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Its full parameter range is resolved. If $p$ is prime, $m$ is positive with $m\\mid p-1$ or $m=2$, $0\\le j<mp$, and $q=\\operatorname{lcm}(2,p-1)$, then the original Euler coefficients are $p$-adically integral and\n\\[\n E_{mpn+qp^r}^{(mp,j)}\\equiv E_{mpn}^{(mp,j)}\\pmod{p^r}\n \\qquad (r\\ge1,\\ n\\ge r).\n\\]\nThe paper states the generating function and the conversion between ordinary degree and packet row before using this formula. The normalized coefficient is $C_{N,j}(n)$ and the original coefficient is $E_{Nn}^{(N,j)}=j!C_{N,j}(n)$. The constant coefficient of the original series is $j!$.","nal edition establishes}\n\\addcontentsline{toc}{section}{What this final edition establishes}\nThe original external problem is Nishibuchi's Conjecture 3.9 for generalized Euler numbers. Its full parameter range is resolved. If $p$ is prime, $m$ is positive with $m\\mid p-1$ or $m=2$, $0\\le j<mp$, and $q=\\operatorname{lcm}(2,p-1)$, then the original Euler coefficients are $p$-adically integral and\n\\[\n E_{mpn+qp^r}^{(mp,j)}\\equiv E_{mpn}^{(mp,j)}\\pmod{p^r}\n \\qquad (r\\ge1,\\ n\\ge r).\n\\]\nThe paper states the generating function and the conversion between ordinary degree and packet row before using this formula. The normalized coefficient is $C_{N,j}(n)$ and the original coefficient is $E_{Nn}^{(N,j)}=j!C_{N,j}(n)$. The constant coefficient of the original series is $j!$.","is positive with $m\\mid p-1$ or $m=2$, $0\\le j<mp$, and $q=\\operatorname{lcm}(2,p-1)$, then the original Euler coefficients are $p$-adically integral and\n\\[\n E_{mpn+qp^r}^{(mp,j)}\\equiv E_{mpn}^{(mp,j)}\\pmod{p^r}\n \\qquad (r\\ge1,\\ n\\ge r).\n\\]\nThe paper states the generating function and the conversion between ordinary degree and packet row before using this formula. The normalized coefficient is $C_{N,j}(n)$ and the original coefficient is $E_{Nn}^{(N,j)}=j!C_{N,j}(n)$. The constant coefficient of the original series is $j!$.","j<mp$, and $q=\\operatorname{lcm}(2,p-1)$, then the original Euler coefficients are $p$-adically integral and\n\\[\n E_{mpn+qp^r}^{(mp,j)}\\equiv E_{mpn}^{(mp,j)}\\pmod{p^r}\n \\qquad (r\\ge1,\\ n\\ge r).\n\\]\nThe paper states the generating function and the conversion between ordinary degree and packet row before using this formula. The normalized coefficient is $C_{N,j}(n)$ and the original coefficient is $E_{Nn}^{(N,j)}=j!C_{N,j}(n)$. The constant coefficient of the original series is $j!$.","e $p$-adically integral and\n\\[\n E_{mpn+qp^r}^{(mp,j)}\\equiv E_{mpn}^{(mp,j)}\\pmod{p^r}\n \\qquad (r\\ge1,\\ n\\ge r).\n\\]\nThe paper states the generating function and the conversion between ordinary degree and packet row before using this formula. The normalized coefficient is $C_{N,j}(n)$ and the original coefficient is $E_{Nn}^{(N,j)}=j!C_{N,j}(n)$. The constant coefficient of the original series is $j!$.","The strongest common-step conclusion in width four is particularly compact. For $n\\ge0$ and $k,H\\ge1$,\n\\begin{equation*}\n \\sum_{i=0}^k(-1)^{k-i}\\binom ki E_{4(n+iH)}^{(4,j)}\n \\equiv0\\pmod{2^{\\,v_2(k!)+kv_2(H)+A_j(k)}},\n\\end{equation*}\nwhere\n\\[\n A_0(k)=A_1(k)=k,\\qquad\n A_2(k)=\\max\\{k+4,\\,2k+\\lfloor k/2\\rfloor+1\\},\\qquad\n A_3(k)=k+2.\n\\]\nThere is no starting-row threshold. The exponent is exact in addresses zero and one for every order and positive step. It is exact in address two at every power-of-two order. The proof chapters also give mixed-step forms, the cancellation criterion in addresses zero and one, and exact first-displacement laws.","\\midrule\nNishibuchi Conjecture 3.9 & Every parameter in the original statement; explicit $n\\ge r$ & Inherited Lean proof","Taylor radii and broader dyadic first differences & Exact radii in addresses $0,1,2$; stated all-rank cases & Written proof and audit","\\[\n(b_0,b_1,b_2,b_3,b_4)=(1,-2,72,-33872,60511872).\n\\]","nge and develop sharper arithmetic consequences. For every prime-power packet width, a charged Hurwitz differential operator transfers integral difference estimates through the reciprocal generating series. In the dyadic case this gives mixed finite-difference bounds at every row, without a starting threshold. For the four addresses of width four, a divided-power argument adds the factorial gain $v_2(k!)$, determines every repeated-step valuation in addresses zero and one, and gives a stronger address-two estimate with exact values at all power-of-two orders. The corresponding first differences yield least periods and bijective isometric continuations onto explicit dyadic residue classes. We also determine exact Taylor radii for three addresses. The manuscript retains the complete earlier arithmetic, lifting, deformation, and analytic-regularity development, with dated historical appendices. A separate appendix states the precise remaining simplicity problem for ordinary Euler tails. Formal coverage is explicit: the external conjecture and several four-address results have inherited …","\\clearpage\n\\appendix\n\\input{../root_scope.tex}\n\\input{parts/70_continuation}\n\\input{parts/71_native}\n\\input{parts/72_recovered}\n\\input{parts/73_supplements}\n\\clearpage\\def\\IXMainAtlas{1}\\def\\AtlasFigDir{../../figures/worked}\\input{../worked_atlas.tex}\\clearpage\n\\begin{thebibliography}{99}\n\\bibitem{Nishibuchi}\nYuta Nishibuchi,\n\\emph{$p$-adic Congruences of Generalized Euler Numbers and Relations to Even\nZeta Values}, arXiv:2605.10677v2, 1 June 2026.\n\\url{https://arxiv.org/abs/2605.10677v2}.\nThe comparison to Conjecture 3.9 uses the source review and fixed-version\nrecord preserved with this edition.","% Generated from references_historical.bib using amsplain, with full repeated authors and DOI links.\n\\bibitem{Amice1964}\nYvette Amice, \\emph{{Interpolation {$p$}-adique}}, Bulletin de la Soci\\'et\\'e\n  Math\\'ematique de France \\textbf{92} (1964), 117--180.\n\\url{https://doi.org/10.24033/bsmf.1606}.","\\bibitem{Gouvea1997}\nFernando~Q. Gouv\\^ea, \\emph{{{$p$}-adic Numbers: An Introduction}}, second ed.,\n  Universitext, Springer, Berlin, 1997.\n\\url{https://doi.org/10.1007/978-3-642-59058-0}.","\\bibitem{Koblitz1984}\nNeal Koblitz, \\emph{{{$p$}-adic Numbers, {$p$}-adic Analysis, and\n  Zeta-Functions}}, second ed., Graduate Texts in Mathematics, vol.~58,\n  Springer, New York, 1984.\n\\url{https://doi.org/10.1007/978-1-4612-1112-9}.","\\bibitem{Mahler1958}\nKurt Mahler, \\emph{{An Interpolation Series for Continuous Functions of a\n  {$p$}-adic Variable}}, Journal f\\\"ur die Reine und Angewandte Mathematik\n  \\textbf{199} (1958), 23--34.\n\\url{https://doi.org/10.1515/crll.1958.199.23}.","\\bibitem{Robert2000}\nAlain~M. Robert, \\emph{{A Course in {$p$}-adic Analysis}}, Graduate Texts in\n  Mathematics, vol. 198, Springer, New York, 2000.\n\\url{https://doi.org/10.1007/978-1-4757-3254-2}.","\\section{Scope and conventions}\n\\label{sec:scope}\nThe central objects are sequences and polynomial equations equipped with\narithmetic coordinates. Residue projectors select addresses in a sequence,\nfinite differences measure compatibility between nearby addresses, and\nvaluation bounds control the passage from finite precision to a $p$-adic\nlimit. These constructions provide a common language for the Euler packet\nresults and the later lifting results. Each theorem nevertheless retains its\nown coefficient ring, topology and nonsingularity hypotheses.","Throughout, $\\N=\\{0,1,2,\\ldots\\}$. When $p$ is prime, $\\Zp$ denotes the\n$p$-adic integers and $\\Qp$ their fraction field. We normalize $v_p(p)=1$.\nFor a rational or $p$-adic quantity, a displayed lower valuation bound is\nunderstood with the mathematical convention $v_p(0)=+\\infty$.\nThe corresponding formal predicates are zero-safe: the quantity is zero or\nits valuation meets the stated bound. An exact finite valuation assertion\nrequires a nonzero quantity, and this is stated where used. Congruence in\n$\\Zp$ means ideal divisibility. Thus\n\\[\n x\\equiv y\\pmod{p^r}\\quad\\Longleftrightarrow\\quad\n x-y\\in p^r\\Zp.\n\\]\nFor vectors, ideal membership and congruence are coordinatewise.","raction field. We normalize $v_p(p)=1$.\nFor a rational or $p$-adic quantity, a displayed lower valuation bound is\nunderstood with the mathematical convention $v_p(0)=+\\infty$.\nThe corresponding formal predicates are zero-safe: the quantity is zero or\nits valuation meets the stated bound. An exact finite valuation assertion\nrequires a nonzero quantity, and this is stated where used. Congruence in\n$\\Zp$ means ideal divisibility. Thus\n\\[\n x\\equiv y\\pmod{p^r}\\quad\\Longleftrightarrow\\quad\n x-y\\in p^r\\Zp.\n\\]\nFor vectors, ideal membership and congruence are coordinatewise.",".\nFor a rational or $p$-adic quantity, a displayed lower valuation bound is\nunderstood with the mathematical convention $v_p(0)=+\\infty$.\nThe corresponding formal predicates are zero-safe: the quantity is zero or\nits valuation meets the stated bound. An exact finite valuation assertion\nrequires a nonzero quantity, and this is stated where used. Congruence in\n$\\Zp$ means ideal divisibility. Thus\n\\[\n x\\equiv y\\pmod{p^r}\\quad\\Longleftrightarrow\\quad\n x-y\\in p^r\\Zp.\n\\]\nFor vectors, ideal membership and congruence are coordinatewise.","Write $\\Delta_u f(n)=f(n+u)-f(n)$ and use $\\Delta_u^k$ for its $k$-fold\niterate. Euler coefficients indexed by ordinary degree and by packet row are\ndistinguished explicitly. A packet displacement by $u$ rows is a degree\ndisplacement by $Nu$ in a packet of width $N$.","The original arithmetic seed notation\n\\[\n\\mathcal S=(A,a_0,\\mathcal O,\\mathcal R,\\{P_{N,j}\\},w,\\tau_p)\n\\]\nrecords a coefficient ring or module, a seed, allowed operators, relations,\nresidue selectors, a depth weight, and a topology. This notation describes the\ningredients of the construction; a category of seed presentations and a\nnotion of equivalence still require separate definitions. We therefore state\neach theorem for a concrete object. Section~\\ref{sec:open} discusses the\nadditional structure needed for the original native conjectures.","Let $R$ be a commutative ring, $a:\\N\\to R$, $N\\geq1$, and $0\\leq j<N$.\nDefine\n\\[\n(P_{N,j}a)(n)=\n\\begin{cases}\na(n),&n\\equiv j\\pmod N,","Let $R$ be a commutative ring, $a:\\N\\to R$, $N\\geq1$, and $0\\leq j<N$.\nDefine\n\\[\n(P_{N,j}a)(n)=\n\\begin{cases}\na(n),&n\\equiv j\\pmod N,\\\\\n0,&\\text{otherwise}.\n\\end{cases}\n\\]\nThis definition keeps the original index set. The compressed subsequence\n$a(Nq+j)$, used below for recurrence packets, is a different representation of\nthe selected coefficients.","0,&\\text{otherwise}.\n\\end{cases}\n\\]\nThis definition keeps the original index set. The compressed subsequence\n$a(Nq+j)$, used below for recurrence packets, is a different representation of\nthe selected coefficients.","\\begin{theorem}[Residue reconstruction]\n\\label{thm:residue-reconstruction}\nThe selectors satisfy\n\\[\nP_{N,j}^2=P_{N,j},\\qquad P_{N,i}P_{N,j}=0\\quad(i\\ne j),\n\\qquad \\sum_{j=0}^{N-1}P_{N,j}=I.\n\\]\nIf $N\\mid M$ and $M>0$, then\n\\[\nP_{N,j}a=\\sum_{\\substack{0\\leq r<M","\\begin{theorem}[Residue reconstruction]\n\\label{thm:residue-reconstruction}\nThe selectors satisfy\n\\[\nP_{N,j}^2=P_{N,j},\\qquad P_{N,i}P_{N,j}=0\\quad(i\\ne j),\n\\qquad \\sum_{j=0}^{N-1}P_{N,j}=I.\n\\]\nIf $N\\mid M$ and $M>0$, then\n\\[\nP_{N,j}a=\\sum_{\\substack{0\\leq r<M\\\\r\\equiv j\\pmod N}}P_{M,r}a.\n\\]\nThe same identities hold coefficientwise for formal power series.\n\\end{theorem}","\\begin{proof}\nAt each index $n$, exactly one standard residue label equals $n\\bmod N$.\nConsequently a repeated selector changes nothing, two different selectors have\nzero composite, and the sum retains $a(n)$ exactly once. If $N\\mid M$, reducing\n$n\\bmod M$ modulo $N$ gives $n\\bmod N$. This proves the refinement identity\nat every coefficient. Equality of all coefficients proves the formal-series\nversions.\n\\end{proof}","Over a characteristic-zero field $K$, put\n\\[\n\\mathcal E(a)=\\sum_{n\\geq0}\\frac{a_n}{n!}t^n,\n\\qquad \\mathcal D(A)_n=n![t^n]A.\n\\]\nThese inverse maps between sequences and $K[[t]]$ provide \\emph{formal}\nexponential coordinates. No analytic convergence is needed for the coefficient\nidentities below.","\\begin{theorem}[Constructed unit-triangular coordinates]\n\\label{thm:egf-coordinates}\nLet $h:\\N\\to K$ satisfy $h_0=1$, write $H=\\mathcal E(h)$ and\n$H^{-1}=\\mathcal E(k)$, and define\n\\[\nU_Ha=\\mathcal D(H\\mathcal E(a)).\n\\]\nThen $U_H$ is an equivalence, with inverse\n$U_H^{-1}a=\\mathcal D(H^{-1}\\mathcal E(a))$. Its coefficient formulas are\n\\[\n(U_Ha)_n=\\sum_{r=0}^{n}\\binom nr h_r a_{n-r},\\qquad\n(U_H^{-1}a)_n=\\sum_{r=0}^{n}\\binom nr k_r a_{n-r}.\n\\]\nThe resulting triangular matrix has diagonal entries $1$. The transported\nselectors\n\\[\n\\widehat P_{N,j}=U_HP_{N,j}U_H^{-1}\n\\]\nare idempotent and pairwise orthogonal, and their sum is the identity.\n\\end{theorem}","satisfy $h_0=1$, write $H=\\mathcal E(h)$ and\n$H^{-1}=\\mathcal E(k)$, and define\n\\[\nU_Ha=\\mathcal D(H\\mathcal E(a)).\n\\]\nThen $U_H$ is an equivalence, with inverse\n$U_H^{-1}a=\\mathcal D(H^{-1}\\mathcal E(a))$. Its coefficient formulas are\n\\[\n(U_Ha)_n=\\sum_{r=0}^{n}\\binom nr h_r a_{n-r},\\qquad\n(U_H^{-1}a)_n=\\sum_{r=0}^{n}\\binom nr k_r a_{n-r}.\n\\]\nThe resulting triangular matrix has diagonal entries $1$. The transported\nselectors\n\\[\n\\widehat P_{N,j}=U_HP_{N,j}U_H^{-1}\n\\]\nare idempotent and pairwise orthogonal, and their sum is the identity.\n\\end{theorem}","cal E(h)$ and\n$H^{-1}=\\mathcal E(k)$, and define\n\\[\nU_Ha=\\mathcal D(H\\mathcal E(a)).\n\\]\nThen $U_H$ is an equivalence, with inverse\n$U_H^{-1}a=\\mathcal D(H^{-1}\\mathcal E(a))$. Its coefficient formulas are\n\\[\n(U_Ha)_n=\\sum_{r=0}^{n}\\binom nr h_r a_{n-r},\\qquad\n(U_H^{-1}a)_n=\\sum_{r=0}^{n}\\binom nr k_r a_{n-r}.\n\\]\nThe resulting triangular matrix has diagonal entries $1$. The transported\nselectors\n\\[\n\\widehat P_{N,j}=U_HP_{N,j}U_H^{-1}\n\\]\nare idempotent and pairwise orthogonal, and their sum is the identity.\n\\end{theorem}","\\begin{proof}\nThe factorials are nonzero in $K$, so encoding and decoding cancel\ncoefficientwise. The constant coefficient of $H$ is $1$; hence its formal\npower-series inverse exists. Multiplication by $H$ and $H^{-1}$ gives the two\ninverse laws. The ordinary product coefficient formula, multiplied by $n!$,\ngives the displayed binomial sums. In row $n$, the coefficient of $a_n$ is\n$h_0=1$ and coefficients of $a_q$ with $q>n$ vanish. Conjugating the identities\nof Theorem~\\ref{thm:residue-reconstruction} by the constructed equivalence proves\nthe transported projector identities.\n\\end{proof}","The factorials are nonzero in $K$, so encoding and decoding cancel\ncoefficientwise. The constant coefficient of $H$ is $1$; hence its formal\npower-series inverse exists. Multiplication by $H$ and $H^{-1}$ gives the two\ninverse laws. The ordinary product coefficient formula, multiplied by $n!$,\ngives the displayed binomial sums. In row $n$, the coefficient of $a_n$ is\n$h_0=1$ and coefficients of $a_q$ with $q>n$ vanish. Conjugating the identities\nof Theorem~\\ref{thm:residue-reconstruction} by the constructed equivalence proves\nthe transported projector identities.\n\\end{proof}","ials are nonzero in $K$, so encoding and decoding cancel\ncoefficientwise. The constant coefficient of $H$ is $1$; hence its formal\npower-series inverse exists. Multiplication by $H$ and $H^{-1}$ gives the two\ninverse laws. The ordinary product coefficient formula, multiplied by $n!$,\ngives the displayed binomial sums. In row $n$, the coefficient of $a_n$ is\n$h_0=1$ and coefficients of $a_q$ with $q>n$ vanish. Conjugating the identities\nof Theorem~\\ref{thm:residue-reconstruction} by the constructed equivalence proves\nthe transported projector identities.\n\\end{proof}","In characteristic $p$, factorial division is unavailable in general, but the\ncoefficient operation remains meaningful over every commutative ring:\n\\[\n(f\\star g)_n=\\sum_{r=0}^{n}\\binom nr f_r g_{n-r},\\qquad\n(D^d f)_n=f_{n+d},\\qquad \\mathbf1_n=\\begin{cases}1,&n=0,","In characteristic $p$, factorial division is unavailable in general, but the\ncoefficient operation remains meaningful over every commutative ring:\n\\[\n(f\\star g)_n=\\sum_{r=0}^{n}\\binom nr f_r g_{n-r},\\qquad\n(D^d f)_n=f_{n+d},\\qquad \\mathbf1_n=\\begin{cases}1,&n=0,\\\\0,&n>0.\\end{cases}\n\\]\nHere $\\star$ is Hurwitz multiplication. The symbol $D^d$ denotes a coefficient\nshift, which agrees with formal differentiation after characteristic-zero\nencoding.","0,&n>0.\\end{cases}\n\\]\nHere $\\star$ is Hurwitz multiplication. The symbol $D^d$ denotes a coefficient\nshift, which agrees with formal differentiation after characteristic-zero\nencoding.","\\begin{theorem}[Hurwitz product rules]\n\\label{thm:hurwitz-product}\nFor $d\\geq0$,\n\\[\nD^d(f\\star g)=\\sum_{r=0}^{d}\\binom dr(D^r f)\\star(D^{d-r}g).\n\\]\nIf $R$ has prime characteristic $p$, then\n\\[\nD^p(f\\star g)=f\\star D^p g+D^p f\\star g.\n\\]\nOver a field, if $g_0=1$, a sequence $b$ is constructed recursively with\n$b\\star g=\\mathbf1$. If also $D^p g=Dg$ in characteristic $p$, then\n$D^p b=Db$.\n\\end{theorem}","\\begin{proof}\nPascal's identity proves the first product rule, and induction proves the\nhigher rule. The intermediate binomial coefficients $\\binom pr$ vanish in\ncharacteristic $p$. For the inverse, solve row $n$ using the coefficient of\n$b_n$, which is $g_0=1$:\n\\[\nb_n=\\mathbf1_n-\\sum_{r<n}\\binom nr b_r g_{n-r}.\n\\]\nSubtract the first and $p$th product rules for $b\\star g=\\mathbf1$.\nThe assumed identity for $g$ leaves\n$(D^p b-Db)\\star g=0$. A second triangular induction shows that convolution\nwith a sequence having constant coefficient $1$ has trivial kernel.\n\\end{proof}","f}\nPascal's identity proves the first product rule, and induction proves the\nhigher rule. The intermediate binomial coefficients $\\binom pr$ vanish in\ncharacteristic $p$. For the inverse, solve row $n$ using the coefficient of\n$b_n$, which is $g_0=1$:\n\\[\nb_n=\\mathbf1_n-\\sum_{r<n}\\binom nr b_r g_{n-r}.\n\\]\nSubtract the first and $p$th product rules for $b\\star g=\\mathbf1$.\nThe assumed identity for $g$ leaves\n$(D^p b-Db)\\star g=0$. A second triangular induction shows that convolution\nwith a sequence having constant coefficient $1$ has trivial kernel.\n\\end{proof}","identity proves the first product rule, and induction proves the\nhigher rule. The intermediate binomial coefficients $\\binom pr$ vanish in\ncharacteristic $p$. For the inverse, solve row $n$ using the coefficient of\n$b_n$, which is $g_0=1$:\n\\[\nb_n=\\mathbf1_n-\\sum_{r<n}\\binom nr b_r g_{n-r}.\n\\]\nSubtract the first and $p$th product rules for $b\\star g=\\mathbf1$.\nThe assumed identity for $g$ leaves\n$(D^p b-Db)\\star g=0$. A second triangular induction shows that convolution\nwith a sequence having constant coefficient $1$ has trivial kernel.\n\\end{proof}","binom pr$ vanish in\ncharacteristic $p$. For the inverse, solve row $n$ using the coefficient of\n$b_n$, which is $g_0=1$:\n\\[\nb_n=\\mathbf1_n-\\sum_{r<n}\\binom nr b_r g_{n-r}.\n\\]\nSubtract the first and $p$th product rules for $b\\star g=\\mathbf1$.\nThe assumed identity for $g$ leaves\n$(D^p b-Db)\\star g=0$. A second triangular induction shows that convolution\nwith a sequence having constant coefficient $1$ has trivial kernel.\n\\end{proof}","\\begingroup\\small\\raggedright\n\\noindent\\emph{Formal correspondence.}\n\\lean{Hurwitz.shift_convolution}, \\lean{Hurwitz.prime_shift_convolution},\n\\lean{Hurwitz.convolution_left_eq_zero},\n\\lean{HurwitzInverse.inverse_convolution}, and\n\\lean{HurwitzInverse.inverse_prime_difference}.\nWorking directly with coefficient sequences avoids the undefined expression\n$1/n!$ in $\\F_p$ when $p\\mid n!$.\n\\par\\endgroup","\\begin{theorem}[Forward substitution and denominator clearing]\n\\label{thm:triangular-foundation}\nLet $T_{nq}$ be coefficients over a field with $T_{nn}\\ne0$ for every $n$.\nFor any sequence $b$, define\n\\[\na_n=\\frac{b_n-\\sum_{q<n}T_{nq}a_q}{T_{nn}}.\n\\]\nThis constructs the unique solution of\n$\\sum_{q=0}^{n}T_{nq}a_q=b_n$ for all $n$. If $T_{nq}$ and $b_n$ are integers\nand the solution is taken over $\\Q$, then\n\\[\n\\left(\\prod_{r=0}^{n}T_{rr}\\right)a_n\\in\\Z.\n\\]\n\\end{theorem}","em}[Forward substitution and denominator clearing]\n\\label{thm:triangular-foundation}\nLet $T_{nq}$ be coefficients over a field with $T_{nn}\\ne0$ for every $n$.\nFor any sequence $b$, define\n\\[\na_n=\\frac{b_n-\\sum_{q<n}T_{nq}a_q}{T_{nn}}.\n\\]\nThis constructs the unique solution of\n$\\sum_{q=0}^{n}T_{nq}a_q=b_n$ for all $n$. If $T_{nq}$ and $b_n$ are integers\nand the solution is taken over $\\Q$, then\n\\[\n\\left(\\prod_{r=0}^{n}T_{rr}\\right)a_n\\in\\Z.\n\\]\n\\end{theorem}","\\begin{proof}\nThe recursion uses only smaller indices, and substitution verifies each row.\nIf a second solution agrees through row $n-1$, subtracting row $n$ and\ncancelling $T_{nn}$ proves equality at $n$. For integrality, use induction and\npromote every earlier clearing product to $\\prod_{r<n}T_{rr}$. Multiplication\nby the additional factor $T_{nn}$ cancels the only new division.\n\\end{proof}","To isolate the filtration argument, let $p$ be any element of a commutative\nring and let $w:\\N\\to\\N\\cup\\{\\infty\\}$. Set $p^{[r]}=p^r$ for finite $r$\nand $p^{[\\infty]}=0$. The latter notation means the coefficient vanishes.","\\begin{theorem}[Filtered finite reconstruction]\n\\label{thm:filtered-foundation}\nFor every $n,L\\in\\N$ and sequences $c,S$ in that ring,\n\\[\n\\begin{split}\n&\\sum_{r=0}^{n}\\binom nr c_rp^{[w(r)]}S_{n-r}","\\begin{theorem}[Filtered finite reconstruction]\n\\label{thm:filtered-foundation}\nFor every $n,L\\in\\N$ and sequences $c,S$ in that ring,\n\\[\n\\begin{split}\n&\\sum_{r=0}^{n}\\binom nr c_rp^{[w(r)]}S_{n-r}\\\\\n&\\hspace{8mm}\\equiv\n\\sum_{\\substack{0\\leq r\\leq n\\\\w(r)<L}}\n\\binom nr c_rp^{[w(r)]}S_{n-r}\\pmod{p^L}.\n\\end{split}\n\\]\nHere congruence means divisibility of the difference by $p^L$ in the\ncoefficient ring. In $\\Zp$, it means membership in $p^L\\Zp$.\n\\end{theorem}","w(r)<L}}\n\\binom nr c_rp^{[w(r)]}S_{n-r}\\pmod{p^L}.\n\\end{split}\n\\]\nHere congruence means divisibility of the difference by $p^L$ in the\ncoefficient ring. In $\\Zp$, it means membership in $p^L\\Zp$.\n\\end{theorem}","\\begin{proof}\nThe difference of the two sums consists exactly of terms with $w(r)\\geq L$.\nEvery such term is divisible by $p^L$, including the zero terms with infinite\nweight. Their finite sum has the same divisibility.\n\\end{proof}","\\begingroup\\small\\raggedright\n\\noindent\\emph{Formal correspondence.}\n\\lean{Filtered.filtered_triangular_reconstruction} and\n\\lean{Filtered.padic_filtered_reconstruction}.\nFor weights $2r$, retaining $r\\leq M$ gives the bound $p^{2M+2}$ through\n\\lean{Filtered.even_weight_tail}. A sharper kernel weight can improve this\nbound, as in the classical dyadic reconstruction later in the paper. Each\ncoefficient row is finite, so no hypothesis $w(r)\\to\\infty$ is needed here.\n\\par\\endgroup","t\\emph{Formal correspondence.}\n\\lean{Filtered.filtered_triangular_reconstruction} and\n\\lean{Filtered.padic_filtered_reconstruction}.\nFor weights $2r$, retaining $r\\leq M$ gives the bound $p^{2M+2}$ through\n\\lean{Filtered.even_weight_tail}. A sharper kernel weight can improve this\nbound, as in the classical dyadic reconstruction later in the paper. Each\ncoefficient row is finite, so no hypothesis $w(r)\\to\\infty$ is needed here.\n\\par\\endgroup","For $f:\\N\\to R$, define the step-$h$ forward difference and its iterates by\n\\[\n\\Delta_h f(n)=f(n+h)-f(n),\\qquad \\Delta_h^0 f=f,\n\\]\nand put $c_r(f)=\\Delta_1^r f(0)$. Thus the coefficients are determined by the\ndifferences of $f$.","\\begin{theorem}[Finite Newton--Mahler identities]\n\\label{thm:mahler-foundation}\nFor every $n,s,h\\in\\N$,\n\\[\n\\begin{aligned}\nf(n)&=\\sum_{r=0}^{n}c_r(f)\\binom nr,","\\begin{theorem}[Finite Newton--Mahler identities]\n\\label{thm:mahler-foundation}\nFor every $n,s,h\\in\\N$,\n\\[\n\\begin{aligned}\nf(n)&=\\sum_{r=0}^{n}c_r(f)\\binom nr,\\\\\nc_r(\\Delta_1^s f)&=c_{s+r}(f),\\\\\n\\Delta_1^s f(n)&=\\sum_{j=0}^{n}c_{s+j}(f)\\binom nj,\\\\\n\\Delta_h^s f(n)&=\\sum_{j=0}^{s}(-1)^{s-j}\\binom sj f(n+jh).\n\\end{aligned}\n\\]\nThe finite interpolant\n$I_Nf(n)=\\sum_{r=0}^{N}c_r(f)\\binom nr$ agrees with $f$ for $n\\leq N$.\nIt has the same order-$s$ unit-step difference at $n$ whenever $n+s\\leq N$.\nFor arbitrary coefficients $c_0,\\ldots,c_N$, the corresponding binomial\ninterpolant has exactly those coefficients and zero coefficients above $N$.\n\\end{theorem}","\\Delta_h^s f(n)&=\\sum_{j=0}^{s}(-1)^{s-j}\\binom sj f(n+jh).\n\\end{aligned}\n\\]\nThe finite interpolant\n$I_Nf(n)=\\sum_{r=0}^{N}c_r(f)\\binom nr$ agrees with $f$ for $n\\leq N$.\nIt has the same order-$s$ unit-step difference at $n$ whenever $n+s\\leq N$.\nFor arbitrary coefficients $c_0,\\ldots,c_N$, the corresponding binomial\ninterpolant has exactly those coefficients and zero coefficients above $N$.\n\\end{theorem}","\\begin{proof}\nWrite $E_h f(n)=f(n+h)$ for the shift operator. The unit shift is\n$I+\\Delta_1$. Expanding its $n$th power and evaluating\nat zero gives reconstruction. Composing difference operators gives the\ncoefficient shift and applying reconstruction to $\\Delta_1^s f$ gives the\nthird identity. The binomial expansion of $(E_h-I)^s$ gives the explicit\nstencil. Binomial coefficients vanish when their lower index exceeds $n$,\nwhich proves the interpolation claim. Equality of every point of the stencil\nproves equality of differences. Repeated finite differences of $\\binom nr$\nat zero give the exact coefficient extraction.\n\\end{proof}","h f(n)=f(n+h)$ for the shift operator. The unit shift is\n$I+\\Delta_1$. Expanding its $n$th power and evaluating\nat zero gives reconstruction. Composing difference operators gives the\ncoefficient shift and applying reconstruction to $\\Delta_1^s f$ gives the\nthird identity. The binomial expansion of $(E_h-I)^s$ gives the explicit\nstencil. Binomial coefficients vanish when their lower index exceeds $n$,\nwhich proves the interpolation claim. Equality of every point of the stencil\nproves equality of differences. Repeated finite differences of $\\binom nr$\nat zero give the exact coefficient extraction.\n\\end{proof}","posing difference operators gives the\ncoefficient shift and applying reconstruction to $\\Delta_1^s f$ gives the\nthird identity. The binomial expansion of $(E_h-I)^s$ gives the explicit\nstencil. Binomial coefficients vanish when their lower index exceeds $n$,\nwhich proves the interpolation claim. Equality of every point of the stencil\nproves equality of differences. Repeated finite differences of $\\binom nr$\nat zero give the exact coefficient extraction.\n\\end{proof}","For $x\\in\\Zp$, extend the valuation to zero by setting\n\\[\n\\operatorname{ord}_p(x)=\n\\begin{cases}v_p(x),&x\\ne0,","For $x\\in\\Zp$, extend the valuation to zero by setting\n\\[\n\\operatorname{ord}_p(x)=\n\\begin{cases}v_p(x),&x\\ne0,\\\\\\infty,&x=0.\\end{cases}\n\\]\nThus $L\\leq\\operatorname{ord}_p(x)$ is equivalent to $x\\in p^L\\Zp$.","infty,&x=0.\\end{cases}\n\\]\nThus $L\\leq\\operatorname{ord}_p(x)$ is equivalent to $x\\in p^L\\Zp$.","\\begin{theorem}[Weighted difference bound]\n\\label{thm:weighted-mahler-foundation}\nLet $f:\\N\\to\\Zp$ and let $w:\\N\\to\\N\\cup\\{\\infty\\}$ satisfy\n$w(s+j)\\leq\\operatorname{ord}_p(c_{s+j}(f))$ for $0\\leq j\\leq n$.\nThen\n\\[\n\\operatorname{ord}_p(\\Delta_1^s f(n))\\geq\n\\min_{0\\leq j\\leq n}\n\\left\\{w(s+j)+v_p\\binom nj\\right\\}.\n\\]\n\\end{theorem}","\\begin{proof}\nApply the nonarchimedean inequality to the finite coefficient formula in\nTheorem~\\ref{thm:mahler-foundation}. For $0\\leq j\\leq n$, the integer\n$\\binom nj$ is positive. The order of its product with $c_{s+j}(f)$ is the\nsum of the two orders, including the case where that coefficient is zero.\n\\end{proof}","\\begingroup\\small\\raggedright\n\\noindent\\emph{Formal correspondence.}\n\\lean{PadicBounds.nat_le_order_iff},\n\\lean{PadicBounds.weighted_difference_bound}, and\n\\lean{PadicBounds.weighted_difference_divisibility}.\nAn expansion on all of $\\Zp$ requires a convergence theorem in addition to\nthese finite algebraic identities and bounds.\n\\par\\endgroup","\\begin{theorem}[Differences of unit modes]\n\\label{thm:unit-modes-foundation}\nOver every commutative ring,\n\\[\n\\Delta_h^s(u^n)=u^n(u^h-1)^s.\n\\]\nFor any prime $p$, units $u_i\\in\\Zp^\\times$, coefficients $a_i\\in\\Zp$, and\na finite index set $I$, the sequence $f(n)=\\sum_{i\\in I}a_i u_i^n$ satisfies\n\\[\n\\Delta_{p-1}^s f(n)\\in p^s\\Zp\n\\qquad(n,s\\in\\N).\n\\]\n\\end{theorem}","\\begin{proof}\nInduction gives the geometric identity. Reduction of a unit to $\\F_p^\\times$\ngives $u_i^{p-1}=1$ modulo $p$. Taking $s$ powers supplies $p^s$ in each\nsummand. The difference operator is linear over finite sums.\n\\end{proof}","\\begingroup\\small\\raggedright\n\\noindent\\emph{Formal correspondence.}\n\\lean{UnitModes.geometric_difference},\n\\lean{UnitModes.unit_fermat_divisibility}, and\n\\lean{UnitModes.finite_unit_modes_divisibility}.\nThe result includes $p=2$, where the displayed step is $1$. Its scope here\nis finite sums of modes.\n\\par\\endgroup","\\begin{theorem}[Finite unit coordinates]\n\\label{thm:finite-unit-foundation}\nLet $p$ be prime, $L\\geq0$, and\n$a\\in(\\Z/p^{L+1}\\Z)^\\times$. Define\n\\[\n\\omega_L(a)=a^{p^L},\\qquad \\langle a\\rangle_L=a\\omega_L(a)^{-1}.\n\\]\nThen\n\\[\na=\\omega_L(a)\\langle a\\rangle_L,\\qquad\n\\omega_L(a)^{p-1}=1,\\qquad \\langle a\\rangle_L\\equiv1\\pmod p.\n\\]\nThe two factors are uniquely determined by these properties and are compatible\nwith reduction to lower precision.\nFor $a\\in(\\Z/2^{L+2}\\Z)^\\times$, there is instead a unique decomposition\n\\[\na=\\epsilon(a)\\langle a\\rangle_4,\n\\qquad \\epsilon(a)\\in\\{1,-1\\},\\qquad\n\\langle a\\rangle_4\\equiv1\\pmod4.\n\\]\n\\end{theorem}","el{thm:finite-unit-foundation}\nLet $p$ be prime, $L\\geq0$, and\n$a\\in(\\Z/p^{L+1}\\Z)^\\times$. Define\n\\[\n\\omega_L(a)=a^{p^L},\\qquad \\langle a\\rangle_L=a\\omega_L(a)^{-1}.\n\\]\nThen\n\\[\na=\\omega_L(a)\\langle a\\rangle_L,\\qquad\n\\omega_L(a)^{p-1}=1,\\qquad \\langle a\\rangle_L\\equiv1\\pmod p.\n\\]\nThe two factors are uniquely determined by these properties and are compatible\nwith reduction to lower precision.\nFor $a\\in(\\Z/2^{L+2}\\Z)^\\times$, there is instead a unique decomposition\n\\[\na=\\epsilon(a)\\langle a\\rangle_4,\n\\qquad \\epsilon(a)\\in\\{1,-1\\},\\qquad\n\\langle a\\rangle_4\\equiv1\\pmod4.\n\\]\n\\end{theorem}","Z/p^{L+1}\\Z)^\\times$. Define\n\\[\n\\omega_L(a)=a^{p^L},\\qquad \\langle a\\rangle_L=a\\omega_L(a)^{-1}.\n\\]\nThen\n\\[\na=\\omega_L(a)\\langle a\\rangle_L,\\qquad\n\\omega_L(a)^{p-1}=1,\\qquad \\langle a\\rangle_L\\equiv1\\pmod p.\n\\]\nThe two factors are uniquely determined by these properties and are compatible\nwith reduction to lower precision.\nFor $a\\in(\\Z/2^{L+2}\\Z)^\\times$, there is instead a unique decomposition\n\\[\na=\\epsilon(a)\\langle a\\rangle_4,\n\\qquad \\epsilon(a)\\in\\{1,-1\\},\\qquad\n\\langle a\\rangle_4\\equiv1\\pmod4.\n\\]\n\\end{theorem}","\\begin{proof}\nThe unit-group exponent $p^L(p-1)$ gives the root identity for $\\omega_L$.\nIts residue is that of $a$, so the remaining factor has residue $1$.\nIf two $(p-1)$st roots have equal residue, raising them to $p^L$ makes them\nequal modulo $p^{L+1}$, while each root is fixed by this exponent. This proves\nuniqueness and then compatibility under reduction. In the dyadic construction,\nan odd residue is exactly $1$ or $-1$ modulo $4$. Choose that sign and multiply\n$a$ by it to obtain the principal factor. The two signs remain distinct\nmodulo $4$, which proves uniqueness.\n\\end{proof}","nit-group exponent $p^L(p-1)$ gives the root identity for $\\omega_L$.\nIts residue is that of $a$, so the remaining factor has residue $1$.\nIf two $(p-1)$st roots have equal residue, raising them to $p^L$ makes them\nequal modulo $p^{L+1}$, while each root is fixed by this exponent. This proves\nuniqueness and then compatibility under reduction. In the dyadic construction,\nan odd residue is exactly $1$ or $-1$ modulo $4$. Choose that sign and multiply\n$a$ by it to obtain the principal factor. The two signs remain distinct\nmodulo $4$, which proves uniqueness.\n\\end{proof}","up exponent $p^L(p-1)$ gives the root identity for $\\omega_L$.\nIts residue is that of $a$, so the remaining factor has residue $1$.\nIf two $(p-1)$st roots have equal residue, raising them to $p^L$ makes them\nequal modulo $p^{L+1}$, while each root is fixed by this exponent. This proves\nuniqueness and then compatibility under reduction. In the dyadic construction,\nan odd residue is exactly $1$ or $-1$ modulo $4$. Choose that sign and multiply\n$a$ by it to obtain the principal factor. The two signs remain distinct\nmodulo $4$, which proves uniqueness.\n\\end{proof}","$p^L(p-1)$ gives the root identity for $\\omega_L$.\nIts residue is that of $a$, so the remaining factor has residue $1$.\nIf two $(p-1)$st roots have equal residue, raising them to $p^L$ makes them\nequal modulo $p^{L+1}$, while each root is fixed by this exponent. This proves\nuniqueness and then compatibility under reduction. In the dyadic construction,\nan odd residue is exactly $1$ or $-1$ modulo $4$. Choose that sign and multiply\n$a$ by it to obtain the principal factor. The two signs remain distinct\nmodulo $4$, which proves uniqueness.\n\\end{proof}","for $\\omega_L$.\nIts residue is that of $a$, so the remaining factor has residue $1$.\nIf two $(p-1)$st roots have equal residue, raising them to $p^L$ makes them\nequal modulo $p^{L+1}$, while each root is fixed by this exponent. This proves\nuniqueness and then compatibility under reduction. In the dyadic construction,\nan odd residue is exactly $1$ or $-1$ modulo $4$. Choose that sign and multiply\n$a$ by it to obtain the principal factor. The two signs remain distinct\nmodulo $4$, which proves uniqueness.\n\\end{proof}","$1$.\nIf two $(p-1)$st roots have equal residue, raising them to $p^L$ makes them\nequal modulo $p^{L+1}$, while each root is fixed by this exponent. This proves\nuniqueness and then compatibility under reduction. In the dyadic construction,\nan odd residue is exactly $1$ or $-1$ modulo $4$. Choose that sign and multiply\n$a$ by it to obtain the principal factor. The two signs remain distinct\nmodulo $4$, which proves uniqueness.\n\\end{proof}","\\begingroup\\small\\raggedright\n\\noindent\\emph{Formal correspondence.}\n\\lean{FiniteUnits.finite_unit_decomposition},\n\\lean{FiniteUnits.decomposition_unique},\n\\lean{UnitRefinement.omega_compatible},\n\\lean{UnitRefinement.principal_compatible}, and\n\\lean{DyadicUnits.dyadic_unit_decomposition} with\n\\lean{DyadicUnits.decomposition_unique}.\nThe finite $p=2$ version of the first construction has trivial root-of-unity\nfactor. The separate modulo-$4$ construction retains the sign that reduction\nmodulo $2$ cannot detect.\n\\par\\endgroup","mph{Formal correspondence.}\n\\lean{FiniteUnits.finite_unit_decomposition},\n\\lean{FiniteUnits.decomposition_unique},\n\\lean{UnitRefinement.omega_compatible},\n\\lean{UnitRefinement.principal_compatible}, and\n\\lean{DyadicUnits.dyadic_unit_decomposition} with\n\\lean{DyadicUnits.decomposition_unique}.\nThe finite $p=2$ version of the first construction has trivial root-of-unity\nfactor. The separate modulo-$4$ construction retains the sign that reduction\nmodulo $2$ cannot detect.\n\\par\\endgroup","t_decomposition},\n\\lean{FiniteUnits.decomposition_unique},\n\\lean{UnitRefinement.omega_compatible},\n\\lean{UnitRefinement.principal_compatible}, and\n\\lean{DyadicUnits.dyadic_unit_decomposition} with\n\\lean{DyadicUnits.decomposition_unique}.\nThe finite $p=2$ version of the first construction has trivial root-of-unity\nfactor. The separate modulo-$4$ construction retains the sign that reduction\nmodulo $2$ cannot detect.\n\\par\\endgroup","\\begin{theorem}[Companion jumps and packet recurrences]\n\\label{thm:companion-foundation}\nLet $C$ be a $d\\times d$ matrix over a commutative ring and\n$x_n=C^n x_0$. For $N,j,k\\in\\N$, put $b_k=x_{Nk+j}$. Then\n\\[\nb_k=C^j(C^N)^kx_0,\\qquad b_{k+1}=C^Nb_k.\n\\]\nIf $\\chi_{C^N}(X)=X^d+\\sum_{r<d}q_rX^r$, then\n\\[\nb_{k+d}=-\\sum_{r=0}^{d-1}q_r b_{k+r}.\n\\]\nEvery coordinate and every linear scalar observable satisfies the same\nrecurrence. No invertibility or diagonalization hypothesis is required.\n\\end{theorem}","\\begin{proof}\nThe jump identity follows from the power law for one matrix. Apply the\nCayley--Hamilton identity $\\chi_{C^N}(C^N)=0$ to the state $b_k$.\nThe leading coefficient is $1$, so the final state can be isolated. A linear\nobservable preserves the resulting finite sum.\n\\end{proof}","\\begingroup\\small\\raggedright\n\\noindent\\emph{Formal correspondence.}\n\\lean{Companion.companion_jump},\n\\lean{Companion.packet_charpoly_recurrence}, and\n\\lean{Companion.packet_observable_recurrence}.\nThe source-sequence versions derive the packet recurrence from the relation\n$x_{n+1}=Cx_n$.\nFor the Fibonacci matrix $C=\\left(\\begin{smallmatrix}1&1","\\begingroup\\small\\raggedright\n\\noindent\\emph{Formal correspondence.}\n\\lean{Companion.companion_jump},\n\\lean{Companion.packet_charpoly_recurrence}, and\n\\lean{Companion.packet_observable_recurrence}.\nThe source-sequence versions derive the packet recurrence from the relation\n$x_{n+1}=Cx_n$.\nFor the Fibonacci matrix $C=\\left(\\begin{smallmatrix}1&1\\\\1&0\\end{smallmatrix}\\right)$,\nits jump trace is the Lucas number $L_N$ and its determinant is $(-1)^N$.\nThus\n\\[\nF_{N(k+2)+j}=L_NF_{N(k+1)+j}-(-1)^NF_{Nk+j}.\n\\]\nThis specialization is\n\\lean{FibonacciPackets.fibonacci_packet_recurrence}; it is derived from the\nFibonacci state.\n\\par\\endgroup","1&0\\end{smallmatrix}\\right)$,\nits jump trace is the Lucas number $L_N$ and its determinant is $(-1)^N$.\nThus\n\\[\nF_{N(k+2)+j}=L_NF_{N(k+1)+j}-(-1)^NF_{Nk+j}.\n\\]\nThis specialization is\n\\lean{FibonacciPackets.fibonacci_packet_recurrence}; it is derived from the\nFibonacci state.\n\\par\\endgroup","\\begin{theorem}[Dyadic--odd packet factorization]\n\\label{thm:crt-foundation}\nLet $a\\geq0$ and let $M$ be a positive odd integer. The Chinese remainder map\nis a ring isomorphism\n\\[\n\\Z/(2^aM)\\Z\\simeq\\Z/2^a\\Z\\times\\Z/M\\Z.\n\\]\nFor $0\\leq r<2^aM$ and any coefficient sequence $f$,\n\\[\nP_{2^aM,r}f\n=P_{2^a,r\\bmod 2^a}\\,P_{M,r\\bmod M}f.\n\\]\nThe same identity holds for formal-series coefficient selectors.\n\\end{theorem}","\\begingroup\\small\\raggedright\n\\noindent\\emph{Formal correspondence.}\n\\lean{CRTPackets.addressEquiv}, \\lean{CRTPackets.residue_equivalence},\n\\lean{CRTPackets.packet_factorization}, and\n\\lean{CRTPackets.series_packet_factorization}.\nThis theorem synchronizes finite arithmetic addresses. The passage from a\nfamily of $p$-adic objects to a global integer object is a separate question,\ndiscussed in Section~\\ref{subsec:native133}.\n\\par\\endgroup","The recovered Annals text develops a broader polyphase vocabulary. We explain\nit through elementary identities with explicit algebraic derivations. These\nidentities have not been separately formalized in Lean for this consolidation.\nFor a sequence $a$ and $M\\geq1$, let $a_r^{[M]}(q)=a_{Mq+r}$ for $0\\leq r<M$.\nIf $A(z)=\\sum_n a_nz^n$ and $A_r(w)=\\sum_q a_{Mq+r}w^q$ are ordinary generating\nfunctions, then unique quotient and remainder give\n\\[\nA(z)=\\sum_{r=0}^{M-1}z^rA_r(z^M),\\qquad\n\\bigl(a_r^{[M]}\\bigr)_s^{[N]}=a_{r+Ms}^{[MN]}.\n\\]\nFor ordinary Cauchy multiplication, the phase-$r$ product has a term without a\ncarry and a term with one carry:\n\\[\nC_r(w)=\\sum_{u+v=r}A_u(w)B_v(w)\n+w\\sum_{u+v=r+M}A_u(w)B_v(w),\\qquad 0\\leq u,v<M.\n\\]\nAdding the two phase addresses gives the formula. It describes ordinary\nCauchy multiplication; Hurwitz convolution has its own binomial weights.","polyphase vocabulary. We explain\nit through elementary identities with explicit algebraic derivations. These\nidentities have not been separately formalized in Lean for this consolidation.\nFor a sequence $a$ and $M\\geq1$, let $a_r^{[M]}(q)=a_{Mq+r}$ for $0\\leq r<M$.\nIf $A(z)=\\sum_n a_nz^n$ and $A_r(w)=\\sum_q a_{Mq+r}w^q$ are ordinary generating\nfunctions, then unique quotient and remainder give\n\\[\nA(z)=\\sum_{r=0}^{M-1}z^rA_r(z^M),\\qquad\n\\bigl(a_r^{[M]}\\bigr)_s^{[N]}=a_{r+Ms}^{[MN]}.\n\\]\nFor ordinary Cauchy multiplication, the phase-$r$ product has a term without a\ncarry and a term with one carry:\n\\[\nC_r(w)=\\sum_{u+v=r}A_u(w)B_v(w)\n+w\\sum_{u+v=r+M}A_u(w)B_v(w),\\qquad 0\\leq u,v<M.\n\\]\nAdding the two phase addresses gives the formula. It describes ordinary\nCauchy multiplication; Hurwitz convolution has its own binomial weights.","arately formalized in Lean for this consolidation.\nFor a sequence $a$ and $M\\geq1$, let $a_r^{[M]}(q)=a_{Mq+r}$ for $0\\leq r<M$.\nIf $A(z)=\\sum_n a_nz^n$ and $A_r(w)=\\sum_q a_{Mq+r}w^q$ are ordinary generating\nfunctions, then unique quotient and remainder give\n\\[\nA(z)=\\sum_{r=0}^{M-1}z^rA_r(z^M),\\qquad\n\\bigl(a_r^{[M]}\\bigr)_s^{[N]}=a_{r+Ms}^{[MN]}.\n\\]\nFor ordinary Cauchy multiplication, the phase-$r$ product has a term without a\ncarry and a term with one carry:\n\\[\nC_r(w)=\\sum_{u+v=r}A_u(w)B_v(w)\n+w\\sum_{u+v=r+M}A_u(w)B_v(w),\\qquad 0\\leq u,v<M.\n\\]\nAdding the two phase addresses gives the formula. It describes ordinary\nCauchy multiplication; Hurwitz convolution has its own binomial weights.","a$ and $M\\geq1$, let $a_r^{[M]}(q)=a_{Mq+r}$ for $0\\leq r<M$.\nIf $A(z)=\\sum_n a_nz^n$ and $A_r(w)=\\sum_q a_{Mq+r}w^q$ are ordinary generating\nfunctions, then unique quotient and remainder give\n\\[\nA(z)=\\sum_{r=0}^{M-1}z^rA_r(z^M),\\qquad\n\\bigl(a_r^{[M]}\\bigr)_s^{[N]}=a_{r+Ms}^{[MN]}.\n\\]\nFor ordinary Cauchy multiplication, the phase-$r$ product has a term without a\ncarry and a term with one carry:\n\\[\nC_r(w)=\\sum_{u+v=r}A_u(w)B_v(w)\n+w\\sum_{u+v=r+M}A_u(w)B_v(w),\\qquad 0\\leq u,v<M.\n\\]\nAdding the two phase addresses gives the formula. It describes ordinary\nCauchy multiplication; Hurwitz convolution has its own binomial weights.","For $M\\geq1$ and $s,r,k\\in\\N$, define the affine Mahler kernel by\n\\[\n\\Theta_{M,s}(r,k)=[z^r](1+z)^s\\bigl((1+z)^M-1\\bigr)^k\n=\\sum_{j=0}^{k}(-1)^{k-j}\\binom kj\\binom{Mj+s}{r}.\n\\]\nIt vanishes unless $k\\leq r\\leq Mk+s$. For a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient ext","For $M\\geq1$ and $s,r,k\\in\\N$, define the affine Mahler kernel by\n\\[\n\\Theta_{M,s}(r,k)=[z^r](1+z)^s\\bigl((1+z)^M-1\\bigr)^k\n=\\sum_{j=0}^{k}(-1)^{k-j}\\binom kj\\binom{Mj+s}{r}.\n\\]\nIt vanishes unless $k\\leq r\\leq Mk+s$. For a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the bi","For $M\\geq1$ and $s,r,k\\in\\N$, define the affine Mahler kernel by\n\\[\n\\Theta_{M,s}(r,k)=[z^r](1+z)^s\\bigl((1+z)^M-1\\bigr)^k\n=\\sum_{j=0}^{k}(-1)^{k-j}\\binom kj\\binom{Mj+s}{r}.\n\\]\nIt vanishes unless $k\\leq r\\leq Mk+s$. For a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the binomial expansion of\n$((1+z)^M-1)^k$ gives the kernel. In the binary zero-address case,\n\\[\n\\Theta_{2,0}(k+j,k)=2^{k-j}\\binom kj\\qquad(0\\leq j\\leq k","For $M\\geq1$ and $s,r,k\\in\\N$, define the affine Mahler kernel by\n\\[\n\\Theta_{M,s}(r,k)=[z^r](1+z)^s\\bigl((1+z)^M-1\\bigr)^k\n=\\sum_{j=0}^{k}(-1)^{k-j}\\binom kj\\binom{Mj+s}{r}.\n\\]\nIt vanishes unless $k\\leq r\\leq Mk+s$. For a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the binomial expansion of\n$((1+z)^M-1)^k$ gives the kernel. In the binary zero-address case,\n\\[\n\\Theta_{2,0}(k+j,k)=2^{k-j}\\binom kj\\qquad(0\\leq j\\leq k),\n\\]\nsince $(2z+z^2)^k=z^k(2+z)^k$.\nEx","For $M\\geq1$ and $s,r,k\\in\\N$, define the affine Mahler kernel by\n\\[\n\\Theta_{M,s}(r,k)=[z^r](1+z)^s\\bigl((1+z)^M-1\\bigr)^k\n=\\sum_{j=0}^{k}(-1)^{k-j}\\binom kj\\binom{Mj+s}{r}.\n\\]\nIt vanishes unless $k\\leq r\\leq Mk+s$. For a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the binomial expansion of\n$((1+z)^M-1)^k$ gives the kernel. In the binary zero-address case,\n\\[\n\\Theta_{2,0}(k+j,k)=2^{k-j}\\binom kj\\qquad(0\\leq j\\leq k),\n\\]\nsince $(2z+z^2)^k=z^k(2+z)^k$.\nExact valuations for the general kernel remain to be determined. The","For $M\\geq1$ and $s,r,k\\in\\N$, define the affine Mahler kernel by\n\\[\n\\Theta_{M,s}(r,k)=[z^r](1+z)^s\\bigl((1+z)^M-1\\bigr)^k\n=\\sum_{j=0}^{k}(-1)^{k-j}\\binom kj\\binom{Mj+s}{r}.\n\\]\nIt vanishes unless $k\\leq r\\leq Mk+s$. For a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the binomial expansion of\n$((1+z)^M-1)^k$ gives the kernel. In the binary zero-address case,\n\\[\n\\Theta_{2,0}(k+j,k)=2^{k-j}\\binom kj\\qquad(0\\leq j\\leq k),\n\\]\nsince $(2z+z^2)^k=z^k(2+z)^k$.\nExact valuations for the general kernel remain to be determined. The\nrecovered","For $M\\geq1$ and $s,r,k\\in\\N$, define the affine Mahler kernel by\n\\[\n\\Theta_{M,s}(r,k)=[z^r](1+z)^s\\bigl((1+z)^M-1\\bigr)^k\n=\\sum_{j=0}^{k}(-1)^{k-j}\\binom kj\\binom{Mj+s}{r}.\n\\]\nIt vanishes unless $k\\leq r\\leq Mk+s$. For a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the binomial expansion of\n$((1+z)^M-1)^k$ gives the kernel. In the binary zero-address case,\n\\[\n\\Theta_{2,0}(k+j,k)=2^{k-j}\\binom kj\\qquad(0\\leq j\\leq k),\n\\]\nsince $(2z+z^2)^k=z^k(2+z)^k$.\nExact valuations for the general kernel remain to be determined. The\nrecovered infinite analytic extensions and stron","For $M\\geq1$ and $s,r,k\\in\\N$, define the affine Mahler kernel by\n\\[\n\\Theta_{M,s}(r,k)=[z^r](1+z)^s\\bigl((1+z)^M-1\\bigr)^k\n=\\sum_{j=0}^{k}(-1)^{k-j}\\binom kj\\binom{Mj+s}{r}.\n\\]\nIt vanishes unless $k\\leq r\\leq Mk+s$. For a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the binomial expansion of\n$((1+z)^M-1)^k$ gives the kernel. In the binary zero-address case,\n\\[\n\\Theta_{2,0}(k+j,k)=2^{k-j}\\binom kj\\qquad(0\\leq j\\leq k),\n\\]\nsince $(2z+z^2)^k=z^k(2+z)^k$.\nExact valuations for the general kernel remain to be determined. The\nrecovered infinite analytic extensions and stronger historical identities\nlie beyond these finite derivations; their sta","affine Mahler kernel by\n\\[\n\\Theta_{M,s}(r,k)=[z^r](1+z)^s\\bigl((1+z)^M-1\\bigr)^k\n=\\sum_{j=0}^{k}(-1)^{k-j}\\binom kj\\binom{Mj+s}{r}.\n\\]\nIt vanishes unless $k\\leq r\\leq Mk+s$. For a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the binomial expansion of\n$((1+z)^M-1)^k$ gives the kernel. In the binary zero-address case,\n\\[\n\\Theta_{2,0}(k+j,k)=2^{k-j}\\binom kj\\qquad(0\\leq j\\leq k),\n\\]\nsince $(2z+z^2)^k=z^k(2+z)^k$.\nExact valuations for the general kernel remain to be determined. The\nrecovered infinite analytic extensions and stronger historical identities\nlie beyond these finite derivations; their status is discussed in\nSection~\\ref{subsec:historical-scope}","binom kj\\binom{Mj+s}{r}.\n\\]\nIt vanishes unless $k\\leq r\\leq Mk+s$. For a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the binomial expansion of\n$((1+z)^M-1)^k$ gives the kernel. In the binary zero-address case,\n\\[\n\\Theta_{2,0}(k+j,k)=2^{k-j}\\binom kj\\qquad(0\\leq j\\leq k),\n\\]\nsince $(2z+z^2)^k=z^k(2+z)^k$.\nExact valuations for the general kernel remain to be determined. The\nrecovered infinite analytic extensions and stronger historical identities\nlie beyond these finite derivations; their status is discussed in\nSection~\\ref{subsec:historical-scope}.","or a finite binomial expansion\n$f(n)=\\sum_{r=0}^{D}c_r\\binom nr$, the $k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the binomial expansion of\n$((1+z)^M-1)^k$ gives the kernel. In the binary zero-address case,\n\\[\n\\Theta_{2,0}(k+j,k)=2^{k-j}\\binom kj\\qquad(0\\leq j\\leq k),\n\\]\nsince $(2z+z^2)^k=z^k(2+z)^k$.\nExact valuations for the general kernel remain to be determined. The\nrecovered infinite analytic extensions and stronger historical identities\nlie beyond these finite derivations; their status is discussed in\nSection~\\ref{subsec:historical-scope}.","$k$th Newton coefficient of\n$g(q)=f(Mq+s)$ is therefore\n\\[\nc_k(g)=\\sum_{r=k}^{\\min(D,Mk+s)}c_r\\Theta_{M,s}(r,k).\n\\]\nIndeed, insert the finite stencil for $\\Delta_1^k g(0)$ and interchange its\ntwo finite sums. Coefficient extraction of the binomial expansion of\n$((1+z)^M-1)^k$ gives the kernel. In the binary zero-address case,\n\\[\n\\Theta_{2,0}(k+j,k)=2^{k-j}\\binom kj\\qquad(0\\leq j\\leq k),\n\\]\nsince $(2z+z^2)^k=z^k(2+z)^k$.\nExact valuations for the general kernel remain to be determined. The\nrecovered infinite analytic extensions and stronger historical identities\nlie beyond these finite derivations; their status is discussed in\nSection~\\ref{subsec:historical-scope}.","Two related constructions run through this section.  The congruential\nEuler family is indexed by a modulus $N$ and an address $j$.  Its\nexceptional dyadic member has $N=4$ and four addresses.  The classical\nsigned Euler numbers arise from $N=2$, $j=0$.  Keeping these indices\nvisible distinguishes the four-packet period theorems from the classical\nedge truncations.","\\begin{definition}[Congruential Euler coefficients]\\label{def:arith-euler}\nFor $N\\ge1$ and $0\\le j<N$, define rational coefficients by the formal identity\n\\begin{equation}\\label{eq:arith-euler-egf}\n \\sum_{d\\ge0}E_d^{(N,j)}\\frac{t^d}{d!}\n =\\left(\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!}\\right)^{-1}.\n\\end{equation}\nThe reciprocal exists because its denominator has constant coefficient\n$1/j!\\ne0$.  Set\n\\[\n U_{N,j}(n)=E_{Nn}^{(N,j)},\\qquad C_{N,j}(n)=\\frac{U_{N,j}(n)}{j!}.\n\\]\nThus $C_{N,j}(0)=1$ and $U_{N,j}(0)=j!$.  All coefficients outside the\nmultiples of $N$ vanish.\n\\end{definition}","or $N\\ge1$ and $0\\le j<N$, define rational coefficients by the formal identity\n\\begin{equation}\\label{eq:arith-euler-egf}\n \\sum_{d\\ge0}E_d^{(N,j)}\\frac{t^d}{d!}\n =\\left(\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!}\\right)^{-1}.\n\\end{equation}\nThe reciprocal exists because its denominator has constant coefficient\n$1/j!\\ne0$.  Set\n\\[\n U_{N,j}(n)=E_{Nn}^{(N,j)},\\qquad C_{N,j}(n)=\\frac{U_{N,j}(n)}{j!}.\n\\]\nThus $C_{N,j}(0)=1$ and $U_{N,j}(0)=j!$.  All coefficients outside the\nmultiples of $N$ vanish.\n\\end{definition}","e j<N$, define rational coefficients by the formal identity\n\\begin{equation}\\label{eq:arith-euler-egf}\n \\sum_{d\\ge0}E_d^{(N,j)}\\frac{t^d}{d!}\n =\\left(\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!}\\right)^{-1}.\n\\end{equation}\nThe reciprocal exists because its denominator has constant coefficient\n$1/j!\\ne0$.  Set\n\\[\n U_{N,j}(n)=E_{Nn}^{(N,j)},\\qquad C_{N,j}(n)=\\frac{U_{N,j}(n)}{j!}.\n\\]\nThus $C_{N,j}(0)=1$ and $U_{N,j}(0)=j!$.  All coefficients outside the\nmultiples of $N$ vanish.\n\\end{definition}","\\begin{equation}\\label{eq:arith-euler-egf}\n \\sum_{d\\ge0}E_d^{(N,j)}\\frac{t^d}{d!}\n =\\left(\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!}\\right)^{-1}.\n\\end{equation}\nThe reciprocal exists because its denominator has constant coefficient\n$1/j!\\ne0$.  Set\n\\[\n U_{N,j}(n)=E_{Nn}^{(N,j)},\\qquad C_{N,j}(n)=\\frac{U_{N,j}(n)}{j!}.\n\\]\nThus $C_{N,j}(0)=1$ and $U_{N,j}(0)=j!$.  All coefficients outside the\nmultiples of $N$ vanish.\n\\end{definition}","The normalized source and its inverse are\n\\begin{equation}\\label{eq:arith-kernel}\n K_{N,j}(t)=j!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n \\mathcal P_{N,j}(t)=\\sum_{n\\ge0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!}.\n\\end{equation}\nTheir product is one.  Clearing factorials in its coefficient of $t^{Nn}$\ngives the triangular system\n\\begin{equation}\\label{eq:arith-triangular}\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0}.\n\\end{equation}\nIn particular, for $n>0$,\n\\[\n C_{N,j}(n)=-\\binom{Nn+j}{Nn}^{-1}\n       \\sum_{q=0}^{n-1}\\binom{Nn+j}{Nq}C_{N,j}(q).\n\\]\nThe diagonal binomial coefficient is positive, so this recurrence determines\nthe sequence uniquely.  The Lean construction begins with this recurrence\nand derives the reciprocal identity from it.  The resulting sequence is\nthen identified with the degree-indexed coefficients of\n\\eqref{eq:arith-euler-egf}, so both descriptions refer to the same famil","The normalized source and its inverse are\n\\begin{equation}\\label{eq:arith-kernel}\n K_{N,j}(t)=j!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n \\mathcal P_{N,j}(t)=\\sum_{n\\ge0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!}.\n\\end{equation}\nTheir product is one.  Clearing factorials in its coefficient of $t^{Nn}$\ngives the triangular system\n\\begin{equation}\\label{eq:arith-triangular}\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0}.\n\\end{equation}\nIn particular, for $n>0$,\n\\[\n C_{N,j}(n)=-\\binom{Nn+j}{Nn}^{-1}\n       \\sum_{q=0}^{n-1}\\binom{Nn+j}{Nq}C_{N,j}(q).\n\\]\nThe diagonal binomial coefficient is positive, so this recurrence determines\nthe sequence uniquely.  The Lean construction begins with this recurrence\nand derives the reciprocal identity from it.  The resulting sequence is\nthen identified with the degree-indexed coefficients of\n\\eqref{eq:arith-euler-egf}, so both descriptions refer to the same family.","ion}\\label{eq:arith-kernel}\n K_{N,j}(t)=j!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n \\mathcal P_{N,j}(t)=\\sum_{n\\ge0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!}.\n\\end{equation}\nTheir product is one.  Clearing factorials in its coefficient of $t^{Nn}$\ngives the triangular system\n\\begin{equation}\\label{eq:arith-triangular}\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0}.\n\\end{equation}\nIn particular, for $n>0$,\n\\[\n C_{N,j}(n)=-\\binom{Nn+j}{Nn}^{-1}\n       \\sum_{q=0}^{n-1}\\binom{Nn+j}{Nq}C_{N,j}(q).\n\\]\nThe diagonal binomial coefficient is positive, so this recurrence determines\nthe sequence uniquely.  The Lean construction begins with this recurrence\nand derives the reciprocal identity from it.  The resulting sequence is\nthen identified with the degree-indexed coefficients of\n\\eqref{eq:arith-euler-egf}, so both descriptions refer to the same family.","el{eq:arith-kernel}\n K_{N,j}(t)=j!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n \\mathcal P_{N,j}(t)=\\sum_{n\\ge0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!}.\n\\end{equation}\nTheir product is one.  Clearing factorials in its coefficient of $t^{Nn}$\ngives the triangular system\n\\begin{equation}\\label{eq:arith-triangular}\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0}.\n\\end{equation}\nIn particular, for $n>0$,\n\\[\n C_{N,j}(n)=-\\binom{Nn+j}{Nn}^{-1}\n       \\sum_{q=0}^{n-1}\\binom{Nn+j}{Nq}C_{N,j}(q).\n\\]\nThe diagonal binomial coefficient is positive, so this recurrence determines\nthe sequence uniquely.  The Lean construction begins with this recurrence\nand derives the reciprocal identity from it.  The resulting sequence is\nthen identified with the degree-indexed coefficients of\n\\eqref{eq:arith-euler-egf}, so both descriptions refer to the same family.","\\begin{proposition}[A denominator bound for every address]\n\\label{prop:arith-denominator}\nFor $N\\ge1$, $0\\le j<N$, and $n\\in\\N$, set\n\\[\n D_{N,j}(n)=\\prod_{r=1}^{n}\\binom{Nr+j}{Nr},\\qquad D_{N,j}(0)=1.\n\\]\nThen $D_{N,j}(n)C_{N,j}(n)\\in\\Z$.\n\\end{proposition}","\\begin{proof}\nProceed by induction in the recurrence \\eqref{eq:arith-triangular}.\nFor every $q<n$, the product $D_{N,j}(q)$ divides $D_{N,j}(n-1)$.\nMultiplying the recurrence by $D_{N,j}(n-1)$ therefore makes every\nprevious term integral.  The diagonal factor completes the product\n$D_{N,j}(n)$, proving the claim.  At $n=0$, both factors are one.\nThis gives a uniform denominator bound without requiring integrality of\nthe normalized coefficients themselves.  The Lean statements are\n\\lean{KomorebiPaperIX.Congruential.clearing_product} and\n\\lean{KomorebiPaperIX.Congruential.universal_denominator_clearing}.\n\\end{proof}","\\begin{example}[The Bernoulli boundary address]\\label{ex:arith-bernoulli}\nAt $N=2$, the address $j=1$ gives a second classical family.  Define the\nBernoulli numbers by\n\\[\n B(z)=\\frac{z}{e^z-1}=\\sum_{r\\ge0}B_r\\frac{z^r}{r!},\n \\qquad B_0=1,\\quad B_1=-\\frac12.\n\\]\nAll quotients here are regular formal series.  The identity\n\\[\n 2B(z)-B(2z)=\\frac{z}{\\sinh z}\n            =\\sum_{n\\ge0}C_{2,1}(n)\\frac{z^{2n}}{(2n)!}\n\\]\nfollows by putting the two fractions on a common denominator and using\n\\eqref{eq:arith-euler-egf}.  Comparing even coefficients yields\n\\[\n C_{2,1}(n)=(2-2^{2n})B_{2n}\\qquad(n\\in\\N).\n\\]\nThis classical derivation restores the boundary example from the original\nmanuscript.  No dedicated Lean bridge for this Bernoulli specialization\nwas identified in the preserved sour","\\begin{example}[The Bernoulli boundary address]\\label{ex:arith-bernoulli}\nAt $N=2$, the address $j=1$ gives a second classical family.  Define the\nBernoulli numbers by\n\\[\n B(z)=\\frac{z}{e^z-1}=\\sum_{r\\ge0}B_r\\frac{z^r}{r!},\n \\qquad B_0=1,\\quad B_1=-\\frac12.\n\\]\nAll quotients here are regular formal series.  The identity\n\\[\n 2B(z)-B(2z)=\\frac{z}{\\sinh z}\n            =\\sum_{n\\ge0}C_{2,1}(n)\\frac{z^{2n}}{(2n)!}\n\\]\nfollows by putting the two fractions on a common denominator and using\n\\eqref{eq:arith-euler-egf}.  Comparing even coefficients yields\n\\[\n C_{2,1}(n)=(2-2^{2n})B_{2n}\\qquad(n\\in\\N).\n\\]\nThis classical derivation restores the boundary example from the original\nmanuscript.  No dedicated Lean bridge for this Bernoulli specialization\nwas identified in the preserved sources.\n\\end{example}","oulli numbers by\n\\[\n B(z)=\\frac{z}{e^z-1}=\\sum_{r\\ge0}B_r\\frac{z^r}{r!},\n \\qquad B_0=1,\\quad B_1=-\\frac12.\n\\]\nAll quotients here are regular formal series.  The identity\n\\[\n 2B(z)-B(2z)=\\frac{z}{\\sinh z}\n            =\\sum_{n\\ge0}C_{2,1}(n)\\frac{z^{2n}}{(2n)!}\n\\]\nfollows by putting the two fractions on a common denominator and using\n\\eqref{eq:arith-euler-egf}.  Comparing even coefficients yields\n\\[\n C_{2,1}(n)=(2-2^{2n})B_{2n}\\qquad(n\\in\\N).\n\\]\nThis classical derivation restores the boundary example from the original\nmanuscript.  No dedicated Lean bridge for this Bernoulli specialization\nwas identified in the preserved sources.\n\\end{example}","\\begin{definition}[Differences and valuation conventions]\\label{def:arith-precision}\nFor a sequence $f$ and $h,k,n\\in\\N$, put\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad\n \\Delta_h^0 f(n)=f(n),\\qquad\n \\Delta_h^{k+1}f(n)=\\Delta_h^k f(n+h)-\\Delta_h^k f(n).\n\\]\nWe use the extended valuation $v_p(0)=+\\infty$ in mathematical statements.\nThus $v_p(x)\\ge L$, including negative integers $L$, means\n$x=0$ or the ordinary valuation of the nonzero rational $x$ is at least $L$.\nFor integral coefficients and $L\\ge0$, it is equivalent to\n$x\\in p^L\\Z_p$.  In expressions involving a step multiplier we write\n$\\nu_p(u)=v_p(u)$ for $u>0$ and set $\\nu_p(0)=0$.  This finite convention\nmakes the displayed exponents meaningful at step zero, where all\npositive-order differences vanish.\n\\end{definition}","ion conventions]\\label{def:arith-precision}\nFor a sequence $f$ and $h,k,n\\in\\N$, put\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad\n \\Delta_h^0 f(n)=f(n),\\qquad\n \\Delta_h^{k+1}f(n)=\\Delta_h^k f(n+h)-\\Delta_h^k f(n).\n\\]\nWe use the extended valuation $v_p(0)=+\\infty$ in mathematical statements.\nThus $v_p(x)\\ge L$, including negative integers $L$, means\n$x=0$ or the ordinary valuation of the nonzero rational $x$ is at least $L$.\nFor integral coefficients and $L\\ge0$, it is equivalent to\n$x\\in p^L\\Z_p$.  In expressions involving a step multiplier we write\n$\\nu_p(u)=v_p(u)$ for $u>0$ and set $\\nu_p(0)=0$.  This finite convention\nmakes the displayed exponents meaningful at step zero, where all\npositive-order differences vanish.\n\\end{definition}","\\label{def:arith-precision}\nFor a sequence $f$ and $h,k,n\\in\\N$, put\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad\n \\Delta_h^0 f(n)=f(n),\\qquad\n \\Delta_h^{k+1}f(n)=\\Delta_h^k f(n+h)-\\Delta_h^k f(n).\n\\]\nWe use the extended valuation $v_p(0)=+\\infty$ in mathematical statements.\nThus $v_p(x)\\ge L$, including negative integers $L$, means\n$x=0$ or the ordinary valuation of the nonzero rational $x$ is at least $L$.\nFor integral coefficients and $L\\ge0$, it is equivalent to\n$x\\in p^L\\Z_p$.  In expressions involving a step multiplier we write\n$\\nu_p(u)=v_p(u)$ for $u>0$ and set $\\nu_p(0)=0$.  This finite convention\nmakes the displayed exponents meaningful at step zero, where all\npositive-order differences vanish.\n\\end{definition}","equence $f$ and $h,k,n\\in\\N$, put\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad\n \\Delta_h^0 f(n)=f(n),\\qquad\n \\Delta_h^{k+1}f(n)=\\Delta_h^k f(n+h)-\\Delta_h^k f(n).\n\\]\nWe use the extended valuation $v_p(0)=+\\infty$ in mathematical statements.\nThus $v_p(x)\\ge L$, including negative integers $L$, means\n$x=0$ or the ordinary valuation of the nonzero rational $x$ is at least $L$.\nFor integral coefficients and $L\\ge0$, it is equivalent to\n$x\\in p^L\\Z_p$.  In expressions involving a step multiplier we write\n$\\nu_p(u)=v_p(u)$ for $u>0$ and set $\\nu_p(0)=0$.  This finite convention\nmakes the displayed exponents meaningful at step zero, where all\npositive-order differences vanish.\n\\end{definition}","put\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad\n \\Delta_h^0 f(n)=f(n),\\qquad\n \\Delta_h^{k+1}f(n)=\\Delta_h^k f(n+h)-\\Delta_h^k f(n).\n\\]\nWe use the extended valuation $v_p(0)=+\\infty$ in mathematical statements.\nThus $v_p(x)\\ge L$, including negative integers $L$, means\n$x=0$ or the ordinary valuation of the nonzero rational $x$ is at least $L$.\nFor integral coefficients and $L\\ge0$, it is equivalent to\n$x\\in p^L\\Z_p$.  In expressions involving a step multiplier we write\n$\\nu_p(u)=v_p(u)$ for $u>0$ and set $\\nu_p(0)=0$.  This finite convention\nmakes the displayed exponents meaningful at step zero, where all\npositive-order differences vanish.\n\\end{definition}","d\n \\Delta_h^{k+1}f(n)=\\Delta_h^k f(n+h)-\\Delta_h^k f(n).\n\\]\nWe use the extended valuation $v_p(0)=+\\infty$ in mathematical statements.\nThus $v_p(x)\\ge L$, including negative integers $L$, means\n$x=0$ or the ordinary valuation of the nonzero rational $x$ is at least $L$.\nFor integral coefficients and $L\\ge0$, it is equivalent to\n$x\\in p^L\\Z_p$.  In expressions involving a step multiplier we write\n$\\nu_p(u)=v_p(u)$ for $u>0$ and set $\\nu_p(0)=0$.  This finite convention\nmakes the displayed exponents meaningful at step zero, where all\npositive-order differences vanish.\n\\end{definition}","\\Delta_h^k f(n+h)-\\Delta_h^k f(n).\n\\]\nWe use the extended valuation $v_p(0)=+\\infty$ in mathematical statements.\nThus $v_p(x)\\ge L$, including negative integers $L$, means\n$x=0$ or the ordinary valuation of the nonzero rational $x$ is at least $L$.\nFor integral coefficients and $L\\ge0$, it is equivalent to\n$x\\in p^L\\Z_p$.  In expressions involving a step multiplier we write\n$\\nu_p(u)=v_p(u)$ for $u>0$ and set $\\nu_p(0)=0$.  This finite convention\nmakes the displayed exponents meaningful at step zero, where all\npositive-order differences vanish.\n\\end{definition}","+h)-\\Delta_h^k f(n).\n\\]\nWe use the extended valuation $v_p(0)=+\\infty$ in mathematical statements.\nThus $v_p(x)\\ge L$, including negative integers $L$, means\n$x=0$ or the ordinary valuation of the nonzero rational $x$ is at least $L$.\nFor integral coefficients and $L\\ge0$, it is equivalent to\n$x\\in p^L\\Z_p$.  In expressions involving a step multiplier we write\n$\\nu_p(u)=v_p(u)$ for $u>0$ and set $\\nu_p(0)=0$.  This finite convention\nmakes the displayed exponents meaningful at step zero, where all\npositive-order differences vanish.\n\\end{definition}","The valuation convention at zero is encoded by\n\\lean{KomorebiPaperIX.GlobalEulerBoundary.precision}.  For $p$-adic\ncongruences, divisibility is interpreted in $\\Z_p$; statements over other\ncoefficient rings specify the ring explicitly.  Field divisibility in\n$\\Q_p$ alone does not encode precision.","\\begin{proposition}[Lucas reduction and address independence]\n\\label{prop:arith-lucas-window}\nLet $p$ be prime, $m>0$, and $N=mp$.  For $0\\le j<p$ and $n,q\\in\\N$,\n\\[\n \\binom{mpn+j}{mpq}\\equiv\\binom{mn}{mq}\\pmod p.\n\\]\nEvery $C_{mp,j}(n)$ belongs to $\\Z_p$.  Its reduction is the unique\nsequence $\\overline C_m:\\N\\to\\F_p$ satisfying\n\\[\n \\sum_{q=0}^{n}\\binom{mn}{mq}\\overline C_m(q)=\\delta_{n0}\n \\quad\\text{in }\\F_p.\n\\]\nConsequently, for any two addresses $0\\le j,k<p$,\n\\[\n C_{mp,j}(n)\\equiv C_{mp,k}(n)\\pmod p\\qquad(n\\in\\N).\n\\]\n\\end{proposition}","ndence]\n\\label{prop:arith-lucas-window}\nLet $p$ be prime, $m>0$, and $N=mp$.  For $0\\le j<p$ and $n,q\\in\\N$,\n\\[\n \\binom{mpn+j}{mpq}\\equiv\\binom{mn}{mq}\\pmod p.\n\\]\nEvery $C_{mp,j}(n)$ belongs to $\\Z_p$.  Its reduction is the unique\nsequence $\\overline C_m:\\N\\to\\F_p$ satisfying\n\\[\n \\sum_{q=0}^{n}\\binom{mn}{mq}\\overline C_m(q)=\\delta_{n0}\n \\quad\\text{in }\\F_p.\n\\]\nConsequently, for any two addresses $0\\le j,k<p$,\n\\[\n C_{mp,j}(n)\\equiv C_{mp,k}(n)\\pmod p\\qquad(n\\in\\N).\n\\]\n\\end{proposition}","prop:arith-lucas-window}\nLet $p$ be prime, $m>0$, and $N=mp$.  For $0\\le j<p$ and $n,q\\in\\N$,\n\\[\n \\binom{mpn+j}{mpq}\\equiv\\binom{mn}{mq}\\pmod p.\n\\]\nEvery $C_{mp,j}(n)$ belongs to $\\Z_p$.  Its reduction is the unique\nsequence $\\overline C_m:\\N\\to\\F_p$ satisfying\n\\[\n \\sum_{q=0}^{n}\\binom{mn}{mq}\\overline C_m(q)=\\delta_{n0}\n \\quad\\text{in }\\F_p.\n\\]\nConsequently, for any two addresses $0\\le j,k<p$,\n\\[\n C_{mp,j}(n)\\equiv C_{mp,k}(n)\\pmod p\\qquad(n\\in\\N).\n\\]\n\\end{proposition}","\\begin{proof}\nLucas reduction separates the lowest base-$p$ digits, which are $j$ and\nzero in the upper and lower arguments.  Their binomial contribution is\none, leaving the displayed reduced row.  On the diagonal the reduced\ncoefficient is one.  The recurrence \\eqref{eq:arith-triangular} therefore\nconstructs an integral packet by successive division by units in $\\Z_p$.\nAfter reduction, every address satisfies the same triangular system over\n$\\F_p$.  Its diagonal entries are one, so its solution is unique.  This\nproves both the common reduction and the congruence between addresses.\nThe row reduction is\n\\lean{KomorebiPaperIX.StandardWindow.lucas_row_reduction}.\nThe integral identification and the two reduction statements are\n\\lean{KomorebiPaperIX.Wi","\\begin{proof}\nLucas reduction separates the lowest base-$p$ digits, which are $j$ and\nzero in the upper and lower arguments.  Their binomial contribution is\none, leaving the displayed reduced row.  On the diagonal the reduced\ncoefficient is one.  The recurrence \\eqref{eq:arith-triangular} therefore\nconstructs an integral packet by successive division by units in $\\Z_p$.\nAfter reduction, every address satisfies the same triangular system over\n$\\F_p$.  Its diagonal entries are one, so its solution is unique.  This\nproves both the common reduction and the congruence between addresses.\nThe row reduction is\n\\lean{KomorebiPaperIX.StandardWindow.lucas_row_reduction}.\nThe integral identification and the two reduction statements are\n\\lean{KomorebiPaperIX.WindowIndependence.stand","\\begin{proof}\nLucas reduction separates the lowest base-$p$ digits, which are $j$ and\nzero in the upper and lower arguments.  Their binomial contribution is\none, leaving the displayed reduced row.  On the diagonal the reduced\ncoefficient is one.  The recurrence \\eqref{eq:arith-triangular} therefore\nconstructs an integral packet by successive division by units in $\\Z_p$.\nAfter reduction, every address satisfies the same triangular system over\n$\\F_p$.  Its diagonal entries are one, so its solution is unique.  This\nproves both the common reduction and the congruence between addresses.\nThe row reduction is\n\\lean{KomorebiPaperIX.StandardWindow.lucas_row_reduction}.\nThe integral identification and the two reduction statements are\n\\lean{KomorebiPaperIX.WindowIndependence.standard_window_integral},\n\\lean{KomorebiPaperIX.WindowIndependence.coeffPadic_coe},\n\\lean{KomorebiPaperIX.WindowIndependence.residue_independence}, and\n\\lean{KomorebiPaperIX.WindowIndependence.two_window_addresses_congruent}.\n\\end{proof}","est base-$p$ digits, which are $j$ and\nzero in the upper and lower arguments.  Their binomial contribution is\none, leaving the displayed reduced row.  On the diagonal the reduced\ncoefficient is one.  The recurrence \\eqref{eq:arith-triangular} therefore\nconstructs an integral packet by successive division by units in $\\Z_p$.\nAfter reduction, every address satisfies the same triangular system over\n$\\F_p$.  Its diagonal entries are one, so its solution is unique.  This\nproves both the common reduction and the congruence between addresses.\nThe row reduction is\n\\lean{KomorebiPaperIX.StandardWindow.lucas_row_reduction}.\nThe integral identification and the two reduction statements are\n\\lean{KomorebiPaperIX.WindowIndependence.standard_window_integral},\n\\lean{KomorebiPaperIX.WindowIndependence.coeffPadic_coe},\n\\lean{KomorebiPaperIX.WindowIndependence.residue_independence}, and\n\\lean{KomorebiPaperIX.WindowIndependence.two_window_addresses_congruent}.\n\\end{proof}","\\begin{theorem}[Nishibuchi congruence]\\label{thm:arith-nishibuchi}\nLet $p$ be prime, let $m>0$, and suppose that $m\\mid p-1$ or $m=2$.\nPut $N=mp$ and $q=\\operatorname{lcm}(2,p-1)$.  For every $0\\le j<N$,\nall $U_{N,j}(n)$ belong to $\\Z_p$.  For $r\\ge1$ and $n\\ge r$,\n\\begin{equation}\\label{eq:arith-nishibuchi}\n E_{Nn+qp^r}^{(N,j)}\\equiv E_{Nn}^{(N,j)}\\pmod{p^r}.\n\\end{equation}\nThus the positive sufficient starting index $n_0=r$ works uniformly.\nIf $m\\mid p-1$, the stronger sufficient condition $r\\le mn$ is valid.\n\\end{theorem}","chi congruence]\\label{thm:arith-nishibuchi}\nLet $p$ be prime, let $m>0$, and suppose that $m\\mid p-1$ or $m=2$.\nPut $N=mp$ and $q=\\operatorname{lcm}(2,p-1)$.  For every $0\\le j<N$,\nall $U_{N,j}(n)$ belong to $\\Z_p$.  For $r\\ge1$ and $n\\ge r$,\n\\begin{equation}\\label{eq:arith-nishibuchi}\n E_{Nn+qp^r}^{(N,j)}\\equiv E_{Nn}^{(N,j)}\\pmod{p^r}.\n\\end{equation}\nThus the positive sufficient starting index $n_0=r$ works uniformly.\nIf $m\\mid p-1$, the stronger sufficient condition $r\\le mn$ is valid.\n\\end{theorem}","ith-nishibuchi}\nLet $p$ be prime, let $m>0$, and suppose that $m\\mid p-1$ or $m=2$.\nPut $N=mp$ and $q=\\operatorname{lcm}(2,p-1)$.  For every $0\\le j<N$,\nall $U_{N,j}(n)$ belong to $\\Z_p$.  For $r\\ge1$ and $n\\ge r$,\n\\begin{equation}\\label{eq:arith-nishibuchi}\n E_{Nn+qp^r}^{(N,j)}\\equiv E_{Nn}^{(N,j)}\\pmod{p^r}.\n\\end{equation}\nThus the positive sufficient starting index $n_0=r$ works uniformly.\nIf $m\\mid p-1$, the stronger sufficient condition $r\\le mn$ is valid.\n\\end{theorem}","d suppose that $m\\mid p-1$ or $m=2$.\nPut $N=mp$ and $q=\\operatorname{lcm}(2,p-1)$.  For every $0\\le j<N$,\nall $U_{N,j}(n)$ belong to $\\Z_p$.  For $r\\ge1$ and $n\\ge r$,\n\\begin{equation}\\label{eq:arith-nishibuchi}\n E_{Nn+qp^r}^{(N,j)}\\equiv E_{Nn}^{(N,j)}\\pmod{p^r}.\n\\end{equation}\nThus the positive sufficient starting index $n_0=r$ works uniformly.\nIf $m\\mid p-1$, the stronger sufficient condition $r\\le mn$ is valid.\n\\end{theorem}","\\begin{theorem}[All-address differences on the divisor branch]\n\\label{thm:arith-divisor-base}\nLet $p$ be prime, $m>0$, $m\\mid p-1$, $N=mp$, and $0\\le j<N$.\nWrite $h=(p-1)/m$.  For every $u,k,n\\in\\N$,\n\\begin{align}\n U_{N,j}(n)&\\in\\Z_p,\\label{eq:arith-divisor-integral}","\\begin{theorem}[All-address differences on the divisor branch]\n\\label{thm:arith-divisor-base}\nLet $p$ be prime, $m>0$, $m\\mid p-1$, $N=mp$, and $0\\le j<N$.\nWrite $h=(p-1)/m$.  For every $u,k,n\\in\\N$,\n\\begin{align}\n U_{N,j}(n)&\\in\\Z_p,\\label{eq:arith-divisor-integral}\\\\\n \\Delta_{uh}^{k}U_{N,j}(n)&\\in p^{\\min(k,mn)}\\Z_p,\\label{eq:arith-divisor-bound}\\\\\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\n   &\\ge\\min(k,mn)-v_p(j!).\\label{eq:arith-divisor-normalized}\n\\end{align}\nThe factorial correction is part of the normalized statement, including\naddresses $j\\ge p$.\n\\end{theorem}","v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\n   &\\ge\\min(k,mn)-v_p(j!).\\label{eq:arith-divisor-normalized}\n\\end{align}\nThe factorial correction is part of the normalized statement, including\naddresses $j\\ge p$.\n\\end{theorem}","\\begin{proof}\nWork in the Hurwitz integral ring of formal series\n\\[\n \\mathcal H_p=\\left\\{F\\in\\Q_p[[t]]:\n            d![t^d]F\\in\\Z_p\\text{ for every }d\\right\\}.\n\\]\nFor $D=d/dt$, let $B_p=(D^{p^2}-D^p)/p$.  Call $F$ flat if\n$B_p^sF\\in\\mathcal H_p$ for every $s\\ge0$.\nWe will prove that $j!\\mathcal P_{N,j}$ is flat at every address.  The main\ndifficulty is to include the addresses at which the usual unit-pivot\nargument fails.","Set\n\\[\n H_c(t)=\\sum_{n\\ge0}\\frac{t^{Nn+c}}{(Nn+c)!},\\qquad\n A_c(t)=\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+c)!},\\qquad\n R_c(t)=\\frac{A_{c-1}(t)}{A_c(t)}\\quad(1\\le c<N).\n\\]\nThe denominator of this last quotient has nonzero constant coefficient.\nThe section identities give\n\\[\n R_cH_c=tH_{c-1},\\qquad R_c(0)=c,\\qquad\n j!\\mathcal P_{N,j}=\\mathcal P_{N,0}\\prod_{c=1}^{j}R_c.\n\\]\nCancellation of consecutive source sections proves the product identity\nand ties the ratios directly to the family under study.","At the singular addresses $c=ap$, $1\\le a<m$, use the weighted ring\n\\[\n \\mathcal W_p=\\left\\{\\sum_{d\\ge0}a_dT^d:\n          p^{\\lfloor d/p\\rfloor}a_d\\in\\Z_p\\right\\}.\n\\]\nThe key estimate is $R_{ap}(\\log(1+T))\\in\\mathcal W_p$.  Its arithmetic\ninput comes from the cyclic group ring $\\Z[\\Z/N\\Z]$.\nFor its canonical element $x$ and $I=(p,x^{p-1}-1)$, the relation\n$(x^{p-1})^p=1$ implies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the","At the singular addresses $c=ap$, $1\\le a<m$, use the weighted ring\n\\[\n \\mathcal W_p=\\left\\{\\sum_{d\\ge0}a_dT^d:\n          p^{\\lfloor d/p\\rfloor}a_d\\in\\Z_p\\right\\}.\n\\]\nThe key estimate is $R_{ap}(\\log(1+T))\\in\\mathcal W_p$.  Its arithmetic\ninput comes from the cyclic group ring $\\Z[\\Z/N\\Z]$.\nFor its canonical element $x$ and $I=(p,x^{p-1}-1)$, the relation\n$(x^{p-1})^p=1$ implies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the weight in $\\","At the singular addresses $c=ap$, $1\\le a<m$, use the weighted ring\n\\[\n \\mathcal W_p=\\left\\{\\sum_{d\\ge0}a_dT^d:\n          p^{\\lfloor d/p\\rfloor}a_d\\in\\Z_p\\right\\}.\n\\]\nThe key estimate is $R_{ap}(\\log(1+T))\\in\\mathcal W_p$.  Its arithmetic\ninput comes from the cyclic group ring $\\Z[\\Z/N\\Z]$.\nFor its canonical element $x$ and $I=(p,x^{p-1}-1)$, the relation\n$(x^{p-1})^p=1$ implies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the weight in $\\mathcal W_p$ after dividing by $d!$.\nComparison of the section and group-ring recurrences identifies these\ncoordinates wi","At the singular addresses $c=ap$, $1\\le a<m$, use the weighted ring\n\\[\n \\mathcal W_p=\\left\\{\\sum_{d\\ge0}a_dT^d:\n          p^{\\lfloor d/p\\rfloor}a_d\\in\\Z_p\\right\\}.\n\\]\nThe key estimate is $R_{ap}(\\log(1+T))\\in\\mathcal W_p$.  Its arithmetic\ninput comes from the cyclic group ring $\\Z[\\Z/N\\Z]$.\nFor its canonical element $x$ and $I=(p,x^{p-1}-1)$, the relation\n$(x^{p-1})^p=1$ implies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the weight in $\\mathcal W_p$ after dividing by $d!$.\nComparison of the section and group-ring recurrences identifies these\ncoordinates with the logarithmically substituted section.","sses $c=ap$, $1\\le a<m$, use the weighted ring\n\\[\n \\mathcal W_p=\\left\\{\\sum_{d\\ge0}a_dT^d:\n          p^{\\lfloor d/p\\rfloor}a_d\\in\\Z_p\\right\\}.\n\\]\nThe key estimate is $R_{ap}(\\log(1+T))\\in\\mathcal W_p$.  Its arithmetic\ninput comes from the cyclic group ring $\\Z[\\Z/N\\Z]$.\nFor its canonical element $x$ and $I=(p,x^{p-1}-1)$, the relation\n$(x^{p-1})^p=1$ implies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the weight in $\\mathcal W_p$ after dividing by $d!$.\nComparison of the section and group-ring recurrences identifies these\ncoordinates with the logarithmically substituted section.","_p=\\left\\{\\sum_{d\\ge0}a_dT^d:\n          p^{\\lfloor d/p\\rfloor}a_d\\in\\Z_p\\right\\}.\n\\]\nThe key estimate is $R_{ap}(\\log(1+T))\\in\\mathcal W_p$.  Its arithmetic\ninput comes from the cyclic group ring $\\Z[\\Z/N\\Z]$.\nFor its canonical element $x$ and $I=(p,x^{p-1}-1)$, the relation\n$(x^{p-1})^p=1$ implies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the weight in $\\mathcal W_p$ after dividing by $d!$.\nComparison of the section and group-ring recurrences identifies these\ncoordinates with the logarithmically substituted section.","\\in\\Z_p\\right\\}.\n\\]\nThe key estimate is $R_{ap}(\\log(1+T))\\in\\mathcal W_p$.  Its arithmetic\ninput comes from the cyclic group ring $\\Z[\\Z/N\\Z]$.\nFor its canonical element $x$ and $I=(p,x^{p-1}-1)$, the relation\n$(x^{p-1})^p=1$ implies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the weight in $\\mathcal W_p$ after dividing by $d!$.\nComparison of the section and group-ring recurrences identifies these\ncoordinates with the logarithmically substituted section.","1+T))\\in\\mathcal W_p$.  Its arithmetic\ninput comes from the cyclic group ring $\\Z[\\Z/N\\Z]$.\nFor its canonical element $x$ and $I=(p,x^{p-1}-1)$, the relation\n$(x^{p-1})^p=1$ implies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the weight in $\\mathcal W_p$ after dividing by $d!$.\nComparison of the section and group-ring recurrences identifies these\ncoordinates with the logarithmically substituted section.","mes from the cyclic group ring $\\Z[\\Z/N\\Z]$.\nFor its canonical element $x$ and $I=(p,x^{p-1}-1)$, the relation\n$(x^{p-1})^p=1$ implies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the weight in $\\mathcal W_p$ after dividing by $d!$.\nComparison of the section and group-ring recurrences identifies these\ncoordinates with the logarithmically substituted section.","$, the relation\n$(x^{p-1})^p=1$ implies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the weight in $\\mathcal W_p$ after dividing by $d!$.\nComparison of the section and group-ring recurrences identifies these\ncoordinates with the logarithmically substituted section.","mplies $I^p\\subseteq(p)I$.  Induction yields\n$I^q\\subseteq(p^{v_p(q!)})$.  Grouping the falling factorial\n$(x)_d=\\prod_{i=0}^{d-1}(x-i)$ into consecutive blocks of length $p$\nthen gives an integral residue-coordinate factor\n$p^{v_p(\\lfloor d/p\\rfloor!)}$.  The identity\n\\[\n v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)\n\\]\naccounts for precisely the weight in $\\mathcal W_p$ after dividing by $d!$.\nComparison of the section and group-ring recurrences identifies these\ncoordinates with the logarithmically substituted section.","To pass from the section to the singular ratio, factor out its initial zero:\nfor $c=ap$, the section has the form\n\\[\n H_c(\\log(1+T))=\\frac{T^c}{c!}\\,V(T),\\qquad V(0)=1,\n \\qquad V\\in\\mathcal W_p.\n\\]\nThe weight of $c!$ compensates for the coefficient shift by $c$.\nThe weighted ring is closed under inversion of series with constant\ncoefficient one and under $L_T=\\log(1+T)(1+T)d/dT$.  With\n$\\ell(T)=\\log(1+T)/T$, interpreted by coefficient shifting, differentiation\nof the displayed factorization gives\n\\[\n R_c(\\log(1+T))=c(1+T)\\ell(T)+(L_TV)V^{-1}.\n\\]\nSince $p\\ell(T)\\in\\mathcal W_p$ and $p\\mid c$, both terms are weighted.\nThis is the singular estimate\n\\lean{KomorebiPaperIX.SingularRatioAmice.paper_singular_amice_estimate}.","the form\n\\[\n H_c(\\log(1+T))=\\frac{T^c}{c!}\\,V(T),\\qquad V(0)=1,\n \\qquad V\\in\\mathcal W_p.\n\\]\nThe weight of $c!$ compensates for the coefficient shift by $c$.\nThe weighted ring is closed under inversion of series with constant\ncoefficient one and under $L_T=\\log(1+T)(1+T)d/dT$.  With\n$\\ell(T)=\\log(1+T)/T$, interpreted by coefficient shifting, differentiation\nof the displayed factorization gives\n\\[\n R_c(\\log(1+T))=c(1+T)\\ell(T)+(L_TV)V^{-1}.\n\\]\nSince $p\\ell(T)\\in\\mathcal W_p$ and $p\\mid c$, both terms are weighted.\nThis is the singular estimate\n\\lean{KomorebiPaperIX.SingularRatioAmice.paper_singular_amice_estimate}.","d V(0)=1,\n \\qquad V\\in\\mathcal W_p.\n\\]\nThe weight of $c!$ compensates for the coefficient shift by $c$.\nThe weighted ring is closed under inversion of series with constant\ncoefficient one and under $L_T=\\log(1+T)(1+T)d/dT$.  With\n$\\ell(T)=\\log(1+T)/T$, interpreted by coefficient shifting, differentiation\nof the displayed factorization gives\n\\[\n R_c(\\log(1+T))=c(1+T)\\ell(T)+(L_TV)V^{-1}.\n\\]\nSince $p\\ell(T)\\in\\mathcal W_p$ and $p\\mid c$, both terms are weighted.\nThis is the singular estimate\n\\lean{KomorebiPaperIX.SingularRatioAmice.paper_singular_amice_estimate}.","W_p.\n\\]\nThe weight of $c!$ compensates for the coefficient shift by $c$.\nThe weighted ring is closed under inversion of series with constant\ncoefficient one and under $L_T=\\log(1+T)(1+T)d/dT$.  With\n$\\ell(T)=\\log(1+T)/T$, interpreted by coefficient shifting, differentiation\nof the displayed factorization gives\n\\[\n R_c(\\log(1+T))=c(1+T)\\ell(T)+(L_TV)V^{-1}.\n\\]\nSince $p\\ell(T)\\in\\mathcal W_p$ and $p\\mid c$, both terms are weighted.\nThis is the singular estimate\n\\lean{KomorebiPaperIX.SingularRatioAmice.paper_singular_amice_estimate}.","To deduce flatness from the weighted estimate, use\n$\\Theta=(1+T)d/dT$.  Both $\\Theta$ and\n$(\\Theta^{p^2}-\\Theta^p)/p$ preserve $\\mathcal W_p$.\nFor the first operator, its coefficient formula is\n$(d+1)a_{d+1}+da_d$; whenever the floor weight rises, the factor $d+1$\nprovides the missing $p$.  For the second operator, the proof uses its\nfull Leibniz correction and the scalar congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intert","To deduce flatness from the weighted estimate, use\n$\\Theta=(1+T)d/dT$.  Both $\\Theta$ and\n$(\\Theta^{p^2}-\\Theta^p)/p$ preserve $\\mathcal W_p$.\nFor the first operator, its coefficient formula is\n$(d+1)a_{d+1}+da_d$; whenever the floor weight rises, the factor $d+1$\nprovides the missing $p$.  For the second operator, the proof uses its\nfull Leibniz correction and the scalar congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $","To deduce flatness from the weighted estimate, use\n$\\Theta=(1+T)d/dT$.  Both $\\Theta$ and\n$(\\Theta^{p^2}-\\Theta^p)/p$ preserve $\\mathcal W_p$.\nFor the first operator, its coefficient formula is\n$(d+1)a_{d+1}+da_d$; whenever the floor weight rises, the factor $d+1$\nprovides the missing $p$.  For the second operator, the proof uses its\nfull Leibniz correction and the scalar congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$","To deduce flatness from the weighted estimate, use\n$\\Theta=(1+T)d/dT$.  Both $\\Theta$ and\n$(\\Theta^{p^2}-\\Theta^p)/p$ preserve $\\mathcal W_p$.\nFor the first operator, its coefficient formula is\n$(d+1)a_{d+1}+da_d$; whenever the floor weight rises, the factor $d+1$\nprovides the missing $p$.  For the second operator, the proof uses its\nfull Leibniz correction and the scalar congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, bec","To deduce flatness from the weighted estimate, use\n$\\Theta=(1+T)d/dT$.  Both $\\Theta$ and\n$(\\Theta^{p^2}-\\Theta^p)/p$ preserve $\\mathcal W_p$.\nFor the first operator, its coefficient formula is\n$(d+1)a_{d+1}+da_d$; whenever the floor weight rises, the factor $d+1$\nprovides the missing $p$.  For the second operator, the proof uses its\nfull Leibniz correction and the scalar congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular","To deduce flatness from the weighted estimate, use\n$\\Theta=(1+T)d/dT$.  Both $\\Theta$ and\n$(\\Theta^{p^2}-\\Theta^p)/p$ preserve $\\mathcal W_p$.\nFor the first operator, its coefficient formula is\n$(d+1)a_{d+1}+da_d$; whenever the floor weight rises, the factor $d+1$\nprovides the missing $p$.  For the second operator, the proof uses its\nfull Leibniz correction and the scalar congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular ratios are flat.  This transfer is proved in\n\\lean","To deduce flatness from the weighted estimate, use\n$\\Theta=(1+T)d/dT$.  Both $\\Theta$ and\n$(\\Theta^{p^2}-\\Theta^p)/p$ preserve $\\mathcal W_p$.\nFor the first operator, its coefficient formula is\n$(d+1)a_{d+1}+da_d$; whenever the floor weight rises, the factor $d+1$\nprovides the missing $p$.  For the second operator, the proof uses its\nfull Leibniz correction and the scalar congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular ratios are flat.  This transfer is proved in\n\\lean{KomorebiPaperIX.Weighted","To deduce flatness from the weighted estimate, use\n$\\Theta=(1+T)d/dT$.  Both $\\Theta$ and\n$(\\Theta^{p^2}-\\Theta^p)/p$ preserve $\\mathcal W_p$.\nFor the first operator, its coefficient formula is\n$(d+1)a_{d+1}+da_d$; whenever the floor weight rises, the factor $d+1$\nprovides the missing $p$.  For the second operator, the proof uses its\nfull Leibniz correction and the scalar congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular ratios are flat.  This transfer is proved in\n\\lean{KomorebiPaperIX.WeightedDworkTransfer.singular_ratio_flat}.","mate, use\n$\\Theta=(1+T)d/dT$.  Both $\\Theta$ and\n$(\\Theta^{p^2}-\\Theta^p)/p$ preserve $\\mathcal W_p$.\nFor the first operator, its coefficient formula is\n$(d+1)a_{d+1}+da_d$; whenever the floor weight rises, the factor $d+1$\nprovides the missing $p$.  For the second operator, the proof uses its\nfull Leibniz correction and the scalar congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular ratios are flat.  This transfer is proved in\n\\lean{KomorebiPaperIX.WeightedDworkTransfer.singular_ratio_flat}.","ection and the scalar congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular ratios are flat.  This transfer is proved in\n\\lean{KomorebiPaperIX.WeightedDworkTransfer.singular_ratio_flat}.","congruence\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular ratios are flat.  This transfer is proved in\n\\lean{KomorebiPaperIX.WeightedDworkTransfer.singular_ratio_flat}.","e\n$p^2\\mid b^{p^2}-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular ratios are flat.  This transfer is proved in\n\\lean{KomorebiPaperIX.WeightedDworkTransfer.singular_ratio_flat}.","-b^p$.  Weighted monomials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular ratios are flat.  This transfer is proved in\n\\lean{KomorebiPaperIX.WeightedDworkTransfer.singular_ratio_flat}.","omials factor as\n\\[\n a_dT^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular ratios are flat.  This transfer is proved in\n\\lean{KomorebiPaperIX.WeightedDworkTransfer.singular_ratio_flat}.","^d=(p^{\\lfloor d/p\\rfloor}a_d)\n          (T^p/p)^{\\lfloor d/p\\rfloor}T^{d\\bmod p}.\n\\]\nThe operator preserves these monomials and hence finite weighted\npolynomials.  Each fixed output coefficient depends on only finitely many\ninput coefficients, which extends the result to the formal series.\nSubstitution $T=e^t-1$ intertwines $D$ with $\\Theta$ and maps\n$\\mathcal W_p$ into $\\mathcal H_p$, because\n\\[\n d![t^d]F(e^t-1)=[T^0]\\Theta^dF.\n\\]\nIt follows that the singular ratios are flat.  This transfer is proved in\n\\lean{KomorebiPaperIX.WeightedDworkTransfer.singular_ratio_flat}.","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On th","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the d","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrali","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the in","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserve","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n S","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two ri","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its se","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the","At a nonsingular address $2\\le c<N$ with $p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteratio","$p\\nmid c$, differentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery…","fferentiation of the section\nidentities gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient o…","s gives\n\\[\n tR_c'=R_c(1+R_{c-1}-R_c).\n\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from…","\\]\nWriting $q=R_{c-1}-(c-1)$ and $y=R_c-c$, this becomes\n\\[\n (tD+c)y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  …","y=cq+yq-y^2.\n\\]\nOn degrees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  The argument is entirely formal.\nThis is the content of\n\\lean{Komor…","ees divisible by $p$, the inverse of $tD+c$ divides the degree-$d$\ncoefficient by the unit $d+c$.  Starting with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  The argument is entirely formal.\nThis is the content of\n\\lean{KomorebiPaperIX.RiccatiFlatP…","ing with $y_0=0$, form\n\\[\n y_{s+1}=(tD+c)^{-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  The argument is entirely formal.\nThis is the content of\n\\lean{KomorebiPaperIX.RiccatiFlatPropagation.riccati_solution_flat}.","-1}(cq+y_sq-y_s^2).\n\\]\nFlatness and support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  The argument is entirely formal.\nThis is the content of\n\\lean{KomorebiPaperIX.RiccatiFlatPropagation.riccati_solution_flat}.","support are preserved. Indeed, put $E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  The argument is entirely formal.\nThis is the content of\n\\lean{KomorebiPaperIX.RiccatiFlatPropagation.riccati_solution_flat}.","$E_c=tD+c$ and $C_p=pD^{p^2}-D^p$.\nThe commutator is $[B_p,E_c]=C_p$, and iteration gives\n\\[\n B_p^rE_cy=E_cB_p^ry+rC_pB_p^{r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  The argument is entirely formal.\nThis is the content of\n\\lean{KomorebiPaperIX.RiccatiFlatPropagation.riccati_solution_flat}.","r-1}y.\n\\]\nFor the inverse equation $E_cy=f$, induction on $r$ makes the\ncorrection term Hurwitz-integral. Both operators preserve support in\ndegrees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  The argument is entirely formal.\nThis is the content of\n\\lean{KomorebiPaperIX.RiccatiFlatPropagation.riccati_solution_flat}.","grees divisible by $p$. On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  The argument is entirely formal.\nThis is the content of\n\\lean{KomorebiPaperIX.RiccatiFlatPropagation.riccati_solution_flat}.",". On that support the degree multiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  The argument is entirely formal.\nThis is the content of\n\\lean{KomorebiPaperIX.RiccatiFlatPropagation.riccati_solution_flat}.","ltiplier $d+c$\nis a unit, so integrality of $E_cB_p^ry$ implies integrality of\n$B_p^ry$. Thus the inverse preserves full flatness, as well as support.\n Since the difference of two right\nsides factors as $(z-y)(q-z-y)$ and its second factor has zero constant\nterm, agreement with the solution gains one coefficient at each iteration.\nConsequently $[t^d]y_s=[t^d]y$ for $d\\le s$.\nEvery coefficient of $B_p^r y$ can therefore be read from a sufficiently\nlong finite iterate.  The argument is entirely formal.\nThis is the content of\n\\lean{KomorebiPaperIX.RiccatiFlatPropagation.riccati_solution_flat}.","The standard window supplies the initial ratio $R_1$.  The weighted\nestimate handles $p\\mid c$, and Riccati propagation handles $p\\nmid c$.\nFlatness is closed under products, so the exact factorization establishes\nflatness of $j!\\mathcal P_{N,j}$.  Extracting its supported Hurwitz coefficients\nand applying the proved saturation and step-multiple lemmas yields\n\\eqref{eq:arith-divisor-integral}--\\eqref{eq:arith-divisor-bound}.","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\D","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalize","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retai","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resul","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements a","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllRe","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiP","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalize","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, an","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.Al","Here is the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_preci","s the coefficient extraction explicitly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","ly. If $a_d=d![t^d]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","]F$,\nthen\n\\[\n d![t^d]B_p^kF=p^{-k}\\Delta_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","_{p(p-1)}^ka_{d+pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","pk}.\n\\]\nFor $F=j!\\mathcal P_{N,j}$ and $k\\le mn$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","n$, take $d=mpn-pk$.\nSupport in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","in multiples of $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","f $mp$ turns this into\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","o\n$p^{-k}\\Delta_h^kU_{N,j}(n)$, so flatness gives divisibility by $p^k$.\nFor $k>mn$, expand $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","d $\\Delta_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","a_h^{k-mn}$ applied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","pplied to\n$\\Delta_h^{mn}U_{N,j}(n)$; every shifted row is at least $n$, so\neach summand is divisible by $p^{mn}$. Finally\n$\\Delta_{uh}=((I+\\Delta_h)^u-I)$, whose $k$th power for $u>0$\nis an integral combination of powers of $\\Delta_h$ of order at least $k$.\nThe case $u=0$ gives zero for $k>0$, and order zero is the original\nintegral coefficient. This proves the bound for all $u$ and $k$.\nFinally $\\Delta_{uh}^kU_{N,j}=j!\\Delta_{uh}^kC_{N,j}$ gives\n\\eqref{eq:arith-divisor-normalized}, with the zero case retained.\nThe resulting statements are\n\\lean{KomorebiPaperIX.AllResidueEuler.all_residue_flat},\n\\lean{KomorebiPaperIX.AllResidueEuler.global_unnormalized_bound}, and\n\\lean{KomorebiPaperIX.AllResidueEuler.global_normalized_precision}.\n\\end{proof}","\\begin{example}[Why the factorial loss is necessary]\n\\label{ex:arith-normalization-loss}\nTake $p=3$, $m=2$, $N=6$, and $j=3$.  The recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exa","\\begin{example}[Why the factorial loss is necessary]\n\\label{ex:arith-normalization-loss}\nTake $p=3$, $m=2$, $N=6$, and $j=3$.  The recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exact valu","\\begin{example}[Why the factorial loss is necessary]\n\\label{ex:arith-normalization-loss}\nTake $p=3$, $m=2$, $N=6$, and $j=3$.  The recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exact values and","\\begin{example}[Why the factorial loss is necessary]\n\\label{ex:arith-normalization-loss}\nTake $p=3$, $m=2$, $N=6$, and $j=3$.  The recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exact values and the refutat","\\begin{example}[Why the factorial loss is necessary]\n\\label{ex:arith-normalization-loss}\nTake $p=3$, $m=2$, $N=6$, and $j=3$.  The recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exact values and the refutation of a universal bound without loss\nare formalized in\n\\lean{KomorebiPaperIX.Normalization.normalized_first},\n\\lean{KomorebiPaperIX.N","\\begin{example}[Why the factorial loss is necessary]\n\\label{ex:arith-normalization-loss}\nTake $p=3$, $m=2$, $N=6$, and $j=3$.  The recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exact values and the refutation of a universal bound without loss\nare formalized in\n\\lean{KomorebiPaperIX.Normalization.normalized_first},\n\\lean{KomorebiPaperIX.Normalization.normalized_secon","\\begin{example}[Why the factorial loss is necessary]\n\\label{ex:arith-normalization-loss}\nTake $p=3$, $m=2$, $N=6$, and $j=3$.  The recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exact values and the refutation of a universal bound without loss\nare formalized in\n\\lean{KomorebiPaperIX.Normalization.normalized_first},\n\\lean{KomorebiPaperIX.Normalization.normalized_second},\n\\lean{KomorebiPaperIX.Normali","\\begin{example}[Why the factorial loss is necessary]\n\\label{ex:arith-normalization-loss}\nTake $p=3$, $m=2$, $N=6$, and $j=3$.  The recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exact values and the refutation of a universal bound without loss\nare formalized in\n\\lean{KomorebiPaperIX.Normalization.normalized_first},\n\\lean{KomorebiPaperIX.Normalization.normalized_second},\n\\lean{KomorebiPaperIX.Normalization.normal","ith-normalization-loss}\nTake $p=3$, $m=2$, $N=6$, and $j=3$.  The recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exact values and the refutation of a universal bound without loss\nare formalized in\n\\lean{KomorebiPaperIX.Normalization.normalized_first},\n\\lean{KomorebiPaperIX.Normalization.normalized_second},\n\\lean{KomorebiPaperIX.Normalization.normalized_difference},\n\\lean{KomorebiPaperIX.Normalization.difference_valuation},\n\\lean{KomorebiPaperIX.Normalization.factorial_loss_is_sh…","ce gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exact values and the refutation of a universal bound without loss\nare formalized in\n\\lean{KomorebiPaperIX.Normalization.normalized_first},\n\\lean{KomorebiPaperIX.Normalization.normalized_second},\n\\lean{KomorebiPaperIX.Normalization.normalized_difference},\n\\lean{KomorebiPaperIX.Normalization.difference_valuation},\n\\lean{KomorebiPaperIX.Normalization.factorial_loss_is_sharp},\n\\lean{KomorebiPaperIX.Normalization.unnormalized_difference},\n\\lean{…","C_{6,3}(1)=-\\frac1{84},\\qquad\n C_{6,3}(2)=\\frac{703}{5460},\\qquad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\nThe last difference has $3$-adic valuation zero.  Here $h=1$, so at\n$n=1$ the order-one bound without factorial loss would require valuation\nat least one.  The corrected normalized bound is sharp, since\n\\[\n v_3\\bigl(C_{6,3}(2)-C_{6,3}(1)\\bigr)=0=1-v_3(3!).\n\\]\nMultiplication by $3!$ gives\n\\[\n U_{6,3}(2)-U_{6,3}(1)=\\frac{384}{455},\\qquad\n v_3\\bigl(U_{6,3}(2)-U_{6,3}(1)\\bigr)=1,\n\\]\nconsistent with the unnormalized theorem.  Thus the factorial correction\nis already necessary at a single step in the all-address parameter range.\nThese exact values and the refutation of a universal bound without loss\nare formalized in\n\\lean{KomorebiPaperIX.Normalization.normalized_first},\n\\lean{KomorebiPaperIX.Normalization.normalized_second},\n\\lean{KomorebiPaperIX.Normalization.normalized_difference},\n\\lean{KomorebiPaperIX.Normalization.difference_valuation},\n\\lean{KomorebiPaperIX.Normalization.factorial_loss_is_sharp},\n\\lean{KomorebiPaperIX.Normalization.unnormalized_difference},\n\\lean{KomorebiPaper…","\\begin{corollary}[A common period in the standard window]\n\\label{cor:arith-window-period}\nLet $p$ be prime, $m>0$, $m\\mid p-1$, and $0\\le j<p$.  Put\n$h=(p-1)/m$.  For every $n\\ge1$ and $u\\in\\N$,\n\\[\n \\overline C_m(n+uh)=\\overline C_m(n),\\qquad\n C_{mp,j}(n+uh)\\equiv C_{mp,j}(n)\\pmod p.\n\\]\nThus $h$ is a sufficient period from row one, common to all addresses\nin the standard window.  It need not be the least period.\n\\end{corollary}","\\begin{proof}\nSince $j<p$, the factorial $j!$ is a unit in $\\Z_p$.  Also $mn\\ge1$.\nThe order-one case of Theorem~\\ref{thm:arith-divisor-base} therefore gives\nvaluation at least one for the normalized difference at every step $uh$.\nThis proves the congruence, and Proposition~\\ref{prop:arith-lucas-window}\nidentifies its reduction with the equality for $\\overline C_m$.\nThe preserved Lean development also proves this period directly from the\nHurwitz inverse in characteristic $p$:\n\\lean{KomorebiPaperIX.WindowPeriod.uniform_standard_window_period},\n\\lean{KomorebiPaperIX.WindowPeriod.multiple_period}, and\n\\lean{KomorebiPaperIX.WindowPeriod.standard_window_period_divisibility}.\n\\end{proof}","ge1$.\nThe order-one case of Theorem~\\ref{thm:arith-divisor-base} therefore gives\nvaluation at least one for the normalized difference at every step $uh$.\nThis proves the congruence, and Proposition~\\ref{prop:arith-lucas-window}\nidentifies its reduction with the equality for $\\overline C_m$.\nThe preserved Lean development also proves this period directly from the\nHurwitz inverse in characteristic $p$:\n\\lean{KomorebiPaperIX.WindowPeriod.uniform_standard_window_period},\n\\lean{KomorebiPaperIX.WindowPeriod.multiple_period}, and\n\\lean{KomorebiPaperIX.WindowPeriod.standard_window_period_divisibility}.\n\\end{proof}","\\begin{lemma}[Prime-power period from finite differences]\n\\label{lem:arith-newton-period}\nLet $R$ be a commutative ring, $p$ a prime, $f:\\N\\to R$, $h,n\\in\\N$,\nand $r\\ge1$.  If\n\\[\n p^{\\min(k,r)}\\mid\\Delta_h^k f(n)\\quad\\text{for every }k\\in\\N,\n\\]\nthen $p^r\\mid f(n+p^{r-1}h)-f(n)$.\n\\end{lemma}","\\begin{proof}\nThe finite Newton expansion gives\n\\[\n f(n+p^{r-1}h)-f(n)\n =\\sum_{k=1}^{p^{r-1}}\\binom{p^{r-1}}k\\Delta_h^k f(n).\n\\]\nTerms with $k\\ge r$ have the required factor from the difference.\nFor $1\\le k<r$, the identity\n\\[\n k\\binom{p^{r-1}}k=p^{r-1}\\binom{p^{r-1}-1}{k-1}\n\\]\nand $v_p(k)<k$ imply\n$p^r\\mid\\binom{p^{r-1}}k p^k$ in $\\Z$.\nMapping that integer divisibility to $R$ proves the claim term by term.\nThe argument includes $r=1$.\nThe declaration is\n\\lean{KomorebiPaperIX.PeriodFromDifferences.prime_power_period}.\n\\end{proof}","pansion gives\n\\[\n f(n+p^{r-1}h)-f(n)\n =\\sum_{k=1}^{p^{r-1}}\\binom{p^{r-1}}k\\Delta_h^k f(n).\n\\]\nTerms with $k\\ge r$ have the required factor from the difference.\nFor $1\\le k<r$, the identity\n\\[\n k\\binom{p^{r-1}}k=p^{r-1}\\binom{p^{r-1}-1}{k-1}\n\\]\nand $v_p(k)<k$ imply\n$p^r\\mid\\binom{p^{r-1}}k p^k$ in $\\Z$.\nMapping that integer divisibility to $R$ proves the claim term by term.\nThe argument includes $r=1$.\nThe declaration is\n\\lean{KomorebiPaperIX.PeriodFromDifferences.prime_power_period}.\n\\end{proof}","\\begin{proof}[Proof of Theorem~\\ref{thm:arith-nishibuchi}]\nOn the divisor branch, $m\\mid q$ and the packet step $q/m$ is an integer\nmultiple of $h=(p-1)/m$.  If $r\\le mn$, Theorem~\\ref{thm:arith-divisor-base}\nsupplies the hypotheses of Lemma~\\ref{lem:arith-newton-period}.\nIts packet displacement is $p^{r-1}q/m$, which converts to the original\ndegree as follows:\n\\[\n mp\\left(n+p^{r-1}\\frac qm\\right)=mpn+qp^r.\n\\]\nThe choice $n\\ge r$ implies $mn\\ge r$ because $m\\ge1$.","orem~\\ref{thm:arith-nishibuchi}]\nOn the divisor branch, $m\\mid q$ and the packet step $q/m$ is an integer\nmultiple of $h=(p-1)/m$.  If $r\\le mn$, Theorem~\\ref{thm:arith-divisor-base}\nsupplies the hypotheses of Lemma~\\ref{lem:arith-newton-period}.\nIts packet displacement is $p^{r-1}q/m$, which converts to the original\ndegree as follows:\n\\[\n mp\\left(n+p^{r-1}\\frac qm\\right)=mpn+qp^r.\n\\]\nThe choice $n\\ge r$ implies $mn\\ge r$ because $m\\ge1$.","th-nishibuchi}]\nOn the divisor branch, $m\\mid q$ and the packet step $q/m$ is an integer\nmultiple of $h=(p-1)/m$.  If $r\\le mn$, Theorem~\\ref{thm:arith-divisor-base}\nsupplies the hypotheses of Lemma~\\ref{lem:arith-newton-period}.\nIts packet displacement is $p^{r-1}q/m$, which converts to the original\ndegree as follows:\n\\[\n mp\\left(n+p^{r-1}\\frac qm\\right)=mpn+qp^r.\n\\]\nThe choice $n\\ge r$ implies $mn\\ge r$ because $m\\ge1$.","the divisor branch, $m\\mid q$ and the packet step $q/m$ is an integer\nmultiple of $h=(p-1)/m$.  If $r\\le mn$, Theorem~\\ref{thm:arith-divisor-base}\nsupplies the hypotheses of Lemma~\\ref{lem:arith-newton-period}.\nIts packet displacement is $p^{r-1}q/m$, which converts to the original\ndegree as follows:\n\\[\n mp\\left(n+p^{r-1}\\frac qm\\right)=mpn+qp^r.\n\\]\nThe choice $n\\ge r$ implies $mn\\ge r$ because $m\\ge1$.","For an odd prime, $2\\mid p-1$, so the option $m=2$ introduces no further\ncase.  The sole remaining branch is $p=2,m=2,N=4$.  The all-row theorem\nproved next gives $2^k\\mid\\Delta_1^kC_{4,j}(n)$ at every row and address.\nAfter multiplying by $j!$, Lemma~\\ref{lem:arith-newton-period} gives packet\nperiod $2^{r-1}$ and degree displacement $4\\cdot2^{r-1}=2^{r+1}=qp^r$.\nIt also gives integrality.  Hence the complete statement follows.\nThe formal degree conversion is\n\\lean{KomorebiPaperIX.NishibuchiCompletion.original_degree_shift};\nthe all-parameter degree theorem is\n\\lean{KomorebiPaperIX.NishibuchiCompletion.full_period_degree}.\n\\end{proof}","For the remainder of the four-packet discussion abbreviate\n$C_j=C_{4,j}$ and $U_j=U_{4,j}$.  Removing the original row cap requires\nan unshifted coefficient operator.","\\begin{theorem}[All-row dyadic higher differences]\\label{thm:arith-dyadic-differences}\nFor $0\\le j<4$ and all $u,k,n\\in\\N$,\n\\begin{align}\n C_j(n)&\\in\\Z_2,","\\begin{theorem}[All-row dyadic higher differences]\\label{thm:arith-dyadic-differences}\nFor $0\\le j<4$ and all $u,k,n\\in\\N$,\n\\begin{align}\n C_j(n)&\\in\\Z_2,\\\\\n \\Delta_u^kC_j(n)&\\in 2^{k(1+\\nu_2(u))}\\Z_2,\\label{eq:arith-allrow-c}\\\\\n \\Delta_u^kU_j(n)&\\in\n 2^{k(1+\\nu_2(u))+v_2(j!)}\\Z_2.\\label{eq:arith-allrow-u}\n\\end{align}\nThere is no restriction $k\\le n$ and no minimum with the row $n$.\n\\end{theorem}","\\Delta_u^kU_j(n)&\\in\n 2^{k(1+\\nu_2(u))+v_2(j!)}\\Z_2.\\label{eq:arith-allrow-u}\n\\end{align}\nThere is no restriction $k\\le n$ and no minimum with the row $n$.\n\\end{theorem}","\\begin{proof}\nThe Hurwitz coefficients of the normalized source $K_{4,j}$ at degrees\n$4n$ are\n\\[\n g_j(n)=\\binom{4n+j}{4n}^{-1}.\n\\]\nTheir explicit expressions are\n\\[\n \\begin{aligned}\n g_0(n)&=1,&\n g_1(n)&=\\frac1{4n+1},","\\begin{proof}\nThe Hurwitz coefficients of the normalized source $K_{4,j}$ at degrees\n$4n$ are\n\\[\n g_j(n)=\\binom{4n+j}{4n}^{-1}.\n\\]\nTheir explicit expressions are\n\\[\n \\begin{aligned}\n g_0(n)&=1,&\n g_1(n)&=\\frac1{4n+1},\\\\\n g_2(n)&=\\frac1{(4n+1)(2n+1)},&\n g_3(n)&=\\frac3{(4n+1)(2n+1)(4n+3)}.\n \\end{aligned}\n\\]\nEvery denominator is odd.  An affine sequence with even slope satisfies\n$2^k\\mid\\Delta_1^k f(n)$ for every $k,n$.  The exact product identity\n\\[\n \\Delta_1^s(fg)(n)=\\sum_{i=0}^s\\binom si\n          \\Delta_1^if(n)\\Delta_1^{s-i}g(n+i)\n\\]\nproves closure of this property under products.  Applying the formula\nwith the factors interchanged to $fg=1$ isolates $f(n+s)\\Delta_1^sg(n)$.\nMultiplication by $f(n+s)^{-1}$ and induction then prove closure under\npointwise unit inverses.\nThus all four $g_j$ have the property.  The explicit formulas and difference\nbounds are formalized in\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_coe} and\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_kummer}.","g_2(n)&=\\frac1{(4n+1)(2n+1)},&\n g_3(n)&=\\frac3{(4n+1)(2n+1)(4n+3)}.\n \\end{aligned}\n\\]\nEvery denominator is odd.  An affine sequence with even slope satisfies\n$2^k\\mid\\Delta_1^k f(n)$ for every $k,n$.  The exact product identity\n\\[\n \\Delta_1^s(fg)(n)=\\sum_{i=0}^s\\binom si\n          \\Delta_1^if(n)\\Delta_1^{s-i}g(n+i)\n\\]\nproves closure of this property under products.  Applying the formula\nwith the factors interchanged to $fg=1$ isolates $f(n+s)\\Delta_1^sg(n)$.\nMultiplication by $f(n+s)^{-1}$ and induction then prove closure under\npointwise unit inverses.\nThus all four $g_j$ have the property.  The explicit formulas and difference\nbounds are formalized in\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_coe} and\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_kummer}.","+1)},&\n g_3(n)&=\\frac3{(4n+1)(2n+1)(4n+3)}.\n \\end{aligned}\n\\]\nEvery denominator is odd.  An affine sequence with even slope satisfies\n$2^k\\mid\\Delta_1^k f(n)$ for every $k,n$.  The exact product identity\n\\[\n \\Delta_1^s(fg)(n)=\\sum_{i=0}^s\\binom si\n          \\Delta_1^if(n)\\Delta_1^{s-i}g(n+i)\n\\]\nproves closure of this property under products.  Applying the formula\nwith the factors interchanged to $fg=1$ isolates $f(n+s)\\Delta_1^sg(n)$.\nMultiplication by $f(n+s)^{-1}$ and induction then prove closure under\npointwise unit inverses.\nThus all four $g_j$ have the property.  The explicit formulas and difference\nbounds are formalized in\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_coe} and\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_kummer}.","\\frac3{(4n+1)(2n+1)(4n+3)}.\n \\end{aligned}\n\\]\nEvery denominator is odd.  An affine sequence with even slope satisfies\n$2^k\\mid\\Delta_1^k f(n)$ for every $k,n$.  The exact product identity\n\\[\n \\Delta_1^s(fg)(n)=\\sum_{i=0}^s\\binom si\n          \\Delta_1^if(n)\\Delta_1^{s-i}g(n+i)\n\\]\nproves closure of this property under products.  Applying the formula\nwith the factors interchanged to $fg=1$ isolates $f(n+s)\\Delta_1^sg(n)$.\nMultiplication by $f(n+s)^{-1}$ and induction then prove closure under\npointwise unit inverses.\nThus all four $g_j$ have the property.  The explicit formulas and difference\nbounds are formalized in\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_coe} and\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_kummer}.","\\]\nEvery denominator is odd.  An affine sequence with even slope satisfies\n$2^k\\mid\\Delta_1^k f(n)$ for every $k,n$.  The exact product identity\n\\[\n \\Delta_1^s(fg)(n)=\\sum_{i=0}^s\\binom si\n          \\Delta_1^if(n)\\Delta_1^{s-i}g(n+i)\n\\]\nproves closure of this property under products.  Applying the formula\nwith the factors interchanged to $fg=1$ isolates $f(n+s)\\Delta_1^sg(n)$.\nMultiplication by $f(n+s)^{-1}$ and induction then prove closure under\npointwise unit inverses.\nThus all four $g_j$ have the property.  The explicit formulas and difference\nbounds are formalized in\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_coe} and\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_kummer}.","^k f(n)$ for every $k,n$.  The exact product identity\n\\[\n \\Delta_1^s(fg)(n)=\\sum_{i=0}^s\\binom si\n          \\Delta_1^if(n)\\Delta_1^{s-i}g(n+i)\n\\]\nproves closure of this property under products.  Applying the formula\nwith the factors interchanged to $fg=1$ isolates $f(n+s)\\Delta_1^sg(n)$.\nMultiplication by $f(n+s)^{-1}$ and induction then prove closure under\npointwise unit inverses.\nThus all four $g_j$ have the property.  The explicit formulas and difference\nbounds are formalized in\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_coe} and\n\\lean{KomorebiPaperIX.DyadicFourKernel.source_kummer}.","For an integer charge $c$, put\n\\[\n A_c=\\frac{D^4-c}{2}.\n\\]\nA series is flat of charge $c$ when applying any power $A_c^s$ to it gives\nan element of $\\mathcal H_2$.  The required product formula is\n\\begin{equation}\\label{eq:arith-charged-product}\n A_{c+d}(FG)=(A_cF)G+F(A_dG)\n       +2F'G'''+3F''G''+2F'''G'.\n\\end{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves t","For an integer charge $c$, put\n\\[\n A_c=\\frac{D^4-c}{2}.\n\\]\nA series is flat of charge $c$ when applying any power $A_c^s$ to it gives\nan element of $\\mathcal H_2$.  The required product formula is\n\\begin{equation}\\label{eq:arith-charged-product}\n A_{c+d}(FG)=(A_cF)G+F(A_dG)\n       +2F'G'''+3F''G''+2F'''G'.\n\\end{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge","For an integer charge $c$, put\n\\[\n A_c=\\frac{D^4-c}{2}.\n\\]\nA series is flat of charge $c$ when applying any power $A_c^s$ to it gives\nan element of $\\mathcal H_2$.  The required product formula is\n\\begin{equation}\\label{eq:arith-charged-product}\n A_{c+d}(FG)=(A_cF)G+F(A_dG)\n       +2F'G'''+3F''G''+2F'''G'.\n\\end{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge.\nFinally $A_{c+2}=A_c-1$, so in","For an integer charge $c$, put\n\\[\n A_c=\\frac{D^4-c}{2}.\n\\]\nA series is flat of charge $c$ when applying any power $A_c^s$ to it gives\nan element of $\\mathcal H_2$.  The required product formula is\n\\begin{equation}\\label{eq:arith-charged-product}\n A_{c+d}(FG)=(A_cF)G+F(A_dG)\n       +2F'G'''+3F''G''+2F'''G'.\n\\end{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge.\nFinally $A_{c+2}=A_c-1$, so increasing a charge by two preserv","For an integer charge $c$, put\n\\[\n A_c=\\frac{D^4-c}{2}.\n\\]\nA series is flat of charge $c$ when applying any power $A_c^s$ to it gives\nan element of $\\mathcal H_2$.  The required product formula is\n\\begin{equation}\\label{eq:arith-charged-product}\n A_{c+d}(FG)=(A_cF)G+F(A_dG)\n       +2F'G'''+3F''G''+2F'''G'.\n\\end{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge.\nFinally $A_{c+2}=A_c-1$, so increasing a charge by two preserves flatness.\nThese are\n\\lean{KomorebiPape","For an integer charge $c$, put\n\\[\n A_c=\\frac{D^4-c}{2}.\n\\]\nA series is flat of charge $c$ when applying any power $A_c^s$ to it gives\nan element of $\\mathcal H_2$.  The required product formula is\n\\begin{equation}\\label{eq:arith-charged-product}\n A_{c+d}(FG)=(A_cF)G+F(A_dG)\n       +2F'G'''+3F''G''+2F'''G'.\n\\end{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge.\nFinally $A_{c+2}=A_c-1$, so increasing a charge by two preserves flatness.\nThese are\n\\lean{KomorebiPaperIX.ChargedDworkClosure.B_mul},\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_inverse}, and\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_charge_","For an integer charge $c$, put\n\\[\n A_c=\\frac{D^4-c}{2}.\n\\]\nA series is flat of charge $c$ when applying any power $A_c^s$ to it gives\nan element of $\\mathcal H_2$.  The required product formula is\n\\begin{equation}\\label{eq:arith-charged-product}\n A_{c+d}(FG)=(A_cF)G+F(A_dG)\n       +2F'G'''+3F''G''+2F'''G'.\n\\end{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge.\nFinally $A_{c+2}=A_c-1$, so increasing a charge by two preserves flatness.\nThese are\n\\lean{KomorebiPaperIX.ChargedDworkClosure.B_mul},\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_inverse}, and\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_charge_shift}.","harge $c$, put\n\\[\n A_c=\\frac{D^4-c}{2}.\n\\]\nA series is flat of charge $c$ when applying any power $A_c^s$ to it gives\nan element of $\\mathcal H_2$.  The required product formula is\n\\begin{equation}\\label{eq:arith-charged-product}\n A_{c+d}(FG)=(A_cF)G+F(A_dG)\n       +2F'G'''+3F''G''+2F'''G'.\n\\end{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge.\nFinally $A_{c+2}=A_c-1$, so increasing a charge by two preserves flatness.\nThese are\n\\lean{KomorebiPaperIX.ChargedDworkClosure.B_mul},\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_inverse}, and\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_charge_shift}.","t of $\\mathcal H_2$.  The required product formula is\n\\begin{equation}\\label{eq:arith-charged-product}\n A_{c+d}(FG)=(A_cF)G+F(A_dG)\n       +2F'G'''+3F''G''+2F'''G'.\n\\end{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge.\nFinally $A_{c+2}=A_c-1$, so increasing a charge by two preserves flatness.\nThese are\n\\lean{KomorebiPaperIX.ChargedDworkClosure.B_mul},\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_inverse}, and\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_charge_shift}.","d product formula is\n\\begin{equation}\\label{eq:arith-charged-product}\n A_{c+d}(FG)=(A_cF)G+F(A_dG)\n       +2F'G'''+3F''G''+2F'''G'.\n\\end{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge.\nFinally $A_{c+2}=A_c-1$, so increasing a charge by two preserves flatness.\nThese are\n\\lean{KomorebiPaperIX.ChargedDworkClosure.B_mul},\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_inverse}, and\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_charge_shift}.","{equation}\nAll middle coefficients are integral.  Derivatives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge.\nFinally $A_{c+2}=A_c-1$, so increasing a charge by two preserves flatness.\nThese are\n\\lean{KomorebiPaperIX.ChargedDworkClosure.B_mul},\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_inverse}, and\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_charge_shift}.","atives preserve\n$\\mathcal H_2$ and commute with $A_c$.  Induction on the number of\niterations in \\eqref{eq:arith-charged-product} proves that multiplication\nadds charges.  If $FG=1$, the same formula yields\n\\[\n A_{-c}G=-G^2A_cF-G(2F'G'''+3F''G''+2F'''G').\n\\]\nInduction, starting from the independently established integrality of\nthe reciprocal $G$, proves that inversion negates the charge.\nFinally $A_{c+2}=A_c-1$, so increasing a charge by two preserves flatness.\nThese are\n\\lean{KomorebiPaperIX.ChargedDworkClosure.B_mul},\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_inverse}, and\n\\lean{KomorebiPaperIX.ChargedDworkClosure.flat_charge_shift}.","Packing $g_j$ into degrees divisible by four gives $K_{4,j}$.\nIts source property is precisely charge-one flatness.  Inversion gives\ncharge $-1$, and the charge shift restores charge one for the\nreciprocal $\\mathcal P_{4,j}$.  If $\\operatorname{dec}(F)(d)=d![t^d]F$, the exact\ncoefficient identity is\n\\[\n \\operatorname{dec}(A_1^sF)(d)\n       =2^{-s}\\Delta_4^s\\operatorname{dec}(F)(d).\n\\]\nAt $d=4n$, packing turns the right side into\n$2^{-s}\\Delta_1^sC_j(n)$.  This extraction requires no negative degree\nand therefore holds at every $n$, including zero.  It proves the unit-step\nbound \\lean{KomorebiPaperIX.DyadicFourAllRows.base_step_bound}.","le by four gives $K_{4,j}$.\nIts source property is precisely charge-one flatness.  Inversion gives\ncharge $-1$, and the charge shift restores charge one for the\nreciprocal $\\mathcal P_{4,j}$.  If $\\operatorname{dec}(F)(d)=d![t^d]F$, the exact\ncoefficient identity is\n\\[\n \\operatorname{dec}(A_1^sF)(d)\n       =2^{-s}\\Delta_4^s\\operatorname{dec}(F)(d).\n\\]\nAt $d=4n$, packing turns the right side into\n$2^{-s}\\Delta_1^sC_j(n)$.  This extraction requires no negative degree\nand therefore holds at every $n$, including zero.  It proves the unit-step\nbound \\lean{KomorebiPaperIX.DyadicFourAllRows.base_step_bound}.","harge $-1$, and the charge shift restores charge one for the\nreciprocal $\\mathcal P_{4,j}$.  If $\\operatorname{dec}(F)(d)=d![t^d]F$, the exact\ncoefficient identity is\n\\[\n \\operatorname{dec}(A_1^sF)(d)\n       =2^{-s}\\Delta_4^s\\operatorname{dec}(F)(d).\n\\]\nAt $d=4n$, packing turns the right side into\n$2^{-s}\\Delta_1^sC_j(n)$.  This extraction requires no negative degree\nand therefore holds at every $n$, including zero.  It proves the unit-step\nbound \\lean{KomorebiPaperIX.DyadicFourAllRows.base_step_bound}.","To incorporate $v_2(u)$, expand a displacement using finite Newton\ncoefficients.  For any prime $p$, $u>0$, and $1\\le a\\le u$,\n\\[\n p^{v_p(u)+1}\\mid\\binom ua p^a,\n\\]\nas follows from $a\\binom ua=u\\binom{u-1}{a-1}$ and $v_p(a)<a$.\nRepeated displacement, retaining the remaining base-difference order\nat each induction step, gives a factor $p^{k(1+v_p(u))}$ from $k$\ndisplacements.  This induction is formalized in\n\\lean{KomorebiPaperIX.StepAmplification.repeated_precision}.\nApplying it with $p=2$ proves \\eqref{eq:arith-allrow-c}; multiplication\nby $j!$ proves \\eqref{eq:arith-allrow-u}.  Zero step and zero order follow\nfrom their definitions.  The final statements are\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.no","To incorporate $v_2(u)$, expand a displacement using finite Newton\ncoefficients.  For any prime $p$, $u>0$, and $1\\le a\\le u$,\n\\[\n p^{v_p(u)+1}\\mid\\binom ua p^a,\n\\]\nas follows from $a\\binom ua=u\\binom{u-1}{a-1}$ and $v_p(a)<a$.\nRepeated displacement, retaining the remaining base-difference order\nat each induction step, gives a factor $p^{k(1+v_p(u))}$ from $k$\ndisplacements.  This induction is formalized in\n\\lean{KomorebiPaperIX.StepAmplification.repeated_precision}.\nApplying it with $p=2$ proves \\eqref{eq:arith-allrow-c}; multiplication\nby $j!$ proves \\eqref{eq:arith-allrow-u}.  Zero step and zero order follow\nfrom their definitions.  The final statements are\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.normalized_step_bound}\nand\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.","To incorporate $v_2(u)$, expand a displacement using finite Newton\ncoefficients.  For any prime $p$, $u>0$, and $1\\le a\\le u$,\n\\[\n p^{v_p(u)+1}\\mid\\binom ua p^a,\n\\]\nas follows from $a\\binom ua=u\\binom{u-1}{a-1}$ and $v_p(a)<a$.\nRepeated displacement, retaining the remaining base-difference order\nat each induction step, gives a factor $p^{k(1+v_p(u))}$ from $k$\ndisplacements.  This induction is formalized in\n\\lean{KomorebiPaperIX.StepAmplification.repeated_precision}.\nApplying it with $p=2$ proves \\eqref{eq:arith-allrow-c}; multiplication\nby $j!$ proves \\eqref{eq:arith-allrow-u}.  Zero step and zero order follow\nfrom their definitions.  The final statements are\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.normalized_step_bound}\nand\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.unnorma","To incorporate $v_2(u)$, expand a displacement using finite Newton\ncoefficients.  For any prime $p$, $u>0$, and $1\\le a\\le u$,\n\\[\n p^{v_p(u)+1}\\mid\\binom ua p^a,\n\\]\nas follows from $a\\binom ua=u\\binom{u-1}{a-1}$ and $v_p(a)<a$.\nRepeated displacement, retaining the remaining base-difference order\nat each induction step, gives a factor $p^{k(1+v_p(u))}$ from $k$\ndisplacements.  This induction is formalized in\n\\lean{KomorebiPaperIX.StepAmplification.repeated_precision}.\nApplying it with $p=2$ proves \\eqref{eq:arith-allrow-c}; multiplication\nby $j!$ proves \\eqref{eq:arith-allrow-u}.  Zero step and zero order follow\nfrom their definitions.  The final statements are\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.normalized_step_bound}\nand\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.unnormalized_step_bound}.","To incorporate $v_2(u)$, expand a displacement using finite Newton\ncoefficients.  For any prime $p$, $u>0$, and $1\\le a\\le u$,\n\\[\n p^{v_p(u)+1}\\mid\\binom ua p^a,\n\\]\nas follows from $a\\binom ua=u\\binom{u-1}{a-1}$ and $v_p(a)<a$.\nRepeated displacement, retaining the remaining base-difference order\nat each induction step, gives a factor $p^{k(1+v_p(u))}$ from $k$\ndisplacements.  This induction is formalized in\n\\lean{KomorebiPaperIX.StepAmplification.repeated_precision}.\nApplying it with $p=2$ proves \\eqref{eq:arith-allrow-c}; multiplication\nby $j!$ proves \\eqref{eq:arith-allrow-u}.  Zero step and zero order follow\nfrom their definitions.  The final statements are\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.normalized_step_bound}\nand\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.unnormalized_step_bound}.\n\\end{proof}","prime $p$, $u>0$, and $1\\le a\\le u$,\n\\[\n p^{v_p(u)+1}\\mid\\binom ua p^a,\n\\]\nas follows from $a\\binom ua=u\\binom{u-1}{a-1}$ and $v_p(a)<a$.\nRepeated displacement, retaining the remaining base-difference order\nat each induction step, gives a factor $p^{k(1+v_p(u))}$ from $k$\ndisplacements.  This induction is formalized in\n\\lean{KomorebiPaperIX.StepAmplification.repeated_precision}.\nApplying it with $p=2$ proves \\eqref{eq:arith-allrow-c}; multiplication\nby $j!$ proves \\eqref{eq:arith-allrow-u}.  Zero step and zero order follow\nfrom their definitions.  The final statements are\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.normalized_step_bound}\nand\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.unnormalized_step_bound}.\n\\end{proof}","binom ua p^a,\n\\]\nas follows from $a\\binom ua=u\\binom{u-1}{a-1}$ and $v_p(a)<a$.\nRepeated displacement, retaining the remaining base-difference order\nat each induction step, gives a factor $p^{k(1+v_p(u))}$ from $k$\ndisplacements.  This induction is formalized in\n\\lean{KomorebiPaperIX.StepAmplification.repeated_precision}.\nApplying it with $p=2$ proves \\eqref{eq:arith-allrow-c}; multiplication\nby $j!$ proves \\eqref{eq:arith-allrow-u}.  Zero step and zero order follow\nfrom their definitions.  The final statements are\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.normalized_step_bound}\nand\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.unnormalized_step_bound}.\n\\end{proof}","\\begin{corollary}[Step-sensitive divisor-branch bound]\\label{cor:arith-amplified}\nUnder the hypotheses of Theorem~\\ref{thm:arith-divisor-base}, for all\n$u,k,n\\in\\N$,\n\\[\n \\Delta_{u(p-1)/m}^{k}U_{mp,j}(n)\n       \\in p^{\\min(k(1+\\nu_p(u)),mn)}\\Z_p.\n\\]\nFor $C_{mp,j}$ the corresponding valuation lower bound is\n$\\min(k(1+\\nu_p(u)),mn)-v_p(j!)$.\n\\end{corollary}","\\begin{proof}\nApply the same repeated-displacement theorem with the fixed cap $mn$ to\n\\eqref{eq:arith-divisor-bound}.  The argument preserves this cap and\nimproves the contribution from the order of the difference.\nDivision by the fixed factorial gives the rational valuation statement,\nwith the zero case included.\nThese are\n\\lean{KomorebiPaperIX.AllResidueStepAmplification.unnormalized_precision}\nand\n\\lean{KomorebiPaperIX.AllResidueStepAmplification.normalized_precision}.\n\\end{proof}","\\begin{corollary}[Dyadic periods from row zero]\\label{cor:arith-allrow-period}\nFor $0\\le j<4$, $r\\ge1$, and $n\\ge0$,\n\\[\n U_j(n+2^{r-1})\\equiv U_j(n)\\pmod{2^r}.\n\\]\nFor $j=2,3$ and $r\\ge2$, $2^{r-2}$ is also a sufficient packet period.\nThe respective original degree displacements are $2^{r+1}$ and $2^r$.\n\\end{corollary}","\\begin{proof}\nUse Lemma~\\ref{lem:arith-newton-period} and the unit-step case of\nTheorem~\\ref{thm:arith-dyadic-differences}.  At $j=2,3$ the factorial\ncontributes an additional power of two.  The original degree versions are\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.all_rows_period_degree}\nand\n\\lean{KomorebiPaperIX.DyadicFourFinalStrengthening.all_rows_half_period_degree}.\nThe next theorem sharpens these sufficient periods to a complete\nclassification.\n\\end{proof}","\\begin{lemma}[Finite Newton dominance]\\label{lem:arith-dominance}\nLet $f:\\N\\to\\Q$ and write $a_k=\\Delta_1^kf(0)$.  Suppose that $c\\in\\N$,\n$a_1\\ne0$, $v_2(a_1)=c$, and\n\\[\n v_2(a_k)\\ge c+1+\\lfloor\\log_2k\\rfloor\\qquad(k\\ge2).\n\\]\nThen for every $n\\ge0$ and $T>0$,\n\\[\n f(n+T)-f(n)\\ne0,\\qquad\n v_2\\bigl(f(n+T)-f(n)\\bigr)=v_2(T)+c.\n\\]\n\\end{lemma}","\\begin{proof}\nFinite Newton reconstruction at an arbitrary row gives\n\\[\n \\Delta_1^kf(n)=\\sum_{i=0}^{n}\\binom ni a_{k+i}.\n\\]\nThe monotonicity of $\\lfloor\\log_2k\\rfloor$ therefore preserves the\nhigher-coefficient bound after translation.  In addition,\n$\\Delta_1f(n)-a_1\\in2^{c+1}\\Z_2$, so the first difference is nonzero\nof exact valuation $c$.","Now expand the displacement:\n\\[\n f(n+T)-f(n)=T\\Delta_1f(n)\n          +\\sum_{k=2}^{T}\\binom Tk\\Delta_1^kf(n).\n\\]\nFor $1\\le k\\le T$, the binomial denominator identity implies\n\\[\n v_2\\binom Tk\\ge v_2(T)-v_2(k)\n            \\ge v_2(T)-\\lfloor\\log_2k\\rfloor.\n\\]\nEvery term of the remaining sum consequently has valuation at least\n$v_2(T)+c+1$.  Its sum cannot cancel the leading term, which has exact\nvaluation $v_2(T)+c$.  This proves nonvanishing and the claimed valuation\nusing only finite sums.\nThe general theorem is\n\\lean{KomorebiPaperIX.NewtonDominance.exact_displacement}.\n\\end{proof}","pand the displacement:\n\\[\n f(n+T)-f(n)=T\\Delta_1f(n)\n          +\\sum_{k=2}^{T}\\binom Tk\\Delta_1^kf(n).\n\\]\nFor $1\\le k\\le T$, the binomial denominator identity implies\n\\[\n v_2\\binom Tk\\ge v_2(T)-v_2(k)\n            \\ge v_2(T)-\\lfloor\\log_2k\\rfloor.\n\\]\nEvery term of the remaining sum consequently has valuation at least\n$v_2(T)+c+1$.  Its sum cannot cancel the leading term, which has exact\nvaluation $v_2(T)+c$.  This proves nonvanishing and the claimed valuation\nusing only finite sums.\nThe general theorem is\n\\lean{KomorebiPaperIX.NewtonDominance.exact_displacement}.\n\\end{proof}","\\begin{theorem}[Exact four-packet displacement law]\\label{thm:arith-exact}\nSet\n\\[\n (c_0,c_1,c_2,c_3)=(1,1,4,2),\\qquad\n (d_0,d_1,d_2,d_3)=(1,1,5,3).\n\\]\nFor $0\\le j<4$, $n\\ge0$, and every positive integer $T$,\n\\begin{align*}\n C_j(n+T)-C_j(n)&\\ne0,&\n v_2\\bigl(C_j(n+T)-C_j(n)\\bigr)&=v_2(T)+c_j,","\\begin{theorem}[Exact four-packet displacement law]\\label{thm:arith-exact}\nSet\n\\[\n (c_0,c_1,c_2,c_3)=(1,1,4,2),\\qquad\n (d_0,d_1,d_2,d_3)=(1,1,5,3).\n\\]\nFor $0\\le j<4$, $n\\ge0$, and every positive integer $T$,\n\\begin{align*}\n C_j(n+T)-C_j(n)&\\ne0,&\n v_2\\bigl(C_j(n+T)-C_j(n)\\bigr)&=v_2(T)+c_j,\\\\\n U_j(n+T)-U_j(n)&\\ne0,&\n v_2\\bigl(U_j(n+T)-U_j(n)\\bigr)&=v_2(T)+d_j.\n\\end{align*}\n\\end{theorem}","\\begin{proof}\nThe first Newton coefficients, derived from\n\\eqref{eq:arith-triangular}, are\n\\[\n a_{0,1}=-2,\\quad a_{1,1}=-\\frac65,\\quad\n a_{2,1}=-\\frac{16}{15},\\quad a_{3,1}=-\\frac{36}{35}.\n\\]\nTheir exact valuations are $c_j$.  Theorem~\\ref{thm:arith-dyadic-differences}\ngives $v_2(a_{j,k})\\ge k$ for all orders.  To obtain the stronger\nlogarithmic envelope required by Lemma~\\ref{lem:arith-dominance}, only\na finite initial segment needs improvement.  The recurrence proves the\nfollowing exact valuations:\n\\begin{center}\n\\begin{tabular}{cll}\\toprule\nAddress $j$ & Further orders $k$ & $v_2(a_{j,k})$ in that order","\\quad a_{3,1}=-\\frac{36}{35}.\n\\]\nTheir exact valuations are $c_j$.  Theorem~\\ref{thm:arith-dyadic-differences}\ngives $v_2(a_{j,k})\\ge k$ for all orders.  To obtain the stronger\nlogarithmic envelope required by Lemma~\\ref{lem:arith-dominance}, only\na finite initial segment needs improvement.  The recurrence proves the\nfollowing exact valuations:\n\\begin{center}\n\\begin{tabular}{cll}\\toprule\nAddress $j$ & Further orders $k$ & $v_2(a_{j,k})$ in that order","36}{35}.\n\\]\nTheir exact valuations are $c_j$.  Theorem~\\ref{thm:arith-dyadic-differences}\ngives $v_2(a_{j,k})\\ge k$ for all orders.  To obtain the stronger\nlogarithmic envelope required by Lemma~\\ref{lem:arith-dominance}, only\na finite initial segment needs improvement.  The recurrence proves the\nfollowing exact valuations:\n\\begin{center}\n\\begin{tabular}{cll}\\toprule\nAddress $j$ & Further orders $k$ & $v_2(a_{j,k})$ in that order","}.\n\\]\nTheir exact valuations are $c_j$.  Theorem~\\ref{thm:arith-dyadic-differences}\ngives $v_2(a_{j,k})\\ge k$ for all orders.  To obtain the stronger\nlogarithmic envelope required by Lemma~\\ref{lem:arith-dominance}, only\na finite initial segment needs improvement.  The recurrence proves the\nfollowing exact valuations:\n\\begin{center}\n\\begin{tabular}{cll}\\toprule\nAddress $j$ & Further orders $k$ & $v_2(a_{j,k})$ in that order","midrule\n$0$ & $2$ & $3$","$1$ & $2$ & $3$","$2$ & $2,3,4,5,6$ & $6,9,13,16,22$","$3$ & $2,3,4$ & $8,8,15$","bottomrule\n\\end{tabular}\n\\end{center}\nFor all subsequent $k$, the elementary inequality\n$c_j+1+\\lfloor\\log_2k\\rfloor\\le k$ completes the envelope.\nThe entries follow by exact rational computation from the defining\nrecurrence, and their valuations are proved in Lean.  Together with the\ninequality for all subsequent orders, they establish the bound at every\norder.\nThe uniform envelope is\n\\lean{KomorebiPaperIX.DyadicFourPrefix.tail_precision}.","Apply Lemma~\\ref{lem:arith-dominance} to $C_j$.  Multiplication by the\nnonzero factorial $j!$ adds $v_2(j!)$ and preserves nonzeroness.\nHence $d_j=c_j+v_2(j!)$, giving the four valuation gains for the original\ncoefficients.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicFourLeastPeriods.normalized_exact_displacement}\nand \\lean{KomorebiPaperIX.DyadicFourLeastPeriods.exact_displacement}.\n\\end{proof}","\\begin{corollary}[All positive periods and their least element]\n\\label{cor:arith-least-period}\nFor $r\\in\\N$ and $0\\le j<4$, a positive integer $T$ is a packet period\nof $U_j$ modulo $2^r$ if and only if\n\\[\n 2^{\\max(0,r-d_j)}\\mid T.\n\\]\nHere a packet period means $U_j(n+T)\\equiv U_j(n)\\pmod{2^r}$ for every\n$n\\ge0$.  Consequently the least positive packet periods are\n\\[\n \\begin{array}{c|cccc}\n j&0&1&2&3","\\begin{corollary}[All positive periods and their least element]\n\\label{cor:arith-least-period}\nFor $r\\in\\N$ and $0\\le j<4$, a positive integer $T$ is a packet period\nof $U_j$ modulo $2^r$ if and only if\n\\[\n 2^{\\max(0,r-d_j)}\\mid T.\n\\]\nHere a packet period means $U_j(n+T)\\equiv U_j(n)\\pmod{2^r}$ for every\n$n\\ge0$.  Consequently the least positive packet periods are\n\\[\n \\begin{array}{c|cccc}\n j&0&1&2&3\\\\\\hline\n \\text{least period}&2^{\\max(0,r-1)}&2^{\\max(0,r-1)}&\n                 2^{\\max(0,r-5)}&2^{\\max(0,r-3)}.\n \\end{array}\n\\]\n\\end{corollary}","\\begin{proof}\nBy the exact displacement law, the congruence at any row is equivalent to\n$v_2(T)+d_j\\ge r$.  This is exactly the displayed integer\ndivisibility criterion.  Its least positive solution is the indicated\npower of two, which also covers $r=0$ and $r\\le d_j$.\nThe all-period and least-element statements are\n\\lean{KomorebiPaperIX.DyadicFourLeastPeriods.period_iff_dvd} and\n\\lean{KomorebiPaperIX.DyadicFourLeastPeriods.least_packet_period}.\n\\end{proof}","For example, $U_2(n)\\equiv2\\pmod{32}$ and $U_3(n)\\equiv6\\pmod8$\nfor every row.  At precision $2^{10}$ the least packet periods are\n$512,512,32,128$.  A packet displacement $T$ has original degree\ndisplacement $4T$, as proved by\n\\lean{KomorebiPaperIX.DyadicFourLeastPeriods.original_degree_exact_displacement}.\nMinimality here concerns packet shifts.  Shifts of the full degree-indexed\nsequence, including its zero coefficients off the packet support, require\na separate analysis.  The new strengthening chapters determine all repeated-step higher differences in addresses zero and one and a power-of-two-order family in address two. The full exact classification of arbitrary higher-order differences\nare also a separate question from t","For example, $U_2(n)\\equiv2\\pmod{32}$ and $U_3(n)\\equiv6\\pmod8$\nfor every row.  At precision $2^{10}$ the least packet periods are\n$512,512,32,128$.  A packet displacement $T$ has original degree\ndisplacement $4T$, as proved by\n\\lean{KomorebiPaperIX.DyadicFourLeastPeriods.original_degree_exact_displacement}.\nMinimality here concerns packet shifts.  Shifts of the full degree-indexed\nsequence, including its zero coefficients off the packet support, require\na separate analysis.  The new strengthening chapters determine all repeated-step higher differences in addresses zero and one and a power-of-two-order family in address two. The full exact classification of arbitrary higher-order differences\nare also a separate question from the first-displacement law.","The classical signed Euler numbers have generating function\n$\\operatorname{sech}t=\\sum_{n\\ge0}E_{2n}t^{2n}/(2n)!$.\nThey are the $N=2$, $j=0$ member of\n\\eqref{eq:arith-euler-egf}.  The coordinate theorem reconstructs\nthem from finite alternating sums of odd powers.","\\begin{definition}[Odd-length source and inverse coefficients]\n\\label{def:arith-odd-source}\nFor a positive odd integer $p$, define\n\\[\n S_d(p)=\\sum_{a=0}^{p-1}(-1)^a(2a+1)^d.\n\\]\nLet $\\gamma_m$ be the integral Hurwitz coefficients of $\\operatorname{sech}^2$:\n\\[\n \\operatorname{sech}^2t=\\sum_{m\\ge0}\\gamma_m\\frac{t^{2m}}{(2m)!},\n \\qquad\n \\gamma_m=\\sum_{b=0}^{2m}\\binom{2m}{b}E_bE_{2m-b},\n\\]\nwhere $E_b=0$ for odd $b$.  Then $\\gamma_0=1$.  Define\n\\[\n T^{\\mathrm{odd}}_{p,M}(n)=\\sum_{m=0}^{\\min(M,n)}\n       \\binom{2n}{2m}\\gamma_m p^{2m}S_{2(n-m)}(p).\n\\]\nHere $p$ need only be odd; primality is needed only to interpret the\nresult as a prime-adic congruence ladder.\n\\end{definition}","nition}[Odd-length source and inverse coefficients]\n\\label{def:arith-odd-source}\nFor a positive odd integer $p$, define\n\\[\n S_d(p)=\\sum_{a=0}^{p-1}(-1)^a(2a+1)^d.\n\\]\nLet $\\gamma_m$ be the integral Hurwitz coefficients of $\\operatorname{sech}^2$:\n\\[\n \\operatorname{sech}^2t=\\sum_{m\\ge0}\\gamma_m\\frac{t^{2m}}{(2m)!},\n \\qquad\n \\gamma_m=\\sum_{b=0}^{2m}\\binom{2m}{b}E_bE_{2m-b},\n\\]\nwhere $E_b=0$ for odd $b$.  Then $\\gamma_0=1$.  Define\n\\[\n T^{\\mathrm{odd}}_{p,M}(n)=\\sum_{m=0}^{\\min(M,n)}\n       \\binom{2n}{2m}\\gamma_m p^{2m}S_{2(n-m)}(p).\n\\]\nHere $p$ need only be odd; primality is needed only to interpret the\nresult as a prime-adic congruence ladder.\n\\end{definition}","-length source and inverse coefficients]\n\\label{def:arith-odd-source}\nFor a positive odd integer $p$, define\n\\[\n S_d(p)=\\sum_{a=0}^{p-1}(-1)^a(2a+1)^d.\n\\]\nLet $\\gamma_m$ be the integral Hurwitz coefficients of $\\operatorname{sech}^2$:\n\\[\n \\operatorname{sech}^2t=\\sum_{m\\ge0}\\gamma_m\\frac{t^{2m}}{(2m)!},\n \\qquad\n \\gamma_m=\\sum_{b=0}^{2m}\\binom{2m}{b}E_bE_{2m-b},\n\\]\nwhere $E_b=0$ for odd $b$.  Then $\\gamma_0=1$.  Define\n\\[\n T^{\\mathrm{odd}}_{p,M}(n)=\\sum_{m=0}^{\\min(M,n)}\n       \\binom{2n}{2m}\\gamma_m p^{2m}S_{2(n-m)}(p).\n\\]\nHere $p$ need only be odd; primality is needed only to interpret the\nresult as a prime-adic congruence ladder.\n\\end{definition}","rse coefficients]\n\\label{def:arith-odd-source}\nFor a positive odd integer $p$, define\n\\[\n S_d(p)=\\sum_{a=0}^{p-1}(-1)^a(2a+1)^d.\n\\]\nLet $\\gamma_m$ be the integral Hurwitz coefficients of $\\operatorname{sech}^2$:\n\\[\n \\operatorname{sech}^2t=\\sum_{m\\ge0}\\gamma_m\\frac{t^{2m}}{(2m)!},\n \\qquad\n \\gamma_m=\\sum_{b=0}^{2m}\\binom{2m}{b}E_bE_{2m-b},\n\\]\nwhere $E_b=0$ for odd $b$.  Then $\\gamma_0=1$.  Define\n\\[\n T^{\\mathrm{odd}}_{p,M}(n)=\\sum_{m=0}^{\\min(M,n)}\n       \\binom{2n}{2m}\\gamma_m p^{2m}S_{2(n-m)}(p).\n\\]\nHere $p$ need only be odd; primality is needed only to interpret the\nresult as a prime-adic congruence ladder.\n\\end{definition}","1}(-1)^a(2a+1)^d.\n\\]\nLet $\\gamma_m$ be the integral Hurwitz coefficients of $\\operatorname{sech}^2$:\n\\[\n \\operatorname{sech}^2t=\\sum_{m\\ge0}\\gamma_m\\frac{t^{2m}}{(2m)!},\n \\qquad\n \\gamma_m=\\sum_{b=0}^{2m}\\binom{2m}{b}E_bE_{2m-b},\n\\]\nwhere $E_b=0$ for odd $b$.  Then $\\gamma_0=1$.  Define\n\\[\n T^{\\mathrm{odd}}_{p,M}(n)=\\sum_{m=0}^{\\min(M,n)}\n       \\binom{2n}{2m}\\gamma_m p^{2m}S_{2(n-m)}(p).\n\\]\nHere $p$ need only be odd; primality is needed only to interpret the\nresult as a prime-adic congruence ladder.\n\\end{definition}","The same inverse coefficients can be written in the signed zigzag notation\nof the original manuscript.  Define $Z_r$ by\n\\[\n \\sec t+\\tan t=\\sum_{r\\ge0}Z_r\\frac{t^r}{r!}.\n\\]\nThen\n\\begin{equation}\\label{eq:arith-signed-zigzag}\n \\gamma_m=(-1)^m Z_{2m+1}\\qquad(m\\in\\N).\n\\end{equation}\nIndeed, the odd coefficients come from $\\tan t$, whose derivative is\n$\\sec^2t$.  Changing the sign of the squared variable in the even series\nfor $\\operatorname{sech}^2t$ gives $\\sec^2t$, so its degree-$2m$\nHurwitz coefficient is $(-1)^m\\gamma_m$.  Differentiation identifies\nthis coefficient with $Z_{2m+1}$, proving the formula.  The Lean bridge is\n\\lean{KomorebiPaperIX.Zigzag.inverse_is_signed_zigzag}; it uses the\nformal generating-function definition of the zigzag numbers.","ts can be written in the signed zigzag notation\nof the original manuscript.  Define $Z_r$ by\n\\[\n \\sec t+\\tan t=\\sum_{r\\ge0}Z_r\\frac{t^r}{r!}.\n\\]\nThen\n\\begin{equation}\\label{eq:arith-signed-zigzag}\n \\gamma_m=(-1)^m Z_{2m+1}\\qquad(m\\in\\N).\n\\end{equation}\nIndeed, the odd coefficients come from $\\tan t$, whose derivative is\n$\\sec^2t$.  Changing the sign of the squared variable in the even series\nfor $\\operatorname{sech}^2t$ gives $\\sec^2t$, so its degree-$2m$\nHurwitz coefficient is $(-1)^m\\gamma_m$.  Differentiation identifies\nthis coefficient with $Z_{2m+1}$, proving the formula.  The Lean bridge is\n\\lean{KomorebiPaperIX.Zigzag.inverse_is_signed_zigzag}; it uses the\nformal generating-function definition of the zigzag numbers.","zigzag notation\nof the original manuscript.  Define $Z_r$ by\n\\[\n \\sec t+\\tan t=\\sum_{r\\ge0}Z_r\\frac{t^r}{r!}.\n\\]\nThen\n\\begin{equation}\\label{eq:arith-signed-zigzag}\n \\gamma_m=(-1)^m Z_{2m+1}\\qquad(m\\in\\N).\n\\end{equation}\nIndeed, the odd coefficients come from $\\tan t$, whose derivative is\n$\\sec^2t$.  Changing the sign of the squared variable in the even series\nfor $\\operatorname{sech}^2t$ gives $\\sec^2t$, so its degree-$2m$\nHurwitz coefficient is $(-1)^m\\gamma_m$.  Differentiation identifies\nthis coefficient with $Z_{2m+1}$, proving the formula.  The Lean bridge is\n\\lean{KomorebiPaperIX.Zigzag.inverse_is_signed_zigzag}; it uses the\nformal generating-function definition of the zigzag numbers.","riginal manuscript.  Define $Z_r$ by\n\\[\n \\sec t+\\tan t=\\sum_{r\\ge0}Z_r\\frac{t^r}{r!}.\n\\]\nThen\n\\begin{equation}\\label{eq:arith-signed-zigzag}\n \\gamma_m=(-1)^m Z_{2m+1}\\qquad(m\\in\\N).\n\\end{equation}\nIndeed, the odd coefficients come from $\\tan t$, whose derivative is\n$\\sec^2t$.  Changing the sign of the squared variable in the even series\nfor $\\operatorname{sech}^2t$ gives $\\sec^2t$, so its degree-$2m$\nHurwitz coefficient is $(-1)^m\\gamma_m$.  Differentiation identifies\nthis coefficient with $Z_{2m+1}$, proving the formula.  The Lean bridge is\n\\lean{KomorebiPaperIX.Zigzag.inverse_is_signed_zigzag}; it uses the\nformal generating-function definition of the zigzag numbers.","$Z_r$ by\n\\[\n \\sec t+\\tan t=\\sum_{r\\ge0}Z_r\\frac{t^r}{r!}.\n\\]\nThen\n\\begin{equation}\\label{eq:arith-signed-zigzag}\n \\gamma_m=(-1)^m Z_{2m+1}\\qquad(m\\in\\N).\n\\end{equation}\nIndeed, the odd coefficients come from $\\tan t$, whose derivative is\n$\\sec^2t$.  Changing the sign of the squared variable in the even series\nfor $\\operatorname{sech}^2t$ gives $\\sec^2t$, so its degree-$2m$\nHurwitz coefficient is $(-1)^m\\gamma_m$.  Differentiation identifies\nthis coefficient with $Z_{2m+1}$, proving the formula.  The Lean bridge is\n\\lean{KomorebiPaperIX.Zigzag.inverse_is_signed_zigzag}; it uses the\nformal generating-function definition of the zigzag numbers.","egin{equation}\\label{eq:arith-signed-zigzag}\n \\gamma_m=(-1)^m Z_{2m+1}\\qquad(m\\in\\N).\n\\end{equation}\nIndeed, the odd coefficients come from $\\tan t$, whose derivative is\n$\\sec^2t$.  Changing the sign of the squared variable in the even series\nfor $\\operatorname{sech}^2t$ gives $\\sec^2t$, so its degree-$2m$\nHurwitz coefficient is $(-1)^m\\gamma_m$.  Differentiation identifies\nthis coefficient with $Z_{2m+1}$, proving the formula.  The Lean bridge is\n\\lean{KomorebiPaperIX.Zigzag.inverse_is_signed_zigzag}; it uses the\nformal generating-function definition of the zigzag numbers.","\\begin{theorem}[Classical odd-length reconstruction]\n\\label{thm:arith-odd-reconstruction}\nFor every positive odd integer $p$ and every $n\\in\\N$,\n\\[\n E_{2n}=\\sum_{m=0}^{n}\\binom{2n}{2m}\\gamma_m p^{2m}S_{2(n-m)}(p).\n\\]\nEquivalently, $E_{2n}=\\sum_{q=0}^n V_{nq}S_{2q}(p)$ for the lower\ntriangular matrix\n\\[\n V_{nq}=\\binom{2n}{2q}\\gamma_{n-q}p^{2(n-q)}\\quad(q\\le n),\n \\qquad V_{nq}=0\\quad(q>n).\n\\]\nIt has $V_{nn}=1$ and $p^{2(n-q)}\\mid V_{nq}$ when $q\\le n$.\n\\end{theorem}","egin{theorem}[Classical odd-length reconstruction]\n\\label{thm:arith-odd-reconstruction}\nFor every positive odd integer $p$ and every $n\\in\\N$,\n\\[\n E_{2n}=\\sum_{m=0}^{n}\\binom{2n}{2m}\\gamma_m p^{2m}S_{2(n-m)}(p).\n\\]\nEquivalently, $E_{2n}=\\sum_{q=0}^n V_{nq}S_{2q}(p)$ for the lower\ntriangular matrix\n\\[\n V_{nq}=\\binom{2n}{2q}\\gamma_{n-q}p^{2(n-q)}\\quad(q\\le n),\n \\qquad V_{nq}=0\\quad(q>n).\n\\]\nIt has $V_{nn}=1$ and $p^{2(n-q)}\\mid V_{nq}$ when $q\\le n$.\n\\end{theorem}","lassical odd-length reconstruction]\n\\label{thm:arith-odd-reconstruction}\nFor every positive odd integer $p$ and every $n\\in\\N$,\n\\[\n E_{2n}=\\sum_{m=0}^{n}\\binom{2n}{2m}\\gamma_m p^{2m}S_{2(n-m)}(p).\n\\]\nEquivalently, $E_{2n}=\\sum_{q=0}^n V_{nq}S_{2q}(p)$ for the lower\ntriangular matrix\n\\[\n V_{nq}=\\binom{2n}{2q}\\gamma_{n-q}p^{2(n-q)}\\quad(q\\le n),\n \\qquad V_{nq}=0\\quad(q>n).\n\\]\nIt has $V_{nn}=1$ and $p^{2(n-q)}\\mid V_{nq}$ when $q\\le n$.\n\\end{theorem}","ction]\n\\label{thm:arith-odd-reconstruction}\nFor every positive odd integer $p$ and every $n\\in\\N$,\n\\[\n E_{2n}=\\sum_{m=0}^{n}\\binom{2n}{2m}\\gamma_m p^{2m}S_{2(n-m)}(p).\n\\]\nEquivalently, $E_{2n}=\\sum_{q=0}^n V_{nq}S_{2q}(p)$ for the lower\ntriangular matrix\n\\[\n V_{nq}=\\binom{2n}{2q}\\gamma_{n-q}p^{2(n-q)}\\quad(q\\le n),\n \\qquad V_{nq}=0\\quad(q>n).\n\\]\nIt has $V_{nn}=1$ and $p^{2(n-q)}\\mid V_{nq}$ when $q\\le n$.\n\\end{theorem}","\\begin{proof}\nThe exponential generating series of the source is the finite sum\n\\[\n \\mathcal S_p(t)=\\sum_{a=0}^{p-1}(-1)^a e^{(2a+1)t}.\n\\]\nMultiplication by $\\cosh t$ telescopes.  Since $p$ is odd,\n\\[\n \\mathcal S_p(t)=\\frac{1+e^{2pt}}{2}\\operatorname{sech}t.\n\\]\nTaking its even part gives\n\\[\n \\sum_{n\\ge0}S_{2n}(p)\\frac{t^{2n}}{(2n)!}\n       =\\cosh^2(pt)\\operatorname{sech}t.\n\\]\nMultiplication by the reciprocal $\\operatorname{sech}^2(pt)$\nyields $\\operatorname{sech}t$.  The EGF product rule now gives the\nreconstruction coefficient by coefficient.  Replacing $m$ by $n-q$\nproduces the displayed lower triangular row.  Its diagonal and weights\nfollow from $\\gamma_0=1$ and the explicit factor $p^{2(n-q)}$.\nThe source-series identity is\n\\lean{KomorebiPaperIX.OddPower.odd_source_even_egf}; the reconstruction\nand row statement are\n\\lea","\\begin{proof}\nThe exponential generating series of the source is the finite sum\n\\[\n \\mathcal S_p(t)=\\sum_{a=0}^{p-1}(-1)^a e^{(2a+1)t}.\n\\]\nMultiplication by $\\cosh t$ telescopes.  Since $p$ is odd,\n\\[\n \\mathcal S_p(t)=\\frac{1+e^{2pt}}{2}\\operatorname{sech}t.\n\\]\nTaking its even part gives\n\\[\n \\sum_{n\\ge0}S_{2n}(p)\\frac{t^{2n}}{(2n)!}\n       =\\cosh^2(pt)\\operatorname{sech}t.\n\\]\nMultiplication by the reciprocal $\\operatorname{sech}^2(pt)$\nyields $\\operatorname{sech}t$.  The EGF product rule now gives the\nreconstruction coefficient by coefficient.  Replacing $m$ by $n-q$\nproduces the displayed lower triangular row.  Its diagonal and weights\nfollow from $\\gamma_0=1$ and the explicit factor $p^{2(n-q)}$.\nThe source-series identity is\n\\lean{KomorebiPaperIX.OddPower.odd_source_even_egf}; the reconstruction\nand row statement are\n\\lean{KomorebiPaperIX.OddEuler.od","\\begin{proof}\nThe exponential generating series of the source is the finite sum\n\\[\n \\mathcal S_p(t)=\\sum_{a=0}^{p-1}(-1)^a e^{(2a+1)t}.\n\\]\nMultiplication by $\\cosh t$ telescopes.  Since $p$ is odd,\n\\[\n \\mathcal S_p(t)=\\frac{1+e^{2pt}}{2}\\operatorname{sech}t.\n\\]\nTaking its even part gives\n\\[\n \\sum_{n\\ge0}S_{2n}(p)\\frac{t^{2n}}{(2n)!}\n       =\\cosh^2(pt)\\operatorname{sech}t.\n\\]\nMultiplication by the reciprocal $\\operatorname{sech}^2(pt)$\nyields $\\operatorname{sech}t$.  The EGF product rule now gives the\nreconstruction coefficient by coefficient.  Replacing $m$ by $n-q$\nproduces the displayed lower triangular row.  Its diagonal and weights\nfollow from $\\gamma_0=1$ and the explicit factor $p^{2(n-q)}$.\nThe source-series identity is\n\\lean{KomorebiPaperIX.OddPower.odd_source_even_egf}; the reconstruction\nand row statement are\n\\lean{KomorebiPaperIX.OddEuler.odd_euler_reconstruction}","\\begin{proof}\nThe exponential generating series of the source is the finite sum\n\\[\n \\mathcal S_p(t)=\\sum_{a=0}^{p-1}(-1)^a e^{(2a+1)t}.\n\\]\nMultiplication by $\\cosh t$ telescopes.  Since $p$ is odd,\n\\[\n \\mathcal S_p(t)=\\frac{1+e^{2pt}}{2}\\operatorname{sech}t.\n\\]\nTaking its even part gives\n\\[\n \\sum_{n\\ge0}S_{2n}(p)\\frac{t^{2n}}{(2n)!}\n       =\\cosh^2(pt)\\operatorname{sech}t.\n\\]\nMultiplication by the reciprocal $\\operatorname{sech}^2(pt)$\nyields $\\operatorname{sech}t$.  The EGF product rule now gives the\nreconstruction coefficient by coefficient.  Replacing $m$ by $n-q$\nproduces the displayed lower triangular row.  Its diagonal and weights\nfollow from $\\gamma_0=1$ and the explicit factor $p^{2(n-q)}$.\nThe source-series identity is\n\\lean{KomorebiPaperIX.OddPower.odd_source_even_egf}; the reconstruction\nand row statement are\n\\lean{KomorebiPaperIX.OddEuler.odd_euler_reconstruction} and\n\\lean{KomorebiPaperIX.OddEuler.row_action}.\n\\end{proof}","The exponential generating series of the source is the finite sum\n\\[\n \\mathcal S_p(t)=\\sum_{a=0}^{p-1}(-1)^a e^{(2a+1)t}.\n\\]\nMultiplication by $\\cosh t$ telescopes.  Since $p$ is odd,\n\\[\n \\mathcal S_p(t)=\\frac{1+e^{2pt}}{2}\\operatorname{sech}t.\n\\]\nTaking its even part gives\n\\[\n \\sum_{n\\ge0}S_{2n}(p)\\frac{t^{2n}}{(2n)!}\n       =\\cosh^2(pt)\\operatorname{sech}t.\n\\]\nMultiplication by the reciprocal $\\operatorname{sech}^2(pt)$\nyields $\\operatorname{sech}t$.  The EGF product rule now gives the\nreconstruction coefficient by coefficient.  Replacing $m$ by $n-q$\nproduces the displayed lower triangular row.  Its diagonal and weights\nfollow from $\\gamma_0=1$ and the explicit factor $p^{2(n-q)}$.\nThe source-series identity is\n\\lean{KomorebiPaperIX.OddPower.odd_source_even_egf}; the reconstruction\nand row statement are\n\\lean{KomorebiPaperIX.OddEuler.odd_euler_reconstruction} and\n\\lean{KomorebiPaperIX.OddEuler.row_action}.\n\\end{proof}","of the source is the finite sum\n\\[\n \\mathcal S_p(t)=\\sum_{a=0}^{p-1}(-1)^a e^{(2a+1)t}.\n\\]\nMultiplication by $\\cosh t$ telescopes.  Since $p$ is odd,\n\\[\n \\mathcal S_p(t)=\\frac{1+e^{2pt}}{2}\\operatorname{sech}t.\n\\]\nTaking its even part gives\n\\[\n \\sum_{n\\ge0}S_{2n}(p)\\frac{t^{2n}}{(2n)!}\n       =\\cosh^2(pt)\\operatorname{sech}t.\n\\]\nMultiplication by the reciprocal $\\operatorname{sech}^2(pt)$\nyields $\\operatorname{sech}t$.  The EGF product rule now gives the\nreconstruction coefficient by coefficient.  Replacing $m$ by $n-q$\nproduces the displayed lower triangular row.  Its diagonal and weights\nfollow from $\\gamma_0=1$ and the explicit factor $p^{2(n-q)}$.\nThe source-series identity is\n\\lean{KomorebiPaperIX.OddPower.odd_source_even_egf}; the reconstruction\nand row statement are\n\\lean{KomorebiPaperIX.OddEuler.odd_euler_reconstruction} and\n\\lean{KomorebiPaperIX.OddEuler.row_action}.\n\\end{proof}","h t$ telescopes.  Since $p$ is odd,\n\\[\n \\mathcal S_p(t)=\\frac{1+e^{2pt}}{2}\\operatorname{sech}t.\n\\]\nTaking its even part gives\n\\[\n \\sum_{n\\ge0}S_{2n}(p)\\frac{t^{2n}}{(2n)!}\n       =\\cosh^2(pt)\\operatorname{sech}t.\n\\]\nMultiplication by the reciprocal $\\operatorname{sech}^2(pt)$\nyields $\\operatorname{sech}t$.  The EGF product rule now gives the\nreconstruction coefficient by coefficient.  Replacing $m$ by $n-q$\nproduces the displayed lower triangular row.  Its diagonal and weights\nfollow from $\\gamma_0=1$ and the explicit factor $p^{2(n-q)}$.\nThe source-series identity is\n\\lean{KomorebiPaperIX.OddPower.odd_source_even_egf}; the reconstruction\nand row statement are\n\\lean{KomorebiPaperIX.OddEuler.odd_euler_reconstruction} and\n\\lean{KomorebiPaperIX.OddEuler.row_action}.\n\\end{proof}","lescopes.  Since $p$ is odd,\n\\[\n \\mathcal S_p(t)=\\frac{1+e^{2pt}}{2}\\operatorname{sech}t.\n\\]\nTaking its even part gives\n\\[\n \\sum_{n\\ge0}S_{2n}(p)\\frac{t^{2n}}{(2n)!}\n       =\\cosh^2(pt)\\operatorname{sech}t.\n\\]\nMultiplication by the reciprocal $\\operatorname{sech}^2(pt)$\nyields $\\operatorname{sech}t$.  The EGF product rule now gives the\nreconstruction coefficient by coefficient.  Replacing $m$ by $n-q$\nproduces the displayed lower triangular row.  Its diagonal and weights\nfollow from $\\gamma_0=1$ and the explicit factor $p^{2(n-q)}$.\nThe source-series identity is\n\\lean{KomorebiPaperIX.OddPower.odd_source_even_egf}; the reconstruction\nand row statement are\n\\lean{KomorebiPaperIX.OddEuler.odd_euler_reconstruction} and\n\\lean{KomorebiPaperIX.OddEuler.row_action}.\n\\end{proof}","\\]\nTaking its even part gives\n\\[\n \\sum_{n\\ge0}S_{2n}(p)\\frac{t^{2n}}{(2n)!}\n       =\\cosh^2(pt)\\operatorname{sech}t.\n\\]\nMultiplication by the reciprocal $\\operatorname{sech}^2(pt)$\nyields $\\operatorname{sech}t$.  The EGF product rule now gives the\nreconstruction coefficient by coefficient.  Replacing $m$ by $n-q$\nproduces the displayed lower triangular row.  Its diagonal and weights\nfollow from $\\gamma_0=1$ and the explicit factor $p^{2(n-q)}$.\nThe source-series identity is\n\\lean{KomorebiPaperIX.OddPower.odd_source_even_egf}; the reconstruction\nand row statement are\n\\lean{KomorebiPaperIX.OddEuler.odd_euler_reconstruction} and\n\\lean{KomorebiPaperIX.OddEuler.row_action}.\n\\end{proof}","um_{n\\ge0}S_{2n}(p)\\frac{t^{2n}}{(2n)!}\n       =\\cosh^2(pt)\\operatorname{sech}t.\n\\]\nMultiplication by the reciprocal $\\operatorname{sech}^2(pt)$\nyields $\\operatorname{sech}t$.  The EGF product rule now gives the\nreconstruction coefficient by coefficient.  Replacing $m$ by $n-q$\nproduces the displayed lower triangular row.  Its diagonal and weights\nfollow from $\\gamma_0=1$ and the explicit factor $p^{2(n-q)}$.\nThe source-series identity is\n\\lean{KomorebiPaperIX.OddPower.odd_source_even_egf}; the reconstruction\nand row statement are\n\\lean{KomorebiPaperIX.OddEuler.odd_euler_reconstruction} and\n\\lean{KomorebiPaperIX.OddEuler.row_action}.\n\\end{proof}","\\begin{corollary}[Odd-prime truncation ladder]\\label{cor:arith-odd-ladder}\nFor every positive odd integer $p$ and all $M,n\\in\\N$,\n\\[\n p^{2M+2}\\mid E_{2n}-T^{\\mathrm{odd}}_{p,M}(n)\\quad\\text{in }\\Z.\n\\]\nIf $M\\ge n$, the truncation equals $E_{2n}$ exactly.  In particular,\nfor an odd prime $p$ this is a congruence modulo $p^{2M+2}$ in $\\Z_p$.\n\\end{corollary}","\\begin{proof}\nEvery omitted summand has $m\\ge M+1$, hence contains the integer factor\n$p^{2m}$ divisible by $p^{2M+2}$.  All its other factors are integral.\nThe finite sum of these omitted terms proves the assertion, with the\nempty-tail case giving exact equality.  These are\n\\lean{KomorebiPaperIX.OddEuler.odd_euler_supercongruence} and\n\\lean{KomorebiPaperIX.OddEuler.full_depth_exact}.\n\\end{proof}","The reconstruction and its truncation bound already hold over the integers.\nFor the dyadic construction we use a different source.  Its original bound\nof $2^{2M+2}$ follows from integral reciprocal coefficients, and the next\nsubsection sharpens that bound to the optimal uniform exponent $4M+3$.","Return now to the classical signed Euler numbers\n\\[\n \\operatorname{sech}t=\\frac1{\\cosh t}\n       =\\sum_{n\\ge0}E_{2n}\\frac{t^{2n}}{(2n)!}.\n\\]\nThus $E_{2n}=E_{2n}^{(2,0)}$ in the convention\n\\eqref{eq:arith-euler-egf}.  In particular $E_0=1$, $E_2=-1$, and $E_4=5$.\nThe following construction concerns these $N=2$ coefficients, not $C_{4,0}$.","\\begin{definition}[Dyadic truncation]\\label{def:arith-truncation}\nDefine $\\beta_m$ by the formal unit-series identity\n\\[\n \\sum_{m\\ge0}\\beta_m z^{2m}=\\left(\\frac{z}{\\sinh z}\\right)^2,\n \\qquad \\beta_0=1,\n\\]\nwhere the quotient means the inverse of the unit series\n$(\\sinh z/z)^2$.  Define\n\\[\n A(n)=\\frac{3^{2n+2}-1}{4(2n+1)(2n+2)},\\qquad\n S_m(n)=\\beta_m2^{2m}\\frac{(2n)!}{(2n-2m)!}A(n-m)\n       \\quad(0\\le m\\le n),\n\\]\nand\n\\begin{equation}\\label{eq:arith-tm}\n T_M(n)=\\sum_{m=0}^{\\min(M,n)}S_m(n)\\qquad(M,n\\in\\N).\n\\end{equation}\n\\end{definition}","n]\\label{def:arith-truncation}\nDefine $\\beta_m$ by the formal unit-series identity\n\\[\n \\sum_{m\\ge0}\\beta_m z^{2m}=\\left(\\frac{z}{\\sinh z}\\right)^2,\n \\qquad \\beta_0=1,\n\\]\nwhere the quotient means the inverse of the unit series\n$(\\sinh z/z)^2$.  Define\n\\[\n A(n)=\\frac{3^{2n+2}-1}{4(2n+1)(2n+2)},\\qquad\n S_m(n)=\\beta_m2^{2m}\\frac{(2n)!}{(2n-2m)!}A(n-m)\n       \\quad(0\\le m\\le n),\n\\]\nand\n\\begin{equation}\\label{eq:arith-tm}\n T_M(n)=\\sum_{m=0}^{\\min(M,n)}S_m(n)\\qquad(M,n\\in\\N).\n\\end{equation}\n\\end{definition}","\\begin{lemma}[Exact reconstruction and source integrality]\n\\label{lem:arith-reconstruction}\nFor every $n\\in\\N$,\n\\[\n E_{2n}=\\sum_{m=0}^{n}S_m(n).\n\\]\nEvery $\\beta_m$ is $2$-adically integral and every $A(n)$ is a nonzero\n$2$-adic unit.  If $M\\ge n$, then $T_M(n)=E_{2n}$.\n\\end{lemma}","\\begin{proof}\nThe ordinary kernel coefficients of $(\\sinh z/z)^2$, in $z^2$, are\n\\[\n \\frac{2^{2m+1}}{(2m+2)!}.\n\\]\nLegendre's strict factorial bound\n$v_2((2m+2)!)<2m+2$ makes them integral.  Their constant coefficient is\none.  The coefficient recurrence for its reciprocal therefore\nproves integrality of every $\\beta_m$.\nFor the source factor, lifting the exponent at the even power $2n+2$\ngives\n\\[\n v_2(3^{2n+2}-1)=2+v_2(2n+2).\n\\]\nIts denominator has the same valuation, since $2n+1$ is odd.  Thus\n$A(n)$ is a nonzero unit.","e\n\\[\n \\frac{2^{2m+1}}{(2m+2)!}.\n\\]\nLegendre's strict factorial bound\n$v_2((2m+2)!)<2m+2$ makes them integral.  Their constant coefficient is\none.  The coefficient recurrence for its reciprocal therefore\nproves integrality of every $\\beta_m$.\nFor the source factor, lifting the exponent at the even power $2n+2$\ngives\n\\[\n v_2(3^{2n+2}-1)=2+v_2(2n+2).\n\\]\nIts denominator has the same valuation, since $2n+1$ is odd.  Thus\n$A(n)$ is a nonzero unit.","{(2m+2)!}.\n\\]\nLegendre's strict factorial bound\n$v_2((2m+2)!)<2m+2$ makes them integral.  Their constant coefficient is\none.  The coefficient recurrence for its reciprocal therefore\nproves integrality of every $\\beta_m$.\nFor the source factor, lifting the exponent at the even power $2n+2$\ngives\n\\[\n v_2(3^{2n+2}-1)=2+v_2(2n+2).\n\\]\nIts denominator has the same valuation, since $2n+1$ is odd.  Thus\n$A(n)$ is a nonzero unit.","Its exponential generating function is\n\\[\n \\mathcal A(t)=\\sum_{n\\ge0}A(n)\\frac{t^{2n}}{(2n)!}\n             =\\frac{\\cosh(3t)-\\cosh t}{4t^2}.\n\\]\nThe quotients by powers of $t$ are well defined because the required\ninitial coefficients vanish.  The hyperbolic identity\n\\[\n \\mathcal A(t)\\cosh t=\\left(\\frac{\\sinh(2t)}{2t}\\right)^2\n\\]\nthen gives the exact formal-series product\n\\[\n \\left(\\sum_{m\\ge0}\\beta_m(2t)^{2m}\\right)\\mathcal A(t)\n       =\\operatorname{sech}t.\n\\]\nTaking the factorial-scaled coefficient of $t^{2n}$ produces the stated\nfinite sum.  Full depth leaves no omitted terms.\nThe relevant declarations are\n\\lean{KomorebiPaperIX.Dyadic.beta_integral},\n\\lean{KomorebiPaperIX.Dyadic.source_valuation},\n\\lean{KomorebiPaperIX.DyadicSeries.dyadic_reconstruction_series}, and\n\\lean{KomorebiPaperIX.DyadicReconstruction.dyadic_reconstruct","Its exponential generating function is\n\\[\n \\mathcal A(t)=\\sum_{n\\ge0}A(n)\\frac{t^{2n}}{(2n)!}\n             =\\frac{\\cosh(3t)-\\cosh t}{4t^2}.\n\\]\nThe quotients by powers of $t$ are well defined because the required\ninitial coefficients vanish.  The hyperbolic identity\n\\[\n \\mathcal A(t)\\cosh t=\\left(\\frac{\\sinh(2t)}{2t}\\right)^2\n\\]\nthen gives the exact formal-series product\n\\[\n \\left(\\sum_{m\\ge0}\\beta_m(2t)^{2m}\\right)\\mathcal A(t)\n       =\\operatorname{sech}t.\n\\]\nTaking the factorial-scaled coefficient of $t^{2n}$ produces the stated\nfinite sum.  Full depth leaves no omitted terms.\nThe relevant declarations are\n\\lean{KomorebiPaperIX.Dyadic.beta_integral},\n\\lean{KomorebiPaperIX.Dyadic.source_valuation},\n\\lean{KomorebiPaperIX.DyadicSeries.dyadic_reconstruction_series}, and\n\\lean{KomorebiPaperIX.DyadicReconstruction.dyadic_reconstruction}.\n\\end{proof}","frac{\\cosh(3t)-\\cosh t}{4t^2}.\n\\]\nThe quotients by powers of $t$ are well defined because the required\ninitial coefficients vanish.  The hyperbolic identity\n\\[\n \\mathcal A(t)\\cosh t=\\left(\\frac{\\sinh(2t)}{2t}\\right)^2\n\\]\nthen gives the exact formal-series product\n\\[\n \\left(\\sum_{m\\ge0}\\beta_m(2t)^{2m}\\right)\\mathcal A(t)\n       =\\operatorname{sech}t.\n\\]\nTaking the factorial-scaled coefficient of $t^{2n}$ produces the stated\nfinite sum.  Full depth leaves no omitted terms.\nThe relevant declarations are\n\\lean{KomorebiPaperIX.Dyadic.beta_integral},\n\\lean{KomorebiPaperIX.Dyadic.source_valuation},\n\\lean{KomorebiPaperIX.DyadicSeries.dyadic_reconstruction_series}, and\n\\lean{KomorebiPaperIX.DyadicReconstruction.dyadic_reconstruction}.\n\\end{proof}","\\begin{lemma}[Exact weighted inverse coefficient]\\label{lem:arith-kappa}\nLet $\\kappa_m=\\beta_m2^{2m}(2m)!$.  For every $m\\ge1$,\n\\[\n \\kappa_m=-2^{4m}(2m-1)B_{2m}\\ne0,\\qquad\n v_2(\\kappa_m)=4m-1,\n\\]\nwhere $B_{2m}$ is the usual even Bernoulli number.\n\\end{lemma}","\\begin{proof}\nLet $Q(z)=z\\coth z$, the regular formal series with constant coefficient\none.  The elementary differential identity\n\\[\n \\left(\\frac z{\\sinh z}\\right)^2=Q(z)-zQ'(z)\n\\]\nand the Bernoulli expansion\n\\[\n Q(z)=1+\\sum_{m\\ge1}\\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}\n\\]\ngive\n$\\beta_m=-2^{2m}(2m-1)B_{2m}/(2m)!$.\nMultiplication by $2^{2m}(2m)!$ proves the identity for $\\kappa_m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli","\\begin{proof}\nLet $Q(z)=z\\coth z$, the regular formal series with constant coefficient\none.  The elementary differential identity\n\\[\n \\left(\\frac z{\\sinh z}\\right)^2=Q(z)-zQ'(z)\n\\]\nand the Bernoulli expansion\n\\[\n Q(z)=1+\\sum_{m\\ge1}\\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}\n\\]\ngive\n$\\beta_m=-2^{2m}(2m-1)B_{2m}/(2m)!$.\nMultiplication by $2^{2m}(2m)!$ proves the identity for $\\kappa_m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli theorem then\ndetermines the valuation of $\\kappa_m$.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_bernoulli} and\n\\lean{Komore","\\begin{proof}\nLet $Q(z)=z\\coth z$, the regular formal series with constant coefficient\none.  The elementary differential identity\n\\[\n \\left(\\frac z{\\sinh z}\\right)^2=Q(z)-zQ'(z)\n\\]\nand the Bernoulli expansion\n\\[\n Q(z)=1+\\sum_{m\\ge1}\\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}\n\\]\ngive\n$\\beta_m=-2^{2m}(2m-1)B_{2m}/(2m)!$.\nMultiplication by $2^{2m}(2m)!$ proves the identity for $\\kappa_m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli theorem then\ndetermines the valuation of $\\kappa_m$.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_bernoulli} and\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_valuation}.\n\\end{proof}","Q(z)=z\\coth z$, the regular formal series with constant coefficient\none.  The elementary differential identity\n\\[\n \\left(\\frac z{\\sinh z}\\right)^2=Q(z)-zQ'(z)\n\\]\nand the Bernoulli expansion\n\\[\n Q(z)=1+\\sum_{m\\ge1}\\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}\n\\]\ngive\n$\\beta_m=-2^{2m}(2m-1)B_{2m}/(2m)!$.\nMultiplication by $2^{2m}(2m)!$ proves the identity for $\\kappa_m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli theorem then\ndetermines the valuation of $\\kappa_m$.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_bernoulli} and\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_valuation}.\n\\end{proof}","egular formal series with constant coefficient\none.  The elementary differential identity\n\\[\n \\left(\\frac z{\\sinh z}\\right)^2=Q(z)-zQ'(z)\n\\]\nand the Bernoulli expansion\n\\[\n Q(z)=1+\\sum_{m\\ge1}\\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}\n\\]\ngive\n$\\beta_m=-2^{2m}(2m-1)B_{2m}/(2m)!$.\nMultiplication by $2^{2m}(2m)!$ proves the identity for $\\kappa_m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli theorem then\ndetermines the valuation of $\\kappa_m$.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_bernoulli} and\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_valuation}.\n\\end{proof}","ent\none.  The elementary differential identity\n\\[\n \\left(\\frac z{\\sinh z}\\right)^2=Q(z)-zQ'(z)\n\\]\nand the Bernoulli expansion\n\\[\n Q(z)=1+\\sum_{m\\ge1}\\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}\n\\]\ngive\n$\\beta_m=-2^{2m}(2m-1)B_{2m}/(2m)!$.\nMultiplication by $2^{2m}(2m)!$ proves the identity for $\\kappa_m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli theorem then\ndetermines the valuation of $\\kappa_m$.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_bernoulli} and\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_valuation}.\n\\end{proof}","The elementary differential identity\n\\[\n \\left(\\frac z{\\sinh z}\\right)^2=Q(z)-zQ'(z)\n\\]\nand the Bernoulli expansion\n\\[\n Q(z)=1+\\sum_{m\\ge1}\\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}\n\\]\ngive\n$\\beta_m=-2^{2m}(2m-1)B_{2m}/(2m)!$.\nMultiplication by $2^{2m}(2m)!$ proves the identity for $\\kappa_m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli theorem then\ndetermines the valuation of $\\kappa_m$.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_bernoulli} and\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_valuation}.\n\\end{proof}","[\n \\left(\\frac z{\\sinh z}\\right)^2=Q(z)-zQ'(z)\n\\]\nand the Bernoulli expansion\n\\[\n Q(z)=1+\\sum_{m\\ge1}\\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}\n\\]\ngive\n$\\beta_m=-2^{2m}(2m-1)B_{2m}/(2m)!$.\nMultiplication by $2^{2m}(2m)!$ proves the identity for $\\kappa_m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli theorem then\ndetermines the valuation of $\\kappa_m$.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_bernoulli} and\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_valuation}.\n\\end{proof}","}\\right)^2=Q(z)-zQ'(z)\n\\]\nand the Bernoulli expansion\n\\[\n Q(z)=1+\\sum_{m\\ge1}\\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}\n\\]\ngive\n$\\beta_m=-2^{2m}(2m-1)B_{2m}/(2m)!$.\nMultiplication by $2^{2m}(2m)!$ proves the identity for $\\kappa_m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli theorem then\ndetermines the valuation of $\\kappa_m$.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_bernoulli} and\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_valuation}.\n\\end{proof}","lli expansion\n\\[\n Q(z)=1+\\sum_{m\\ge1}\\frac{2^{2m}B_{2m}}{(2m)!}z^{2m}\n\\]\ngive\n$\\beta_m=-2^{2m}(2m-1)B_{2m}/(2m)!$.\nMultiplication by $2^{2m}(2m)!$ proves the identity for $\\kappa_m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli theorem then\ndetermines the valuation of $\\kappa_m$.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_bernoulli} and\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_valuation}.\n\\end{proof}","m$.\nThe Bernoulli denominator theorem gives $v_2(B_{2m})=-1$ for $m>0$: its denominator contains each prime $q$ with $(q-1)\\mid 2m$ exactly once, including $2$\\cite{DLMF24}.\nSince $2m-1$ is odd, the valuation is exactly $4m-1$.\nThis also proves nonvanishing.  In Lean, the hyperbolic identity is proved\nafter clearing removable zeros.  The library Bernoulli theorem then\ndetermines the valuation of $\\kappa_m$.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_bernoulli} and\n\\lean{KomorebiPaperIX.DyadicSharpTower.kappa_valuation}.\n\\end{proof}","\\begin{theorem}[Optimal uniform dyadic tower]\\label{thm:arith-sharp-tower}\nFor all $M,n\\in\\N$,\n\\begin{equation}\\label{eq:arith-sharp-tail}\n E_{2n}-T_M(n)\\in2^{4M+3}\\Z_2.\n\\end{equation}\nAt the first omitted row $n=M+1$, the error is nonzero and\n\\[\n v_2\\bigl(E_{2M+2}-T_M(M+1)\\bigr)=4M+3.\n\\]\nThus the exponent $4M+3$ cannot be increased uniformly in the Euler row.\n\\end{theorem}","\\begin{proof}\nRewrite the summand exactly as\n\\[\n S_m(n)=\\kappa_m\\binom{2n}{2m}A(n-m).\n\\]\nFor $m>M$, its first factor is divisible by $2^{4m-1}$, the binomial\ncoefficient is integral, and $A(n-m)$ is a unit.  Hence every omitted\nsummand is divisible by $2^{4M+3}$.  Summing the finite tail proves\n\\eqref{eq:arith-sharp-tail}.  If $M\\ge n$, the tail is empty and the\nstatement includes the resulting zero error.","At $n=M+1$ there is exactly one omitted summand.  Here\n$\\binom{2n}{2n}=1$ and $A(0)=1$, so the error is $\\kappa_{M+1}$.\nLemma~\\ref{lem:arith-kappa} gives its exact valuation $4M+3$ and\nnonvanishing.  The statements are\n\\lean{KomorebiPaperIX.DyadicSharpTower.dyadic_supercongruence_sharp},\n\\lean{KomorebiPaperIX.DyadicSharpTower.first_omitted_error}, and\n\\lean{KomorebiPaperIX.DyadicSharpTower.first_omitted_error_valuation}.\n\\end{proof}","\\begin{proposition}[Row-dependent summand weights]\\label{prop:arith-row-weight}\nFor $1\\le m\\le n$, the summand is nonzero and\n\\[\n v_2(S_m(n))=4m-1+v_2\\binom{2n}{2m}.\n\\]\nMore generally, if $L\\in\\N$ satisfies\n\\[\n L\\le4m-1+v_2\\binom{2n}{2m}\\qquad\\text{for every }M<m\\le n,\n\\]\nthen $E_{2n}-T_M(n)\\in2^L\\Z_2$.\n\\end{proposition}","\\begin{proof}\nUse the same factorization of $S_m(n)$, the nonzero binomial coefficient\nfor $m\\le n$, the exact valuation of $\\kappa_m$, and the unit property\nof $A(n-m)$.  The tail conclusion follows by addition in $2^L\\Z_2$.\nThese are \\lean{KomorebiPaperIX.DyadicSharpTower.summand_valuation} and\n\\lean{KomorebiPaperIX.DyadicSharpTower.dyadic_supercongruence_row_weight}.\n\\end{proof}","The minimum summand weight bounds the tail valuation from below, but\ncancellation may raise that valuation. Appendix~\\ref{app:root:scope} gives a later exact factorization and finite-polynomial reduction for each fixed depth. A universal explicit root and simplicity classification for that factor remains open.  Uniform optimality follows\nfrom the equality already established at $n=M+1$.","\\begin{theorem}[Isometric packet bijections]\\label{thm:arith-isometry}\nFor $0\\le j<4$, define on natural numbers\n\\[\n f_j(n)=\\frac{U_j(n)-j!}{2^{d_j}},\\qquad (d_0,d_1,d_2,d_3)=(1,1,5,3).\n\\]\nThere is a unique continuous map $\\Phi_j:\\Z_2\\to\\Z_2$ whose value at\nevery $n\\in\\N$ equals the embedded rational $f_j(n)$.\nIt is an isometric bijection and $\\Phi_j(0)=0$:\n\\[\n |\\Phi_j(x)-\\Phi_j(y)|_2=|x-y|_2\\qquad(x,y\\in\\Z_2).\n\\]\nIn particular, every element of $\\Z_2$ has exactly one preimage.\n\\end{theorem}","thm:arith-isometry}\nFor $0\\le j<4$, define on natural numbers\n\\[\n f_j(n)=\\frac{U_j(n)-j!}{2^{d_j}},\\qquad (d_0,d_1,d_2,d_3)=(1,1,5,3).\n\\]\nThere is a unique continuous map $\\Phi_j:\\Z_2\\to\\Z_2$ whose value at\nevery $n\\in\\N$ equals the embedded rational $f_j(n)$.\nIt is an isometric bijection and $\\Phi_j(0)=0$:\n\\[\n |\\Phi_j(x)-\\Phi_j(y)|_2=|x-y|_2\\qquad(x,y\\in\\Z_2).\n\\]\nIn particular, every element of $\\Z_2$ has exactly one preimage.\n\\end{theorem}","\\begin{proof}\nFor $n>0$, the displacement law at row zero gives\n$v_2(U_j(n)-j!)=v_2(n)+d_j$, so $f_j(n)\\in\\Z_2$.\nThe value at zero is zero.  Applying the same law to the smaller of two\nunequal natural numbers shows\n\\[\n |f_j(n)-f_j(m)|_2=|n-m|_2.\n\\]\nThe natural numbers are dense in $\\Z_2$, and the target is complete.\nThe map therefore extends continuously.  The distance identity extends\non the dense product of natural-number points, making the extension\nan isometry.  Density also proves uniqueness among all continuous maps\nwith these prescribed values.","To prove surjectivity, use compactness.  For an isometry $g$ of a compact\nmetric space into itself, the sequence of iterates $g^n(x)$ has a\nconvergent subsequence.  Two sufficiently close iterates, with exponents\n$a<b$, yield $d(x,g^{b-a}(x))$ equally small after removing $a$ isometric\niterations.  Since $b-a>0$, the latter point is in the image of $g$.\nThus the image is dense.  It is also compact and hence closed, so it is\nthe whole space.  Apply this to $\\Phi_j$ on $\\Z_2$.","of a compact\nmetric space into itself, the sequence of iterates $g^n(x)$ has a\nconvergent subsequence.  Two sufficiently close iterates, with exponents\n$a<b$, yield $d(x,g^{b-a}(x))$ equally small after removing $a$ isometric\niterations.  Since $b-a>0$, the latter point is in the image of $g$.\nThus the image is dense.  It is also compact and hence closed, so it is\nthe whole space.  Apply this to $\\Phi_j$ on $\\Z_2$.","metric space into itself, the sequence of iterates $g^n(x)$ has a\nconvergent subsequence.  Two sufficiently close iterates, with exponents\n$a<b$, yield $d(x,g^{b-a}(x))$ equally small after removing $a$ isometric\niterations.  Since $b-a>0$, the latter point is in the image of $g$.\nThus the image is dense.  It is also compact and hence closed, so it is\nthe whole space.  Apply this to $\\Phi_j$ on $\\Z_2$.","\\begin{corollary}[Exact image of the original coefficients]\n\\label{cor:arith-affine-image}\nThe continuous extension of the original sequence is\n\\[\n \\widetilde U_j(x)=j!+2^{d_j}\\Phi_j(x).\n\\]\nIt is injective, its image is exactly $j!+2^{d_j}\\Z_2$, and\n\\[\n |\\widetilde U_j(x)-\\widetilde U_j(y)|_2\n     =2^{-d_j}|x-y|_2.\n\\]\nEvery point of that affine image has exactly one preimage in $\\Z_2$.\n\\end{corollary}","\\begin{proof}\nThe formula agrees with $U_j$ on every natural number.  Continuity and\nthe distance equality follow from the corresponding properties of\n$\\Phi_j$.  Its surjectivity gives precisely the displayed affine image,\nand multiplication by the nonzero scalar $2^{d_j}$ preserves injectivity.\nThe declarations are\n\\lean{KomorebiPaperIX.DyadicFourIsometry.originalExtension_nat},\n\\lean{KomorebiPaperIX.DyadicFourIsometry.originalExtension_dist}, and\n\\lean{KomorebiPaperIX.DyadicFourIsometry.originalExtension_range}.\n\\end{proof}","Unless a different coefficient ring is specified, a polynomial system is a\nfinite list of integer polynomials. All divisibility conditions on vectors are\ncoordinatewise. For a positive integer $q$, write $S_q=\\Z/q\\Z$ and use an\noverline for reduction modulo $q$. For a prime $p$, write\n$\\F_p=S_p$, normalize $|p|_p=p^{-1}$, and put\n\\[\n B_L(x)=x+p^L\\Zp^d.\n\\]\nAn integer center is included in this notation by its natural image in\n$\\Zp^d$. All dimensions below are finite and may be zero. A statement over\n$S_q$ allows composite $q$ and $q=1$ when explicitly stated. A cardinality\nformula in terms of vector-space rank uses prime $p$.","ring is specified, a polynomial system is a\nfinite list of integer polynomials. All divisibility conditions on vectors are\ncoordinatewise. For a positive integer $q$, write $S_q=\\Z/q\\Z$ and use an\noverline for reduction modulo $q$. For a prime $p$, write\n$\\F_p=S_p$, normalize $|p|_p=p^{-1}$, and put\n\\[\n B_L(x)=x+p^L\\Zp^d.\n\\]\nAn integer center is included in this notation by its natural image in\n$\\Zp^d$. All dimensions below are finite and may be zero. A statement over\n$S_q$ allows composite $q$ and $q=1$ when explicitly stated. A cardinality\nformula in terms of vector-space rank uses prime $p$.","ials. All divisibility conditions on vectors are\ncoordinatewise. For a positive integer $q$, write $S_q=\\Z/q\\Z$ and use an\noverline for reduction modulo $q$. For a prime $p$, write\n$\\F_p=S_p$, normalize $|p|_p=p^{-1}$, and put\n\\[\n B_L(x)=x+p^L\\Zp^d.\n\\]\nAn integer center is included in this notation by its natural image in\n$\\Zp^d$. All dimensions below are finite and may be zero. A statement over\n$S_q$ allows composite $q$ and $q=1$ when explicitly stated. A cardinality\nformula in terms of vector-space rank uses prime $p$.","conditions on vectors are\ncoordinatewise. For a positive integer $q$, write $S_q=\\Z/q\\Z$ and use an\noverline for reduction modulo $q$. For a prime $p$, write\n$\\F_p=S_p$, normalize $|p|_p=p^{-1}$, and put\n\\[\n B_L(x)=x+p^L\\Zp^d.\n\\]\nAn integer center is included in this notation by its natural image in\n$\\Zp^d$. All dimensions below are finite and may be zero. A statement over\n$S_q$ allows composite $q$ and $q=1$ when explicitly stated. A cardinality\nformula in terms of vector-space rank uses prime $p$.","ns on vectors are\ncoordinatewise. For a positive integer $q$, write $S_q=\\Z/q\\Z$ and use an\noverline for reduction modulo $q$. For a prime $p$, write\n$\\F_p=S_p$, normalize $|p|_p=p^{-1}$, and put\n\\[\n B_L(x)=x+p^L\\Zp^d.\n\\]\nAn integer center is included in this notation by its natural image in\n$\\Zp^d$. All dimensions below are finite and may be zero. A statement over\n$S_q$ allows composite $q$ and $q=1$ when explicitly stated. A cardinality\nformula in terms of vector-space rank uses prime $p$.","verline for reduction modulo $q$. For a prime $p$, write\n$\\F_p=S_p$, normalize $|p|_p=p^{-1}$, and put\n\\[\n B_L(x)=x+p^L\\Zp^d.\n\\]\nAn integer center is included in this notation by its natural image in\n$\\Zp^d$. All dimensions below are finite and may be zero. A statement over\n$S_q$ allows composite $q$ and $q=1$ when explicitly stated. A cardinality\nformula in terms of vector-space rank uses prime $p$.","For $F=(F_1,\\ldots,F_e)\\in\\Z[X_1,\\ldots,X_d]^e$, define its evaluated\nJacobian by the formal partial derivatives:\n\\[\n DF(x)_{ij}=(\\partial_jF_i)(x),\\qquad\n J_x=DF(x)\\bmod q:S_q^d\\longrightarrow S_q^e.\n\\]\nThe obstruction therefore uses the Jacobian of the given polynomials. The\nfollowing algebraic Taylor identity relates this linear map to polynomial\nevaluation.","\\begin{lemma}[Polynomial Taylor remainder]\\label{lem:polynomial-taylor}\nLet $R$ be a commutative ring, $P\\in R[X_1,\\ldots,X_d]$, $a\\in R^d$, and\n$c\\in R$. There is a polynomial $Q\\in R[U_1,\\ldots,U_d]$, with $Q(0)=0$,\nsuch that for every $u\\in R^d$,\n\\begin{equation}\\label{eq:polynomial-taylor}\n P(a+cu)=P(a)+c\\sum_{j=1}^d(\\partial_jP)(a)u_j+c^2Q(u).\n\\end{equation}\nIn particular, when $R=\\Z$ and $a,u,c$ are integral, subtracting the first\ntwo terms on the right from $P(a+cu)$ gives a multiple of $c^2$.\n\\end{lemma}","}\nLet $R$ be a commutative ring, $P\\in R[X_1,\\ldots,X_d]$, $a\\in R^d$, and\n$c\\in R$. There is a polynomial $Q\\in R[U_1,\\ldots,U_d]$, with $Q(0)=0$,\nsuch that for every $u\\in R^d$,\n\\begin{equation}\\label{eq:polynomial-taylor}\n P(a+cu)=P(a)+c\\sum_{j=1}^d(\\partial_jP)(a)u_j+c^2Q(u).\n\\end{equation}\nIn particular, when $R=\\Z$ and $a,u,c$ are integral, subtracting the first\ntwo terms on the right from $P(a+cu)$ gives a multiple of $c^2$.\n\\end{lemma}","g, $P\\in R[X_1,\\ldots,X_d]$, $a\\in R^d$, and\n$c\\in R$. There is a polynomial $Q\\in R[U_1,\\ldots,U_d]$, with $Q(0)=0$,\nsuch that for every $u\\in R^d$,\n\\begin{equation}\\label{eq:polynomial-taylor}\n P(a+cu)=P(a)+c\\sum_{j=1}^d(\\partial_jP)(a)u_j+c^2Q(u).\n\\end{equation}\nIn particular, when $R=\\Z$ and $a,u,c$ are integral, subtracting the first\ntwo terms on the right from $P(a+cu)$ gives a multiple of $c^2$.\n\\end{lemma}","\\begin{proof}\nApply polynomial induction using constants, addition, and multiplication by a\nvariable. Constants have zero remainder, and remainders add. For the\nmultiplication step, suppose the assertion is known for $P$ and write\n$D_P(U)=\\sum_j(\\partial_jP)(a)U_j$. Multiplication of its expansion by\n$a_i+cU_i$ gives constant term $P(a)a_i$, linear term\n$c(a_iD_P(U)+P(a)U_i)$, and remainder\n\\[\n c^2\\bigl(Q(U)(a_i+cU_i)+D_P(U)U_i\\bigr).\n\\]\nThe product rule identifies the linear term with the derivative of\n$PX_i$. The displayed remainder vanishes at $U=0$. This constructs the\npolynomial remainder before evaluation at $u$, even over rings where\ndivision by $c$ would be invalid.\n\\end{proof}","s have zero remainder, and remainders add. For the\nmultiplication step, suppose the assertion is known for $P$ and write\n$D_P(U)=\\sum_j(\\partial_jP)(a)U_j$. Multiplication of its expansion by\n$a_i+cU_i$ gives constant term $P(a)a_i$, linear term\n$c(a_iD_P(U)+P(a)U_i)$, and remainder\n\\[\n c^2\\bigl(Q(U)(a_i+cU_i)+D_P(U)U_i\\bigr).\n\\]\nThe product rule identifies the linear term with the derivative of\n$PX_i$. The displayed remainder vanishes at $U=0$. This constructs the\npolynomial remainder before evaluation at $u$, even over rings where\ndivision by $c$ would be invalid.\n\\end{proof}","or the\nmultiplication step, suppose the assertion is known for $P$ and write\n$D_P(U)=\\sum_j(\\partial_jP)(a)U_j$. Multiplication of its expansion by\n$a_i+cU_i$ gives constant term $P(a)a_i$, linear term\n$c(a_iD_P(U)+P(a)U_i)$, and remainder\n\\[\n c^2\\bigl(Q(U)(a_i+cU_i)+D_P(U)U_i\\bigr).\n\\]\nThe product rule identifies the linear term with the derivative of\n$PX_i$. The displayed remainder vanishes at $U=0$. This constructs the\npolynomial remainder before evaluation at $u$, even over rings where\ndivision by $c$ would be invalid.\n\\end{proof}","d write\n$D_P(U)=\\sum_j(\\partial_jP)(a)U_j$. Multiplication of its expansion by\n$a_i+cU_i$ gives constant term $P(a)a_i$, linear term\n$c(a_iD_P(U)+P(a)U_i)$, and remainder\n\\[\n c^2\\bigl(Q(U)(a_i+cU_i)+D_P(U)U_i\\bigr).\n\\]\nThe product rule identifies the linear term with the derivative of\n$PX_i$. The displayed remainder vanishes at $U=0$. This constructs the\npolynomial remainder before evaluation at $u$, even over rings where\ndivision by $c$ would be invalid.\n\\end{proof}","_j$. Multiplication of its expansion by\n$a_i+cU_i$ gives constant term $P(a)a_i$, linear term\n$c(a_iD_P(U)+P(a)U_i)$, and remainder\n\\[\n c^2\\bigl(Q(U)(a_i+cU_i)+D_P(U)U_i\\bigr).\n\\]\nThe product rule identifies the linear term with the derivative of\n$PX_i$. The displayed remainder vanishes at $U=0$. This constructs the\npolynomial remainder before evaluation at $u$, even over rings where\ndivision by $c$ would be invalid.\n\\end{proof}","\\begin{theorem}[Finite polynomial obstruction]\\label{thm:finite-polynomial-obstruction}\nLet $q\\ge1$, $L\\ge1$, $x\\in\\Z^d$, and suppose\n$F(x)\\in q^L\\Z^e$. Define the exact integer residual $r_x$ by\n$F(x)=q^Lr_x$, and put\n\\[\n \\mathcal O_x=S_q^e/\\im J_x,\\qquad o_x=[\\overline{r_x}].\n\\]\nThen $\\mathcal O_x$ is finite and\n\\begin{equation}\\label{eq:finite-obstruction}\n o_x=0\\quad\\Longleftrightarrow\\quad\n \\exists y\\in\\Z^d:\\quad\n y\\equiv x\\pmod{q^L},\\qquad F(y)\\equiv0\\pmod{q^{L+1}}.\n\\end{equation}\nNo surjectivity or nonsingularity assumption is required.\n\\end{theorem}","\\begin{proof}\nEvery point in the lower residue class has the form $y=x+q^Lt$ with\n$t\\in\\Z^d$. By Lemma~\\ref{lem:polynomial-taylor} and $2L\\ge L+1$,\n\\[\n F(x+q^Lt)\\equiv q^L\\bigl(r_x+DF(x)t\\bigr)\\pmod{q^{L+1}}.\n\\]\nThe integer $q^L$ is nonzero. Cancelling it in the corresponding divisibility\ncondition proves\n\\begin{equation}\\label{eq:digit-correction}\n F(x+q^Lt)\\equiv0\\pmod{q^{L+1}}\n \\quad\\Longleftrightarrow\\quad\n \\overline{r_x}+J_x\\overline t=0.\n\\end{equation}\nInteger vectors reduce onto $S_q^d$. The equation on the right is solvable\nexactly when $\\overline{r_x}$ belongs to $\\im J_x$, since that image is\nclosed under negation. This is precisely $o_x=0$. Finiteness follows because\n$\\mathcal O_x$ is a quotient of the finite module $S_q^e$.\n\\end{proof}",". By Lemma~\\ref{lem:polynomial-taylor} and $2L\\ge L+1$,\n\\[\n F(x+q^Lt)\\equiv q^L\\bigl(r_x+DF(x)t\\bigr)\\pmod{q^{L+1}}.\n\\]\nThe integer $q^L$ is nonzero. Cancelling it in the corresponding divisibility\ncondition proves\n\\begin{equation}\\label{eq:digit-correction}\n F(x+q^Lt)\\equiv0\\pmod{q^{L+1}}\n \\quad\\Longleftrightarrow\\quad\n \\overline{r_x}+J_x\\overline t=0.\n\\end{equation}\nInteger vectors reduce onto $S_q^d$. The equation on the right is solvable\nexactly when $\\overline{r_x}$ belongs to $\\im J_x$, since that image is\nclosed under negation. This is precisely $o_x=0$. Finiteness follows because\n$\\mathcal O_x$ is a quotient of the finite module $S_q^e$.\n\\end{proof}","F(x+q^Lt)\\equiv q^L\\bigl(r_x+DF(x)t\\bigr)\\pmod{q^{L+1}}.\n\\]\nThe integer $q^L$ is nonzero. Cancelling it in the corresponding divisibility\ncondition proves\n\\begin{equation}\\label{eq:digit-correction}\n F(x+q^Lt)\\equiv0\\pmod{q^{L+1}}\n \\quad\\Longleftrightarrow\\quad\n \\overline{r_x}+J_x\\overline t=0.\n\\end{equation}\nInteger vectors reduce onto $S_q^d$. The equation on the right is solvable\nexactly when $\\overline{r_x}$ belongs to $\\im J_x$, since that image is\nclosed under negation. This is precisely $o_x=0$. Finiteness follows because\n$\\mathcal O_x$ is a quotient of the finite module $S_q^e$.\n\\end{proof}","+DF(x)t\\bigr)\\pmod{q^{L+1}}.\n\\]\nThe integer $q^L$ is nonzero. Cancelling it in the corresponding divisibility\ncondition proves\n\\begin{equation}\\label{eq:digit-correction}\n F(x+q^Lt)\\equiv0\\pmod{q^{L+1}}\n \\quad\\Longleftrightarrow\\quad\n \\overline{r_x}+J_x\\overline t=0.\n\\end{equation}\nInteger vectors reduce onto $S_q^d$. The equation on the right is solvable\nexactly when $\\overline{r_x}$ belongs to $\\im J_x$, since that image is\nclosed under negation. This is precisely $o_x=0$. Finiteness follows because\n$\\mathcal O_x$ is a quotient of the finite module $S_q^e$.\n\\end{proof}","it in the corresponding divisibility\ncondition proves\n\\begin{equation}\\label{eq:digit-correction}\n F(x+q^Lt)\\equiv0\\pmod{q^{L+1}}\n \\quad\\Longleftrightarrow\\quad\n \\overline{r_x}+J_x\\overline t=0.\n\\end{equation}\nInteger vectors reduce onto $S_q^d$. The equation on the right is solvable\nexactly when $\\overline{r_x}$ belongs to $\\im J_x$, since that image is\nclosed under negation. This is precisely $o_x=0$. Finiteness follows because\n$\\mathcal O_x$ is a quotient of the finite module $S_q^e$.\n\\end{proof}","y\ncondition proves\n\\begin{equation}\\label{eq:digit-correction}\n F(x+q^Lt)\\equiv0\\pmod{q^{L+1}}\n \\quad\\Longleftrightarrow\\quad\n \\overline{r_x}+J_x\\overline t=0.\n\\end{equation}\nInteger vectors reduce onto $S_q^d$. The equation on the right is solvable\nexactly when $\\overline{r_x}$ belongs to $\\im J_x$, since that image is\nclosed under negation. This is precisely $o_x=0$. Finiteness follows because\n$\\mathcal O_x$ is a quotient of the finite module $S_q^e$.\n\\end{proof}","-correction}\n F(x+q^Lt)\\equiv0\\pmod{q^{L+1}}\n \\quad\\Longleftrightarrow\\quad\n \\overline{r_x}+J_x\\overline t=0.\n\\end{equation}\nInteger vectors reduce onto $S_q^d$. The equation on the right is solvable\nexactly when $\\overline{r_x}$ belongs to $\\im J_x$, since that image is\nclosed under negation. This is precisely $o_x=0$. Finiteness follows because\n$\\mathcal O_x$ is a quotient of the finite module $S_q^e$.\n\\end{proof}","The formal theorem\n\\lean{Komorebi.PolynomialObstruction.finite_polynomial_obstruction}\npresents the image using the integer-domain map\n$\\Z^d\\xrightarrow{DF(x)}\\Z^e\\to S_q^e$. Reduction onto $S_q^d$ identifies\nthat image with $\\im J_x$. The class-preserving comparison is explicit in\n\\lean{Komorebi.PolynomialLiftCounts.obstructionEquiv}; the change of\npresentation does not change the distinguished residual class.","\\begin{proposition}[Affine depth zero]\\label{prop:affine-obstruction}\nLet $F(x)=Ax-b$ for an integer linear map $A:\\Z^d\\to\\Z^e$. The criterion\nof Theorem~\\ref{thm:finite-polynomial-obstruction} holds for every $L\\ge0$,\nwith $J=A\\bmod q$.\n\\end{proposition}","\\begin{proof}\nThe affine identity\n$F(x+q^Lt)=F(x)+q^LAt$ has no quadratic remainder. The proof of\n\\eqref{eq:digit-correction} therefore needs no inequality involving $2L$.\nThe residual and the finite quotient are obtained as above.\n\\end{proof}","The affine statement is formalized by\n\\lean{Komorebi.NativeAffine.finite_affine_obstruction}.\nPositive depth is essential in the nonlinear theorem: for $F(X)=X^2-1$,\n$q=2$, $L=0$, and $x=0$, the Jacobian is zero and the residual class is\nnonzero, yet $y=1$ is a root modulo two in the same residue modulo one.","\\begin{proposition}[Independence of the integer representative]\n\\label{prop:representative-independence}\nUnder the hypotheses of Theorem~\\ref{thm:finite-polynomial-obstruction},\nlet $x'=x+q^Ls$ and write $F(x')=q^Lr_{x'}$. Then\n\\[\n J_{x'}=J_x,\\qquad\n \\overline{r_{x'}}=\\overline{r_x}+J_x\\overline s.\n\\]\nConsequently the two residual classes coincide in the common quotient\n$S_q^e/\\im J_x$.\n\\end{proposition}","\\begin{proof}\nPolynomial evaluation respects coordinatewise congruence. Apply this first\nto every formal partial derivative: $x'\\equiv x\\pmod q$ gives\n$J_{x'}=J_x$. Next apply \\eqref{eq:polynomial-taylor} to $F$ at scale $q^L$.\nSubtract the two exact residual identities and cancel $q^L$ to obtain the\ndisplayed relation modulo $q$. Its additional term belongs to the common\nJacobian image, so it vanishes in the quotient.\n\\end{proof}","This comparison is formalized by\n\\lean{Komorebi.PolynomialRepresentative.representative_independence}.\nThis proves independence of the integer representative within a fixed\npolynomial presentation. It makes no identification of the raw cokernels\nof different equation lists. For example, adjoining a zero equation leaves\nthe roots unchanged while adding an $S_q$ summand to the raw cokernel.","The next obstruction concerns compatibility between individually attainable\ncorrections. Its algebraic form works over an arbitrary commutative ring.\nLet $U,V,W$ be $R$-modules and let $J:U\\to V$, $K:U\\to W$ be linear.\nDefine\n\\begin{equation}\\label{eq:common-quotient}\n C(J,K)=U/(\\ker J+\\ker K).\n\\end{equation}\nAn attainable value $Ju$ has label $[u]\\in C(J,K)$; this is independent of\nthe choice of its preimage because $\\ker J$ is killed in the quotient.\nSimilarly, $Kv$ has label $[v]$. Denote these maps by\n$\\lambda_J:\\im J\\to C(J,K)$ and $\\lambda_K:\\im K\\to C(J,K)$.","oncerns compatibility between individually attainable\ncorrections. Its algebraic form works over an arbitrary commutative ring.\nLet $U,V,W$ be $R$-modules and let $J:U\\to V$, $K:U\\to W$ be linear.\nDefine\n\\begin{equation}\\label{eq:common-quotient}\n C(J,K)=U/(\\ker J+\\ker K).\n\\end{equation}\nAn attainable value $Ju$ has label $[u]\\in C(J,K)$; this is independent of\nthe choice of its preimage because $\\ker J$ is killed in the quotient.\nSimilarly, $Kv$ has label $[v]$. Denote these maps by\n$\\lambda_J:\\im J\\to C(J,K)$ and $\\lambda_K:\\im K\\to C(J,K)$.","inable\ncorrections. Its algebraic form works over an arbitrary commutative ring.\nLet $U,V,W$ be $R$-modules and let $J:U\\to V$, $K:U\\to W$ be linear.\nDefine\n\\begin{equation}\\label{eq:common-quotient}\n C(J,K)=U/(\\ker J+\\ker K).\n\\end{equation}\nAn attainable value $Ju$ has label $[u]\\in C(J,K)$; this is independent of\nthe choice of its preimage because $\\ker J$ is killed in the quotient.\nSimilarly, $Kv$ has label $[v]$. Denote these maps by\n$\\lambda_J:\\im J\\to C(J,K)$ and $\\lambda_K:\\im K\\to C(J,K)$.","ons. Its algebraic form works over an arbitrary commutative ring.\nLet $U,V,W$ be $R$-modules and let $J:U\\to V$, $K:U\\to W$ be linear.\nDefine\n\\begin{equation}\\label{eq:common-quotient}\n C(J,K)=U/(\\ker J+\\ker K).\n\\end{equation}\nAn attainable value $Ju$ has label $[u]\\in C(J,K)$; this is independent of\nthe choice of its preimage because $\\ker J$ is killed in the quotient.\nSimilarly, $Kv$ has label $[v]$. Denote these maps by\n$\\lambda_J:\\im J\\to C(J,K)$ and $\\lambda_K:\\im K\\to C(J,K)$.","s over an arbitrary commutative ring.\nLet $U,V,W$ be $R$-modules and let $J:U\\to V$, $K:U\\to W$ be linear.\nDefine\n\\begin{equation}\\label{eq:common-quotient}\n C(J,K)=U/(\\ker J+\\ker K).\n\\end{equation}\nAn attainable value $Ju$ has label $[u]\\in C(J,K)$; this is independent of\nthe choice of its preimage because $\\ker J$ is killed in the quotient.\nSimilarly, $Kv$ has label $[v]$. Denote these maps by\n$\\lambda_J:\\im J\\to C(J,K)$ and $\\lambda_K:\\im K\\to C(J,K)$.","ring.\nLet $U,V,W$ be $R$-modules and let $J:U\\to V$, $K:U\\to W$ be linear.\nDefine\n\\begin{equation}\\label{eq:common-quotient}\n C(J,K)=U/(\\ker J+\\ker K).\n\\end{equation}\nAn attainable value $Ju$ has label $[u]\\in C(J,K)$; this is independent of\nthe choice of its preimage because $\\ker J$ is killed in the quotient.\nSimilarly, $Kv$ has label $[v]$. Denote these maps by\n$\\lambda_J:\\im J\\to C(J,K)$ and $\\lambda_K:\\im K\\to C(J,K)$.","\\begin{theorem}[Common correction and fiber product]\\label{thm:common-correction}\nFor every $u,v\\in U$,\n\\begin{equation}\\label{eq:common-correction}\n [u]=[v]\\text{ in }C(J,K)\n \\quad\\Longleftrightarrow\\quad\n \\exists w\\in U:\\ Jw=Ju,\\quad Kw=Kv.\n\\end{equation}\nBoth label maps are surjective, and the image of the joint map is exactly\nthe fiber product:\n\\begin{equation}\\label{eq:joint-fiber-product}\n \\im\\bigl(U\\to\\im J\\times\\im K,\\ u\\mapsto(Ju,Ku)\\bigr)\n =\\im J\\mathbin{\\times}_{C(J,K)}\\im K.\n\\end{equation}\nIf either $V$ or $W$ is finite, then $C(J,K)$ is finite.\n\\end{theorem}","n}\\label{eq:common-correction}\n [u]=[v]\\text{ in }C(J,K)\n \\quad\\Longleftrightarrow\\quad\n \\exists w\\in U:\\ Jw=Ju,\\quad Kw=Kv.\n\\end{equation}\nBoth label maps are surjective, and the image of the joint map is exactly\nthe fiber product:\n\\begin{equation}\\label{eq:joint-fiber-product}\n \\im\\bigl(U\\to\\im J\\times\\im K,\\ u\\mapsto(Ju,Ku)\\bigr)\n =\\im J\\mathbin{\\times}_{C(J,K)}\\im K.\n\\end{equation}\nIf either $V$ or $W$ is finite, then $C(J,K)$ is finite.\n\\end{theorem}","l{eq:common-correction}\n [u]=[v]\\text{ in }C(J,K)\n \\quad\\Longleftrightarrow\\quad\n \\exists w\\in U:\\ Jw=Ju,\\quad Kw=Kv.\n\\end{equation}\nBoth label maps are surjective, and the image of the joint map is exactly\nthe fiber product:\n\\begin{equation}\\label{eq:joint-fiber-product}\n \\im\\bigl(U\\to\\im J\\times\\im K,\\ u\\mapsto(Ju,Ku)\\bigr)\n =\\im J\\mathbin{\\times}_{C(J,K)}\\im K.\n\\end{equation}\nIf either $V$ or $W$ is finite, then $C(J,K)$ is finite.\n\\end{theorem}","on}\n [u]=[v]\\text{ in }C(J,K)\n \\quad\\Longleftrightarrow\\quad\n \\exists w\\in U:\\ Jw=Ju,\\quad Kw=Kv.\n\\end{equation}\nBoth label maps are surjective, and the image of the joint map is exactly\nthe fiber product:\n\\begin{equation}\\label{eq:joint-fiber-product}\n \\im\\bigl(U\\to\\im J\\times\\im K,\\ u\\mapsto(Ju,Ku)\\bigr)\n =\\im J\\mathbin{\\times}_{C(J,K)}\\im K.\n\\end{equation}\nIf either $V$ or $W$ is finite, then $C(J,K)$ is finite.\n\\end{theorem}","\\begin{proof}\nIf $[u]=[v]$, write $u-v=s+t$ with $s\\in\\ker J$ and $t\\in\\ker K$.\nThen $w=u-s=v+t$ satisfies both required equalities. Conversely, a common\n$w$ gives\n\\[\n u-v=(u-w)+(w-v)\\in\\ker J+\\ker K.\n\\]\nEvery class $[u]$ is the label of $Ju$ and of $Ku$, proving surjectivity.\nFor attainable $a=Ju$ and $b=Kv$, the first equivalence says exactly that\nequal labels are equivalent to a simultaneous preimage. This proves\n\\eqref{eq:joint-fiber-product}. Finally, a finite target has finite image,\nand the associated surjective label map makes $C(J,K)$ finite.\n\\end{proof}","y, a common\n$w$ gives\n\\[\n u-v=(u-w)+(w-v)\\in\\ker J+\\ker K.\n\\]\nEvery class $[u]$ is the label of $Ju$ and of $Ku$, proving surjectivity.\nFor attainable $a=Ju$ and $b=Kv$, the first equivalence says exactly that\nequal labels are equivalent to a simultaneous preimage. This proves\n\\eqref{eq:joint-fiber-product}. Finally, a finite target has finite image,\nand the associated surjective label map makes $C(J,K)$ finite.\n\\end{proof}","These statements are\n\\lean{Komorebi.SharedCorrection.quotient_mk_eq_iff_common} and\n\\lean{Komorebi.SharedCorrection.joint_range_eq_fiber}, together with the\nfinite-target corollaries in the same module. The quotient also has the\nexpected universal property: every linear map from $U$ annihilating both\nkernels factors uniquely through $C(J,K)$. Equivalently, compatible linear\nobservations on $\\im J$ and $\\im K$ factor uniquely through their label\nmaps. This is formalized by\n\\lean{Komorebi.SharedCorrection.common_paired_universal}.","statements are\n\\lean{Komorebi.SharedCorrection.quotient_mk_eq_iff_common} and\n\\lean{Komorebi.SharedCorrection.joint_range_eq_fiber}, together with the\nfinite-target corollaries in the same module. The quotient also has the\nexpected universal property: every linear map from $U$ annihilating both\nkernels factors uniquely through $C(J,K)$. Equivalently, compatible linear\nobservations on $\\im J$ and $\\im K$ factor uniquely through their label\nmaps. This is formalized by\n\\lean{Komorebi.SharedCorrection.common_paired_universal}.","\\begin{corollary}[Simultaneous polynomial lifting]\\label{cor:shared-polynomial}\nLet $F\\in\\Z[X_1,\\ldots,X_d]^e$ and\n$G\\in\\Z[X_1,\\ldots,X_d]^f$ have a common lower root $x$ modulo $q^L$,\nwhere $q\\ge1$ and $L\\ge1$. Use the integer-domain reduced Jacobians\n$J:\\Z^d\\to S_q^e$ and $K:\\Z^d\\to S_q^f$. Suppose $u,v\\in\\Z^d$ are\nseparately valid corrections for $F$ and $G$, respectively. Then\n\\[\n [u]=[v]\\text{ in }\\Z^d/(\\ker J+\\ker K)\n\\]\nif and only if a single $y\\equiv x\\pmod{q^L}$ solves both systems modulo\n$q^{L+1}$. Each separate label is independent of the chosen valid\ncorrection. The common quotient is finite.\n\\end{corollary}","l{cor:shared-polynomial}\nLet $F\\in\\Z[X_1,\\ldots,X_d]^e$ and\n$G\\in\\Z[X_1,\\ldots,X_d]^f$ have a common lower root $x$ modulo $q^L$,\nwhere $q\\ge1$ and $L\\ge1$. Use the integer-domain reduced Jacobians\n$J:\\Z^d\\to S_q^e$ and $K:\\Z^d\\to S_q^f$. Suppose $u,v\\in\\Z^d$ are\nseparately valid corrections for $F$ and $G$, respectively. Then\n\\[\n [u]=[v]\\text{ in }\\Z^d/(\\ker J+\\ker K)\n\\]\nif and only if a single $y\\equiv x\\pmod{q^L}$ solves both systems modulo\n$q^{L+1}$. Each separate label is independent of the chosen valid\ncorrection. The common quotient is finite.\n\\end{corollary}","_d]^e$ and\n$G\\in\\Z[X_1,\\ldots,X_d]^f$ have a common lower root $x$ modulo $q^L$,\nwhere $q\\ge1$ and $L\\ge1$. Use the integer-domain reduced Jacobians\n$J:\\Z^d\\to S_q^e$ and $K:\\Z^d\\to S_q^f$. Suppose $u,v\\in\\Z^d$ are\nseparately valid corrections for $F$ and $G$, respectively. Then\n\\[\n [u]=[v]\\text{ in }\\Z^d/(\\ker J+\\ker K)\n\\]\nif and only if a single $y\\equiv x\\pmod{q^L}$ solves both systems modulo\n$q^{L+1}$. Each separate label is independent of the chosen valid\ncorrection. The common quotient is finite.\n\\end{corollary}","\\begin{proof}\nFor each system, \\eqref{eq:digit-correction} translates a valid next root\ninto a linear correction equation. Once $u$ corrects $F$, another digit\n$w$ corrects $F$ precisely when $Jw=Ju$. The analogous condition for $G$\nis $Kw=Kv$. Theorem~\\ref{thm:common-correction} now gives the criterion.\nTwo valid corrections of one block differ by a vector in its Jacobian\nkernel, so they have equal labels. Finiteness follows from either finite\nresidue target. An unconditional formulation quantifies over separately\nvalid $u,v$ with equal labels, thereby including their existence in the\ncriterion rather than assuming it.\n\\end{proof}","$u$ corrects $F$, another digit\n$w$ corrects $F$ precisely when $Jw=Ju$. The analogous condition for $G$\nis $Kw=Kv$. Theorem~\\ref{thm:common-correction} now gives the criterion.\nTwo valid corrections of one block differ by a vector in its Jacobian\nkernel, so they have equal labels. Finiteness follows from either finite\nresidue target. An unconditional formulation quantifies over separately\nvalid $u,v$ with equal labels, thereby including their existence in the\ncriterion rather than assuming it.\n\\end{proof}","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged.","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstate","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lif","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komoreb","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffineP","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.fini","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.finite_affine_pair_compatibili","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.finite_affine_pair_compatibility}.\nSep","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.finite_affine_pair_compatibility}.\nSeparate liftability a","For affine integer systems $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.finite_affine_pair_compatibility}.\nSeparate liftability alone is insufficient. At $x=0$ modulo two,","ms $F(X)=AX-b$ and $G(X)=BX-c$, where\n$A:\\Z^d\\to\\Z^e$ and $B:\\Z^d\\to\\Z^f$ are linear,\nCorollary~\\ref{cor:shared-polynomial} holds for $q\\ge1$ and every\n$L\\ge0$, including depth zero. Here $x$ is a common root modulo $q^L$,\nand $J$ and $K$ are the reductions of $A$ and $B$ with domain $\\Z^d$.\nGiven separately valid corrections $u,v$, a simultaneous next lift exists\nif and only if their classes in $\\Z^d/(\\ker J+\\ker K)$ agree. The quotient\nis finite, and each label is independent of the chosen valid correction.\nIndeed, Proposition~\\ref{prop:affine-obstruction} supplies the correction\nequation without a positive-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.finite_affine_pair_compatibility}.\nSeparate liftability alone is insufficient. At $x=0$ modulo two,\n$F(X)=X$ and $G(X)=X-2$ both lift to modulus four, but their correc…","-depth restriction, and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.finite_affine_pair_compatibility}.\nSeparate liftability alone is insufficient. At $x=0$ modulo two,\n$F(X)=X$ and $G(X)=X-2$ both lift to modulus four, but their correcting\ndigits are respectively $0$ and $1$ modulo two. No single residue solves\nboth equations modulo four.",", and the proof of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.finite_affine_pair_compatibility}.\nSeparate liftability alone is insufficient. At $x=0$ modulo two,\n$F(X)=X$ and $G(X)=X-2$ both lift to modulus four, but their correcting\ndigits are respectively $0$ and $1$ modulo two. No single residue solves\nboth equations modulo four.","of of\nTheorem~\\ref{thm:common-correction} then applies unchanged. The formal\nstatement is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.finite_affine_pair_compatibility}.\nSeparate liftability alone is insufficient. At $x=0$ modulo two,\n$F(X)=X$ and $G(X)=X-2$ both lift to modulus four, but their correcting\ndigits are respectively $0$ and $1$ modulo two. No single residue solves\nboth equations modulo four.","ent is \\lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.finite_affine_pair_compatibility}.\nSeparate liftability alone is insufficient. At $x=0$ modulo two,\n$F(X)=X$ and $G(X)=X-2$ both lift to modulus four, but their correcting\ndigits are respectively $0$ and $1$ modulo two. No single residue solves\nboth equations modulo four.","lean{Komorebi.AffinePair.joint_lift_iff_class_eq}.\nQuantifying over separately valid corrections with equal labels also\nincludes their existence in the criterion, as in\n\\lean{Komorebi.AffinePair.finite_affine_pair_compatibility}.\nSeparate liftability alone is insufficient. At $x=0$ modulo two,\n$F(X)=X$ and $G(X)=X-2$ both lift to modulus four, but their correcting\ndigits are respectively $0$ and $1$ modulo two. No single residue solves\nboth equations modulo four.","Integer representatives and residue classes at the next level must also be\ndistinguished. For $q\\ge1$ and $L\\ge0$, the map\n\\[\n S_q^d\\longrightarrow\n \\{y\\in(\\Z/q^{L+1}\\Z)^d:y\\equiv x\\pmod{q^L}\\},\n \\qquad t\\longmapsto x+q^L\\widetilde t\n\\]\nis a bijection, where $\\widetilde t$ is any integer representative.\nIndependence of the representative and injectivity both follow by cancelling\n$q^L$ in integer divisibility; surjectivity follows by expressing the lower\ncongruence as a multiple of $q^L$. This is the equivalence\n\\lean{Komorebi.ResidueLiftDigits.digitEquivResidue}.","d. For $q\\ge1$ and $L\\ge0$, the map\n\\[\n S_q^d\\longrightarrow\n \\{y\\in(\\Z/q^{L+1}\\Z)^d:y\\equiv x\\pmod{q^L}\\},\n \\qquad t\\longmapsto x+q^L\\widetilde t\n\\]\nis a bijection, where $\\widetilde t$ is any integer representative.\nIndependence of the representative and injectivity both follow by cancelling\n$q^L$ in integer divisibility; surjectivity follows by expressing the lower\ncongruence as a multiple of $q^L$. This is the equivalence\n\\lean{Komorebi.ResidueLiftDigits.digitEquivResidue}.","At positive polynomial depth, restrict it to the correction fiber\n\\[\n \\mathcal T_x=\\{t\\in S_q^d:\\overline{r_x}+J_xt=0\\}\n\\]\nand the lift set\n\\[\n \\mathcal L_{x,L}=\n \\{y\\in(\\Z/q^{L+1}\\Z)^d:y\\equiv x\\pmod{q^L},\\ F(y)=0\\}.\n\\]\nThe polynomial predicate in the second display is evaluation in the residue\nring. Evaluation on canonical integer representatives gives the same\npredicate, because integer polynomial evaluation respects congruence.","\\begin{theorem}[Torsor and lift cardinalities]\\label{thm:lift-torsor-count}\nUnder the hypotheses of Theorem~\\ref{thm:finite-polynomial-obstruction},\nthere is an equivalence $\\mathcal T_x\\simeq\\mathcal L_{x,L}$.\nEvery nonempty fiber is a torsor for $\\ker J_x$ under addition of\ncorrection digits. If $q=p$ is prime and\n$\\rho_x=\\rank_{\\F_p}J_x$, then\n\\begin{equation}\\label{eq:lift-count}\n \\#\\mathcal L_{x,L}=\n \\begin{cases}p^{d-\\rho_x},&o_x=0,","\\begin{theorem}[Torsor and lift cardinalities]\\label{thm:lift-torsor-count}\nUnder the hypotheses of Theorem~\\ref{thm:finite-polynomial-obstruction},\nthere is an equivalence $\\mathcal T_x\\simeq\\mathcal L_{x,L}$.\nEvery nonempty fiber is a torsor for $\\ker J_x$ under addition of\ncorrection digits. If $q=p$ is prime and\n$\\rho_x=\\rank_{\\F_p}J_x$, then\n\\begin{equation}\\label{eq:lift-count}\n \\#\\mathcal L_{x,L}=\n \\begin{cases}p^{d-\\rho_x},&o_x=0,\\\\0,&o_x\\ne0,\\end{cases}\n \\qquad\n \\#\\mathcal O_x=p^{e-\\rho_x}.\n\\end{equation}\n\\end{theorem}","\\begin{proof}\nEquation~\\eqref{eq:digit-correction} identifies exactly which points of the\nresidue fiber satisfy the equations, proving the restricted equivalence.\nIf $u,t\\in\\mathcal T_x$, subtraction gives $J_x(t-u)=0$. Conversely,\nadding any element of $\\ker J_x$ preserves the correction equation. The\nunique kernel vector carrying $u$ to $t$ is $t-u$, so the action is free\nand transitive. Choosing one valid digit identifies the fiber with the\nkernel, without making that choice part of its translation action.","For prime $p$, rank-nullity gives\n$\\dim\\ker J_x=d-\\rho_x$ and\n$\\dim(\\F_p^e/\\im J_x)=e-\\rho_x$. The cardinality of an\n$n$-dimensional $\\F_p$-space is $p^n$. Combine these identities with the\nobstruction criterion and the residue equivalence.\n\\end{proof}","The cardinality formula is formalized by\n\\lean{Komorebi.PolynomialLiftCounts.finite_polynomial_lift_count}.\nThe explicit torsor and chosen-point equivalence are developed in\n\\lean{Komorebi.FiniteFiberCounts.translate_exists_unique} and\n\\lean{Komorebi.PolynomialLiftCounts.liftEquivKernel}.\nFor composite $q$, the torsor and kernel cardinality remain valid, while the\nvector-space rank formula is unavailable. Multiplication by two on\n$\\Z/4\\Z$ already has a kernel of size two.","ed by\n\\lean{Komorebi.PolynomialLiftCounts.finite_polynomial_lift_count}.\nThe explicit torsor and chosen-point equivalence are developed in\n\\lean{Komorebi.FiniteFiberCounts.translate_exists_unique} and\n\\lean{Komorebi.PolynomialLiftCounts.liftEquivKernel}.\nFor composite $q$, the torsor and kernel cardinality remain valid, while the\nvector-space rank formula is unavailable. Multiplication by two on\n$\\Z/4\\Z$ already has a kernel of size two.","\\begin{theorem}[Compatible polynomial tower]\\label{thm:smooth-tower}\nLet $q\\ge1$, $L\\ge1$, $F\\in\\Z[X_1,\\ldots,X_d]^e$, and\n$x\\in\\Z^d$ satisfy $F(x)\\in q^L\\Z^e$. Suppose $J_x:S_q^d\\to S_q^e$\nis surjective. There is a single sequence $z_n\\in\\Z^d$ such that\n\\begin{align*}\n z_0&=x,& F(z_n)&\\in q^{L+n}\\Z^e,","\\begin{theorem}[Compatible polynomial tower]\\label{thm:smooth-tower}\nLet $q\\ge1$, $L\\ge1$, $F\\in\\Z[X_1,\\ldots,X_d]^e$, and\n$x\\in\\Z^d$ satisfy $F(x)\\in q^L\\Z^e$. Suppose $J_x:S_q^d\\to S_q^e$\nis surjective. There is a single sequence $z_n\\in\\Z^d$ such that\n\\begin{align*}\n z_0&=x,& F(z_n)&\\in q^{L+n}\\Z^e,\\\\\n z_{n+1}-z_n&\\in q^{L+n}\\Z^d,& J_{z_n}&=J_x.\n\\end{align*}\nFor $a\\le b$, it also satisfies $z_b-z_a\\in q^{L+a}\\Z^d$.\n\\end{theorem}","z_{n+1}-z_n&\\in q^{L+n}\\Z^d,& J_{z_n}&=J_x.\n\\end{align*}\nFor $a\\le b$, it also satisfies $z_b-z_a\\in q^{L+a}\\Z^d$.\n\\end{theorem}","\\begin{proof}\nAt level $N=L+n$, divide the residual by $q^N$. Surjectivity supplies\n$t\\in S_q^d$ with $J_{z_n}t=-\\overline{F(z_n)/q^N}$. Choose an integer\nrepresentative and apply \\eqref{eq:digit-correction} to obtain $z_{n+1}$.\nThe new point is congruent to $z_n$ modulo $q$. Evaluation of each formal\npartial derivative therefore gives $J_{z_{n+1}}=J_{z_n}$.","Construct the sequence recursively, retaining the root condition and the\nfixed Jacobian at each step. Surjectivity at every later stage then follows\nfrom the single initial assumption. The cross-level congruence follows by\ninduction, adding consecutive differences whose divisibilities all imply divisibility\nby $q^{L+a}$.\n\\end{proof}","\\begin{theorem}[Completion and smooth branching]\\label{thm:smooth-padic-root}\nIn Theorem~\\ref{thm:smooth-tower}, let $q=p$ be prime. The tower can be\nchosen with a limit $a\\in\\Zp^d$ such that\n\\[\n F(a)=0,\\qquad z_n\\longrightarrow a,\\qquad\n a-z_n\\in p^{L+n}\\Zp^d\\quad(n\\ge0).\n\\]\nAt each stage, the number of distinct next-level residue roots over $z_n$\nis exactly $p^{d-e}$, and the finite obstruction module has one element.\nIf $d=e$, there is exactly one root of $F$ in $B_L(x)$.\n\\end{theorem}","nd smooth branching]\\label{thm:smooth-padic-root}\nIn Theorem~\\ref{thm:smooth-tower}, let $q=p$ be prime. The tower can be\nchosen with a limit $a\\in\\Zp^d$ such that\n\\[\n F(a)=0,\\qquad z_n\\longrightarrow a,\\qquad\n a-z_n\\in p^{L+n}\\Zp^d\\quad(n\\ge0).\n\\]\nAt each stage, the number of distinct next-level residue roots over $z_n$\nis exactly $p^{d-e}$, and the finite obstruction module has one element.\nIf $d=e$, there is exactly one root of $F$ in $B_L(x)$.\n\\end{theorem}","h-padic-root}\nIn Theorem~\\ref{thm:smooth-tower}, let $q=p$ be prime. The tower can be\nchosen with a limit $a\\in\\Zp^d$ such that\n\\[\n F(a)=0,\\qquad z_n\\longrightarrow a,\\qquad\n a-z_n\\in p^{L+n}\\Zp^d\\quad(n\\ge0).\n\\]\nAt each stage, the number of distinct next-level residue roots over $z_n$\nis exactly $p^{d-e}$, and the finite obstruction module has one element.\nIf $d=e$, there is exactly one root of $F$ in $B_L(x)$.\n\\end{theorem}","-root}\nIn Theorem~\\ref{thm:smooth-tower}, let $q=p$ be prime. The tower can be\nchosen with a limit $a\\in\\Zp^d$ such that\n\\[\n F(a)=0,\\qquad z_n\\longrightarrow a,\\qquad\n a-z_n\\in p^{L+n}\\Zp^d\\quad(n\\ge0).\n\\]\nAt each stage, the number of distinct next-level residue roots over $z_n$\nis exactly $p^{d-e}$, and the finite obstruction module has one element.\nIf $d=e$, there is exactly one root of $F$ in $B_L(x)$.\n\\end{theorem}","\\begin{proof}\nCross-level compatibility bounds the supremum distance between later terms\nby $p^{-(L+n)}$. Completeness gives a limit, and closed norm bounds retain\nthe congruence to every $z_n$. View the coefficients of the original integer\npolynomials in $\\Zp$. Their evaluations satisfy\nboth $F_i(z_n)\\to F_i(a)$, by continuity, and $F_i(z_n)\\to0$, by their\ngrowing divisibility. Uniqueness of limits gives exact vanishing.","Surjectivity gives $\\rank J_x=e\\le d$. Theorem~\\ref{thm:lift-torsor-count}\nthen gives $p^{d-e}$ next residue roots and obstruction cardinality one at\nevery stage.","For square uniqueness, let $a,b\\in B_L(x)$ be arbitrary exact roots.\nTheir canonical integer truncations are genuine roots in the respective\nfinite residue rings, since evaluation commutes with reduction. They agree\nat depth $L$. If they agree at depth $N\\ge L$, both next residues belong\nto the same lift set, which has cardinality $p^{d-d}=1$. Induction gives\nagreement at every depth. The residue maps separate $p$-adic integers, so\n$a=b$. Existence was already supplied by the tower.\n\\end{proof}","ueness, let $a,b\\in B_L(x)$ be arbitrary exact roots.\nTheir canonical integer truncations are genuine roots in the respective\nfinite residue rings, since evaluation commutes with reduction. They agree\nat depth $L$. If they agree at depth $N\\ge L$, both next residues belong\nto the same lift set, which has cardinality $p^{d-d}=1$. Induction gives\nagreement at every depth. The residue maps separate $p$-adic integers, so\n$a=b$. Existence was already supplied by the tower.\n\\end{proof}","L(x)$ be arbitrary exact roots.\nTheir canonical integer truncations are genuine roots in the respective\nfinite residue rings, since evaluation commutes with reduction. They agree\nat depth $L$. If they agree at depth $N\\ge L$, both next residues belong\nto the same lift set, which has cardinality $p^{d-d}=1$. Induction gives\nagreement at every depth. The residue maps separate $p$-adic integers, so\n$a=b$. Existence was already supplied by the tower.\n\\end{proof}","\\begin{remark}\\label{rem:singular-tower-boundary}\nThe one-step singular criterion and the smooth infinite theorem have\ndifferent hypotheses. The polynomial $X^2-5$ has odd roots modulo two and\nfour but has no root modulo eight. Thus a previous zero obstruction does\nnot force the next one to vanish. Also, rectangular surjectivity alone\ndoes not imply uniqueness: $F(X,Y)=X$ has the distinct dyadic roots\n$(0,0)$ and $(0,2)$ in the same residue ball modulo two. These examples show\nwhy the hypotheses of the tower theorem and its uniqueness clause are needed.\n\\end{remark}","{remark}\\label{rem:singular-tower-boundary}\nThe one-step singular criterion and the smooth infinite theorem have\ndifferent hypotheses. The polynomial $X^2-5$ has odd roots modulo two and\nfour but has no root modulo eight. Thus a previous zero obstruction does\nnot force the next one to vanish. Also, rectangular surjectivity alone\ndoes not imply uniqueness: $F(X,Y)=X$ has the distinct dyadic roots\n$(0,0)$ and $(0,2)$ in the same residue ball modulo two. These examples show\nwhy the hypotheses of the tower theorem and its uniqueness clause are needed.\n\\end{remark}","el{rem:singular-tower-boundary}\nThe one-step singular criterion and the smooth infinite theorem have\ndifferent hypotheses. The polynomial $X^2-5$ has odd roots modulo two and\nfour but has no root modulo eight. Thus a previous zero obstruction does\nnot force the next one to vanish. Also, rectangular surjectivity alone\ndoes not imply uniqueness: $F(X,Y)=X$ has the distinct dyadic roots\n$(0,0)$ and $(0,2)$ in the same residue ball modulo two. These examples show\nwhy the hypotheses of the tower theorem and its uniqueness clause are needed.\n\\end{remark}","\\begin{proposition}[Smooth coefficient stability]\\label{prop:smooth-coefficient-stability}\nLet $p$ be prime, $M\\ge1$, $F,H\\in\\Z[X_1,\\ldots,X_d]^e$, and\n$c\\in p^M\\Z$. Suppose $DF(x)\\bmod p$ is surjective for an integer\n$x\\in\\Z^d$. If $a\\in\\Zp^d$ satisfies $F(a)=0$ and\n$a\\equiv x\\pmod p$, then there is a root $b$ of $G=F+cH$ with\n$b-a\\in p^M\\Zp^d$.","If $d=e$ and $M\\ge L\\ge1$, every pair of roots $a,b\\in B_L(x)$ of\n$F$ and $G$, respectively, satisfies $b-a\\in p^M\\Zp^d$. The perturbed\nroot is unique in that ball whenever it exists.\n\\end{proposition}","\\begin{proof}\nSince $p\\mid c$, differentiation gives\n$DG(x)\\equiv DF(x)\\pmod p$. Truncate $a$ to an integer vector $y$ at\ndepth $M$. Then $F(y)\\equiv0\\pmod{p^M}$, hence also\n$G(y)\\equiv0\\pmod{p^M}$. Congruence of $y$ and $x$ modulo $p$ preserves\nthe reduced Jacobian. The smooth root theorem gives\n$b\\equiv y\\equiv a\\pmod{p^M}$.","In the square case this constructed root lies in $B_L(x)$ because $M\\ge L$.\nSquare uniqueness for $G$ identifies it with every competing root in that\nball. The construction therefore gives the precision assertion for an\narbitrary root there.\n\\end{proof}","See \\lean{Komorebi.PolynomialRootStability.exists_nearby_perturbed_root}\nand \\lean{Komorebi.PolynomialRootStability.perturbed_roots_congruent}.\nThe scalar $c$ may be negative or zero. The weighted theorems below allow\narbitrary integral $p$-adic coefficients and give an explicit first-order\ndisplacement.","\\begin{theorem}[Continuous charts with a selected Jacobian minor]\n\\label{thm:continuous-parameter-chart}\nLet $p$ be prime, $L\\ge1$,\n$F\\in\\Z[X_1,\\ldots,X_{e+k}]^e$, and $x\\in\\Z^{e+k}$ satisfy\n$F(x)\\in p^L\\Z^e$. Suppose the full reduced Jacobian is surjective.\nThere is a permutation $\\sigma$ of the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv","\\begin{theorem}[Continuous charts with a selected Jacobian minor]\n\\label{thm:continuous-parameter-chart}\nLet $p$ be prime, $L\\ge1$,\n$F\\in\\Z[X_1,\\ldots,X_{e+k}]^e$, and $x\\in\\Z^{e+k}$ satisfy\n$F(x)\\in p^L\\Z^e$. Suppose the full reduced Jacobian is surjective.\nThere is a permutation $\\sigma$ of the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end","\\begin{theorem}[Continuous charts with a selected Jacobian minor]\n\\label{thm:continuous-parameter-chart}\nLet $p$ be prime, $L\\ge1$,\n$F\\in\\Z[X_1,\\ldots,X_{e+k}]^e$, and $x\\in\\Z^{e+k}$ satisfy\n$F(x)\\in p^L\\Z^e$. Suppose the full reduced Jacobian is surjective.\nThere is a permutation $\\sigma$ of the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma","\\begin{theorem}[Continuous charts with a selected Jacobian minor]\n\\label{thm:continuous-parameter-chart}\nLet $p$ be prime, $L\\ge1$,\n$F\\in\\Z[X_1,\\ldots,X_{e+k}]^e$, and $x\\in\\Z^{e+k}$ satisfy\n$F(x)\\in p^L\\Z^e$. Suppose the full reduced Jacobian is surjective.\nThere is a permutation $\\sigma$ of the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ f","\\begin{theorem}[Continuous charts with a selected Jacobian minor]\n\\label{thm:continuous-parameter-chart}\nLet $p$ be prime, $L\\ge1$,\n$F\\in\\Z[X_1,\\ldots,X_{e+k}]^e$, and $x\\in\\Z^{e+k}$ satisfy\n$F(x)\\in p^L\\Z^e$. Suppose the full reduced Jacobian is surjective.\nThere is a permutation $\\sigma$ of the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ for a continuous\n$\\phi:T\\to","\\begin{theorem}[Continuous charts with a selected Jacobian minor]\n\\label{thm:continuous-parameter-chart}\nLet $p$ be prime, $L\\ge1$,\n$F\\in\\Z[X_1,\\ldots,X_{e+k}]^e$, and $x\\in\\Z^{e+k}$ satisfy\n$F(x)\\in p^L\\Z^e$. Suppose the full reduced Jacobian is surjective.\nThere is a permutation $\\sigma$ of the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ for a continuous\n$\\phi:T\\to y_0+p^L\\Zp^e$ whose graph is the entire reordered root locus.\n\\end{theorem}","arts with a selected Jacobian minor]\n\\label{thm:continuous-parameter-chart}\nLet $p$ be prime, $L\\ge1$,\n$F\\in\\Z[X_1,\\ldots,X_{e+k}]^e$, and $x\\in\\Z^{e+k}$ satisfy\n$F(x)\\in p^L\\Z^e$. Suppose the full reduced Jacobian is surjective.\nThere is a permutation $\\sigma$ of the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ for a continuous\n$\\phi:T\\to y_0+p^L\\Zp^e$ whose graph is the entire reordered root locus.\n\\end{theorem}","parameter-chart}\nLet $p$ be prime, $L\\ge1$,\n$F\\in\\Z[X_1,\\ldots,X_{e+k}]^e$, and $x\\in\\Z^{e+k}$ satisfy\n$F(x)\\in p^L\\Z^e$. Suppose the full reduced Jacobian is surjective.\nThere is a permutation $\\sigma$ of the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ for a continuous\n$\\phi:T\\to y_0+p^L\\Zp^e$ whose graph is the entire reordered root locus.\n\\end{theorem}","\\in p^L\\Z^e$. Suppose the full reduced Jacobian is surjective.\nThere is a permutation $\\sigma$ of the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ for a continuous\n$\\phi:T\\to y_0+p^L\\Zp^e$ whose graph is the entire reordered root locus.\n\\end{theorem}","ian is surjective.\nThere is a permutation $\\sigma$ of the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ for a continuous\n$\\phi:T\\to y_0+p^L\\Zp^e$ whose graph is the entire reordered root locus.\n\\end{theorem}","the coordinates such that, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ for a continuous\n$\\phi:T\\to y_0+p^L\\Zp^e$ whose graph is the entire reordered root locus.\n\\end{theorem}","t, writing\n$\\sigma x=(y_0,t_0)$ with $e$ solved coordinates and $k$ parameter\ncoordinates, the parameter ball\n$T=t_0+p^L\\Zp^k$ is homeomorphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ for a continuous\n$\\phi:T\\to y_0+p^L\\Zp^e$ whose graph is the entire reordered root locus.\n\\end{theorem}","rphic to the entire root locus\n\\[\n \\mathcal R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ for a continuous\n$\\phi:T\\to y_0+p^L\\Zp^e$ whose graph is the entire reordered root locus.\n\\end{theorem}","R=\\{z\\in B_L(x):F(z)=0\\}.\n\\]\nThe inverse homeomorphism is the actual free-coordinate projection\n$z\\mapsto\\operatorname{free}(\\sigma z)$. If $h:T\\to\\mathcal R$ is the\nforward map, then for every $N\\ge0$,\n\\begin{equation}\\label{eq:chart-congruence}\n h(t)\\equiv h(u)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad t\\equiv u\\pmod{p^N}.\n\\end{equation}\nEquivalently, $\\sigma h(t)=(\\phi(t),t)$ for a continuous\n$\\phi:T\\to y_0+p^L\\Zp^e$ whose graph is the entire reordered root locus.\n\\end{theorem}","\\begin{proof}\nFirst consider a square integer polynomial map $P$ whose reduced Jacobian\nat $x$ is surjective. Its Jacobian is invertible over $\\F_p$. A digit\ncomparison using Taylor expansion shows that equal outputs modulo\n$p^{N+1}$ force equal inputs modulo $p^{N+1}$ whenever the inputs already\nagree modulo $p^N$ and belong to $B_1(x)$. Induction and integer truncation\ngive, for arbitrary $a,b\\in B_1(x)$,\n\\[\n P(a)\\equiv P(b)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad a\\equiv b\\pmod{p^N}.\n\\]","First consider a square integer polynomial map $P$ whose reduced Jacobian\nat $x$ is surjective. Its Jacobian is invertible over $\\F_p$. A digit\ncomparison using Taylor expansion shows that equal outputs modulo\n$p^{N+1}$ force equal inputs modulo $p^{N+1}$ whenever the inputs already\nagree modulo $p^N$ and belong to $B_1(x)$. Induction and integer truncation\ngive, for arbitrary $a,b\\in B_1(x)$,\n\\[\n P(a)\\equiv P(b)\\pmod{p^N}\n \\quad\\Longleftrightarrow\\quad a\\equiv b\\pmod{p^N}.\n\\]","More generally, for a rectangular map with surjective reduced Jacobian,\n\\begin{equation}\\label{eq:polynomial-ball-image}\n P(B_L(x))=P(x)+p^L\\Zp^e.\n\\end{equation}\nOne inclusion follows from polynomial congruence. For integer targets in\nthe right-hand ball, subtract the target from the constant terms and use\nthe smooth root theorem. For arbitrary $p$-adic targets, approximate by\ninteger truncations. The input ball is compact, so its continuous image is\nclosed. It contains all these approximants and hence their limit. This\nargument does not assume that independently chosen preimages form a\ncompatible sequence.","For square $P$, the resulting bijection between the two compact residue\nballs is continuous, and its target is Hausdorff. It is therefore a\nhomeomorphism, with the same congruence preservation for its inverse.","Now suppose a chosen set of $e$ columns of $DF(x)\\bmod p$ is independent,\nand put those coordinates first. Apply the square result to the augmented\npolynomial map\n\\[\n \\Psi(Y,T)=(F(Y,T),T).\n\\]\nIts reduced Jacobian has upper blocks $(D_YF,D_TF)$ and lower blocks\n$(0,I_k)$, so it is invertible. For $t\\in T$, the target $(0,t)$ belongs\nto $\\Psi(x)+p^L\\Zp^{e+k}$ because $F(x)\\equiv0\\pmod{p^L}$.\nThe inverse of $\\Psi$ at that target gives the unique root with parameters\n$t$. The inverse congruence property gives one direction of\n\\eqref{eq:chart-congruence}; projection onto the parameter coordinates\ngives the other.","se a chosen set of $e$ columns of $DF(x)\\bmod p$ is independent,\nand put those coordinates first. Apply the square result to the augmented\npolynomial map\n\\[\n \\Psi(Y,T)=(F(Y,T),T).\n\\]\nIts reduced Jacobian has upper blocks $(D_YF,D_TF)$ and lower blocks\n$(0,I_k)$, so it is invertible. For $t\\in T$, the target $(0,t)$ belongs\nto $\\Psi(x)+p^L\\Zp^{e+k}$ because $F(x)\\equiv0\\pmod{p^L}$.\nThe inverse of $\\Psi$ at that target gives the unique root with parameters\n$t$. The inverse congruence property gives one direction of\n\\eqref{eq:chart-congruence}; projection onto the parameter coordinates\ngives the other.","ent,\nand put those coordinates first. Apply the square result to the augmented\npolynomial map\n\\[\n \\Psi(Y,T)=(F(Y,T),T).\n\\]\nIts reduced Jacobian has upper blocks $(D_YF,D_TF)$ and lower blocks\n$(0,I_k)$, so it is invertible. For $t\\in T$, the target $(0,t)$ belongs\nto $\\Psi(x)+p^L\\Zp^{e+k}$ because $F(x)\\equiv0\\pmod{p^L}$.\nThe inverse of $\\Psi$ at that target gives the unique root with parameters\n$t$. The inverse congruence property gives one direction of\n\\eqref{eq:chart-congruence}; projection onto the parameter coordinates\ngives the other.","Finally, surjectivity of the original rectangular Jacobian guarantees a\nbasis among its columns. Extend the corresponding injection of coordinate\nindices to a permutation $\\sigma$. Renaming the variables by\n$\\sigma^{-1}$ transports polynomial evaluation and every formal partial\nderivative. The fixed-coordinate construction therefore applies to the\noriginal system after undoing the permutation.\n\\end{proof}","The main theorem is\n\\lean{Komorebi.PolynomialGeneralChart.exists_parameter_homeomorph}.\nThe square local inverse and surjectivity for arbitrary targets are\n\\lean{Komorebi.PolynomialLocalInverse.localHomeomorph} and\n\\lean{Komorebi.PolynomialLocalSurjective.exists_padic_preimage}.\nThe center $x$ is only an approximate root: the chart need not satisfy\n$\\phi(t_0)=y_0$ unless $F(x)=0$ exactly. The assertion selects suitable\ncoordinates; it does not permit an arbitrary prescribed solved block.","Suppose now that the split into solved variables and parameter variables\nis fixed in advance. Let\n\\[\n F\\in\\Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\\qquad x=(y_0,t_0),\n\\]\nand assume $F(x)=q^Lr_x$ with $q\\ge1$ and $L\\ge1$. Write the reduced\nJacobian as\n\\[\n DF(x)\\bmod q=(J_x\\ K_x),\\qquad\n J_x:S_q^m\\to S_q^e,\\quad K_x:S_q^k\\to S_q^e.\n\\]\nFor a prescribed parameter digit $v\\in S_q^k$, let $\\mathcal L_{x,L,v}$\nbe the set of next residue roots over $x$ whose parameter coordinates are\n$t_0+q^L\\widetilde v$ modulo $q^{L+1}$.","riables and parameter variables\nis fixed in advance. Let\n\\[\n F\\in\\Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\\qquad x=(y_0,t_0),\n\\]\nand assume $F(x)=q^Lr_x$ with $q\\ge1$ and $L\\ge1$. Write the reduced\nJacobian as\n\\[\n DF(x)\\bmod q=(J_x\\ K_x),\\qquad\n J_x:S_q^m\\to S_q^e,\\quad K_x:S_q^k\\to S_q^e.\n\\]\nFor a prescribed parameter digit $v\\in S_q^k$, let $\\mathcal L_{x,L,v}$\nbe the set of next residue roots over $x$ whose parameter coordinates are\n$t_0+q^L\\widetilde v$ modulo $q^{L+1}$.","ixed in advance. Let\n\\[\n F\\in\\Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\\qquad x=(y_0,t_0),\n\\]\nand assume $F(x)=q^Lr_x$ with $q\\ge1$ and $L\\ge1$. Write the reduced\nJacobian as\n\\[\n DF(x)\\bmod q=(J_x\\ K_x),\\qquad\n J_x:S_q^m\\to S_q^e,\\quad K_x:S_q^k\\to S_q^e.\n\\]\nFor a prescribed parameter digit $v\\in S_q^k$, let $\\mathcal L_{x,L,v}$\nbe the set of next residue roots over $x$ whose parameter coordinates are\n$t_0+q^L\\widetilde v$ modulo $q^{L+1}$.","\\Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\\qquad x=(y_0,t_0),\n\\]\nand assume $F(x)=q^Lr_x$ with $q\\ge1$ and $L\\ge1$. Write the reduced\nJacobian as\n\\[\n DF(x)\\bmod q=(J_x\\ K_x),\\qquad\n J_x:S_q^m\\to S_q^e,\\quad K_x:S_q^k\\to S_q^e.\n\\]\nFor a prescribed parameter digit $v\\in S_q^k$, let $\\mathcal L_{x,L,v}$\nbe the set of next residue roots over $x$ whose parameter coordinates are\n$t_0+q^L\\widetilde v$ modulo $q^{L+1}$.","\\begin{theorem}[Prescribed-parameter obstruction and count]\n\\label{thm:prescribed-obstruction}\nDefine\n\\[\n o_x(v)=[\\overline{r_x}+K_xv]\\in S_q^e/\\im J_x.\n\\]\nThen $\\mathcal L_{x,L,v}$ is nonempty if and only if $o_x(v)=0$.\nWhen nonempty, it is a torsor for $\\ker J_x$. If $q=p$ is prime and\n$\\rho_x=\\rank_{\\F_p}J_x$, then\n\\begin{equation}\\label{eq:prescribed-count}\n \\#\\mathcal L_{x,L,v}\n =\\begin{cases}p^{m-\\rho_x},&o_x(v)=0,","\\begin{theorem}[Prescribed-parameter obstruction and count]\n\\label{thm:prescribed-obstruction}\nDefine\n\\[\n o_x(v)=[\\overline{r_x}+K_xv]\\in S_q^e/\\im J_x.\n\\]\nThen $\\mathcal L_{x,L,v}$ is nonempty if and only if $o_x(v)=0$.\nWhen nonempty, it is a torsor for $\\ker J_x$. If $q=p$ is prime and\n$\\rho_x=\\rank_{\\F_p}J_x$, then\n\\begin{equation}\\label{eq:prescribed-count}\n \\#\\mathcal L_{x,L,v}\n =\\begin{cases}p^{m-\\rho_x},&o_x(v)=0,\\\\0,&o_x(v)\\ne0,\\end{cases}\n \\qquad \\#(\\F_p^e/\\im J_x)=p^{e-\\rho_x}.\n\\end{equation}\nThe criterion holds uniformly for every prescribed $v$ without a\nsurjectivity hypothesis.\n\\end{theorem}","0,&o_x(v)\\ne0,\\end{cases}\n \\qquad \\#(\\F_p^e/\\im J_x)=p^{e-\\rho_x}.\n\\end{equation}\nThe criterion holds uniformly for every prescribed $v$ without a\nsurjectivity hypothesis.\n\\end{theorem}","\\begin{proof}\nA candidate for the next lift has the form\n$x+q^L(\\widetilde u,\\widetilde v)$, with solved digit $u\\in S_q^m$.\nIts Taylor correction equation is\n\\[\n \\overline{r_x}+J_xu+K_xv=0.\n\\]\nSolvability is exactly vanishing of the displayed cokernel class. Restrict\nthe residue-digit equivalence to the prescribed parameter coordinates.\nThis identifies the lift fiber with the solved-digit fiber.\nTranslation by $\\ker J_x$ is free and transitive on any nonempty such\nfiber. The prime-field count follows from rank-nullity as before.\n\\end{proof}","candidate for the next lift has the form\n$x+q^L(\\widetilde u,\\widetilde v)$, with solved digit $u\\in S_q^m$.\nIts Taylor correction equation is\n\\[\n \\overline{r_x}+J_xu+K_xv=0.\n\\]\nSolvability is exactly vanishing of the displayed cokernel class. Restrict\nthe residue-digit equivalence to the prescribed parameter coordinates.\nThis identifies the lift fiber with the solved-digit fiber.\nTranslation by $\\ker J_x$ is free and transitive on any nonempty such\nfiber. The prime-field count follows from rank-nullity as before.\n\\end{proof}","The formal statements start from the lower root and construct its residual:\n\\lean{Komorebi.PolynomialPrescribedObstruction.finite_prescribed_obstruction}\nand \\lean{Komorebi.PolynomialPrescribedObstruction.finite_prescribed_lift_count}.\nFor $F(Y,T)=T$ at the dyadic origin, the full Jacobian $(0\\ 1)$ is\nsurjective but $J_x=0$. Prescribing $v=1$ forces $T=2\\pmod4$, so no root\nexists; prescribing $v=0$ gives two choices of $Y$ modulo four over the\ninitial zero residue. This example explains why the criterion depends on\nthe solved block, rather than merely on the full Jacobian.","art from the lower root and construct its residual:\n\\lean{Komorebi.PolynomialPrescribedObstruction.finite_prescribed_obstruction}\nand \\lean{Komorebi.PolynomialPrescribedObstruction.finite_prescribed_lift_count}.\nFor $F(Y,T)=T$ at the dyadic origin, the full Jacobian $(0\\ 1)$ is\nsurjective but $J_x=0$. Prescribing $v=1$ forces $T=2\\pmod4$, so no root\nexists; prescribing $v=0$ gives two choices of $Y$ modulo four over the\ninitial zero residue. This example explains why the criterion depends on\nthe solved block, rather than merely on the full Jacobian.","\\begin{theorem}[Every compatible integer parameter path]\n\\label{thm:prescribed-integer-tower}\nIn the preceding notation, assume $J_x$ is surjective. For every supplied\ninteger path $t_n\\in\\Z^k$ satisfying\n\\[\n t_0=\\operatorname{free}(x),\\qquad\n t_{n+1}-t_n\\in q^{L+n}\\Z^k,\n\\]\nthere is a compatible polynomial tower $z_n\\in\\Z^{m+k}$ with\n\\[\n z_0=x,\\quad \\operatorname{free}(z_n)=t_n,\\quad\n F(z_n)\\in q^{L+n}\\Z^e,\\quad\n z_{n+1}-z_n\\in q^{L+n}\\Z^{m+k}.\n\\]\nThe free-coordinate equality is exact in integers. Both reduced Jacobian\nblocks remain fixed, and $z_b-z_a\\in q^{L+a}\\Z^{m+k}$ whenever $a\\le b$.\n\\end{theorem}","r every supplied\ninteger path $t_n\\in\\Z^k$ satisfying\n\\[\n t_0=\\operatorname{free}(x),\\qquad\n t_{n+1}-t_n\\in q^{L+n}\\Z^k,\n\\]\nthere is a compatible polynomial tower $z_n\\in\\Z^{m+k}$ with\n\\[\n z_0=x,\\quad \\operatorname{free}(z_n)=t_n,\\quad\n F(z_n)\\in q^{L+n}\\Z^e,\\quad\n z_{n+1}-z_n\\in q^{L+n}\\Z^{m+k}.\n\\]\nThe free-coordinate equality is exact in integers. Both reduced Jacobian\nblocks remain fixed, and $z_b-z_a\\in q^{L+a}\\Z^{m+k}$ whenever $a\\le b$.\n\\end{theorem}","Z^k$ satisfying\n\\[\n t_0=\\operatorname{free}(x),\\qquad\n t_{n+1}-t_n\\in q^{L+n}\\Z^k,\n\\]\nthere is a compatible polynomial tower $z_n\\in\\Z^{m+k}$ with\n\\[\n z_0=x,\\quad \\operatorname{free}(z_n)=t_n,\\quad\n F(z_n)\\in q^{L+n}\\Z^e,\\quad\n z_{n+1}-z_n\\in q^{L+n}\\Z^{m+k}.\n\\]\nThe free-coordinate equality is exact in integers. Both reduced Jacobian\nblocks remain fixed, and $z_b-z_a\\in q^{L+a}\\Z^{m+k}$ whenever $a\\le b$.\n\\end{theorem}","\\begin{proof}\nAt level $N=L+n$, write the supplied parameter increment exactly as\n$t_{n+1}-t_n=q^Nw$ with $w\\in\\Z^k$. Surjectivity of the solved block\ngives a digit $u$ satisfying\n$J_{z_n}u=-\\overline{F(z_n)/q^N}-K_{z_n}\\overline w$.\nChoose an integer representative $\\widetilde u$ and set\n$z_{n+1}=z_n+q^N(\\widetilde u,w)$. Its parameter coordinates equal the\nsupplied target exactly. Taylor expansion proves the next root condition,\nwhile congruence modulo $q$ preserves the full reduced Jacobian.\nConstruct the sequence recursively, preserving the exact parameter\ncoordinates and the fixed solved Jacobian. The earlier cross-level\nargument proves all the stated integer congruences.\n\\end{proof}","increment exactly as\n$t_{n+1}-t_n=q^Nw$ with $w\\in\\Z^k$. Surjectivity of the solved block\ngives a digit $u$ satisfying\n$J_{z_n}u=-\\overline{F(z_n)/q^N}-K_{z_n}\\overline w$.\nChoose an integer representative $\\widetilde u$ and set\n$z_{n+1}=z_n+q^N(\\widetilde u,w)$. Its parameter coordinates equal the\nsupplied target exactly. Taylor expansion proves the next root condition,\nwhile congruence modulo $q$ preserves the full reduced Jacobian.\nConstruct the sequence recursively, preserving the exact parameter\ncoordinates and the fixed solved Jacobian. The earlier cross-level\nargument proves all the stated integer congruences.\n\\end{proof}","\\begin{theorem}[Arbitrary $p$-adic parameter endpoints]\n\\label{thm:prescribed-padic-tower}\nLet $q=p$ be prime in Theorem~\\ref{thm:prescribed-integer-tower}.\nFor every $\\tau\\in t_0+p^L\\Zp^k$, there are an integer tower $z_n$ and\n$b\\in\\Zp^{m+k}$ such that\n\\[\n F(b)=0,\\qquad\\operatorname{free}(b)=\\tau,\\qquad\n z_n\\longrightarrow b,\\qquad b-z_n\\in p^{L+n}\\Zp^{m+k}.\n\\]\nThe initial value is $z_0=x$. At every stage and for every proposed\nparameter digit $v\\in\\F_p^k$, the corresponding next residue fiber has\nexactly $p^{m-e}$ elements. If $m=e$, then for each $\\tau$ there is exactly\none such root in $B_L(x)$, independently of all choices of integer\napproximants and correcting digits.\n\\end{theorem}","ints]\n\\label{thm:prescribed-padic-tower}\nLet $q=p$ be prime in Theorem~\\ref{thm:prescribed-integer-tower}.\nFor every $\\tau\\in t_0+p^L\\Zp^k$, there are an integer tower $z_n$ and\n$b\\in\\Zp^{m+k}$ such that\n\\[\n F(b)=0,\\qquad\\operatorname{free}(b)=\\tau,\\qquad\n z_n\\longrightarrow b,\\qquad b-z_n\\in p^{L+n}\\Zp^{m+k}.\n\\]\nThe initial value is $z_0=x$. At every stage and for every proposed\nparameter digit $v\\in\\F_p^k$, the corresponding next residue fiber has\nexactly $p^{m-e}$ elements. If $m=e$, then for each $\\tau$ there is exactly\none such root in $B_L(x)$, independently of all choices of integer\napproximants and correcting digits.\n\\end{theorem}","Theorem~\\ref{thm:prescribed-integer-tower}.\nFor every $\\tau\\in t_0+p^L\\Zp^k$, there are an integer tower $z_n$ and\n$b\\in\\Zp^{m+k}$ such that\n\\[\n F(b)=0,\\qquad\\operatorname{free}(b)=\\tau,\\qquad\n z_n\\longrightarrow b,\\qquad b-z_n\\in p^{L+n}\\Zp^{m+k}.\n\\]\nThe initial value is $z_0=x$. At every stage and for every proposed\nparameter digit $v\\in\\F_p^k$, the corresponding next residue fiber has\nexactly $p^{m-e}$ elements. If $m=e$, then for each $\\tau$ there is exactly\none such root in $B_L(x)$, independently of all choices of integer\napproximants and correcting digits.\n\\end{theorem}","ibed-integer-tower}.\nFor every $\\tau\\in t_0+p^L\\Zp^k$, there are an integer tower $z_n$ and\n$b\\in\\Zp^{m+k}$ such that\n\\[\n F(b)=0,\\qquad\\operatorname{free}(b)=\\tau,\\qquad\n z_n\\longrightarrow b,\\qquad b-z_n\\in p^{L+n}\\Zp^{m+k}.\n\\]\nThe initial value is $z_0=x$. At every stage and for every proposed\nparameter digit $v\\in\\F_p^k$, the corresponding next residue fiber has\nexactly $p^{m-e}$ elements. If $m=e$, then for each $\\tau$ there is exactly\none such root in $B_L(x)$, independently of all choices of integer\napproximants and correcting digits.\n\\end{theorem}","For every $\\tau\\in t_0+p^L\\Zp^k$, there are an integer tower $z_n$ and\n$b\\in\\Zp^{m+k}$ such that\n\\[\n F(b)=0,\\qquad\\operatorname{free}(b)=\\tau,\\qquad\n z_n\\longrightarrow b,\\qquad b-z_n\\in p^{L+n}\\Zp^{m+k}.\n\\]\nThe initial value is $z_0=x$. At every stage and for every proposed\nparameter digit $v\\in\\F_p^k$, the corresponding next residue fiber has\nexactly $p^{m-e}$ elements. If $m=e$, then for each $\\tau$ there is exactly\none such root in $B_L(x)$, independently of all choices of integer\napproximants and correcting digits.\n\\end{theorem}","re an integer tower $z_n$ and\n$b\\in\\Zp^{m+k}$ such that\n\\[\n F(b)=0,\\qquad\\operatorname{free}(b)=\\tau,\\qquad\n z_n\\longrightarrow b,\\qquad b-z_n\\in p^{L+n}\\Zp^{m+k}.\n\\]\nThe initial value is $z_0=x$. At every stage and for every proposed\nparameter digit $v\\in\\F_p^k$, the corresponding next residue fiber has\nexactly $p^{m-e}$ elements. If $m=e$, then for each $\\tau$ there is exactly\none such root in $B_L(x)$, independently of all choices of integer\napproximants and correcting digits.\n\\end{theorem}","\\begin{proof}\nSet the first integer parameter approximation equal to the prescribed\n$t_0$. For $n\\ge1$, use canonical representatives of\n$\\tau\\bmod p^{L+n}$. The initial congruence and truncation precision give\n\\[\n \\tau-t_n\\in p^{L+n}\\Zp^k.\n\\]\nSubtracting consecutive relations gives compatibility in $\\Zp$.\nFaithfulness of integer divisibility under $\\Z\\hookrightarrow\\Zp$\ngives compatibility in $\\Z$, and the same norm bounds give $t_n\\to\\tau$.\nTheorem~\\ref{thm:prescribed-integer-tower} and completion now produce $b$\nwith exact polynomial vanishing and all retained precisions. Each free\ncoordinate has both the corresponding coordinate of $b$ and that of\n$\\tau$ as its limit, proving exact parameter equality.","r parameter approximation equal to the prescribed\n$t_0$. For $n\\ge1$, use canonical representatives of\n$\\tau\\bmod p^{L+n}$. The initial congruence and truncation precision give\n\\[\n \\tau-t_n\\in p^{L+n}\\Zp^k.\n\\]\nSubtracting consecutive relations gives compatibility in $\\Zp$.\nFaithfulness of integer divisibility under $\\Z\\hookrightarrow\\Zp$\ngives compatibility in $\\Z$, and the same norm bounds give $t_n\\to\\tau$.\nTheorem~\\ref{thm:prescribed-integer-tower} and completion now produce $b$\nwith exact polynomial vanishing and all retained precisions. Each free\ncoordinate has both the corresponding coordinate of $b$ and that of\n$\\tau$ as its limit, proving exact parameter equality.","representatives of\n$\\tau\\bmod p^{L+n}$. The initial congruence and truncation precision give\n\\[\n \\tau-t_n\\in p^{L+n}\\Zp^k.\n\\]\nSubtracting consecutive relations gives compatibility in $\\Zp$.\nFaithfulness of integer divisibility under $\\Z\\hookrightarrow\\Zp$\ngives compatibility in $\\Z$, and the same norm bounds give $t_n\\to\\tau$.\nTheorem~\\ref{thm:prescribed-integer-tower} and completion now produce $b$\nwith exact polynomial vanishing and all retained precisions. Each free\ncoordinate has both the corresponding coordinate of $b$ and that of\n$\\tau$ as its limit, proving exact parameter equality.","ubtracting consecutive relations gives compatibility in $\\Zp$.\nFaithfulness of integer divisibility under $\\Z\\hookrightarrow\\Zp$\ngives compatibility in $\\Z$, and the same norm bounds give $t_n\\to\\tau$.\nTheorem~\\ref{thm:prescribed-integer-tower} and completion now produce $b$\nwith exact polynomial vanishing and all retained precisions. Each free\ncoordinate has both the corresponding coordinate of $b$ and that of\n$\\tau$ as its limit, proving exact parameter equality.","cutive relations gives compatibility in $\\Zp$.\nFaithfulness of integer divisibility under $\\Z\\hookrightarrow\\Zp$\ngives compatibility in $\\Z$, and the same norm bounds give $t_n\\to\\tau$.\nTheorem~\\ref{thm:prescribed-integer-tower} and completion now produce $b$\nwith exact polynomial vanishing and all retained precisions. Each free\ncoordinate has both the corresponding coordinate of $b$ and that of\n$\\tau$ as its limit, proving exact parameter equality.","The solved Jacobian remains surjective, so every shifted obstruction\n$[\\overline r+K_{z_n}v]$ vanishes. Formula~\\eqref{eq:prescribed-count}\ntherefore gives $p^{m-e}$ choices for every parameter digit at every stage.\nWhen $m=e$, the fixed-coordinate chart in the proof of\nTheorem~\\ref{thm:continuous-parameter-chart} contains every root in the\ninitial ball with the specified parameters. Its uniqueness proves the\nendpoint assertion. This is uniqueness of the actual endpoint, not of the\ninteger representatives used to approach it.\n\\end{proof}","The endpoint theorems are\n\\lean{Komorebi.PolynomialPrescribedPadicTower.exists_prescribed_padic_root_with_tower}\nand \\lean{Komorebi.PolynomialPrescribedPadicTower.exists_unique_prescribed_padic_root}.\nThe separate supplied-path theorem\n\\lean{Komorebi.PolynomialPrescribedPadicTower.exists_root_with_prescribed_path}\naccepts any compatible integer parameter path with limit $\\tau$.\nFor $m>e$, endpoint uniqueness is false in general: the system\n$F(Y_1,Y_2,T)=Y_1$ has two distinct roots $(0,0,0)$ and $(0,2,0)$ in the\nsame dyadic ball with identical parameter value.","nomialPrescribedPadicTower.exists_prescribed_padic_root_with_tower}\nand \\lean{Komorebi.PolynomialPrescribedPadicTower.exists_unique_prescribed_padic_root}.\nThe separate supplied-path theorem\n\\lean{Komorebi.PolynomialPrescribedPadicTower.exists_root_with_prescribed_path}\naccepts any compatible integer parameter path with limit $\\tau$.\nFor $m>e$, endpoint uniqueness is false in general: the system\n$F(Y_1,Y_2,T)=Y_1$ has two distinct roots $(0,0,0)$ and $(0,2,0)$ in the\nsame dyadic ball with identical parameter value.","ed_padic_root_with_tower}\nand \\lean{Komorebi.PolynomialPrescribedPadicTower.exists_unique_prescribed_padic_root}.\nThe separate supplied-path theorem\n\\lean{Komorebi.PolynomialPrescribedPadicTower.exists_root_with_prescribed_path}\naccepts any compatible integer parameter path with limit $\\tau$.\nFor $m>e$, endpoint uniqueness is false in general: the system\n$F(Y_1,Y_2,T)=Y_1$ has two distinct roots $(0,0,0)$ and $(0,2,0)$ in the\nsame dyadic ball with identical parameter value.","with_tower}\nand \\lean{Komorebi.PolynomialPrescribedPadicTower.exists_unique_prescribed_padic_root}.\nThe separate supplied-path theorem\n\\lean{Komorebi.PolynomialPrescribedPadicTower.exists_root_with_prescribed_path}\naccepts any compatible integer parameter path with limit $\\tau$.\nFor $m>e$, endpoint uniqueness is false in general: the system\n$F(Y_1,Y_2,T)=Y_1$ has two distinct roots $(0,0,0)$ and $(0,2,0)$ in the\nsame dyadic ball with identical parameter value.","The remaining theorems in this section allow arbitrary coefficients and\ncenters in $R=\\Zp$. They concern square solved Jacobians invertible over\n$\\Qp$, even when their reductions modulo $p$ are singular. The exponent\ncontrolling an integral inverse is retained explicitly.","\\begin{theorem}[Weighted nonlinear lifting]\\label{thm:weighted-lifting}\nLet $p$ be prime, $d,s,r\\in\\N$, $F\\in R[X_1,\\ldots,X_d]^d$, and\n$a\\in R^d$. Set $J=DF(a)$. Suppose\n\\begin{equation}\\label{eq:weighted-hypotheses}\n J\\in\\operatorname{GL}_d(\\Qp),\\qquad\n p^sJ^{-1}\\in M_d(R),\\qquad F(a)\\in p^rR^d,\\qquad r>2s.\n\\end{equation}\nThen there is exactly one root $b$ of $F$ in $a+p^{s+1}R^d$, and it\nsatisfies\n\\begin{equation}\\label{eq:weighted-displacement}\n b-a\\in p^{r-s}R^d.\n\\end{equation}\nUniqueness holds throughout the full ball of depth $s+1$.\n\\end{theorem}","hted nonlinear lifting]\\label{thm:weighted-lifting}\nLet $p$ be prime, $d,s,r\\in\\N$, $F\\in R[X_1,\\ldots,X_d]^d$, and\n$a\\in R^d$. Set $J=DF(a)$. Suppose\n\\begin{equation}\\label{eq:weighted-hypotheses}\n J\\in\\operatorname{GL}_d(\\Qp),\\qquad\n p^sJ^{-1}\\in M_d(R),\\qquad F(a)\\in p^rR^d,\\qquad r>2s.\n\\end{equation}\nThen there is exactly one root $b$ of $F$ in $a+p^{s+1}R^d$, and it\nsatisfies\n\\begin{equation}\\label{eq:weighted-displacement}\n b-a\\in p^{r-s}R^d.\n\\end{equation}\nUniqueness holds throughout the full ball of depth $s+1$.\n\\end{theorem}","n R^d$. Set $J=DF(a)$. Suppose\n\\begin{equation}\\label{eq:weighted-hypotheses}\n J\\in\\operatorname{GL}_d(\\Qp),\\qquad\n p^sJ^{-1}\\in M_d(R),\\qquad F(a)\\in p^rR^d,\\qquad r>2s.\n\\end{equation}\nThen there is exactly one root $b$ of $F$ in $a+p^{s+1}R^d$, and it\nsatisfies\n\\begin{equation}\\label{eq:weighted-displacement}\n b-a\\in p^{r-s}R^d.\n\\end{equation}\nUniqueness holds throughout the full ball of depth $s+1$.\n\\end{theorem}","\\begin{proof}\nLet $A=p^sJ^{-1}\\in M_d(R)$. Inclusion into $\\Qp$ and its injectivity\ngive the two integral identities\n\\[\n AJ=JA=p^sI_d.\n\\]\nThese identities are also an equivalent integral certificate for the\nfirst two hypotheses of \\eqref{eq:weighted-hypotheses}. In particular,\n$A$ is injective: $Aw=0$ implies $p^sw=JAw=0$, hence $w=0$.","Precondition the polynomial system by $A$ and set $c=p^{s+1}$.\nIts derivative at $a$ is $AJ=p^sI_d$. Lemma~\\ref{lem:polynomial-taylor}\ngives a polynomial vector $Q$ with $Q(0)=0$ and\n\\[\n (AF)(a+cu)=(AF)(a)+p^{2s+1}u+p^{2s+2}Q(u).\n\\]\nSince $A$ is integral, $(AF)(a)\\in p^rR^d$. The strict integer inequality\n$r>2s$ permits $k=r-(2s+1)\\ge0$ and a vector $C\\in p^kR^d$ such that\n$(AF)(a)=p^{2s+1}C$. The normalized root equation is therefore\n\\begin{equation}\\label{eq:weighted-fixed-point}\n u=-C-pQ(u).\n\\end{equation}","{s+1}$.\nIts derivative at $a$ is $AJ=p^sI_d$. Lemma~\\ref{lem:polynomial-taylor}\ngives a polynomial vector $Q$ with $Q(0)=0$ and\n\\[\n (AF)(a+cu)=(AF)(a)+p^{2s+1}u+p^{2s+2}Q(u).\n\\]\nSince $A$ is integral, $(AF)(a)\\in p^rR^d$. The strict integer inequality\n$r>2s$ permits $k=r-(2s+1)\\ge0$ and a vector $C\\in p^kR^d$ such that\n$(AF)(a)=p^{2s+1}C$. The normalized root equation is therefore\n\\begin{equation}\\label{eq:weighted-fixed-point}\n u=-C-pQ(u).\n\\end{equation}","Integral polynomial evaluation is $1$-Lipschitz for the supremum\n$p$-adic metric: a common divisibility of input differences is preserved\nunder addition and multiplication by integral elements. Consequently the\nright side of \\eqref{eq:weighted-fixed-point} has contraction constant\n$|p|_p<1$ on the complete space $R^d$. It has a unique fixed point $u$.","To obtain the sharper displacement, begin with $u\\in R^d$. If\n$u\\in p^nR^d$, then $Q(0)=0$ and polynomial congruence give\n$Q(u)\\in p^nR^d$. For $n<k$, both $C$ and $pQ(u)$ are divisible by\n$p^{n+1}$. Equation~\\eqref{eq:weighted-fixed-point} therefore improves\nthe divisibility of $u$ by one digit. Induction gives $u\\in p^kR^d$.\nThus $b=a+p^{s+1}u$ satisfies \\eqref{eq:weighted-displacement}, because\n$s+1+k=r-s$. Its normalized preconditioned value vanishes, and injectivity\nof $A$ gives $F(b)=0$.","obtain the sharper displacement, begin with $u\\in R^d$. If\n$u\\in p^nR^d$, then $Q(0)=0$ and polynomial congruence give\n$Q(u)\\in p^nR^d$. For $n<k$, both $C$ and $pQ(u)$ are divisible by\n$p^{n+1}$. Equation~\\eqref{eq:weighted-fixed-point} therefore improves\nthe divisibility of $u$ by one digit. Induction gives $u\\in p^kR^d$.\nThus $b=a+p^{s+1}u$ satisfies \\eqref{eq:weighted-displacement}, because\n$s+1+k=r-s$. Its normalized preconditioned value vanishes, and injectivity\nof $A$ gives $F(b)=0$.","n $Q(0)=0$ and polynomial congruence give\n$Q(u)\\in p^nR^d$. For $n<k$, both $C$ and $pQ(u)$ are divisible by\n$p^{n+1}$. Equation~\\eqref{eq:weighted-fixed-point} therefore improves\nthe divisibility of $u$ by one digit. Induction gives $u\\in p^kR^d$.\nThus $b=a+p^{s+1}u$ satisfies \\eqref{eq:weighted-displacement}, because\n$s+1+k=r-s$. Its normalized preconditioned value vanishes, and injectivity\nof $A$ gives $F(b)=0$.","and polynomial congruence give\n$Q(u)\\in p^nR^d$. For $n<k$, both $C$ and $pQ(u)$ are divisible by\n$p^{n+1}$. Equation~\\eqref{eq:weighted-fixed-point} therefore improves\nthe divisibility of $u$ by one digit. Induction gives $u\\in p^kR^d$.\nThus $b=a+p^{s+1}u$ satisfies \\eqref{eq:weighted-displacement}, because\n$s+1+k=r-s$. Its normalized preconditioned value vanishes, and injectivity\nof $A$ gives $F(b)=0$.","Every competing root $z\\in a+p^{s+1}R^d$ can be written\n$z=a+p^{s+1}w$. Its root equation implies the same fixed-point\nequation after cancellation of the nonzero scalar $p^{2s+1}$.\nContraction uniqueness gives $w=u$, proving uniqueness on the entire\nstated ball.\n\\end{proof}","The existence and uniqueness statements are\n\\lean{Komorebi.PolynomialWeightedLift.exists_weighted_root} and\n\\lean{Komorebi.PolynomialWeightedLift.exists_unique_weighted_root}.\nThe inverse hypothesis in Lean specifies an integral matrix whose image\nis exactly $p^sJ^{-1}$; it does not substitute a potentially larger\ndeterminant valuation for $s$.","\\begin{example}[Strict threshold and displacement boundary]\n\\label{ex:weighted-boundary}\nFor $p=2$, $F(X)=X^2+3$, and $a=1$, one has $F(a)=4$ and $DF(a)=2$.\nThus $s=1$ and $r=2$ satisfy every displayed hypothesis except the strict\ninequality. No root exists in $\\Z_2$, since none exists modulo eight.\nHence $r\\ge2s$ cannot replace $r>2s$ uniformly. Also,\n$F(X)=p^sX+p^r$ has root $-p^{r-s}$, whose displacement is not divisible\nby $p^{r-s+1}$. The guaranteed displacement cannot uniformly gain a digit.\n\\end{example}","and displacement boundary]\n\\label{ex:weighted-boundary}\nFor $p=2$, $F(X)=X^2+3$, and $a=1$, one has $F(a)=4$ and $DF(a)=2$.\nThus $s=1$ and $r=2$ satisfy every displayed hypothesis except the strict\ninequality. No root exists in $\\Z_2$, since none exists modulo eight.\nHence $r\\ge2s$ cannot replace $r>2s$ uniformly. Also,\n$F(X)=p^sX+p^r$ has root $-p^{r-s}$, whose displacement is not divisible\nby $p^{r-s+1}$. The guaranteed displacement cannot uniformly gain a digit.\n\\end{example}","\\begin{theorem}[Weighted coefficient deformation]\\label{thm:coefficient-motion}\nLet $F,H\\in R[X_1,\\ldots,X_d]^d$, $F(a)=0$, and $J=DF(a)$. Suppose\n$J\\in\\operatorname{GL}_d(\\Qp)$, $p^sJ^{-1}\\in M_d(R)$, and $M>2s$.\nFor $G=F+p^MH$, there is exactly one root $b\\in a+p^{s+1}R^d$.\nIt satisfies $b-a\\in p^{M-s}R^d$ and, in $\\Qp^d$,\n\\begin{equation}\\label{eq:coefficient-first-order}\n b-a=-p^MJ^{-1}H(a)+\\varepsilon,\n \\qquad\\varepsilon\\in p^{2M-3s}R^d.\n\\end{equation}\nOnly the original Jacobian $DF(a)$ is subject to an inverse hypothesis.\n\\end{theorem}","n R[X_1,\\ldots,X_d]^d$, $F(a)=0$, and $J=DF(a)$. Suppose\n$J\\in\\operatorname{GL}_d(\\Qp)$, $p^sJ^{-1}\\in M_d(R)$, and $M>2s$.\nFor $G=F+p^MH$, there is exactly one root $b\\in a+p^{s+1}R^d$.\nIt satisfies $b-a\\in p^{M-s}R^d$ and, in $\\Qp^d$,\n\\begin{equation}\\label{eq:coefficient-first-order}\n b-a=-p^MJ^{-1}H(a)+\\varepsilon,\n \\qquad\\varepsilon\\in p^{2M-3s}R^d.\n\\end{equation}\nOnly the original Jacobian $DF(a)$ is subject to an inverse hypothesis.\n\\end{theorem}","\\begin{proof}\nRetain the original certificate $A=p^sJ^{-1}$ and put $c=p^{s+1}$.\nTaylor expansion and a polynomial difference identity yield polynomial\nvectors $Q_F,Q_H$, both zero at the origin, such that\n\\[\n (AF)(a+cu)=p^{2s+1}u+p^{2s+2}Q_F(u),\\qquad\n (AH)(a+cu)=AH(a)+cQ_H(u).\n\\]\nConsequently\n\\[\n (AG)(a+cu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\]\nwhere\n\\[\n C=p^{M-(2s+1)}AH(a),\\qquad\n Q=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\]\nAll exponents are nonnegative. The same contraction and divisibility\nargument as in Theorem~\\ref{thm:weighted-lifting} gives the root,\nthe displacement $p^{M-s}$, and full-ball uniqueness. This constructs the\nroot without introducing an inverse hypothesis for $DG(a)$.","th zero at the origin, such that\n\\[\n (AF)(a+cu)=p^{2s+1}u+p^{2s+2}Q_F(u),\\qquad\n (AH)(a+cu)=AH(a)+cQ_H(u).\n\\]\nConsequently\n\\[\n (AG)(a+cu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\]\nwhere\n\\[\n C=p^{M-(2s+1)}AH(a),\\qquad\n Q=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\]\nAll exponents are nonnegative. The same contraction and divisibility\nargument as in Theorem~\\ref{thm:weighted-lifting} gives the root,\nthe displacement $p^{M-s}$, and full-ball uniqueness. This constructs the\nroot without introducing an inverse hypothesis for $DG(a)$.","]\nConsequently\n\\[\n (AG)(a+cu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\]\nwhere\n\\[\n C=p^{M-(2s+1)}AH(a),\\qquad\n Q=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\]\nAll exponents are nonnegative. The same contraction and divisibility\nargument as in Theorem~\\ref{thm:weighted-lifting} gives the root,\nthe displacement $p^{M-s}$, and full-ball uniqueness. This constructs the\nroot without introducing an inverse hypothesis for $DG(a)$.","For the error estimate, let $\\delta=b-a$ and $D=M-s$. Taylor's theorem\nfor the preconditioned polynomial gives\n\\[\n E_F=(AF)(b)-(AF)(a)-p^s\\delta\\in p^{2D}R^d.\n\\]\nPolynomial congruence gives\n$E_H=(AH)(b)-(AH)(a)\\in p^DR^d$.\nThe equations $F(a)=0$ and $G(b)=0$ imply\n\\[\n p^s\\delta+p^MAH(a)=-E_F-p^ME_H\\in p^{2D}R^d,\n\\]\nbecause $M+D\\ge2D$. Cancel $p^s$ in the integral domain $R$ to obtain\n\\[\n \\delta+p^{M-s}AH(a)\\in p^{2M-3s}R^d.\n\\]\nThis integral vector is $\\varepsilon$. After inclusion into $\\Qp$,\n$p^{M-s}A=p^MJ^{-1}$ gives \\eqref{eq:coefficient-first-order}.\n\\end{proof}","ylor's theorem\nfor the preconditioned polynomial gives\n\\[\n E_F=(AF)(b)-(AF)(a)-p^s\\delta\\in p^{2D}R^d.\n\\]\nPolynomial congruence gives\n$E_H=(AH)(b)-(AH)(a)\\in p^DR^d$.\nThe equations $F(a)=0$ and $G(b)=0$ imply\n\\[\n p^s\\delta+p^MAH(a)=-E_F-p^ME_H\\in p^{2D}R^d,\n\\]\nbecause $M+D\\ge2D$. Cancel $p^s$ in the integral domain $R$ to obtain\n\\[\n \\delta+p^{M-s}AH(a)\\in p^{2M-3s}R^d.\n\\]\nThis integral vector is $\\varepsilon$. After inclusion into $\\Qp$,\n$p^{M-s}A=p^MJ^{-1}$ gives \\eqref{eq:coefficient-first-order}.\n\\end{proof}","d polynomial gives\n\\[\n E_F=(AF)(b)-(AF)(a)-p^s\\delta\\in p^{2D}R^d.\n\\]\nPolynomial congruence gives\n$E_H=(AH)(b)-(AH)(a)\\in p^DR^d$.\nThe equations $F(a)=0$ and $G(b)=0$ imply\n\\[\n p^s\\delta+p^MAH(a)=-E_F-p^ME_H\\in p^{2D}R^d,\n\\]\nbecause $M+D\\ge2D$. Cancel $p^s$ in the integral domain $R$ to obtain\n\\[\n \\delta+p^{M-s}AH(a)\\in p^{2M-3s}R^d.\n\\]\nThis integral vector is $\\varepsilon$. After inclusion into $\\Qp$,\n$p^{M-s}A=p^MJ^{-1}$ gives \\eqref{eq:coefficient-first-order}.\n\\end{proof}","omial gives\n\\[\n E_F=(AF)(b)-(AF)(a)-p^s\\delta\\in p^{2D}R^d.\n\\]\nPolynomial congruence gives\n$E_H=(AH)(b)-(AH)(a)\\in p^DR^d$.\nThe equations $F(a)=0$ and $G(b)=0$ imply\n\\[\n p^s\\delta+p^MAH(a)=-E_F-p^ME_H\\in p^{2D}R^d,\n\\]\nbecause $M+D\\ge2D$. Cancel $p^s$ in the integral domain $R$ to obtain\n\\[\n \\delta+p^{M-s}AH(a)\\in p^{2M-3s}R^d.\n\\]\nThis integral vector is $\\varepsilon$. After inclusion into $\\Qp$,\n$p^{M-s}A=p^MJ^{-1}$ gives \\eqref{eq:coefficient-first-order}.\n\\end{proof}","The complete theorem is\n\\lean{Komorebi.PolynomialRootMotion.exists_unique_root_motion}.\nThe error depth exceeds the displacement depth by $M-2s\\ge1$. It is a\nguaranteed divisibility, not an assertion of exact valuation for every\nsystem.","\\begin{theorem}[Joint coefficient and parameter deformation]\n\\label{thm:joint-deformation}\\leavevmode\\par\nLet $F,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e$ and\n$a=(y_0,t_0)\\in R^{e+k}$ satisfy $F(a)=0$. Define the actual blocks\n\\[\n J=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\]\nSuppose $J\\in\\operatorname{GL}_e(\\Qp)$,\n$p^sJ^{-1}\\in M_e(R)$, and $M>2s$. For every $v\\in R^k$, set\n\\[\n \\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\]\nThere is exactly one $y\\in y_0+p^{s+1}R^e$ with $G(y,\\tau)=0$.\nIt satisfies\n\\begin{equation}\\label{eq:joint-first-order}\n \\begin{aligned}\n  y-y_0&\\in p^{M-s}R^e,","rmation]\n\\label{thm:joint-deformation}\\leavevmode\\par\nLet $F,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e$ and\n$a=(y_0,t_0)\\in R^{e+k}$ satisfy $F(a)=0$. Define the actual blocks\n\\[\n J=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\]\nSuppose $J\\in\\operatorname{GL}_e(\\Qp)$,\n$p^sJ^{-1}\\in M_e(R)$, and $M>2s$. For every $v\\in R^k$, set\n\\[\n \\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\]\nThere is exactly one $y\\in y_0+p^{s+1}R^e$ with $G(y,\\tau)=0$.\nIt satisfies\n\\begin{equation}\\label{eq:joint-first-order}\n \\begin{aligned}\n  y-y_0&\\in p^{M-s}R^e,","eformation}\\leavevmode\\par\nLet $F,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e$ and\n$a=(y_0,t_0)\\in R^{e+k}$ satisfy $F(a)=0$. Define the actual blocks\n\\[\n J=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\]\nSuppose $J\\in\\operatorname{GL}_e(\\Qp)$,\n$p^sJ^{-1}\\in M_e(R)$, and $M>2s$. For every $v\\in R^k$, set\n\\[\n \\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\]\nThere is exactly one $y\\in y_0+p^{s+1}R^e$ with $G(y,\\tau)=0$.\nIt satisfies\n\\begin{equation}\\label{eq:joint-first-order}\n \\begin{aligned}\n  y-y_0&\\in p^{M-s}R^e,","1,\\ldots,Y_e,T_1,\\ldots,T_k]^e$ and\n$a=(y_0,t_0)\\in R^{e+k}$ satisfy $F(a)=0$. Define the actual blocks\n\\[\n J=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\]\nSuppose $J\\in\\operatorname{GL}_e(\\Qp)$,\n$p^sJ^{-1}\\in M_e(R)$, and $M>2s$. For every $v\\in R^k$, set\n\\[\n \\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\]\nThere is exactly one $y\\in y_0+p^{s+1}R^e$ with $G(y,\\tau)=0$.\nIt satisfies\n\\begin{equation}\\label{eq:joint-first-order}\n \\begin{aligned}\n  y-y_0&\\in p^{M-s}R^e,\\\\\n  y-y_0&=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon,\n  \\qquad\\varepsilon\\in p^{2M-3s}R^e.\n \\end{aligned}\n\\end{equation}\nThe equality involving $J^{-1}$ is in $\\Qp^e$, with an integral error\nvector. The parameter equality is exact.\n\\end{theorem}","y-y_0&=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon,\n  \\qquad\\varepsilon\\in p^{2M-3s}R^e.\n \\end{aligned}\n\\end{equation}\nThe equality involving $J^{-1}$ is in $\\Qp^e$, with an integral error\nvector. The parameter equality is exact.\n\\end{theorem}","\\begin{proof}\nUse the augmented polynomial systems\n\\[\n \\mathcal F(Y,T)=(F(Y,T),T-t_0),\\qquad\n \\mathcal H(Y,T)=(H(Y,T),-v).\n\\]\nThen $\\mathcal F(a)=0$, and its derivative is\n\\[\n B=D\\mathcal F(a)=\\begin{pmatrix}J&K","\\begin{proof}\nUse the augmented polynomial systems\n\\[\n \\mathcal F(Y,T)=(F(Y,T),T-t_0),\\qquad\n \\mathcal H(Y,T)=(H(Y,T),-v).\n\\]\nThen $\\mathcal F(a)=0$, and its derivative is\n\\[\n B=D\\mathcal F(a)=\\begin{pmatrix}J&K\\\\0&I_k\\end{pmatrix}.\n\\]\nFor $A=p^sJ^{-1}\\in M_e(R)$ and $c=p^s$, define\n\\[\n C=\\begin{pmatrix}A&-AK","0&I_k\\end{pmatrix}.\n\\]\nFor $A=p^sJ^{-1}\\in M_e(R)$ and $c=p^s$, define\n\\[\n C=\\begin{pmatrix}A&-AK","0&I_k\\end{pmatrix}.\n\\]\nFor $A=p^sJ^{-1}\\in M_e(R)$ and $c=p^s$, define\n\\[\n C=\\begin{pmatrix}A&-AK\\\\0&cI_k\\end{pmatrix}.\n\\]\nThe identities $AJ=JA=cI_e$ give $CB=BC=cI_{e+k}$. In the second\nproduct, the upper right block is $-JAK+cK=0$; in the first, it is\n$AK-AK=0$. Thus augmentation preserves the original inverse bound $s$.","0&cI_k\\end{pmatrix}.\n\\]\nThe identities $AJ=JA=cI_e$ give $CB=BC=cI_{e+k}$. In the second\nproduct, the upper right block is $-JAK+cK=0$; in the first, it is\n$AK-AK=0$. Thus augmentation preserves the original inverse bound $s$.","Theorem~\\ref{thm:coefficient-motion} gives a unique root $b$ of\n$\\mathcal F+p^M\\mathcal H$ in $a+p^{s+1}R^{e+k}$, with displacement\ndepth $M-s$. Its last equations say exactly\n$b_T=t_0+p^Mv=\\tau$. Its first equations are $G(b)=0$.\nThe integral first-order expression uses\n\\[\n C\\binom{H(a)}{-v}\n =\\binom{A(H(a)+Kv)}{-p^sv}.\n\\]\nProjection onto the solved coordinates gives\n\\[\n b_Y-y_0+p^{M-s}A(H(a)+Kv)\\in p^{2M-3s}R^e.\n\\]\nThe same inclusion identity as before converts this to\n\\eqref{eq:joint-first-order}.","Finally, if $z\\in y_0+p^{s+1}R^e$ is any competing solved root with\nparameters $\\tau$, then $(z,\\tau)$ belongs to the full augmented ball.\nIndeed, $M>2s$ implies $M\\ge s+1$. Full-point uniqueness therefore gives\n$(z,\\tau)=b$. This comparison proves uniqueness on exactly the solved\nball stated in the theorem.\n\\end{proof}","The exact solved-coordinate statement is\n\\lean{Komorebi.PolynomialJointDeformation.exists_joint_deformation},\nwith its unique-existence interface\n\\lean{Komorebi.PolynomialJointDeformation.exists_unique_joint_deformation}.\nThe triangular inverse identities hold over arbitrary commutative rings,\nso they apply in particular over the $p$-adic integers.","\\begin{example}[First-order sign and error precision]\\label{ex:motion-boundary}\nFor $p=2$, $s=0$, $M=2$, $F(Y,T)=Y+T$, $H=1$, $a=0$, and $v=1$,\nthe prescribed parameter is $\\tau=4$ and the solved root is $y=-8$.\nHere $J=K=1$, and the forcing term is $H(a)+Kv=2$. Omitting the parameter\nterm would predict $-4$, and reversing its sign would predict $0$.\nThe corresponding errors $-4$ and $-8$ are both outside\n$2^{2M}\\Z_2=16\\Z_2$. Thus either change would violate the stated error\nbound.","mple}[First-order sign and error precision]\\label{ex:motion-boundary}\nFor $p=2$, $s=0$, $M=2$, $F(Y,T)=Y+T$, $H=1$, $a=0$, and $v=1$,\nthe prescribed parameter is $\\tau=4$ and the solved root is $y=-8$.\nHere $J=K=1$, and the forcing term is $H(a)+Kv=2$. Omitting the parameter\nterm would predict $-4$, and reversing its sign would predict $0$.\nThe corresponding errors $-4$ and $-8$ are both outside\n$2^{2M}\\Z_2=16\\Z_2$. Thus either change would violate the stated error\nbound.","The error exponent cannot uniformly be increased. For\n$F(Y,T)=Y+Y^2+T$, $H=0$, $a=0$, $p=2$, $s=0$, $M=1$, and $v=1$,\nthere is an even root $y$ of $y+y^2+2=0$. Its equation forces\n$y\\equiv2\\pmod4$ and $y+2=-y^2$, so\n\\[\n 4\\mid y+2,\\qquad 8\\nmid y+2.\n\\]\nThus the error depth $2M-3s=2$ is attained even for parameter-only motion.\nThe same quadratic with a constant coefficient perturbation gives the\ncoefficient-only boundary. These examples refute a uniform extra error\ndigit; they do not assert optimality for every individual choice of\n$p,s,M$.\n\\end{example}","ot $y$ of $y+y^2+2=0$. Its equation forces\n$y\\equiv2\\pmod4$ and $y+2=-y^2$, so\n\\[\n 4\\mid y+2,\\qquad 8\\nmid y+2.\n\\]\nThus the error depth $2M-3s=2$ is attained even for parameter-only motion.\nThe same quadratic with a constant coefficient perturbation gives the\ncoefficient-only boundary. These examples refute a uniform extra error\ndigit; they do not assert optimality for every individual choice of\n$p,s,M$.\n\\end{example}","\\section{Analytic regularity of the local solution map}\n\\label{sec:regularity}\nThe deformation estimates control how far an exact root moves under a small\nperturbation. We now show that its local dependence on the parameters is\nanalytic, and compute the derivative from the polynomial Jacobian. The\nanalytic argument takes place over a complete normed field. For integral\n$p$-adic data, we first include the coefficients and base point in $\\Qp$;\n$\\Zp$ itself is not a vector space over $\\Qp$.","l solution map}\n\\label{sec:regularity}\nThe deformation estimates control how far an exact root moves under a small\nperturbation. We now show that its local dependence on the parameters is\nanalytic, and compute the derivative from the polynomial Jacobian. The\nanalytic argument takes place over a complete normed field. For integral\n$p$-adic data, we first include the coefficients and base point in $\\Qp$;\n$\\Zp$ itself is not a vector space over $\\Qp$.","ion map}\n\\label{sec:regularity}\nThe deformation estimates control how far an exact root moves under a small\nperturbation. We now show that its local dependence on the parameters is\nanalytic, and compute the derivative from the polynomial Jacobian. The\nanalytic argument takes place over a complete normed field. For integral\n$p$-adic data, we first include the coefficients and base point in $\\Qp$;\n$\\Zp$ itself is not a vector space over $\\Qp$.","rmation estimates control how far an exact root moves under a small\nperturbation. We now show that its local dependence on the parameters is\nanalytic, and compute the derivative from the polynomial Jacobian. The\nanalytic argument takes place over a complete normed field. For integral\n$p$-adic data, we first include the coefficients and base point in $\\Qp$;\n$\\Zp$ itself is not a vector space over $\\Qp$.","Let $\\mathbb K$ be a complete nontrivially normed field. Let $e,k\\in\\N$, and\nwrite the variables as $(Y,T)\\in\\mathbb K^e\\times\\mathbb K^k$. For\n\\[\n F\\in\\mathbb K[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n \\qquad a=(y_0,t_0),\n\\]\ndefine the matrices of evaluated formal partial derivatives\n\\begin{equation}\n J=D_YF(a)\\in M_e(\\mathbb K),\\qquad\n K=D_TF(a)\\in M_{e\\times k}(\\mathbb K).\n \\label{eq:regularity-blocks}\n\\end{equation}\nRows index equations and columns index variables. We first establish that\nthese formal derivatives are the derivatives of the evaluation map.","\\begin{lemma}[Derivative of polynomial evaluation]\n\\label{lem:actual-polynomial-derivative}\nFor $P\\in\\mathbb K[X_1,\\ldots,X_n]$, polynomial evaluation is analytic at\nevery point. At $a\\in\\mathbb K^n$ its strict Fr\\'echet derivative is\n\\[\n v\\longmapsto\\sum_{j=1}^n (\\partial_jP)(a)v_j.\n\\]\nFor a finite system, its strict derivative is the continuous linear map\nrepresented by the matrix of evaluated formal partial derivatives.\n\\end{lemma}\n\\begin{proof}\nFor the derivative statement, use polynomial induction on constants,\naddition, and multiplication by a variable. Constants have zero derivative.\nThe sum rule agrees with the additivity of formal partial differentiation.\nFor $PX_j$, the product rule gives\n\\[\n d(PX_j)_a(v)=a_j\\,dP_a(v)+P(a)v_j,\n\\]\nwhich agrees with the evaluated formal product rule for every coordinate.\nTh","\\begin{lemma}[Derivative of polynomial evaluation]\n\\label{lem:actual-polynomial-derivative}\nFor $P\\in\\mathbb K[X_1,\\ldots,X_n]$, polynomial evaluation is analytic at\nevery point. At $a\\in\\mathbb K^n$ its strict Fr\\'echet derivative is\n\\[\n v\\longmapsto\\sum_{j=1}^n (\\partial_jP)(a)v_j.\n\\]\nFor a finite system, its strict derivative is the continuous linear map\nrepresented by the matrix of evaluated formal partial derivatives.\n\\end{lemma}\n\\begin{proof}\nFor the derivative statement, use polynomial induction on constants,\naddition, and multiplication by a variable. Constants have zero derivative.\nThe sum rule agrees with the additivity of formal partial differentiation.\nFor $PX_j$, the product rule gives\n\\[\n d(PX_j)_a(v)=a_j\\,dP_a(v)+P(a)v_j,\n\\]\nwhich agrees with the evaluated formal product rule for every coordinate.\nThe finite product rule for maps to coordinate spaces then assembles the\ns","\\begin{lemma}[Derivative of polynomial evaluation]\n\\label{lem:actual-polynomial-derivative}\nFor $P\\in\\mathbb K[X_1,\\ldots,X_n]$, polynomial evaluation is analytic at\nevery point. At $a\\in\\mathbb K^n$ its strict Fr\\'echet derivative is\n\\[\n v\\longmapsto\\sum_{j=1}^n (\\partial_jP)(a)v_j.\n\\]\nFor a finite system, its strict derivative is the continuous linear map\nrepresented by the matrix of evaluated formal partial derivatives.\n\\end{lemma}\n\\begin{proof}\nFor the derivative statement, use polynomial induction on constants,\naddition, and multiplication by a variable. Constants have zero derivative.\nThe sum rule agrees with the additivity of formal partial differentiation.\nFor $PX_j$, the product rule gives\n\\[\n d(PX_j)_a(v)=a_j\\,dP_a(v)+P(a)v_j,\n\\]\nwhich agrees with the evaluated formal product rule for every coordinate.\nThe finite product rule for maps to coordinate spaces then assembles the\nsystem derivative. Each matrix defines a continuous linear map because the\ncoordinate sp","a)v_j.\n\\]\nFor a finite system, its strict derivative is the continuous linear map\nrepresented by the matrix of evaluated formal partial derivatives.\n\\end{lemma}\n\\begin{proof}\nFor the derivative statement, use polynomial induction on constants,\naddition, and multiplication by a variable. Constants have zero derivative.\nThe sum rule agrees with the additivity of formal partial differentiation.\nFor $PX_j$, the product rule gives\n\\[\n d(PX_j)_a(v)=a_j\\,dP_a(v)+P(a)v_j,\n\\]\nwhich agrees with the evaluated formal product rule for every coordinate.\nThe finite product rule for maps to coordinate spaces then assembles the\nsystem derivative. Each matrix defines a continuous linear map because the\ncoordinate spaces are finite-dimensional over the complete field.\nPolynomial analyticity follows from the corresponding polynomial induction\nin Mathlib and its finite-coordinate assembly.\n\\end{proof}\nThe derivative bridge and analytic statements are\n\\lean{Komorebi.PolynomialAnalyticEvaluation.hasStrictFDerivAt_evalSystem},\n\\lean{Komorebi.PolynomialAnalyticEvaluation.analyticAt_evalSystem}, and\n\\lean{Kom…","ts strict derivative is the continuous linear map\nrepresented by the matrix of evaluated formal partial derivatives.\n\\end{lemma}\n\\begin{proof}\nFor the derivative statement, use polynomial induction on constants,\naddition, and multiplication by a variable. Constants have zero derivative.\nThe sum rule agrees with the additivity of formal partial differentiation.\nFor $PX_j$, the product rule gives\n\\[\n d(PX_j)_a(v)=a_j\\,dP_a(v)+P(a)v_j,\n\\]\nwhich agrees with the evaluated formal product rule for every coordinate.\nThe finite product rule for maps to coordinate spaces then assembles the\nsystem derivative. Each matrix defines a continuous linear map because the\ncoordinate spaces are finite-dimensional over the complete field.\nPolynomial analyticity follows from the corresponding polynomial induction\nin Mathlib and its finite-coordinate assembly.\n\\end{proof}\nThe derivative bridge and analytic statements are\n\\lean{Komorebi.PolynomialAnalyticEvaluation.hasStrictFDerivAt_evalSystem},\n\\lean{Komorebi.PolynomialAnalyticEvaluation.analyticAt_evalSystem}, and\n\\lean{Komorebi.PolynomialAnalyticEvaluati…","\\begin{lemma}[The augmented derivative]\n\\label{lem:augmented-analytic-inverse}\nIf $J\\in\\operatorname{GL}_e(\\mathbb K)$, the polynomial map\n\\[\n \\Phi(Y,T)=(F(Y,T),T)\n\\]\nhas invertible strict derivative at $a$, with\n\\begin{equation}\n D\\Phi(a)=\\begin{pmatrix}J&K","\\begin{lemma}[The augmented derivative]\n\\label{lem:augmented-analytic-inverse}\nIf $J\\in\\operatorname{GL}_e(\\mathbb K)$, the polynomial map\n\\[\n \\Phi(Y,T)=(F(Y,T),T)\n\\]\nhas invertible strict derivative at $a$, with\n\\begin{equation}\n D\\Phi(a)=\\begin{pmatrix}J&K\\\\0&I_k\\end{pmatrix},\\qquad\n D\\Phi(a)^{-1}=\\begin{pmatrix}J^{-1}&-J^{-1}K\\\\0&I_k\\end{pmatrix}.\n \\label{eq:analytic-block-inverse}\n\\end{equation}\nThe local inverse supplied by this derivative is analytic near $\\Phi(a)$.\n\\end{lemma}\n\\begin{proof}\nThe lower coordinates of $\\Phi$ are the parameter variables. Differentiating\nthem gives the blocks $0$ and $I_k$. Lemma~\\ref{lem:actual-polynomial-derivative}\nidentifies the remaining blocks with~\\eqref{eq:regularity-blocks}.\nMultiplication of the two matrices in~\\eqref{eq:analytic-block-inverse},\nin both orders, gives the identity. The resulting linear equivalence is\ncontinuous, as is its inverse. Apply the strict inverse-function theorem to\n$\\Phi$. The analytic inverse theorem applies to the same open partial\nhomeomorphism, because $\\Phi$ is analytic and its derivative is that","0&I_k\\end{pmatrix}.\n \\label{eq:analytic-block-inverse}\n\\end{equation}\nThe local inverse supplied by this derivative is analytic near $\\Phi(a)$.\n\\end{lemma}\n\\begin{proof}\nThe lower coordinates of $\\Phi$ are the parameter variables. Differentiating\nthem gives the blocks $0$ and $I_k$. Lemma~\\ref{lem:actual-polynomial-derivative}\nidentifies the remaining blocks with~\\eqref{eq:regularity-blocks}.\nMultiplication of the two matrices in~\\eqref{eq:analytic-block-inverse},\nin both orders, gives the identity. The resulting linear equivalence is\ncontinuous, as is its inverse. Apply the strict inverse-function theorem to\n$\\Phi$. The analytic inverse theorem applies to the same open partial\nhomeomorphism, because $\\Phi$ is analytic and its derivative is that\ncontinuous linear equivalence. It supplies a convergent local power-series\nrepresenta","0&I_k\\end{pmatrix}.\n \\label{eq:analytic-block-inverse}\n\\end{equation}\nThe local inverse supplied by this derivative is analytic near $\\Phi(a)$.\n\\end{lemma}\n\\begin{proof}\nThe lower coordinates of $\\Phi$ are the parameter variables. Differentiating\nthem gives the blocks $0$ and $I_k$. Lemma~\\ref{lem:actual-polynomial-derivative}\nidentifies the remaining blocks with~\\eqref{eq:regularity-blocks}.\nMultiplication of the two matrices in~\\eqref{eq:analytic-block-inverse},\nin both orders, gives the identity. The resulting linear equivalence is\ncontinuous, as is its inverse. Apply the strict inverse-function theorem to\n$\\Phi$. The analytic inverse theorem applies to the same open partial\nhomeomorphism, because $\\Phi$ is analytic and its derivative is that\ncontinuous linear equivalence. It supplies a convergent local power-series\nrepresentation of this inverse.\n\\end{proof}\nThe augmented Jacobian is","0&I_k\\end{pmatrix}.\n \\label{eq:analytic-block-inverse}\n\\end{equation}\nThe local inverse supplied by this derivative is analytic near $\\Phi(a)$.\n\\end{lemma}\n\\begin{proof}\nThe lower coordinates of $\\Phi$ are the parameter variables. Differentiating\nthem gives the blocks $0$ and $I_k$. Lemma~\\ref{lem:actual-polynomial-derivative}\nidentifies the remaining blocks with~\\eqref{eq:regularity-blocks}.\nMultiplication of the two matrices in~\\eqref{eq:analytic-block-inverse},\nin both orders, gives the identity. The resulting linear equivalence is\ncontinuous, as is its inverse. Apply the strict inverse-function theorem to\n$\\Phi$. The analytic inverse theorem applies to the same open partial\nhomeomorphism, because $\\Phi$ is analytic and its derivative is that\ncontinuous linear equivalence. It supplies a convergent local power-series\nrepresentation of this inverse.\n\\end{proof}\nThe augmented Jacobian is identified in\n\\lean{Komorebi.PolynomialJointSystem.jacobian_augmented}","0&I_k\\end{pmatrix}.\n \\label{eq:analytic-block-inverse}\n\\end{equation}\nThe local inverse supplied by this derivative is analytic near $\\Phi(a)$.\n\\end{lemma}\n\\begin{proof}\nThe lower coordinates of $\\Phi$ are the parameter variables. Differentiating\nthem gives the blocks $0$ and $I_k$. Lemma~\\ref{lem:actual-polynomial-derivative}\nidentifies the remaining blocks with~\\eqref{eq:regularity-blocks}.\nMultiplication of the two matrices in~\\eqref{eq:analytic-block-inverse},\nin both orders, gives the identity. The resulting linear equivalence is\ncontinuous, as is its inverse. Apply the strict inverse-function theorem to\n$\\Phi$. The analytic inverse theorem applies to the same open partial\nhomeomorphism, because $\\Phi$ is analytic and its derivative is that\ncontinuous linear equivalence. It supplies a convergent local power-series\nrepresentation of this inverse.\n\\end{proof}\nThe augmented Jacobian is identified in\n\\lean{Komorebi.PolynomialJointSystem.jacobian_augmented}.\nThe cont","r variables. Differentiating\nthem gives the blocks $0$ and $I_k$. Lemma~\\ref{lem:actual-polynomial-derivative}\nidentifies the remaining blocks with~\\eqref{eq:regularity-blocks}.\nMultiplication of the two matrices in~\\eqref{eq:analytic-block-inverse},\nin both orders, gives the identity. The resulting linear equivalence is\ncontinuous, as is its inverse. Apply the strict inverse-function theorem to\n$\\Phi$. The analytic inverse theorem applies to the same open partial\nhomeomorphism, because $\\Phi$ is analytic and its derivative is that\ncontinuous linear equivalence. It supplies a convergent local power-series\nrepresentation of this inverse.\n\\end{proof}\nThe augmented Jacobian is identified in\n\\lean{Komorebi.PolynomialJointSystem.jacobian_augmented}.\nThe continuous inverse block and its restriction are\n\\lean{Komorebi.PolynomialAnalyticBlocks.upperEquiv} and\n\\lean{Komorebi.PolynomialAnalyticBlocks.inverse_solved_zero}.\nAnalyticity of the same constructed inverse is\n\\lean{Komorebi.PolynomialAnalyticInverse.analyticAt_inverse}.","omial-derivative}\nidentifies the remaining blocks with~\\eqref{eq:regularity-blocks}.\nMultiplication of the two matrices in~\\eqref{eq:analytic-block-inverse},\nin both orders, gives the identity. The resulting linear equivalence is\ncontinuous, as is its inverse. Apply the strict inverse-function theorem to\n$\\Phi$. The analytic inverse theorem applies to the same open partial\nhomeomorphism, because $\\Phi$ is analytic and its derivative is that\ncontinuous linear equivalence. It supplies a convergent local power-series\nrepresentation of this inverse.\n\\end{proof}\nThe augmented Jacobian is identified in\n\\lean{Komorebi.PolynomialJointSystem.jacobian_augmented}.\nThe continuous inverse block and its restriction are\n\\lean{Komorebi.PolynomialAnalyticBlocks.upperEquiv} and\n\\lean{Komorebi.PolynomialAnalyticBlocks.inverse_solved_zero}.\nAnalyticity of the same constructed inverse is\n\\lean{Komorebi.PolynomialAnalyticInverse.analyticAt_inverse}.","\\begin{theorem}[Analytic local root graph]\n\\label{thm:analytic-local-graph}\nSuppose $F(a)=0$ and $J\\in\\operatorname{GL}_e(\\mathbb K)$. There are open\nneighborhoods $t_0\\in U\\subseteq\\mathbb K^k$ and\n$y_0\\in V\\subseteq\\mathbb K^e$, and a map\n$\\psi:\\mathbb K^k\\to\\mathbb K^e$, such that\n\\begin{gather}\n \\psi(t_0)=y_0,\\qquad \\psi(U)\\subseteq V,\\qquad\n \\psi\\text{ is analytic at every point of }U,\\label{eq:analytic-psi}","\\begin{theorem}[Analytic local root graph]\n\\label{thm:analytic-local-graph}\nSuppose $F(a)=0$ and $J\\in\\operatorname{GL}_e(\\mathbb K)$. There are open\nneighborhoods $t_0\\in U\\subseteq\\mathbb K^k$ and\n$y_0\\in V\\subseteq\\mathbb K^e$, and a map\n$\\psi:\\mathbb K^k\\to\\mathbb K^e$, such that\n\\begin{gather}\n \\psi(t_0)=y_0,\\qquad \\psi(U)\\subseteq V,\\qquad\n \\psi\\text{ is analytic at every point of }U,\\label{eq:analytic-psi}\\\\\n F(y,t)=0\\quad\\Longleftrightarrow\\quad y=\\psi(t)\n \\qquad(t\\in U,\\ y\\in V).\\label{eq:analytic-root-graph}\n\\end{gather}\nAt the base parameter, the strict Fr\\'echet derivative is\n\\begin{equation}\n \\boxed{D\\psi(t_0)=-J^{-1}K.}\n \\label{eq:implicit-derivative}\n\\end{equation}\nIn particular, $\\psi$ is differentiable on $U$.\n\\end{theorem}\n\\begin{proof}\nLet $g$ be the analytic local inverse of $\\Phi$ supplied by\nLemma~\\ref{lem:augmented-analytic-inverse}. Since $F(a)=0$,\n$\\Phi(a)=(0,t_0)$ and $g(0,t_0)=a$. Define\n\\[\n \\psi(t)=\\pi_Yg(0,t),\n\\]\nwhere $\\pi_Y$ projects to the solved coordinates. This gives the base value\nand analyticity at $t_0$. The chain rule and\n\\eqref{eq:analytic-bloc…","F(y,t)=0\\quad\\Longleftrightarrow\\quad y=\\psi(t)\n \\qquad(t\\in U,\\ y\\in V).\\label{eq:analytic-root-graph}\n\\end{gather}\nAt the base parameter, the strict Fr\\'echet derivative is\n\\begin{equation}\n \\boxed{D\\psi(t_0)=-J^{-1}K.}\n \\label{eq:implicit-derivative}\n\\end{equation}\nIn particular, $\\psi$ is differentiable on $U$.\n\\end{theorem}\n\\begin{proof}\nLet $g$ be the analytic local inverse of $\\Phi$ supplied by\nLemma~\\ref{lem:augmented-analytic-inverse}. Since $F(a)=0$,\n$\\Phi(a)=(0,t_0)$ and $g(0,t_0)=a$. Define\n\\[\n \\psi(t)=\\pi_Yg(0,t),\n\\]\nwhere $\\pi_Y$ projects to the solved coordinates. This gives the base value\nand analyticity at $t_0$. The chain rule and\n\\eqref{eq:analytic-block-inverse} give\n\\[\n D\\psi(t_0)v\n =\\pi_Y\\begin{pmatrix}J^{-1}&-J^{-1}K","t\\in U,\\ y\\in V).\\label{eq:analytic-root-graph}\n\\end{gather}\nAt the base parameter, the strict Fr\\'echet derivative is\n\\begin{equation}\n \\boxed{D\\psi(t_0)=-J^{-1}K.}\n \\label{eq:implicit-derivative}\n\\end{equation}\nIn particular, $\\psi$ is differentiable on $U$.\n\\end{theorem}\n\\begin{proof}\nLet $g$ be the analytic local inverse of $\\Phi$ supplied by\nLemma~\\ref{lem:augmented-analytic-inverse}. Since $F(a)=0$,\n$\\Phi(a)=(0,t_0)$ and $g(0,t_0)=a$. Define\n\\[\n \\psi(t)=\\pi_Yg(0,t),\n\\]\nwhere $\\pi_Y$ projects to the solved coordinates. This gives the base value\nand analyticity at $t_0$. The chain rule and\n\\eqref{eq:analytic-block-inverse} give\n\\[\n D\\psi(t_0)v\n =\\pi_Y\\begin{pmatrix}J^{-1}&-J^{-1}K","\\in V).\\label{eq:analytic-root-graph}\n\\end{gather}\nAt the base parameter, the strict Fr\\'echet derivative is\n\\begin{equation}\n \\boxed{D\\psi(t_0)=-J^{-1}K.}\n \\label{eq:implicit-derivative}\n\\end{equation}\nIn particular, $\\psi$ is differentiable on $U$.\n\\end{theorem}\n\\begin{proof}\nLet $g$ be the analytic local inverse of $\\Phi$ supplied by\nLemma~\\ref{lem:augmented-analytic-inverse}. Since $F(a)=0$,\n$\\Phi(a)=(0,t_0)$ and $g(0,t_0)=a$. Define\n\\[\n \\psi(t)=\\pi_Yg(0,t),\n\\]\nwhere $\\pi_Y$ projects to the solved coordinates. This gives the base value\nand analyticity at $t_0$. The chain rule and\n\\eqref{eq:analytic-block-inverse} give\n\\[\n D\\psi(t_0)v\n =\\pi_Y\\begin{pmatrix}J^{-1}&-J^{-1}K","tic-root-graph}\n\\end{gather}\nAt the base parameter, the strict Fr\\'echet derivative is\n\\begin{equation}\n \\boxed{D\\psi(t_0)=-J^{-1}K.}\n \\label{eq:implicit-derivative}\n\\end{equation}\nIn particular, $\\psi$ is differentiable on $U$.\n\\end{theorem}\n\\begin{proof}\nLet $g$ be the analytic local inverse of $\\Phi$ supplied by\nLemma~\\ref{lem:augmented-analytic-inverse}. Since $F(a)=0$,\n$\\Phi(a)=(0,t_0)$ and $g(0,t_0)=a$. Define\n\\[\n \\psi(t)=\\pi_Yg(0,t),\n\\]\nwhere $\\pi_Y$ projects to the solved coordinates. This gives the base value\nand analyticity at $t_0$. The chain rule and\n\\eqref{eq:analytic-block-inverse} give\n\\[\n D\\psi(t_0)v\n =\\pi_Y\\begin{pmatrix}J^{-1}&-J^{-1}K","ather}\nAt the base parameter, the strict Fr\\'echet derivative is\n\\begin{equation}\n \\boxed{D\\psi(t_0)=-J^{-1}K.}\n \\label{eq:implicit-derivative}\n\\end{equation}\nIn particular, $\\psi$ is differentiable on $U$.\n\\end{theorem}\n\\begin{proof}\nLet $g$ be the analytic local inverse of $\\Phi$ supplied by\nLemma~\\ref{lem:augmented-analytic-inverse}. Since $F(a)=0$,\n$\\Phi(a)=(0,t_0)$ and $g(0,t_0)=a$. Define\n\\[\n \\psi(t)=\\pi_Yg(0,t),\n\\]\nwhere $\\pi_Y$ projects to the solved coordinates. This gives the base value\nand analyticity at $t_0$. The chain rule and\n\\eqref{eq:analytic-block-inverse} give\n\\[\n D\\psi(t_0)v\n =\\pi_Y\\begin{pmatrix}J^{-1}&-J^{-1}K","strict Fr\\'echet derivative is\n\\begin{equation}\n \\boxed{D\\psi(t_0)=-J^{-1}K.}\n \\label{eq:implicit-derivative}\n\\end{equation}\nIn particular, $\\psi$ is differentiable on $U$.\n\\end{theorem}\n\\begin{proof}\nLet $g$ be the analytic local inverse of $\\Phi$ supplied by\nLemma~\\ref{lem:augmented-analytic-inverse}. Since $F(a)=0$,\n$\\Phi(a)=(0,t_0)$ and $g(0,t_0)=a$. Define\n\\[\n \\psi(t)=\\pi_Yg(0,t),\n\\]\nwhere $\\pi_Y$ projects to the solved coordinates. This gives the base value\nand analyticity at $t_0$. The chain rule and\n\\eqref{eq:analytic-block-inverse} give\n\\[\n D\\psi(t_0)v\n =\\pi_Y\\begin{pmatrix}J^{-1}&-J^{-1}K",":implicit-derivative}\n\\end{equation}\nIn particular, $\\psi$ is differentiable on $U$.\n\\end{theorem}\n\\begin{proof}\nLet $g$ be the analytic local inverse of $\\Phi$ supplied by\nLemma~\\ref{lem:augmented-analytic-inverse}. Since $F(a)=0$,\n$\\Phi(a)=(0,t_0)$ and $g(0,t_0)=a$. Define\n\\[\n \\psi(t)=\\pi_Yg(0,t),\n\\]\nwhere $\\pi_Y$ projects to the solved coordinates. This gives the base value\nand analyticity at $t_0$. The chain rule and\n\\eqref{eq:analytic-block-inverse} give\n\\[\n D\\psi(t_0)v\n =\\pi_Y\\begin{pmatrix}J^{-1}&-J^{-1}K","erentiable on $U$.\n\\end{theorem}\n\\begin{proof}\nLet $g$ be the analytic local inverse of $\\Phi$ supplied by\nLemma~\\ref{lem:augmented-analytic-inverse}. Since $F(a)=0$,\n$\\Phi(a)=(0,t_0)$ and $g(0,t_0)=a$. Define\n\\[\n \\psi(t)=\\pi_Yg(0,t),\n\\]\nwhere $\\pi_Y$ projects to the solved coordinates. This gives the base value\nand analyticity at $t_0$. The chain rule and\n\\eqref{eq:analytic-block-inverse} give\n\\[\n D\\psi(t_0)v\n =\\pi_Y\\begin{pmatrix}J^{-1}&-J^{-1}K\\\\0&I_k\\end{pmatrix}\n       \\binom{0}{v}\n =-J^{-1}Kv.\n\\]","To obtain the root equivalence on a product neighborhood, we choose the\nneighborhoods as follows. The source of the open partial homeomorphism defined by\n$\\Phi$ is open around $a$; its inverse is $g$. Choose open sets $V$ and $U_0$\ncontaining $y_0$ and $t_0$ such that\n$V\\times U_0$ lies in that source. Restrict $U_0$ further by the preimage of\nan open analytic inverse neighborhood under $t\\mapsto(0,t)$, the open locus\nwhere this same $\\psi$ is analytic, and the interior of $\\psi^{-1}(V)$.\nAll of these sets contain $t_0$. For the last set, use the continuity of\n$\\psi$ at $t_0$ and $\\psi(t_0)=y_0\\in V$. Their intersection is an open\nneighborhood $U$ with $\\psi(U)\\subseteq V$.","duct neighborhood, we choose the\nneighborhoods as follows. The source of the open partial homeomorphism defined by\n$\\Phi$ is open around $a$; its inverse is $g$. Choose open sets $V$ and $U_0$\ncontaining $y_0$ and $t_0$ such that\n$V\\times U_0$ lies in that source. Restrict $U_0$ further by the preimage of\nan open analytic inverse neighborhood under $t\\mapsto(0,t)$, the open locus\nwhere this same $\\psi$ is analytic, and the interior of $\\psi^{-1}(V)$.\nAll of these sets contain $t_0$. For the last set, use the continuity of\n$\\psi$ at $t_0$ and $\\psi(t_0)=y_0\\in V$. Their intersection is an open\nneighborhood $U$ with $\\psi(U)\\subseteq V$.","rhoods as follows. The source of the open partial homeomorphism defined by\n$\\Phi$ is open around $a$; its inverse is $g$. Choose open sets $V$ and $U_0$\ncontaining $y_0$ and $t_0$ such that\n$V\\times U_0$ lies in that source. Restrict $U_0$ further by the preimage of\nan open analytic inverse neighborhood under $t\\mapsto(0,t)$, the open locus\nwhere this same $\\psi$ is analytic, and the interior of $\\psi^{-1}(V)$.\nAll of these sets contain $t_0$. For the last set, use the continuity of\n$\\psi$ at $t_0$ and $\\psi(t_0)=y_0\\in V$. Their intersection is an open\nneighborhood $U$ with $\\psi(U)\\subseteq V$.","partial homeomorphism defined by\n$\\Phi$ is open around $a$; its inverse is $g$. Choose open sets $V$ and $U_0$\ncontaining $y_0$ and $t_0$ such that\n$V\\times U_0$ lies in that source. Restrict $U_0$ further by the preimage of\nan open analytic inverse neighborhood under $t\\mapsto(0,t)$, the open locus\nwhere this same $\\psi$ is analytic, and the interior of $\\psi^{-1}(V)$.\nAll of these sets contain $t_0$. For the last set, use the continuity of\n$\\psi$ at $t_0$ and $\\psi(t_0)=y_0\\in V$. Their intersection is an open\nneighborhood $U$ with $\\psi(U)\\subseteq V$.","around $a$; its inverse is $g$. Choose open sets $V$ and $U_0$\ncontaining $y_0$ and $t_0$ such that\n$V\\times U_0$ lies in that source. Restrict $U_0$ further by the preimage of\nan open analytic inverse neighborhood under $t\\mapsto(0,t)$, the open locus\nwhere this same $\\psi$ is analytic, and the interior of $\\psi^{-1}(V)$.\nAll of these sets contain $t_0$. For the last set, use the continuity of\n$\\psi$ at $t_0$ and $\\psi(t_0)=y_0\\in V$. Their intersection is an open\nneighborhood $U$ with $\\psi(U)\\subseteq V$.","$; its inverse is $g$. Choose open sets $V$ and $U_0$\ncontaining $y_0$ and $t_0$ such that\n$V\\times U_0$ lies in that source. Restrict $U_0$ further by the preimage of\nan open analytic inverse neighborhood under $t\\mapsto(0,t)$, the open locus\nwhere this same $\\psi$ is analytic, and the interior of $\\psi^{-1}(V)$.\nAll of these sets contain $t_0$. For the last set, use the continuity of\n$\\psi$ at $t_0$ and $\\psi(t_0)=y_0\\in V$. Their intersection is an open\nneighborhood $U$ with $\\psi(U)\\subseteq V$.","erse is $g$. Choose open sets $V$ and $U_0$\ncontaining $y_0$ and $t_0$ such that\n$V\\times U_0$ lies in that source. Restrict $U_0$ further by the preimage of\nan open analytic inverse neighborhood under $t\\mapsto(0,t)$, the open locus\nwhere this same $\\psi$ is analytic, and the interior of $\\psi^{-1}(V)$.\nAll of these sets contain $t_0$. For the last set, use the continuity of\n$\\psi$ at $t_0$ and $\\psi(t_0)=y_0\\in V$. Their intersection is an open\nneighborhood $U$ with $\\psi(U)\\subseteq V$.","$t_0$ such that\n$V\\times U_0$ lies in that source. Restrict $U_0$ further by the preimage of\nan open analytic inverse neighborhood under $t\\mapsto(0,t)$, the open locus\nwhere this same $\\psi$ is analytic, and the interior of $\\psi^{-1}(V)$.\nAll of these sets contain $t_0$. For the last set, use the continuity of\n$\\psi$ at $t_0$ and $\\psi(t_0)=y_0\\in V$. Their intersection is an open\nneighborhood $U$ with $\\psi(U)\\subseteq V$.","h that\n$V\\times U_0$ lies in that source. Restrict $U_0$ further by the preimage of\nan open analytic inverse neighborhood under $t\\mapsto(0,t)$, the open locus\nwhere this same $\\psi$ is analytic, and the interior of $\\psi^{-1}(V)$.\nAll of these sets contain $t_0$. For the last set, use the continuity of\n$\\psi$ at $t_0$ and $\\psi(t_0)=y_0\\in V$. Their intersection is an open\nneighborhood $U$ with $\\psi(U)\\subseteq V$.","If $t\\in U$, $y\\in V$ and $F(y,t)=0$, then\n$\\Phi(y,t)=(0,t)$. The left inverse identity forces\n$g(0,t)=(y,t)$, hence $y=\\psi(t)$. Conversely, the right inverse identity\n$\\Phi(g(0,t))=(0,t)$ says both that its solved equations vanish and that its\nparameter coordinates are exactly $t$. Thus\n$g(0,t)=(\\psi(t),t)$ and $F(\\psi(t),t)=0$.\nAnalyticity at every point of $U$ implies differentiability there.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialAnalyticImplicit.exists_analytic_implicit}.\nDifferentiability is also included explicitly in the statement contract\n\\lean{Komorebi.PolynomialRegularityContracts.actual_padic_analytic_root_graph}.","\\begin{theorem}[Regularity of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.Pol","\\begin{theorem}[Regularity of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelectio","\\begin{theorem}[Regularity of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_select","\\begin{theorem}[Regularity of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts fo","\\begin{theorem}[Regularity of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation deri","\\begin{theorem}[Regularity of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equ","\\begin{theorem}[Regularity of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near","\\begin{theorem}[Regularity of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near the base\np","\\begin{theorem}[Regularity of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near the base\nparameter and the specified base v","\\begin{theorem}[Regularity of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near the base\nparameter and the specified base value.","ty of a supplied root selection]\n\\label{thm:selection-regularity}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near the base\nparameter and the specified base value.","ty}\nUnder the hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near the base\nparameter and the specified base value.","hypotheses of Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near the base\nparameter and the specified base value.","Theorem~\\ref{thm:analytic-local-graph}, let\n$h:\\mathbb K^k\\to\\mathbb K^e$ be continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near the base\nparameter and the specified base value.","e continuous at $t_0$, with\n$h(t_0)=y_0$. Suppose $F(h(t),t)=0$ for every $t$ in some neighborhood of\n$t_0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near the base\nparameter and the specified base value.","0$. Then $h$ and $\\psi$ agree in a neighborhood of $t_0$. Consequently\n$h$ is analytic there as a germ and has strict derivative\n$Dh(t_0)=-J^{-1}K$.\n\\end{theorem}\n\\begin{proof}\nContinuity at $t_0$ places $h(t)$ inside $V$ for all sufficiently close\nparameters. Intersect this neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near the base\nparameter and the specified base value.","neighborhood with $U$ and the neighborhood on\nwhich the root equation holds. Equation~\\eqref{eq:analytic-root-graph} gives\n$h(t)=\\psi(t)$ throughout the intersection. Analyticity and strict\ndifferentiability at a point are unchanged by agreement on a neighborhood.\n\\end{proof}\nThe formal result is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_regularity}.\nIts formulation derives $F(a)=0$ from the root equation near the base\nparameter and the specified base value.","\\begin{corollary}[Regularity throughout an open parameter set]\n\\label{cor:selection-open-regularity}\nLet $U\\subseteq\\mathbb K^k$ be open, and let $h$ be continuous on $U$.\nSuppose that for every $t\\in U$,\n\\[\n F(h(t),t)=0,\n \\qquad J_t=D_YF(h(t),t)\\in\\operatorname{GL}_e(\\mathbb K).\n\\]\nThen $h$ is analytic at every point of $U$ and its strict derivative is\n\\[\n Dh(t)=-J_t^{-1}D_TF(h(t),t)\\qquad(t\\in U).\n\\]\n\\end{corollary}\n\\begin{proof}\nAt each $t\\in U$, openness turns continuity on $U$ into continuity at $t$\nand makes the root equation hold on a neighborhood of $t$. Apply\nTheorem~\\ref{thm:selection-regularity} at $(h(t),t)$.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_analyticOnNhd}.\nThe pointwise invertibility assumption is part of this corollary.","en parameter set]\n\\label{cor:selection-open-regularity}\nLet $U\\subseteq\\mathbb K^k$ be open, and let $h$ be continuous on $U$.\nSuppose that for every $t\\in U$,\n\\[\n F(h(t),t)=0,\n \\qquad J_t=D_YF(h(t),t)\\in\\operatorname{GL}_e(\\mathbb K).\n\\]\nThen $h$ is analytic at every point of $U$ and its strict derivative is\n\\[\n Dh(t)=-J_t^{-1}D_TF(h(t),t)\\qquad(t\\in U).\n\\]\n\\end{corollary}\n\\begin{proof}\nAt each $t\\in U$, openness turns continuity on $U$ into continuity at $t$\nand makes the root equation hold on a neighborhood of $t$. Apply\nTheorem~\\ref{thm:selection-regularity} at $(h(t),t)$.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_analyticOnNhd}.\nThe pointwise invertibility assumption is part of this corollary.","open-regularity}\nLet $U\\subseteq\\mathbb K^k$ be open, and let $h$ be continuous on $U$.\nSuppose that for every $t\\in U$,\n\\[\n F(h(t),t)=0,\n \\qquad J_t=D_YF(h(t),t)\\in\\operatorname{GL}_e(\\mathbb K).\n\\]\nThen $h$ is analytic at every point of $U$ and its strict derivative is\n\\[\n Dh(t)=-J_t^{-1}D_TF(h(t),t)\\qquad(t\\in U).\n\\]\n\\end{corollary}\n\\begin{proof}\nAt each $t\\in U$, openness turns continuity on $U$ into continuity at $t$\nand makes the root equation hold on a neighborhood of $t$. Apply\nTheorem~\\ref{thm:selection-regularity} at $(h(t),t)$.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_analyticOnNhd}.\nThe pointwise invertibility assumption is part of this corollary.","\\subseteq\\mathbb K^k$ be open, and let $h$ be continuous on $U$.\nSuppose that for every $t\\in U$,\n\\[\n F(h(t),t)=0,\n \\qquad J_t=D_YF(h(t),t)\\in\\operatorname{GL}_e(\\mathbb K).\n\\]\nThen $h$ is analytic at every point of $U$ and its strict derivative is\n\\[\n Dh(t)=-J_t^{-1}D_TF(h(t),t)\\qquad(t\\in U).\n\\]\n\\end{corollary}\n\\begin{proof}\nAt each $t\\in U$, openness turns continuity on $U$ into continuity at $t$\nand makes the root equation hold on a neighborhood of $t$. Apply\nTheorem~\\ref{thm:selection-regularity} at $(h(t),t)$.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_analyticOnNhd}.\nThe pointwise invertibility assumption is part of this corollary.","n $U$.\nSuppose that for every $t\\in U$,\n\\[\n F(h(t),t)=0,\n \\qquad J_t=D_YF(h(t),t)\\in\\operatorname{GL}_e(\\mathbb K).\n\\]\nThen $h$ is analytic at every point of $U$ and its strict derivative is\n\\[\n Dh(t)=-J_t^{-1}D_TF(h(t),t)\\qquad(t\\in U).\n\\]\n\\end{corollary}\n\\begin{proof}\nAt each $t\\in U$, openness turns continuity on $U$ into continuity at $t$\nand makes the root equation hold on a neighborhood of $t$. Apply\nTheorem~\\ref{thm:selection-regularity} at $(h(t),t)$.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_analyticOnNhd}.\nThe pointwise invertibility assumption is part of this corollary.","=0,\n \\qquad J_t=D_YF(h(t),t)\\in\\operatorname{GL}_e(\\mathbb K).\n\\]\nThen $h$ is analytic at every point of $U$ and its strict derivative is\n\\[\n Dh(t)=-J_t^{-1}D_TF(h(t),t)\\qquad(t\\in U).\n\\]\n\\end{corollary}\n\\begin{proof}\nAt each $t\\in U$, openness turns continuity on $U$ into continuity at $t$\nand makes the root equation hold on a neighborhood of $t$. Apply\nTheorem~\\ref{thm:selection-regularity} at $(h(t),t)$.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialAnalyticSelection.continuous_selection_analyticOnNhd}.\nThe pointwise invertibility assumption is part of this corollary.","\\begin{corollary}[Integral coefficients and centers]\n\\label{cor:integral-analytic-graph}\nLet $p$ be any prime. Suppose $F\\in\\Zp[Y,T]^e$, $a\\in\\Zp^{e+k}$,\nand $F(a)=0$. Let $\\iota:\\Zp\\hookrightarrow\\Qp$ be the inclusion.\nIf $\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\Qp)$, the polynomial system\nobtained by including the original coefficients in $\\Qp$ has the analytic\nlocal root graph of Theorem~\\ref{thm:analytic-local-graph} at $\\iota(a)$.\nThe strict derivative at the base is\n\\[\n -\\iota(D_YF(a))^{-1}\\iota(D_TF(a)).\n\\]\nThe open neighborhoods and the output map are in the ambient spaces over\n$\\Qp$.\n\\end{corollary}\n\\begin{proof}\nCoefficient inclusion commutes with evaluation and formal partial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of th","\\begin{corollary}[Integral coefficients and centers]\n\\label{cor:integral-analytic-graph}\nLet $p$ be any prime. Suppose $F\\in\\Zp[Y,T]^e$, $a\\in\\Zp^{e+k}$,\nand $F(a)=0$. Let $\\iota:\\Zp\\hookrightarrow\\Qp$ be the inclusion.\nIf $\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\Qp)$, the polynomial system\nobtained by including the original coefficients in $\\Qp$ has the analytic\nlocal root graph of Theorem~\\ref{thm:analytic-local-graph} at $\\iota(a)$.\nThe strict derivative at the base is\n\\[\n -\\iota(D_YF(a))^{-1}\\iota(D_TF(a)).\n\\]\nThe open neighborhoods and the output map are in the ambient spaces over\n$\\Qp$.\n\\end{corollary}\n\\begin{proof}\nCoefficient inclusion commutes with evaluation and formal partial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of the original blocks, and the original exa","\\begin{corollary}[Integral coefficients and centers]\n\\label{cor:integral-analytic-graph}\nLet $p$ be any prime. Suppose $F\\in\\Zp[Y,T]^e$, $a\\in\\Zp^{e+k}$,\nand $F(a)=0$. Let $\\iota:\\Zp\\hookrightarrow\\Qp$ be the inclusion.\nIf $\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\Qp)$, the polynomial system\nobtained by including the original coefficients in $\\Qp$ has the analytic\nlocal root graph of Theorem~\\ref{thm:analytic-local-graph} at $\\iota(a)$.\nThe strict derivative at the base is\n\\[\n -\\iota(D_YF(a))^{-1}\\iota(D_TF(a)).\n\\]\nThe open neighborhoods and the output map are in the ambient spaces over\n$\\Qp$.\n\\end{corollary}\n\\begin{proof}\nCoefficient inclusion commutes with evaluation and formal partial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of the original blocks, and the original exact root\nremains a","\\begin{corollary}[Integral coefficients and centers]\n\\label{cor:integral-analytic-graph}\nLet $p$ be any prime. Suppose $F\\in\\Zp[Y,T]^e$, $a\\in\\Zp^{e+k}$,\nand $F(a)=0$. Let $\\iota:\\Zp\\hookrightarrow\\Qp$ be the inclusion.\nIf $\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\Qp)$, the polynomial system\nobtained by including the original coefficients in $\\Qp$ has the analytic\nlocal root graph of Theorem~\\ref{thm:analytic-local-graph} at $\\iota(a)$.\nThe strict derivative at the base is\n\\[\n -\\iota(D_YF(a))^{-1}\\iota(D_TF(a)).\n\\]\nThe open neighborhoods and the output map are in the ambient spaces over\n$\\Qp$.\n\\end{corollary}\n\\begin{proof}\nCoefficient inclusion commutes with evaluation and formal partial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of the original blocks, and the original exact root\nremains an exact root.","\\begin{corollary}[Integral coefficients and centers]\n\\label{cor:integral-analytic-graph}\nLet $p$ be any prime. Suppose $F\\in\\Zp[Y,T]^e$, $a\\in\\Zp^{e+k}$,\nand $F(a)=0$. Let $\\iota:\\Zp\\hookrightarrow\\Qp$ be the inclusion.\nIf $\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\Qp)$, the polynomial system\nobtained by including the original coefficients in $\\Qp$ has the analytic\nlocal root graph of Theorem~\\ref{thm:analytic-local-graph} at $\\iota(a)$.\nThe strict derivative at the base is\n\\[\n -\\iota(D_YF(a))^{-1}\\iota(D_TF(a)).\n\\]\nThe open neighborhoods and the output map are in the ambient spaces over\n$\\Qp$.\n\\end{corollary}\n\\begin{proof}\nCoefficient inclusion commutes with evaluation and formal partial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of the original blocks, and the original exact root\nremains an exact root. Apply Theorem~\\ref{thm:analytic-loc","\\begin{corollary}[Integral coefficients and centers]\n\\label{cor:integral-analytic-graph}\nLet $p$ be any prime. Suppose $F\\in\\Zp[Y,T]^e$, $a\\in\\Zp^{e+k}$,\nand $F(a)=0$. Let $\\iota:\\Zp\\hookrightarrow\\Qp$ be the inclusion.\nIf $\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\Qp)$, the polynomial system\nobtained by including the original coefficients in $\\Qp$ has the analytic\nlocal root graph of Theorem~\\ref{thm:analytic-local-graph} at $\\iota(a)$.\nThe strict derivative at the base is\n\\[\n -\\iota(D_YF(a))^{-1}\\iota(D_TF(a)).\n\\]\nThe open neighborhoods and the output map are in the ambient spaces over\n$\\Qp$.\n\\end{corollary}\n\\begin{proof}\nCoefficient inclusion commutes with evaluation and formal partial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of the original blocks, and the original exact root\nremains an exact root. Apply Theorem~\\ref{thm:analytic-local-graph} to the system with included\ncoefficients.\n\\end{proof}\nT","\\begin{corollary}[Integral coefficients and centers]\n\\label{cor:integral-analytic-graph}\nLet $p$ be any prime. Suppose $F\\in\\Zp[Y,T]^e$, $a\\in\\Zp^{e+k}$,\nand $F(a)=0$. Let $\\iota:\\Zp\\hookrightarrow\\Qp$ be the inclusion.\nIf $\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\Qp)$, the polynomial system\nobtained by including the original coefficients in $\\Qp$ has the analytic\nlocal root graph of Theorem~\\ref{thm:analytic-local-graph} at $\\iota(a)$.\nThe strict derivative at the base is\n\\[\n -\\iota(D_YF(a))^{-1}\\iota(D_TF(a)).\n\\]\nThe open neighborhoods and the output map are in the ambient spaces over\n$\\Qp$.\n\\end{corollary}\n\\begin{proof}\nCoefficient inclusion commutes with evaluation and formal partial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of the original blocks, and the original exact root\nremains an exact root. Apply Theorem~\\ref{thm:analytic-local-graph} to the system with included\ncoefficients.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialRegularityIntegral.exists_analytic_integral_impl","coefficients and centers]\n\\label{cor:integral-analytic-graph}\nLet $p$ be any prime. Suppose $F\\in\\Zp[Y,T]^e$, $a\\in\\Zp^{e+k}$,\nand $F(a)=0$. Let $\\iota:\\Zp\\hookrightarrow\\Qp$ be the inclusion.\nIf $\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\Qp)$, the polynomial system\nobtained by including the original coefficients in $\\Qp$ has the analytic\nlocal root graph of Theorem~\\ref{thm:analytic-local-graph} at $\\iota(a)$.\nThe strict derivative at the base is\n\\[\n -\\iota(D_YF(a))^{-1}\\iota(D_TF(a)).\n\\]\nThe open neighborhoods and the output map are in the ambient spaces over\n$\\Qp$.\n\\end{corollary}\n\\begin{proof}\nCoefficient inclusion commutes with evaluation and formal partial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of the original blocks, and the original exact root\nremains an exact root. Apply Theorem~\\ref{thm:analytic-local-graph} to the system with included\ncoefficients.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialRegularityIntegral.exists_analytic_integral_implicit}.\nIt requires no invertibility modulo $p$. In particular, it includes solved\n…","ytic-graph}\nLet $p$ be any prime. Suppose $F\\in\\Zp[Y,T]^e$, $a\\in\\Zp^{e+k}$,\nand $F(a)=0$. Let $\\iota:\\Zp\\hookrightarrow\\Qp$ be the inclusion.\nIf $\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\Qp)$, the polynomial system\nobtained by including the original coefficients in $\\Qp$ has the analytic\nlocal root graph of Theorem~\\ref{thm:analytic-local-graph} at $\\iota(a)$.\nThe strict derivative at the base is\n\\[\n -\\iota(D_YF(a))^{-1}\\iota(D_TF(a)).\n\\]\nThe open neighborhoods and the output map are in the ambient spaces over\n$\\Qp$.\n\\end{corollary}\n\\begin{proof}\nCoefficient inclusion commutes with evaluation and formal partial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of the original blocks, and the original exact root\nremains an exact root. Apply Theorem~\\ref{thm:analytic-local-graph} to the system with included\ncoefficients.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialRegularityIntegral.exists_analytic_integral_implicit}.\nIt requires no invertibility modulo $p$. In particular, it includes solved\nJacobians singular modulo $p$ but invertible over …","arrow\\Qp$ be the inclusion.\nIf $\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\Qp)$, the polynomial system\nobtained by including the original coefficients in $\\Qp$ has the analytic\nlocal root graph of Theorem~\\ref{thm:analytic-local-graph} at $\\iota(a)$.\nThe strict derivative at the base is\n\\[\n -\\iota(D_YF(a))^{-1}\\iota(D_TF(a)).\n\\]\nThe open neighborhoods and the output map are in the ambient spaces over\n$\\Qp$.\n\\end{corollary}\n\\begin{proof}\nCoefficient inclusion commutes with evaluation and formal partial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of the original blocks, and the original exact root\nremains an exact root. Apply Theorem~\\ref{thm:analytic-local-graph} to the system with included\ncoefficients.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialRegularityIntegral.exists_analytic_integral_implicit}.\nIt requires no invertibility modulo $p$. In particular, it includes solved\nJacobians singular modulo $p$ but invertible over the fraction field.","tial\ndifferentiation. Consequently both Jacobian blocks after inclusion are the\nentrywise inclusions of the original blocks, and the original exact root\nremains an exact root. Apply Theorem~\\ref{thm:analytic-local-graph} to the system with included\ncoefficients.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialRegularityIntegral.exists_analytic_integral_implicit}.\nIt requires no invertibility modulo $p$. In particular, it includes solved\nJacobians singular modulo $p$ but invertible over the fraction field.","lusion are the\nentrywise inclusions of the original blocks, and the original exact root\nremains an exact root. Apply Theorem~\\ref{thm:analytic-local-graph} to the system with included\ncoefficients.\n\\end{proof}\nThis is\n\\lean{Komorebi.PolynomialRegularityIntegral.exists_analytic_integral_implicit}.\nIt requires no invertibility modulo $p$. In particular, it includes solved\nJacobians singular modulo $p$ but invertible over the fraction field.","\\begin{example}[The derivative sign and a nonlinear control]\nFor $F(Y,T)=Y+T$, the exact branch is $\\psi(T)=-T$ and $J=K=1$.\nThe derivative is $-1$. Uniqueness of the derivative rules out $+1$ over\n$\\Qp$, including $p=2$. The nonlinear polynomial $F(Y,T)=Y+Y^2+T$\nhas $F(0,0)=0$ and $J=K=1$ there. Applying the formal theorem to this polynomial constructs an analytic\nlocal root graph with derivative $-1$. Thus the example checks the existence\nconclusion as well as the derivative formula.\nThe declarations and the deliberately rejected positive-sign statement are\nin the regularity control and statement-contract files.\n\\end{example}","egin{example}[The derivative sign and a nonlinear control]\nFor $F(Y,T)=Y+T$, the exact branch is $\\psi(T)=-T$ and $J=K=1$.\nThe derivative is $-1$. Uniqueness of the derivative rules out $+1$ over\n$\\Qp$, including $p=2$. The nonlinear polynomial $F(Y,T)=Y+Y^2+T$\nhas $F(0,0)=0$ and $J=K=1$ there. Applying the formal theorem to this polynomial constructs an analytic\nlocal root graph with derivative $-1$. Thus the example checks the existence\nconclusion as well as the derivative formula.\nThe declarations and the deliberately rejected positive-sign statement are\nin the regularity control and statement-contract files.\n\\end{example}","The results cover arbitrary finite $e$ and $k$, including zero. The local\ninverse machinery explicitly handles a zero-dimensional domain. The analytic theorem does not assign a uniform numerical radius to its\nneighborhoods. The earlier\nweighted lifting and joint deformation theorems retain their quantitative\nradii and error bounds, while the present statements establish regularity\nof the local solution germ. A direct Lean identification with each\nhistorical subtype-valued chart is an additional interface task. General\npersistence a","The results cover arbitrary finite $e$ and $k$, including zero. The local\ninverse machinery explicitly handles a zero-dimensional domain. The analytic theorem does not assign a uniform numerical radius to its\nneighborhoods. The earlier\nweighted lifting and joint deformation theorems retain their quantitative\nradii and error bounds, while the present statements establish regularity\nof the local solution germ. A direct Lean identification with each\nhistorical subtype-valued chart is an additional interface task. General\npersistence at a frac","The weighted lifting theorem allows the solved Jacobian to be noninvertible\nmodulo $p$ provided that it is invertible over $\\Qp$ with a controlled inverse.\nThis includes a substantial class of lifts often called singular modulo $p$.\nThe next problem is to understand Jacobians that remain rank-deficient over\n$\\Qp$, where the inverse construction is no longer available.","For a fixed polynomial system $F\\in\\Z[X_1,\\ldots,X_d]^e$, a precision $L\\geq1$,\nand a root $\\bar x$ modulo $p^L$, let\n\\[\n\\mathcal L_{F,L}(\\bar x)\n=\\{x'\\in(\\Z/p^{L+1}\\Z)^d:\nF(x')=0,\\ x'\\equiv\\bar x\\pmod{p^L}\\}.\n\\]\nThe finite obstruction theorem decides whether this set is empty from the\nresidual digit and reduced Jacobian. When it is nonempty, the theorem also\ndetermines its cardinality and affine kernel structure. The conclusion\nconcerns one level: vanishing of the obstruction permits the next step,\nbut does not guarantee that subsequent steps are possible.","For example, $X^2-5$ has the root $1$ modulo $2$, and both $1$ and $3$ lift it\nto roots modulo $4$. Every odd square is $1$ modulo $8$, so neither branch\ncontinues to a root modulo $8$. Thus even a successful first correction can\nlead to branches that stop at the following level.","The original conjecture asks for a canonical finite obstruction packet whose\nvanishing is necessary and sufficient for a one-level lift in a finitely\npresented arithmetic seed system. For a fixed finite polynomial presentation,\nthe paper now supplies the precise obstruction\n\\[\no_{F,\\bar x}=[r]\\in\n\\coker\\!\\left(J_{F,\\bar x}:\\F_p^d\\longrightarrow\\F_p^e\\right),\n\\]\nwhere $r$ is the residual digit. We have proved independence from the chosen\ninteger representative. Invariance under a change of equations or presentation\nrequires more.","The raw cokernel already reveals the difficulty.\nThe presentations $F=(X)$ and $G=(X,0)$ describe the same zero locus over every\ncoefficient ring. At zero their Jacobians are, respectively,\n\\[\n\\F_p\\xrightarrow{\\,1\\,}\\F_p,\n\\qquad\n\\F_p\\xrightarrow{\\,(1,0)\\,}\\F_p^2.\n\\]\nTheir raw cokernels are $0$ and $\\F_p$. Both distinguished obstruction classes\nvanish, but the ambient obstruction groups are not isomorphic. The example\nrules out the raw cokernel as a presentation-independent object while leaving\nopen the possibility of an intrinsic obstruction construction.","The original proposed form is\n\\[\nX\\simeq X_2^{\\mathrm{wt}}\\times_C\\prod_{p>2}'X_p.\n\\]\nIt leaves $X$, the distinguished subsets defining the restricted product,\nthe maps to $C$, the ambient category, and the universal meaning of\n``smallest'' unspecified. In particular, one must decide whether $X$ is a\nglobal object or a completed object.","The linear compatibility theorem provides one concrete part of this program.\nIf two local observations\n$J:U\\to V$ and $K:U\\to W$ are linear maps from the \\emph{same} correction\nmodule, define\n\\[\nC_{J,K}=U/(\\ker J+\\ker K).\n\\]\nThe attainable observations $\\im J$ and $\\im K$ have canonical labels in this\nquotient. Equality of labels is equivalent to simultaneous realization by a\nsingle correction. The joint image is exactly the fiber product over\n$C_{J,K}$, with the proved linear universal property.\nThe declarations are\n\\begingroup\\small\\raggedright\n\\lean{Komorebi.SharedCorrection.labels_eq_iff_common},\n\\lean{Komorebi.SharedCorrection.joint_range_eq_fiber}, and\n\\lean{Komorebi.SharedCorrection.common_universal}.\n\\par\\endgroup\nTo apply this result arithmetically, one must construct the common correction\nmo","The linear compatibility theorem provides one concrete part of this program.\nIf two local observations\n$J:U\\to V$ and $K:U\\to W$ are linear maps from the \\emph{same} correction\nmodule, define\n\\[\nC_{J,K}=U/(\\ker J+\\ker K).\n\\]\nThe attainable observations $\\im J$ and $\\im K$ have canonical labels in this\nquotient. Equality of labels is equivalent to simultaneous realization by a\nsingle correction. The joint image is exactly the fiber product over\n$C_{J,K}$, with the proved linear universal property.\nThe declarations are\n\\begingroup\\small\\raggedright\n\\lean{Komorebi.SharedCorrection.labels_eq_iff_common},\n\\lean{Komorebi.SharedCorrection.joint_range_eq_fiber}, and\n\\lean{Komorebi.SharedCorrection.common_universal}.\n\\par\\endgroup\nTo apply this result arithmetically, one must construct the common correction\nmodule and the ob","The linear compatibility theorem provides one concrete part of this program.\nIf two local observations\n$J:U\\to V$ and $K:U\\to W$ are linear maps from the \\emph{same} correction\nmodule, define\n\\[\nC_{J,K}=U/(\\ker J+\\ker K).\n\\]\nThe attainable observations $\\im J$ and $\\im K$ have canonical labels in this\nquotient. Equality of labels is equivalent to simultaneous realization by a\nsingle correction. The joint image is exactly the fiber product over\n$C_{J,K}$, with the proved linear universal property.\nThe declarations are\n\\begingroup\\small\\raggedright\n\\lean{Komorebi.SharedCorrection.labels_eq_iff_common},\n\\lean{Komorebi.SharedCorrection.joint_range_eq_fiber}, and\n\\lean{Komorebi.SharedCorrection.common_universal}.\n\\par\\endgroup\nTo apply this result arithmetically, one must construct the common correction\nmodule and the observation maps, then show that the resulting compatibility\ncondition agrees with the intended a","The linear compatibility theorem provides one concrete part of this program.\nIf two local observations\n$J:U\\to V$ and $K:U\\to W$ are linear maps from the \\emph{same} correction\nmodule, define\n\\[\nC_{J,K}=U/(\\ker J+\\ker K).\n\\]\nThe attainable observations $\\im J$ and $\\im K$ have canonical labels in this\nquotient. Equality of labels is equivalent to simultaneous realization by a\nsingle correction. The joint image is exactly the fiber product over\n$C_{J,K}$, with the proved linear universal property.\nThe declarations are\n\\begingroup\\small\\raggedright\n\\lean{Komorebi.SharedCorrection.labels_eq_iff_common},\n\\lean{Komorebi.SharedCorrection.joint_range_eq_fiber}, and\n\\lean{Komorebi.SharedCorrection.common_universal}.\n\\par\\endgroup\nTo apply this result arithmetically, one must construct the common correction\nmodule and the observation maps, then show that the resulting compatibility\ncondition agrees with the intended arithmetic one.","crete part of this program.\nIf two local observations\n$J:U\\to V$ and $K:U\\to W$ are linear maps from the \\emph{same} correction\nmodule, define\n\\[\nC_{J,K}=U/(\\ker J+\\ker K).\n\\]\nThe attainable observations $\\im J$ and $\\im K$ have canonical labels in this\nquotient. Equality of labels is equivalent to simultaneous realization by a\nsingle correction. The joint image is exactly the fiber product over\n$C_{J,K}$, with the proved linear universal property.\nThe declarations are\n\\begingroup\\small\\raggedright\n\\lean{Komorebi.SharedCorrection.labels_eq_iff_common},\n\\lean{Komorebi.SharedCorrection.joint_range_eq_fiber}, and\n\\lean{Komorebi.SharedCorrection.common_universal}.\n\\par\\endgroup\nTo apply this result arithmetically, one must construct the common correction\nmodule and the observation maps, then show that the resulting compatibility\ncondition agrees with the intended arithmetic one.","\\paragraph{Why the global/completed distinction is necessary.}\nConsider constant sequences with integer initial value. Their global parameter\nspace is $\\Z$, while their local parameter spaces are $\\Z_p$. Under the\ncanonical local maps, the image of the global objects is the diagonal\n\\[\nD=\\{(n)_p:n\\in\\Z\\}\\subseteq\\prod_p\\Z_p.\n\\]\nThis diagonal is dense: a basic open set specifies finitely many congruences,\nand the Chinese remainder theorem provides an integer satisfying them. It is\nproper: the tuple with $2$-adic coordinate zero and all odd-adic coordinates\none is not the image of an integer. Indeed an integer zero in $\\Z_2$ must be\nzero, contradicting its $3$-adic coordinate.","al value. Their global parameter\nspace is $\\Z$, while their local parameter spaces are $\\Z_p$. Under the\ncanonical local maps, the image of the global objects is the diagonal\n\\[\nD=\\{(n)_p:n\\in\\Z\\}\\subseteq\\prod_p\\Z_p.\n\\]\nThis diagonal is dense: a basic open set specifies finitely many congruences,\nand the Chinese remainder theorem provides an integer satisfying them. It is\nproper: the tuple with $2$-adic coordinate zero and all odd-adic coordinates\none is not the image of an integer. Indeed an integer zero in $\\Z_2$ must be\nzero, contradicting its $3$-adic coordinate.","ocal maps, the image of the global objects is the diagonal\n\\[\nD=\\{(n)_p:n\\in\\Z\\}\\subseteq\\prod_p\\Z_p.\n\\]\nThis diagonal is dense: a basic open set specifies finitely many congruences,\nand the Chinese remainder theorem provides an integer satisfying them. It is\nproper: the tuple with $2$-adic coordinate zero and all odd-adic coordinates\none is not the image of an integer. Indeed an integer zero in $\\Z_2$ must be\nzero, contradicting its $3$-adic coordinate.","cts is the diagonal\n\\[\nD=\\{(n)_p:n\\in\\Z\\}\\subseteq\\prod_p\\Z_p.\n\\]\nThis diagonal is dense: a basic open set specifies finitely many congruences,\nand the Chinese remainder theorem provides an integer satisfying them. It is\nproper: the tuple with $2$-adic coordinate zero and all odd-adic coordinates\none is not the image of an integer. Indeed an integer zero in $\\Z_2$ must be\nzero, contradicting its $3$-adic coordinate.","Consequently this canonical global diagonal cannot be an equality locus\n\\[\n\\{(a,b):f(a)=g(b)\\}\n\\subseteq \\Z_2\\times\\prod_{p>2}\\Z_p\n\\]\nfor continuous maps to a Hausdorff space $C$. Such an equality locus is closed,\nas the inverse image of the closed diagonal in $C\\times C$, whereas $D$ is\nproper and dense. This topological argument, retained from the native progress\nnote, concerns the stated topological interpretation. It has not been\nformalized in Lean for this manuscript.","Finite CRT reconstruction leads to the familiar completed object\n$\\widehat\\Z=\\prod_p\\Z_p$, but does not identify which local tuples come from a\nsingle global integer. A complete form of Native 13.3 must therefore specify\nfinite-level compatibility, transition maps and completion, and any additional\nglobal descent condition. If recurrence, spectral, or Frobenius information\nis to supply that condition, it must be expressed through defined maps or\nconstraints on the local data, with a precise account of how they enter $C$.","l tuples come from a\nsingle global integer. A complete form of Native 13.3 must therefore specify\nfinite-level compatibility, transition maps and completion, and any additional\nglobal descent condition. If recurrence, spectral, or Frobenius information\nis to supply that condition, it must be expressed through defined maps or\nconstraints on the local data, with a precise account of how they enter $C$.","The accompanying \\nolinkurl{native/Native_Conjectures_Progress.md} records\nfurther results in restricted settings: an intrinsic finite\n$\\operatorname{Ext}^1$ obstruction for finitely presented commutative integer\nalgebras, finite CRT reconstruction for polynomial solution sets, and a\ncharacterization of rectangular finite relations by equality of common\nlabels. The obstruction result is a stated specialization of a classical\nexternal theorem; the CRT and rectangularity results have prose proofs in\nthe note. These are preserved mathematical progress, with the restrictions\nand source records given there, rather than Lean completions of the general\nNative 13.2 and 13.3 statements.","Write $U_{N,j}(n)=j!C_{N,j}(n)$ and $U_j=U_{4,j}$, and use\n$v_p(0)=\\infty$ when discussing valuation bounds. The following classification\nquestions remain open in this program.","\\begin{enumerate}\n\\item \\textbf{The remaining higher-difference classification.}\nThe new strengthening chapters determine every repeated-step valuation in addresses $j=0,1$, give the exact criterion for equality in the common mixed-step lower bound there, and determine address-two valuations at all power-of-two orders. What remains is a complete exact formula in address two at the other orders, a full classification in address three, and the amount of additional cancellation in mixed-step cases where the leading term vanishes. Numerical patterns through a finite range do not settle these questions.","\\item \\textbf{Upper-half addresses at larger dyadic rank.}\nThe exact first-difference laws now include $N=2,4,8$ and every lower-half address for $N=2^a\\ge4$. Larger upper-half addresses need a finer classification. Explicit counterexamples show that extrapolating one low-order Newton coefficient to a universal metric law is false. A least period is measured in packet rows; its Euler-degree displacement is $NP$.","bf{Upper-half addresses at larger dyadic rank.}\nThe exact first-difference laws now include $N=2,4,8$ and every lower-half address for $N=2^a\\ge4$. Larger upper-half addresses need a finer classification. Explicit counterexamples show that extrapolating one low-order Newton coefficient to a universal metric law is false. A least period is measured in packet rows; its Euler-degree displacement is $NP$.","\\item \\textbf{Exact affine Mahler-kernel valuations.}\nFor the kernel of Section~\\ref{subsec:polyphase-exposition}, determine exact\nvalues of\n\\[\nv_p\\bigl(\\Theta_{M,s}(r,k)\\bigr),\\qquad\n\\Theta_{M,s}(r,k)=[z^r](1+z)^s((1+z)^M-1)^k,\n\\]\nbeyond the explicit special families. The minimum valuation of the summands\ngives a lower bound. Equality requires a separate argument excluding\ncancellation.","One further task is to connect the older chart constructions to the new\nregularity theorem. That theorem applies to polynomials with arbitrary\ncoefficients in the ambient field $\\Qp$ and identifies the analytic\ngerm of every continuous ambient-field root selection under its hypotheses.\nSeveral older rectangular charts were defined using integer coefficients,\ninteger centers, or subtype-valued maps. Comparing each such historical\ninterface explicitly with the new ambient formulation remains to be done.\nThese comparisons would connect the existing constructions to the proved\nanalytic implicit-function theorem. General singular persistence remains the\nseparate problem describ","\\midrule\nOriginal completion baseline & 79 & $79+37+3$ & 1000","retained logs. In particular, the following interfaces distinguish\nthe principal completed results from generic auxiliary lemmas:\n\\begin{itemize}\n\\item The external conjecture is assembled by\n\\lean{NishibuchiCompletion.conjecture_3_9} in the original completion\nbaseline. Its coefficient identification, integrality, parameter split and\nexplicit threshold are part of that proof chain.\n\\item Exact $N=4$ packet periods are supplied by\n\\lean{DyadicFourLeastPeriods.least_packet_period}; the isometric extension\nis supplied by \\lean{DyadicFourIsometry.existsUnique_isometry}. The\nnormalization of the original coefficients is an explicit part of these\ninterfaces.\n\\item The classical sharp edge tower is supplied by\n\\lean{DyadicSharpTower.dyadic_supercongruence_sharp}, together with its\nfirst omitted error valuation. Its optimality concerns the stated uniform\ntruncation bound.\n\\item Prescribed parameter lifting uses the actual residual and actual\npartial Jacobians in\n\\lean{PolynomialPrescribedObstruction.prescribed_lift_iff_obstruction_zero}.\nThe quotient criterion and exact lift counts are not …","ion.prescribed_lift_iff_obstruction_zero}.\nThe quotient criterion and exact lift counts are not assumptions of the\nlater tower results.\n\\item Joint deformation is supplied by\n\\lean{PolynomialJointDeformation.exists_unique_joint_deformation}.\nLocal regularity is supplied by\n\\lean{PolynomialAnalyticImplicit.exists_analytic_implicit}, with the\nactual polynomial derivative and the derivative formula $-J^{-1}K$.\n\\item \\lean{PolynomialAnalyticSelection.continuous_selection_regularity}\ntransfers the analytic germ and strict derivative to any continuous local\nroot selection satisfying the stated root and base-point conditions.\n\\end{itemize}\nHere arithmetic declaration names have namespace prefix\n\\texttt{KomorebiPaperIX.}; polynomial and native declarations have prefix\n\\texttt{Komorebi.}. The section maps and CSV retain the full names.","The original conjecture correspondence was reviewed against Nishibuchi's\narXiv version \\texttt{2605.10677v2}, dated 1 June 2026. The retained source\nrecord identifies the official version, the full title, the PDF checksum and\nthe extracted-text checksum. The detailed review compares Definition~2.1,\nNotation~3.2 and Conjecture~3.9 with the actual reciprocal EGF coefficients,\nthe factor $j!$, the residue range, the degree displacement, and the threshold\nin the formal capstone. Nishibuchi's conjecture is the conclusion to be\nproved; it is not an axiom imported into Lean.","texttt{Classical.choice} and \\texttt{Quot.sound}.\nThe reports do not describe a rebuild of Mathlib from source. The local\nregularity audit separately records the applicable Mathlib analytic inverse\nand strict derivative interfaces and their hypotheses. In particular, the\nfield-valued analysis is performed over a complete nontrivially normed field;\nits integral specialization embeds the data into $\\Qp$. It does not use\n$\\Zp$ as a vector space over $\\Qp$.","} and \\texttt{Quot.sound}.\nThe reports do not describe a rebuild of Mathlib from source. The local\nregularity audit separately records the applicable Mathlib analytic inverse\nand strict derivative interfaces and their hypotheses. In particular, the\nfield-valued analysis is performed over a complete nontrivially normed field;\nits integral specialization embeds the data into $\\Qp$. It does not use\n$\\Zp$ as a vector space over $\\Qp$.","e reports do not describe a rebuild of Mathlib from source. The local\nregularity audit separately records the applicable Mathlib analytic inverse\nand strict derivative interfaces and their hypotheses. In particular, the\nfield-valued analysis is performed over a complete nontrivially normed field;\nits integral specialization embeds the data into $\\Qp$. It does not use\n$\\Zp$ as a vector space over $\\Qp$.","ly-row script separately reproduced 312 finite period\ncases and its odd-prime row-zero boundary control. Other research-script\nreplays, unchanged baseline regression gates and recovered-text controls\nare recorded separately in the same audit directory; their totals are\nnot added to the independent gate. Lean and Lake were unavailable in this\nexecution environment. No fresh compiler run, infinite $p$-adic computation,\nanalytic-continuation computation or hardware benchmark is asserted.","The dedicated arithmetic experiment computed the four reciprocal rows through index 512 modulo $2^{4096}$, with independent exact-rational inversion through index 16. It checked 525,312 first differences, 1,141,424 repeated-step factorial bounds, 139,000 mixed-step bounds, and 2,048 Newton coefficients. The supplemental experiments cover additional mixed-step and sharpness cases and retain their separate tested ranges. Their outputs and source hashes are preserved under \\nolinkurl{preserved/threshold_free/computation}. All these are bounded implementation checks; the written arguments establish the universal claims.","\\textbf{Scope.} Fix a prime \\(p\\), an integer \\(L\\ge1\\), finite natural\nnumbers \\(d,r\\), and polynomials\n\\(F_1,\\ldots,F_r\\in\\mathbb Z[X_1,\\ldots,X_d]\\). Let\n\\(\\bar x\\in(\\mathbb Z/p^L\\mathbb Z)^d\\) satisfy every equation\n\\(F_i(\\bar x)=0\\). Choose any integer representative\n\\(x\\in\\mathbb Z^d\\). Set","\\[e_i=F_i(x)/p^L\\pmod p,\\qquad J_{ij}=\\frac{\\partial F_i}{\\partial X_j}(x)\\pmod p.\\]","The divisions are exact integer divisions. Regard \\(J\\) as a linear map\n\\(\\mathbb F_p^d\\to\\mathbb F_p^r\\). Define","\\[O_{F,\\bar x}=\\mathbb F_p^r/\\operatorname{im}J,\\qquad o_{F,\\bar x}=[e].\\]","\\textbf{Theorem.} For this fixed polynomial presentation, the group and\ndistinguished class are independent of the chosen integer\nrepresentative. The class vanishes if and only if \\(\\bar x\\) lifts to a\nsolution modulo \\(p^{L+1}\\). The obstruction group has\n\\(p^{r-\\operatorname{rank}J}\\) elements. When lifts exist, their set is\na torsor for \\(\\ker J\\) and contains exactly\n\\(p^{d-\\operatorname{rank}J}\\) elements.","\\textbf{Proof.} For every integer correction vector \\(t\\), expansion of\neach monomial gives","\\[F(x+p^Lt)\\equiv F(x)+p^LJ(x)t\\pmod {p^{L+1}}.\\tag{1}\\]","Indeed, each term involving at least two correction factors is divisible\nby \\(p^{2L}\\), and \\(2L\\ge L+1\\). Constant and linear correction terms\ngive the two displayed summands. Adding the monomial expansions proves\n(1) for every polynomial, with no denominator or analytic convergence\nissue.","Every element reducing to \\(\\bar x\\) is uniquely expressible modulo\n\\(p^{L+1}\\) as \\(x+p^Lt\\) with \\(t\\in\\mathbb F_p^d\\). By (1), it solves\nthe equations exactly when","\\[e+Jt=0\\quad\\text{in }\\mathbb F_p^r.\\tag{2}\\]","Equation (2) is soluble exactly when \\(e\\in\\operatorname{im}J\\): the\nsign makes no difference because the image is a subspace. This is\nequivalent to \\([e]=0\\).","Another representative has the form \\(x'=x+p^Ls\\). Since \\(L\\ge1\\), its\nJacobian is the same modulo \\(p\\). Dividing (1) by \\(p^L\\) shows that\nits residual digit is \\(e'=e+Js\\), so it gives exactly the same quotient\nclass. If \\(t_0\\) solves (2), all its solutions are \\(t_0+\\ker J\\), and\nno others. The finite cardinalities follow from rank-nullity. This also\ncovers no variables, no equations, zero Jacobian and singular solutions.\n\\(\\square\\)","m \\(x'=x+p^Ls\\). Since \\(L\\ge1\\), its\nJacobian is the same modulo \\(p\\). Dividing (1) by \\(p^L\\) shows that\nits residual digit is \\(e'=e+Js\\), so it gives exactly the same quotient\nclass. If \\(t_0\\) solves (2), all its solutions are \\(t_0+\\ker J\\), and\nno others. The finite cardinalities follow from rank-nullity. This also\ncovers no variables, no equations, zero Jacobian and singular solutions.\n\\(\\square\\)","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruc","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at t","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step.","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infin","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\i","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solutio","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system.","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their ra","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\ma","\\begin{itemize}\n\\tightlist\n\\item\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\)","em\n  \\textbf{A failed singular lift:} \\(p=3\\), \\(L=1\\), \\(F(X)=X^2-3\\),\n  \\(x=0\\). Here \\(e=-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\). Both actual distinguished\n  o…","-1\\ne0\\) in \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\). Both actual distinguished\n  obstructions are zero. Thus one cannot claim a canonical isomorphism\n  of these two raw group…","n \\(\\mathbb F_3\\) and \\(J=0\\). There are\n  no lifts of zero modulo 3 to a root modulo 9.\n\\item\n  \\textbf{Representative independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\). Both actual distinguished\n  obstructions are zero. Thus one cannot claim a canonical isomorphism\n  of these two raw groups merely b…","tive independence is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\). Both actual distinguished\n  obstructions are zero. Thus one cannot claim a canonical isomorphism\n  of these two raw groups merely because the equations define the same\n  seed.\n\\end{itemize}","e is a quotient assertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\). Both actual distinguished\n  obstructions are zero. Thus one cannot claim a canonical isomorphism\n  of these two raw groups merely because the equations define the same\n  seed.\n\\end{itemize}","sertion:} for\n  \\(F(X)=X^2-1\\), \\(p=3\\), \\(L=1\\), representatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\). Both actual distinguished\n  obstructions are zero. Thus one cannot claim a canonical isomorphism\n  of these two raw groups merely because the equations define the same\n  seed.\n\\end{itemize}","epresentatives \\(1\\) and \\(4\\) have\n  residual digits \\(0\\) and \\(2\\). Their common Jacobian is \\(2\\), with\n  full image, and both give zero obstruction. Their corrected\n  representatives \\(1\\) and \\(10\\) are the same root modulo 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\). Both actual distinguished\n  obstructions are zero. Thus one cannot claim a canonical isomorphism\n  of these two raw groups merely because the equations define the same\n  seed.\n\\end{itemize}","o 9.\n\\item\n  \\textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\). Both actual distinguished\n  obstructions are zero. Thus one cannot claim a canonical isomorphism\n  of these two raw groups merely because the equations define the same\n  seed.\n\\end{itemize}","textbf{A successful step need not give indefinite continuation:} for\n  \\(F(X)=X^2-5\\), \\(p=2\\), the root \\(1\\) modulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\). Both actual distinguished\n  obstructions are zero. Thus one cannot claim a canonical isomorphism\n  of these two raw groups merely because the equations define the same\n  seed.\n\\end{itemize}","odulo 2 has zero obstruction\n  and lifts to \\(1,3\\) modulo 4. Both have nonzero obstruction at the\n  next step. There is no root modulo 8 and no compatible infinite lift.\n\\item\n  \\textbf{The raw Jacobian cokernel need not be presentation\n  independent:} the presentations \\(F=(X)\\) and \\(G=(X,0)\\) define the\n  same solution system. Their raw obstruction groups at zero are\n  respectively zero and \\(\\mathbb F_p\\). Both actual distinguished\n  obstructions are zero. Thus one cannot claim a canonical isomorphism\n  of these two raw groups merely because the equations define the same\n  seed.\n\\end{itemize}","Let \\(B\\) be a finitely presented commutative \\(\\mathbb Z\\)-algebra,\n\\(p\\) prime and \\(L\\ge1\\). For a ring map\n\\(f:B\\to\\mathbb Z/p^L\\mathbb Z\\), use its reduction to make\n\\(\\mathbb F_p\\) a \\(B\\)-module. There is a canonical class","\\[o(f)\\in\\operatorname{Ext}^1_B(\\mathrm{NL}_{B/\\mathbb Z},\\mathbb F_p)\\]","whose vanishing is equivalent to a lift\n\\(B\\to\\mathbb Z/p^{L+1}\\mathbb Z\\). This follows from the ring-map\nlifting theorem in\n\\href{https://stacks.math.columbia.edu/tag/08S5}{Stacks Project, Lemma\n91.2.1, tag 08S5}. Specialize the source extension to\n\\(0\\to0\\to B\\to B\\) and the target extension to the kernel\n\\(p^L/p^{L+1}\\cong\\mathbb F_p\\). The base maps are identities of\n\\(\\mathbb Z\\). The kernel is square zero because \\(L\\ge1\\).","ent to a lift\n\\(B\\to\\mathbb Z/p^{L+1}\\mathbb Z\\). This follows from the ring-map\nlifting theorem in\n\\href{https://stacks.math.columbia.edu/tag/08S5}{Stacks Project, Lemma\n91.2.1, tag 08S5}. Specialize the source extension to\n\\(0\\to0\\to B\\to B\\) and the target extension to the kernel\n\\(p^L/p^{L+1}\\cong\\mathbb F_p\\). The base maps are identities of\n\\(\\mathbb Z\\). The kernel is square zero because \\(L\\ge1\\).","The group is finite. Choose a finite presentation\n\\(P=\\mathbb Z[X_1,\\ldots,X_d]\\twoheadrightarrow B\\) with finitely\ngenerated kernel \\(I\\). The presentation description in\n\\href{https://stacks.math.columbia.edu/tag/0GPU}{Stacks Project, Remark\n91.2.4, tag 0GPU} identifies the group as","\\[\\operatorname{coker}\\left(\\operatorname{Hom}_B(\\Omega_{P/\\mathbb Z}\\otimes_P B,\\mathbb F_p)\\longrightarrow\\operatorname{Hom}_B(I/I^2,\\mathbb F_p)\\right).\\]","If \\(I\\) has \\(r\\) chosen generators, evaluation on their classes\ninjects the latter Hom module into \\(\\mathbb F_p^r\\). Hence its quotient\nis finite. The relation module \\(I/I^2\\) retains dependencies between\nequations, including redundant zero equations, which the raw target\n\\(\\mathbb F_p^r\\) does not.","Consider the recurrence \\(a_{n+1}=a_n\\) with integer initial value. Its\nglobal parameter space is \\(X=\\mathbb Z\\). Its local parameter spaces\nare \\(X_p=\\mathbb Z_p\\), and every local realization has the same\nprime-independent recurrence. Give the local factors their usual\n\\(p\\)-adic topologies and the product its product topology. For these\nintegral local factors, the restricted product relative to all\n\\(\\mathbb Z_p\\) is the ordinary product. Write","the recurrence \\(a_{n+1}=a_n\\) with integer initial value. Its\nglobal parameter space is \\(X=\\mathbb Z\\). Its local parameter spaces\nare \\(X_p=\\mathbb Z_p\\), and every local realization has the same\nprime-independent recurrence. Give the local factors their usual\n\\(p\\)-adic topologies and the product its product topology. For these\nintegral local factors, the restricted product relative to all\n\\(\\mathbb Z_p\\) is the ordinary product. Write","\\[A=\\mathbb Z_2,\\qquad B=\\prod_{p>2}\\mathbb Z_p,\\qquad D=\\{(n,(n)_{p>2}):n\\in\\mathbb Z\\}\\subset A\\times B.\\]","\\textbf{Theorem.} There are no Hausdorff space \\(C\\) and continuous maps\n\\(f:A\\to C\\), \\(g:B\\to C\\) for which the equality locus","\\[\\{(a,b):f(a)=g(b)\\}\\]","is exactly \\(D\\). Thus the canonical local realizations of global\ninteger constant sequences cannot satisfy the manuscript's fiber-product\nformula in this Hausdorff topological interpretation.","\\textbf{Proof.} The equality locus is closed: it is the inverse image of\nthe diagonal of \\(C\\times C\\) under the continuous map \\(f\\times g\\),\nand a Hausdorff space has closed diagonal.","The set \\(D\\) is dense in \\(A\\times B\\). A nonempty basic open set\nprescribes congruences at only finitely many primes and finite precision\nat each. The moduli are pairwise coprime. The finite Chinese remainder\ntheorem supplies an integer meeting all those congruences, so every such\nopen set meets \\(D\\).","However, \\(D\\) is not the whole product. The element with 2-adic\ncoordinate zero and every odd-adic coordinate one is not diagonal. If an\ninteger had zero 2-adic coordinate, it would be divisible by every power\nof two, forcing it to be zero. Its 3-adic coordinate would then be zero,\nnot one. Consequently \\(D\\) is a proper dense subset and is not closed.\nIt cannot be the equality locus above. \\(\\square\\)","For a finite set \\(S\\) of primes, positive precisions \\(L_p\\), and a\nfixed integer polynomial system \\(F\\), put","\\[N=\\prod_{p\\in S}p^{L_p},\\qquad\\operatorname{Sol}_F(R)=\\{x\\in R^d:F(x)=0\\}.\\]","\\[\\operatorname{Sol}_F(\\mathbb Z/N\\mathbb Z)\\cong\\prod_{p\\in S}\\operatorname{Sol}_F(\\mathbb Z/p^{L_p}\\mathbb Z).\\]","The reason is exact: the CRT ring isomorphism preserves polynomial\nevaluation, and zero in a finite product is equivalent to zero in every\ncomponent. For \\(F\\) empty, passing to all compatible finite congruences\nidentifies the profinite completion \\(\\widehat{\\mathbb Z}\\) with\n\\(\\prod_p\\mathbb Z_p\\). The same finite reconstruction is compatible\nwith coefficient projections and integer recurrence matrices because\nthese operations commute with reduction and CRT.","Here the polynomial equations or recurrence matrices are fixed common\ndata. The theorem reconstructs finite or profinite objects. It does not\nidentify a ``smallest'' spine, impose additional spectral descent, or\nrecover global \\(\\mathbb Z\\) from arbitrary tuples of local points. A\ncompleted version of 13.3 must say exactly which object is completed and\nspecify its universal property.","Fix finite sets \\(A,B\\) of local packets and an explicitly given\nrelation \\(R\\subseteq A\\times B\\). Say that \\(R\\) is rectangular when","\\[R(a,b),\\ R(a',b),\\ R(a',b')\\quad\\Longrightarrow\\quad R(a,b')\\]","for all \\(a,a'\\in A\\) and \\(b,b'\\in B\\).","\\textbf{Theorem.} The relation \\(R\\) can be represented exactly as","\\[R(a,b)\\quad\\Longleftrightarrow\\quad f(a)=g(b)\\]","for maps from \\(A\\) and \\(B\\) to a common set if and only if \\(R\\) is\nrectangular. In that case a canonical finite choice of common set is the\nconnected-component set \\(C_R\\) of the bipartite graph with vertex set\n\\(A\\sqcup B\\) and edges given by \\(R\\).","The component maps \\(f_R:A\\to C_R\\) and \\(g_R:B\\to C_R\\) have the\nfollowing universal property. For every set \\(Y\\) and maps \\(h:A\\to Y\\),\n\\(k:B\\to Y\\) satisfying \\(h(a)=k(b)\\) whenever \\(R(a,b)\\), there is a\nunique map \\(u:C_R\\to Y\\) with \\(u\\circ f_R=h\\) and \\(u\\circ g_R=k\\).","\\textbf{Proof.} Any equality representation is rectangular, since the\nthree assumed equalities imply \\(f(a)=g(b)=f(a')=g(b')\\).","Conversely, suppose \\(R\\) is rectangular. An edge always gives equal\ncomponent labels. If \\(a\\in A\\) and \\(b\\in B\\) have equal component\nlabels, there is a finite path between them. Its length is odd. A path\nof length one is already an \\(R\\)-edge. A longer path begins with three\nedges \\(a-b_1-a_1-b_2\\). Rectangularity gives the edge \\(a-b_2\\).\nReplacing these three edges by that edge shortens the path by two.\nInduction therefore yields \\(R(a,b)\\). This proves the exact equality\nrepresentation. Finiteness follows because the component set is a\nquotient of the finite set \\(A\\sqcup B\\).","An edge always gives equal\ncomponent labels. If \\(a\\in A\\) and \\(b\\in B\\) have equal component\nlabels, there is a finite path between them. Its length is odd. A path\nof length one is already an \\(R\\)-edge. A longer path begins with three\nedges \\(a-b_1-a_1-b_2\\). Rectangularity gives the edge \\(a-b_2\\).\nReplacing these three edges by that edge shortens the path by two.\nInduction therefore yields \\(R(a,b)\\). This proves the exact equality\nrepresentation. Finiteness follows because the component set is a\nquotient of the finite set \\(A\\sqcup B\\).","em. Its length is odd. A path\nof length one is already an \\(R\\)-edge. A longer path begins with three\nedges \\(a-b_1-a_1-b_2\\). Rectangularity gives the edge \\(a-b_2\\).\nReplacing these three edges by that edge shortens the path by two.\nInduction therefore yields \\(R(a,b)\\). This proves the exact equality\nrepresentation. Finiteness follows because the component set is a\nquotient of the finite set \\(A\\sqcup B\\).","For the universal property, combine \\(h\\) and \\(k\\) into a function on\n\\(A\\sqcup B\\). It is constant across every edge by hypothesis, hence\nalong every finite path. It therefore descends to a well-defined\nfunction on components. The two required factorization identities hold\nby construction. They determine the value on every component, proving\nuniqueness. \\(\\square\\)","For example, the three pairs \\((a_0,b_0),(a_1,b_0),(a_1,b_1)\\) without\n\\((a_0,b_1)\\) fail rectangularity. Every equality-label representation\nagreeing on those three pairs also admits the fourth pair, so it cannot\nrepresent that relation exactly.","This is a precise finite gluing obligation: future arithmetic\napplications must define \\(A\\), \\(B\\) and \\(R\\) and prove\nrectangularity. The universal property holds for the component quotient\neven when rectangularity fails, but then the equality locus strictly\nenlarges \\(R\\). The theorem establishes neither a minimum number of\nlabels nor a smallest spine in the manuscript's unspecified category. It\ncarries no automatic topology, spectral information or Frobenius\ncompatibility. \\textbf{Evidence: prose proof in this note, not a Lean\nformalization.}","reconstruction lacked all 14 original figure files. This\nreading edition supplies new reconstructions at their retained labels and\ncaptions. These are explanatory schematics and exact finite examples, not\nrecovered original artwork. The reproducible program\n\\nolinkurl{tools/reconstructed_figures.py} checks the displayed finite data.\nThe modulo-seven inverse plot uses the corrected positive tail\n$6,5,2,6,5,2,\\ldots$. A reconstructed illustration does not upgrade the\nhistorical proof or formalization status. The original historical source\npackage, including all 14 vector figures and the bibliography, has now been\nrecovered and is preserved unchanged in the accompanying release. The current\nreading edition retains its explicitly identified reconstructions and corrected\nfinite data. The 22 historical references now have recovered bibliographic\nentries and checked identities. Superseded claims about the divisor-branch\nfinite-difference problem have local notices directing the reader to the\ncurrent theorem. The remaining higher kernel-valuation and unrestricted\nlocal-to-global questions re…","This article begins with the Hurwitz product\n\\[\n(a\\starh b)_n\n=\n\\sum_{r=0}^{n}\n\\binom nr a_rb_{n-r}.\n\\]\nOver a field of characteristic zero it is multiplication of exponential generating functions. Over a finite field it realizes the additive group algebra through the exponential modes \\(\\epsilon_u=(u^n)_{n\\geq0}\\). This finite algebra gives an exact proof of the standard window period. Hurwitz series rings are treated in \\cite{Keigher1997,BarberoCerrutiMurru2018}. The finite difference and umbral background may be found in \\cite{Roman1984,Comtet1974,GrahamKnuthPatashnik1994}.","Fix an odd prime \\(p\\), a divisor \\(m\\mid p-1\\), and put\n\\[\nh=\\frac{p-1}{m}.\n\\]\nThe reduced standard window inverse is defined by\n\\[\n\\sum_{q=0}^{n}\n\\binom{mn}{mq}e_q^{(m,p)}\n=\n\\delta_{n0}\n\\quad\\text{in }\\FF_p.\n\\]\nWe prove that\n\\[\ne_{n+h}^{(m,p)}=e_n^{(m,p)}\n\\qquad(n\\geq1).\n\\]\nThe result first follows from subgroup symmetry in the finite Hurwitz algebra. A second proof shows that the standard window operator preserves the finite space of sequences that are \\(h\\) periodic after their constant term. The two arguments expose different parts of the same finite character structure.","\\frac{p-1}{m}.\n\\]\nThe reduced standard window inverse is defined by\n\\[\n\\sum_{q=0}^{n}\n\\binom{mn}{mq}e_q^{(m,p)}\n=\n\\delta_{n0}\n\\quad\\text{in }\\FF_p.\n\\]\nWe prove that\n\\[\ne_{n+h}^{(m,p)}=e_n^{(m,p)}\n\\qquad(n\\geq1).\n\\]\nThe result first follows from subgroup symmetry in the finite Hurwitz algebra. A second proof shows that the standard window operator preserves the finite space of sequences that are \\(h\\) periodic after their constant term. The two arguments expose different parts of the same finite character structure.","The central change in this edition is the treatment of the odd and dyadic Euler formulas. They are no longer presented as unrelated reconstruction devices. For each prime \\(p\\) we define an even source series \\(H_p\\) and an inverse kernel \\(K_p\\) satisfying\n\\[\nK_p(t)H_p(t)=\\operatorname{sech}t.\n\\]\nAt an odd prime,\n\\[\nH_p(t)=\\cosh^2(pt)\\operatorname{sech}t,\n\\qquad\nK_p(t)=\\operatorname{sech}^2(pt).\n\\]\nAt the dyadic prime,\n\\[\nH_2(t)=\\frac{\\sinh(2t)\\sinh t}{2t^2},\n\\qquad\nK_2(t)=\\frac{4t^2}{\\sinh^2(2t)}.\n\\]\nThe coefficient convolution is identical in form, but the kernels have different integral weights. This is the correct sense in which the local theories unify.","edition is the treatment of the odd and dyadic Euler formulas. They are no longer presented as unrelated reconstruction devices. For each prime \\(p\\) we define an even source series \\(H_p\\) and an inverse kernel \\(K_p\\) satisfying\n\\[\nK_p(t)H_p(t)=\\operatorname{sech}t.\n\\]\nAt an odd prime,\n\\[\nH_p(t)=\\cosh^2(pt)\\operatorname{sech}t,\n\\qquad\nK_p(t)=\\operatorname{sech}^2(pt).\n\\]\nAt the dyadic prime,\n\\[\nH_2(t)=\\frac{\\sinh(2t)\\sinh t}{2t^2},\n\\qquad\nK_2(t)=\\frac{4t^2}{\\sinh^2(2t)}.\n\\]\nThe coefficient convolution is identical in form, but the kernels have different integral weights. This is the correct sense in which the local theories unify.","The dyadic kernel is substantially sharper than the separate coefficient bound alone suggests. If\n\\[\n\\frac{z^2}{\\sinh^2z}\n=\n\\sum_{r\\geq0}\\beta_rz^{2r}\n\\]\nand\n\\[\n\\kappa_{2,r}=\\beta_r2^{2r}(2r)!,\n\\]\nthen\n\\[\n\\vTwo(\\kappa_{2,r})=4r-1\n\\qquad(r\\geq1).\n\\]\nConsequently, truncation after depth \\(M\\) is uniformly valid modulo\n\\[\n2^{4M+3},\n\\]\nand this exponent is optimal. The rowwise estimate records the additional binary carries in the binomial coefficient. Classical congruence theory for Euler families is extensive \\cite{CarlitzLevine1960,Young1999,Young2001,Sun2008,Chen2004}. The formulas here are stated as exact coordinate identities. Historical priority requires a separate theorem by theorem comparison.","The second expansion concerns local interpolation. The exponent of the finite unit group moves every unit mode into a principal unit neighborhood. Binomial expansion then gives a Mahler series whose depth \\(r\\) coefficient gains at least \\(Lr\\) digits when the chosen jump is the identity modulo \\(p^L\\). The same argument applies to polynomial exponential modes and to invertible matrix state systems. Mahler interpolation and the Amice transform are classical \\cite{Mahler1958,Amice1964,Robert2000}. The local unit structure and valuation tools used here are standard \\cite{Koblitz1984,Serre1979,Washington1997,Gouvea1997}.","The odd Euler source contains one nonunit base. Removing that base gives an analytic unit source and an interpolation of a depleted Euler sequence on every even character branch. The dyadic source is already analytic on \\(\\Zp\\) after a removable singularity is filled. It yields a direct analytic interpolation of \\(E_{2n}\\) on \\(\\mathbb Z_2\\). These branches are then refined by the substitution\n\\[\nx=Mq+s.\n\\]\nThe induced Mahler coefficient transform is finite and explicit. Its kernels compose exactly. At prime power radix they become sparse modulo the residue prime. Every monotone valuation ladder survives the refinement.","The remaining sections treat arithmetic lift obstructions, recurrence packets, and Chinese remainder reconstruction. The lifting obstruction is the residual defect modulo the image of the declared correction map. A recurrence packet is governed by a jump matrix \\(C^N\\). If one integral recurrence is fixed, all of its local realizations have one common source. These statements do not prove that arbitrary unrelated local objects have a global origin.","Four boundaries remain explicit. A period modulo \\(p\\) does not by itself lift to all powers of \\(p\\). The full finite difference estimate for every congruential reciprocal packet still requires control of the triangular inverse. Exact higher valuations of the affine polyphase kernel are not yet known in general. Chinese remainder reconstruction and finite adelic synchronization do not prove a global source for arbitrary local systems.\n\\subsection{Hurwitz series and finite field subgroup inversion}\\label{hist:sec:hurwitz}\n\\par\\smallskip\\noindent\\textbf{Current status.} This dated statement is superseded for the stated divisor-branch family by Theorem~\\ref{thm:arith-divisor-base}, with the $j!$ normalization and factorial valuation loss explic","Four boundaries remain explicit. A period modulo \\(p\\) does not by itself lift to all powers of \\(p\\). The full finite difference estimate for every congruential reciprocal packet still requires control of the triangular inverse. Exact higher valuations of the affine polyphase kernel are not yet known in general. Chinese remainder reconstruction and finite adelic synchronization do not prove a global source for arbitrary local systems.\n\\subsection{Hurwitz series and finite field subgroup inversion}\\label{hist:sec:hurwitz}\n\\par\\smallskip\\noindent\\textbf{Current status.} This dated statement is superseded for the stated divisor-branch family by Theorem~\\ref{thm:arith-divisor-base}, with the $j!$ normalization and factorial valuation loss explicitly retained. It makes no claim for unrestrict","own in general. Chinese remainder reconstruction and finite adelic synchronization do not prove a global source for arbitrary local systems.\n\\subsection{Hurwitz series and finite field subgroup inversion}\\label{hist:sec:hurwitz}\n\\par\\smallskip\\noindent\\textbf{Current status.} This dated statement is superseded for the stated divisor-branch family by Theorem~\\ref{thm:arith-divisor-base}, with the $j!$ normalization and factorial valuation loss explicitly retained. It makes no claim for unrestricted families.\\par","\\begin{histdefinition}[Hurwitz series ring]\\label{hist:def:hurwitz_ring}\nLet \\(R\\) be a commutative ring with identity. The Hurwitz series ring \\(\\Hcal(R)\\) is the set \\(R^{\\NN}\\) with componentwise addition and product\n\\[\n(a\\starh b)_n\n=\n\\sum_{r=0}^{n}\n\\binom nr a_r b_{n-r}.\n\\]\nIts identity is\n\\[\n\\delta=(1,0,0,\\ldots).\n\\]\n\\end{histdefinition}","When the factorials are invertible, the sequence \\(a\\) may be written as the exponential generating series\n\\[\nA(t)=\\sum_{n\\geq0}a_n\\frac{t^n}{n!},\n\\]\nand the Hurwitz product becomes ordinary multiplication. The sequence definition remains valid in every characteristic.","\\begin{histdefinition}[Exponential mode]\\label{hist:def:exponential_mode}\nFor \\(u\\in R\\), define\n\\[\n\\epsilon_u=(u^n)_{n\\geq0},\n\\]\nwhere \\(0^0=1\\).\n\\end{histdefinition}","\\begin{histproposition}[Addition law for modes]\\label{hist:prop:mode_addition}\nFor every \\(u,v\\in R\\),\n\\[\n\\boxed{\n\\epsilon_u\\starh\\epsilon_v\n=\n\\epsilon_{u+v}.\n}\n\\]\n\\end{histproposition}","\\begin{proof}\nThe \\(n\\)th coefficient is\n\\[\n\\sum_{r=0}^{n}\n\\binom nr u^rv^{n-r}\n=\n(u+v)^n\n\\]\nby the binomial theorem.\n\\end{proof}","Let \\(p\\) be prime. Write \\(\\FF_p^+\\) for the additive group of the finite field.","\\begin{histtheorem}[Finite field mode algebra]\\label{hist:thm:mode_group_algebra}\nThe map\n\\[\n\\Theta:\n\\FF_p[\\FF_p^+]\n\\longrightarrow\n\\Hcal(\\FF_p),\n\\qquad\n[u]\\longmapsto\\epsilon_u,\n\\]\nis an injective algebra homomorphism. Its image is the \\(p\\) dimensional subalgebra spanned by the modes \\(\\epsilon_u\\).\n\\end{histtheorem}","\\begin{proof}\nThe product law follows from \\cref{hist:prop:mode_addition}. To prove injectivity, suppose\n\\[\n\\sum_{u\\in\\FF_p}c_u\\epsilon_u=0.\n\\]\nLooking at the coordinates \\(n=0,1,\\ldots,p-1\\) gives a Vandermonde system with matrix\n\\[\n(u^n)_{0\\leq n\\leq p-1,\\ u\\in\\FF_p}.\n\\]\nIts determinant is the product of all differences between distinct field elements, so it is nonzero. Hence every \\(c_u\\) vanishes.\n\\end{proof}","\\begin{histproposition}[Augmentation criterion]\\label{hist:prop:augmentation_criterion}\nLet\n\\[\nx=\\sum_{u\\in\\FF_p}c_u[u]\n\\in\n\\FF_p[\\FF_p^+].\n\\]\nThen \\(x\\) is a unit if and only if\n\\[\n\\operatorname{aug}(x)\n=\n\\sum_uc_u\n\\]\nis nonzero. If \\(J\\) is the augmentation ideal, then\n\\[\nJ^p=0.\n\\]\n\\end{histproposition}","\\begin{proof}\nChoose a generator \\(g\\) of the cyclic additive group. Then\n\\[\n\\FF_p[\\FF_p^+]\n\\cong\n\\FF_p[y]/(y^p-1)\n=\n\\FF_p[y]/(y-1)^p.\n\\]\nThe augmentation ideal is generated by \\(y-1\\), so its \\(p\\)th power is zero. Every element has the form \\(a+n\\) with \\(a\\in\\FF_p\\) and \\(n\\in J\\). It is invertible exactly when \\(a\\neq0\\), which is the augmentation condition.\n\\end{proof}","\\begin{histcorollary}[Finite inverse formula]\\label{hist:cor:finite_inverse_formula}\nIf \\(x\\) has augmentation one and \\(n=x-1\\), then\n\\[\n\\boxed{\nx^{-1}\n=\n\\sum_{r=0}^{p-1}(-n)^r.\n}\n\\]\n\\end{histcorollary}","\\begin{proof}\nBy \\cref{hist:prop:augmentation_criterion}, \\(n^p=0\\). The finite geometric sum gives the inverse.\n\\end{proof}","Fix a divisor \\(m\\) of \\(p-1\\). Let\n\\[\nH_m\n=\n\\{\\zeta\\in\\FF_p^\\times:\\zeta^m=1\\}.\n\\]\nThis group has order \\(m\\).","\\begin{histdefinition}[Multiple selector]\\label{hist:def:multiple_selector}\nDefine \\(a^{(m)}\\in\\Hcal(\\FF_p)\\) by\n\\[\na_n^{(m)}\n=\n\\begin{cases}\n1,&m\\mid n,","\\begin{histdefinition}[Multiple selector]\\label{hist:def:multiple_selector}\nDefine \\(a^{(m)}\\in\\Hcal(\\FF_p)\\) by\n\\[\na_n^{(m)}\n=\n\\begin{cases}\n1,&m\\mid n,\\\\\n0,&m\\nmid n.\n\\end{cases}\n\\]\n\\end{histdefinition}","\\begin{histlemma}[Subgroup average]\\label{hist:lem:subgroup_average}\nThe multiple selector has the mode expansion\n\\[\n\\boxed{\na^{(m)}\n=\n\\frac1m\n\\sum_{\\zeta\\in H_m}\n\\epsilon_\\zeta.\n}\n\\]\n\\end{histlemma}","\\begin{proof}\nThe \\(n\\)th coefficient on the right is\n\\[\n\\frac1m\n\\sum_{\\zeta\\in H_m}\\zeta^n.\n\\]\nThe character sum is one when \\(m\\mid n\\) and zero otherwise.\n\\end{proof}","Since the subgroup average has augmentation one, it has a unique Hurwitz inverse. Write\n\\[\nb^{(m,p)}\n=\n(a^{(m)})^{-1}.\n\\]","\\begin{histtheorem}[Support and positive degree period]\\label{hist:thm:support_period}\nThe inverse \\(b^{(m,p)}\\) has the following properties.\n\\begin{enumerate}\n\\item \\(b_n^{(m,p)}=0\\) unless \\(m\\mid n\\).\n\\item For every positive \\(n\\),\n\\[\n\\boxed{\nb_{n+p-1}^{(m,p)}\n=\nb_n^{(m,p)}.\n}\n\\]\n\\end{enumerate}\n\\end{histtheorem}","\\begin{proof}\nMultiplication by an element \\(\\eta\\in H_m\\) gives an automorphism of the additive group and therefore an automorphism\n\\[\n\\epsilon_u\\longmapsto\\epsilon_{\\eta u}\n\\]\nof the mode algebra. The selector \\(a^{(m)}\\) is fixed by every such automorphism, so its unique inverse is fixed as well.","Write\n\\[\nb^{(m,p)}\n=\n\\sum_{u\\in\\FF_p}c_u\\epsilon_u.\n\\]\nInvariance gives \\(c_{\\eta u}=c_u\\). After the change of variable \\(u\\mapsto\\eta u\\),\n\\[\nb_n^{(m,p)}\n=\n\\eta^n b_n^{(m,p)}\n\\]\nfor every \\(\\eta\\in H_m\\). If \\(m\\nmid n\\), some \\(\\eta\\) satisfies \\(\\eta^n\\neq1\\), so the coefficient vanishes.","For positive \\(n\\), the zero mode contributes nothing. Every nonzero mode satisfies \\(u^{p-1}=1\\). Hence\n\\[\nb_{n+p-1}^{(m,p)}\n=\n\\sum_{u\\neq0}c_uu^{n+p-1}\n=\n\\sum_{u\\neq0}c_uu^n\n=\nb_n^{(m,p)}.\n\\]\n\\end{proof}","\\begin{histdefinition}[Compressed inverse]\\label{hist:def:compressed_inverse}\nSet\n\\[\ne_n^{(m,p)}\n=\nb_{mn}^{(m,p)}.\n\\]\n\\end{histdefinition}","\\begin{histproposition}[Compressed inverse recurrence]\\label{hist:prop:compressed_inverse_recurrence}\nThe compressed coefficients satisfy\n\\[\n\\boxed{\n\\sum_{q=0}^{n}\n\\binom{mn}{mq}\ne_q^{(m,p)}\n=\n\\delta_{n0}.\n}\n\\]\nIn particular, \\(e_0^{(m,p)}=1\\), and the positive coefficients are determined recursively because the diagonal is one.\n\\end{histproposition}","\\begin{proof}\nEvaluate the Hurwitz identity\n\\[\na^{(m)}\\starh b^{(m,p)}=\\delta\n\\]\nat degree \\(mn\\). Both factors vanish outside degrees divisible by \\(m\\). Reindexing the surviving terms gives the displayed recurrence.\n\\end{proof}","\\begin{histcorollary}[Uniform compressed period]\\label{hist:cor:compressed_period}\nLet\n\\[\nh=\\frac{p-1}{m}.\n\\]\nThen\n\\[\n\\boxed{\ne_{n+h}^{(m,p)}\n=\ne_n^{(m,p)}\n\\qquad(n\\geq1).\n}\n\\]\n\\end{histcorollary}","\\begin{proof}\nSince \\(m(n+h)=mn+p-1\\), the conclusion follows from \\cref{hist:thm:support_period}.\n\\end{proof}","\\begin{histcorollary}[Rational ordinary generating function]\\label{hist:cor:periodic_ogf}\nIn \\(\\FF_p[[z]]\\),\n\\[\n\\boxed{\n\\sum_{n\\geq0}e_n^{(m,p)}z^n\n=\n1+\n\\frac{\n\\sum_{r=1}^{h}e_r^{(m,p)}z^r\n}{1-z^h}.\n}\n\\]\n\\end{histcorollary}","\\begin{proof}\nSeparate the constant coefficient and group the positive indices into residue classes modulo \\(h\\).\n\\end{proof}","\\begin{histtheorem}[Exponential form and minimal period]\\label{hist:thm:minimal_period}\nLet\n\\[\nG_m=(\\FF_p^\\times)^m.\n\\]\nThis cyclic group has order \\(h\\). There are uniquely determined coefficients \\(d_v\\in\\FF_p\\), indexed by \\(v\\in G_m\\), such that\n\\[\n\\boxed{\ne_n^{(m,p)}\n=\n\\sum_{v\\in G_m}d_vv^n\n\\qquad(n\\geq1).\n}\n\\]\nThe minimal positive period is the least common multiple of the orders of those \\(v\\) for which \\(d_v\\neq0\\). In particular, it divides \\(h\\).\n\\end{histtheorem}","em}[Exponential form and minimal period]\\label{hist:thm:minimal_period}\nLet\n\\[\nG_m=(\\FF_p^\\times)^m.\n\\]\nThis cyclic group has order \\(h\\). There are uniquely determined coefficients \\(d_v\\in\\FF_p\\), indexed by \\(v\\in G_m\\), such that\n\\[\n\\boxed{\ne_n^{(m,p)}\n=\n\\sum_{v\\in G_m}d_vv^n\n\\qquad(n\\geq1).\n}\n\\]\nThe minimal positive period is the least common multiple of the orders of those \\(v\\) for which \\(d_v\\neq0\\). In particular, it divides \\(h\\).\n\\end{histtheorem}","\\label{hist:thm:minimal_period}\nLet\n\\[\nG_m=(\\FF_p^\\times)^m.\n\\]\nThis cyclic group has order \\(h\\). There are uniquely determined coefficients \\(d_v\\in\\FF_p\\), indexed by \\(v\\in G_m\\), such that\n\\[\n\\boxed{\ne_n^{(m,p)}\n=\n\\sum_{v\\in G_m}d_vv^n\n\\qquad(n\\geq1).\n}\n\\]\nThe minimal positive period is the least common multiple of the orders of those \\(v\\) for which \\(d_v\\neq0\\). In particular, it divides \\(h\\).\n\\end{histtheorem}","\\begin{proof}\nThe coefficients \\(c_u\\) in the proof of \\cref{hist:thm:support_period} are constant on the cosets of \\(H_m\\). For \\(u\\neq0\\), the value \\(u^m\\) depends only on the coset, and every element of \\(G_m\\) arises. Grouping the modes by cosets gives the displayed expansion.","Distinct elements of \\(G_m\\) define linearly independent functions on one full period, by the Vandermonde determinant. Therefore a positive integer \\(r\\) is a period exactly when \\(v^r=1\\) for every active mode. The least such \\(r\\) is the stated least common multiple.\n\\end{proof}","\\begin{histexample}[A compressed inverse modulo seven]\\label{hist:ex:compressed_mod_seven}\n\\par\\noindent\\textbf{Editorial numerical correction.} The positive-index standard-window tail printed below is erroneous. The defining recurrence gives $e_1=-1\\equiv6$, $e_2=5$ and $e_3=-61\\equiv2\\pmod7$, so the correct positive tail is $6,5,2,6,5,2,\\ldots$. The period-three conclusion remains correct. The original line is retained only as a corrected historical record.\\par\\medskip\nTake \\(p=7\\) and \\(m=2\\), so \\(h=3\\). The recurrence gives\n\\[\ne_n^{(2,7)}\n=\n1,6,4,1,6,4,\\ldots\n\\qquad(n\\geq1).\n\\]\nThe positive tail has exact period three.\n\\end{histexample}","_seven}\n\\par\\noindent\\textbf{Editorial numerical correction.} The positive-index standard-window tail printed below is erroneous. The defining recurrence gives $e_1=-1\\equiv6$, $e_2=5$ and $e_3=-61\\equiv2\\pmod7$, so the correct positive tail is $6,5,2,6,5,2,\\ldots$. The period-three conclusion remains correct. The original line is retained only as a corrected historical record.\\par\\medskip\nTake \\(p=7\\) and \\(m=2\\), so \\(h=3\\). The recurrence gives\n\\[\ne_n^{(2,7)}\n=\n1,6,4,1,6,4,\\ldots\n\\qquad(n\\geq1).\n\\]\nThe positive tail has exact period three.\n\\end{histexample}","\\noindent\\textbf{Editorial numerical correction.} The positive-index standard-window tail printed below is erroneous. The defining recurrence gives $e_1=-1\\equiv6$, $e_2=5$ and $e_3=-61\\equiv2\\pmod7$, so the correct positive tail is $6,5,2,6,5,2,\\ldots$. The period-three conclusion remains correct. The original line is retained only as a corrected historical record.\\par\\medskip\nTake \\(p=7\\) and \\(m=2\\), so \\(h=3\\). The recurrence gives\n\\[\ne_n^{(2,7)}\n=\n1,6,4,1,6,4,\\ldots\n\\qquad(n\\geq1).\n\\]\nThe positive tail has exact period three.\n\\end{histexample}","xtbf{Editorial numerical correction.} The positive-index standard-window tail printed below is erroneous. The defining recurrence gives $e_1=-1\\equiv6$, $e_2=5$ and $e_3=-61\\equiv2\\pmod7$, so the correct positive tail is $6,5,2,6,5,2,\\ldots$. The period-three conclusion remains correct. The original line is retained only as a corrected historical record.\\par\\medskip\nTake \\(p=7\\) and \\(m=2\\), so \\(h=3\\). The recurrence gives\n\\[\ne_n^{(2,7)}\n=\n1,6,4,1,6,4,\\ldots\n\\qquad(n\\geq1).\n\\]\nThe positive tail has exact period three.\n\\end{histexample}","ction.} The positive-index standard-window tail printed below is erroneous. The defining recurrence gives $e_1=-1\\equiv6$, $e_2=5$ and $e_3=-61\\equiv2\\pmod7$, so the correct positive tail is $6,5,2,6,5,2,\\ldots$. The period-three conclusion remains correct. The original line is retained only as a corrected historical record.\\par\\medskip\nTake \\(p=7\\) and \\(m=2\\), so \\(h=3\\). The recurrence gives\n\\[\ne_n^{(2,7)}\n=\n1,6,4,1,6,4,\\ldots\n\\qquad(n\\geq1).\n\\]\nThe positive tail has exact period three.\n\\end{histexample}","The examples later in the paper use different kernels, but their coefficient algebra has one common form. This section records that form once. Let \\(R\\) be a commutative algebra over \\(\\QQ\\), and take all coefficients in \\(R\\).","Let\n\\[\nH(t)=\\sum_{r\\geq0}h_r\\frac{t^r}{r!},\n\\qquad\nh_0=1.\n\\]\nThere is a unique inverse\n\\[\nK(t)=H(t)^{-1}\n=\\sum_{r\\geq0}k_r\\frac{t^r}{r!}.\n\\]","\\begin{histtheorem}[Exact coefficient transport]\\label{hist:thm:exact_coefficient_transport}\nLet\n\\[\nA(t)=\\sum_{n\\geq0}A_n\\frac{t^n}{n!},\n\\qquad\nS(t)=H(t)A(t).\n\\]\nThen\n\\[\n\\boxed{\nS_n\n=\n\\sum_{r=0}^{n}\n\\binom nr h_rA_{n-r}.\n}\n\\]\nThe inverse coordinate formula is\n\\[\n\\boxed{\nA_n\n=\n\\sum_{r=0}^{n}\n\\binom nr k_rS_{n-r}.\n}\n\\]\nBoth transforms are lower triangular with diagonal one.\n\\end{histtheorem}","\\begin{proof}\nThe first identity is the coefficient rule for the product of exponential generating functions. Multiply by \\(K(t)\\) and use \\(K(t)H(t)=1\\) for the inverse identity. The diagonal coefficient in each transform is the constant coefficient of its kernel, which is one.\n\\end{proof}","\\begin{histcorollary}[Composition of coefficient transports]\\label{hist:cor:transport_composition}\nIf \\(U_H\\) and \\(U_G\\) are the triangular transforms associated with unit kernels \\(H\\) and \\(G\\), then\n\\[\n\\boxed{\nU_HU_G=U_{HG}.\n}\n\\]\nIn particular, the inverse of \\(U_H\\) is \\(U_{H^{-1}}\\).\n\\end{histcorollary}","\\begin{proof}\nMultiplication by \\(G\\), followed by multiplication by \\(H\\), is multiplication by \\(HG\\).\n\\end{proof}","\\begin{histtheorem}[Filtered triangular reconstruction]\\label{hist:thm:filtered_triangular_reconstruction}\nLet \\(p\\) be prime. Suppose\n\\[\nA_n\n=\n\\sum_{r=0}^{n}\n\\binom nr\nc_rp^{w(r)}S_{n-r},\n\\]\nwhere \\(c_r\\in\\Zp\\), \\(S_n\\in\\Zp\\), and\n\\[\nw(r)\\in\\NN.\n\\]\nFor every precision \\(L\\geq1\\),\n\\[\n\\boxed{\nA_n\n\\equiv\n\\sum_{\\substack{0\\leq r\\leq n","\\begin{histtheorem}[Filtered triangular reconstruction]\\label{hist:thm:filtered_triangular_reconstruction}\nLet \\(p\\) be prime. Suppose\n\\[\nA_n\n=\n\\sum_{r=0}^{n}\n\\binom nr\nc_rp^{w(r)}S_{n-r},\n\\]\nwhere \\(c_r\\in\\Zp\\), \\(S_n\\in\\Zp\\), and\n\\[\nw(r)\\in\\NN.\n\\]\nFor every precision \\(L\\geq1\\),\n\\[\n\\boxed{\nA_n\n\\equiv\n\\sum_{\\substack{0\\leq r\\leq n\\\\w(r)<L}}\n\\binom nr\nc_rp^{w(r)}S_{n-r}\n\\pmod {p^L}.\n}\n\\]\nWhen the omitted set is nonempty, the tail has valuation at least\n\\[\n\\boxed{\n\\min_{\\substack{0\\leq r\\leq n","w(r)<L}}\n\\binom nr\nc_rp^{w(r)}S_{n-r}\n\\pmod {p^L}.\n}\n\\]\nWhen the omitted set is nonempty, the tail has valuation at least\n\\[\n\\boxed{\n\\min_{\\substack{0\\leq r\\leq n\\\\w(r)\\geq L}}\n\\left(\nw(r)+\\vp\\binom nr+\\vp(S_{n-r})\n\\right).\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nEvery omitted summand belongs to \\(p^L\\Zp\\). The sharper statement follows by taking the minimum of the exact lower bounds for the summands and applying the nonarchimedean triangle inequality.\n\\end{proof}","For integers \\(N\\geq1\\) and \\(0\\leq j<N\\), define\n\\[\n\\Ecal_{N,j}(t)\n=\n\\sum_{n\\geq0}\n\\frac{t^{Nn+j}}{(Nn+j)!}.\n\\]\nThe forced factor \\(t^j\\) may be removed without changing the residue address.","\\begin{histdefinition}[Congruential reciprocal packet]\\label{hist:def:congruential_packet}\nDefine\n\\[\n\\boxed{\n\\Ccal_{N,j}(t)\n=\n\\frac{t^j/j!}{\\Ecal_{N,j}(t)}\n=\n\\sum_{n\\geq0}\nC_{N,j}(n)\n\\frac{t^{Nn}}{(Nn)!}.\n}\n\\]\n\\end{histdefinition}","\\begin{histtheorem}[Triangular inverse system]\\label{hist:thm:triangular_inverse}\nThe coefficients satisfy \\(C_{N,j}(0)=1\\) and\n\\[\n\\boxed{\n\\sum_{q=0}^{n}\n\\binom{Nn+j}{Nq}\nC_{N,j}(q)\n=\n\\delta_{n0}.\n}\n\\]\nFor \\(n\\geq1\\),\n\\[\n\\boxed{\nC_{N,j}(n)\n=\n-\n\\frac{\n\\sum_{q=0}^{n-1}\n\\binom{Nn+j}{Nq}C_{N,j}(q)\n}{\n\\binom{Nn+j}{Nn}\n}.\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nMultiply \\(\\Ccal_{N,j}(t)\\) by \\(\\Ecal_{N,j}(t)\\) and compare the coefficient of \\(t^{Nn+j}/(Nn+j)!\\). The last summand is the diagonal term, which is nonzero over \\(\\QQ\\).\n\\end{proof}","\\begin{histproposition}[A denominator bound]\\label{hist:prop:denominator_bound}\nLet\n\\[\nD_{N,j}(n)\n=\n\\prod_{r=1}^{n}\n\\binom{Nr+j}{Nr},\n\\qquad\nD_{N,j}(0)=1.\n\\]\nThen\n\\[\n\\boxed{\nD_{N,j}(n)C_{N,j}(n)\n\\in\n\\ZZ.\n}\n\\]\nFor every prime \\(p\\),\n\\[\n\\vp(C_{N,j}(n))\n\\geq\n-\n\\sum_{r=1}^{n}\n\\vp\\binom{Nr+j}{Nr}.\n\\]\n\\end{histproposition}","\\begin{proof}\nUse induction on \\(n\\) in the recurrence of \\cref{hist:thm:triangular_inverse}. Every earlier denominator divides \\(D_{N,j}(n-1)\\), while the new diagonal denominator is the final factor of \\(D_{N,j}(n)\\). The valuation statement follows.\n\\end{proof}","\\begin{histexample}[Euler and Bernoulli boundaries]\\label{hist:ex:euler_bernoulli_boundaries}\nFor \\((N,j)=(2,0)\\),\n\\[\n\\Ccal_{2,0}(t)=\\operatorname{sech}t,\n\\]\nso\n\\[\nC_{2,0}(n)=E_{2n}.\n\\]\nFor \\((N,j)=(2,1)\\),\n\\[\n\\Ccal_{2,1}(t)=\\frac{t}{\\sinh t},\n\\]\nand\n\\[\nC_{2,1}(n)\n=\n2\\left(1-2^{2n-1}\\right)B_{2n}.\n\\]\nThe usual convention \\(B_1=-1/2\\) is used. Euler and Bernoulli number conventions are reviewed in \\cite{Comtet1974,Stanley2012}.\n\\end{histexample}","Fix a prime \\(p\\), an integer \\(m\\geq1\\), and an address \\(0\\leq j<p\\). Put\n\\[\nN=mp.\n\\]\nFor arbitrary \\(m\\), let \\(\\widetilde e_n^{(m,p)}\\in\\FF_p\\) be the unique sequence satisfying\n\\[\n\\boxed{\n\\sum_{q=0}^{n}\n\\binom{mn}{mq}\n\\widetilde e_q^{(m,p)}\n=\n\\delta_{n0}.\n}\n\\]\nWhen \\(m\\mid p-1\\), this sequence agrees with the compressed Hurwitz inverse in \\cref{hist:prop:compressed_inverse_recurrence}.","\\begin{histstandardfact}[Lucas congruence]\\label{hist:fact:lucas}\nIf \\(a=a_0+a_1p+\\cdots\\) and \\(b=b_0+b_1p+\\cdots\\) are base \\(p\\) expansions, then\n\\[\n\\binom ab\n\\equiv\n\\prod_r\\binom{a_r}{b_r}\n\\pmod p.\n\\]\nThis is the classical theorem of Lucas \\cite{Lucas1878a}.\n\\end{histstandardfact}","\\begin{histtheorem}[Standard window reduction]\\label{hist:thm:standard_window_reduction}\nFor \\(0\\leq q\\leq n\\),\n\\[\n\\boxed{\n\\binom{mpn+j}{mpq}\n\\equiv\n\\binom{mn}{mq}\n\\pmod p.\n}\n\\]\nThe diagonal coefficient is a unit modulo \\(p\\):\n\\[\n\\binom{mpn+j}{mpn}\n\\equiv1\n\\pmod p.\n\\]\n\\end{histtheorem}","\\begin{proof}\nWrite\n\\[\nmpn+j=p(mn)+j,\n\\qquad\nmpq=p(mq).\n\\]\nLucas reduction separates the lowest base \\(p\\) digit and gives\n\\[\n\\binom{mpn+j}{mpq}\n\\equiv\n\\binom j0\n\\binom{mn}{mq}\n\\pmod p.\n\\]\nThe diagonal case is obtained by setting \\(q=n\\).\n\\end{proof}","\\begin{histcorollary}[Standard window integrality]\\label{hist:cor:standard_integrality}\nEvery coefficient \\(C_{mp,j}(n)\\) belongs to \\(\\Zp\\).\n\\end{histcorollary}","\\begin{proof}\nThe recurrence in \\cref{hist:thm:triangular_inverse} starts with a \\(p\\) integral constant term. The binomial coefficients are integers, and every diagonal term is a \\(p\\) adic unit by \\cref{hist:thm:standard_window_reduction}. Induction gives the result.\n\\end{proof}","\\begin{histcorollary}[Address independence modulo the prime]\\label{hist:cor:address_independence}\nLet \\(\\widetilde e_n^{(m,p)}\\) be the reduced standard inverse above. Then\n\\[\n\\boxed{\nC_{mp,j}(n)\n\\equiv\n\\widetilde e_n^{(m,p)}\n\\pmod p\n}\n\\]\nfor every \\(0\\leq j<p\\).\n\\end{histcorollary}","\\begin{proof}\nModulo \\(p\\), the recurrence in \\cref{hist:thm:triangular_inverse} becomes\n\\[\n\\sum_{q=0}^{n}\n\\binom{mn}{mq}\nC_{mp,j}(q)\n=\n\\delta_{n0}.\n\\]\nThis is the defining recurrence for \\(\\widetilde e_n^{(m,p)}\\). The diagonal is one, so the solution is unique.\n\\end{proof}","\\begin{histtheorem}[Uniform standard window period]\\label{hist:thm:standard_window_period}\nAssume that \\(p\\) is odd and \\(m\\mid p-1\\). Put\n\\[\nh=\\frac{p-1}{m}.\n\\]\nThen for every standard window address \\(0\\leq j<p\\),\n\\[\n\\boxed{\nC_{mp,j}(n+h)\n\\equiv\nC_{mp,j}(n)\n\\pmod p\n\\qquad(n\\geq1).\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nBy \\cref{hist:cor:address_independence}, every address reduces to the same reduced standard inverse. Under the divisibility assumption, that inverse agrees with the compressed Hurwitz inverse by \\cref{hist:prop:compressed_inverse_recurrence}. Its positive tail has period \\(h\\) by \\cref{hist:cor:compressed_period}.\n\\end{proof}","\\begin{histcorollary}[Rational residual generating function]\\label{hist:cor:standard_rational_ogf}\nUnder the hypotheses of \\cref{hist:thm:standard_window_period}, for every standard address \\(j\\),\n\\[\n\\sum_{n\\geq0}\n\\left(C_{mp,j}(n)\\bmod p\\right)z^n\n=\n1+\n\\frac{\n\\sum_{r=1}^{h}\\widetilde e_r^{(m,p)}z^r\n}{1-z^h}.\n\\]\n\\end{histcorollary}","\\begin{histtheorem}[Fourier invariance of the standard window operator]\\label{hist:thm:standard_window_fourier}\nLet \\(p\\) be odd, let \\(m\\mid p-1\\), and let \\(h=(p-1)/m\\). Define\n\\[\n(T_{m,p}f)_n\n=\n\\sum_{q=0}^{n}\\binom{mn}{mq}f_q\n\\quad\\text{in }\\FF_p.\n\\]\nLet \\(V_h\\) be the vector space of sequences satisfying\n\\[\nf_{n+h}=f_n\n\\qquad(n\\geq1).\n\\]\nThen\n\\[\n\\boxed{T_{m,p}(V_h)=V_h.}\n\\]\nThe restriction of \\(T_{m,p}\\) to \\(V_h\\) is an automorphism.\n\\end{histtheorem}","egin{histtheorem}[Fourier invariance of the standard window operator]\\label{hist:thm:standard_window_fourier}\nLet \\(p\\) be odd, let \\(m\\mid p-1\\), and let \\(h=(p-1)/m\\). Define\n\\[\n(T_{m,p}f)_n\n=\n\\sum_{q=0}^{n}\\binom{mn}{mq}f_q\n\\quad\\text{in }\\FF_p.\n\\]\nLet \\(V_h\\) be the vector space of sequences satisfying\n\\[\nf_{n+h}=f_n\n\\qquad(n\\geq1).\n\\]\nThen\n\\[\n\\boxed{T_{m,p}(V_h)=V_h.}\n\\]\nThe restriction of \\(T_{m,p}\\) to \\(V_h\\) is an automorphism.\n\\end{histtheorem}","m}[Fourier invariance of the standard window operator]\\label{hist:thm:standard_window_fourier}\nLet \\(p\\) be odd, let \\(m\\mid p-1\\), and let \\(h=(p-1)/m\\). Define\n\\[\n(T_{m,p}f)_n\n=\n\\sum_{q=0}^{n}\\binom{mn}{mq}f_q\n\\quad\\text{in }\\FF_p.\n\\]\nLet \\(V_h\\) be the vector space of sequences satisfying\n\\[\nf_{n+h}=f_n\n\\qquad(n\\geq1).\n\\]\nThen\n\\[\n\\boxed{T_{m,p}(V_h)=V_h.}\n\\]\nThe restriction of \\(T_{m,p}\\) to \\(V_h\\) is an automorphism.\n\\end{histtheorem}","\\begin{proof}\nChoose \\(d_0=f_h\\) and \\(d_n=f_n\\) for \\(n\\geq1\\). Then \\(d\\) is fully \\(h\\) periodic and \\(f=d+c\\delta_0\\) for \\(c=f_0-f_h\\). Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower tr","\\begin{proof}\nChoose \\(d_0=f_h\\) and \\(d_n=f_n\\) for \\(n\\geq1\\). Then \\(d\\) is fully \\(h\\) periodic and \\(f=d+c\\delta_0\\) for \\(c=f_0-f_h\\). Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on ever","\\begin{proof}\nChoose \\(d_0=f_h\\) and \\(d_n=f_n\\) for \\(n\\geq1\\). Then \\(d\\) is fully \\(h\\) periodic and \\(f=d+c\\delta_0\\) for \\(c=f_0-f_h\\). Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initia","\\begin{proof}\nChoose \\(d_0=f_h\\) and \\(d_n=f_n\\) for \\(n\\geq1\\). Then \\(d\\) is fully \\(h\\) periodic and \\(f=d+c\\delta_0\\) for \\(c=f_0-f_h\\). Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. I","\\begin{proof}\nChoose \\(d_0=f_h\\) and \\(d_n=f_n\\) for \\(n\\geq1\\). Then \\(d\\) is fully \\(h\\) periodic and \\(f=d+c\\delta_0\\) for \\(c=f_0-f_h\\). Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction","\\begin{proof}\nChoose \\(d_0=f_h\\) and \\(d_n=f_n\\) for \\(n\\geq1\\). Then \\(d\\) is fully \\(h\\) periodic and \\(f=d+c\\delta_0\\) for \\(c=f_0-f_h\\). Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space","\\begin{proof}\nChoose \\(d_0=f_h\\) and \\(d_n=f_n\\) for \\(n\\geq1\\). Then \\(d\\) is fully \\(h\\) periodic and \\(f=d+c\\delta_0\\) for \\(c=f_0-f_h\\). Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f","\\begin{proof}\nChoose \\(d_0=f_h\\) and \\(d_n=f_n\\) for \\(n\\geq1\\). Then \\(d\\) is fully \\(h\\) periodic and \\(f=d+c\\delta_0\\) for \\(c=f_0-f_h\\). Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f_0,f_1,\\ldots,f_h\\) is inje","\\begin{proof}\nChoose \\(d_0=f_h\\) and \\(d_n=f_n\\) for \\(n\\geq1\\). Then \\(d\\) is fully \\(h\\) periodic and \\(f=d+c\\delta_0\\) for \\(c=f_0-f_h\\). Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f_0,f_1,\\ldots,f_h\\) is injective and hence bijective.\n\\end{proof}","s fully \\(h\\) periodic and \\(f=d+c\\delta_0\\) for \\(c=f_0-f_h\\). Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f_0,f_1,\\ldots,f_h\\) is injective and hence bijective.\n\\end{proof}",". Since \\(T_{m,p}\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f_0,f_1,\\ldots,f_h\\) is injective and hence bijective.\n\\end{proof}","\\delta_0\\) is the constant sequence one, it is enough to treat \\(d\\). Expand\n\\[\nd_q=\\sum_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f_0,f_1,\\ldots,f_h\\) is injective and hence bijective.\n\\end{proof}","um_{\\chi\\in\\mu_h}c_\\chi\\chi^q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f_0,f_1,\\ldots,f_h\\) is injective and hence bijective.\n\\end{proof}","q.\n\\]\nThe \\(m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f_0,f_1,\\ldots,f_h\\) is injective and hence bijective.\n\\end{proof}","m\\)th power map on \\(\\FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f_0,f_1,\\ldots,f_h\\) is injective and hence bijective.\n\\end{proof}","FF_p^\\times\\) has image \\(\\mu_h\\). Choose \\(y\\) with \\(y^m=\\chi\\). The root filter gives\n\\[\nT_{m,p}(\\chi^\\bullet)_n\n=\n\\frac1m\n\\sum_{\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f_0,f_1,\\ldots,f_h\\) is injective and hence bijective.\n\\end{proof}","\\zeta\\in\\mu_m}\n(1+\\zeta y)^{mn}.\n\\]\nIf \\(1+\\zeta y=0\\), the corresponding term vanishes for positive \\(n\\). Otherwise\n\\[\n\\left((1+\\zeta y)^m\\right)^h\n=\n(1+\\zeta y)^{p-1}\n=1.\n\\]\nEvery summand is therefore \\(h\\) periodic for positive \\(n\\). Thus \\(T_{m,p}\\) preserves \\(V_h\\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \\(f_0,f_1,\\ldots,f_h\\) is injective and hence bijective.\n\\end{proof}","\\begin{histcorollary}[A second proof of the standard window period]\\label{hist:cor:standard_window_fourier_period}\nThe reduced inverse \\(e^{(m,p)}\\) and every standard address satisfy the period in \\cref{hist:thm:standard_window_period}.\n\\end{histcorollary}","\\begin{proof}\nThe sequence \\(\\delta=(1,0,0,\\ldots)\\) belongs to \\(V_h\\). Its inverse image under the automorphism in \\cref{hist:thm:standard_window_fourier} also belongs to \\(V_h\\). Address independence modulo \\(p\\) gives the statement for every standard address.\n\\end{proof}","\\begin{figure}[t]\n\\centering\n\\includegraphics[width=.95\\linewidth,height=3.15in,keepaspectratio]{../../figures/reconstructed/standard_fourier.pdf}\n\\caption{New reconstruction. Finite field Fourier transfer for a standard window. Each character becomes a finite sum of positive index modes with period dividing \\(h\\).}\n\\label{hist:fig:standard_fourier}\n\\end{figure}\n\\begin{figure}[t]\n\\centering\n\\includegraphics[width=.95\\linewidth,height=3.15in,keepaspectratio]{../../figures/reconstructed/standard_matrix.pdf}\n\\caption{New reconstruction. The standard window matrix for $p=7$, $m=2$, and $0\\le j<7$. Reduction removes the address and retains diagonal one.}\n\\label{hist:fig:standard_matrix}\n\\end{figure}","on. Finite field Fourier transfer for a standard window. Each character becomes a finite sum of positive index modes with period dividing \\(h\\).}\n\\label{hist:fig:standard_fourier}\n\\end{figure}\n\\begin{figure}[t]\n\\centering\n\\includegraphics[width=.95\\linewidth,height=3.15in,keepaspectratio]{../../figures/reconstructed/standard_matrix.pdf}\n\\caption{New reconstruction. The standard window matrix for $p=7$, $m=2$, and $0\\le j<7$. Reduction removes the address and retains diagonal one.}\n\\label{hist:fig:standard_matrix}\n\\end{figure}","ite field Fourier transfer for a standard window. Each character becomes a finite sum of positive index modes with period dividing \\(h\\).}\n\\label{hist:fig:standard_fourier}\n\\end{figure}\n\\begin{figure}[t]\n\\centering\n\\includegraphics[width=.95\\linewidth,height=3.15in,keepaspectratio]{../../figures/reconstructed/standard_matrix.pdf}\n\\caption{New reconstruction. The standard window matrix for $p=7$, $m=2$, and $0\\le j<7$. Reduction removes the address and retains diagonal one.}\n\\label{hist:fig:standard_matrix}\n\\end{figure}","ourier transfer for a standard window. Each character becomes a finite sum of positive index modes with period dividing \\(h\\).}\n\\label{hist:fig:standard_fourier}\n\\end{figure}\n\\begin{figure}[t]\n\\centering\n\\includegraphics[width=.95\\linewidth,height=3.15in,keepaspectratio]{../../figures/reconstructed/standard_matrix.pdf}\n\\caption{New reconstruction. The standard window matrix for $p=7$, $m=2$, and $0\\le j<7$. Reduction removes the address and retains diagonal one.}\n\\label{hist:fig:standard_matrix}\n\\end{figure}","\\begin{histremark}\\label{hist:rem:period_boundary}\nThe theorem gives a period modulo \\(p\\). A period modulo higher powers requires additional control of the triangular inverse and cannot be inferred from the finite field argument alone.","Let \\(q\\) be a positive odd integer. Define\n\\[\nS_n(q)\n=\n\\sum_{a=0}^{q-1}\n(-1)^a(2a+1)^n.\n\\]\nLet the signed Euler numbers and Euler zigzag numbers be defined by\n\\[\n\\operatorname{sech}t\n=\n\\sum_{n\\geq0}E_n\\frac{t^n}{n!}\n\\]\nand\n\\[\n\\sec t+\\tan t\n=\n\\sum_{n\\geq0}Z_n\\frac{t^n}{n!}.\n\\]","\\begin{histlemma}[Odd source generating function]\\label{hist:lem:odd_source_egf}\nFor every positive odd \\(q\\),\n\\[\n\\boxed{\n\\sum_{n\\geq0}S_n(q)\\frac{t^n}{n!}\n=\n\\e^{qt}\\cosh(qt)\\operatorname{sech}t.\n}\n\\]\nIts even part is\n\\[\n\\boxed{\n\\sum_{n\\geq0}S_{2n}(q)\\frac{t^{2n}}{(2n)!}\n=\n\\cosh^2(qt)\\operatorname{sech}t.\n}\n\\]\n\\end{histlemma}","\\begin{proof}\nThe finite geometric sum gives\n\\[\n\\sum_{a=0}^{q-1}(-1)^a\\e^{(2a+1)t}\n=\n\\e^t\\frac{1+\\e^{2qt}}{1+\\e^{2t}}\n=\n\\e^{qt}\\cosh(qt)\\operatorname{sech}t.\n\\]\nAveraging at \\(t\\) and \\(-t\\) gives the even part.\n\\end{proof}","\\begin{histlemma}[Zigzag inverse kernel]\\label{hist:lem:local_zigzag_kernel}\n\\[\n\\boxed{\n\\operatorname{sech}^2t\n=\n\\sum_{r\\geq0}\n(-1)^rZ_{2r+1}\n\\frac{t^{2r}}{(2r)!}.\n}\n\\]\n\\end{histlemma}","\\begin{proof}\nThe identity \\(\\tanh t=-i\\tan(it)\\) gives the odd zigzag expansion of \\(\\tanh t\\). Differentiate.\n\\end{proof}","\\begin{histtheorem}[Odd modulus Euler coordinate pair]\\label{hist:thm:odd_coordinate_pair}\nFor every positive odd \\(q\\) and every even \\(n\\geq0\\),\n\\[\n\\boxed{\nE_n\n=\n\\sum_{r=0}^{n/2}\n(-1)^rZ_{2r+1}\n\\binom n{2r}\nq^{2r}S_{n-2r}(q).\n}\n\\]\nThe inverse transform is\n\\[\n\\boxed{\nS_n(q)\n=\nE_n\n+\n\\sum_{r=1}^{n/2}\n2^{2r-1}\n\\binom n{2r}\nq^{2r}E_{n-2r}.\n}\n\\]\nBoth transforms are lower triangular with diagonal one.\n\\end{histtheorem}","\\begin{proof}\nThe even source identity gives\n\\[\n\\operatorname{sech}t\n=\n\\operatorname{sech}^2(qt)\n\\sum_{n\\geq0}S_{2n}(q)\\frac{t^{2n}}{(2n)!}.\n\\]\nInsert \\cref{hist:lem:local_zigzag_kernel} and compare coefficients. For the inverse, multiply \\(\\operatorname{sech}t\\) by\n\\[\n\\cosh^2(qt)\n=\n1+\n\\sum_{r\\geq1}2^{2r-1}q^{2r}\\frac{t^{2r}}{(2r)!}.\n\\]\n\\end{proof}","\\begin{histcorollary}[Odd modulus truncation]\\label{hist:cor:odd_modulus_truncation}\nFor every \\(M\\geq0\\), every positive odd \\(q\\), and every even \\(n\\),\n\\[\n\\boxed{\nE_n\n\\equiv\n\\sum_{r=0}^{\\min(M,n/2)}\n(-1)^rZ_{2r+1}\n\\binom n{2r}q^{2r}S_{n-2r}(q)\n\\pmod {q^{2M+2}}.\n}\n\\]\nIf \\(p\\mid q\\) and \\(a=\\vp(q)\\), the same truncation is valid modulo \\(p^{2a(M+1)}\\).\n\\end{histcorollary}","\\begin{proof}\nEvery omitted term contains \\(q^{2r}\\) with \\(r\\geq M+1\\). The local statement follows by taking the \\(p\\) adic valuation.\n\\end{proof}","\\begin{histproposition}[Odd row weight]\\label{hist:prop:odd_row_weight}\nFor an odd prime \\(p\\), the \\(r\\)th nonzero term in the reconstruction has valuation\n\\[\n\\boxed{\nW_p(n,r)\n=\n2r+\\vp(Z_{2r+1})\n+\\vp\\binom n{2r}\n+\\vp\\bigl(S_{n-2r}(p)\\bigr).\n}\n\\]\nThe binomial contribution is\n\\[\n\\boxed{\n\\vp\\binom n{2r}\n=\n\\frac{s_p(2r)+s_p(n-2r)-s_p(n)}{p-1},\n}\n\\]\nwhere \\(s_p\\) is the base \\(p\\) digit sum.\n\\end{histproposition}","For even \\(n\\geq0\\), define\n\\[\n\\boxed{\nR_n\n=\n\\frac{3^{n+2}-1}{4(n+1)(n+2)}.\n}\n\\]\nIts even exponential generating function is\n\\[\n\\boxed{\n\\sum_{n\\text{ even}}R_n\\frac{t^n}{n!}\n=\n\\frac{\\sinh(2t)\\sinh t}{2t^2}.\n}\n\\]\nDefine \\(\\beta_r\\) by\n\\[\n\\boxed{\n\\frac{z^2}{\\sinh^2z}\n=\n\\sum_{r\\geq0}\\beta_rz^{2r}.\n}\n\\]","\\begin{histproposition}[Dyadic inverse coefficients]\\label{hist:prop:dyadic_beta_unified}\nFor \\(r\\geq1\\),\n\\[\n\\boxed{\n\\beta_r\n=\n-\n\\frac{2^{2r}(2r-1)B_{2r}}{(2r)!}.\n}\n\\]\nConsequently,\n\\[\n\\boxed{\n\\vTwo(\\beta_r)=\\sTwo(r)-1.\n}\n\\]\nEvery \\(R_n\\) is a dyadic unit.\n\\end{histproposition}","\\begin{proof}\nDifferentiate the Bernoulli expansion of \\(\\coth z\\) and use \\(-(\\coth z)'=\\sinh(z)^{-2}\\). The von Staudt and Clausen theorem gives \\(\\vTwo(B_{2r})=-1\\). Legendre's formula gives\n\\[\n\\vTwo((2r)!)=2r-\\sTwo(r).\n\\]\nFor \\(R_n\\), put \\(k=n+2\\). Since \\(k\\) is even,\n\\[\n\\vTwo(3^k-1)=2+\\vTwo(k).\n\\]\nThe denominator has the same valuation because \\(n+1\\) is odd.\n\\end{proof}","\\begin{histtheorem}[Dyadic Euler coordinate pair]\\label{hist:thm:dyadic_coordinate_pair}\nFor every even \\(n\\geq0\\),\n\\[\n\\boxed{\nE_n\n=\n\\sum_{r=0}^{n/2}\n\\beta_r2^{2r}\n\\fall n{2r}\nR_{n-2r}.\n}\n\\]\nThe inverse transform is\n\\[\n\\boxed{\nR_n\n=\n\\sum_{r=0}^{n/2}\n\\frac{2^{4r+1}}{(2r+2)!}\n\\fall n{2r}\nE_{n-2r}.\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nSubstitute \\(z=2t\\) in the inverse series and multiply\n\\[\n\\frac{4t^2}{\\sinh^2(2t)}\n\\frac{\\sinh(2t)\\sinh t}{2t^2}\n=\n\\operatorname{sech}t.\n\\]\nCoefficient extraction gives the forward transform. The inverse follows from\n\\[\n\\left(\\frac{\\sinh(2t)}{2t}\\right)^2\n=\n\\sum_{r\\geq0}\n\\frac{2^{4r+1}}{(2r+2)!}t^{2r}.\n\\]\n\\end{proof}","For a prime \\(p\\), define\n\\[\nA_{p,n}\n=\n\\begin{cases}\nS_n(p),&p\\text{ odd},","For a prime \\(p\\), define\n\\[\nA_{p,n}\n=\n\\begin{cases}\nS_n(p),&p\\text{ odd},\\\\\nR_n,&p=2\\text{ and }n\\text{ even},\\\\\n0,&p=2\\text{ and }n\\text{ odd},\n\\end{cases}\n\\]\nand\n\\[\n\\kappa_{p,r}\n=\n\\begin{cases}\n(-1)^rZ_{2r+1}p^{2r},&p\\text{ odd},","0,&p=2\\text{ and }n\\text{ odd},\n\\end{cases}\n\\]\nand\n\\[\n\\kappa_{p,r}\n=\n\\begin{cases}\n(-1)^rZ_{2r+1}p^{2r},&p\\text{ odd},\\\\\n\\beta_r2^{2r}(2r)!,&p=2.\n\\end{cases}\n\\]","\\begin{histtheorem}[Unified local Euler transfer]\\label{hist:thm:unified_local_euler}\nFor every prime \\(p\\) and every even \\(n\\geq0\\),\n\\[\n\\boxed{\nE_n\n=\n\\sum_{r=0}^{n/2}\n\\binom n{2r}\n\\kappa_{p,r}\nA_{p,n-2r}.\n}\n\\]\nIf\n\\[\nH_p(t)\n=\n\\sum_{m\\geq0}A_{p,2m}\\frac{t^{2m}}{(2m)!}\n\\]\nand\n\\[\nK_p(t)\n=\n\\sum_{r\\geq0}\\kappa_{p,r}\\frac{t^{2r}}{(2r)!},\n\\]\nthen\n\\[\n\\boxed{K_p(t)H_p(t)=\\operatorname{sech}t.}\n\\]\nFor odd \\(p\\),\n\\[\nH_p(t)=\\cosh^2(pt)\\operatorname{sech}t,\n\\qquad\nK_p(t)=\\operatorname{sech}^2(pt).\n\\]\nFor \\(p=2\\),\n\\[\nH_2(t)=\\frac{\\sinh(2t)\\sinh t}{2t^2},\n\\qquad\nK_2(t)=\\frac{4t^2}{\\sinh^2(2t)}.\n\\]\n\\end{histtheorem}","egin{histtheorem}[Unified local Euler transfer]\\label{hist:thm:unified_local_euler}\nFor every prime \\(p\\) and every even \\(n\\geq0\\),\n\\[\n\\boxed{\nE_n\n=\n\\sum_{r=0}^{n/2}\n\\binom n{2r}\n\\kappa_{p,r}\nA_{p,n-2r}.\n}\n\\]\nIf\n\\[\nH_p(t)\n=\n\\sum_{m\\geq0}A_{p,2m}\\frac{t^{2m}}{(2m)!}\n\\]\nand\n\\[\nK_p(t)\n=\n\\sum_{r\\geq0}\\kappa_{p,r}\\frac{t^{2r}}{(2r)!},\n\\]\nthen\n\\[\n\\boxed{K_p(t)H_p(t)=\\operatorname{sech}t.}\n\\]\nFor odd \\(p\\),\n\\[\nH_p(t)=\\cosh^2(pt)\\operatorname{sech}t,\n\\qquad\nK_p(t)=\\operatorname{sech}^2(pt).\n\\]\nFor \\(p=2\\),\n\\[\nH_2(t)=\\frac{\\sinh(2t)\\sinh t}{2t^2},\n\\qquad\nK_2(t)=\\frac{4t^2}{\\sinh^2(2t)}.\n\\]\n\\end{histtheorem}","sttheorem}[Unified local Euler transfer]\\label{hist:thm:unified_local_euler}\nFor every prime \\(p\\) and every even \\(n\\geq0\\),\n\\[\n\\boxed{\nE_n\n=\n\\sum_{r=0}^{n/2}\n\\binom n{2r}\n\\kappa_{p,r}\nA_{p,n-2r}.\n}\n\\]\nIf\n\\[\nH_p(t)\n=\n\\sum_{m\\geq0}A_{p,2m}\\frac{t^{2m}}{(2m)!}\n\\]\nand\n\\[\nK_p(t)\n=\n\\sum_{r\\geq0}\\kappa_{p,r}\\frac{t^{2r}}{(2r)!},\n\\]\nthen\n\\[\n\\boxed{K_p(t)H_p(t)=\\operatorname{sech}t.}\n\\]\nFor odd \\(p\\),\n\\[\nH_p(t)=\\cosh^2(pt)\\operatorname{sech}t,\n\\qquad\nK_p(t)=\\operatorname{sech}^2(pt).\n\\]\nFor \\(p=2\\),\n\\[\nH_2(t)=\\frac{\\sinh(2t)\\sinh t}{2t^2},\n\\qquad\nK_2(t)=\\frac{4t^2}{\\sinh^2(2t)}.\n\\]\n\\end{histtheorem}","e \\(p\\) and every even \\(n\\geq0\\),\n\\[\n\\boxed{\nE_n\n=\n\\sum_{r=0}^{n/2}\n\\binom n{2r}\n\\kappa_{p,r}\nA_{p,n-2r}.\n}\n\\]\nIf\n\\[\nH_p(t)\n=\n\\sum_{m\\geq0}A_{p,2m}\\frac{t^{2m}}{(2m)!}\n\\]\nand\n\\[\nK_p(t)\n=\n\\sum_{r\\geq0}\\kappa_{p,r}\\frac{t^{2r}}{(2r)!},\n\\]\nthen\n\\[\n\\boxed{K_p(t)H_p(t)=\\operatorname{sech}t.}\n\\]\nFor odd \\(p\\),\n\\[\nH_p(t)=\\cosh^2(pt)\\operatorname{sech}t,\n\\qquad\nK_p(t)=\\operatorname{sech}^2(pt).\n\\]\nFor \\(p=2\\),\n\\[\nH_2(t)=\\frac{\\sinh(2t)\\sinh t}{2t^2},\n\\qquad\nK_2(t)=\\frac{4t^2}{\\sinh^2(2t)}.\n\\]\n\\end{histtheorem}","nd every even \\(n\\geq0\\),\n\\[\n\\boxed{\nE_n\n=\n\\sum_{r=0}^{n/2}\n\\binom n{2r}\n\\kappa_{p,r}\nA_{p,n-2r}.\n}\n\\]\nIf\n\\[\nH_p(t)\n=\n\\sum_{m\\geq0}A_{p,2m}\\frac{t^{2m}}{(2m)!}\n\\]\nand\n\\[\nK_p(t)\n=\n\\sum_{r\\geq0}\\kappa_{p,r}\\frac{t^{2r}}{(2r)!},\n\\]\nthen\n\\[\n\\boxed{K_p(t)H_p(t)=\\operatorname{sech}t.}\n\\]\nFor odd \\(p\\),\n\\[\nH_p(t)=\\cosh^2(pt)\\operatorname{sech}t,\n\\qquad\nK_p(t)=\\operatorname{sech}^2(pt).\n\\]\nFor \\(p=2\\),\n\\[\nH_2(t)=\\frac{\\sinh(2t)\\sinh t}{2t^2},\n\\qquad\nK_2(t)=\\frac{4t^2}{\\sinh^2(2t)}.\n\\]\n\\end{histtheorem}","\\begin{histtheorem}[Unified inverse local transfer]\\label{hist:thm:unified_inverse_local_euler}\nDefine\n\\[\n\\lambda_{p,r}\n=\n\\begin{cases}\n1,&p\\text{ odd and }r=0,\\\\\n2^{2r-1}p^{2r},&p\\text{ odd and }r\\geq1,\\\\\n\\dfrac{2^{4r+1}}{(2r+1)(2r+2)},&p=2.\n\\end{cases}\n\\]\nThen for every prime \\(p\\) and every even \\(n\\geq0\\),\n\\[\n\\boxed{\nA_{p,n}\n=\n\\sum_{r=0}^{n/2}\n\\binom n{2r}\n\\lambda_{p,r}\nE_{n-2r}.\n}\n\\]\nEquivalently, if\n\\[\nL_p(t)=K_p(t)^{-1},\n\\]\nthen\n\\[\n\\boxed{H_p(t)=L_p(t)\\operatorname{sech}t,}\n\\]\nwhere\n\\[\nL_p(t)=\\cosh^2(pt)\n\\]\nfor odd \\(p\\), and\n\\[\nL_2(t)=\\left(\\frac{\\sinh(2t)}{2t}\\right)^2\n\\]\nfor \\(p=2\\).\n\\end{histtheorem}","\\dfrac{2^{4r+1}}{(2r+1)(2r+2)},&p=2.\n\\end{cases}\n\\]\nThen for every prime \\(p\\) and every even \\(n\\geq0\\),\n\\[\n\\boxed{\nA_{p,n}\n=\n\\sum_{r=0}^{n/2}\n\\binom n{2r}\n\\lambda_{p,r}\nE_{n-2r}.\n}\n\\]\nEquivalently, if\n\\[\nL_p(t)=K_p(t)^{-1},\n\\]\nthen\n\\[\n\\boxed{H_p(t)=L_p(t)\\operatorname{sech}t,}\n\\]\nwhere\n\\[\nL_p(t)=\\cosh^2(pt)\n\\]\nfor odd \\(p\\), and\n\\[\nL_2(t)=\\left(\\frac{\\sinh(2t)}{2t}\\right)^2\n\\]\nfor \\(p=2\\).\n\\end{histtheorem}","\\begin{proof}\nFor odd \\(p\\), this is the inverse identity in \\cref{hist:thm:odd_coordinate_pair}. For \\(p=2\\), rewrite the inverse identity in \\cref{hist:thm:dyadic_coordinate_pair} by using\n\\[\n\\fall n{2r}=(2r)!\\binom n{2r}.\n\\]\nThe coefficient becomes\n\\[\n\\frac{2^{4r+1}(2r)!}{(2r+2)!}\n=\n\\frac{2^{4r+1}}{(2r+1)(2r+2)}.\n\\]\nThe generating functions are the inverses of the kernels in \\cref{hist:thm:unified_local_euler}.\n\\end{proof}","\\begin{histcorollary}[Exact change between local source charts]\\label{hist:cor:local_source_change}\n\\par\\noindent\\textbf{Editorial range correction.} This comparison holds for the even source subsequences, equivalently the evenized generating functions $H_p$. It does not identify their full odd-degree source values. For $p=3$, the original odd source has $A_{3,1}=S_1(3)=3$, whereas every odd dyadic source coefficient is zero; the even kernel $L_3K_2$ cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity $H_p=L_pK_\\ell H_\\ell$ and its proof remain valid.\\par\\medskip\nFor any two primes \\(p\\) and \\(\\ell\\), the source sequences \\(A_{p,n}\\) and \\(A_{\\ell,n}\\) are related by an exact unit lower triangular transformation obtained by composing the inverse kernel \\(L_p\\) with the reconstruction kernel \\(K_\\ell\\).\n\\end{histcorollary}","l source charts]\\label{hist:cor:local_source_change}\n\\par\\noindent\\textbf{Editorial range correction.} This comparison holds for the even source subsequences, equivalently the evenized generating functions $H_p$. It does not identify their full odd-degree source values. For $p=3$, the original odd source has $A_{3,1}=S_1(3)=3$, whereas every odd dyadic source coefficient is zero; the even kernel $L_3K_2$ cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity $H_p=L_pK_\\ell H_\\ell$ and its proof remain valid.\\par\\medskip\nFor any two primes \\(p\\) and \\(\\ell\\), the source sequences \\(A_{p,n}\\) and \\(A_{\\ell,n}\\) are related by an exact unit lower triangular transformation obtained by composing the inverse kernel \\(L_p\\) with the reconstruction kernel \\(K_\\ell\\).\n\\end{histcorollary}","the even source subsequences, equivalently the evenized generating functions $H_p$. It does not identify their full odd-degree source values. For $p=3$, the original odd source has $A_{3,1}=S_1(3)=3$, whereas every odd dyadic source coefficient is zero; the even kernel $L_3K_2$ cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity $H_p=L_pK_\\ell H_\\ell$ and its proof remain valid.\\par\\medskip\nFor any two primes \\(p\\) and \\(\\ell\\), the source sequences \\(A_{p,n}\\) and \\(A_{\\ell,n}\\) are related by an exact unit lower triangular transformation obtained by composing the inverse kernel \\(L_p\\) with the reconstruction kernel \\(K_\\ell\\).\n\\end{histcorollary}","It does not identify their full odd-degree source values. For $p=3$, the original odd source has $A_{3,1}=S_1(3)=3$, whereas every odd dyadic source coefficient is zero; the even kernel $L_3K_2$ cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity $H_p=L_pK_\\ell H_\\ell$ and its proof remain valid.\\par\\medskip\nFor any two primes \\(p\\) and \\(\\ell\\), the source sequences \\(A_{p,n}\\) and \\(A_{\\ell,n}\\) are related by an exact unit lower triangular transformation obtained by composing the inverse kernel \\(L_p\\) with the reconstruction kernel \\(K_\\ell\\).\n\\end{histcorollary}","ot identify their full odd-degree source values. For $p=3$, the original odd source has $A_{3,1}=S_1(3)=3$, whereas every odd dyadic source coefficient is zero; the even kernel $L_3K_2$ cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity $H_p=L_pK_\\ell H_\\ell$ and its proof remain valid.\\par\\medskip\nFor any two primes \\(p\\) and \\(\\ell\\), the source sequences \\(A_{p,n}\\) and \\(A_{\\ell,n}\\) are related by an exact unit lower triangular transformation obtained by composing the inverse kernel \\(L_p\\) with the reconstruction kernel \\(K_\\ell\\).\n\\end{histcorollary}","ee source values. For $p=3$, the original odd source has $A_{3,1}=S_1(3)=3$, whereas every odd dyadic source coefficient is zero; the even kernel $L_3K_2$ cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity $H_p=L_pK_\\ell H_\\ell$ and its proof remain valid.\\par\\medskip\nFor any two primes \\(p\\) and \\(\\ell\\), the source sequences \\(A_{p,n}\\) and \\(A_{\\ell,n}\\) are related by an exact unit lower triangular transformation obtained by composing the inverse kernel \\(L_p\\) with the reconstruction kernel \\(K_\\ell\\).\n\\end{histcorollary}",". For $p=3$, the original odd source has $A_{3,1}=S_1(3)=3$, whereas every odd dyadic source coefficient is zero; the even kernel $L_3K_2$ cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity $H_p=L_pK_\\ell H_\\ell$ and its proof remain valid.\\par\\medskip\nFor any two primes \\(p\\) and \\(\\ell\\), the source sequences \\(A_{p,n}\\) and \\(A_{\\ell,n}\\) are related by an exact unit lower triangular transformation obtained by composing the inverse kernel \\(L_p\\) with the reconstruction kernel \\(K_\\ell\\).\n\\end{histcorollary}","even kernel $L_3K_2$ cannot change that odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity $H_p=L_pK_\\ell H_\\ell$ and its proof remain valid.\\par\\medskip\nFor any two primes \\(p\\) and \\(\\ell\\), the source sequences \\(A_{p,n}\\) and \\(A_{\\ell,n}\\) are related by an exact unit lower triangular transformation obtained by composing the inverse kernel \\(L_p\\) with the reconstruction kernel \\(K_\\ell\\).\n\\end{histcorollary}","t odd coefficient. Read the coefficient consequence below only at even indices. The displayed identity $H_p=L_pK_\\ell H_\\ell$ and its proof remain valid.\\par\\medskip\nFor any two primes \\(p\\) and \\(\\ell\\), the source sequences \\(A_{p,n}\\) and \\(A_{\\ell,n}\\) are related by an exact unit lower triangular transformation obtained by composing the inverse kernel \\(L_p\\) with the reconstruction kernel \\(K_\\ell\\).\n\\end{histcorollary}","\\begin{proof}\nBoth source series reconstruct \\(\\operatorname{sech}t\\). Hence\n\\[\nH_p(t)=L_p(t)K_\\ell(t)H_\\ell(t).\n\\]\nThe multiplier has constant coefficient one, so \\cref{hist:thm:exact_coefficient_transport} gives an exact unit lower triangular coordinate change.\n\\end{proof}","\\midrule\nFinite torsion address & Modulo \\(p-1\\) & Sign address modulo two","Principal units & \\(1+p\\mathbb Z_p\\) & \\(1+4\\mathbb Z_2\\)","Source & Alternating odd power sum \\(S_n(p)\\) & Weighted source \\(R_n\\)","Kernel & \\(\\operatorname{sech}^2(pt)\\) & \\(4t^2/\\sinh^2(2t)\\)","Positive depth weight & At least \\(2r\\), with zigzag and row factors & Exactly \\(4r-1\\), before the binomial carry","Natural refined step & \\((p-1)p^a\\) & \\(2^{a+1}\\)","\\begin{histexample}[The first odd reconstruction]\\label{hist:ex:first_odd_reconstruction}\nFor \\(p=3\\) and \\(n=2\\),\n\\[\nS_2(3)=1-9+25=17.\n\\]\nSince \\(Z_1=1\\), \\(Z_3=2\\), and \\(S_0(3)=1\\), the odd coordinate formula gives\n\\[\nE_2\n=17-2\\cdot9\n=-1.\n\\]\nThe large source value and the depth one correction are both necessary.\n\\end{histexample}","\\begin{histexample}[The first nontrivial dyadic row]\\label{hist:ex:first_dyadic_reconstruction}\nFor \\(n=4\\),\n\\[\nR_4=\\frac{91}{15},\n\\qquad\nR_2=\\frac53,\n\\qquad\nR_0=1.\n\\]\nThe exact kernel coefficients are\n\\[\n\\kappa_{2,1}=-\\frac83,\n\\qquad\n\\kappa_{2,2}=\\frac{128}{5}.\n\\]\nTherefore\n\\[\nE_4\n=\n\\frac{91}{15}\n-\\frac83\\binom42\\frac53\n+\\frac{128}{5}\n=5.\n\\]\nThe rational source values are dyadic units, while the transfer depth is carried by the kernel.\n\\end{histexample}","\\begin{histremark}\\label{hist:rem:no_dyadic_substitution}\nThe common convolution does not identify the integral charts. At an odd prime the finite character coordinate has order \\(p-1\\). At the dyadic prime it begins with the sign coordinate, and the principal unit subgroup begins at \\(1+4\\mathbb Z_2\\).\n\\end{histremark}","For \\(r\\geq1\\), write\n\\[\n\\kappa_{2,r}\n=\n\\beta_r2^{2r}(2r)!\n=\n-2^{4r}(2r-1)B_{2r}.\n\\]","\\begin{histtheorem}[Exact dyadic kernel weight]\\label{hist:thm:exact_dyadic_kernel_weight}\n\\par\\noindent\\textbf{Editorial index conventions.} The standalone kernel valuation holds for all $r\\ge1$. Its reconstruction-term formula requires even $n\\ge0$ and $1\\le r\\le n/2$. In the following truncation formulas only terms with $r\\le n/2$ occur; terms that would use $R_{n-2r}$ at a negative index are omitted. An empty tail is exactly zero, and the minimum over an empty set of term valuations is $+\\infty$. These conventions include $M\\ge n/2$.\\par\\medskip\nFor every \\(r\\geq1\\),\n\\[\n\\boxed{\\vTwo(\\kappa_{2,r})=4r-1.}\n\\]\nThe \\(r\\)th reconstruction term has exact valuation\n\\[\n\\boxed{\n\\vTwo\\left(\n\\kappa_{2,r}\\binom n{2r}R_{n-2r}\n\\right)\n=\n4r-1+\\vTwo\\binom n{2r}.\n}\n\\]\nEquivalently,\n\\[\n\\boxed{\nW_2(n,r)\n=\n4r-1+\\sTwo(r)+\\sTwo(n-2r)-\\sTwo(n).\n}\n\\]\n\\end{histtheorem}","noindent\\textbf{Editorial index conventions.} The standalone kernel valuation holds for all $r\\ge1$. Its reconstruction-term formula requires even $n\\ge0$ and $1\\le r\\le n/2$. In the following truncation formulas only terms with $r\\le n/2$ occur; terms that would use $R_{n-2r}$ at a negative index are omitted. An empty tail is exactly zero, and the minimum over an empty set of term valuations is $+\\infty$. These conventions include $M\\ge n/2$.\\par\\medskip\nFor every \\(r\\geq1\\),\n\\[\n\\boxed{\\vTwo(\\kappa_{2,r})=4r-1.}\n\\]\nThe \\(r\\)th reconstruction term has exact valuation\n\\[\n\\boxed{\n\\vTwo\\left(\n\\kappa_{2,r}\\binom n{2r}R_{n-2r}\n\\right)\n=\n4r-1+\\vTwo\\binom n{2r}.\n}\n\\]\nEquivalently,\n\\[\n\\boxed{\nW_2(n,r)\n=\n4r-1+\\sTwo(r)+\\sTwo(n-2r)-\\sTwo(n).\n}\n\\]\n\\end{histtheorem}","ntions.} The standalone kernel valuation holds for all $r\\ge1$. Its reconstruction-term formula requires even $n\\ge0$ and $1\\le r\\le n/2$. In the following truncation formulas only terms with $r\\le n/2$ occur; terms that would use $R_{n-2r}$ at a negative index are omitted. An empty tail is exactly zero, and the minimum over an empty set of term valuations is $+\\infty$. These conventions include $M\\ge n/2$.\\par\\medskip\nFor every \\(r\\geq1\\),\n\\[\n\\boxed{\\vTwo(\\kappa_{2,r})=4r-1.}\n\\]\nThe \\(r\\)th reconstruction term has exact valuation\n\\[\n\\boxed{\n\\vTwo\\left(\n\\kappa_{2,r}\\binom n{2r}R_{n-2r}\n\\right)\n=\n4r-1+\\vTwo\\binom n{2r}.\n}\n\\]\nEquivalently,\n\\[\n\\boxed{\nW_2(n,r)\n=\n4r-1+\\sTwo(r)+\\sTwo(n-2r)-\\sTwo(n).\n}\n\\]\n\\end{histtheorem}","ation holds for all $r\\ge1$. Its reconstruction-term formula requires even $n\\ge0$ and $1\\le r\\le n/2$. In the following truncation formulas only terms with $r\\le n/2$ occur; terms that would use $R_{n-2r}$ at a negative index are omitted. An empty tail is exactly zero, and the minimum over an empty set of term valuations is $+\\infty$. These conventions include $M\\ge n/2$.\\par\\medskip\nFor every \\(r\\geq1\\),\n\\[\n\\boxed{\\vTwo(\\kappa_{2,r})=4r-1.}\n\\]\nThe \\(r\\)th reconstruction term has exact valuation\n\\[\n\\boxed{\n\\vTwo\\left(\n\\kappa_{2,r}\\binom n{2r}R_{n-2r}\n\\right)\n=\n4r-1+\\vTwo\\binom n{2r}.\n}\n\\]\nEquivalently,\n\\[\n\\boxed{\nW_2(n,r)\n=\n4r-1+\\sTwo(r)+\\sTwo(n-2r)-\\sTwo(n).\n}\n\\]\n\\end{histtheorem}","for all $r\\ge1$. Its reconstruction-term formula requires even $n\\ge0$ and $1\\le r\\le n/2$. In the following truncation formulas only terms with $r\\le n/2$ occur; terms that would use $R_{n-2r}$ at a negative index are omitted. An empty tail is exactly zero, and the minimum over an empty set of term valuations is $+\\infty$. These conventions include $M\\ge n/2$.\\par\\medskip\nFor every \\(r\\geq1\\),\n\\[\n\\boxed{\\vTwo(\\kappa_{2,r})=4r-1.}\n\\]\nThe \\(r\\)th reconstruction term has exact valuation\n\\[\n\\boxed{\n\\vTwo\\left(\n\\kappa_{2,r}\\binom n{2r}R_{n-2r}\n\\right)\n=\n4r-1+\\vTwo\\binom n{2r}.\n}\n\\]\nEquivalently,\n\\[\n\\boxed{\nW_2(n,r)\n=\n4r-1+\\sTwo(r)+\\sTwo(n-2r)-\\sTwo(n).\n}\n\\]\n\\end{histtheorem}","mula requires even $n\\ge0$ and $1\\le r\\le n/2$. In the following truncation formulas only terms with $r\\le n/2$ occur; terms that would use $R_{n-2r}$ at a negative index are omitted. An empty tail is exactly zero, and the minimum over an empty set of term valuations is $+\\infty$. These conventions include $M\\ge n/2$.\\par\\medskip\nFor every \\(r\\geq1\\),\n\\[\n\\boxed{\\vTwo(\\kappa_{2,r})=4r-1.}\n\\]\nThe \\(r\\)th reconstruction term has exact valuation\n\\[\n\\boxed{\n\\vTwo\\left(\n\\kappa_{2,r}\\binom n{2r}R_{n-2r}\n\\right)\n=\n4r-1+\\vTwo\\binom n{2r}.\n}\n\\]\nEquivalently,\n\\[\n\\boxed{\nW_2(n,r)\n=\n4r-1+\\sTwo(r)+\\sTwo(n-2r)-\\sTwo(n).\n}\n\\]\n\\end{histtheorem}","In the following truncation formulas only terms with $r\\le n/2$ occur; terms that would use $R_{n-2r}$ at a negative index are omitted. An empty tail is exactly zero, and the minimum over an empty set of term valuations is $+\\infty$. These conventions include $M\\ge n/2$.\\par\\medskip\nFor every \\(r\\geq1\\),\n\\[\n\\boxed{\\vTwo(\\kappa_{2,r})=4r-1.}\n\\]\nThe \\(r\\)th reconstruction term has exact valuation\n\\[\n\\boxed{\n\\vTwo\\left(\n\\kappa_{2,r}\\binom n{2r}R_{n-2r}\n\\right)\n=\n4r-1+\\vTwo\\binom n{2r}.\n}\n\\]\nEquivalently,\n\\[\n\\boxed{\nW_2(n,r)\n=\n4r-1+\\sTwo(r)+\\sTwo(n-2r)-\\sTwo(n).\n}\n\\]\n\\end{histtheorem}","egative index are omitted. An empty tail is exactly zero, and the minimum over an empty set of term valuations is $+\\infty$. These conventions include $M\\ge n/2$.\\par\\medskip\nFor every \\(r\\geq1\\),\n\\[\n\\boxed{\\vTwo(\\kappa_{2,r})=4r-1.}\n\\]\nThe \\(r\\)th reconstruction term has exact valuation\n\\[\n\\boxed{\n\\vTwo\\left(\n\\kappa_{2,r}\\binom n{2r}R_{n-2r}\n\\right)\n=\n4r-1+\\vTwo\\binom n{2r}.\n}\n\\]\nEquivalently,\n\\[\n\\boxed{\nW_2(n,r)\n=\n4r-1+\\sTwo(r)+\\sTwo(n-2r)-\\sTwo(n).\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nThe Bernoulli formula gives \\(\\vTwo(B_{2r})=-1\\). The source value is a unit. Kummer's binary carry formula gives the last expression.\n\\end{proof}","\\begin{histtheorem}[Optimal dyadic truncation]\\label{hist:thm:optimal_dyadic_truncation}\nLet\n\\[\nT_{2,M}(n)\n=\n\\sum_{r=0}^{\\min(M,n/2)}\n\\kappa_{2,r}\\binom n{2r}R_{n-2r}.\n\\]\nThen\n\\[\n\\boxed{\nE_n\\equiv T_{2,M}(n)\\pmod {2^{4M+3}}.\n}\n\\]\nMore precisely,\n\\[\n\\boxed{\n\\vTwo\\bigl(E_n-T_{2,M}(n)\\bigr)\n\\geq\n\\min_{M+1\\leq r\\leq n/2}\n\\left(\n4r-1+\\vTwo\\binom n{2r}\n\\right).\n}\n\\]\nThe uniform exponent \\(4M+3\\) is best possible.\n\\end{histtheorem}","\\begin{proof}\nEvery omitted term has valuation at least \\(4(M+1)-1\\). For optimality, take \\(n=2M+2\\). There is one omitted term. Its binomial coefficient and source value are one, so its valuation is exactly \\(4M+3\\).\n\\end{proof}","The first three sharp truncations are\n\\[\n\\begin{aligned}\nE_n&\\equiv R_n&&\\pmod 8,\\\\\nE_n&\\equiv R_n-\\frac83\\binom n2R_{n-2}&&\\pmod {128},\\\\\nE_n&\\equiv R_n-\\frac83\\binom n2R_{n-2}\n+\\frac{128}{5}\\binom n4R_{n-4}&&\\pmod {2048}.\n\\end{aligned}\n\\]","For a function \\(f\\) on the nonnegative integers, define\n\\[\n\\Delta f(n)=f(n+1)-f(n).\n\\]","\\begin{histtheorem}[Finite Mahler reconstruction]\\label{hist:thm:finite_mahler_unified}\nGiven \\(f(0),\\ldots,f(N)\\), put \\(c_r=\\Delta^rf(0)\\). Then\n\\[\n\\boxed{\nf(n)=\\sum_{r=0}^{n}c_r\\binom nr\n\\qquad(0\\leq n\\leq N).\n}\n\\]\nFor every \\(s\\) with \\(0\\leq s\\leq N\\),\n\\[\n\\boxed{\n\\Delta^sf(n)\n=\n\\sum_{r=s}^{N}c_r\\binom n{r-s}\n\\qquad(0\\leq n\\leq N-s).\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nLet\n\\[\nP(x)=\\sum_{r=0}^{N}c_r\\binom xr.\n\\]\nNewton interpolation gives \\(P(n)=f(n)\\) for \\(0\\leq n\\leq N\\). Iterating\n\\[\n\\Delta\\binom xr=\\binom x{r-1}\n\\]\ngives\n\\[\n\\Delta^sP(n)=\\sum_{r=s}^{N}c_r\\binom n{r-s}.\n\\]\nWhen \\(0\\leq n\\leq N-s\\), every value used in \\(\\Delta^sf(n)\\) lies among the prescribed data, so \\(\\Delta^sf(n)=\\Delta^sP(n)\\).\n\\end{proof}","\\begin{histcorollary}[Mahler valuation ledger]\\label{hist:cor:mahler_valuation_ledger}\nIf \\(c_r\\in\\Zp\\) and \\(\\vp(c_r)\\geq w(r)\\), then\n\\[\n\\boxed{\n\\vp(\\Delta^sf(n))\n\\geq\n\\min_{r\\geq s}\n\\left(\nw(r)+\\vp\\binom n{r-s}\n\\right).\n}\n\\]\n\\end{histcorollary}","The infinite interpolation theorem states that every continuous function on \\(\\Zp\\) has a unique convergent Mahler expansion \\cite{Mahler1958,Robert2000}. We use that theorem as a standard input when the local variable is allowed to range over all of \\(\\Zp\\).","\\begin{histstandardfact}[Local unit decomposition]\\label{hist:fact:local_unit_decomposition}\nFor odd \\(p\\), every unit has a unique factorization\n\\[\nu=\\omega(u)\\langle u\\rangle,\n\\qquad\n\\omega(u)^{p-1}=1,\n\\qquad\n\\langle u\\rangle\\in1+p\\mathbb Z_p.\n\\]\nEvery dyadic unit has a unique factorization\n\\[\nu=(-1)^{\\epsilon(u)}\\langle u\\rangle,\n\\qquad\n\\epsilon(u)\\in\\{0,1\\},\n\\qquad\n\\langle u\\rangle\\in1+4\\mathbb Z_2.\n\\]\nThe first factor gives the finite character address. The second factor admits logarithmic and Mahler coordinates.\n\\end{histstandardfact}","For odd \\(p\\) and \\(L\\geq1\\), put\n\\[\n\\Lambda_{p,L}=(p-1)p^{L-1}.\n\\]\nFor \\(p=2\\) and \\(L\\geq3\\), put\n\\[\n\\Lambda_{2,L}=2^{L-2}.\n\\]\nThese are exponents of the unit groups modulo the declared prime powers.","\\begin{histtheorem}[One unit mode]\\label{hist:thm:carmichael_mahler_scalar}\nLet \\(u\\in\\Zp^\\times\\), let \\(H=\\Lambda_{p,L}\\), and put\n\\[\nq_{u,L}=u^H-1\\in p^L\\Zp.\n\\]\nFor every integer address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\nu^{Hx+j}\n=\nu^j\\sum_{r\\geq0}q_{u,L}^{\\,r}\\binom xr.\n}\n\\]\nThe \\(r\\)th Mahler coefficient has valuation at least \\(Lr\\).\n\\end{histtheorem}","\\begin{proof}\nWrite \\(u^{Hx+j}=u^j(1+q_{u,L})^x\\) and use the convergent binomial series on \\(\\Zp\\).\n\\end{proof}","\\begin{histexample}[Paired principal unit expansions]\\label{hist:ex:paired_unit_expansions}\nAt the odd prime three,\n\\[\n2^{2x+j}\n=\n2^j4^x\n=\n2^j\\sum_{r\\geq0}3^r\\binom xr.\n\\]\nAt the dyadic prime,\n\\[\n3^{2x+j}\n=\n3^j9^x\n=\n3^j\\sum_{r\\geq0}8^r\\binom xr.\n\\]\nThe formal expansions are identical. Their first local slopes are one and three because \\(4-1=3\\) while \\(9-1=8\\).\n\\end{histexample}","\\begin{histcorollary}[Basic odd and dyadic difference gains]\\label{hist:cor:basic_local_difference_gains}\nFor odd \\(p\\),\n\\[\n\\boxed{\n\\Delta_{(p-1)p^a}^{\\,r}u^n\n=\nu^n\\left(u^{(p-1)p^a}-1\\right)^r,\n\\qquad\n\\vp\\left(\\Delta_{(p-1)p^a}^{\\,r}u^n\\right)\\geq(a+1)r.\n}\n\\]\nFor odd \\(u\\in\\mathbb Z_2^\\times\\),\n\\[\n\\boxed{\n\\Delta_{2^{a+1}}^{\\,r}u^n\n=\nu^n\\left(u^{2^{a+1}}-1\\right)^r,\n\\qquad\n\\vTwo\\left(\\Delta_{2^{a+1}}^{\\,r}u^n\\right)\\geq(a+3)r.\n}\n\\]\n\\end{histcorollary}","\\begin{proof}\nOne step difference multiplies a unit mode by \\(u^H-1\\). Fermat reduction and lifting the exponent give the odd estimate. For odd \\(u\\),\n\\[\n\\vTwo\\left(u^{2^{a+1}}-1\\right)\n=\n\\vTwo(u^2-1)+a\n\\geq a+3.\n\\]\n\\end{proof}","\\begin{histtheorem}[Polynomial exponential Mahler transfer]\\label{hist:thm:polynomial_exponential_mahler}\nLet\n\\[\nf(n)=P(n)u^n\n\\]\nwith \\(P\\in\\Zp[n]\\) of degree \\(d\\). Let \\(H=\\Lambda_{p,L}\\), fix an address \\(j\\), and write\n\\[\nQ_j(x)=P(Hx+j),\n\\qquad\ny=u^H,\n\\qquad q=y-1.\n\\]\nThen\n\\[\n\\boxed{\nf(Hx+j)=\\sum_{r\\geq0}c_{j,r}\\binom xr,\n}\n\\]\nwhere\n\\[\n\\boxed{\nc_{j,r}\n=\nu^j\n\\sum_{k=0}^{\\min(d,r)}\n\\binom rk\n\\Delta^kQ_j(0)y^kq^{r-k}.\n}\n\\]\nConsequently,\n\\[\n\\boxed{\n\\vp(c_{j,r})\\geq L\\max(r-d,0).\n}\n\\]\nFor \\(R\\geq d\\), truncation at depth \\(R\\) is valid modulo \\(p^{L(R+1-d)}\\) uniformly on \\(\\Zp\\).\n\\end{histtheorem}","ahler transfer]\\label{hist:thm:polynomial_exponential_mahler}\nLet\n\\[\nf(n)=P(n)u^n\n\\]\nwith \\(P\\in\\Zp[n]\\) of degree \\(d\\). Let \\(H=\\Lambda_{p,L}\\), fix an address \\(j\\), and write\n\\[\nQ_j(x)=P(Hx+j),\n\\qquad\ny=u^H,\n\\qquad q=y-1.\n\\]\nThen\n\\[\n\\boxed{\nf(Hx+j)=\\sum_{r\\geq0}c_{j,r}\\binom xr,\n}\n\\]\nwhere\n\\[\n\\boxed{\nc_{j,r}\n=\nu^j\n\\sum_{k=0}^{\\min(d,r)}\n\\binom rk\n\\Delta^kQ_j(0)y^kq^{r-k}.\n}\n\\]\nConsequently,\n\\[\n\\boxed{\n\\vp(c_{j,r})\\geq L\\max(r-d,0).\n}\n\\]\nFor \\(R\\geq d\\), truncation at depth \\(R\\) is valid modulo \\(p^{L(R+1-d)}\\) uniformly on \\(\\Zp\\).\n\\end{histtheorem}","al_mahler}\nLet\n\\[\nf(n)=P(n)u^n\n\\]\nwith \\(P\\in\\Zp[n]\\) of degree \\(d\\). Let \\(H=\\Lambda_{p,L}\\), fix an address \\(j\\), and write\n\\[\nQ_j(x)=P(Hx+j),\n\\qquad\ny=u^H,\n\\qquad q=y-1.\n\\]\nThen\n\\[\n\\boxed{\nf(Hx+j)=\\sum_{r\\geq0}c_{j,r}\\binom xr,\n}\n\\]\nwhere\n\\[\n\\boxed{\nc_{j,r}\n=\nu^j\n\\sum_{k=0}^{\\min(d,r)}\n\\binom rk\n\\Delta^kQ_j(0)y^kq^{r-k}.\n}\n\\]\nConsequently,\n\\[\n\\boxed{\n\\vp(c_{j,r})\\geq L\\max(r-d,0).\n}\n\\]\nFor \\(R\\geq d\\), truncation at depth \\(R\\) is valid modulo \\(p^{L(R+1-d)}\\) uniformly on \\(\\Zp\\).\n\\end{histtheorem}","with \\(P\\in\\Zp[n]\\) of degree \\(d\\). Let \\(H=\\Lambda_{p,L}\\), fix an address \\(j\\), and write\n\\[\nQ_j(x)=P(Hx+j),\n\\qquad\ny=u^H,\n\\qquad q=y-1.\n\\]\nThen\n\\[\n\\boxed{\nf(Hx+j)=\\sum_{r\\geq0}c_{j,r}\\binom xr,\n}\n\\]\nwhere\n\\[\n\\boxed{\nc_{j,r}\n=\nu^j\n\\sum_{k=0}^{\\min(d,r)}\n\\binom rk\n\\Delta^kQ_j(0)y^kq^{r-k}.\n}\n\\]\nConsequently,\n\\[\n\\boxed{\n\\vp(c_{j,r})\\geq L\\max(r-d,0).\n}\n\\]\nFor \\(R\\geq d\\), truncation at depth \\(R\\) is valid modulo \\(p^{L(R+1-d)}\\) uniformly on \\(\\Zp\\).\n\\end{histtheorem}","f degree \\(d\\). Let \\(H=\\Lambda_{p,L}\\), fix an address \\(j\\), and write\n\\[\nQ_j(x)=P(Hx+j),\n\\qquad\ny=u^H,\n\\qquad q=y-1.\n\\]\nThen\n\\[\n\\boxed{\nf(Hx+j)=\\sum_{r\\geq0}c_{j,r}\\binom xr,\n}\n\\]\nwhere\n\\[\n\\boxed{\nc_{j,r}\n=\nu^j\n\\sum_{k=0}^{\\min(d,r)}\n\\binom rk\n\\Delta^kQ_j(0)y^kq^{r-k}.\n}\n\\]\nConsequently,\n\\[\n\\boxed{\n\\vp(c_{j,r})\\geq L\\max(r-d,0).\n}\n\\]\nFor \\(R\\geq d\\), truncation at depth \\(R\\) is valid modulo \\(p^{L(R+1-d)}\\) uniformly on \\(\\Zp\\).\n\\end{histtheorem}","a_{p,L}\\), fix an address \\(j\\), and write\n\\[\nQ_j(x)=P(Hx+j),\n\\qquad\ny=u^H,\n\\qquad q=y-1.\n\\]\nThen\n\\[\n\\boxed{\nf(Hx+j)=\\sum_{r\\geq0}c_{j,r}\\binom xr,\n}\n\\]\nwhere\n\\[\n\\boxed{\nc_{j,r}\n=\nu^j\n\\sum_{k=0}^{\\min(d,r)}\n\\binom rk\n\\Delta^kQ_j(0)y^kq^{r-k}.\n}\n\\]\nConsequently,\n\\[\n\\boxed{\n\\vp(c_{j,r})\\geq L\\max(r-d,0).\n}\n\\]\nFor \\(R\\geq d\\), truncation at depth \\(R\\) is valid modulo \\(p^{L(R+1-d)}\\) uniformly on \\(\\Zp\\).\n\\end{histtheorem}","\\begin{proof}\nOn the polynomial factor the difference operator acts as \\(q+y\\Delta\\). The two terms commute. Expand \\((q+y\\Delta)^r\\) and evaluate at zero. Since \\(\\Delta^kQ_j=0\\) for \\(k>d\\), every surviving term contains at least \\(q^{r-d}\\).\n\\end{proof}","Let\n\\[\nv_{n+1}=Cv_n,\n\\qquad\na_n=w^TC^nv_0,\n\\]\nwith \\(C\\in M_d(\\Zp)\\) invertible modulo \\(p\\).","\\begin{histtheorem}[Matrix Carmichael and Mahler transfer]\\label{hist:thm:matrix_carmichael_mahler}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The original displayed statement leaves the vectors untyped. Its integrality conclusion requires $v_0,w\\in\\Z_p^d$, $L\\ge1$, and $H,j,n\\in\\N$ with $H>0$; the value at $Hx+j$ for $x\\in\\Z_p$ means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","in{histtheorem}[Matrix Carmichael and Mahler transfer]\\label{hist:thm:matrix_carmichael_mahler}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The original displayed statement leaves the vectors untyped. Its integrality conclusion requires $v_0,w\\in\\Z_p^d$, $L\\ge1$, and $H,j,n\\in\\N$ with $H>0$; the value at $Hx+j$ for $x\\in\\Z_p$ means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","theorem}[Matrix Carmichael and Mahler transfer]\\label{hist:thm:matrix_carmichael_mahler}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The original displayed statement leaves the vectors untyped. Its integrality conclusion requires $v_0,w\\in\\Z_p^d$, $L\\ge1$, and $H,j,n\\in\\N$ with $H>0$; the value at $Hx+j$ for $x\\in\\Z_p$ means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","atrix Carmichael and Mahler transfer]\\label{hist:thm:matrix_carmichael_mahler}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The original displayed statement leaves the vectors untyped. Its integrality conclusion requires $v_0,w\\in\\Z_p^d$, $L\\ge1$, and $H,j,n\\in\\N$ with $H>0$; the value at $Hx+j$ for $x\\in\\Z_p$ means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","d Mahler transfer]\\label{hist:thm:matrix_carmichael_mahler}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The original displayed statement leaves the vectors untyped. Its integrality conclusion requires $v_0,w\\in\\Z_p^d$, $L\\ge1$, and $H,j,n\\in\\N$ with $H>0$; the value at $Hx+j$ for $x\\in\\Z_p$ means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}",":matrix_carmichael_mahler}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The original displayed statement leaves the vectors untyped. Its integrality conclusion requires $v_0,w\\in\\Z_p^d$, $L\\ge1$, and $H,j,n\\in\\N$ with $H>0$; the value at $Hx+j$ for $x\\in\\Z_p$ means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","0,w\\in\\Z_p^d$, $L\\ge1$, and $H,j,n\\in\\N$ with $H>0$; the value at $Hx+j$ for $x\\in\\Z_p$ means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}",", and $H,j,n\\in\\N$ with $H>0$; the value at $Hx+j$ for $x\\in\\Z_p$ means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","lue at $Hx+j$ for $x\\in\\Z_p$ means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","r $x\\in\\Z_p$ means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","means the matrix-binomial interpolation. With $p=3$, $C=4$, $L=1$, $w=1$ and $v_0=1/3$, one has $\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","$\\Delta_1a_0=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","=1$, which is not in $3\\Z_3$. The printed claim without these restrictions is therefore not an as-written certificate. The full original proof follows under the corrected hypotheses.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\). For every address \\(j\\) and every \\(x\\in\\Zp\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r\\geq0}\np^{Lr}\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nThe series converges on \\(\\Zp\\), and\n\\[\n\\boxed{\n\\Delta_H^ra_n\n\\equiv0\n\\pmod {p^{Lr}}.\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nUse\n\\[\nC^{Hx+j}=C^j(I+p^LB)^x\n\\]\nand the matrix binomial series. One difference in the coarse variable multiplies the series by \\(p^LB\\).\n\\end{proof}","\\begin{histtheorem}[Finite adelic transfer]\\label{hist:thm:finite_adelic_transfer}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} Read this statement with $v_0,w\\in\\Z^d$, $H>0$ and $j,n,r\\in\\N$, and positive integer $L_p$. These restrictions make the integral congruences and powers well typed. Allowing $v_0=1/3$ in the one-prime example above loses the advertised digit; invertibility of $C$ alone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","f{Editorial hypothesis correction.} Read this statement with $v_0,w\\in\\Z^d$, $H>0$ and $j,n,r\\in\\N$, and positive integer $L_p$. These restrictions make the integral congruences and powers well typed. Allowing $v_0=1/3$ in the one-prime example above loses the advertised digit; invertibility of $C$ alone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","v_0,w\\in\\Z^d$, $H>0$ and $j,n,r\\in\\N$, and positive integer $L_p$. These restrictions make the integral congruences and powers well typed. Allowing $v_0=1/3$ in the one-prime example above loses the advertised digit; invertibility of $C$ alone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","and positive integer $L_p$. These restrictions make the integral congruences and powers well typed. Allowing $v_0=1/3$ in the one-prime example above loses the advertised digit; invertibility of $C$ alone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","er $L_p$. These restrictions make the integral congruences and powers well typed. Allowing $v_0=1/3$ in the one-prime example above loses the advertised digit; invertibility of $C$ alone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","restrictions make the integral congruences and powers well typed. Allowing $v_0=1/3$ in the one-prime example above loses the advertised digit; invertibility of $C$ alone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","e integral congruences and powers well typed. Allowing $v_0=1/3$ in the one-prime example above loses the advertised digit; invertibility of $C$ alone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","$v_0=1/3$ in the one-prime example above loses the advertised digit; invertibility of $C$ alone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","/3$ in the one-prime example above loses the advertised digit; invertibility of $C$ alone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","f $C$ alone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","lone does not restore it. The original proof is retained under these hypotheses.\\par\\medskip\nLet\n\\[\nQ=\\prod_{p\\in S}p^{L_p}\n\\]\nfor a finite set of primes. Let \\(C\\in M_d(\\ZZ)\\) be invertible modulo \\(Q\\), and choose \\(H\\) such that\n\\[\nC^H=I+QB\n\\]\nwith \\(B\\in M_d(\\ZZ)\\). Then for every nonnegative integer \\(x\\),\n\\[\n\\boxed{\na_{Hx+j}\n=\n\\sum_{r=0}^{x}Q^r\\binom xr\nw^TC^jB^rv_0.\n}\n\\]\nFor every \\(r\\),\n\\[\n\\boxed{\n\\Delta_H^ra_n\\equiv0\\pmod {Q^r}.\n}\n\\]\n\\end{histtheorem}","\\begin{histexample}[A simultaneous Fibonacci jump]\\label{hist:ex:fibonacci_finite_adelic}\nLet\n\\[\nC=\n\\begin{pmatrix}\n1&1\\\\\n1&0\n\\end{pmatrix}.\n\\]\nThe matrix \\(C\\) is invertible modulo twenty four, and\n\\[\nC^{24}-I\n=\n24\n\\begin{pmatrix}\n3126&1932","1&0\n\\end{pmatrix}.\n\\]\nThe matrix \\(C\\) is invertible modulo twenty four, and\n\\[\nC^{24}-I\n=\n24\n\\begin{pmatrix}\n3126&1932","1&0\n\\end{pmatrix}.\n\\]\nThe matrix \\(C\\) is invertible modulo twenty four, and\n\\[\nC^{24}-I\n=\n24\n\\begin{pmatrix}\n3126&1932\\\\\n1932&1194\n\\end{pmatrix}.\n\\]\nFor every scalar coordinate of the state sequence \\(C^nv_0\\),\n\\[\n\\boxed{\n\\Delta_{24}^{\\,r}a_n\n\\equiv0\n\\pmod {24^r}.\n}\n\\]\nOne integral jump therefore gives simultaneous congruence ladders at the primes two and three.\n\\end{histexample}","1932&1194\n\\end{pmatrix}.\n\\]\nFor every scalar coordinate of the state sequence \\(C^nv_0\\),\n\\[\n\\boxed{\n\\Delta_{24}^{\\,r}a_n\n\\equiv0\n\\pmod {24^r}.\n}\n\\]\nOne integral jump therefore gives simultaneous congruence ladders at the primes two and three.\n\\end{histexample}","\\begin{histdefinition}[Amice pullback]\\label{hist:def:amice_pullback_unified}\nFor a formal exponential generating series\n\\[\nF(z)=\\sum_{n\\geq0}a_n\\frac{z^n}{n!},\n\\]\ndefine\n\\[\n\\widehat F(T)=F(\\log(1+T)).\n\\]\n\\end{histdefinition}","\\begin{histtheorem}[Derivative conjugacy]\\label{hist:thm:amice_derivative_unified}\n\\[\n\\boxed{\n\\widehat{F'}(T)\n=\n(1+T)\\frac{d}{dT}\\widehat F(T).\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nDifferentiate the composition \\(F(\\log(1+T))\\).\n\\end{proof}","\\begin{histproposition}[Weighted convergence]\\label{hist:prop:amice_weighted_convergence_unified}\nIf\n\\[\n\\widehat F(T)=\\sum_{r\\geq0}c_rT^r\n\\]\nand\n\\[\n\\vp(c_r)+r\\alpha\\longrightarrow+\\infty,\n\\]\nthen the series converges whenever \\(\\vp(T)\\geq\\alpha\\).\n\\end{histproposition}","\\begin{proof}\nThe valuation of the \\(r\\)th term is \\(\\vp(c_r)+r\\vp(T)\\), which tends to positive infinity.\n\\end{proof}","For an odd prime \\(p\\), the source \\(S_n(p)\\) contains one nonunit base. Put\n\\[\n\\sigma_p=(-1)^{(p-1)/2}\n\\]\nand\n\\[\nS_n^\\times(p)=S_n(p)-\\sigma_pp^n.\n\\]","\\begin{histtheorem}[Depleted odd prime reconstruction]\\label{hist:thm:depleted_odd_reconstruction}\nFor every even \\(n\\),\n\\[\n\\boxed{\n\\left(1-\\sigma_pp^n\\right)E_n\n=\n\\sum_{r=0}^{n/2}\n(-1)^rZ_{2r+1}\n\\binom n{2r}p^{2r}S_{n-2r}^\\times(p).\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nThe removed source mode contributes \\(\\sigma_p\\cosh(pt)\\). Multiplication by \\(\\operatorname{sech}^2(pt)\\) gives \\(\\sigma_p\\operatorname{sech}(pt)\\). Subtract it from the full source and kernel product.\n\\end{proof}","Fix an even address \\(j\\in\\{0,2,\\ldots,p-3\\}\\). For \\(x\\in\\Zp\\), define\n\\[\n\\begin{aligned}\n\\mathscr E_{p,j}^\\times(x)\n=\n\\sum_{r\\geq0}\n&(-1)^rZ_{2r+1}p^{2r}\n\\binom{(p-1)x+j}{2r}","Fix an even address \\(j\\in\\{0,2,\\ldots,p-3\\}\\). For \\(x\\in\\Zp\\), define\n\\[\n\\begin{aligned}\n\\mathscr E_{p,j}^\\times(x)\n=\n\\sum_{r\\geq0}\n&(-1)^rZ_{2r+1}p^{2r}\n\\binom{(p-1)x+j}{2r}\\\\\n&\\times\n\\sum_{\\substack{0\\leq a<p\\\\2a+1\\neq p}}\n(-1)^a\n(2a+1)^{j-2r}\n\\left((2a+1)^{p-1}\\right)^x.\n\\end{aligned}\n\\]","\\begin{histtheorem}[Odd prime analytic Euler branch]\\label{hist:thm:odd_analytic_branch}\nThe series defining \\(\\mathscr E_{p,j}^\\times\\) converges in the Tate algebra on \\(\\Zp\\). For every nonnegative integer \\(x\\),\n\\[\n\\boxed{\n\\mathscr E_{p,j}^\\times(x)\n=\n\\left(1-\\sigma_pp^{(p-1)x+j}\\right)\nE_{(p-1)x+j}.\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nEvery retained base is a unit, and its \\((p-1)\\)st power lies in \\(1+p\\Zp\\). The polynomial\n\\[\np^{2r}\\binom{(p-1)x+j}{2r}\n\\]\nhas Gauss norm at most\n\\[\np^{-2r+\\vp((2r)!)}.\n\\]\nThis tends to zero because \\(p\\) is odd. The remaining factors have bounded Gauss norm. Thus the series converges in the Tate algebra. At a nonnegative integer \\(x\\), the binomial factor vanishes for large \\(r\\), and \\cref{hist:thm:depleted_odd_reconstruction} gives the value.\n\\end{proof}","\\begin{histremark}\\label{hist:rem:depleted_factor_boundary}\nThe analytic function is defined by its unit source series. The expression \\(p^{(p-1)x+j}\\) in the interpolation identity is evaluated only at nonnegative integers. No analytic function \\(x\\mapsto p^x\\) on all of \\(\\Zp\\) is used.\n\\end{histremark}","For \\(x\\in\\mathbb Z_2\\), define\n\\[\n\\boxed{\n\\mathscr R(x)\n=\n\\frac{9^{x+1}-1}{8(x+1)(2x+1)}.\n}\n\\]\nThe apparent singularity at \\(x=-1\\) is removable, with value \\(-\\log(9)/8\\).","\\begin{histlemma}[Analytic dyadic source]\\label{hist:lem:analytic_dyadic_source}\nThe function \\(\\mathscr R\\) belongs to the dyadic Tate algebra and takes values in \\(\\mathbb Z_2\\). For every nonnegative integer \\(n\\),\n\\[\n\\boxed{\\mathscr R(n)=R_{2n}.}\n\\]\n\\end{histlemma}","\\begin{proof}\nPut \\(y=x+1\\) and let \\(\\ell=\\log 9\\). Since \\(\\vTwo(\\ell)=3\\),\n\\[\n\\frac{9^y-1}{8y}\n=\n\\frac{\\ell}{8}\n\\sum_{m\\geq0}\n\\frac{\\ell^m}{(m+1)!}y^m.\n\\]\nThe coefficient of \\(y^m\\) has valuation\n\\[\n3m-\\vTwo((m+1)!),\n\\]\nwhich is nonnegative and tends to positive infinity. The quotient therefore belongs to \\(\\mathbb Z_2\\langle y\\rangle\\). The factor \\(2x+1=1+2x\\) is a unit in \\(\\mathbb Z_2\\langle x\\rangle\\), with inverse \\(\\sum_{m\\geq0}(-2x)^m\\). Hence \\(\\mathscr R\\) is integral and analytic. The integer values follow from the definition of \\(R_{2n}\\).\n\\end{proof}","nd let \\(\\ell=\\log 9\\). Since \\(\\vTwo(\\ell)=3\\),\n\\[\n\\frac{9^y-1}{8y}\n=\n\\frac{\\ell}{8}\n\\sum_{m\\geq0}\n\\frac{\\ell^m}{(m+1)!}y^m.\n\\]\nThe coefficient of \\(y^m\\) has valuation\n\\[\n3m-\\vTwo((m+1)!),\n\\]\nwhich is nonnegative and tends to positive infinity. The quotient therefore belongs to \\(\\mathbb Z_2\\langle y\\rangle\\). The factor \\(2x+1=1+2x\\) is a unit in \\(\\mathbb Z_2\\langle x\\rangle\\), with inverse \\(\\sum_{m\\geq0}(-2x)^m\\). Hence \\(\\mathscr R\\) is integral and analytic. The integer values follow from the definition of \\(R_{2n}\\).\n\\end{proof}","\\vTwo(\\ell)=3\\),\n\\[\n\\frac{9^y-1}{8y}\n=\n\\frac{\\ell}{8}\n\\sum_{m\\geq0}\n\\frac{\\ell^m}{(m+1)!}y^m.\n\\]\nThe coefficient of \\(y^m\\) has valuation\n\\[\n3m-\\vTwo((m+1)!),\n\\]\nwhich is nonnegative and tends to positive infinity. The quotient therefore belongs to \\(\\mathbb Z_2\\langle y\\rangle\\). The factor \\(2x+1=1+2x\\) is a unit in \\(\\mathbb Z_2\\langle x\\rangle\\), with inverse \\(\\sum_{m\\geq0}(-2x)^m\\). Hence \\(\\mathscr R\\) is integral and analytic. The integer values follow from the definition of \\(R_{2n}\\).\n\\end{proof}","^m.\n\\]\nThe coefficient of \\(y^m\\) has valuation\n\\[\n3m-\\vTwo((m+1)!),\n\\]\nwhich is nonnegative and tends to positive infinity. The quotient therefore belongs to \\(\\mathbb Z_2\\langle y\\rangle\\). The factor \\(2x+1=1+2x\\) is a unit in \\(\\mathbb Z_2\\langle x\\rangle\\), with inverse \\(\\sum_{m\\geq0}(-2x)^m\\). Hence \\(\\mathscr R\\) is integral and analytic. The integer values follow from the definition of \\(R_{2n}\\).\n\\end{proof}","Define\n\\[\n\\boxed{\n\\mathscr E_2(x)\n=\n\\sum_{r\\geq0}\n\\kappa_{2,r}\n\\binom{2x}{2r}\n\\mathscr R(x-r).\n}\n\\]","\\begin{histtheorem}[Direct dyadic Euler interpolation]\\label{hist:thm:direct_dyadic_interpolation}\nThe series defining \\(\\mathscr E_2\\) converges in the dyadic Tate algebra. For every nonnegative integer \\(n\\),\n\\[\n\\boxed{\\mathscr E_2(n)=E_{2n}.}\n\\]\nTruncation after depth \\(M\\) is uniformly valid modulo \\(2^{4M+3}\\) on \\(\\mathbb Z_2\\).\n\\end{histtheorem}","\\begin{proof}\nThe monomial Gauss norm of \\(\\binom{2x}{2r}\\) is at most \\(2^{\\vTwo((2r)!)}\\). Since\n\\[\n\\vTwo(\\kappa_{2,r})=4r-1,\n\\]\nthe product tends to zero in Gauss norm. Translation by the integer \\(r\\) preserves the unit disc, so \\(\\mathscr R(x-r)\\) remains an integral analytic function with uniformly bounded supremum norm. The series therefore converges in the Tate algebra.","At a nonnegative integer \\(n\\), every term with \\(r>n\\) vanishes and \\cref{hist:thm:dyadic_coordinate_pair} remains. For arbitrary \\(x\\in\\mathbb Z_2\\), the integer valued polynomial \\(\\binom{2x}{2r}\\) and the source value \\(\\mathscr R(x-r)\\) lie in \\(\\mathbb Z_2\\). Every omitted term after depth \\(M\\) is therefore divisible by \\(2^{4M+3}\\). This proves the uniform truncation statement.\n\\end{proof}","\\begin{histtheorem}[Unified local interpolation statement]\\label{hist:thm:unified_local_interpolation}\nThe odd branch \\(\\mathscr E_{p,j}^\\times\\) and the dyadic branch \\(\\mathscr E_2\\) are analytic local realizations of the source and kernel law in \\cref{hist:thm:unified_local_euler}. At integer arguments they recover, respectively,\n\\[\n\\left(1-\\sigma_pp^n\\right)E_n\n\\]\non each even odd prime character branch and\n\\[\nE_{2n}\n\\]\non the dyadic even branch. The two realizations have distinct source corrections and distinct kernel valuation slopes.\n\\end{histtheorem}","m}[Unified local interpolation statement]\\label{hist:thm:unified_local_interpolation}\nThe odd branch \\(\\mathscr E_{p,j}^\\times\\) and the dyadic branch \\(\\mathscr E_2\\) are analytic local realizations of the source and kernel law in \\cref{hist:thm:unified_local_euler}. At integer arguments they recover, respectively,\n\\[\n\\left(1-\\sigma_pp^n\\right)E_n\n\\]\non each even odd prime character branch and\n\\[\nE_{2n}\n\\]\non the dyadic even branch. The two realizations have distinct source corrections and distinct kernel valuation slopes.\n\\end{histtheorem}","Polyphase refinement replaces one local branch by finitely many digit branches while preserving exact synthesis. For sequences it is residue analysis. For local functions it is the digit tree of \\(\\mathbb Z_p\\). In Mahler coordinates it becomes a finite triangular kernel.","For a sequence \\(a=(a_n)_{n\\geq0}\\) and \\(M\\geq2\\), define\n\\[\na_r^{[M]}(q)=a_{Mq+r}\n\\qquad(0\\leq r<M).\n\\]\nThe analysis map is\n\\[\n\\mathcal P_Ma\n=\n\\left(a_0^{[M]},\\ldots,a_{M-1}^{[M]}\\right).\n\\]\nSynthesis interleaves the branches by the rule\n\\[\na_{Mq+r}=a_r^{[M]}(q).\n\\]","\\begin{histtheorem}[Exact discrete polyphase reconstruction]\\label{hist:thm:discrete_polyphase_reconstruction}\nThe analysis map \\(\\mathcal P_M\\) is a bijection between sequences and ordered families of \\(M\\) sequences. If\n\\[\nA(z)=\\sum_{n\\geq0}a_nz^n\n\\]\nand\n\\[\nA_r^{[M]}(w)=\\sum_{q\\geq0}a_{Mq+r}w^q,\n\\]\nthen\n\\[\n\\boxed{\nA(z)=\\sum_{r=0}^{M-1}z^rA_r^{[M]}(z^M).\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nEvery nonnegative integer has a unique expression \\(Mq+r\\) with \\(0\\leq r<M\\). Group the terms of the ordinary generating function by this residue.\n\\end{proof}","\\begin{histtheorem}[Nested discrete refinement]\\label{hist:thm:nested_discrete_polyphase}\nFor positive integers \\(M,N\\),\n\\[\n\\boxed{\n\\left(a_r^{[M]}\\right)_s^{[N]}\n=\na_{r+Ms}^{[MN]}.\n}\n\\]\nFor a mixed radix path \\((M_1,\\ldots,M_\\ell)\\) with digits \\(0\\leq s_i<M_i\\), the final address is\n\\[\n\\boxed{\nr(\\boldsymbol s)\n=\ns_1+M_1s_2+\\cdots+\n\\left(\\prod_{i=1}^{\\ell-1}M_i\\right)s_\\ell.\n}\n\\]\nOnly the product radix and accumulated address determine the final branch.\n\\end{histtheorem}","\\begin{proof}\nDirect substitution gives\n\\[\n\\left(a_r^{[M]}\\right)_s^{[N]}(q)\n=\na_{M(Nq+s)+r}.\n\\]\nIterate this identity along the mixed radix path.\n\\end{proof}","Let \\(c=a*b\\) be the ordinary Cauchy convolution. Write \\(A_r\\), \\(B_r\\), and \\(C_r\\) for the phase ordinary generating functions.","\\begin{histproposition}[One carry convolution law]\\label{hist:prop:polyphase_carry_convolution}\nFor every residue \\(r\\),\n\\[\n\\boxed{\nC_r(w)\n=\n\\sum_{u+v=r}A_u(w)B_v(w)\n+\nw\\sum_{u+v=r+M}A_u(w)B_v(w).\n}\n\\]\nThe first sum has no phase carry. The second has one coarse index carry.\n\\end{histproposition}","\\begin{proof}\nWrite the input indices as \\(Mq+u\\) and \\(Mq'+v\\). Since \\(0\\leq u,v<M\\), their sum has residue \\(u+v\\) when \\(u+v<M\\), and residue \\(u+v-M\\) with one additional coarse index when \\(u+v\\geq M\\).\n\\end{proof}","\\begin{histtheorem}[Polyphase shift matrix]\\label{hist:thm:polyphase_shift_matrix}\nLet \\(E\\) be the scalar shift \\((Ea)_n=a_{n+1}\\), and let \\(E_q\\) shift the coarse index. Then\n\\[\n\\boxed{\n\\mathcal P_M(Ea)\n=\n\\mathsf S_M(E_q)\\mathcal P_Ma,\n}\n\\]\nwhere\n\\[\n\\mathsf S_M(E_q)\n=\n\\begin{pmatrix}\n0&1&0&\\cdots&0","\\begin{histtheorem}[Polyphase shift matrix]\\label{hist:thm:polyphase_shift_matrix}\nLet \\(E\\) be the scalar shift \\((Ea)_n=a_{n+1}\\), and let \\(E_q\\) shift the coarse index. Then\n\\[\n\\boxed{\n\\mathcal P_M(Ea)\n=\n\\mathsf S_M(E_q)\\mathcal P_Ma,\n}\n\\]\nwhere\n\\[\n\\mathsf S_M(E_q)\n=\n\\begin{pmatrix}\n0&1&0&\\cdots&0\\\\\n0&0&1&\\cdots&0\\\\\n\\vdots&&&\\ddots&\\vdots\\\\\n0&0&0&\\cdots&1\\\\\nE_q&0&0&\\cdots&0\n\\end{pmatrix}.\n\\]\nIt also satisfies\n\\[\n\\boxed{\\mathsf S_M(E_q)^M=E_qI_M.}\n\\]\n\\end{histtheorem}","E_q&0&0&\\cdots&0\n\\end{pmatrix}.\n\\]\nIt also satisfies\n\\[\n\\boxed{\\mathsf S_M(E_q)^M=E_qI_M.}\n\\]\n\\end{histtheorem}","\\begin{proof}\nFor \\(r<M-1\\),\n\\[\n(Ea)_{Mq+r}=a_{Mq+r+1}=a_{r+1}^{[M]}(q).\n\\]\nFor \\(r=M-1\\),\n\\[\n(Ea)_{Mq+M-1}=a_{M(q+1)}=E_qa_0^{[M]}(q).\n\\]\nAfter \\(M\\) shifts every phase has advanced by one coarse index.\n\\end{proof}","For \\(a\\geq1\\) and \\(0\\leq s<p^a\\), define\n\\[\nf_{a,s}(x)=f(p^ax+s).\n\\]","\\begin{histtheorem}[Exact local branch reconstruction]\\label{hist:thm:local_branch_reconstruction}\nThe family \\(\\{f_{a,s}\\}_{0\\leq s<p^a}\\) determines \\(f\\) uniquely on \\(\\mathbb Z_p\\). More precisely, if \\(y\\in s+p^a\\mathbb Z_p\\), then\n\\[\n\\boxed{\nf(y)=f_{a,s}\\left(\\frac{y-s}{p^a}\\right).\n}\n\\]\nFor \\(a,b\\geq1\\),\n\\[\n\\boxed{\n(f_{a,s})_{b,t}\n=\nf_{a+b,s+p^at}.\n}\n\\]\nThus local refinement is exactly the \\(p\\) adic digit tree.\n\\end{histtheorem}","egin{histtheorem}[Exact local branch reconstruction]\\label{hist:thm:local_branch_reconstruction}\nThe family \\(\\{f_{a,s}\\}_{0\\leq s<p^a}\\) determines \\(f\\) uniquely on \\(\\mathbb Z_p\\). More precisely, if \\(y\\in s+p^a\\mathbb Z_p\\), then\n\\[\n\\boxed{\nf(y)=f_{a,s}\\left(\\frac{y-s}{p^a}\\right).\n}\n\\]\nFor \\(a,b\\geq1\\),\n\\[\n\\boxed{\n(f_{a,s})_{b,t}\n=\nf_{a+b,s+p^at}.\n}\n\\]\nThus local refinement is exactly the \\(p\\) adic digit tree.\n\\end{histtheorem}","\\begin{proof}\nThe cosets \\(s+p^a\\mathbb Z_p\\) form a disjoint partition of \\(\\mathbb Z_p\\). The first formula is the inverse affine change of variables on one coset. The second follows from\n\\[\np^a(p^bx+t)+s\n=\np^{a+b}x+p^at+s.\n\\]\n\\end{proof}","Let\n\\[\n\\Phi_M(z)=(1+z)^M-1.\n\\]\nFor \\(M\\geq1\\), \\(s\\geq0\\), and \\(r,k\\geq0\\), define\n\\[\n\\boxed{\n\\Theta_{M,s}(r,k)\n=\n[z^r](1+z)^s\\Phi_M(z)^k.\n}\n\\]\nEquivalently,\n\\[\n\\boxed{\n\\Theta_{M,s}(r,k)\n=\n\\sum_{j=0}^{k}\n(-1)^{k-j}\\binom kj\\binom{Mj+s}{r}.\n}\n\\]\nIt vanishes unless\n\\[\nk\\leq r\\leq Mk+s.\n\\]","\\begin{histproposition}[Edge coefficients of the affine kernel]\\label{hist:prop:affine_kernel_edges}\nFor every \\(M\\geq1\\), \\(s\\geq0\\), and \\(k\\geq0\\),\n\\[\n\\boxed{\n\\Theta_{M,s}(k,k)=M^k,\n\\qquad\n\\Theta_{M,s}(Mk+s,k)=1.\n}\n\\]\nFor binary radix and zero address,\n\\[\n\\boxed{\n\\Theta_{2,0}(k+j,k)\n=\n2^{k-j}\\binom kj\n\\qquad(0\\leq j\\leq k).\n}\n\\]\nHence\n\\[\n\\boxed{\n\\vTwo\\left(\\Theta_{2,0}(k+j,k)\\right)\n=\nk-j+\\vTwo\\binom kj.\n}\n\\]\n\\end{histproposition}","\\begin{proof}\nThe lowest degree term of \\(\\Phi_M(z)\\) is \\(Mz\\), and its highest degree term is \\(z^M\\). Multiplication by \\((1+z)^s\\) gives the two edge coefficients. When \\(M=2\\) and \\(s=0\\),\n\\[\n\\Phi_2(z)^k\n=\n(2z+z^2)^k\n=\nz^k(2+z)^k.\n\\]\nTaking the coefficient of \\(z^{k+j}\\) gives the binary formula.\n\\end{proof}","\\begin{histtheorem}[Mahler polyphase transfer]\\label{hist:thm:mahler_polyphase_transfer}\nLet\n\\[\nf(x)=\\sum_{r\\geq0}c_r\\binom xr\n\\]\nbe a convergent Mahler expansion. For\n\\[\ng_s(q)=f(Mq+s),\n\\]\none has\n\\[\n\\boxed{\ng_s(q)=\\sum_{k\\geq0}d_{s,k}\\binom qk,}\n\\]\nwhere\n\\[\n\\boxed{\nd_{s,k}\n=\n\\sum_{r=k}^{Mk+s}\nc_r\\Theta_{M,s}(r,k).\n}\n\\]\nEvery refined coefficient uses a finite window of the original Mahler coefficients.\n\\end{histtheorem}","\\begin{proof}\nUse\n\\[\n(1+z)^{Mq+s}\n=\n(1+z)^s\\left(1+\\Phi_M(z)\\right)^q\n=\n\\sum_{k\\geq0}\\binom qk(1+z)^s\\Phi_M(z)^k.\n\\]\nTaking the coefficient of \\(z^r\\) gives the affine expansion of \\(\\binom{Mq+s}{r}\\). Since \\(c_r\\to0\\) and every \\(d_{s,k}\\) uses indices \\(r\\geq k\\), the transformed coefficients also tend to zero.\n\\end{proof}","\\begin{histtheorem}[Kernel composition]\\label{hist:thm:polyphase_kernel_composition}\nFor positive integers \\(M,N\\),\n\\[\n\\boxed{\n\\sum_{k\\geq0}\n\\Theta_{M,s}(r,k)\n\\Theta_{N,t}(k,\\ell)\n=\n\\Theta_{MN,Mt+s}(r,\\ell).\n}\n\\]\nThe sum is finite.\n\\end{histtheorem}","\\begin{proof}\nApply \\cref{hist:thm:mahler_polyphase_transfer} first to \\(x=Mq+s\\) and then to \\(q=Ny+t\\). The composite substitution is \\(x=MNy+Mt+s\\). Uniqueness of Mahler coefficients gives the identity.\n\\end{proof}","\\begin{histtheorem}[Polyphase valuation ledger]\\label{hist:thm:polyphase_valuation_ledger}\nFor every refined coefficient,\n\\[\n\\boxed{\n\\vp(d_{s,k})\n\\geq\n\\min_{k\\leq r\\leq Mk+s}\n\\left(\n\\vp(c_r)+\\vp(\\Theta_{M,s}(r,k))\n\\right).\n}\n\\]\nIf \\(\\vp(c_r)\\geq w(r)\\) for a nondecreasing function \\(w\\), then\n\\[\n\\boxed{\\vp(d_{s,k})\\geq w(k).}\n\\]\nThus every monotone Mahler precision ladder survives arbitrary polyphase refinement.\n\\end{histtheorem}","\\begin{proof}\nApply the nonarchimedean triangle inequality to the finite coefficient formula. Since \\(r\\geq k\\), monotonicity gives \\(w(r)\\geq w(k)\\), and the affine kernel is integral.\n\\end{proof}","\\begin{histtheorem}[Prime power sparsification]\\label{hist:thm:prime_power_sparsification}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The affine-kernel congruence below needs no change, but its consequence for transformed Mahler coefficients requires $c_r\\in\\Z_p$ for every $r$. Convergence of a $\\Q_p$-valued Mahler expansion alone is insufficient. For $p=2$, $M=2$, $s=0$ and $f(x)=x/2$, one has $c_1=1/2$, $c_2=0$, while $g(q)=q$ has $d_1=1$; thus $d_1\\not\\equiv c_2\\pmod2$. The coefficient consequence is read with integral original coefficients.\\par\\medskip\nLet \\(M=p^a\\). Then\n\\[\n\\boxed{\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p,\n}\n\\]\nwhere the binomial coefficient is zero outside its natural range. Consequently,\n\\[\n\\boxed{\nd_{s,k}\n\\equiv\n\\sum_{u=0}^{s}\n\\binom su\nc_{p^ak+u}\n\\pmod p.\n}\n\\]\nIn particular,\n\\[\n\\boxed{d_{0,k}\\equiv c_{p^ak}\\pmod p.}\n\\]\n\\end{histtheorem}","theorem}[Prime power sparsification]\\label{hist:thm:prime_power_sparsification}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The affine-kernel congruence below needs no change, but its consequence for transformed Mahler coefficients requires $c_r\\in\\Z_p$ for every $r$. Convergence of a $\\Q_p$-valued Mahler expansion alone is insufficient. For $p=2$, $M=2$, $s=0$ and $f(x)=x/2$, one has $c_1=1/2$, $c_2=0$, while $g(q)=q$ has $d_1=1$; thus $d_1\\not\\equiv c_2\\pmod2$. The coefficient consequence is read with integral original coefficients.\\par\\medskip\nLet \\(M=p^a\\). Then\n\\[\n\\boxed{\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p,\n}\n\\]\nwhere the binomial coefficient is zero outside its natural range. Consequently,\n\\[\n\\boxed{\nd_{s,k}\n\\equiv\n\\sum_{u=0}^{s}\n\\binom su\nc_{p^ak+u}\n\\pmod p.\n}\n\\]\nIn particular,\n\\[\n\\boxed{d_{0,k}\\equiv c_{p^ak}\\pmod p.}\n\\]\n\\end{histtheorem}","ime power sparsification]\\label{hist:thm:prime_power_sparsification}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The affine-kernel congruence below needs no change, but its consequence for transformed Mahler coefficients requires $c_r\\in\\Z_p$ for every $r$. Convergence of a $\\Q_p$-valued Mahler expansion alone is insufficient. For $p=2$, $M=2$, $s=0$ and $f(x)=x/2$, one has $c_1=1/2$, $c_2=0$, while $g(q)=q$ has $d_1=1$; thus $d_1\\not\\equiv c_2\\pmod2$. The coefficient consequence is read with integral original coefficients.\\par\\medskip\nLet \\(M=p^a\\). Then\n\\[\n\\boxed{\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p,\n}\n\\]\nwhere the binomial coefficient is zero outside its natural range. Consequently,\n\\[\n\\boxed{\nd_{s,k}\n\\equiv\n\\sum_{u=0}^{s}\n\\binom su\nc_{p^ak+u}\n\\pmod p.\n}\n\\]\nIn particular,\n\\[\n\\boxed{d_{0,k}\\equiv c_{p^ak}\\pmod p.}\n\\]\n\\end{histtheorem}","ification]\\label{hist:thm:prime_power_sparsification}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The affine-kernel congruence below needs no change, but its consequence for transformed Mahler coefficients requires $c_r\\in\\Z_p$ for every $r$. Convergence of a $\\Q_p$-valued Mahler expansion alone is insufficient. For $p=2$, $M=2$, $s=0$ and $f(x)=x/2$, one has $c_1=1/2$, $c_2=0$, while $g(q)=q$ has $d_1=1$; thus $d_1\\not\\equiv c_2\\pmod2$. The coefficient consequence is read with integral original coefficients.\\par\\medskip\nLet \\(M=p^a\\). Then\n\\[\n\\boxed{\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p,\n}\n\\]\nwhere the binomial coefficient is zero outside its natural range. Consequently,\n\\[\n\\boxed{\nd_{s,k}\n\\equiv\n\\sum_{u=0}^{s}\n\\binom su\nc_{p^ak+u}\n\\pmod p.\n}\n\\]\nIn particular,\n\\[\n\\boxed{d_{0,k}\\equiv c_{p^ak}\\pmod p.}\n\\]\n\\end{histtheorem}","bel{hist:thm:prime_power_sparsification}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The affine-kernel congruence below needs no change, but its consequence for transformed Mahler coefficients requires $c_r\\in\\Z_p$ for every $r$. Convergence of a $\\Q_p$-valued Mahler expansion alone is insufficient. For $p=2$, $M=2$, $s=0$ and $f(x)=x/2$, one has $c_1=1/2$, $c_2=0$, while $g(q)=q$ has $d_1=1$; thus $d_1\\not\\equiv c_2\\pmod2$. The coefficient consequence is read with integral original coefficients.\\par\\medskip\nLet \\(M=p^a\\). Then\n\\[\n\\boxed{\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p,\n}\n\\]\nwhere the binomial coefficient is zero outside its natural range. Consequently,\n\\[\n\\boxed{\nd_{s,k}\n\\equiv\n\\sum_{u=0}^{s}\n\\binom su\nc_{p^ak+u}\n\\pmod p.\n}\n\\]\nIn particular,\n\\[\n\\boxed{d_{0,k}\\equiv c_{p^ak}\\pmod p.}\n\\]\n\\end{histtheorem}","rime_power_sparsification}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} The affine-kernel congruence below needs no change, but its consequence for transformed Mahler coefficients requires $c_r\\in\\Z_p$ for every $r$. Convergence of a $\\Q_p$-valued Mahler expansion alone is insufficient. For $p=2$, $M=2$, $s=0$ and $f(x)=x/2$, one has $c_1=1/2$, $c_2=0$, while $g(q)=q$ has $d_1=1$; thus $d_1\\not\\equiv c_2\\pmod2$. The coefficient consequence is read with integral original coefficients.\\par\\medskip\nLet \\(M=p^a\\). Then\n\\[\n\\boxed{\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p,\n}\n\\]\nwhere the binomial coefficient is zero outside its natural range. Consequently,\n\\[\n\\boxed{\nd_{s,k}\n\\equiv\n\\sum_{u=0}^{s}\n\\binom su\nc_{p^ak+u}\n\\pmod p.\n}\n\\]\nIn particular,\n\\[\n\\boxed{d_{0,k}\\equiv c_{p^ak}\\pmod p.}\n\\]\n\\end{histtheorem}","needs no change, but its consequence for transformed Mahler coefficients requires $c_r\\in\\Z_p$ for every $r$. Convergence of a $\\Q_p$-valued Mahler expansion alone is insufficient. For $p=2$, $M=2$, $s=0$ and $f(x)=x/2$, one has $c_1=1/2$, $c_2=0$, while $g(q)=q$ has $d_1=1$; thus $d_1\\not\\equiv c_2\\pmod2$. The coefficient consequence is read with integral original coefficients.\\par\\medskip\nLet \\(M=p^a\\). Then\n\\[\n\\boxed{\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p,\n}\n\\]\nwhere the binomial coefficient is zero outside its natural range. Consequently,\n\\[\n\\boxed{\nd_{s,k}\n\\equiv\n\\sum_{u=0}^{s}\n\\binom su\nc_{p^ak+u}\n\\pmod p.\n}\n\\]\nIn particular,\n\\[\n\\boxed{d_{0,k}\\equiv c_{p^ak}\\pmod p.}\n\\]\n\\end{histtheorem}","but its consequence for transformed Mahler coefficients requires $c_r\\in\\Z_p$ for every $r$. Convergence of a $\\Q_p$-valued Mahler expansion alone is insufficient. For $p=2$, $M=2$, $s=0$ and $f(x)=x/2$, one has $c_1=1/2$, $c_2=0$, while $g(q)=q$ has $d_1=1$; thus $d_1\\not\\equiv c_2\\pmod2$. The coefficient consequence is read with integral original coefficients.\\par\\medskip\nLet \\(M=p^a\\). Then\n\\[\n\\boxed{\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p,\n}\n\\]\nwhere the binomial coefficient is zero outside its natural range. Consequently,\n\\[\n\\boxed{\nd_{s,k}\n\\equiv\n\\sum_{u=0}^{s}\n\\binom su\nc_{p^ak+u}\n\\pmod p.\n}\n\\]\nIn particular,\n\\[\n\\boxed{d_{0,k}\\equiv c_{p^ak}\\pmod p.}\n\\]\n\\end{histtheorem}","insufficient. For $p=2$, $M=2$, $s=0$ and $f(x)=x/2$, one has $c_1=1/2$, $c_2=0$, while $g(q)=q$ has $d_1=1$; thus $d_1\\not\\equiv c_2\\pmod2$. The coefficient consequence is read with integral original coefficients.\\par\\medskip\nLet \\(M=p^a\\). Then\n\\[\n\\boxed{\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p,\n}\n\\]\nwhere the binomial coefficient is zero outside its natural range. Consequently,\n\\[\n\\boxed{\nd_{s,k}\n\\equiv\n\\sum_{u=0}^{s}\n\\binom su\nc_{p^ak+u}\n\\pmod p.\n}\n\\]\nIn particular,\n\\[\n\\boxed{d_{0,k}\\equiv c_{p^ak}\\pmod p.}\n\\]\n\\end{histtheorem}","(q)=q$ has $d_1=1$; thus $d_1\\not\\equiv c_2\\pmod2$. The coefficient consequence is read with integral original coefficients.\\par\\medskip\nLet \\(M=p^a\\). Then\n\\[\n\\boxed{\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p,\n}\n\\]\nwhere the binomial coefficient is zero outside its natural range. Consequently,\n\\[\n\\boxed{\nd_{s,k}\n\\equiv\n\\sum_{u=0}^{s}\n\\binom su\nc_{p^ak+u}\n\\pmod p.\n}\n\\]\nIn particular,\n\\[\n\\boxed{d_{0,k}\\equiv c_{p^ak}\\pmod p.}\n\\]\n\\end{histtheorem}","\\begin{proof}\nIn characteristic \\(p\\),\n\\[\n(1+z)^{p^a}-1\\equiv z^{p^a}.\n\\]\nSubstitute this congruence in the definition of \\(\\Theta_{p^a,s}\\).\n\\end{proof}","\\begin{histtheorem}[Scalar polyphase precision amplification]\\label{hist:thm:scalar_polyphase_precision}\nLet \\(u\\in\\mathbb Z_p^\\times\\), let \\(a,r\\geq0\\), and put\n\\[\nH_a=\n\\begin{cases}\n(p-1)p^a,&p\\text{ odd},","\\begin{histtheorem}[Scalar polyphase precision amplification]\\label{hist:thm:scalar_polyphase_precision}\nLet \\(u\\in\\mathbb Z_p^\\times\\), let \\(a,r\\geq0\\), and put\n\\[\nH_a=\n\\begin{cases}\n(p-1)p^a,&p\\text{ odd},\\\\\n2^{a+1},&p=2.\n\\end{cases}\n\\]\nThen\n\\[\n\\boxed{\n\\Delta_{H_a}^{\\,r}u^n\n=\nu^n\\left(u^{H_a}-1\\right)^r.\n}\n\\]\nFor odd \\(p\\),\n\\[\n\\boxed{\n\\vp\\left(\\Delta_{H_a}^{\\,r}u^n\\right)\n\\geq(a+1)r.\n}\n\\]\nFor \\(p=2\\) and odd \\(u\\),\n\\[\n\\boxed{\n\\vTwo\\left(\\Delta_{H_a}^{\\,r}u^n\\right)\n\\geq(a+3)r.\n}\n\\]\nEquivalently, every additional prime power digit increases the certified Mahler slope by one.\n\\end{histtheorem}","2^{a+1},&p=2.\n\\end{cases}\n\\]\nThen\n\\[\n\\boxed{\n\\Delta_{H_a}^{\\,r}u^n\n=\nu^n\\left(u^{H_a}-1\\right)^r.\n}\n\\]\nFor odd \\(p\\),\n\\[\n\\boxed{\n\\vp\\left(\\Delta_{H_a}^{\\,r}u^n\\right)\n\\geq(a+1)r.\n}\n\\]\nFor \\(p=2\\) and odd \\(u\\),\n\\[\n\\boxed{\n\\vTwo\\left(\\Delta_{H_a}^{\\,r}u^n\\right)\n\\geq(a+3)r.\n}\n\\]\nEquivalently, every additional prime power digit increases the certified Mahler slope by one.\n\\end{histtheorem}","\\begin{proof}\nOne difference with step \\(H_a\\) multiplies the unit mode by \\(u^{H_a}-1\\). Iterate. The valuation estimates are those of \\cref{hist:cor:basic_local_difference_gains}.\n\\end{proof}","\\begin{histlemma}[Matrix power amplification]\\label{hist:lem:matrix_power_amplification}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} Require $L\\ge1$. At $L=0$ the standalone claim is false: for $p=3$, $d=1$ and $X=2=1+B$, one has $X^3-1=7\\notin3\\Z_3$. With $L\\ge1$, every higher binomial term in the retained proof has the required valuation, including $p=2$.\\par\\medskip\nIf\n\\[\nX=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\), then for every \\(a\\geq0\\) there is \\(B_a\\in M_d(\\Zp)\\) such that\n\\[\n\\boxed{X^{p^a}=I+p^{L+a}B_a.}\n\\]\n\\end{histlemma}","istlemma}[Matrix power amplification]\\label{hist:lem:matrix_power_amplification}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} Require $L\\ge1$. At $L=0$ the standalone claim is false: for $p=3$, $d=1$ and $X=2=1+B$, one has $X^3-1=7\\notin3\\Z_3$. With $L\\ge1$, every higher binomial term in the retained proof has the required valuation, including $p=2$.\\par\\medskip\nIf\n\\[\nX=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\), then for every \\(a\\geq0\\) there is \\(B_a\\in M_d(\\Zp)\\) such that\n\\[\n\\boxed{X^{p^a}=I+p^{L+a}B_a.}\n\\]\n\\end{histlemma}","on]\\label{hist:lem:matrix_power_amplification}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} Require $L\\ge1$. At $L=0$ the standalone claim is false: for $p=3$, $d=1$ and $X=2=1+B$, one has $X^3-1=7\\notin3\\Z_3$. With $L\\ge1$, every higher binomial term in the retained proof has the required valuation, including $p=2$.\\par\\medskip\nIf\n\\[\nX=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\), then for every \\(a\\geq0\\) there is \\(B_a\\in M_d(\\Zp)\\) such that\n\\[\n\\boxed{X^{p^a}=I+p^{L+a}B_a.}\n\\]\n\\end{histlemma}","atrix_power_amplification}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} Require $L\\ge1$. At $L=0$ the standalone claim is false: for $p=3$, $d=1$ and $X=2=1+B$, one has $X^3-1=7\\notin3\\Z_3$. With $L\\ge1$, every higher binomial term in the retained proof has the required valuation, including $p=2$.\\par\\medskip\nIf\n\\[\nX=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\), then for every \\(a\\geq0\\) there is \\(B_a\\in M_d(\\Zp)\\) such that\n\\[\n\\boxed{X^{p^a}=I+p^{L+a}B_a.}\n\\]\n\\end{histlemma}","r\\noindent\\textbf{Editorial hypothesis correction.} Require $L\\ge1$. At $L=0$ the standalone claim is false: for $p=3$, $d=1$ and $X=2=1+B$, one has $X^3-1=7\\notin3\\Z_3$. With $L\\ge1$, every higher binomial term in the retained proof has the required valuation, including $p=2$.\\par\\medskip\nIf\n\\[\nX=I+p^LB\n\\]\nwith \\(B\\in M_d(\\Zp)\\), then for every \\(a\\geq0\\) there is \\(B_a\\in M_d(\\Zp)\\) such that\n\\[\n\\boxed{X^{p^a}=I+p^{L+a}B_a.}\n\\]\n\\end{histlemma}","\\begin{proof}\nInduct on \\(a\\). Expand the \\(p\\)th power of \\(I+p^{L+a}B_a\\). The linear term has valuation at least \\(L+a+1\\), and every higher term has at least that valuation.\n\\end{proof}","\\begin{histtheorem}[Matrix polyphase precision amplification]\\label{hist:thm:matrix_polyphase_precision}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} Require $L\\ge1$, $C,B\\in M_d(\\Z_p)$, $v_0,w\\in\\Z_p^d$, $H>0$, and $a,j,n,r\\in\\N$. These are the hypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive $L$, the preceding counterexamples invalidate the claimed precision.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB.\n\\]\nPut \\(H_a=Hp^a\\). Then\n\\[\nC^{H_a}=I+p^{L+a}B_a\n\\]\nfor an integral matrix \\(B_a\\). Every refined packet\n\\[\nb_{a,j}(x)=a_{H_ax+j}\n\\]\nhas Mahler coefficients satisfying\n\\[\n\\boxed{\\vp(c_{a,j,r})\\geq(L+a)r.}\n\\]\nEquivalently,\n\\[\n\\boxed{\n\\Delta_{H_a}^{\\,r}a_n\n\\equiv0\n\\pmod {p^{(L+a)r}}.\n}\n\\]\n\\end{histtheorem}","thm:matrix_polyphase_precision}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} Require $L\\ge1$, $C,B\\in M_d(\\Z_p)$, $v_0,w\\in\\Z_p^d$, $H>0$, and $a,j,n,r\\in\\N$. These are the hypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive $L$, the preceding counterexamples invalidate the claimed precision.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB.\n\\]\nPut \\(H_a=Hp^a\\). Then\n\\[\nC^{H_a}=I+p^{L+a}B_a\n\\]\nfor an integral matrix \\(B_a\\). Every refined packet\n\\[\nb_{a,j}(x)=a_{H_ax+j}\n\\]\nhas Mahler coefficients satisfying\n\\[\n\\boxed{\\vp(c_{a,j,r})\\geq(L+a)r.}\n\\]\nEquivalently,\n\\[\n\\boxed{\n\\Delta_{H_a}^{\\,r}a_n\n\\equiv0\n\\pmod {p^{(L+a)r}}.\n}\n\\]\n\\end{histtheorem}","recision}\n\\par\\noindent\\textbf{Editorial hypothesis correction.} Require $L\\ge1$, $C,B\\in M_d(\\Z_p)$, $v_0,w\\in\\Z_p^d$, $H>0$, and $a,j,n,r\\in\\N$. These are the hypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive $L$, the preceding counterexamples invalidate the claimed precision.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB.\n\\]\nPut \\(H_a=Hp^a\\). Then\n\\[\nC^{H_a}=I+p^{L+a}B_a\n\\]\nfor an integral matrix \\(B_a\\). Every refined packet\n\\[\nb_{a,j}(x)=a_{H_ax+j}\n\\]\nhas Mahler coefficients satisfying\n\\[\n\\boxed{\\vp(c_{a,j,r})\\geq(L+a)r.}\n\\]\nEquivalently,\n\\[\n\\boxed{\n\\Delta_{H_a}^{\\,r}a_n\n\\equiv0\n\\pmod {p^{(L+a)r}}.\n}\n\\]\n\\end{histtheorem}","dent\\textbf{Editorial hypothesis correction.} Require $L\\ge1$, $C,B\\in M_d(\\Z_p)$, $v_0,w\\in\\Z_p^d$, $H>0$, and $a,j,n,r\\in\\N$. These are the hypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive $L$, the preceding counterexamples invalidate the claimed precision.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB.\n\\]\nPut \\(H_a=Hp^a\\). Then\n\\[\nC^{H_a}=I+p^{L+a}B_a\n\\]\nfor an integral matrix \\(B_a\\). Every refined packet\n\\[\nb_{a,j}(x)=a_{H_ax+j}\n\\]\nhas Mahler coefficients satisfying\n\\[\n\\boxed{\\vp(c_{a,j,r})\\geq(L+a)r.}\n\\]\nEquivalently,\n\\[\n\\boxed{\n\\Delta_{H_a}^{\\,r}a_n\n\\equiv0\n\\pmod {p^{(L+a)r}}.\n}\n\\]\n\\end{histtheorem}","ire $L\\ge1$, $C,B\\in M_d(\\Z_p)$, $v_0,w\\in\\Z_p^d$, $H>0$, and $a,j,n,r\\in\\N$. These are the hypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive $L$, the preceding counterexamples invalidate the claimed precision.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB.\n\\]\nPut \\(H_a=Hp^a\\). Then\n\\[\nC^{H_a}=I+p^{L+a}B_a\n\\]\nfor an integral matrix \\(B_a\\). Every refined packet\n\\[\nb_{a,j}(x)=a_{H_ax+j}\n\\]\nhas Mahler coefficients satisfying\n\\[\n\\boxed{\\vp(c_{a,j,r})\\geq(L+a)r.}\n\\]\nEquivalently,\n\\[\n\\boxed{\n\\Delta_{H_a}^{\\,r}a_n\n\\equiv0\n\\pmod {p^{(L+a)r}}.\n}\n\\]\n\\end{histtheorem}","$, $v_0,w\\in\\Z_p^d$, $H>0$, and $a,j,n,r\\in\\N$. These are the hypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive $L$, the preceding counterexamples invalidate the claimed precision.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB.\n\\]\nPut \\(H_a=Hp^a\\). Then\n\\[\nC^{H_a}=I+p^{L+a}B_a\n\\]\nfor an integral matrix \\(B_a\\). Every refined packet\n\\[\nb_{a,j}(x)=a_{H_ax+j}\n\\]\nhas Mahler coefficients satisfying\n\\[\n\\boxed{\\vp(c_{a,j,r})\\geq(L+a)r.}\n\\]\nEquivalently,\n\\[\n\\boxed{\n\\Delta_{H_a}^{\\,r}a_n\n\\equiv0\n\\pmod {p^{(L+a)r}}.\n}\n\\]\n\\end{histtheorem}","ypotheses of the two corrected matrix statements used by the retained proof. Without integral observation vectors or positive $L$, the preceding counterexamples invalidate the claimed precision.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB.\n\\]\nPut \\(H_a=Hp^a\\). Then\n\\[\nC^{H_a}=I+p^{L+a}B_a\n\\]\nfor an integral matrix \\(B_a\\). Every refined packet\n\\[\nb_{a,j}(x)=a_{H_ax+j}\n\\]\nhas Mahler coefficients satisfying\n\\[\n\\boxed{\\vp(c_{a,j,r})\\geq(L+a)r.}\n\\]\nEquivalently,\n\\[\n\\boxed{\n\\Delta_{H_a}^{\\,r}a_n\n\\equiv0\n\\pmod {p^{(L+a)r}}.\n}\n\\]\n\\end{histtheorem}","by the retained proof. Without integral observation vectors or positive $L$, the preceding counterexamples invalidate the claimed precision.\\par\\medskip\nAssume\n\\[\nC^H=I+p^LB.\n\\]\nPut \\(H_a=Hp^a\\). Then\n\\[\nC^{H_a}=I+p^{L+a}B_a\n\\]\nfor an integral matrix \\(B_a\\). Every refined packet\n\\[\nb_{a,j}(x)=a_{H_ax+j}\n\\]\nhas Mahler coefficients satisfying\n\\[\n\\boxed{\\vp(c_{a,j,r})\\geq(L+a)r.}\n\\]\nEquivalently,\n\\[\n\\boxed{\n\\Delta_{H_a}^{\\,r}a_n\n\\equiv0\n\\pmod {p^{(L+a)r}}.\n}\n\\]\n\\end{histtheorem}","\\begin{histexample}[Refined dyadic Fibonacci ladder]\\label{hist:ex:refined_fibonacci_ladder}\nFor the Fibonacci companion matrix\n\\[\nC=\n\\begin{pmatrix}\n1&1\\\\\n1&0\n\\end{pmatrix},\n\\]\none has\n\\[\nC^{12}\n=I+8\n\\begin{pmatrix}\n29&18","1&0\n\\end{pmatrix},\n\\]\none has\n\\[\nC^{12}\n=I+8\n\\begin{pmatrix}\n29&18\\\\\n18&11\n\\end{pmatrix}.\n\\]\nTherefore\n\\[\n\\boxed{\n\\Delta_{12\\cdot2^a}^{\\,r}F_n\n\\equiv0\n\\pmod {2^{(a+3)r}}.\n}\n\\]\nEvery dyadic refinement doubles the step and adds one guaranteed binary digit at each Mahler depth.\n\\end{histexample}","18&11\n\\end{pmatrix}.\n\\]\nTherefore\n\\[\n\\boxed{\n\\Delta_{12\\cdot2^a}^{\\,r}F_n\n\\equiv0\n\\pmod {2^{(a+3)r}}.\n}\n\\]\nEvery dyadic refinement doubles the step and adds one guaranteed binary digit at each Mahler depth.\n\\end{histexample}","For odd \\(p\\), write\n\\[\n\\boxed{\nn=j+(p-1)(s+p^aq),\n}\n\\]\nwith \\(0\\leq j<p-1\\) and \\(0\\leq s<p^a\\). The address \\(j\\) records the finite character. The address \\(s\\) records the principal unit digit. The variable \\(q\\) carries the remaining Mahler expansion.","At the dyadic prime, write\n\\[\n\\boxed{\nn=j+2(s+2^aq),\n}\n\\]\nwith \\(j\\in\\{0,1\\}\\) and \\(0\\leq s<2^a\\). The sign coordinate replaces the odd prime character coordinate.","\\begin{histproposition}[Local phase lattice]\\label{hist:prop:local_phase_lattice}\nFor a unit mode \\(u^n\\), the odd prime coordinate gives\n\\[\n\\boxed{\nu^n\n=\nu^{j+(p-1)s}\n\\left(u^{(p-1)p^a}\\right)^q.\n}\n\\]\nThe dyadic coordinate gives\n\\[\n\\boxed{\nu^n\n=\nu^{j+2s}\n\\left(u^{2^{a+1}}\\right)^q.\n}\n\\]\nThus finite character phase, local digit address, and Mahler depth are separate coordinates.\n\\end{histproposition}","\\begin{proof}\nSubstitute the two displayed decompositions of \\(n\\).\n\\end{proof}","\\begin{histcorollary}[Polyphase refinement of the Euler branches]\\label{hist:cor:euler_polyphase_refinement}\nThe analytic branches in \\cref{hist:thm:odd_analytic_branch,hist:thm:direct_dyadic_interpolation} admit exact refinements under \\(x=p^aq+s\\). Their refined Mahler coefficients are obtained from \\(\\Theta_{p^a,s}\\). The original monotone valuation ladders are preserved, and unit mode or invertible matrix representations gain one certified digit per refinement level.\n\\end{histcorollary}","Let \\(p\\) be a prime. Let\n\\[\nM\\in M_{r,d}(\\Zp),\n\\qquad\nb\\in\\Zp^r.\n\\]\nSuppose that \\(x_L\\in\\Zp^d\\) satisfies\n\\[\nMx_L\\equiv b\\pmod {p^L}\n\\]\nfor some \\(L\\geq1\\). Define the reduced residual\n\\[\n\\rho_L(x_L)\n=\n\\frac{Mx_L-b}{p^L}\n\\pmod p\n\\in\\FF_p^r.\n\\]\nWrite \\(\\overline M\\) for the reduction of \\(M\\) modulo \\(p\\).","\\begin{histdefinition}[Linear lift obstruction]\\label{hist:def:linear_lift_obstruction}\nThe linear obstruction at depth \\(L\\) is\n\\[\n\\boxed{\n\\operatorname{Ob}^{\\mathrm{lin}}_L(x_L)\n=\n[\\rho_L(x_L)]\n\\in\n\\coker(\\overline M).\n}\n\\]\n\\end{histdefinition}","\\begin{histtheorem}[One level linear lifting]\\label{hist:thm:linear_lifting}\nA class represented by \\(x_L\\) lifts to a solution modulo \\(p^{L+1}\\) if and only if\n\\[\n\\boxed{\n\\operatorname{Ob}^{\\mathrm{lin}}_L(x_L)=0.\n}\n\\]\nMore precisely, a correction \\(y\\in\\FF_p^d\\) gives a lift\n\\[\nx_{L+1}=x_L+p^Ly\n\\pmod {p^{L+1}}\n\\]\nif and only if\n\\[\n\\boxed{\n\\overline My=-\\rho_L(x_L).\n}\n\\]\nWhen one correction exists, the set of all correction classes is a coset of \\(\\ker\\overline M\\).\n\\end{histtheorem}","st:thm:linear_lifting}\nA class represented by \\(x_L\\) lifts to a solution modulo \\(p^{L+1}\\) if and only if\n\\[\n\\boxed{\n\\operatorname{Ob}^{\\mathrm{lin}}_L(x_L)=0.\n}\n\\]\nMore precisely, a correction \\(y\\in\\FF_p^d\\) gives a lift\n\\[\nx_{L+1}=x_L+p^Ly\n\\pmod {p^{L+1}}\n\\]\nif and only if\n\\[\n\\boxed{\n\\overline My=-\\rho_L(x_L).\n}\n\\]\nWhen one correction exists, the set of all correction classes is a coset of \\(\\ker\\overline M\\).\n\\end{histtheorem}","\\begin{proof}\nModulo \\(p^{L+1}\\),\n\\[\nM(x_L+p^Ly)-b\n=\nMx_L-b+p^LMy.\n\\]\nAfter division by \\(p^L\\) and reduction modulo \\(p\\), the lift condition becomes\n\\[\n\\rho_L(x_L)+\\overline My=0.\n\\]\nThis equation is solvable exactly when the residual lies in the image of \\(\\overline M\\). If \\(y_0\\) is one solution, every solution is \\(y_0+z\\) with \\(z\\in\\ker\\overline M\\).\n\\end{proof}","\\begin{histproposition}[Representative independence]\\label{hist:prop:linear_obstruction_independence}\nThe class \\(\\operatorname{Ob}^{\\mathrm{lin}}_L(x_L)\\) depends only on the solution class of \\(x_L\\) modulo \\(p^L\\).\n\\end{histproposition}","\\begin{proof}\nReplace \\(x_L\\) by \\(x_L+p^Lz\\). The reduced residual changes by \\(\\overline Mz\\). Its class in the cokernel is unchanged.\n\\end{proof}","\\begin{histcorollary}[Exact number of linear children]\\label{hist:cor:linear_child_count}\nIf \\(\\rho=\\rank_{\\FF_p}\\overline M\\), then a solvable depth \\(L\\) class has exactly\n\\[\n\\boxed{\np^{d-\\rho}\n}\n\\]\nchildren modulo \\(p^{L+1}\\).\n\\end{histcorollary}","\\begin{proof}\nThe correction set is a coset of a vector space of dimension \\(d-\\rho\\).\n\\end{proof}","\\begin{histexample}[A visible linear obstruction]\\label{hist:ex:linear_lift_obstruction}\nTake \\(p=3\\), \\(L=1\\), and\n\\[\nM=\n\\begin{pmatrix}\n1&1","\\begin{histexample}[A visible linear obstruction]\\label{hist:ex:linear_lift_obstruction}\nTake \\(p=3\\), \\(L=1\\), and\n\\[\nM=\n\\begin{pmatrix}\n1&1\\\\\n0&0\n\\end{pmatrix}.\n\\]\nThe class \\(x_1=(0,0)\\) solves\n\\[\nMx_1\\equiv b\\pmod3\n\\]\nfor both\n\\[\nb_1=(-3,0)\n\\qquad\\text{and}\\qquad\nb_2=(0,-3).\n\\]\nFor \\(b_1\\), the reduced residual is \\((1,0)\\), which belongs to the image of \\(\\overline M\\). There are three correction classes, given by\n\\[\ny_1+y_2=2\n\\pmod3.\n\\]\nFor \\(b_2\\), the residual is \\((0,1)\\), whose cokernel class is nonzero. No lift modulo nine exists because the second row of \\(M\\) remains zero.\n\\end{histexample}","0&0\n\\end{pmatrix}.\n\\]\nThe class \\(x_1=(0,0)\\) solves\n\\[\nMx_1\\equiv b\\pmod3\n\\]\nfor both\n\\[\nb_1=(-3,0)\n\\qquad\\text{and}\\qquad\nb_2=(0,-3).\n\\]\nFor \\(b_1\\), the reduced residual is \\((1,0)\\), which belongs to the image of \\(\\overline M\\). There are three correction classes, given by\n\\[\ny_1+y_2=2\n\\pmod3.\n\\]\nFor \\(b_2\\), the residual is \\((0,1)\\), whose cokernel class is nonzero. No lift modulo nine exists because the second row of \\(M\\) remains zero.\n\\end{histexample}","\\(x_1=(0,0)\\) solves\n\\[\nMx_1\\equiv b\\pmod3\n\\]\nfor both\n\\[\nb_1=(-3,0)\n\\qquad\\text{and}\\qquad\nb_2=(0,-3).\n\\]\nFor \\(b_1\\), the reduced residual is \\((1,0)\\), which belongs to the image of \\(\\overline M\\). There are three correction classes, given by\n\\[\ny_1+y_2=2\n\\pmod3.\n\\]\nFor \\(b_2\\), the residual is \\((0,1)\\), whose cokernel class is nonzero. No lift modulo nine exists because the second row of \\(M\\) remains zero.\n\\end{histexample}","Let\n\\[\nF=(F_1,\\ldots,F_r):\\Zp^d\\longrightarrow\\Zp^r\n\\]\nbe a polynomial map.\\par\\noindent\\textbf{Editorial coefficient clarification.} Here polynomial means $F_i\\in\\Z_p[X_1,\\ldots,X_d]$, not merely an integer-valued polynomial with coefficients in $\\Q_p$. This makes the Jacobian and Taylor remainder integral and their reduction well typed.\\par Suppose\n\\[\nF(x_L)\\in p^L\\Zp^r.\n\\]\nDefine\n\\[\n\\rho_L^F(x_L)\n=\n\\frac{F(x_L)}{p^L}\n\\pmod p.\n\\]\nLet \\(J_F(x_L)\\) denote the Jacobian matrix and write \\(\\overline J_F(x_L)\\) for its reduction modulo \\(p\\).","rrow\\Zp^r\n\\]\nbe a polynomial map.\\par\\noindent\\textbf{Editorial coefficient clarification.} Here polynomial means $F_i\\in\\Z_p[X_1,\\ldots,X_d]$, not merely an integer-valued polynomial with coefficients in $\\Q_p$. This makes the Jacobian and Taylor remainder integral and their reduction well typed.\\par Suppose\n\\[\nF(x_L)\\in p^L\\Zp^r.\n\\]\nDefine\n\\[\n\\rho_L^F(x_L)\n=\n\\frac{F(x_L)}{p^L}\n\\pmod p.\n\\]\nLet \\(J_F(x_L)\\) denote the Jacobian matrix and write \\(\\overline J_F(x_L)\\) for its reduction modulo \\(p\\).","tbf{Editorial coefficient clarification.} Here polynomial means $F_i\\in\\Z_p[X_1,\\ldots,X_d]$, not merely an integer-valued polynomial with coefficients in $\\Q_p$. This makes the Jacobian and Taylor remainder integral and their reduction well typed.\\par Suppose\n\\[\nF(x_L)\\in p^L\\Zp^r.\n\\]\nDefine\n\\[\n\\rho_L^F(x_L)\n=\n\\frac{F(x_L)}{p^L}\n\\pmod p.\n\\]\nLet \\(J_F(x_L)\\) denote the Jacobian matrix and write \\(\\overline J_F(x_L)\\) for its reduction modulo \\(p\\).","olynomial means $F_i\\in\\Z_p[X_1,\\ldots,X_d]$, not merely an integer-valued polynomial with coefficients in $\\Q_p$. This makes the Jacobian and Taylor remainder integral and their reduction well typed.\\par Suppose\n\\[\nF(x_L)\\in p^L\\Zp^r.\n\\]\nDefine\n\\[\n\\rho_L^F(x_L)\n=\n\\frac{F(x_L)}{p^L}\n\\pmod p.\n\\]\nLet \\(J_F(x_L)\\) denote the Jacobian matrix and write \\(\\overline J_F(x_L)\\) for its reduction modulo \\(p\\).","\\begin{histdefinition}[Polynomial lift obstruction]\\label{hist:def:polynomial_lift_obstruction}\nThe polynomial obstruction at depth \\(L\\) is\n\\[\n\\boxed{\n\\operatorname{Ob}^{F}_L(x_L)\n=\n[\\rho_L^F(x_L)]\n\\in\n\\coker\\bigl(\\overline J_F(x_L)\\bigr).\n}\n\\]\n\\end{histdefinition}","\\begin{histtheorem}[One level polynomial lifting]\\label{hist:thm:polynomial_lifting}\nLet \\(L\\geq1\\). A class represented by \\(x_L\\) lifts to a zero of \\(F\\) modulo \\(p^{L+1}\\) if and only if\n\\[\n\\boxed{\n\\operatorname{Ob}^{F}_L(x_L)=0.\n}\n\\]\nA correction \\(y\\in\\FF_p^d\\) gives a lift if and only if\n\\[\n\\boxed{\n\\overline J_F(x_L)y\n=\n-\\rho_L^F(x_L).\n}\n\\]\nWhen a correction exists, the correction classes form a coset of\n\\[\n\\ker\\overline J_F(x_L).\n\\]\n\\end{histtheorem}","m}[One level polynomial lifting]\\label{hist:thm:polynomial_lifting}\nLet \\(L\\geq1\\). A class represented by \\(x_L\\) lifts to a zero of \\(F\\) modulo \\(p^{L+1}\\) if and only if\n\\[\n\\boxed{\n\\operatorname{Ob}^{F}_L(x_L)=0.\n}\n\\]\nA correction \\(y\\in\\FF_p^d\\) gives a lift if and only if\n\\[\n\\boxed{\n\\overline J_F(x_L)y\n=\n-\\rho_L^F(x_L).\n}\n\\]\nWhen a correction exists, the correction classes form a coset of\n\\[\n\\ker\\overline J_F(x_L).\n\\]\n\\end{histtheorem}","\\begin{proof}\nThe polynomial Taylor expansion gives\n\\[\nF(x_L+p^Ly)\n\\equiv\nF(x_L)+p^LJ_F(x_L)y\n\\pmod {p^{L+1}}.\n\\]\nEvery term of degree at least two in the increment contains \\(p^{2L}\\), which is divisible by \\(p^{L+1}\\). Divide by \\(p^L\\), reduce modulo \\(p\\), and apply the same cokernel criterion as in \\cref{hist:thm:linear_lifting}.\n\\end{proof}","\\begin{histproposition}[Polynomial representative independence]\\label{hist:prop:polynomial_obstruction_independence}\nThe class \\(\\operatorname{Ob}^{F}_L(x_L)\\) depends only on the class of \\(x_L\\) modulo \\(p^L\\).\n\\end{histproposition}","\\begin{proof}\nIf \\(x_L\\) is replaced by \\(x_L+p^Lz\\), Taylor expansion shows that the reduced residual changes by an element of the image of \\(\\overline J_F(x_L)\\). The reduced Jacobian itself is unchanged because \\(L\\geq1\\).\n\\end{proof}","\\begin{histcorollary}[Simple root lifting]\\label{hist:cor:simple_root_lifting}\nAssume \\(r=d\\) and\n\\[\n\\det J_F(x_L)\\not\\equiv0\\pmod p.\n\\]\nThen the depth \\(L\\) solution has exactly one child modulo \\(p^{L+1}\\). Iterating gives a unique \\(p\\) adic zero above the initial class.\n\\end{histcorollary}","\\begin{proof}\nThe reduced Jacobian is invertible, so every residual has one correction. The resulting compatible sequence is Cauchy in \\(\\Zp^d\\), and continuity of \\(F\\) gives a zero in the limit. Uniqueness holds at every depth.\n\\end{proof}","\\begin{histexample}[A singular polynomial obstruction]\\label{hist:ex:singular_polynomial_obstruction}\nLet \\(p=3\\) and consider the class \\(x_1=0\\) modulo three. For\n\\[\nF_0(x)=x^2,\n\\]\nthe residual and the reduced derivative both vanish. Every correction \\(y\\in\\FF_3\\) gives\n\\[\nF_0(3y)\\equiv0\\pmod9,\n\\]\nso the class has three children. For\n\\[\nF_1(x)=x^2+3,\n\\]\nthe reduced derivative is still zero, but the reduced residual is one. Its class in the cokernel is nonzero, so no lift modulo nine exists.\n\\end{histexample}","\\begin{histtheorem}[Packet decomposition of a lift obstruction]\\label{hist:thm:packet_lift_obstruction}\nSuppose\n\\[\n\\FF_p^d\n=\n\\bigoplus_{a\\in A}U_a,\n\\qquad\n\\FF_p^r\n=\n\\bigoplus_{a\\in A}V_a,\n\\]\nand suppose a linear correction map \\(D:\\FF_p^d\\to\\FF_p^r\\) satisfies\n\\[\nD(U_a)\\subseteq V_a\n\\]\nfor every \\(a\\). Write \\(D_a=D|_{U_a}\\). Then\n\\[\n\\boxed{\n\\coker D\n\\cong\n\\bigoplus_{a\\in A}\\coker D_a.\n}\n\\]\nA residual lifts if and only if every packet obstruction vanishes.\n\\end{histtheorem}","\\begin{proof}\nThe image of \\(D\\) is the direct sum of the images of the maps \\(D_a\\). Taking the quotient of the codomain packet decomposition gives the stated cokernel decomposition.\n\\end{proof}","Let \\(R\\) be a commutative ring. Let\n\\[\nv_{n+1}=Cv_n,\n\\qquad\nC\\in M_d(R).\n\\]\nFor a positive integer \\(N\\) and an address \\(j\\in\\{0,\\ldots,N-1\\}\\), define the residue state packet\n\\[\nu_k^{(j)}=v_{Nk+j}.\n\\]","\\begin{histtheorem}[Companion jump recurrence]\\label{hist:thm:companion_jump}\nEvery residue state packet satisfies\n\\[\n\\boxed{\nu_{k+1}^{(j)}=C^Nu_k^{(j)}.}\n\\]\nMore explicitly,\n\\[\n\\boxed{\nu_k^{(j)}=C^j(C^N)^kv_0.}\n\\]\n\\end{histtheorem}","\\begin{proof}\nThe state relation gives \\(v_n=C^nv_0\\). Substitute \\(n=Nk+j\\).\n\\end{proof}","\\begin{figure}[t]\n\\centering\n\\includegraphics[width=.95\\linewidth,height=3.15in,keepaspectratio]{../../figures/reconstructed/recurrence_packets.pdf}\n\\caption{New reconstruction. Fibonacci residues $F_{3k+j}$, with $F_0=0$ and $F_1=1$. Each packet evolves by the same jump matrix \\(C^3\\) and has a different initial address.}\n\\label{hist:fig:recurrence_packets}\n\\end{figure}","We use the resultant convention\n\\[\n\\Res_y(f,g)\n=\n\\prod_{f(\\alpha)=0}g(\\alpha)\n\\]\nwhen \\(f\\) is monic, with roots counted with multiplicity.","\\begin{histtheorem}[Characteristic polynomial of the jump]\\label{hist:thm:jump_characteristic}\nLet\n\\[\n\\chi_C(y)=\\det(yI-C).\n\\]\nThen\n\\[\n\\boxed{\n\\chi_{C^N}(x)\n=\n\\Res_y\\bigl(\\chi_C(y),x-y^N\\bigr).\n}\n\\]\nThe identity holds over the coefficient ring after interpreting the resultant universally.\n\\end{histtheorem}","\\begin{proof}\nAfter passage to an algebraic closure, let \\(\\alpha_1,\\ldots,\\alpha_d\\) be the eigenvalues of \\(C\\), counted with algebraic multiplicity. The resultant is\n\\[\n\\prod_{i=1}^{d}(x-\\alpha_i^N),\n\\]\nwhich is the characteristic polynomial of \\(C^N\\). Both sides are polynomial expressions with integer coefficients in the entries of \\(C\\), so the identity descends to every commutative coefficient ring.\n\\end{proof}","\\begin{figure}[t]\n\\centering\n\\includegraphics[width=.95\\linewidth,height=3.15in,keepaspectratio]{../../figures/reconstructed/recurrence_root_power.pdf}\n\\caption{New reconstruction. Passing to every \\(N\\)th term raises each characteristic root to its \\(N\\)th power.}\n\\label{hist:fig:recurrence_root_power}\n\\end{figure}","\\begin{histtheorem}[Residue subsequences of a second order recurrence]\\label{hist:thm:second_order_jump}\nSuppose\n\\[\nx_{n+2}=Px_{n+1}-Qx_n\n\\]\nover a commutative ring. Define \\(V_0=2\\), \\(V_1=P\\), and\n\\[\nV_{n+2}=PV_{n+1}-QV_n.\n\\]\nFor every positive integer \\(N\\) and every address \\(j\\), the subsequence\n\\[\nu_k=x_{Nk+j}\n\\]\nsatisfies\n\\[\n\\boxed{\nu_{k+2}=V_Nu_{k+1}-Q^Nu_k.}\n\\]\n\\end{histtheorem}","\\begin{proof}\nThe companion matrix\n\\[\nC=\n\\begin{pmatrix}\nP&-Q\\\\\n1&0\n\\end{pmatrix}\n\\]\nhas trace \\(P\\) and determinant \\(Q\\). The trace of \\(C^N\\) is \\(V_N\\), while its determinant is \\(Q^N\\). Cayley Hamilton applied to \\(C^N\\) gives\n\\[\n(C^N)^2-V_NC^N+Q^NI=0.\n\\]\nApply this identity to the residue state packet and take its scalar coordinate.\n\\end{proof}","1&0\n\\end{pmatrix}\n\\]\nhas trace \\(P\\) and determinant \\(Q\\). The trace of \\(C^N\\) is \\(V_N\\), while its determinant is \\(Q^N\\). Cayley Hamilton applied to \\(C^N\\) gives\n\\[\n(C^N)^2-V_NC^N+Q^NI=0.\n\\]\nApply this identity to the residue state packet and take its scalar coordinate.\n\\end{proof}","\\begin{histcorollary}[Fibonacci residue packets]\\label{hist:cor:fibonacci_packets}\nLet \\(F_0=0\\), \\(F_1=1\\), and \\(F_{n+2}=F_{n+1}+F_n\\). Let \\(L_N\\) denote the Lucas number. For every \\(N\\geq1\\) and every address \\(j\\),\n\\[\n\\boxed{\nF_{N(k+2)+j}\n=\nL_NF_{N(k+1)+j}\n-(-1)^NF_{Nk+j}.\n}\n\\]\n\\end{histcorollary}","\\begin{proof}\nApply \\cref{hist:thm:second_order_jump} with \\(P=1\\) and \\(Q=-1\\). Then \\(V_N=L_N\\).\n\\end{proof}","\\begin{histtheorem}[Periodicity of an invertible recurrence modulo a prime power]\\label{hist:thm:recurrence_periodicity}\nLet\n\\[\nC\\in M_d(\\ZZ),\n\\qquad\np\\nmid\\det C.\n\\]\nFor every \\(L\\geq1\\), every state orbit modulo \\(p^L\\) is purely periodic. Its period divides the order of \\(C\\) in\n\\[\n\\operatorname{GL}_d(\\ZZ/p^L\\ZZ).\n\\]\nIf \\(p\\mid\\det C\\), every orbit is still eventually periodic.\n\\end{histtheorem}","\\begin{proof}\nWhen \\(p\\nmid\\det C\\), multiplication by \\(C\\) is a permutation of the finite state set \\((\\ZZ/p^L\\ZZ)^d\\). Every orbit of a permutation is periodic, and applying the group order relation for \\(C\\) gives the divisibility statement. Without invertibility, repeated application of a map on a finite set eventually repeats a state, after which the orbit is periodic.\n\\end{proof}","\\begin{histcorollary}[Residue packet periods]\\label{hist:cor:residue_packet_periods}\nUnder the invertibility hypothesis of \\cref{hist:thm:recurrence_periodicity}, every residue packet with jump \\(N\\) has period dividing the order of \\(C^N\\) modulo \\(p^L\\).\n\\end{histcorollary}","\\begin{histtheorem}[Base change of an integral recurrence]\\label{hist:thm:integral_recurrence_spine}\nLet\n\\[\nC\\in M_d(\\ZZ),\n\\qquad\nv_0\\in\\ZZ^d.\n\\]\nThe integral orbit\n\\[\nv_n=C^nv_0\n\\]\ndetermines, by base change, compatible orbits in every ring \\(R\\) equipped with a homomorphism \\(\\ZZ\\to R\\). In particular, the reductions modulo every positive integer and the orbits in every \\(\\Zp^d\\) share the same companion matrix, characteristic polynomial, and integer recurrence relations.\n\\end{histtheorem}","\\begin{proof}\nEvery coordinate of \\(C^nv_0\\) is an integer polynomial in the entries of \\(C\\) and \\(v_0\\). Applying a ring homomorphism commutes with matrix multiplication and preserves every polynomial relation with integer coefficients.\n\\end{proof}","\\begin{histdefinition}[Residue projector]\\label{hist:def:residue_projector}\nFor a sequence \\(a=(a_n)_{n\\geq0}\\), a positive integer \\(N\\), and an address \\(r\\) modulo \\(N\\), define\n\\[\n(P_{N,r}a)_n\n=\n\\begin{cases}\na_n,&n\\equiv r\\pmod N,","\\begin{histdefinition}[Residue projector]\\label{hist:def:residue_projector}\nFor a sequence \\(a=(a_n)_{n\\geq0}\\), a positive integer \\(N\\), and an address \\(r\\) modulo \\(N\\), define\n\\[\n(P_{N,r}a)_n\n=\n\\begin{cases}\na_n,&n\\equiv r\\pmod N,\\\\\n0,&n\\not\\equiv r\\pmod N.\n\\end{cases}\n\\]\n\\end{histdefinition}","\\begin{histproposition}[Complete residue packet system]\\label{hist:prop:residue_packet_system}\nFor fixed \\(N\\), the projectors \\(P_{N,r}\\) satisfy\n\\[\nP_{N,r}P_{N,s}\n=\n\\delta_{rs}P_{N,r}\n\\]\nand\n\\[\n\\boxed{\n\\sum_{r\\bmod N}P_{N,r}=I.\n}\n\\]\n\\end{histproposition}","\\begin{proof}\nEvery nonnegative integer belongs to exactly one residue class modulo \\(N\\).\n\\end{proof}","\\begin{histtheorem}[Coprime packet factorization]\\label{hist:thm:crt_packet_factorization}\nLet \\(A\\) and \\(B\\) be coprime positive integers. Let \\(r_A\\) be an address modulo \\(A\\) and \\(r_B\\) an address modulo \\(B\\). Let \\(r\\) be the unique address modulo \\(AB\\) satisfying\n\\[\nr\\equiv r_A\\pmod A,\n\\qquad\nr\\equiv r_B\\pmod B.\n\\]\nThen\n\\[\n\\boxed{\nP_{AB,r}\n=\nP_{A,r_A}P_{B,r_B}\n=\nP_{B,r_B}P_{A,r_A}.\n}\n\\]\n\\end{histtheorem}","\\begin{proof}\nThe product retains exactly the indices satisfying both congruences. By the Chinese remainder theorem, those indices form one residue class modulo \\(AB\\) \\cite{IrelandRosen1990}.\n\\end{proof}","\\begin{histcorollary}[Several coprime moduli]\\label{hist:cor:many_crt_packets}\nLet \\(N_1,\\ldots,N_s\\) be pairwise coprime and put\n\\[\nN=\\prod_{j=1}^{s}N_j.\n\\]\nFor compatible addresses \\(r_j\\),\n\\[\n\\boxed{\nP_{N,r}\n=\n\\prod_{j=1}^{s}P_{N_j,r_j},\n}\n\\]\nwhere \\(r\\) is the unique combined address modulo \\(N\\).\n\\end{histcorollary}","\\begin{histstandardfact}[Finite local reconstruction]\\label{hist:fact:finite_local_reconstruction}\nLet\n\\[\nN=\\prod_{j=1}^{s}p_j^{a_j}\n\\]\nbe the prime power factorization of a positive integer. The map\n\\[\n\\ZZ/N\\ZZ\n\\longrightarrow\n\\prod_{j=1}^{s}\\ZZ/p_j^{a_j}\\ZZ\n\\]\nis a ring isomorphism. Thus a compatible collection of prime power residues reconstructs one residue modulo \\(N\\) \\cite{IrelandRosen1990}.\n\\end{histstandardfact}","\\begin{histproposition}[Dyadic and odd residue refinement]\\label{hist:prop:dyadic_odd_refinement}\nLet \\(N=2^am\\) with \\(m\\) odd. Every residue packet modulo \\(N\\) factors uniquely as the product of one dyadic packet modulo \\(2^a\\) and one odd packet modulo \\(m\\).\n\\end{histproposition}","\\begin{proof}\nThe moduli \\(2^a\\) and \\(m\\) are coprime. Apply \\cref{hist:thm:crt_packet_factorization}.\n\\end{proof}","Let \\(C_{mp,j}(n)\\) be a congruential reciprocal coefficient and let\n\\[\nh=\\frac{p-1}{m}.\n\\]\nThe unit mode calculations suggest bounds for\n\\[\n\\vp\\left(\\Delta_h^sC_{mp,j}(n)\\right).\n\\]\nThe theorems of \\cref{hist:sec:mahler,hist:sec:polyphase} determine the gain before the congruential triangular inverse is applied. They do not yet determine every valuation loss introduced by that inverse. The full family bound therefore remains open.","The residual formula\n\\[\n\\Theta_{p^a,s}(r,k)\n\\equiv\n\\binom{s}{r-p^ak}\n\\pmod p\n\\]\nis exact. A closed formula for\n\\[\n\\vp\\left(\\Theta_{p^a,s}(r,k)\\right)\n\\]\nin terms of base \\(p\\) carries and the three digit addresses is not proved here. Such a formula would sharpen the min plus valuation ledger and could distinguish optimal refinement trees.","The odd and dyadic cases agree at the level of convolution and differ at the level of integral geometry. At odd primes the source must be depleted before it becomes a unit mode interpolation. At the dyadic prime the source extends directly to an analytic integral function. The dyadic kernel has exact weight \\(4r-1\\), giving the optimal uniform modulus \\(2^{4M+3}\\) after depth \\(M\\).","\\par\\smallskip\\noindent\\textbf{Current status.} This dated statement is superseded for the stated divisor-branch family by Theorem~\\ref{thm:arith-divisor-base}, with the $j!$ normalization and factorial valuation loss explicitly retained. It makes no claim for unrestricted families.\\par","Let \\[\nF=(F_1,\\ldots,F_e)\\in\\mathbb Z[X_1,\\ldots,X_d]^e,\n\\quad q\\ge1,\\quad L\\ge1,\n\\quad x\\in\\mathbb Z^d,\n\\quad F(x)\\equiv0\\pmod{q^L}.\n\\] Write \\(R_q=\\mathbb Z/q\\mathbb Z\\), let \\(\\rho_q\\) denote\ncoordinatewise reduction, and put \\[\nr_x=F(x)/q^L\\in\\mathbb Z^e,\n\\qquad A_x=DF(x):\\mathbb Z^d\\longrightarrow\\mathbb Z^e,\n\\qquad J_x=\\rho_q\\circ A_x.\n\\] Here \\(A_x\\) is computed from the actual formal partial derivatives\nof the input polynomials. The Lean map \\(J_x\\) has domain\n\\(\\mathbb Z^d\\) and codomain \\(R_q^e\\), viewed as integer modules. It\nfactors through reduction to \\(R_q^d\\); the cokernel below is presented\ndirectly using its integer-domain image.","\\quad q\\ge1,\\quad L\\ge1,\n\\quad x\\in\\mathbb Z^d,\n\\quad F(x)\\equiv0\\pmod{q^L}.\n\\] Write \\(R_q=\\mathbb Z/q\\mathbb Z\\), let \\(\\rho_q\\) denote\ncoordinatewise reduction, and put \\[\nr_x=F(x)/q^L\\in\\mathbb Z^e,\n\\qquad A_x=DF(x):\\mathbb Z^d\\longrightarrow\\mathbb Z^e,\n\\qquad J_x=\\rho_q\\circ A_x.\n\\] Here \\(A_x\\) is computed from the actual formal partial derivatives\nof the input polynomials. The Lean map \\(J_x\\) has domain\n\\(\\mathbb Z^d\\) and codomain \\(R_q^e\\), viewed as integer modules. It\nfactors through reduction to \\(R_q^d\\); the cokernel below is presented\ndirectly using its integer-domain image.","L\\ge1,\n\\quad x\\in\\mathbb Z^d,\n\\quad F(x)\\equiv0\\pmod{q^L}.\n\\] Write \\(R_q=\\mathbb Z/q\\mathbb Z\\), let \\(\\rho_q\\) denote\ncoordinatewise reduction, and put \\[\nr_x=F(x)/q^L\\in\\mathbb Z^e,\n\\qquad A_x=DF(x):\\mathbb Z^d\\longrightarrow\\mathbb Z^e,\n\\qquad J_x=\\rho_q\\circ A_x.\n\\] Here \\(A_x\\) is computed from the actual formal partial derivatives\nof the input polynomials. The Lean map \\(J_x\\) has domain\n\\(\\mathbb Z^d\\) and codomain \\(R_q^e\\), viewed as integer modules. It\nfactors through reduction to \\(R_q^d\\); the cokernel below is presented\ndirectly using its integer-domain image.","\\quad F(x)\\equiv0\\pmod{q^L}.\n\\] Write \\(R_q=\\mathbb Z/q\\mathbb Z\\), let \\(\\rho_q\\) denote\ncoordinatewise reduction, and put \\[\nr_x=F(x)/q^L\\in\\mathbb Z^e,\n\\qquad A_x=DF(x):\\mathbb Z^d\\longrightarrow\\mathbb Z^e,\n\\qquad J_x=\\rho_q\\circ A_x.\n\\] Here \\(A_x\\) is computed from the actual formal partial derivatives\nof the input polynomials. The Lean map \\(J_x\\) has domain\n\\(\\mathbb Z^d\\) and codomain \\(R_q^e\\), viewed as integer modules. It\nfactors through reduction to \\(R_q^d\\); the cokernel below is presented\ndirectly using its integer-domain image.","et \\(\\rho_q\\) denote\ncoordinatewise reduction, and put \\[\nr_x=F(x)/q^L\\in\\mathbb Z^e,\n\\qquad A_x=DF(x):\\mathbb Z^d\\longrightarrow\\mathbb Z^e,\n\\qquad J_x=\\rho_q\\circ A_x.\n\\] Here \\(A_x\\) is computed from the actual formal partial derivatives\nof the input polynomials. The Lean map \\(J_x\\) has domain\n\\(\\mathbb Z^d\\) and codomain \\(R_q^e\\), viewed as integer modules. It\nfactors through reduction to \\(R_q^d\\); the cokernel below is presented\ndirectly using its integer-domain image.","Define the finite obstruction module and its distinguished class by \\[\n\\mathcal O_x=R_q^e/\\operatorname{im}J_x,\n\\qquad o_x=[\\rho_q(r_x)]\\in\\mathcal O_x.\n\\] The checked criterion is \\[\n\\boxed{\n o_x=0\n \\iff\n \\exists y\\in\\mathbb Z^d:\\quad\n y\\equiv x\\pmod{q^L},\\qquad F(y)\\equiv0\\pmod{q^{L+1}}.\n}\n\\] The residual exists by the lower congruence and is unique because\n\\(q>0\\). The quotient is finite since its ambient module \\(R_q^e\\) is\nfinite. The theorem permits zero variables or zero equations as well.","truction module and its distinguished class by \\[\n\\mathcal O_x=R_q^e/\\operatorname{im}J_x,\n\\qquad o_x=[\\rho_q(r_x)]\\in\\mathcal O_x.\n\\] The checked criterion is \\[\n\\boxed{\n o_x=0\n \\iff\n \\exists y\\in\\mathbb Z^d:\\quad\n y\\equiv x\\pmod{q^L},\\qquad F(y)\\equiv0\\pmod{q^{L+1}}.\n}\n\\] The residual exists by the lower congruence and is unique because\n\\(q>0\\). The quotient is finite since its ambient module \\(R_q^e\\) is\nfinite. The theorem permits zero variables or zero equations as well.","For every \\(P\\in\\mathbb Z[X_1,\\ldots,X_d]\\), integer \\(a\\), and\n\\(x,t\\in\\mathbb Z^d\\), Lean proves \\[\n\\boxed{\n a^2\\mid P(x+at)-P(x)-a\\sum_{i=1}^{d}(\\partial_iP)(x)t_i.\n}\n\\] The proof uses polynomial induction: constants, addition, and\nmultiplication by a variable. The product rule for the actual partial\nderivatives controls the multiplication step. It does not assume a\nTaylor identity or a liftability condition.","Taking \\(a=q^L\\), the inequality \\(2L\\ge L+1\\) gives \\[\nF(x+q^Lt)\\equiv q^L(r_x+A_xt)\\pmod{q^{L+1}}.\n\\] Since \\(q^L\\ne0\\) in the integers, cancellation in divisibility\nyields \\[\nF(x+q^Lt)\\equiv0\\pmod{q^{L+1}}\n\\iff \\rho_q(r_x)+J_xt=0.\n\\] Every integer representative of the lower residue has the form\n\\(x+q^Lt\\), and this correction equation has a solution exactly when\n\\(o_x=0\\).","The positive-depth hypothesis is necessary for the general nonlinear\ntheorem. For \\(F(X)=X^2-1\\), \\(q=2\\), \\(L=0\\), \\(x=0\\), the Jacobian is\nzero and the residual class is nonzero, yet \\(y=1\\) is a solution modulo\ntwo in the same residue modulo one. Affine systems retain the previously\nchecked \\(L\\ge0\\) theorem.","Suppose \\(x'=x+q^Ls\\) represents the same lower residue, with exact\nresidual \\(r_{x'}\\). Polynomial evaluation respects coordinatewise\ncongruences, including when applied to each partial derivative. Hence \\[\nJ_{x'}=J_x,\n\\qquad\n\\rho_q(r_{x'})=\\rho_q(r_x)+J_xs.\n\\] The two cokernels therefore have the same ambient module and the same\ndefining submodule, and \\[\n[\\rho_q(r_{x'})]=[\\rho_q(r_x)].\n\\] Lean states this using one explicit common reduced map, avoiding an\nimplicit comparison between differently indexed quotient types. This\nestablishes independence of the integer representative within the fixed\npolynomial presentation.","Changing the polynomial presentation is a separate question. Appending a\nzero equation can change the raw cokernel by adding a copy of \\(R_q\\),\neven though it does not change the solutions.","Let \\(G\\in\\mathbb Z[X_1,\\ldots,X_d]^f\\) be a second fixed system with\nthe same variables and the same lower root \\(x\\). Let\n\\(K_x=\\rho_q\\circ DG(x)\\), and define \\[\n\\mathcal C_x=\\mathbb Z^d/(\\ker J_x+\\ker K_x).\n\\] This quotient is finite: it is a quotient of the finite image of\n\\(J_x\\), or equivalently of the finite image of \\(K_x\\).","Suppose \\(u,v\\in\\mathbb Z^d\\) are separately valid corrections: \\[\nF(x+q^Lu)\\equiv0\\pmod{q^{L+1}},\n\\qquad\nG(x+q^Lv)\\equiv0\\pmod{q^{L+1}}.\n\\] Then \\[\n\\boxed{\n\\begin{gathered}\n[u]=[v]\\text{ in }\\mathcal C_x\n\\iff\n\\exists y\\equiv x\\pmod{q^L}:","Suppose \\(u,v\\in\\mathbb Z^d\\) are separately valid corrections: \\[\nF(x+q^Lu)\\equiv0\\pmod{q^{L+1}},\n\\qquad\nG(x+q^Lv)\\equiv0\\pmod{q^{L+1}}.\n\\] Then \\[\n\\boxed{\n\\begin{gathered}\n[u]=[v]\\text{ in }\\mathcal C_x\n\\iff\n\\exists y\\equiv x\\pmod{q^L}:\\\\\nF(y)\\equiv0\\pmod{q^{L+1}},\\quad G(y)\\equiv0\\pmod{q^{L+1}}.\n\\end{gathered}\n}\n\\] The labels are independent of the choices of valid \\(u\\) and \\(v\\).\nAn unconditional version asks for separately valid corrections with\nequal labels; it does not assume that each block is liftable.","F(y)\\equiv0\\pmod{q^{L+1}},\\quad G(y)\\equiv0\\pmod{q^{L+1}}.\n\\end{gathered}\n}\n\\] The labels are independent of the choices of valid \\(u\\) and \\(v\\).\nAn unconditional version asks for separately valid corrections with\nequal labels; it does not assume that each block is liftable.","Taylor transport turns a common correction \\(t\\) into the simultaneous\nequations \\(J_xt=J_xu\\) and \\(K_xt=K_xv\\). The previously checked\ncommon-quotient theorem gives precisely the displayed equivalence.\nSeparate solvability alone does not imply joint solvability; the\ncontrols include \\(F(X)=X\\), \\(G(X)=X-2\\), starting at \\(x=0\\) modulo\ntwo and lifting to modulus four.","Let \\(p\\) be prime, \\(L\\ge1\\), and \\[\nF\\in\\mathbb Z[X_1,\\ldots,X_d]^e,\n\\qquad x\\in\\mathbb Z^d,\n\\qquad F(x)\\equiv0\\pmod{p^L}.\n\\] Write \\[\nr_x=F(x)/p^L,\\qquad b_x=r_x\\bmod p,\n\\qquad A_x=DF(x)\\bmod p:\\mathbb F_p^d\\longrightarrow\\mathbb F_p^e,\n\\qquad \\rho_x=\\operatorname{rank}_{\\mathbb F_p}A_x.\n\\] The actual lift set is \\[\n\\mathcal L_{x,L}=\n\\left\\{y\\in(\\mathbb Z/p^{L+1}\\mathbb Z)^d:\n y\\equiv x\\pmod{p^L},\\quad F(y)=0\\right\\},\n\\] where polynomial evaluation in the displayed set is modulo\n\\(p^{L+1}\\). Define \\[\n\\mathcal O_x=\\mathbb F_p^e/\\operatorname{im}A_x,\n\\qquad o_x=[b_x].\n\\] Lean proves \\[\n\\boxed{\n\\#\\mathcal L_{x,L}=\n\\begin{cases}\np^{d-\\rho_x},&o_x=0,","Z^d,\n\\qquad F(x)\\equiv0\\pmod{p^L}.\n\\] Write \\[\nr_x=F(x)/p^L,\\qquad b_x=r_x\\bmod p,\n\\qquad A_x=DF(x)\\bmod p:\\mathbb F_p^d\\longrightarrow\\mathbb F_p^e,\n\\qquad \\rho_x=\\operatorname{rank}_{\\mathbb F_p}A_x.\n\\] The actual lift set is \\[\n\\mathcal L_{x,L}=\n\\left\\{y\\in(\\mathbb Z/p^{L+1}\\mathbb Z)^d:\n y\\equiv x\\pmod{p^L},\\quad F(y)=0\\right\\},\n\\] where polynomial evaluation in the displayed set is modulo\n\\(p^{L+1}\\). Define \\[\n\\mathcal O_x=\\mathbb F_p^e/\\operatorname{im}A_x,\n\\qquad o_x=[b_x].\n\\] Lean proves \\[\n\\boxed{\n\\#\\mathcal L_{x,L}=\n\\begin{cases}\np^{d-\\rho_x},&o_x=0,","uiv0\\pmod{p^L}.\n\\] Write \\[\nr_x=F(x)/p^L,\\qquad b_x=r_x\\bmod p,\n\\qquad A_x=DF(x)\\bmod p:\\mathbb F_p^d\\longrightarrow\\mathbb F_p^e,\n\\qquad \\rho_x=\\operatorname{rank}_{\\mathbb F_p}A_x.\n\\] The actual lift set is \\[\n\\mathcal L_{x,L}=\n\\left\\{y\\in(\\mathbb Z/p^{L+1}\\mathbb Z)^d:\n y\\equiv x\\pmod{p^L},\\quad F(y)=0\\right\\},\n\\] where polynomial evaluation in the displayed set is modulo\n\\(p^{L+1}\\). Define \\[\n\\mathcal O_x=\\mathbb F_p^e/\\operatorname{im}A_x,\n\\qquad o_x=[b_x].\n\\] Lean proves \\[\n\\boxed{\n\\#\\mathcal L_{x,L}=\n\\begin{cases}\np^{d-\\rho_x},&o_x=0,","p:\\mathbb F_p^d\\longrightarrow\\mathbb F_p^e,\n\\qquad \\rho_x=\\operatorname{rank}_{\\mathbb F_p}A_x.\n\\] The actual lift set is \\[\n\\mathcal L_{x,L}=\n\\left\\{y\\in(\\mathbb Z/p^{L+1}\\mathbb Z)^d:\n y\\equiv x\\pmod{p^L},\\quad F(y)=0\\right\\},\n\\] where polynomial evaluation in the displayed set is modulo\n\\(p^{L+1}\\). Define \\[\n\\mathcal O_x=\\mathbb F_p^e/\\operatorname{im}A_x,\n\\qquad o_x=[b_x].\n\\] Lean proves \\[\n\\boxed{\n\\#\\mathcal L_{x,L}=\n\\begin{cases}\np^{d-\\rho_x},&o_x=0,\\\\\n0,&o_x\\ne0,\n\\end{cases}\n\\qquad\n\\#\\mathcal O_x=p^{e-\\rho_x}.\n}\n\\] These formulas permit singular Jacobians and zero variables or zero\nequations. Nonsingularity is not assumed. In particular, rank alone\ngives the size of a nonempty lift fiber; the residual obstruction\ndetermines whether it is empty.","For every positive integer \\(q\\), including composite moduli and modulus\none, and every \\(L\\ge0\\), the checked map \\[\n\\Phi_{x,L}:(\\mathbb Z/q\\mathbb Z)^d\\longrightarrow\n\\{y\\bmod q^{L+1}:y\\equiv x\\pmod{q^L}\\},\n\\qquad t\\longmapsto x+q^L\\widetilde t\n\\] is a bijection. Here \\(\\widetilde t\\) is the canonical integer\nrepresentative of the digit vector. The entire residue fiber therefore\nhas \\(q^d\\) elements. Injectivity is proved by cancelling \\(q^L\\) in the\ncorresponding integer divisibility statement; surjectivity comes from\nthe lower-residue congruence.","composite moduli and modulus\none, and every \\(L\\ge0\\), the checked map \\[\n\\Phi_{x,L}:(\\mathbb Z/q\\mathbb Z)^d\\longrightarrow\n\\{y\\bmod q^{L+1}:y\\equiv x\\pmod{q^L}\\},\n\\qquad t\\longmapsto x+q^L\\widetilde t\n\\] is a bijection. Here \\(\\widetilde t\\) is the canonical integer\nrepresentative of the digit vector. The entire residue fiber therefore\nhas \\(q^d\\) elements. Injectivity is proved by cancelling \\(q^L\\) in the\ncorresponding integer divisibility statement; surjectivity comes from\nthe lower-residue congruence.","The formal type \\mcode{ActualLifts} is a subtype of these genuine\n\\texttt{ZMod\\ (q\\^{}(L+1))} residue vectors. Its polynomial predicate is\nevaluated on canonical integer representatives and tested for\ndivisibility by \\(q^{L+1}\\). The checked polynomial congruence theorem\nmakes that predicate independent of the integer representative. Thus\nusing canonical representatives does not discard or duplicate residue\nclasses.","At positive depth, the preceding actual multivariable Taylor theorem\ngives \\[\nF(x+p^L\\widetilde t)\\equiv\np^L(r_x+DF(x)\\widetilde t)\\pmod{p^{L+1}}.\n\\] The new bridge proves that the evaluated integer Jacobian reduces to\nthe actual \\(\\mathbb F_p\\)-linear map \\(A_x\\). Restricting\n\\(\\Phi_{x,L}\\) therefore gives an explicit equivalence \\[\n\\boxed{\\{t\\in\\mathbb F_p^d:b_x+A_xt=0\\}\\simeq\\mathcal L_{x,L}.}\n\\] The proof uses the actual formal partial derivatives of the original\npolynomial list.","Let \\(\\mathcal T_x=\\{t:b_x+A_xt=0\\}\\). Its natural kernel action is \\[\nk\\cdot t=k+t,\\qquad k\\in\\ker A_x.\n\\] For every pair \\(u,t\\in\\mathcal T_x\\), exactly one kernel vector\nsends \\(u\\) to \\(t\\), namely \\(t-u\\). Thus every nonempty correction\nfiber is a torsor for \\(\\ker A_x\\). Choosing one valid digit \\(u\\) gives\n\\[\n\\mathcal T_x\\simeq\\ker A_x,\n\\qquad t\\longmapsto t-u.\n\\] Lean constructs an explicit \\mcode{AddTorsor} structure for the\ndigit fiber, proves the unique translation property, and supplies a\nchosen-point equivalence from the actual residue lift set to the kernel.\nA basepoint is required for this equivalence, but the translation action\nitself does not depend on choosing one.","The earlier obstruction used the integer-domain map \\[\nJ_x=\\rho_p\\circ DF(x):\\mathbb Z^d\\to\\mathbb F_p^e.\n\\] The new count uses \\(A_x:\\mathbb F_p^d\\to\\mathbb F_p^e\\). Reduction\nof integer vectors onto finite vectors is surjective, so the two maps\nhave the same image after restricting scalars. Lean constructs a\nclass-preserving integer-linear equivalence \\[\n\\mathbb F_p^e/\\operatorname{im}J_x\n\\simeq\n\\mathbb F_p^e/\\operatorname{im}A_x.\n\\] Consequently the cardinality theorem counts the original obstruction\ngroup, preserving its distinguished residual class. It is not merely a\nformula for a separately introduced comparison object.","Rank-nullity gives \\(\\dim\\ker A_x=d-\\rho_x\\), and the quotient dimension\nis \\(e-\\rho_x\\). Cardinalities of finite vector spaces complete both\nformulas.","Let \\(p\\) be prime, \\(L\\ge1\\), and \\[\nF\\in\\mathbb Z[X_1,\\ldots,X_d]^e,\\qquad x\\in\\mathbb Z^d,\n\\qquad F(x)\\equiv0\\pmod{p^L}.\n\\] Assume the actual reduced Jacobian \\[\nA_x=DF(x)\\bmod p:\\mathbb F_p^d\\longrightarrow\\mathbb F_p^e\n\\] is surjective. Then Lean proves that there is a point \\[\n\\boxed{a\\in\\mathbb Z_p^d,\\qquad F(a)=0,\\qquad\n a\\equiv x\\pmod{p^L}.}\n\\] Here the coefficients of the original integer polynomials are cast\ninto \\(\\mathbb Z_p\\), and their multivariable evaluation is exactly\nzero. This is an existence theorem for a fixed polynomial presentation\nwith the stated Jacobian hypothesis.","and \\[\nF\\in\\mathbb Z[X_1,\\ldots,X_d]^e,\\qquad x\\in\\mathbb Z^d,\n\\qquad F(x)\\equiv0\\pmod{p^L}.\n\\] Assume the actual reduced Jacobian \\[\nA_x=DF(x)\\bmod p:\\mathbb F_p^d\\longrightarrow\\mathbb F_p^e\n\\] is surjective. Then Lean proves that there is a point \\[\n\\boxed{a\\in\\mathbb Z_p^d,\\qquad F(a)=0,\\qquad\n a\\equiv x\\pmod{p^L}.}\n\\] Here the coefficients of the original integer polynomials are cast\ninto \\(\\mathbb Z_p\\), and their multivariable evaluation is exactly\nzero. This is an existence theorem for a fixed polynomial presentation\nwith the stated Jacobian hypothesis.","The stronger checked statement constructs \\(z_n\\in\\mathbb Z^d\\),\n\\(n\\ge0\\), with \\[\n\\begin{aligned}\nz_0&=x,","The stronger checked statement constructs \\(z_n\\in\\mathbb Z^d\\),\n\\(n\\ge0\\), with \\[\n\\begin{aligned}\nz_0&=x,\\\\\nF(z_n)&\\equiv0\\pmod{p^{L+n}},\\\\\nz_{n+1}&\\equiv z_n\\pmod{p^{L+n}},\\\\\nDF(z_n)\\bmod p&=A_x,\\\\\nz_n&\\longrightarrow a\\quad\\text{in }\\mathbb Z_p^d,\\\\\na-z_n&\\in p^{L+n}\\mathbb Z_p^d.\n\\end{aligned}\n\\] The last inclusion retains every prescribed precision. It is a\ndivisibility lower bound, not an assertion that the error has exactly\nthat valuation. An approximation can already equal the limiting root.","The one-step Taylor theorem already supplies the exact correction\nequation \\[\nb_n+A_{z_n}t=0,\\qquad\nb_n=F(z_n)/p^{L+n}\\bmod p.\n\\] Surjectivity supplies a correcting digit for every residual \\(b_n\\).\nThe chosen correction gives an actual new integer point \\(z_{n+1}\\),\nusing the proved nonlinear Taylor remainder.","Every coordinate change is divisible by \\(p^{L+n}\\), hence by \\(p\\).\nEach formal partial derivative is itself an integer polynomial, so its\nvalue is unchanged modulo \\(p\\). Therefore \\[\nA_{z_{n+1}}=A_{z_n}=A_x.\n\\] The proof assumes no separate surjectivity or solvability statement\nat later levels. A dependent recursion chooses each next point from the\nproved one-step existence theorem. It produces one compatible sequence,\nrather than unrelated witnesses at different precisions.","The integer tower construction works more generally for every positive\nmodulus \\(q\\), including composite \\(q\\) and \\(q=1\\), with a surjective\nmap over \\(\\mathbb Z/q\\mathbb Z\\). The p-adic completion and the\nrank-based branching formulas in this block use prime \\(p\\).","Compatibility implies, for \\(m\\ge n\\), \\[\np^{L+n}\\mid z_m-z_n.\n\\] After casting to \\(\\mathbb Z_p\\), this bounds each coordinate\ndistance by \\(p^{-(L+n)}\\), so the sequence is Cauchy. Completeness\nsupplies its limit. Closed norm bounds show that the limit retains the\ncongruence at every stage.","The generic convergence theorem is proved independently of the\npolynomial application. It also proves \\[\np^{L+n}\\mid v_n\\ \\text{for all }n\n\\quad\\Longrightarrow\\quad v_n\\longrightarrow0\\text{ in }\\mathbb Z_p.\n\\] Apply this to each integer sequence \\(F_i(z_n)\\). Continuity of\nmultivariable polynomial evaluation also gives \\(F_i(z_n)\\to F_i(a)\\).\nUniqueness of limits then yields \\(F_i(a)=0\\).","Surjectivity implies \\[\n\\operatorname{rank}_{\\mathbb F_p}A_x=e,\\qquad e\\le d.\n\\] Combining this with the preceding exact residue-lift count gives \\[\n\\boxed{\\#\\mathcal L_{z_n,L+n}=p^{d-e}\\quad\\text{for every }n.}\n\\] These are distinct solutions modulo \\(p^{L+n+1}\\) lying above the\nselected point modulo \\(p^{L+n}\\). The original finite obstruction group\nat each smooth point has cardinality one.","The theorem constructs one path through these finite choices. When\n\\(d>e\\), it does not claim that all paths have the same limit. The\ngeneric uniqueness theorem concerns the limit of a fixed sequence only.","\\begin{enumerate}\n\\def\\labelenumi{\\arabic{enumi}.}\n\\tightlist\n\\item\n  \\textbf{An initial lift does not guarantee an infinite tower without\n  the Jacobian condition.} \\(X^2-5\\) has the odd lower root modulo two\n  and a lift modulo four, but no root modulo eight. Lean proves that no\n  integer tower can continue through all precisions in this example.\n\\item\n  \\textbf{Surjective rectangular Jacobians do not give uniqueness of all\n  roots.} The system \\(F(X,Y)=X\\) has a surjective Jacobian and distinct\n  dyadic roots \\((0,0)\\) and \\((0,2)\\), both in the same residue modulo\n  two.\n\\item\n  \\textbf{The retained initial precision cannot be increased without\n  justification.} For \\(F(X)=X-2\\), zero is a lower root modulo two with\n  surjective derivative, but its exact root two does not retain zero\n  modulo four.\n\\end{enumerate}","ightlist\n\\item\n  \\textbf{An initial lift does not guarantee an infinite tower without\n  the Jacobian condition.} \\(X^2-5\\) has the odd lower root modulo two\n  and a lift modulo four, but no root modulo eight. Lean proves that no\n  integer tower can continue through all precisions in this example.\n\\item\n  \\textbf{Surjective rectangular Jacobians do not give uniqueness of all\n  roots.} The system \\(F(X,Y)=X\\) has a surjective Jacobian and distinct\n  dyadic roots \\((0,0)\\) and \\((0,2)\\), both in the same residue modulo\n  two.\n\\item\n  \\textbf{The retained initial precision cannot be increased without\n  justification.} For \\(F(X)=X-2\\), zero is a lower root modulo two with\n  surjective derivative, but its exact root two does not retain zero\n  modulo four.\n\\end{enumerate}","nite tower without\n  the Jacobian condition.} \\(X^2-5\\) has the odd lower root modulo two\n  and a lift modulo four, but no root modulo eight. Lean proves that no\n  integer tower can continue through all precisions in this example.\n\\item\n  \\textbf{Surjective rectangular Jacobians do not give uniqueness of all\n  roots.} The system \\(F(X,Y)=X\\) has a surjective Jacobian and distinct\n  dyadic roots \\((0,0)\\) and \\((0,2)\\), both in the same residue modulo\n  two.\n\\item\n  \\textbf{The retained initial precision cannot be increased without\n  justification.} For \\(F(X)=X-2\\), zero is a lower root modulo two with\n  surjective derivative, but its exact root two does not retain zero\n  modulo four.\n\\end{enumerate}","hout\n  the Jacobian condition.} \\(X^2-5\\) has the odd lower root modulo two\n  and a lift modulo four, but no root modulo eight. Lean proves that no\n  integer tower can continue through all precisions in this example.\n\\item\n  \\textbf{Surjective rectangular Jacobians do not give uniqueness of all\n  roots.} The system \\(F(X,Y)=X\\) has a surjective Jacobian and distinct\n  dyadic roots \\((0,0)\\) and \\((0,2)\\), both in the same residue modulo\n  two.\n\\item\n  \\textbf{The retained initial precision cannot be increased without\n  justification.} For \\(F(X)=X-2\\), zero is a lower root modulo two with\n  surjective derivative, but its exact root two does not retain zero\n  modulo four.\n\\end{enumerate}","integer tower can continue through all precisions in this example.\n\\item\n  \\textbf{Surjective rectangular Jacobians do not give uniqueness of all\n  roots.} The system \\(F(X,Y)=X\\) has a surjective Jacobian and distinct\n  dyadic roots \\((0,0)\\) and \\((0,2)\\), both in the same residue modulo\n  two.\n\\item\n  \\textbf{The retained initial precision cannot be increased without\n  justification.} For \\(F(X)=X-2\\), zero is a lower root modulo two with\n  surjective derivative, but its exact root two does not retain zero\n  modulo four.\n\\end{enumerate}","Let \\(p\\) be prime, \\(L\\ge1\\), and \\[\nF,H\\in\\mathbb Z[X_1,\\ldots,X_d]^e,\n\\qquad x\\in\\mathbb Z^d.\n\\] Write \\[\nA_x=DF(x)\\bmod p:\\mathbb F_p^d\\longrightarrow\\mathbb F_p^e,\n\\qquad B_L(x)=x+p^L\\mathbb Z_p^d.\n\\] For an integer \\(c\\), the perturbed system is the actual polynomial\nlist \\[\nG=F+cH.\n\\] There is no restriction on the sign of \\(c\\), and \\(c=0\\) is allowed.\nDivisibility conclusions give lower bounds on precision and allow zero\ndifferences.","If \\(d=e\\), \\(A_x\\) is surjective, and \\(F(x)\\equiv0\\pmod{p^L}\\), then\n\\[\n\\boxed{\\exists!\\,a\\in B_L(x):F(a)=0.}\n\\] The stronger comparison lemma proves that any two exact roots of\n\\(F\\) in \\(B_L(x)\\) coincide, without needing a separate lower-root\nhypothesis.","\\textbf{Proof implemented.} Map every arbitrary p-adic root into the\nactual residue rings \\(\\mathbb Z/p^n\\mathbb Z\\), and choose its\ncanonical integer representative. Evaluation commutes with reduction, so\nthese representatives are genuine finite polynomial roots. The preceding\ncount theorem gives exactly \\(p^{d-d}=1\\) next-level residue lift.\nInductively any two roots agree at every precision. The p-adic residue\nmaps separate points, so the roots are equal. Existence is the preceding\nchecked tower theorem.","If \\(p^L\\mid c\\), the actual identities give \\[\nG(x)=F(x)+cH(x),\\qquad DG(x)\\equiv DF(x)\\pmod p.\n\\] Thus if \\(F(x)\\equiv0\\pmod{p^L}\\) and \\(A_x\\) is surjective, then \\[\n\\exists b\\in B_L(x):G(b)=0.\n\\] This theorem allows rectangular systems. The evaluation, derivative\nand Jacobian identities are proved from the polynomial definition\n\\(G_i=F_i+C(c)H_i\\), rather than supplied as hypotheses. Algebraic\ncongruence helpers also permit composite moduli; exact p-adic root\nexistence uses prime \\(p\\).","od p.\n\\] Thus if \\(F(x)\\equiv0\\pmod{p^L}\\) and \\(A_x\\) is surjective, then \\[\n\\exists b\\in B_L(x):G(b)=0.\n\\] This theorem allows rectangular systems. The evaluation, derivative\nand Jacobian identities are proved from the polynomial definition\n\\(G_i=F_i+C(c)H_i\\), rather than supplied as hypotheses. Algebraic\ncongruence helpers also permit composite moduli; exact p-adic root\nexistence uses prime \\(p\\).","Let \\(M\\ge1\\) and \\(p^M\\mid c\\). Suppose \\(A_x\\) is surjective and\n\\(a\\in\\mathbb Z_p^d\\) is an exact root of \\(F\\) with\n\\(a\\equiv x\\pmod p\\). Then \\[\n\\boxed{\\exists b\\in\\mathbb Z_p^d:\\quad G(b)=0,\n\\qquad b\\equiv a\\pmod{p^M}.}\n\\] No uniqueness is claimed when \\(d>e\\).","\\textbf{Proof implemented.} Truncate \\(a\\) to an integer vector \\(y\\)\nmodulo \\(p^M\\). It is an actual lower root of \\(F\\), and its reduced\nJacobian agrees with \\(A_x\\). Apply the coefficient-perturbed existence\ntheorem at precision \\(M\\), then combine the two congruences through\n\\(y\\).","Suppose \\(d=e\\), \\(M\\ge L\\ge1\\), \\(p^M\\mid c\\), and \\(A_x\\) is\nsurjective. If \\[\na,b\\in B_L(x),\\qquad F(a)=0,\\qquad G(b)=0,\n\\] then \\[\n\\boxed{b-a\\in p^M\\mathbb Z_p^d.}\n\\] In particular, for \\(G=F+p^MH\\), no p-adic digits below depth \\(M\\)\nare changed in the selected root.","The formalization also proves a combined existence theorem. Starting\nonly with \\(F(x)\\equiv0\\pmod{p^L}\\), it constructs an original root\n\\(a\\in B_L(x)\\) and proves that there is exactly one \\(b\\in B_L(x)\\)\nsatisfying \\(G(b)=0\\) and \\(b\\equiv a\\pmod{p^M}\\). The comparison\ntheorem shows that the precision bound holds for every root of \\(G\\) in\nthat ball, so uniqueness is not obtained merely by adding a restrictive\ndisplacement condition.","\\textbf{Proof implemented.} Section 3 produces a nearby root \\(b'\\).\nSince \\(M\\ge L\\), it belongs to \\(B_L(x)\\). The perturbed Jacobian\nremains surjective, and square uniqueness identifies \\(b'\\) with the\narbitrary root \\(b\\).","Let \\(p\\) be prime, \\(L\\ge1\\), \\(e,q\\ge0\\), and \\[\nF\\in\\mathbb Z[X_1,\\ldots,X_{e+q}]^e,\\qquad\nx\\in\\mathbb Z^{e+q},\\qquad F(x)\\equiv0\\pmod{p^L}.\n\\] Assume the actual reduced Jacobian \\[\nA_x=DF(x)\\bmod p:\\mathbb F_p^{e+q}\\longrightarrow\\mathbb F_p^e\n\\] is surjective. All vector congruences are coordinatewise.","There is a coordinate permutation \\(\\sigma\\), with\n\\((\\sigma z)_i=z_{\\sigma(i)}\\). Write \\(\\sigma z=(y,t)\\), using \\(e\\)\nsolved coordinates and \\(q\\) free coordinates. Define \\[\n(y_0,t_0)=\\sigma x,\\qquad\nB=x+p^L\\mathbb Z_p^{e+q},\\qquad\nT=t_0+p^L\\mathbb Z_p^q,\\qquad\n\\mathcal R=\\{z\\in B:F(z)=0\\}.\n\\] There is a homeomorphism \\[\nh:T\\xrightarrow{\\ \\sim\\ }\\mathcal R,\\qquad\nh^{-1}(z)=\\operatorname{free}(\\sigma z).\n\\] Equivalently, a continuous function\n\\(\\phi:T\\to y_0+p^L\\mathbb Z_p^e\\) gives the entire root locus through\n\\(\\sigma h(t)=(\\phi(t),t)\\). Every parameter has exactly one\ncorresponding root in \\(B\\). For every \\(N\\ge0\\), \\[\n\\boxed{h(t)\\equiv h(u)\\pmod{p^N}\\iff t\\equiv u\\pmod{p^N}.}\n\\]","((\\sigma z)_i=z_{\\sigma(i)}\\). Write \\(\\sigma z=(y,t)\\), using \\(e\\)\nsolved coordinates and \\(q\\) free coordinates. Define \\[\n(y_0,t_0)=\\sigma x,\\qquad\nB=x+p^L\\mathbb Z_p^{e+q},\\qquad\nT=t_0+p^L\\mathbb Z_p^q,\\qquad\n\\mathcal R=\\{z\\in B:F(z)=0\\}.\n\\] There is a homeomorphism \\[\nh:T\\xrightarrow{\\ \\sim\\ }\\mathcal R,\\qquad\nh^{-1}(z)=\\operatorname{free}(\\sigma z).\n\\] Equivalently, a continuous function\n\\(\\phi:T\\to y_0+p^L\\mathbb Z_p^e\\) gives the entire root locus through\n\\(\\sigma h(t)=(\\phi(t),t)\\). Every parameter has exactly one\ncorresponding root in \\(B\\). For every \\(N\\ge0\\), \\[\n\\boxed{h(t)\\equiv h(u)\\pmod{p^N}\\iff t\\equiv u\\pmod{p^N}.}\n\\]","lved coordinates and \\(q\\) free coordinates. Define \\[\n(y_0,t_0)=\\sigma x,\\qquad\nB=x+p^L\\mathbb Z_p^{e+q},\\qquad\nT=t_0+p^L\\mathbb Z_p^q,\\qquad\n\\mathcal R=\\{z\\in B:F(z)=0\\}.\n\\] There is a homeomorphism \\[\nh:T\\xrightarrow{\\ \\sim\\ }\\mathcal R,\\qquad\nh^{-1}(z)=\\operatorname{free}(\\sigma z).\n\\] Equivalently, a continuous function\n\\(\\phi:T\\to y_0+p^L\\mathbb Z_p^e\\) gives the entire root locus through\n\\(\\sigma h(t)=(\\phi(t),t)\\). Every parameter has exactly one\ncorresponding root in \\(B\\). For every \\(N\\ge0\\), \\[\n\\boxed{h(t)\\equiv h(u)\\pmod{p^N}\\iff t\\equiv u\\pmod{p^N}.}\n\\]","x+p^L\\mathbb Z_p^{e+q},\\qquad\nT=t_0+p^L\\mathbb Z_p^q,\\qquad\n\\mathcal R=\\{z\\in B:F(z)=0\\}.\n\\] There is a homeomorphism \\[\nh:T\\xrightarrow{\\ \\sim\\ }\\mathcal R,\\qquad\nh^{-1}(z)=\\operatorname{free}(\\sigma z).\n\\] Equivalently, a continuous function\n\\(\\phi:T\\to y_0+p^L\\mathbb Z_p^e\\) gives the entire root locus through\n\\(\\sigma h(t)=(\\phi(t),t)\\). Every parameter has exactly one\ncorresponding root in \\(B\\). For every \\(N\\ge0\\), \\[\n\\boxed{h(t)\\equiv h(u)\\pmod{p^N}\\iff t\\equiv u\\pmod{p^N}.}\n\\]","{e+q},\\qquad\nT=t_0+p^L\\mathbb Z_p^q,\\qquad\n\\mathcal R=\\{z\\in B:F(z)=0\\}.\n\\] There is a homeomorphism \\[\nh:T\\xrightarrow{\\ \\sim\\ }\\mathcal R,\\qquad\nh^{-1}(z)=\\operatorname{free}(\\sigma z).\n\\] Equivalently, a continuous function\n\\(\\phi:T\\to y_0+p^L\\mathbb Z_p^e\\) gives the entire root locus through\n\\(\\sigma h(t)=(\\phi(t),t)\\). Every parameter has exactly one\ncorresponding root in \\(B\\). For every \\(N\\ge0\\), \\[\n\\boxed{h(t)\\equiv h(u)\\pmod{p^N}\\iff t\\equiv u\\pmod{p^N}.}\n\\]","d\nT=t_0+p^L\\mathbb Z_p^q,\\qquad\n\\mathcal R=\\{z\\in B:F(z)=0\\}.\n\\] There is a homeomorphism \\[\nh:T\\xrightarrow{\\ \\sim\\ }\\mathcal R,\\qquad\nh^{-1}(z)=\\operatorname{free}(\\sigma z).\n\\] Equivalently, a continuous function\n\\(\\phi:T\\to y_0+p^L\\mathbb Z_p^e\\) gives the entire root locus through\n\\(\\sigma h(t)=(\\phi(t),t)\\). Every parameter has exactly one\ncorresponding root in \\(B\\). For every \\(N\\ge0\\), \\[\n\\boxed{h(t)\\equiv h(u)\\pmod{p^N}\\iff t\\equiv u\\pmod{p^N}.}\n\\]","The coordinate choice is a conclusion. A suitable minor and a\npreexisting chart are not hypotheses. The theorem covers the original\nroot locus after undoing the coordinate reordering. Since \\(x\\) is an\napproximate root, it does not assert \\(\\phi(t_0)=y_0\\) unless \\(F(x)=0\\)\nexactly. The cases \\(p=2\\), \\(L=1\\), \\(e=0\\), and \\(q=0\\) are included.","For a rectangular integer polynomial system with a surjective reduced\nJacobian at \\(x\\), and \\(L\\ge1\\), \\[\nF(x+p^L\\mathbb Z_p^d)=F(x)+p^L\\mathbb Z_p^e.\n\\] This covers arbitrary p-adic output targets. No lower-root hypothesis\nis needed.","When \\(d=e\\), the actual restricted polynomial map is a homeomorphism.\nMoreover, for arbitrary \\(a,b\\in x+p\\mathbb Z_p^d\\) and every \\(N\\ge0\\),\n\\[\nF(a)\\equiv F(b)\\pmod{p^N}\\iff a\\equiv b\\pmod{p^N}.\n\\] Its inverse therefore preserves every congruence depth.","}.}\n\\tightlist\n\\item\n  The actual integer Taylor formula gives one additional input digit\n  from one additional output digit when the reduced square Jacobian is\n  surjective. Its finite domain and codomain have the same size, so it\n  is injective. Induction gives every integer precision.\n\\item\n  Canonical integer truncations of arbitrary p-adic vectors preserve\n  evaluation modulo each power of \\(p\\). Applying the integer result to\n  truncations proves the p-adic congruence equivalence.\n\\item\n  Subtracting an integer target from the constant terms leaves the\n  actual Jacobian unchanged. The previous root-lifting theorem supplies\n  a preimage. The image of the compact input ball is closed and contains\n  the convergent truncations of every admissible p-adic target, hence\n  contains the target. No compatibility of arbitrarily chosen preimages\n  is assumed.\n\\item\n  In the square case, the resulting continuous bijection from a compact\n  ball to a Hausdorff ball is a homeomorphism. Its forward map is actual\n  polynomial evaluation, and its inverse solves the output equation.\n\\item\n  For a …","compatibility of arbitrarily chosen preimages\n  is assumed.\n\\item\n  In the square case, the resulting continuous bijection from a compact\n  ball to a Hausdorff ball is a homeomorphism. Its forward map is actual\n  polynomial evaluation, and its inverse solves the output equation.\n\\item\n  For a specified surjective solved-variable block, apply this inverse\n  to the actual augmented polynomial map \\[\n  \\Psi(y,t)=(F(y,t),t).\n  \\] Its derivative has an identity free-coordinate block. The inverse\n  at \\((0,t)\\) gives the root. The lower-root congruence places this\n  target in the output ball. Injectivity proves uniqueness and full\n  graph coverage. Projection identifies the inverse chart and reflects\n  congruences.\n\\item\n  The columns of a surjective finite-dimensional linear map contain a\n  basis. Select those columns and extend their injection to a coordinate\n  permutation. Polynomial renaming by \\(\\sigma^{-1}\\) transports actual\n  evaluations and partial derivatives. The fixed-coordinate chart then\n  gives the main theorem on the entire original root locus.\n\\end{enumerate}","ontinuous bijection from a compact\n  ball to a Hausdorff ball is a homeomorphism. Its forward map is actual\n  polynomial evaluation, and its inverse solves the output equation.\n\\item\n  For a specified surjective solved-variable block, apply this inverse\n  to the actual augmented polynomial map \\[\n  \\Psi(y,t)=(F(y,t),t).\n  \\] Its derivative has an identity free-coordinate block. The inverse\n  at \\((0,t)\\) gives the root. The lower-root congruence places this\n  target in the output ball. Injectivity proves uniqueness and full\n  graph coverage. Projection identifies the inverse chart and reflects\n  congruences.\n\\item\n  The columns of a surjective finite-dimensional linear map contain a\n  basis. Select those columns and extend their injection to a coordinate\n  permutation. Polynomial renaming by \\(\\sigma^{-1}\\) transports actual\n  evaluations and partial derivatives. The fixed-coordinate chart then\n  gives the main theorem on the entire original root locus.\n\\end{enumerate}","\\) gives the root. The lower-root congruence places this\n  target in the output ball. Injectivity proves uniqueness and full\n  graph coverage. Projection identifies the inverse chart and reflects\n  congruences.\n\\item\n  The columns of a surjective finite-dimensional linear map contain a\n  basis. Select those columns and extend their injection to a coordinate\n  permutation. Polynomial renaming by \\(\\sigma^{-1}\\) transports actual\n  evaluations and partial derivatives. The fixed-coordinate chart then\n  gives the main theorem on the entire original root locus.\n\\end{enumerate}","Full Jacobian surjectivity need not allow an arbitrary prescribed\nparameter choice. At the dyadic origin, \\(F(X,T)=X^2-T\\) has a\nsurjective full Jacobian but zero \\(X\\)-derivative. Prescribing \\(T=2\\)\ngives no root even modulo \\(4\\). Suitable coordinates instead solve for\n\\(T\\).","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,r\\in\\mathbb N\\). Let\n\\(F=(F_1,\\ldots,F_d)\\in R[X_1,\\ldots,X_d]^d\\), \\(a\\in R^d\\), and let \\[\nJ=DF(a)=\\bigl(\\partial_jF_i(a)\\bigr)_{i,j}.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad\nF(a)\\in p^rR^d,\\qquad r>2s.\n\\] Then \\[\n\\boxed{\\exists!\\,b\\in a+p^{s+1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,r\\in\\mathbb N\\). Let\n\\(F=(F_1,\\ldots,F_d)\\in R[X_1,\\ldots,X_d]^d\\), \\(a\\in R^d\\), and let \\[\nJ=DF(a)=\\bigl(\\partial_jF_i(a)\\bigr)_{i,j}.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad\nF(a)\\in p^rR^d,\\qquad r>2s.\n\\] Then \\[\n\\boxed{\\exists!\\,b\\in a+p^{s+1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,r\\in\\mathbb N\\). Let\n\\(F=(F_1,\\ldots,F_d)\\in R[X_1,\\ldots,X_d]^d\\), \\(a\\in R^d\\), and let \\[\nJ=DF(a)=\\bigl(\\partial_jF_i(a)\\bigr)_{i,j}.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad\nF(a)\\in p^rR^d,\\qquad r>2s.\n\\] Then \\[\n\\boxed{\\exists!\\,b\\in a+p^{s+1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); th","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,r\\in\\mathbb N\\). Let\n\\(F=(F_1,\\ldots,F_d)\\in R[X_1,\\ldots,X_d]^d\\), \\(a\\in R^d\\), and let \\[\nJ=DF(a)=\\bigl(\\partial_jF_i(a)\\bigr)_{i,j}.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad\nF(a)\\in p^rR^d,\\qquad r>2s.\n\\] Then \\[\n\\boxed{\\exists!\\,b\\in a+p^{s+1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); there is no replacement by a","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,r\\in\\mathbb N\\). Let\n\\(F=(F_1,\\ldots,F_d)\\in R[X_1,\\ldots,X_d]^d\\), \\(a\\in R^d\\), and let \\[\nJ=DF(a)=\\bigl(\\partial_jF_i(a)\\bigr)_{i,j}.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad\nF(a)\\in p^rR^d,\\qquad r>2s.\n\\] Then \\[\n\\boxed{\\exists!\\,b\\in a+p^{s+1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); there is no replacement by a determinant valuation. Integrality\nof \\(p^sJ^{-1}\\)","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,r\\in\\mathbb N\\). Let\n\\(F=(F_1,\\ldots,F_d)\\in R[X_1,\\ldots,X_d]^d\\), \\(a\\in R^d\\), and let \\[\nJ=DF(a)=\\bigl(\\partial_jF_i(a)\\bigr)_{i,j}.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad\nF(a)\\in p^rR^d,\\qquad r>2s.\n\\] Then \\[\n\\boxed{\\exists!\\,b\\in a+p^{s+1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); there is no replacement by a determinant valuation. Integrality\nof \\(p^sJ^{-1}\\) is expressed","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,r\\in\\mathbb N\\). Let\n\\(F=(F_1,\\ldots,F_d)\\in R[X_1,\\ldots,X_d]^d\\), \\(a\\in R^d\\), and let \\[\nJ=DF(a)=\\bigl(\\partial_jF_i(a)\\bigr)_{i,j}.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad\nF(a)\\in p^rR^d,\\qquad r>2s.\n\\] Then \\[\n\\boxed{\\exists!\\,b\\in a+p^{s+1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); there is no replacement by a determinant valuation. Integrality\nof \\(p^sJ^{-1}\\) is expressed in Lean by existence of an integral matrix\nwith exactly tha","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,r\\in\\mathbb N\\). Let\n\\(F=(F_1,\\ldots,F_d)\\in R[X_1,\\ldots,X_d]^d\\), \\(a\\in R^d\\), and let \\[\nJ=DF(a)=\\bigl(\\partial_jF_i(a)\\bigr)_{i,j}.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad\nF(a)\\in p^rR^d,\\qquad r>2s.\n\\] Then \\[\n\\boxed{\\exists!\\,b\\in a+p^{s+1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); there is no replacement by a determinant valuation. Integrality\nof \\(p^sJ^{-1}\\) is expressed in Lean by existence of an integral matrix\nwith exactly that image under the coefficient inclusion\n\\(R\\hookrightarrow K\\).","n R[X_1,\\ldots,X_d]^d\\), \\(a\\in R^d\\), and let \\[\nJ=DF(a)=\\bigl(\\partial_jF_i(a)\\bigr)_{i,j}.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad\nF(a)\\in p^rR^d,\\qquad r>2s.\n\\] Then \\[\n\\boxed{\\exists!\\,b\\in a+p^{s+1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); there is no replacement by a determinant valuation. Integrality\nof \\(p^sJ^{-1}\\) is expressed in Lean by existence of an integral matrix\nwith exactly that image under the coefficient inclusion\n\\(R\\hookrightarrow K\\).","a)=\\bigl(\\partial_jF_i(a)\\bigr)_{i,j}.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad\nF(a)\\in p^rR^d,\\qquad r>2s.\n\\] Then \\[\n\\boxed{\\exists!\\,b\\in a+p^{s+1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); there is no replacement by a determinant valuation. Integrality\nof \\(p^sJ^{-1}\\) is expressed in Lean by existence of an integral matrix\nwith exactly that image under the coefficient inclusion\n\\(R\\hookrightarrow K\\).","1}R^d:\\ F(b)=0,}\n\\qquad\n\\boxed{b-a\\in p^{r-s}R^d.}\n\\] All vector divisibility assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); there is no replacement by a determinant valuation. Integrality\nof \\(p^sJ^{-1}\\) is expressed in Lean by existence of an integral matrix\nwith exactly that image under the coefficient inclusion\n\\(R\\hookrightarrow K\\).","ity assertions are coordinatewise. The uniqueness\npredicate is the entire ball of depth \\(s+1\\): competitors need not be\nassumed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); there is no replacement by a determinant valuation. Integrality\nof \\(p^sJ^{-1}\\) is expressed in Lean by existence of an integral matrix\nwith exactly that image under the coefficient inclusion\n\\(R\\hookrightarrow K\\).","ed to have depth \\(r-s\\). The theorem permits arbitrary integral\np-adic coefficients and centers, every prime including two, and\ndimension zero. It retains the supplied inverse-denominator exponent\n\\(s\\); there is no replacement by a determinant valuation. Integrality\nof \\(p^sJ^{-1}\\) is expressed in Lean by existence of an integral matrix\nwith exactly that image under the coefficient inclusion\n\\(R\\hookrightarrow K\\).","Write \\(A=p^sJ^{-1}\\in M_d(R)\\). Then \\[\nAJ=JA=p^sI.\n\\] The formal matrix bridge proves the equivalence with the stated\nfraction-field hypotheses, including nonsingularity. In particular,\nmultiplication by \\(A\\) is injective, since multiplication by \\(J\\)\nsends \\(Av=0\\) to \\(p^sv=0\\).","Put \\(G=AF\\), \\(t=p^{s+1}\\), and \\(k=r-(2s+1)\\). Taylor expansion of the\nactual preconditioned polynomials gives one polynomial system \\(Q\\) such\nthat \\[\nG(a+tu)=G(a)+p^{2s+1}u+p^{2s+2}Q(u),\\qquad Q(0)=0.\n\\] This expansion is proved first over every commutative ring. Its\nremainder is a polynomial valid for every \\(u\\), not a pointwise\nquotient depending on \\(u\\). Since \\(G(a)\\in p^rR^d\\), it can be written\n\\(p^{2s+1}C\\) with \\(C\\in p^kR^d\\). Thus \\[\nG(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr).\n\\]",", \\(t=p^{s+1}\\), and \\(k=r-(2s+1)\\). Taylor expansion of the\nactual preconditioned polynomials gives one polynomial system \\(Q\\) such\nthat \\[\nG(a+tu)=G(a)+p^{2s+1}u+p^{2s+2}Q(u),\\qquad Q(0)=0.\n\\] This expansion is proved first over every commutative ring. Its\nremainder is a polynomial valid for every \\(u\\), not a pointwise\nquotient depending on \\(u\\). Since \\(G(a)\\in p^rR^d\\), it can be written\n\\(p^{2s+1}C\\) with \\(C\\in p^kR^d\\). Thus \\[\nG(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr).\n\\]","d \\(k=r-(2s+1)\\). Taylor expansion of the\nactual preconditioned polynomials gives one polynomial system \\(Q\\) such\nthat \\[\nG(a+tu)=G(a)+p^{2s+1}u+p^{2s+2}Q(u),\\qquad Q(0)=0.\n\\] This expansion is proved first over every commutative ring. Its\nremainder is a polynomial valid for every \\(u\\), not a pointwise\nquotient depending on \\(u\\). Since \\(G(a)\\in p^rR^d\\), it can be written\n\\(p^{2s+1}C\\) with \\(C\\in p^kR^d\\). Thus \\[\nG(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr).\n\\]","or expansion of the\nactual preconditioned polynomials gives one polynomial system \\(Q\\) such\nthat \\[\nG(a+tu)=G(a)+p^{2s+1}u+p^{2s+2}Q(u),\\qquad Q(0)=0.\n\\] This expansion is proved first over every commutative ring. Its\nremainder is a polynomial valid for every \\(u\\), not a pointwise\nquotient depending on \\(u\\). Since \\(G(a)\\in p^rR^d\\), it can be written\n\\(p^{2s+1}C\\) with \\(C\\in p^kR^d\\). Thus \\[\nG(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr).\n\\]","Integral polynomial evaluation is 1-Lipschitz in the supremum p-adic\nmetric. Consequently \\(T(u)=-C-pQ(u)\\) has contraction constant\n\\(|p|_p<1\\) on the complete space \\(R^d\\), and has exactly one fixed\npoint \\(u\\).","For completeness, the sharp precision is additional to this fixed-point\nexistence. Inductively, if \\(u\\in p^nR^d\\), then polynomial congruence\nand \\(Q(0)=0\\) imply \\(Q(u)\\in p^nR^d\\). For \\(n<k\\), the equation\n\\(u=-C-pQ(u)\\) therefore gives \\(u\\in p^{n+1}R^d\\). Starting at \\(n=0\\),\nthis proves \\(u\\in p^kR^d\\). Hence \\(b=a+tu\\) satisfies \\[\nb-a\\in p^{s+1+k}R^d=p^{r-s}R^d.\n\\] The normalization and injectivity of \\(A\\) give \\(F(b)=0\\).",", the sharp precision is additional to this fixed-point\nexistence. Inductively, if \\(u\\in p^nR^d\\), then polynomial congruence\nand \\(Q(0)=0\\) imply \\(Q(u)\\in p^nR^d\\). For \\(n<k\\), the equation\n\\(u=-C-pQ(u)\\) therefore gives \\(u\\in p^{n+1}R^d\\). Starting at \\(n=0\\),\nthis proves \\(u\\in p^kR^d\\). Hence \\(b=a+tu\\) satisfies \\[\nb-a\\in p^{s+1+k}R^d=p^{r-s}R^d.\n\\] The normalization and injectivity of \\(A\\) give \\(F(b)=0\\).","precision is additional to this fixed-point\nexistence. Inductively, if \\(u\\in p^nR^d\\), then polynomial congruence\nand \\(Q(0)=0\\) imply \\(Q(u)\\in p^nR^d\\). For \\(n<k\\), the equation\n\\(u=-C-pQ(u)\\) therefore gives \\(u\\in p^{n+1}R^d\\). Starting at \\(n=0\\),\nthis proves \\(u\\in p^kR^d\\). Hence \\(b=a+tu\\) satisfies \\[\nb-a\\in p^{s+1+k}R^d=p^{r-s}R^d.\n\\] The normalization and injectivity of \\(A\\) give \\(F(b)=0\\).","Every other \\(z\\in a+p^{s+1}R^d\\) has the form \\(a+tw\\). If \\(F(z)=0\\),\nthe same normalization, with cancellation of the nonzero scalar\n\\(p^{2s+1}\\), gives \\(w=T(w)\\). Fixed-point uniqueness implies \\(w=u\\),\nhence \\(z=b\\).","The strict inequality cannot in general be weakened to \\(r\\ge2s\\): for\n\\(p=2\\), \\(F(X)=X^2+3\\), and \\(a=1\\), we have \\(F(a)=4\\), \\(DF(a)=2\\),\nand may take \\(s=1,r=2\\). Yet no root exists in \\(\\mathbb Z_2\\), since\nthere is none modulo eight. The auxiliary Lean controls prove this\nobstruction and the exact residual depth.","The displacement exponent cannot uniformly be increased by one. For\n\\(F(X)=p^sX+p^r\\) at \\(a=0\\), with \\(r>2s\\), its root is \\(b=-p^{r-s}\\),\nwhich is not divisible by \\(p^{r-s+1}\\). The controls prove the linear\nroot and the absence of that extra digit for all primes.","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Ja","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis o","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and c","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic elements. Every\np","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic elements. Every\nprime and finite squ","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic elements. Every\nprime and finite square dimension are included, including \\(p","Let \\(p\\) be prime, \\(R=\\mathbb Z_p\\), \\(K=\\mathbb Q_p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic elements. Every\nprime and finite square dimension are included, including \\(p=2\\) and\ndimension zero.","mathbb Q_p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic elements. Every\nprime and finite square dimension are included, including \\(p=2\\) and\ndimension zero.","p\\), and\n\\(d,s,M\\in\\mathbb N\\). Let \\(F,H\\in R[X_1,\\ldots,X_d]^d\\), let\n\\(a\\in R^d\\) satisfy \\(F(a)=0\\), and set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic elements. Every\nprime and finite square dimension are included, including \\(p=2\\) and\ndimension zero.","set \\[\nJ=DF(a),\\qquad G=F+p^M H.\n\\] Assume \\[\nJ\\in\\operatorname{GL}_d(K),\\qquad p^sJ^{-1}\\in M_d(R),\\qquad M>2s.\n\\] Then there is exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic elements. Every\nprime and finite square dimension are included, including \\(p=2\\) and\ndimension zero.","exactly one root \\(b\\) of \\(G\\) in \\(a+p^{s+1}R^d\\). It\nsatisfies \\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic elements. Every\nprime and finite square dimension are included, including \\(p=2\\) and\ndimension zero.","\\[\nb-a\\in p^{M-s}R^d,\n\\] and there is \\(\\varepsilon\\in p^{2M-3s}R^d\\) such that, in \\(K^d\\),\n\\[\n\\boxed{b-a=-p^M J^{-1}H(a)+\\varepsilon.}\n\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic elements. Every\nprime and finite square dimension are included, including \\(p=2\\) and\ndimension zero.","\\] The inverse in this formula belongs to the original Jacobian\n\\(DF(a)\\). The theorem constructs the perturbed root from these\nhypotheses and proves uniqueness throughout the full ball of depth\n\\(s+1\\). It does not require an additional Jacobian hypothesis on \\(G\\).\nCoefficients and centers are arbitrary integral p-adic elements. Every\nprime and finite square dimension are included, including \\(p=2\\) and\ndimension zero.","The error depth exceeds the displacement depth by \\[\n(2M-3s)-(M-s)=M-2s\\ge1.\n\\] Thus the first-order term determines strictly more digits than the\ndisplacement bound alone. These are guaranteed divisibility bounds;\nequality of valuations is not asserted for every system.","Write \\(A=p^sJ^{-1}\\in M_d(R)\\), so \\(AJ=JA=p^sI\\). Let \\(t=p^{s+1}\\)\nand \\(k=M-(2s+1)\\ge0\\). The actual polynomial Taylor and difference\nidentities give polynomial systems \\(Q_F,Q_H\\), both vanishing at zero,\nsuch that \\[\n(AF)(a+tu)=p^{2s+1}u+p^{2s+2}Q_F(u),\n\\] \\[\n(AH)(a+tu)=(AH)(a)+tQ_H(u).\n\\] Consequently \\[\n(AG)(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\] where \\[\nC=p^kAH(a),\\qquad\nQ=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\] All exponents are nonnegative under \\(M>2s\\). Actual integral\npolynomial evaluation is 1-Lipschitz, so \\(u\\mapsto-C-pQ(u)\\) is a\nstrict contraction on the complete space \\(R^d\\). Its unique fixed point\ninherits the depth \\(k\\) of \\(C\\), because \\(Q(0)=0\\). Thus \\(b=a+tu\\)\nhas displacement depth \\(s+1+k=M-s","Write \\(A=p^sJ^{-1}\\in M_d(R)\\), so \\(AJ=JA=p^sI\\). Let \\(t=p^{s+1}\\)\nand \\(k=M-(2s+1)\\ge0\\). The actual polynomial Taylor and difference\nidentities give polynomial systems \\(Q_F,Q_H\\), both vanishing at zero,\nsuch that \\[\n(AF)(a+tu)=p^{2s+1}u+p^{2s+2}Q_F(u),\n\\] \\[\n(AH)(a+tu)=(AH)(a)+tQ_H(u).\n\\] Consequently \\[\n(AG)(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\] where \\[\nC=p^kAH(a),\\qquad\nQ=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\] All exponents are nonnegative under \\(M>2s\\). Actual integral\npolynomial evaluation is 1-Lipschitz, so \\(u\\mapsto-C-pQ(u)\\) is a\nstrict contraction on the complete space \\(R^d\\). Its unique fixed point\ninherits the depth \\(k\\) of \\(C\\), because \\(Q(0)=0\\). Thus \\(b=a+tu\\)\nhas displacement depth \\(s+1+k=M-s\\).","p^{s+1}\\)\nand \\(k=M-(2s+1)\\ge0\\). The actual polynomial Taylor and difference\nidentities give polynomial systems \\(Q_F,Q_H\\), both vanishing at zero,\nsuch that \\[\n(AF)(a+tu)=p^{2s+1}u+p^{2s+2}Q_F(u),\n\\] \\[\n(AH)(a+tu)=(AH)(a)+tQ_H(u).\n\\] Consequently \\[\n(AG)(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\] where \\[\nC=p^kAH(a),\\qquad\nQ=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\] All exponents are nonnegative under \\(M>2s\\). Actual integral\npolynomial evaluation is 1-Lipschitz, so \\(u\\mapsto-C-pQ(u)\\) is a\nstrict contraction on the complete space \\(R^d\\). Its unique fixed point\ninherits the depth \\(k\\) of \\(C\\), because \\(Q(0)=0\\). Thus \\(b=a+tu\\)\nhas displacement depth \\(s+1+k=M-s\\).","difference\nidentities give polynomial systems \\(Q_F,Q_H\\), both vanishing at zero,\nsuch that \\[\n(AF)(a+tu)=p^{2s+1}u+p^{2s+2}Q_F(u),\n\\] \\[\n(AH)(a+tu)=(AH)(a)+tQ_H(u).\n\\] Consequently \\[\n(AG)(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\] where \\[\nC=p^kAH(a),\\qquad\nQ=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\] All exponents are nonnegative under \\(M>2s\\). Actual integral\npolynomial evaluation is 1-Lipschitz, so \\(u\\mapsto-C-pQ(u)\\) is a\nstrict contraction on the complete space \\(R^d\\). Its unique fixed point\ninherits the depth \\(k\\) of \\(C\\), because \\(Q(0)=0\\). Thus \\(b=a+tu\\)\nhas displacement depth \\(s+1+k=M-s\\).","ishing at zero,\nsuch that \\[\n(AF)(a+tu)=p^{2s+1}u+p^{2s+2}Q_F(u),\n\\] \\[\n(AH)(a+tu)=(AH)(a)+tQ_H(u).\n\\] Consequently \\[\n(AG)(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\] where \\[\nC=p^kAH(a),\\qquad\nQ=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\] All exponents are nonnegative under \\(M>2s\\). Actual integral\npolynomial evaluation is 1-Lipschitz, so \\(u\\mapsto-C-pQ(u)\\) is a\nstrict contraction on the complete space \\(R^d\\). Its unique fixed point\ninherits the depth \\(k\\) of \\(C\\), because \\(Q(0)=0\\). Thus \\(b=a+tu\\)\nhas displacement depth \\(s+1+k=M-s\\).","^{2s+2}Q_F(u),\n\\] \\[\n(AH)(a+tu)=(AH)(a)+tQ_H(u).\n\\] Consequently \\[\n(AG)(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\] where \\[\nC=p^kAH(a),\\qquad\nQ=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\] All exponents are nonnegative under \\(M>2s\\). Actual integral\npolynomial evaluation is 1-Lipschitz, so \\(u\\mapsto-C-pQ(u)\\) is a\nstrict contraction on the complete space \\(R^d\\). Its unique fixed point\ninherits the depth \\(k\\) of \\(C\\), because \\(Q(0)=0\\). Thus \\(b=a+tu\\)\nhas displacement depth \\(s+1+k=M-s\\).","F(u),\n\\] \\[\n(AH)(a+tu)=(AH)(a)+tQ_H(u).\n\\] Consequently \\[\n(AG)(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\] where \\[\nC=p^kAH(a),\\qquad\nQ=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\] All exponents are nonnegative under \\(M>2s\\). Actual integral\npolynomial evaluation is 1-Lipschitz, so \\(u\\mapsto-C-pQ(u)\\) is a\nstrict contraction on the complete space \\(R^d\\). Its unique fixed point\ninherits the depth \\(k\\) of \\(C\\), because \\(Q(0)=0\\). Thus \\(b=a+tu\\)\nhas displacement depth \\(s+1+k=M-s\\).",")(a+tu)=(AH)(a)+tQ_H(u).\n\\] Consequently \\[\n(AG)(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\] where \\[\nC=p^kAH(a),\\qquad\nQ=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\] All exponents are nonnegative under \\(M>2s\\). Actual integral\npolynomial evaluation is 1-Lipschitz, so \\(u\\mapsto-C-pQ(u)\\) is a\nstrict contraction on the complete space \\(R^d\\). Its unique fixed point\ninherits the depth \\(k\\) of \\(C\\), because \\(Q(0)=0\\). Thus \\(b=a+tu\\)\nhas displacement depth \\(s+1+k=M-s\\).","Q_H(u).\n\\] Consequently \\[\n(AG)(a+tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\] where \\[\nC=p^kAH(a),\\qquad\nQ=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\] All exponents are nonnegative under \\(M>2s\\). Actual integral\npolynomial evaluation is 1-Lipschitz, so \\(u\\mapsto-C-pQ(u)\\) is a\nstrict contraction on the complete space \\(R^d\\). Its unique fixed point\ninherits the depth \\(k\\) of \\(C\\), because \\(Q(0)=0\\). Thus \\(b=a+tu\\)\nhas displacement depth \\(s+1+k=M-s\\).","tu)=p^{2s+1}\\bigl(C+u+pQ(u)\\bigr),\n\\] where \\[\nC=p^kAH(a),\\qquad\nQ=Q_F+p^{M-s-1}Q_H,\\qquad Q(0)=0.\n\\] All exponents are nonnegative under \\(M>2s\\). Actual integral\npolynomial evaluation is 1-Lipschitz, so \\(u\\mapsto-C-pQ(u)\\) is a\nstrict contraction on the complete space \\(R^d\\). Its unique fixed point\ninherits the depth \\(k\\) of \\(C\\), because \\(Q(0)=0\\). Thus \\(b=a+tu\\)\nhas displacement depth \\(s+1+k=M-s\\).","The certificate \\(JA=p^sI\\) makes multiplication by \\(A\\) injective, so\nthe normalized zero is an actual root of \\(G\\). Every root in \\(a+tR^d\\)\nhas the form \\(a+tw\\) and satisfies the same fixed-point equation. This\nproves uniqueness on the entire stated ball using the original matrix\ncertificate.","Put \\(\\delta=b-a\\) and \\(L=M-s\\). The displacement form of Taylor's\ntheorem, proved over any commutative ring, gives \\[\nE_F=(AF)(b)-(AF)(a)-p^s\\delta\\in p^{2L}R^d.\n\\] Polynomial congruence gives \\[\nE_H=(AH)(b)-(AH)(a)\\in p^LR^d.\n\\] Since \\(F(a)=0\\) and \\(G(b)=0\\), \\[\np^s\\delta+p^MAH(a)=-E_F-p^ME_H.\n\\] We have \\(M+L\\ge2L\\), so the right side lies in \\(p^{2L}R^d\\).\nCancelling the nonzero scalar \\(p^s\\) proves the integral estimate \\[\n\\delta+p^{M-s}AH(a)\\in p^{2M-3s}R^d.\n\\] Define this integral vector to be \\(\\varepsilon\\). Under\n\\(R\\hookrightarrow K\\), the exact identity \\(p^{M-s}A=p^MJ^{-1}\\) gives\nthe displayed first-order formula. The Lean bridge checks matrix\nevaluation and all coefficient inclusions expl","Put \\(\\delta=b-a\\) and \\(L=M-s\\). The displacement form of Taylor's\ntheorem, proved over any commutative ring, gives \\[\nE_F=(AF)(b)-(AF)(a)-p^s\\delta\\in p^{2L}R^d.\n\\] Polynomial congruence gives \\[\nE_H=(AH)(b)-(AH)(a)\\in p^LR^d.\n\\] Since \\(F(a)=0\\) and \\(G(b)=0\\), \\[\np^s\\delta+p^MAH(a)=-E_F-p^ME_H.\n\\] We have \\(M+L\\ge2L\\), so the right side lies in \\(p^{2L}R^d\\).\nCancelling the nonzero scalar \\(p^s\\) proves the integral estimate \\[\n\\delta+p^{M-s}AH(a)\\in p^{2M-3s}R^d.\n\\] Define this integral vector to be \\(\\varepsilon\\). Under\n\\(R\\hookrightarrow K\\), the exact identity \\(p^{M-s}A=p^MJ^{-1}\\) gives\nthe displayed first-order formula. The Lean bridge checks matrix\nevaluation and all coefficient inclusions explicitly.","he displacement form of Taylor's\ntheorem, proved over any commutative ring, gives \\[\nE_F=(AF)(b)-(AF)(a)-p^s\\delta\\in p^{2L}R^d.\n\\] Polynomial congruence gives \\[\nE_H=(AH)(b)-(AH)(a)\\in p^LR^d.\n\\] Since \\(F(a)=0\\) and \\(G(b)=0\\), \\[\np^s\\delta+p^MAH(a)=-E_F-p^ME_H.\n\\] We have \\(M+L\\ge2L\\), so the right side lies in \\(p^{2L}R^d\\).\nCancelling the nonzero scalar \\(p^s\\) proves the integral estimate \\[\n\\delta+p^{M-s}AH(a)\\in p^{2M-3s}R^d.\n\\] Define this integral vector to be \\(\\varepsilon\\). Under\n\\(R\\hookrightarrow K\\), the exact identity \\(p^{M-s}A=p^MJ^{-1}\\) gives\nthe displayed first-order formula. The Lean bridge checks matrix\nevaluation and all coefficient inclusions explicitly.","ives \\[\nE_F=(AF)(b)-(AF)(a)-p^s\\delta\\in p^{2L}R^d.\n\\] Polynomial congruence gives \\[\nE_H=(AH)(b)-(AH)(a)\\in p^LR^d.\n\\] Since \\(F(a)=0\\) and \\(G(b)=0\\), \\[\np^s\\delta+p^MAH(a)=-E_F-p^ME_H.\n\\] We have \\(M+L\\ge2L\\), so the right side lies in \\(p^{2L}R^d\\).\nCancelling the nonzero scalar \\(p^s\\) proves the integral estimate \\[\n\\delta+p^{M-s}AH(a)\\in p^{2M-3s}R^d.\n\\] Define this integral vector to be \\(\\varepsilon\\). Under\n\\(R\\hookrightarrow K\\), the exact identity \\(p^{M-s}A=p^MJ^{-1}\\) gives\nthe displayed first-order formula. The Lean bridge checks matrix\nevaluation and all coefficient inclusions explicitly.",")(a)-p^s\\delta\\in p^{2L}R^d.\n\\] Polynomial congruence gives \\[\nE_H=(AH)(b)-(AH)(a)\\in p^LR^d.\n\\] Since \\(F(a)=0\\) and \\(G(b)=0\\), \\[\np^s\\delta+p^MAH(a)=-E_F-p^ME_H.\n\\] We have \\(M+L\\ge2L\\), so the right side lies in \\(p^{2L}R^d\\).\nCancelling the nonzero scalar \\(p^s\\) proves the integral estimate \\[\n\\delta+p^{M-s}AH(a)\\in p^{2M-3s}R^d.\n\\] Define this integral vector to be \\(\\varepsilon\\). Under\n\\(R\\hookrightarrow K\\), the exact identity \\(p^{M-s}A=p^MJ^{-1}\\) gives\nthe displayed first-order formula. The Lean bridge checks matrix\nevaluation and all coefficient inclusions explicitly.","congruence gives \\[\nE_H=(AH)(b)-(AH)(a)\\in p^LR^d.\n\\] Since \\(F(a)=0\\) and \\(G(b)=0\\), \\[\np^s\\delta+p^MAH(a)=-E_F-p^ME_H.\n\\] We have \\(M+L\\ge2L\\), so the right side lies in \\(p^{2L}R^d\\).\nCancelling the nonzero scalar \\(p^s\\) proves the integral estimate \\[\n\\delta+p^{M-s}AH(a)\\in p^{2M-3s}R^d.\n\\] Define this integral vector to be \\(\\varepsilon\\). Under\n\\(R\\hookrightarrow K\\), the exact identity \\(p^{M-s}A=p^MJ^{-1}\\) gives\nthe displayed first-order formula. The Lean bridge checks matrix\nevaluation and all coefficient inclusions explicitly.","For \\(p=2\\), take \\(F(X)=X+X^2\\), \\(H(X)=1\\), \\(a=0\\), \\(s=0\\), \\(M=1\\).\nThere is a unique root \\(b\\in2\\mathbb Z_2\\) of \\(X+X^2+2\\). Its equation\ngives \\(b+2=-b^2\\). It also forces \\(b\\equiv2\\pmod4\\). Therefore \\[\nv_2(b+2)=2=2M-3s.\n\\] The accompanying Lean controls show that the error is divisible by\nfour and is not divisible by eight. This refutes a uniform replacement\nof \\(2M-3s\\) by \\(2M-3s+1\\), even for a one-variable unit Jacobian.","The prior strict-threshold example also applies to perturbations: take\n\\(F(X)=X^2-1\\), \\(H(X)=1\\), \\(a=1\\), \\(p=2\\), \\(s=1\\), \\(M=2\\). Then\n\\(F(a)=0\\), \\(DF(a)=2\\), and \\(F+2^2H=X^2+3\\) has no root modulo eight,\nhence none in \\(\\mathbb Z_2\\). The new explanatory identification uses\nthe previously checked no-root polynomial. The strict inequality\n\\(M>2s\\) cannot uniformly be replaced by \\(M\\ge2s\\).","Let \\(q>0\\), \\(L\\ge1\\), and \\(m,k,e\\ge0\\). Write\n\\(S=\\mathbb Z/q\\mathbb Z\\). Let \\[\nF\\in\\mathbb Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\n\\qquad x=(y_0,t_0)\\in\\mathbb Z^{m+k},\n\\qquad F(x)\\equiv0\\pmod{q^L}.\n\\] The actual lower root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equi","Let \\(q>0\\), \\(L\\ge1\\), and \\(m,k,e\\ge0\\). Write\n\\(S=\\mathbb Z/q\\mathbb Z\\). Let \\[\nF\\in\\mathbb Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\n\\qquad x=(y_0,t_0)\\in\\mathbb Z^{m+k},\n\\qquad F(x)\\equiv0\\pmod{q^L}.\n\\] The actual lower root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{","Let \\(q>0\\), \\(L\\ge1\\), and \\(m,k,e\\ge0\\). Write\n\\(S=\\mathbb Z/q\\mathbb Z\\). Let \\[\nF\\in\\mathbb Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\n\\qquad x=(y_0,t_0)\\in\\mathbb Z^{m+k},\n\\qquad F(x)\\equiv0\\pmod{q^L}.\n\\] The actual lower root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\eq","Let \\(q>0\\), \\(L\\ge1\\), and \\(m,k,e\\ge0\\). Write\n\\(S=\\mathbb Z/q\\mathbb Z\\). Let \\[\nF\\in\\mathbb Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\n\\qquad x=(y_0,t_0)\\in\\mathbb Z^{m+k},\n\\qquad F(x)\\equiv0\\pmod{q^L}.\n\\] The actual lower root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\equiv t_0+q^L\\widehat v\\pmod{q^{L+1}","Let \\(q>0\\), \\(L\\ge1\\), and \\(m,k,e\\ge0\\). Write\n\\(S=\\mathbb Z/q\\mathbb Z\\). Let \\[\nF\\in\\mathbb Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\n\\qquad x=(y_0,t_0)\\in\\mathbb Z^{m+k},\n\\qquad F(x)\\equiv0\\pmod{q^L}.\n\\] The actual lower root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\equiv t_0+q^L\\widehat v\\pmod{q^{L+1}}.\n\\] The formal predicate uses canonical integer representatives for\nevaluation and the lower congruence. The previously check","Let \\(q>0\\), \\(L\\ge1\\), and \\(m,k,e\\ge0\\). Write\n\\(S=\\mathbb Z/q\\mathbb Z\\). Let \\[\nF\\in\\mathbb Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\n\\qquad x=(y_0,t_0)\\in\\mathbb Z^{m+k},\n\\qquad F(x)\\equiv0\\pmod{q^L}.\n\\] The actual lower root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\equiv t_0+q^L\\widehat v\\pmod{q^{L+1}}.\n\\] The formal predicate uses canonical integer representatives for\nevaluation and the lower congruence. The previously checked polynomial\ncongruence theorem and digit-to-residue equivalence establish the actual\nresidue-class interpretation. Changin","Let \\(q>0\\), \\(L\\ge1\\), and \\(m,k,e\\ge0\\). Write\n\\(S=\\mathbb Z/q\\mathbb Z\\). Let \\[\nF\\in\\mathbb Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\n\\qquad x=(y_0,t_0)\\in\\mathbb Z^{m+k},\n\\qquad F(x)\\equiv0\\pmod{q^L}.\n\\] The actual lower root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\equiv t_0+q^L\\widehat v\\pmod{q^{L+1}}.\n\\] The formal predicate uses canonical integer representatives for\nevaluation and the lower congruence. The previously checked polynomial\ncongruence theorem and digit-to-residue equivalence establish the actual\nresidue-class interpretation. Changing an integral representative of a\nparameter digit by a multiple of \\(q\\) doe…","t \\[\nF\\in\\mathbb Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\n\\qquad x=(y_0,t_0)\\in\\mathbb Z^{m+k},\n\\qquad F(x)\\equiv0\\pmod{q^L}.\n\\] The actual lower root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\equiv t_0+q^L\\widehat v\\pmod{q^{L+1}}.\n\\] The formal predicate uses canonical integer representatives for\nevaluation and the lower congruence. The previously checked polynomial\ncongruence theorem and digit-to-residue equivalence establish the actual\nresidue-class interpretation. Changing an integral representative of a\nparameter digit by a multiple of \\(q\\) does not change its prescribed\nresidue modulo \\(q^{L+1}\\).","x=(y_0,t_0)\\in\\mathbb Z^{m+k},\n\\qquad F(x)\\equiv0\\pmod{q^L}.\n\\] The actual lower root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\equiv t_0+q^L\\widehat v\\pmod{q^{L+1}}.\n\\] The formal predicate uses canonical integer representatives for\nevaluation and the lower congruence. The previously checked polynomial\ncongruence theorem and digit-to-residue equivalence establish the actual\nresidue-class interpretation. Changing an integral representative of a\nparameter digit by a multiple of \\(q\\) does not change its prescribed\nresidue modulo \\(q^{L+1}\\).","Z^{m+k},\n\\qquad F(x)\\equiv0\\pmod{q^L}.\n\\] The actual lower root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\equiv t_0+q^L\\widehat v\\pmod{q^{L+1}}.\n\\] The formal predicate uses canonical integer representatives for\nevaluation and the lower congruence. The previously checked polynomial\ncongruence theorem and digit-to-residue equivalence establish the actual\nresidue-class interpretation. Changing an integral representative of a\nparameter digit by a multiple of \\(q\\) does not change its prescribed\nresidue modulo \\(q^{L+1}\\).","r root supplies the unique integral residual \\[\nr=q^{-L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\equiv t_0+q^L\\widehat v\\pmod{q^{L+1}}.\n\\] The formal predicate uses canonical integer representatives for\nevaluation and the lower congruence. The previously checked polynomial\ncongruence theorem and digit-to-residue equivalence establish the actual\nresidue-class interpretation. Changing an integral representative of a\nparameter digit by a multiple of \\(q\\) does not change its prescribed\nresidue modulo \\(q^{L+1}\\).","L}F(x)\\in\\mathbb Z^e,\\qquad \\bar r=r\\bmod q\\in S^e.\n\\] Let the actual reduced Jacobian blocks be \\[\nJ=(D_YF)(x)\\bmod q:S^m\\to S^e,\n\\qquad K=(D_TF)(x)\\bmod q:S^k\\to S^e.\n\\] For a prescribed parameter digit \\(v\\in S^k\\), choose its canonical\nintegral representative \\(\\widehat v\\). Define \\(\\mathcal L_v\\) to be\nthe set of actual residue vectors\n\\(z\\in(\\mathbb Z/q^{L+1}\\mathbb Z)^{m+k}\\) satisfying \\[\nz\\equiv x\\pmod{q^L},\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\equiv t_0+q^L\\widehat v\\pmod{q^{L+1}}.\n\\] The formal predicate uses canonical integer representatives for\nevaluation and the lower congruence. The previously checked polynomial\ncongruence theorem and digit-to-residue equivalence establish the actual\nresidue-class interpretation. Changing an integral representative of a\nparameter digit by a multiple of \\(q\\) does not change its prescribed\nresidue modulo \\(q^{L+1}\\).","\\qquad F(z)\\equiv0\\pmod{q^{L+1}},\\qquad\nz_T\\equiv t_0+q^L\\widehat v\\pmod{q^{L+1}}.\n\\] The formal predicate uses canonical integer representatives for\nevaluation and the lower congruence. The previously checked polynomial\ncongruence theorem and digit-to-residue equivalence establish the actual\nresidue-class interpretation. Changing an integral representative of a\nparameter digit by a multiple of \\(q\\) does not change its prescribed\nresidue modulo \\(q^{L+1}\\).","0+q^L\\widehat v\\pmod{q^{L+1}}.\n\\] The formal predicate uses canonical integer representatives for\nevaluation and the lower congruence. The previously checked polynomial\ncongruence theorem and digit-to-residue equivalence establish the actual\nresidue-class interpretation. Changing an integral representative of a\nparameter digit by a multiple of \\(q\\) does not change its prescribed\nresidue modulo \\(q^{L+1}\\).","Define the finite solved-block cokernel and its parameter-dependent\nclass by \\[\n\\mathcal C_x=S^e/\\operatorname{im}J,\n\\qquad o_x(v)=[\\bar r+Kv]\\in\\mathcal C_x.\n\\] Then \\[\n\\boxed{\\mathcal L_v\\ne\\varnothing\\iff o_x(v)=0.}\n\\] More precisely, the checked digit-to-lift equivalence is \\[\n\\boxed{\\mathcal L_v\\cong\\{u\\in S^m:Ju=-\\bar r-Kv\\}.}\n\\] It sends a solved digit \\(u\\) to the actual next residue represented\nby \\(x+q^L(\\widehat u,\\widehat v)\\). The obstruction depends on the\nsolved block \\(J\\), not on the range of the full map \\((J\\ K)\\).","inite solved-block cokernel and its parameter-dependent\nclass by \\[\n\\mathcal C_x=S^e/\\operatorname{im}J,\n\\qquad o_x(v)=[\\bar r+Kv]\\in\\mathcal C_x.\n\\] Then \\[\n\\boxed{\\mathcal L_v\\ne\\varnothing\\iff o_x(v)=0.}\n\\] More precisely, the checked digit-to-lift equivalence is \\[\n\\boxed{\\mathcal L_v\\cong\\{u\\in S^m:Ju=-\\bar r-Kv\\}.}\n\\] It sends a solved digit \\(u\\) to the actual next residue represented\nby \\(x+q^L(\\widehat u,\\widehat v)\\). The obstruction depends on the\nsolved block \\(J\\), not on the range of the full map \\((J\\ K)\\).","_x=S^e/\\operatorname{im}J,\n\\qquad o_x(v)=[\\bar r+Kv]\\in\\mathcal C_x.\n\\] Then \\[\n\\boxed{\\mathcal L_v\\ne\\varnothing\\iff o_x(v)=0.}\n\\] More precisely, the checked digit-to-lift equivalence is \\[\n\\boxed{\\mathcal L_v\\cong\\{u\\in S^m:Ju=-\\bar r-Kv\\}.}\n\\] It sends a solved digit \\(u\\) to the actual next residue represented\nby \\(x+q^L(\\widehat u,\\widehat v)\\). The obstruction depends on the\nsolved block \\(J\\), not on the range of the full map \\((J\\ K)\\).","=[\\bar r+Kv]\\in\\mathcal C_x.\n\\] Then \\[\n\\boxed{\\mathcal L_v\\ne\\varnothing\\iff o_x(v)=0.}\n\\] More precisely, the checked digit-to-lift equivalence is \\[\n\\boxed{\\mathcal L_v\\cong\\{u\\in S^m:Ju=-\\bar r-Kv\\}.}\n\\] It sends a solved digit \\(u\\) to the actual next residue represented\nby \\(x+q^L(\\widehat u,\\widehat v)\\). The obstruction depends on the\nsolved block \\(J\\), not on the range of the full map \\((J\\ K)\\).","Every nonempty \\(\\mathcal L_v\\) has a checked free and transitive action\nof \\(\\ker J\\). The action is defined without choosing a base lift.\nChoosing one valid solved digit then gives an explicit bijection\n\\(\\mathcal L_v\\cong\\ker J\\). In particular, for every positive modulus,\n\\[\no_x(v)=0\\quad\\Longrightarrow\\quad\\#\\mathcal L_v=\\#\\ker J.\n\\] These results permit composite \\(q\\), the degenerate modulus \\(q=1\\),\nand zero dimensions. Neither \\(J\\) nor the full reduced Jacobian is\nassumed surjective.","y nonempty \\(\\mathcal L_v\\) has a checked free and transitive action\nof \\(\\ker J\\). The action is defined without choosing a base lift.\nChoosing one valid solved digit then gives an explicit bijection\n\\(\\mathcal L_v\\cong\\ker J\\). In particular, for every positive modulus,\n\\[\no_x(v)=0\\quad\\Longrightarrow\\quad\\#\\mathcal L_v=\\#\\ker J.\n\\] These results permit composite \\(q\\), the degenerate modulus \\(q=1\\),\nand zero dimensions. Neither \\(J\\) nor the full reduced Jacobian is\nassumed surjective.","ked free and transitive action\nof \\(\\ker J\\). The action is defined without choosing a base lift.\nChoosing one valid solved digit then gives an explicit bijection\n\\(\\mathcal L_v\\cong\\ker J\\). In particular, for every positive modulus,\n\\[\no_x(v)=0\\quad\\Longrightarrow\\quad\\#\\mathcal L_v=\\#\\ker J.\n\\] These results permit composite \\(q\\), the degenerate modulus \\(q=1\\),\nand zero dimensions. Neither \\(J\\) nor the full reduced Jacobian is\nassumed surjective.","When \\(q=p\\) is prime, put \\(\\rho=\\operatorname{rank}_{\\mathbb F_p}J\\).\nThen \\[\n\\boxed{\\#\\mathcal L_v=\n\\begin{cases}\np^{m-\\rho},&o_x(v)=0,","When \\(q=p\\) is prime, put \\(\\rho=\\operatorname{rank}_{\\mathbb F_p}J\\).\nThen \\[\n\\boxed{\\#\\mathcal L_v=\n\\begin{cases}\np^{m-\\rho},&o_x(v)=0,\\\\\n0,&o_x(v)\\ne0,\n\\end{cases}}\n\\qquad\n\\boxed{\\#\\mathcal C_x=p^{e-\\rho}.}\n\\] The count is of distinct actual residue classes at level \\(L+1\\). The\nsolved dimension \\(m\\), rather than the total number \\(m+k\\) of\nvariables, appears in the exponent because the parameter digit is fixed.\nThese rank formulas are asserted only over prime fields.","0,&o_x(v)\\ne0,\n\\end{cases}}\n\\qquad\n\\boxed{\\#\\mathcal C_x=p^{e-\\rho}.}\n\\] The count is of distinct actual residue classes at level \\(L+1\\). The\nsolved dimension \\(m\\), rather than the total number \\(m+k\\) of\nvariables, appears in the exponent because the parameter digit is fixed.\nThese rank formulas are asserted only over prime fields.","The main wrapper assumes only the original lower polynomial congruence.\nIt constructs one residual \\(r\\) and supplies the criterion and count\nuniformly for every \\(v\\), together with the exact obstruction\ncardinality.","The actual derivative calculation gives \\[\n(DF(x)\\bmod q)(u,v)=Ju+Kv.\n\\] Taylor expansion at depth \\(L\\ge1\\) makes the quadratic remainder\ndivisible by \\(q^{L+1}\\). Therefore the digit \\((u,v)\\) is a polynomial\nlifting correction exactly when \\[\n\\bar r+Ju+Kv=0.\n\\] The quotient class \\([\\bar r+Kv]\\) vanishes exactly when this\nequation admits \\(u\\). This uses additive inverses in the image of\n\\(J\\), so the sign agrees with the actual Taylor equation.","The earlier unrestricted correction-digit equivalence is then restricted\nto the predicate that the parameter digit equals \\(v\\). Taking free\ncoordinates commutes with forming digit residues, and digit residues are\ninjective in the digit. Consequently the restricted equivalence enforces\nexactly the actual parameter residue condition on \\(\\mathcal L_v\\).","Translation by \\(\\ker J\\) acts freely and transitively on the\nsolved-digit fiber. Transporting this explicit action through the\nresidue equivalence yields the torsor on \\(\\mathcal L_v\\). Rank-nullity\nover \\(\\mathbb F_p\\) then gives the two displayed cardinalities. No\nnumerical enumeration is used to establish the general theorem.","Take \\(F(Y,T)=T\\), \\(q=2\\), \\(L=1\\), and \\(x=(0,0)\\). The full Jacobian\n\\((0\\ 1)\\) is surjective, but \\(J=0\\). Prescribing \\(v=1\\) forces\n\\(T\\equiv2\\pmod4\\), which cannot satisfy \\(F=0\\pmod4\\). Prescribing\n\\(v=0\\) gives exactly two lifts, corresponding to \\(Y\\equiv0,2\\pmod4\\).","Let \\(q>0\\), \\(L\\ge1\\), and \\[\nF\\in\\mathbb Z[Y_1,\\ldots,Y_m,T_1,\\ldots,T_k]^e,\n\\qquad x=(y_0,t_0)\\in\\mathbb Z^{m+k},\n\\qquad F(x)\\in q^L\\mathbb Z^e.\n\\] Write \\(S_q=\\mathbb Z/q\\mathbb Z\\), and let the actual reduced\nJacobian blocks be \\[\nJ_x=D_YF(x)\\bmod q:S_q^m\\longrightarrow S_q^e,\n\\qquad K_x=D_TF(x)\\bmod q:S_q^k\\longrightarrow S_q^e.\n\\] Assume \\(J_x\\) is surjective. A vector divisibility condition below\nis coordinatewise.","\\textbf{Integer prescribed-path theorem.} For every supplied path\n\\(t:\\mathbb N\\to\\mathbb Z^k\\) satisfying \\[\nt(0)=t_0,\\qquad t(n+1)-t(n)\\in q^{L+n}\\mathbb Z^k,\n\\] there is a single tower \\(z:\\mathbb N\\to\\mathbb Z^{m+k}\\) with \\[\nz(0)=x,\\qquad \\operatorname{free}(z(n))=t(n),\n\\] \\[\nF(z(n))\\in q^{L+n}\\mathbb Z^e,\\qquad\nz(n+1)-z(n)\\in q^{L+n}\\mathbb Z^{m+k}.\n\\] The parameter equality is exact in integers. For \\(a\\le b\\), \\[\nz(b)-z(a)\\in q^{L+a}\\mathbb Z^{m+k},\\qquad\nDF(z(n))\\bmod q=DF(x)\\bmod q.\n\\] Consequently \\(J_{z(n)}=J_x\\) at every level. No later surjectivity\nor per-step solvability is assumed. This theorem includes positive\ncomposite bases and \\(q=1\\).","ger prescribed-path theorem.} For every supplied path\n\\(t:\\mathbb N\\to\\mathbb Z^k\\) satisfying \\[\nt(0)=t_0,\\qquad t(n+1)-t(n)\\in q^{L+n}\\mathbb Z^k,\n\\] there is a single tower \\(z:\\mathbb N\\to\\mathbb Z^{m+k}\\) with \\[\nz(0)=x,\\qquad \\operatorname{free}(z(n))=t(n),\n\\] \\[\nF(z(n))\\in q^{L+n}\\mathbb Z^e,\\qquad\nz(n+1)-z(n)\\in q^{L+n}\\mathbb Z^{m+k}.\n\\] The parameter equality is exact in integers. For \\(a\\le b\\), \\[\nz(b)-z(a)\\in q^{L+a}\\mathbb Z^{m+k},\\qquad\nDF(z(n))\\bmod q=DF(x)\\bmod q.\n\\] Consequently \\(J_{z(n)}=J_x\\) at every level. No later surjectivity\nor per-step solvability is assumed. This theorem includes positive\ncomposite bases and \\(q=1\\).","ed-path theorem.} For every supplied path\n\\(t:\\mathbb N\\to\\mathbb Z^k\\) satisfying \\[\nt(0)=t_0,\\qquad t(n+1)-t(n)\\in q^{L+n}\\mathbb Z^k,\n\\] there is a single tower \\(z:\\mathbb N\\to\\mathbb Z^{m+k}\\) with \\[\nz(0)=x,\\qquad \\operatorname{free}(z(n))=t(n),\n\\] \\[\nF(z(n))\\in q^{L+n}\\mathbb Z^e,\\qquad\nz(n+1)-z(n)\\in q^{L+n}\\mathbb Z^{m+k}.\n\\] The parameter equality is exact in integers. For \\(a\\le b\\), \\[\nz(b)-z(a)\\in q^{L+a}\\mathbb Z^{m+k},\\qquad\nDF(z(n))\\bmod q=DF(x)\\bmod q.\n\\] Consequently \\(J_{z(n)}=J_x\\) at every level. No later surjectivity\nor per-step solvability is assumed. This theorem includes positive\ncomposite bases and \\(q=1\\).","_0,\\qquad t(n+1)-t(n)\\in q^{L+n}\\mathbb Z^k,\n\\] there is a single tower \\(z:\\mathbb N\\to\\mathbb Z^{m+k}\\) with \\[\nz(0)=x,\\qquad \\operatorname{free}(z(n))=t(n),\n\\] \\[\nF(z(n))\\in q^{L+n}\\mathbb Z^e,\\qquad\nz(n+1)-z(n)\\in q^{L+n}\\mathbb Z^{m+k}.\n\\] The parameter equality is exact in integers. For \\(a\\le b\\), \\[\nz(b)-z(a)\\in q^{L+a}\\mathbb Z^{m+k},\\qquad\nDF(z(n))\\bmod q=DF(x)\\bmod q.\n\\] Consequently \\(J_{z(n)}=J_x\\) at every level. No later surjectivity\nor per-step solvability is assumed. This theorem includes positive\ncomposite bases and \\(q=1\\).","free}(z(n))=t(n),\n\\] \\[\nF(z(n))\\in q^{L+n}\\mathbb Z^e,\\qquad\nz(n+1)-z(n)\\in q^{L+n}\\mathbb Z^{m+k}.\n\\] The parameter equality is exact in integers. For \\(a\\le b\\), \\[\nz(b)-z(a)\\in q^{L+a}\\mathbb Z^{m+k},\\qquad\nDF(z(n))\\bmod q=DF(x)\\bmod q.\n\\] Consequently \\(J_{z(n)}=J_x\\) at every level. No later surjectivity\nor per-step solvability is assumed. This theorem includes positive\ncomposite bases and \\(q=1\\).","\\textbf{Arbitrary p-adic target theorem.} Now let \\(q=p\\) be prime. For\nevery \\[\n\\tau\\in t_0+p^L\\mathbb Z_p^k,\n\\] there are a compatible integer tower \\(z\\) as above and\n\\(b\\in\\mathbb Z_p^{m+k}\\) such that \\[\nz(n)\\longrightarrow b,\\qquad F(b)=0,\\qquad\n\\operatorname{free}(b)=\\tau,\n\\qquad b-z(n)\\in p^{L+n}\\mathbb Z_p^{m+k}\\quad(n\\ge0).\n\\] A compatible integer parameter path converging to \\(\\tau\\) is\nconstructed in the proof, with its initial representative fixed exactly.\nA separate theorem accepts every supplied compatible integer path with\nlimit \\(\\tau\\) and retains its exact coordinates in the constructed\ntower.","z\\) as above and\n\\(b\\in\\mathbb Z_p^{m+k}\\) such that \\[\nz(n)\\longrightarrow b,\\qquad F(b)=0,\\qquad\n\\operatorname{free}(b)=\\tau,\n\\qquad b-z(n)\\in p^{L+n}\\mathbb Z_p^{m+k}\\quad(n\\ge0).\n\\] A compatible integer parameter path converging to \\(\\tau\\) is\nconstructed in the proof, with its initial representative fixed exactly.\nA separate theorem accepts every supplied compatible integer path with\nlimit \\(\\tau\\) and retains its exact coordinates in the constructed\ntower.","\\textbf{Exact branching.} At any stage \\(n\\), put \\(N=L+n\\). For every\nparameter digit \\(v\\in\\mathbb F_p^k\\), let \\[\n\\mathcal L_{n,v}=\\left\\{\\bar w\\in(\\mathbb Z/p^{N+1}\\mathbb Z)^{m+k}:\n\\begin{array}{l}\n\\bar w\\equiv z(n)\\pmod{p^N},","\\textbf{Exact branching.} At any stage \\(n\\), put \\(N=L+n\\). For every\nparameter digit \\(v\\in\\mathbb F_p^k\\), let \\[\n\\mathcal L_{n,v}=\\left\\{\\bar w\\in(\\mathbb Z/p^{N+1}\\mathbb Z)^{m+k}:\n\\begin{array}{l}\n\\bar w\\equiv z(n)\\pmod{p^N},\\\\\nF(\\bar w)=0,\\\\\n\\operatorname{free}(\\bar w)=t(n)+p^Nv\\pmod{p^{N+1}}\n\\end{array}\\right\\}.\n\\] Here \\(p^Nv\\) means any integer representative of \\(v\\), multiplied\nby \\(p^N\\); the residue class does not depend on the representative.\nThen \\[\n\\#\\mathcal L_{n,v}=p^{m-e}\\qquad\\text{for every }n\\text{ and every }v.\n\\] Surjectivity implies \\(e\\le m\\). This is a count of distinct actual\nnext-level residue classes. It does not count integer representatives or\nentire infinite towers.","\\operatorname{free}(\\bar w)=t(n)+p^Nv\\pmod{p^{N+1}}\n\\end{array}\\right\\}.\n\\] Here \\(p^Nv\\) means any integer representative of \\(v\\), multiplied\nby \\(p^N\\); the residue class does not depend on the representative.\nThen \\[\n\\#\\mathcal L_{n,v}=p^{m-e}\\qquad\\text{for every }n\\text{ and every }v.\n\\] Surjectivity implies \\(e\\le m\\). This is a count of distinct actual\nnext-level residue classes. It does not count integer representatives or\nentire infinite towers.","\\textbf{Square solved-block uniqueness.} If \\(m=e\\), every prescribed\n\\(\\tau\\) has exactly one root in the original ball: \\[\n\\exists!\\,b\\in x+p^L\\mathbb Z_p^{e+k}:\\quad\nF(b)=0,\\qquad\\operatorname{free}(b)=\\tau.\n\\] Thus all valid integer parameter approximations and all lifting\nchoices with the same target have the same endpoint. This does not\nassert uniqueness of the integer representatives in the tower. All prime\nstatements include \\(p=2\\).","} If \\(m=e\\), every prescribed\n\\(\\tau\\) has exactly one root in the original ball: \\[\n\\exists!\\,b\\in x+p^L\\mathbb Z_p^{e+k}:\\quad\nF(b)=0,\\qquad\\operatorname{free}(b)=\\tau.\n\\] Thus all valid integer parameter approximations and all lifting\nchoices with the same target have the same endpoint. This does not\nassert uniqueness of the integer representatives in the tower. All prime\nstatements include \\(p=2\\).","At depth \\(N\\ge1\\), suppose \\(F(a)=q^Nr\\) and a desired next integer\nparameter target satisfies \\(t'-\\operatorname{free}(a)=q^Nw\\), with\n\\(w\\in\\mathbb Z^k\\). Surjectivity supplies a solved digit \\(u\\in S_q^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative g","At depth \\(N\\ge1\\), suppose \\(F(a)=q^Nr\\) and a desired next integer\nparameter target satisfies \\(t'-\\operatorname{free}(a)=q^Nw\\), with\n\\(w\\in\\mathbb Z^k\\). Surjectivity supplies a solved digit \\(u\\in S_q^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced J","At depth \\(N\\ge1\\), suppose \\(F(a)=q^Nr\\) and a desired next integer\nparameter target satisfies \\(t'-\\operatorname{free}(a)=q^Nw\\), with\n\\(w\\in\\mathbb Z^k\\). Surjectivity supplies a solved digit \\(u\\in S_q^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJa","At depth \\(N\\ge1\\), suppose \\(F(a)=q^Nr\\) and a desired next integer\nparameter target satisfies \\(t'-\\operatorname{free}(a)=q^Nw\\), with\n\\(w\\in\\mathbb Z^k\\). Surjectivity supplies a solved digit \\(u\\in S_q^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJacobian as its state, const","At depth \\(N\\ge1\\), suppose \\(F(a)=q^Nr\\) and a desired next integer\nparameter target satisfies \\(t'-\\operatorname{free}(a)=q^Nw\\), with\n\\(w\\in\\mathbb Z^k\\). Surjectivity supplies a solved digit \\(u\\in S_q^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJacobian as its state, constructs the entire integer tower. The\nexistence proof u","At depth \\(N\\ge1\\), suppose \\(F(a)=q^Nr\\) and a desired next integer\nparameter target satisfies \\(t'-\\operatorname{free}(a)=q^Nw\\), with\n\\(w\\in\\mathbb Z^k\\). Surjectivity supplies a solved digit \\(u\\in S_q^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJacobian as its state, constructs the entire integer tower. The\nexistence proof uses ordinary classical choice; it does not","At depth \\(N\\ge1\\), suppose \\(F(a)=q^Nr\\) and a desired next integer\nparameter target satisfies \\(t'-\\operatorname{free}(a)=q^Nw\\), with\n\\(w\\in\\mathbb Z^k\\). Surjectivity supplies a solved digit \\(u\\in S_q^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJacobian as its state, constructs the entire integer tower. The\nexistence proof uses ordinary classical choice; it does not supply an\noptimized computational lifting algorit","At depth \\(N\\ge1\\), suppose \\(F(a)=q^Nr\\) and a desired next integer\nparameter target satisfies \\(t'-\\operatorname{free}(a)=q^Nw\\), with\n\\(w\\in\\mathbb Z^k\\). Surjectivity supplies a solved digit \\(u\\in S_q^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJacobian as its state, constructs the entire integer tower. The\nexistence proof uses ordinary classical choice; it does not supply an\noptimized computational lifting algorithm.","\\(N\\ge1\\), suppose \\(F(a)=q^Nr\\) and a desired next integer\nparameter target satisfies \\(t'-\\operatorname{free}(a)=q^Nw\\), with\n\\(w\\in\\mathbb Z^k\\). Surjectivity supplies a solved digit \\(u\\in S_q^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJacobian as its state, constructs the entire integer tower. The\nexistence proof uses ordinary classical choice; it does not supply an\noptimized computational lifting algorithm.","s \\(t'-\\operatorname{free}(a)=q^Nw\\), with\n\\(w\\in\\mathbb Z^k\\). Surjectivity supplies a solved digit \\(u\\in S_q^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJacobian as its state, constructs the entire integer tower. The\nexistence proof uses ordinary classical choice; it does not supply an\noptimized computational lifting algorithm.","^m\\)\nsatisfying \\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJacobian as its state, constructs the entire integer tower. The\nexistence proof uses ordinary classical choice; it does not supply an\noptimized computational lifting algorithm.","\\[\nJ_a u=-\\bar r-K_a\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJacobian as its state, constructs the entire integer tower. The\nexistence proof uses ordinary classical choice; it does not supply an\noptimized computational lifting algorithm.","\\bar w.\n\\] Choose an integer representative \\(\\widetilde u\\), and set \\[\na'=a+q^N(\\widetilde u,w).\n\\] The parameter quotient is the supplied integer \\(w\\). Therefore\n\\(\\operatorname{free}(a')=t'\\) exactly. The checked polynomial Taylor\nidentity gives \\[\nF(a')\\equiv F(a)+q^NDF(a)(\\widetilde u,w)\n\\equiv0\\pmod{q^{N+1}}.\n\\] Its quadratic remainder is divisible by \\(q^{2N}\\), and\n\\(2N\\ge N+1\\). Also \\(a'\\equiv a\\pmod q\\), so evaluation of every\npolynomial partial derivative gives the same reduced Jacobian. Induction\non the level, carrying the root, exact parameters, and fixed solved\nJacobian as its state, constructs the entire integer tower. The\nexistence proof uses ordinary classical choice; it does not supply an\noptimized computational lifting algorithm.","For a p-adic target \\(\\tau\\), use \\(t(0)=t_0\\) and, for \\(n\\ge1\\), the\ncanonical integer representatives of \\(\\tau\\bmod p^{L+n}\\). The initial\ncongruence hypothesis and the truncation precision imply \\[\n\\tau-t(n)\\in p^{L+n}\\mathbb Z_p^k.\n\\] Subtracting successive precision relations gives compatibility in\n\\(\\mathbb Z_p\\); the checked faithful integer-divisibility theorem gives\ncompatibility in \\(\\mathbb Z\\). These precision relations also give\nconvergence to \\(\\tau\\).",", the\ncanonical integer representatives of \\(\\tau\\bmod p^{L+n}\\). The initial\ncongruence hypothesis and the truncation precision imply \\[\n\\tau-t(n)\\in p^{L+n}\\mathbb Z_p^k.\n\\] Subtracting successive precision relations gives compatibility in\n\\(\\mathbb Z_p\\); the checked faithful integer-divisibility theorem gives\ncompatibility in \\(\\mathbb Z\\). These precision relations also give\nconvergence to \\(\\tau\\).","The compatible vector-tower completion theorem constructs \\(b\\), proves\nconvergence, and retains every stage precision. Continuous polynomial\nevaluation and the increasing root precision give \\(F(b)=0\\). Each\nparameter coordinate has both the corresponding coordinate of \\(b\\) and\n\\(\\tau\\) as its limit, so uniqueness of limits gives exact parameter\nequality.","For every proposed next parameter digit, the previously checked\nobstruction is \\([\\bar r+K_av]\\in S_p^e/\\operatorname{im}J_a\\). The\nfixed solved Jacobian is surjective, so this class vanishes for every\ndigit at every stage. The actual lift fiber is a torsor for\n\\(\\ker J_a\\). Rank-nullity gives its cardinality \\(p^{m-e}\\).","When \\(m=e\\), the new solved-Jacobian definition is definitionally the\nexisting square block used by the checked continuous parameter chart.\nEvery root in the initial ball with the given parameters is that chart\npoint. Applying this to any two roots proves equality, including\nendpoints obtained by different constructions.","The inherited checked example \\(F(Y,T)=T\\), \\(p=2\\), \\(x=0\\), \\(L=1\\)\nhas a surjective full Jacobian but zero solved block. Prescribing\nparameter digit \\(1\\) forces \\(T=2\\pmod4\\), so no next root exists.\nFull-Jacobian surjectivity cannot replace the hypothesis on the\nprescribed solved block.","The new checked example \\(F(Y_1,Y_2,T)=Y_1\\) has a surjective solved\nblock \\((1,0)\\). Nevertheless \\((0,0,0)\\) and \\((0,2,0)\\) are distinct\nexact roots in the same modulo-two ball with identical parameters.\nUniqueness therefore requires the square solved case. The corresponding\nnegative control deliberately attempts the stronger false conclusion and\nmust be rejected by Lean.","Depth \\(L\\ge1\\) is essential for the Taylor step used here. The general\nsingular case can have a nonzero next obstruction after an earlier zero\nobstruction. Neither the finite one-step criterion nor this smooth\ntheorem establishes unrestricted singular persistence.","Let \\(p\\) be any prime, \\(R=\\mathbb Z_p\\), and \\(Q=\\mathbb Q_p\\). Let \\[\nF,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in R^{e+k},\\qquad F(a)=0.\n\\] Here \\(e,k\\ge0\\) are arbitrary finite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral v","Let \\(p\\) be any prime, \\(R=\\mathbb Z_p\\), and \\(Q=\\mathbb Q_p\\). Let \\[\nF,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in R^{e+k},\\qquad F(a)=0.\n\\] Here \\(e,k\\ge0\\) are arbitrary finite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}","Let \\(p\\) be any prime, \\(R=\\mathbb Z_p\\), and \\(Q=\\mathbb Q_p\\). Let \\[\nF,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in R^{e+k},\\qquad F(a)=0.\n\\] Here \\(e,k\\ge0\\) are arbitrary finite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q","Let \\(p\\) be any prime, \\(R=\\mathbb Z_p\\), and \\(Q=\\mathbb Q_p\\). Let \\[\nF,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in R^{e+k},\\qquad F(a)=0.\n\\] Here \\(e,k\\ge0\\) are arbitrary finite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same origi","Let \\(p\\) be any prime, \\(R=\\mathbb Z_p\\), and \\(Q=\\mathbb Q_p\\). Let \\[\nF,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in R^{e+k},\\qquad F(a)=0.\n\\] Here \\(e,k\\ge0\\) are arbitrary finite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\","Let \\(p\\) be any prime, \\(R=\\mathbb Z_p\\), and \\(Q=\\mathbb Q_p\\). Let \\[\nF,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in R^{e+k},\\qquad F(a)=0.\n\\] Here \\(e,k\\ge0\\) are arbitrary finite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.","rime, \\(R=\\mathbb Z_p\\), and \\(Q=\\mathbb Q_p\\). Let \\[\nF,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in R^{e+k},\\qquad F(a)=0.\n\\] Here \\(e,k\\ge0\\) are arbitrary finite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.","hbb Z_p\\), and \\(Q=\\mathbb Q_p\\). Let \\[\nF,H\\in R[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in R^{e+k},\\qquad F(a)=0.\n\\] Here \\(e,k\\ge0\\) are arbitrary finite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.","\\qquad a=(y_0,t_0)\\in R^{e+k},\\qquad F(a)=0.\n\\] Here \\(e,k\\ge0\\) are arbitrary finite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.",")\\in R^{e+k},\\qquad F(a)=0.\n\\] Here \\(e,k\\ge0\\) are arbitrary finite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.","inite dimensions. There are as many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.","s many\nsolved coordinates as equations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.","ations. Define the actual derivative blocks at\nthe original root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.","l root: \\[\nJ=D_YF(a)\\in M_e(R),\\qquad K=D_TF(a)\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.",")\\in M_{e\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.","\\times k}(R).\n\\] Suppose, for natural numbers \\(s,M\\), that \\[\nJ\\in\\operatorname{GL}_e(Q),\\qquad p^sJ^{-1}\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.","\\in M_e(R),\\qquad M>2s.\n\\] For every \\(v\\in R^k\\), set \\[\n\\tau=t_0+p^Mv,\\qquad G=F+p^MH.\n\\] Then there is exactly one \\(y\\in y_0+p^{s+1}R^e\\) with \\[\nG(y,\\tau)=0.\n\\] It satisfies \\[\ny-y_0\\in p^{M-s}R^e\n\\] and there is an integral vector \\(\\varepsilon\\in p^{2M-3s}R^e\\) such\nthat, in \\(Q^e\\), \\[\n\\boxed{\\displaystyle\n y-y_0=-p^MJ^{-1}\\bigl(H(a)+Kv\\bigr)+\\varepsilon.\n}\n\\] The same original exponent \\(s\\) appears throughout. No inverse or\nrank assumption on the perturbed Jacobian is imposed. The statement\nincludes arbitrary integral p-adic coefficients and centers, the dyadic\nprime, and zero dimensions.","The full-point theorem constructs \\(b=(y,\\tau)\\) and also proves \\[\nb-a\\in p^{M-s}R^{e+k}\\cap p^{s+1}R^{e+k}.\n\\] Parameter equality \\(b_T=\\tau\\) is exact. Uniqueness concerns every\nroot in the stated solved-coordinate ball with these parameters, not\nonly roots obtained by the construction. No global uniqueness is\nclaimed.","Define polynomial systems with \\(e+k\\) equations and variables by \\[\n\\mathcal F(Y,T)=\\bigl(F(Y,T),T-t_0\\bigr),\\qquad\n\\mathcal H(Y,T)=\\bigl(H(Y,T),-v\\bigr).\n\\] Both are actual polynomials over \\(R\\). Direct evaluation proves \\[\n\\mathcal F(a)=0,\n\\qquad\n(\\mathcal F+p^M\\mathcal H)(b)=0\n\\iff F(b)+p^MH(b)=0\\ \\text{and}\\ b_T=t_0+p^Mv.\n\\] Actual polynomial partial differentiation gives \\[\nD\\mathcal F(a)=B=\\begin{pmatrix}J&K","Define polynomial systems with \\(e+k\\) equations and variables by \\[\n\\mathcal F(Y,T)=\\bigl(F(Y,T),T-t_0\\bigr),\\qquad\n\\mathcal H(Y,T)=\\bigl(H(Y,T),-v\\bigr).\n\\] Both are actual polynomials over \\(R\\). Direct evaluation proves \\[\n\\mathcal F(a)=0,\n\\qquad\n(\\mathcal F+p^M\\mathcal H)(b)=0\n\\iff F(b)+p^MH(b)=0\\ \\text{and}\\ b_T=t_0+p^Mv.\n\\] Actual polynomial partial differentiation gives \\[\nD\\mathcal F(a)=B=\\begin{pmatrix}J&K\\\\0&I_k\\end{pmatrix}.\n\\] These identities are proved from evaluation and partial derivatives,\nrather than supplied as additional hypotheses.","Choose the integral matrix \\(A=p^sJ^{-1}\\). The existing\ninverse-certificate equivalence gives \\[\nAJ=JA=p^sI_e.\n\\] Put \\(c=p^s\\) and define an integral block matrix \\[\nC=\\begin{pmatrix}A&-AK","Choose the integral matrix \\(A=p^sJ^{-1}\\). The existing\ninverse-certificate equivalence gives \\[\nAJ=JA=p^sI_e.\n\\] Put \\(c=p^s\\) and define an integral block matrix \\[\nC=\\begin{pmatrix}A&-AK\\\\0&cI_k\\end{pmatrix}.\n\\] Then \\[\nCB=\\begin{pmatrix}AJ&AK-AK","0&cI_k\\end{pmatrix}.\n\\] Then \\[\nCB=\\begin{pmatrix}AJ&AK-AK\\\\0&cI_k\\end{pmatrix}=cI_{e+k},\n\\] \\[\nBC=\\begin{pmatrix}JA&-JAK+cK","0&cI_k\\end{pmatrix}=cI_{e+k},\n\\] \\[\nBC=\\begin{pmatrix}JA&-JAK+cK\\\\0&cI_k\\end{pmatrix}=cI_{e+k}.\n\\] The second equality uses \\(JA=cI_e\\). The generic block proof works\nover any commutative ring. The p-adic specialization also explicitly\nproves that \\(B\\) is invertible over \\(Q\\) and that \\(p^sB^{-1}\\) is\nintegral. There is no denominator loss from appending the parameter\nequations.","0&cI_k\\end{pmatrix}=cI_{e+k}.\n\\] The second equality uses \\(JA=cI_e\\). The generic block proof works\nover any commutative ring. The p-adic specialization also explicitly\nproves that \\(B\\) is invertible over \\(Q\\) and that \\(p^sB^{-1}\\) is\nintegral. There is no denominator loss from appending the parameter\nequations.","Apply the previously checked weighted coefficient-perturbation\nconstruction to \\(\\mathcal F,\\mathcal H,a,C\\). Its hypotheses follow\nfrom the actual augmented identities and the two-sided certificate. It\nconstructs \\(b\\), proves the full displacement bound \\(p^{M-s}\\), and\nproves uniqueness throughout \\(a+p^{s+1}R^{e+k}\\). The augmented root\nequivalence gives the required original equations and exact parameter\nvalues.","The checked integral first-order estimate gives \\[\nb-a+p^{M-s}C\\mathcal H(a)\\in p^{2M-3s}R^{e+k}.\n\\] The decisive block action is \\[\nC\\binom{H(a)}{-v}\n=\n\\binom{A\\bigl(H(a)+Kv\\bigr)}{-p^sv}.\n\\] Projecting the solved coordinates therefore gives \\[\ny-y_0+p^{M-s}A\\bigl(H(a)+Kv\\bigr)\n\\in p^{2M-3s}R^e.\n\\] The existing rational-conversion theorem turns \\(p^{M-s}A\\) into\n\\(p^MJ^{-1}\\) and retains an integral error vector. This proves the\ndisplayed first-order formula with the original derivative blocks.","For any competing solved-coordinate root \\(z\\in y_0+p^{s+1}R^e\\), the\njoined point \\((z,\\tau)\\) lies in the augmented uniqueness ball:\n\\(M>2s\\) implies \\(M\\ge s+1\\), so \\(\\tau-t_0=p^Mv\\in p^{s+1}R^k\\).\nFull-point uniqueness gives \\((z,\\tau)=b\\), hence \\(z=y\\). This is the\nadditional comparison needed to obtain uniqueness in exactly the ball\nstated in the solved-coordinate theorem.","For \\(p=2\\), \\(s=0\\), \\(F(Y,T)=Y+T\\), \\(H=1\\), \\(a=0\\), and \\(v=1\\),\nboth original blocks satisfy \\(J=K=1\\). At \\(M=1\\), the exact parameter\nis \\(\\tau=2\\), and the unique solved value is \\(y=-4\\). The combined\ncorrection is \\(-2(1+1)=-4\\), with zero error.","The sign controls use \\(M=2\\): the actual augmented root is\n\\((y,\\tau)=(-8,4)\\), and the required error ideal is \\(16\\mathbb Z_2\\).\nOmitting \\(Kv\\) predicts \\(-4\\), with error \\(-4\\notin16\\mathbb Z_2\\).\nReversing its sign predicts \\(0\\), with error \\(-8\\notin16\\mathbb Z_2\\).\nBoth failures are checked. At binary depth \\(M=1\\), the reversed-sign\ndifference would lie in the allowed error ideal, so that shallower\nexample alone would not distinguish the sign. This control-design\nlimitation was identified and corrected before the release gate.","For parameter-only motion, use \\[\nF(Y,T)=Y+Y^2+T,\\quad H=0,\\quad a=0,\\quad v=1,\\quad p=2,\\ s=0,\\ M=1.\n\\] A checked transfer from the prior scalar quadratic root constructs an\neven \\(y\\) with \\[\ny+y^2+2=0,\\qquad 4\\mid y+2,\\qquad 8\\nmid y+2.\n\\] Thus the error exponent \\(2M-3s=2\\) is attained even when only\nparameters move. No uniform extra error digit can be added to the\ngeneral theorem. This example establishes that limitation; it does not\nassert optimality at every choice of \\(p,s,M\\).","ad p=2,\\ s=0,\\ M=1.\n\\] A checked transfer from the prior scalar quadratic root constructs an\neven \\(y\\) with \\[\ny+y^2+2=0,\\qquad 4\\mid y+2,\\qquad 8\\nmid y+2.\n\\] Thus the error exponent \\(2M-3s=2\\) is attained even when only\nparameters move. No uniform extra error digit can be added to the\ngeneral theorem. This example establishes that limitation; it does not\nassert optimality at every choice of \\(p,s,M\\).","This block establishes both analyticity and differentiability of the\nactual local polynomial solution map at a root with an invertible solved\nJacobian. The statements apply to every prime, including \\(p=2\\). They\nalso hold over any complete nontrivially normed field.","Let \\(\\mathbb K\\) be such a field, let \\(e,k\\in\\mathbb N\\), and let \\[\nF\\in\\mathbb K[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in\\mathbb K^e\\times\\mathbb K^k.\n\\] Evaluation of \\(F\\) means coordinatewise evaluation of its actual\npolynomials. Define the actual derivative blocks \\[\nJ=D_YF(a)\\in M_e(\\mathbb K),\n\\qquad K=D_TF(a)\\in M_{e\\times k}(\\mathbb K),\n\\] where each matrix entry is the corresponding formal partial\nderivative evaluated at \\(a\\). The hypotheses are exactly \\[\nF(a)=0,\n\\qquad J\\in\\operatorname{GL}_e(\\mathbb K).\n\\] Neither analyticity of a proposed solution nor a derivative formula\nis assumed.","N\\), and let \\[\nF\\in\\mathbb K[Y_1,\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in\\mathbb K^e\\times\\mathbb K^k.\n\\] Evaluation of \\(F\\) means coordinatewise evaluation of its actual\npolynomials. Define the actual derivative blocks \\[\nJ=D_YF(a)\\in M_e(\\mathbb K),\n\\qquad K=D_TF(a)\\in M_{e\\times k}(\\mathbb K),\n\\] where each matrix entry is the corresponding formal partial\nderivative evaluated at \\(a\\). The hypotheses are exactly \\[\nF(a)=0,\n\\qquad J\\in\\operatorname{GL}_e(\\mathbb K).\n\\] Neither analyticity of a proposed solution nor a derivative formula\nis assumed.","\\ldots,Y_e,T_1,\\ldots,T_k]^e,\n\\qquad a=(y_0,t_0)\\in\\mathbb K^e\\times\\mathbb K^k.\n\\] Evaluation of \\(F\\) means coordinatewise evaluation of its actual\npolynomials. Define the actual derivative blocks \\[\nJ=D_YF(a)\\in M_e(\\mathbb K),\n\\qquad K=D_TF(a)\\in M_{e\\times k}(\\mathbb K),\n\\] where each matrix entry is the corresponding formal partial\nderivative evaluated at \\(a\\). The hypotheses are exactly \\[\nF(a)=0,\n\\qquad J\\in\\operatorname{GL}_e(\\mathbb K).\n\\] Neither analyticity of a proposed solution nor a derivative formula\nis assumed.","\\textbf{Local analytic graph theorem.} There exist a map \\[\n\\psi:\\mathbb K^k\\longrightarrow\\mathbb K^e\n\\] and open sets \\(U\\subseteq\\mathbb K^k\\), \\(V\\subseteq\\mathbb K^e\\)\nsuch that \\[\nt_0\\in U,\\qquad y_0\\in V,\\qquad \\psi(t_0)=y_0,\n\\qquad \\psi(U)\\subseteq V.\n\\] The map is analytic at every point of \\(U\\), and its graph describes\nall roots in the product neighborhood: \\[\n\\forall t\\in U\\;\\forall y\\in V,\n\\qquad F(y,t)=0\\quad\\Longleftrightarrow\\quad y=\\psi(t).\n\\] At the base parameter it has the strict Fréchet derivative \\[\n\\boxed{D\\psi(t_0)=-J^{-1}K.}\n\\] Consequently it is differentiable on \\(U\\). The map is defined on the\nambient parameter space, but its root, analyticity and uniqueness\nconclusions are restricted to the specified neighborhood. Uniqueness is\namong solved coordinates in \\(V\\).","bseteq\\mathbb K^k\\), \\(V\\subseteq\\mathbb K^e\\)\nsuch that \\[\nt_0\\in U,\\qquad y_0\\in V,\\qquad \\psi(t_0)=y_0,\n\\qquad \\psi(U)\\subseteq V.\n\\] The map is analytic at every point of \\(U\\), and its graph describes\nall roots in the product neighborhood: \\[\n\\forall t\\in U\\;\\forall y\\in V,\n\\qquad F(y,t)=0\\quad\\Longleftrightarrow\\quad y=\\psi(t).\n\\] At the base parameter it has the strict Fréchet derivative \\[\n\\boxed{D\\psi(t_0)=-J^{-1}K.}\n\\] Consequently it is differentiable on \\(U\\). The map is defined on the\nambient parameter space, but its root, analyticity and uniqueness\nconclusions are restricted to the specified neighborhood. Uniqueness is\namong solved coordinates in \\(V\\).","uad y_0\\in V,\\qquad \\psi(t_0)=y_0,\n\\qquad \\psi(U)\\subseteq V.\n\\] The map is analytic at every point of \\(U\\), and its graph describes\nall roots in the product neighborhood: \\[\n\\forall t\\in U\\;\\forall y\\in V,\n\\qquad F(y,t)=0\\quad\\Longleftrightarrow\\quad y=\\psi(t).\n\\] At the base parameter it has the strict Fréchet derivative \\[\n\\boxed{D\\psi(t_0)=-J^{-1}K.}\n\\] Consequently it is differentiable on \\(U\\). The map is defined on the\nambient parameter space, but its root, analyticity and uniqueness\nconclusions are restricted to the specified neighborhood. Uniqueness is\namong solved coordinates in \\(V\\).","y\\in V,\n\\qquad F(y,t)=0\\quad\\Longleftrightarrow\\quad y=\\psi(t).\n\\] At the base parameter it has the strict Fréchet derivative \\[\n\\boxed{D\\psi(t_0)=-J^{-1}K.}\n\\] Consequently it is differentiable on \\(U\\). The map is defined on the\nambient parameter space, but its root, analyticity and uniqueness\nconclusions are restricted to the specified neighborhood. Uniqueness is\namong solved coordinates in \\(V\\).","\\textbf{Proof chain.} The scalar polynomial derivative bridge is proved\nby induction on actual multivariate polynomials: constants, sums, and\nmultiplication by a variable. For a scalar polynomial \\(P\\), its strict\nderivative is \\[\nv\\longmapsto\\sum_j \\frac{\\partial P}{\\partial x_j}(a)v_j.\n\\] Assembling the coordinates gives the strict derivative of the\nevaluated system as its actual Jacobian matrix. Polynomial analyticity\nis supplied by the pinned Mathlib theorem for multivariate polynomial\nevaluation, followed by finite coordinate assembly.","Consider the actual augmented polynomial map \\[\n\\Phi(y,t)=(F(y,t),t).\n\\] Its derivative and inverse derivative at \\(a\\) are \\[\nD\\Phi(a)=\n\\begin{pmatrix}J&K","Consider the actual augmented polynomial map \\[\n\\Phi(y,t)=(F(y,t),t).\n\\] Its derivative and inverse derivative at \\(a\\) are \\[\nD\\Phi(a)=\n\\begin{pmatrix}J&K\\\\0&I_k\\end{pmatrix},\n\\qquad\nD\\Phi(a)^{-1}=\n\\begin{pmatrix}J^{-1}&-J^{-1}K\\\\0&I_k\\end{pmatrix}.\n\\] The proof constructs the continuous linear equivalence from these\nexplicit matrices and verifies both inverse identities. The strict\ninverse theorem gives a local inverse \\(g\\) of \\(\\Phi\\), and the\nanalytic inverse result shows that this same inverse is analytic near\n\\(\\Phi(a)=(0,t_0)\\). Set \\[\n\\psi(t)=\\pi_Y g(0,t).\n\\] The two inverse identities and a sufficiently small product\nneighborhood give the exact local graph statement. Composing the\nderivative of \\(g\\) with parameter insertion and solved-coordinate\nprojection yields \\(-J^{-1}K\\). The chosen parameter neighborhood is\nalso restricted so that the inverse identities, analyticity and\ninclusion \\(\\psi(U)\\subseteq V\\) hold there simulta","0&I_k\\end{pmatrix}.\n\\] The proof constructs the continuous linear equivalence from these\nexplicit matrices and verifies both inverse identities. The strict\ninverse theorem gives a local inverse \\(g\\) of \\(\\Phi\\), and the\nanalytic inverse result shows that this same inverse is analytic near\n\\(\\Phi(a)=(0,t_0)\\). Set \\[\n\\psi(t)=\\pi_Y g(0,t).\n\\] The two inverse identities and a sufficiently small product\nneighborhood give the exact local graph statement. Composing the\nderivative of \\(g\\) with parameter insertion and solved-coordinate\nprojection yields \\(-J^{-1}K\\). The chosen parameter neighborhood is\nalso restricted so that the inverse identities, analyticity and\ninclusion \\(\\psi(U)\\subseteq V\\) hold there simultaneously.","ese\nexplicit matrices and verifies both inverse identities. The strict\ninverse theorem gives a local inverse \\(g\\) of \\(\\Phi\\), and the\nanalytic inverse result shows that this same inverse is analytic near\n\\(\\Phi(a)=(0,t_0)\\). Set \\[\n\\psi(t)=\\pi_Y g(0,t).\n\\] The two inverse identities and a sufficiently small product\nneighborhood give the exact local graph statement. Composing the\nderivative of \\(g\\) with parameter insertion and solved-coordinate\nprojection yields \\(-J^{-1}K\\). The chosen parameter neighborhood is\nalso restricted so that the inverse identities, analyticity and\ninclusion \\(\\psi(U)\\subseteq V\\) hold there simultaneously.","inverse theorem gives a local inverse \\(g\\) of \\(\\Phi\\), and the\nanalytic inverse result shows that this same inverse is analytic near\n\\(\\Phi(a)=(0,t_0)\\). Set \\[\n\\psi(t)=\\pi_Y g(0,t).\n\\] The two inverse identities and a sufficiently small product\nneighborhood give the exact local graph statement. Composing the\nderivative of \\(g\\) with parameter insertion and solved-coordinate\nprojection yields \\(-J^{-1}K\\). The chosen parameter neighborhood is\nalso restricted so that the inverse identities, analyticity and\ninclusion \\(\\psi(U)\\subseteq V\\) hold there simultaneously.","ytic near\n\\(\\Phi(a)=(0,t_0)\\). Set \\[\n\\psi(t)=\\pi_Y g(0,t).\n\\] The two inverse identities and a sufficiently small product\nneighborhood give the exact local graph statement. Composing the\nderivative of \\(g\\) with parameter insertion and solved-coordinate\nprojection yields \\(-J^{-1}K\\). The chosen parameter neighborhood is\nalso restricted so that the inverse identities, analyticity and\ninclusion \\(\\psi(U)\\subseteq V\\) hold there simultaneously.","\\textbf{Transfer to a supplied continuous solution.} Suppose\n\\(g:\\mathbb K^k\\to\\mathbb K^e\\) is continuous at \\(t_0\\),\n\\(g(t_0)=y_0\\), and \\[\nF(g(t),t)=0\n\\] for all \\(t\\) in some neighborhood of \\(t_0\\). Under the same\ninvertibility hypothesis at \\(a\\), continuity places \\(g(t)\\) in the\nuniqueness neighborhood \\(V\\) for all sufficiently close parameters.\nHence \\(g\\) and \\(\\psi\\) agree on a neighborhood of \\(t_0\\). The theorem\n\\mcode{continuous_selection_regularity} transfers both analyticity\nand the strict derivative formula to \\(g\\).","nsfer to a supplied continuous solution.} Suppose\n\\(g:\\mathbb K^k\\to\\mathbb K^e\\) is continuous at \\(t_0\\),\n\\(g(t_0)=y_0\\), and \\[\nF(g(t),t)=0\n\\] for all \\(t\\) in some neighborhood of \\(t_0\\). Under the same\ninvertibility hypothesis at \\(a\\), continuity places \\(g(t)\\) in the\nuniqueness neighborhood \\(V\\) for all sufficiently close parameters.\nHence \\(g\\) and \\(\\psi\\) agree on a neighborhood of \\(t_0\\). The theorem\n\\mcode{continuous_selection_regularity} transfers both analyticity\nand the strict derivative formula to \\(g\\).","nd \\[\nF(g(t),t)=0\n\\] for all \\(t\\) in some neighborhood of \\(t_0\\). Under the same\ninvertibility hypothesis at \\(a\\), continuity places \\(g(t)\\) in the\nuniqueness neighborhood \\(V\\) for all sufficiently close parameters.\nHence \\(g\\) and \\(\\psi\\) agree on a neighborhood of \\(t_0\\). The theorem\n\\mcode{continuous_selection_regularity} transfers both analyticity\nand the strict derivative formula to \\(g\\).","There is also an explicit open-set version. Let \\(U\\) be open, let \\(g\\)\nbe continuous on \\(U\\), and suppose for every \\(t\\in U\\) that \\[\nF(g(t),t)=0,\n\\qquad\nJ_t:=D_YF(g(t),t)\\in\\operatorname{GL}_e(\\mathbb K).\n\\] Writing \\(K_t=D_TF(g(t),t)\\), the theorem\n\\mcode{continuous_selection_analyticOnNhd} proves \\[\ng\\text{ is analytic at every }t\\in U,\n\\qquad Dg(t)=-J_t^{-1}K_t\\quad(t\\in U),\n\\] with a strict Fréchet derivative at each point. The per-point\ninvertibility hypothesis is part of this theorem. This is a transfer\nstatement for continuous local root selections satisfying its\nhypotheses; it does not assert equality of arbitrary global branch\nchoices.","\\textbf{Integral p-adic specialization.} For any prime \\(p\\), let \\[\nF\\in\\mathbb Z_p[Y,T]^e,\n\\qquad a\\in\\mathbb Z_p^{e+k},\n\\qquad F(a)=0.\n\\] Let \\(\\iota:\\mathbb Z_p\\hookrightarrow\\mathbb Q_p\\) be the\ncoefficient inclusion, applied entrywise to points and matrices. Assume\nonly that \\[\n\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\mathbb Q_p).\n\\] The theorem\n\\mcode{PolynomialRegularityIntegral.exists_analytic_integral_implicit}\nconstructs the preceding analytic graph in the ambient spaces\n\\(\\mathbb Q_p^k\\) and \\(\\mathbb Q_p^e\\). Its root equation evaluates the\noriginal integral polynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferenti","\\textbf{Integral p-adic specialization.} For any prime \\(p\\), let \\[\nF\\in\\mathbb Z_p[Y,T]^e,\n\\qquad a\\in\\mathbb Z_p^{e+k},\n\\qquad F(a)=0.\n\\] Let \\(\\iota:\\mathbb Z_p\\hookrightarrow\\mathbb Q_p\\) be the\ncoefficient inclusion, applied entrywise to points and matrices. Assume\nonly that \\[\n\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\mathbb Q_p).\n\\] The theorem\n\\mcode{PolynomialRegularityIntegral.exists_analytic_integral_implicit}\nconstructs the preceding analytic graph in the ambient spaces\n\\(\\mathbb Q_p^k\\) and \\(\\mathbb Q_p^e\\). Its root equation evaluates the\noriginal integral polynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility mod","\\textbf{Integral p-adic specialization.} For any prime \\(p\\), let \\[\nF\\in\\mathbb Z_p[Y,T]^e,\n\\qquad a\\in\\mathbb Z_p^{e+k},\n\\qquad F(a)=0.\n\\] Let \\(\\iota:\\mathbb Z_p\\hookrightarrow\\mathbb Q_p\\) be the\ncoefficient inclusion, applied entrywise to points and matrices. Assume\nonly that \\[\n\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\mathbb Q_p).\n\\] The theorem\n\\mcode{PolynomialRegularityIntegral.exists_analytic_integral_implicit}\nconstructs the preceding analytic graph in the ambient spaces\n\\(\\mathbb Q_p^k\\) and \\(\\mathbb Q_p^e\\). Its root equation evaluates the\noriginal integral polynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This perm","\\textbf{Integral p-adic specialization.} For any prime \\(p\\), let \\[\nF\\in\\mathbb Z_p[Y,T]^e,\n\\qquad a\\in\\mathbb Z_p^{e+k},\n\\qquad F(a)=0.\n\\] Let \\(\\iota:\\mathbb Z_p\\hookrightarrow\\mathbb Q_p\\) be the\ncoefficient inclusion, applied entrywise to points and matrices. Assume\nonly that \\[\n\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\mathbb Q_p).\n\\] The theorem\n\\mcode{PolynomialRegularityIntegral.exists_analytic_integral_implicit}\nconstructs the preceding analytic graph in the ambient spaces\n\\(\\mathbb Q_p^k\\) and \\(\\mathbb Q_p^e\\). Its root equation evaluates the\noriginal integral polynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This permits\na solved Jacobian that is singular modulo \\(p\\) while remaining\ninvertible over \\(\\mathbb Q_p\\). The new specialization is an\nambient-field th",",T]^e,\n\\qquad a\\in\\mathbb Z_p^{e+k},\n\\qquad F(a)=0.\n\\] Let \\(\\iota:\\mathbb Z_p\\hookrightarrow\\mathbb Q_p\\) be the\ncoefficient inclusion, applied entrywise to points and matrices. Assume\nonly that \\[\n\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\mathbb Q_p).\n\\] The theorem\n\\mcode{PolynomialRegularityIntegral.exists_analytic_integral_implicit}\nconstructs the preceding analytic graph in the ambient spaces\n\\(\\mathbb Q_p^k\\) and \\(\\mathbb Q_p^e\\). Its root equation evaluates the\noriginal integral polynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This permits\na solved Jacobian that is singular modulo \\(p\\) while remaining\ninvertible over \\(\\mathbb Q_p\\). The new specialization is an\nambient-field theorem; it does not treat \\(\\mathbb Z_p\\) as a vector\nspace over \\(\\mathbb Q_p\\) or assert a new explicit integral radius.","hbb Z_p^{e+k},\n\\qquad F(a)=0.\n\\] Let \\(\\iota:\\mathbb Z_p\\hookrightarrow\\mathbb Q_p\\) be the\ncoefficient inclusion, applied entrywise to points and matrices. Assume\nonly that \\[\n\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\mathbb Q_p).\n\\] The theorem\n\\mcode{PolynomialRegularityIntegral.exists_analytic_integral_implicit}\nconstructs the preceding analytic graph in the ambient spaces\n\\(\\mathbb Q_p^k\\) and \\(\\mathbb Q_p^e\\). Its root equation evaluates the\noriginal integral polynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This permits\na solved Jacobian that is singular modulo \\(p\\) while remaining\ninvertible over \\(\\mathbb Q_p\\). The new specialization is an\nambient-field theorem; it does not treat \\(\\mathbb Z_p\\) as a vector\nspace over \\(\\mathbb Q_p\\) or assert a new explicit integral radius.","ient inclusion, applied entrywise to points and matrices. Assume\nonly that \\[\n\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\mathbb Q_p).\n\\] The theorem\n\\mcode{PolynomialRegularityIntegral.exists_analytic_integral_implicit}\nconstructs the preceding analytic graph in the ambient spaces\n\\(\\mathbb Q_p^k\\) and \\(\\mathbb Q_p^e\\). Its root equation evaluates the\noriginal integral polynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This permits\na solved Jacobian that is singular modulo \\(p\\) while remaining\ninvertible over \\(\\mathbb Q_p\\). The new specialization is an\nambient-field theorem; it does not treat \\(\\mathbb Z_p\\) as a vector\nspace over \\(\\mathbb Q_p\\) or assert a new explicit integral radius.","oints and matrices. Assume\nonly that \\[\n\\iota(D_YF(a))\\in\\operatorname{GL}_e(\\mathbb Q_p).\n\\] The theorem\n\\mcode{PolynomialRegularityIntegral.exists_analytic_integral_implicit}\nconstructs the preceding analytic graph in the ambient spaces\n\\(\\mathbb Q_p^k\\) and \\(\\mathbb Q_p^e\\). Its root equation evaluates the\noriginal integral polynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This permits\na solved Jacobian that is singular modulo \\(p\\) while remaining\ninvertible over \\(\\mathbb Q_p\\). The new specialization is an\nambient-field theorem; it does not treat \\(\\mathbb Z_p\\) as a vector\nspace over \\(\\mathbb Q_p\\) or assert a new explicit integral radius.","ng analytic graph in the ambient spaces\n\\(\\mathbb Q_p^k\\) and \\(\\mathbb Q_p^e\\). Its root equation evaluates the\noriginal integral polynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This permits\na solved Jacobian that is singular modulo \\(p\\) while remaining\ninvertible over \\(\\mathbb Q_p\\). The new specialization is an\nambient-field theorem; it does not treat \\(\\mathbb Z_p\\) as a vector\nspace over \\(\\mathbb Q_p\\) or assert a new explicit integral radius.","tion evaluates the\noriginal integral polynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This permits\na solved Jacobian that is singular modulo \\(p\\) while remaining\ninvertible over \\(\\mathbb Q_p\\). The new specialization is an\nambient-field theorem; it does not treat \\(\\mathbb Z_p\\) as a vector\nspace over \\(\\mathbb Q_p\\) or assert a new explicit integral radius.","olynomial coefficients using \\(\\iota\\), and its base\nderivative is \\[\n-\\iota(D_YF(a))^{-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This permits\na solved Jacobian that is singular modulo \\(p\\) while remaining\ninvertible over \\(\\mathbb Q_p\\). The new specialization is an\nambient-field theorem; it does not treat \\(\\mathbb Z_p\\) as a vector\nspace over \\(\\mathbb Q_p\\) or assert a new explicit integral radius.","-1}\\,\\iota(D_TF(a)).\n\\] The source proves that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This permits\na solved Jacobian that is singular modulo \\(p\\) while remaining\ninvertible over \\(\\mathbb Q_p\\). The new specialization is an\nambient-field theorem; it does not treat \\(\\mathbb Z_p\\) as a vector\nspace over \\(\\mathbb Q_p\\) or assert a new explicit integral radius.","es that polynomial evaluation, formal partial\ndifferentiation and both Jacobian blocks commute with this inclusion. No\ninvertibility modulo \\(p\\) or integral inverse is required. This permits\na solved Jacobian that is singular modulo \\(p\\) while remaining\ninvertible over \\(\\mathbb Q_p\\). The new specialization is an\nambient-field theorem; it does not treat \\(\\mathbb Z_p\\) as a vector\nspace over \\(\\mathbb Q_p\\) or assert a new explicit integral radius.","2.2 & Filtered finite coefficient reconstruction, including $\\Z_p$","\\subsubsection*{The correction is preserved}\nThe normalized packet is not the same object as the packet multiplied by $j!$. The checked recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad C_{6,3}(2)=\\frac{703}{5460},\n\\]\n\\[\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455},\\qquad\n v_3\\!\\left(\\frac{64}{455}\\right)=0.\n\\]\nThus the unrestricted normalized no-loss claim fails already at a valid one-step parameter instance. Multiplication by $3!$ restores one factor of three in this instance. The general corrected finite-difference theorem is \\textbf{not} inferred from this calculation and is not yet formalized here.","ion is preserved}\nThe normalized packet is not the same object as the packet multiplied by $j!$. The checked recurrence gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\qquad C_{6,3}(2)=\\frac{703}{5460},\n\\]\n\\[\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455},\\qquad\n v_3\\!\\left(\\frac{64}{455}\\right)=0.\n\\]\nThus the unrestricted normalized no-loss claim fails already at a valid one-step parameter instance. Multiplication by $3!$ restores one factor of three in this instance. The general corrected finite-difference theorem is \\textbf{not} inferred from this calculation and is not yet formalized here.","theorems and all 22 explicit definitions. Their transitive axiom dependencies are confined to \\mcode{propext}, \\mcode{Classical.choice} and \\mcode{Quot.sound}, or subsets. Three invalid-proof controls were rejected. The proof, scope, and numerical-regression records remain distinct.\n\\par\\medskip\n\\pageheading{1. Residue packets and actual EGF coordinates}{Source items 1.2 and 2.1}\nFor a sequence $a:\\N\\to R$, define the actual selector\n\\[\n (P_{N,j}a)(n)=\\begin{cases}a(n),&n\\bmod N=j,",". Their transitive axiom dependencies are confined to \\mcode{propext}, \\mcode{Classical.choice} and \\mcode{Quot.sound}, or subsets. Three invalid-proof controls were rejected. The proof, scope, and numerical-regression records remain distinct.\n\\par\\medskip\n\\pageheading{1. Residue packets and actual EGF coordinates}{Source items 1.2 and 2.1}\nFor a sequence $a:\\N\\to R$, define the actual selector\n\\[\n (P_{N,j}a)(n)=\\begin{cases}a(n),&n\\bmod N=j,\\\\0,&\\text{otherwise.}\\end{cases}\n\\]\nThe source proves idempotence for every selector, orthogonality for distinct standard residue labels, and, when $N>0$,\n\\[\n \\sum_{j=0}^{N-1}P_{N,j}a=a.\n\\]\nIf $N\\mid M$ and $M>0$, it also proves\n\\[\n P_{N,j}a=\\sum_{\\substack{0\\le r<M","0,&\\text{otherwise.}\\end{cases}\n\\]\nThe source proves idempotence for every selector, orthogonality for distinct standard residue labels, and, when $N>0$,\n\\[\n \\sum_{j=0}^{N-1}P_{N,j}a=a.\n\\]\nIf $N\\mid M$ and $M>0$, it also proves\n\\[\n P_{N,j}a=\\sum_{\\substack{0\\le r<M","0,&\\text{otherwise.}\\end{cases}\n\\]\nThe source proves idempotence for every selector, orthogonality for distinct standard residue labels, and, when $N>0$,\n\\[\n \\sum_{j=0}^{N-1}P_{N,j}a=a.\n\\]\nIf $N\\mid M$ and $M>0$, it also proves\n\\[\n P_{N,j}a=\\sum_{\\substack{0\\le r<M\\\\r\\bmod N=j}}P_{M,r}a.\n\\]\nThe formal power-series versions are proved coefficientwise for actual \\mcode{PowerSeries} objects. There is no finite truncation of the sequence domain in these theorems.","\\subsection*{A constructed inverse, not an inverse hypothesis}\nLet $K$ be a characteristic-zero field. Define\n\\[\n \\mathcal E(a)=\\sum_{n\\ge0}\\frac{a_n}{n!}t^n,\\qquad\n \\mathcal D(A)_n=n![t^n]A.\n\\]\nBoth inverse laws $\\mathcal D\\mathcal E=I$ and $\\mathcal E\\mathcal D=I$ are proved. The ordinary power-series product then yields the genuine binomial convolution\n\\[\n \\mathcal D\\bigl(\\mathcal E(h)\\mathcal E(a)\\bigr)_n\n =\\sum_{r=0}^{n}\\binom nr h_r a_{n-r}.\n\\]\nFor $h_0=1$, the formal inverse of $H=\\mathcal E(h)$ is used to construct a sequence equivalence\n\\[\n U_H a=\\mathcal D(H\\mathcal E(a)),\\qquad\n U_H^{-1}a=\\mathcal D(H^{-1}\\mathcal E(a)).\n\\]\nThe equivalence has proved left and right inverse laws. The inverse's constant coefficient is proved to be one. The associated","\\subsection*{A constructed inverse, not an inverse hypothesis}\nLet $K$ be a characteristic-zero field. Define\n\\[\n \\mathcal E(a)=\\sum_{n\\ge0}\\frac{a_n}{n!}t^n,\\qquad\n \\mathcal D(A)_n=n![t^n]A.\n\\]\nBoth inverse laws $\\mathcal D\\mathcal E=I$ and $\\mathcal E\\mathcal D=I$ are proved. The ordinary power-series product then yields the genuine binomial convolution\n\\[\n \\mathcal D\\bigl(\\mathcal E(h)\\mathcal E(a)\\bigr)_n\n =\\sum_{r=0}^{n}\\binom nr h_r a_{n-r}.\n\\]\nFor $h_0=1$, the formal inverse of $H=\\mathcal E(h)$ is used to construct a sequence equivalence\n\\[\n U_H a=\\mathcal D(H\\mathcal E(a)),\\qquad\n U_H^{-1}a=\\mathcal D(H^{-1}\\mathcal E(a)).\n\\]\nThe equivalence has proved left and right inverse laws. The inverse's constant coefficient is proved to be one. The associated lower-triangular entries have diagonal one and vanish above the diagonal.","s}\nLet $K$ be a characteristic-zero field. Define\n\\[\n \\mathcal E(a)=\\sum_{n\\ge0}\\frac{a_n}{n!}t^n,\\qquad\n \\mathcal D(A)_n=n![t^n]A.\n\\]\nBoth inverse laws $\\mathcal D\\mathcal E=I$ and $\\mathcal E\\mathcal D=I$ are proved. The ordinary power-series product then yields the genuine binomial convolution\n\\[\n \\mathcal D\\bigl(\\mathcal E(h)\\mathcal E(a)\\bigr)_n\n =\\sum_{r=0}^{n}\\binom nr h_r a_{n-r}.\n\\]\nFor $h_0=1$, the formal inverse of $H=\\mathcal E(h)$ is used to construct a sequence equivalence\n\\[\n U_H a=\\mathcal D(H\\mathcal E(a)),\\qquad\n U_H^{-1}a=\\mathcal D(H^{-1}\\mathcal E(a)).\n\\]\nThe equivalence has proved left and right inverse laws. The inverse's constant coefficient is proved to be one. The associated lower-triangular entries have diagonal one and vanish above the diagonal.","zero field. Define\n\\[\n \\mathcal E(a)=\\sum_{n\\ge0}\\frac{a_n}{n!}t^n,\\qquad\n \\mathcal D(A)_n=n![t^n]A.\n\\]\nBoth inverse laws $\\mathcal D\\mathcal E=I$ and $\\mathcal E\\mathcal D=I$ are proved. The ordinary power-series product then yields the genuine binomial convolution\n\\[\n \\mathcal D\\bigl(\\mathcal E(h)\\mathcal E(a)\\bigr)_n\n =\\sum_{r=0}^{n}\\binom nr h_r a_{n-r}.\n\\]\nFor $h_0=1$, the formal inverse of $H=\\mathcal E(h)$ is used to construct a sequence equivalence\n\\[\n U_H a=\\mathcal D(H\\mathcal E(a)),\\qquad\n U_H^{-1}a=\\mathcal D(H^{-1}\\mathcal E(a)).\n\\]\nThe equivalence has proved left and right inverse laws. The inverse's constant coefficient is proved to be one. The associated lower-triangular entries have diagonal one and vanish above the diagonal.","t^n,\\qquad\n \\mathcal D(A)_n=n![t^n]A.\n\\]\nBoth inverse laws $\\mathcal D\\mathcal E=I$ and $\\mathcal E\\mathcal D=I$ are proved. The ordinary power-series product then yields the genuine binomial convolution\n\\[\n \\mathcal D\\bigl(\\mathcal E(h)\\mathcal E(a)\\bigr)_n\n =\\sum_{r=0}^{n}\\binom nr h_r a_{n-r}.\n\\]\nFor $h_0=1$, the formal inverse of $H=\\mathcal E(h)$ is used to construct a sequence equivalence\n\\[\n U_H a=\\mathcal D(H\\mathcal E(a)),\\qquad\n U_H^{-1}a=\\mathcal D(H^{-1}\\mathcal E(a)).\n\\]\nThe equivalence has proved left and right inverse laws. The inverse's constant coefficient is proved to be one. The associated lower-triangular entries have diagonal one and vanish above the diagonal.","For the actual conjugated selectors\n\\[\n \\widehat P_{N,j}=U_HP_{N,j}U_H^{-1},\n\\]\nidempotence, orthogonality, and finite reconstruction are proved from the constructed maps. No projector identity for the conjugates is inserted as an assumption.","\\textbf{Principal targets.}\\par\n\\target{ResiduePackets.packet_reconstruction}\\par\n\\target{ResiduePackets.packet_refinement}\\par\n\\target{EGF.coordinateEquiv, EGF.coordinate_forward, EGF.coordinate_inverse}\\par\n\\target{EGF.transported_reconstruction}\n\\par\\medskip\n\\pageheading{2. Triangular inversion and denominator control}{Source items 3.2 and 3.3}\nFor a general lower-triangular coefficient array $T$ over a field with nonzero diagonal, the code defines forward substitution by well-founded recursion:\n\\[\n a_n=\\frac{b_n-\\sum_{q<n}T_{nq}a_q}{T_{nn}}.\n\\]\nIt proves that this constructed sequence solves every row and that every sequence solving those rows equals it. No existence or uniqueness premise is used.","acket_refinement}\\par\n\\target{EGF.coordinateEquiv, EGF.coordinate_forward, EGF.coordinate_inverse}\\par\n\\target{EGF.transported_reconstruction}\n\\par\\medskip\n\\pageheading{2. Triangular inversion and denominator control}{Source items 3.2 and 3.3}\nFor a general lower-triangular coefficient array $T$ over a field with nonzero diagonal, the code defines forward substitution by well-founded recursion:\n\\[\n a_n=\\frac{b_n-\\sum_{q<n}T_{nq}a_q}{T_{nn}}.\n\\]\nIt proves that this constructed sequence solves every row and that every sequence solving those rows equals it. No existence or uniqueness premise is used.","\\subsection*{The congruential coefficients}\nOver $\\Q$, specialize to\n\\[\n T_{nq}=\\binom{Nn+j}{Nq},\\qquad b_n=\\delta_{n0}.\n\\]\nThe diagonal is positive as an integer, hence nonzero. The resulting sequence $C_{N,j}(n)$ satisfies\n\\[\n C_{N,j}(0)=1,\\qquad\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0}.\n\\]\nThe displayed recurrence in the manuscript and the unique-solution theorem are consequences of this definition.","The formalization does not stop at a recurrence. In the variable $z=t^N$, construct\n\\[\n K_{N,j}(z)=\\sum_{n\\ge0}\\frac{j!}{(Nn+j)!}z^n,\\qquad\n F_{N,j}(z)=\\sum_{n\\ge0}\\frac{C_{N,j}(n)}{(Nn)!}z^n.\n\\]\nIt proves the factorial coefficient identity, then $F_{N,j}K_{N,j}=1$ and therefore $F_{N,j}=K_{N,j}^{-1}$.","For $N>0$ and $0\\le j<N$, the independently defined original exponential section is\n\\[\n E_{N,j}(t)=\\sum_{d\\equiv j\\ ({\\rm mod}\\ N)}\\frac{t^d}{d!}.\n\\]\nUsing the actual power-series expansion homomorphism $z\\mapsto t^N$, the code proves\n\\[\n E_{N,j}(t)=\\frac{t^j}{j!}K_{N,j}(t^N),\\qquad\n \\boxed{F_{N,j}(t^N)E_{N,j}(t)=\\frac{t^j}{j!}.}\n\\]\nThus the recurrence is connected back to the original normalized EGF definition. The section vanishes at zero when $j>0$; no inverse of that nonunit section is assumed.","original exponential section is\n\\[\n E_{N,j}(t)=\\sum_{d\\equiv j\\ ({\\rm mod}\\ N)}\\frac{t^d}{d!}.\n\\]\nUsing the actual power-series expansion homomorphism $z\\mapsto t^N$, the code proves\n\\[\n E_{N,j}(t)=\\frac{t^j}{j!}K_{N,j}(t^N),\\qquad\n \\boxed{F_{N,j}(t^N)E_{N,j}(t)=\\frac{t^j}{j!}.}\n\\]\nThus the recurrence is connected back to the original normalized EGF definition. The section vanishes at zero when $j>0$; no inverse of that nonunit section is assumed.","\\subsection*{Universal clearing}\nDefine\n\\[\n D_{N,j}(n)=\\prod_{r=1}^{n}\\binom{Nr+j}{Nr},\\qquad D_{N,j}(0)=1.\n\\]\nFor every $n$, the code proves $D_{N,j}(n)C_{N,j}(n)\\in\\Z$. The general proof promotes each earlier clearing denominator to the common product, then cancels the new nonzero diagonal. This is a denominator bound, not a claim that every normalized coefficient is integral.\n\\par\\medskip\n\\pageheading{3. Filtered tails and the standard window}{Source items 2.2 and 8.2}\n\\subsection*{Filtration at a finite coefficient row}\nLet $R$ be any commutative ring. For $w\\in\\N\\cup\\{\\infty\\}$ define\n\\[\n p^{[w]}=\\begin{cases}p^w,&w<\\infty,","proves $D_{N,j}(n)C_{N,j}(n)\\in\\Z$. The general proof promotes each earlier clearing denominator to the common product, then cancels the new nonzero diagonal. This is a denominator bound, not a claim that every normalized coefficient is integral.\n\\par\\medskip\n\\pageheading{3. Filtered tails and the standard window}{Source items 2.2 and 8.2}\n\\subsection*{Filtration at a finite coefficient row}\nLet $R$ be any commutative ring. For $w\\in\\N\\cup\\{\\infty\\}$ define\n\\[\n p^{[w]}=\\begin{cases}p^w,&w<\\infty,","$. The general proof promotes each earlier clearing denominator to the common product, then cancels the new nonzero diagonal. This is a denominator bound, not a claim that every normalized coefficient is integral.\n\\par\\medskip\n\\pageheading{3. Filtered tails and the standard window}{Source items 2.2 and 8.2}\n\\subsection*{Filtration at a finite coefficient row}\nLet $R$ be any commutative ring. For $w\\in\\N\\cup\\{\\infty\\}$ define\n\\[\n p^{[w]}=\\begin{cases}p^w,&w<\\infty,","ach earlier clearing denominator to the common product, then cancels the new nonzero diagonal. This is a denominator bound, not a claim that every normalized coefficient is integral.\n\\par\\medskip\n\\pageheading{3. Filtered tails and the standard window}{Source items 2.2 and 8.2}\n\\subsection*{Filtration at a finite coefficient row}\nLet $R$ be any commutative ring. For $w\\in\\N\\cup\\{\\infty\\}$ define\n\\[\n p^{[w]}=\\begin{cases}p^w,&w<\\infty,\\\\0,&w=\\infty.\\end{cases}\n\\]\nThe infinity convention expresses a vanishing coefficient, not an infinite exponent operation. For any finite index set, every term with weight at least $L$ is divisible by $p^L$. A finite-sum proof yields\n\\[\n \\boxed{p^L\\ \\bigm|\\ \n \\sum_{r=0}^{n}\\binom nr c_rp^{[w(r)]}S_{n-r}\n -\\sum_{\\substack{0\\le r\\le n","0,&w=\\infty.\\end{cases}\n\\]\nThe infinity convention expresses a vanishing coefficient, not an infinite exponent operation. For any finite index set, every term with weight at least $L$ is divisible by $p^L$. A finite-sum proof yields\n\\[\n \\boxed{p^L\\ \\bigm|\\ \n \\sum_{r=0}^{n}\\binom nr c_rp^{[w(r)]}S_{n-r}\n -\\sum_{\\substack{0\\le r\\le n","0,&w=\\infty.\\end{cases}\n\\]\nThe infinity convention expresses a vanishing coefficient, not an infinite exponent operation. For any finite index set, every term with weight at least $L$ is divisible by $p^L$. A finite-sum proof yields\n\\[\n \\boxed{p^L\\ \\bigm|\\ \n \\sum_{r=0}^{n}\\binom nr c_rp^{[w(r)]}S_{n-r}\n -\\sum_{\\substack{0\\le r\\le n\\\\w(r)<L}}\n     \\binom nr c_rp^{[w(r)]}S_{n-r}.}\n\\]\nThe result is specialized to actual $p$-adic integers $R=\\Z_p$ for prime $p$. Divisibility there is precisely the asserted membership in $p^L\\Z_p$.","w(r)<L}}\n     \\binom nr c_rp^{[w(r)]}S_{n-r}.}\n\\]\nThe result is specialized to actual $p$-adic integers $R=\\Z_p$ for prime $p$. Divisibility there is precisely the asserted membership in $p^L\\Z_p$.","For an individual row the sum is finite, so the proof does not require $w(r)\\to\\infty$. This is a stronger coefficient-row scope, not a theorem about an infinite limiting process. A supporting lemma also proves a $p^{2M+2}$ tail bound for a sum with visible factors $p^{2m}$. It is not counted as either Euler supercongruence theorem: their generating-function reconstruction identities remain pending.","\\subsection*{Actual Lucas reduction}\nFor prime $p$, all $m,n,q\\in\\N$, and $0\\le j<p$, the checked binomial congruence is\n\\[\n \\boxed{\\binom{mpn+j}{mpq}\\equiv\\binom{mn}{mq}\\pmod p.}\n\\]\nThe proof applies Mathlib's Lucas low-digit reduction to the actual natural binomial coefficients. It does not assume a pre-reduced row. The statement also permits $q>n$. If $m>0$, both binomial coefficients then vanish.","Taking $q=n$ gives\n\\[\n \\binom{mpn+j}{mpn}\\equiv1\\pmod p.\n\\]\nNondivisibility by $p$ and invertibility of that diagonal in $\\Z/p\\Z$ are proved. The separate supporting lemma $p\\nmid j!$ for $j<p$ establishes the standard-window factorial-unit condition.","\\textbf{Principal targets.}\\par\n\\target{Filtered.padic_filtered_reconstruction}\\par\n\\target{StandardWindow.lucas_row_reduction}\\par\n\\target{StandardWindow.standard_diagonal_unit}\\par\n\\target{StandardWindow.factorial_valuation_zero}\n\\par\\medskip\n\\pageheading{4. Sharp normalization boundary}{Source item 13.1}\nThe construction fixes the two coefficient conventions:\n\\[\n C_{N,j}(n),\\qquad \\widetilde C_{N,j}(n)=j!\\,C_{N,j}(n).\n\\]\nThe witness uses $p=3$, $m=2$, $N=mp=6$, $j=3$, $n=1$ and difference step $h=(p-1)/m=1$. The parameter conditions are checked explicitly, including odd primality, $m\\mid p-1$ and $j<N$.","ed.padic_filtered_reconstruction}\\par\n\\target{StandardWindow.lucas_row_reduction}\\par\n\\target{StandardWindow.standard_diagonal_unit}\\par\n\\target{StandardWindow.factorial_valuation_zero}\n\\par\\medskip\n\\pageheading{4. Sharp normalization boundary}{Source item 13.1}\nThe construction fixes the two coefficient conventions:\n\\[\n C_{N,j}(n),\\qquad \\widetilde C_{N,j}(n)=j!\\,C_{N,j}(n).\n\\]\nThe witness uses $p=3$, $m=2$, $N=mp=6$, $j=3$, $n=1$ and difference step $h=(p-1)/m=1$. The parameter conditions are checked explicitly, including odd primality, $m\\mid p-1$ and $j<N$.","c_filtered_reconstruction}\\par\n\\target{StandardWindow.lucas_row_reduction}\\par\n\\target{StandardWindow.standard_diagonal_unit}\\par\n\\target{StandardWindow.factorial_valuation_zero}\n\\par\\medskip\n\\pageheading{4. Sharp normalization boundary}{Source item 13.1}\nThe construction fixes the two coefficient conventions:\n\\[\n C_{N,j}(n),\\qquad \\widetilde C_{N,j}(n)=j!\\,C_{N,j}(n).\n\\]\nThe witness uses $p=3$, $m=2$, $N=mp=6$, $j=3$, $n=1$ and difference step $h=(p-1)/m=1$. The parameter conditions are checked explicitly, including odd primality, $m\\mid p-1$ and $j<N$.","red_reconstruction}\\par\n\\target{StandardWindow.lucas_row_reduction}\\par\n\\target{StandardWindow.standard_diagonal_unit}\\par\n\\target{StandardWindow.factorial_valuation_zero}\n\\par\\medskip\n\\pageheading{4. Sharp normalization boundary}{Source item 13.1}\nThe construction fixes the two coefficient conventions:\n\\[\n C_{N,j}(n),\\qquad \\widetilde C_{N,j}(n)=j!\\,C_{N,j}(n).\n\\]\nThe witness uses $p=3$, $m=2$, $N=mp=6$, $j=3$, $n=1$ and difference step $h=(p-1)/m=1$. The parameter conditions are checked explicitly, including odd primality, $m\\mid p-1$ and $j<N$.","truction}\\par\n\\target{StandardWindow.lucas_row_reduction}\\par\n\\target{StandardWindow.standard_diagonal_unit}\\par\n\\target{StandardWindow.factorial_valuation_zero}\n\\par\\medskip\n\\pageheading{4. Sharp normalization boundary}{Source item 13.1}\nThe construction fixes the two coefficient conventions:\n\\[\n C_{N,j}(n),\\qquad \\widetilde C_{N,j}(n)=j!\\,C_{N,j}(n).\n\\]\nThe witness uses $p=3$, $m=2$, $N=mp=6$, $j=3$, $n=1$ and difference step $h=(p-1)/m=1$. The parameter conditions are checked explicitly, including odd primality, $m\\mid p-1$ and $j<N$.","n}\\par\n\\target{StandardWindow.lucas_row_reduction}\\par\n\\target{StandardWindow.standard_diagonal_unit}\\par\n\\target{StandardWindow.factorial_valuation_zero}\n\\par\\medskip\n\\pageheading{4. Sharp normalization boundary}{Source item 13.1}\nThe construction fixes the two coefficient conventions:\n\\[\n C_{N,j}(n),\\qquad \\widetilde C_{N,j}(n)=j!\\,C_{N,j}(n).\n\\]\nThe witness uses $p=3$, $m=2$, $N=mp=6$, $j=3$, $n=1$ and difference step $h=(p-1)/m=1$. The parameter conditions are checked explicitly, including odd primality, $m\\mid p-1$ and $j<N$.","dow.lucas_row_reduction}\\par\n\\target{StandardWindow.standard_diagonal_unit}\\par\n\\target{StandardWindow.factorial_valuation_zero}\n\\par\\medskip\n\\pageheading{4. Sharp normalization boundary}{Source item 13.1}\nThe construction fixes the two coefficient conventions:\n\\[\n C_{N,j}(n),\\qquad \\widetilde C_{N,j}(n)=j!\\,C_{N,j}(n).\n\\]\nThe witness uses $p=3$, $m=2$, $N=mp=6$, $j=3$, $n=1$ and difference step $h=(p-1)/m=1$. The parameter conditions are checked explicitly, including odd primality, $m\\mid p-1$ and $j<N$.","tandardWindow.factorial_valuation_zero}\n\\par\\medskip\n\\pageheading{4. Sharp normalization boundary}{Source item 13.1}\nThe construction fixes the two coefficient conventions:\n\\[\n C_{N,j}(n),\\qquad \\widetilde C_{N,j}(n)=j!\\,C_{N,j}(n).\n\\]\nThe witness uses $p=3$, $m=2$, $N=mp=6$, $j=3$, $n=1$ and difference step $h=(p-1)/m=1$. The parameter conditions are checked explicitly, including odd primality, $m\\mid p-1$ and $j<N$.","ctorial_valuation_zero}\n\\par\\medskip\n\\pageheading{4. Sharp normalization boundary}{Source item 13.1}\nThe construction fixes the two coefficient conventions:\n\\[\n C_{N,j}(n),\\qquad \\widetilde C_{N,j}(n)=j!\\,C_{N,j}(n).\n\\]\nThe witness uses $p=3$, $m=2$, $N=mp=6$, $j=3$, $n=1$ and difference step $h=(p-1)/m=1$. The parameter conditions are checked explicitly, including odd primality, $m\\mid p-1$ and $j<N$.","The first row gives $84C_{6,3}(1)+1=0$. The second row, evaluated by the same recursive definition, gives $C_{6,3}(2)=703/5460$. Therefore\n\\[\n \\Delta_1 C_{6,3}(1)=\\frac{64}{455}\\ne0.\n\\]\nNeither numerator nor denominator is divisible by three, so its actual rational 3-adic valuation is zero. A further theorem negates the universal one-step no-loss statement by applying it to this valid tuple and deriving $1\\le0$.","t row gives $84C_{6,3}(1)+1=0$. The second row, evaluated by the same recursive definition, gives $C_{6,3}(2)=703/5460$. Therefore\n\\[\n \\Delta_1 C_{6,3}(1)=\\frac{64}{455}\\ne0.\n\\]\nNeither numerator nor denominator is divisible by three, so its actual rational 3-adic valuation is zero. A further theorem negates the universal one-step no-loss statement by applying it to this valid tuple and deriving $1\\le0$.","For the unnormalized convention,\n\\[\n \\Delta_1\\widetilde C_{6,3}(1)=3!\\,\\frac{64}{455}=\\frac{384}{455},\n \\qquad v_3\\!\\left(\\frac{384}{455}\\right)=1.\n\\]\nThe factorial's valuation is also proved:\n\\[\n v_3(3!)=1,\\qquad\n v_3\\bigl(\\Delta_1C_{6,3}(1)\\bigr)\n =1-v_3(3!)=0.\n\\]\nThis proves sharpness at the displayed normalization boundary. It does not establish the corrected global inequality for all parameters, and does not assert that every nonunit-factorial parameter tuple must violate every no-loss specialization. All valuation calculations here concern nonzero rationals.","\\subsubsection*{A useful scope strengthening}\nThe algebraic proof needs \\emph{oddness}, not primality. The exact reconstruction and the integer divisibility statement are therefore checked for every positive odd length $p$, including $p=1$ and odd composites. The manuscript's odd-prime statement follows as a specialization. This is a sharpening of assumptions in this development, not a historical-priority claim. Even lengths are not included; a checked row-zero counterexample shows why.","-prime finite-difference theorem remain pending. No continuum convergence or analytic summability theorem is inferred from the formal power-series equalities in this block. The historical numerical study is not being relabelled as a fresh experiment.\n\\par\\medskip\n\\pageheading{1. The actual odd-power generating function}{Lemma 4.1; manuscript page 10}\nWork in the rational formal power-series ring $\\Q[[t]]$. Define\n\\[\n S_n(p)=\\sum_{a=0}^{p-1}(-1)^a(2a+1)^n\\in\\Z,\n \\qquad F_p(t)=\\sum_{n\\ge0}S_n(p)\\frac{t^n}{n!}.\n\\]\nThe formal exponential is Mathlib's \\mcode{PowerSeries.exp}; scalar substitution is its actual \\mcode{rescale} homomorphism. We define\n\\[\n \\exp_a(t)=e^{at},\\qquad\n \\cosh_a(t)=\\frac{e^{at}+e^{-at}}2,\\qquad\n \\sech(t)=\\cosh_1(t)^{-1}.\n\\]\nThe constant term of $\\cosh_a$ is proved to be one, so the reciprocal used here is a genuine unit inverse.","rence theorem remain pending. No continuum convergence or analytic summability theorem is inferred from the formal power-series equalities in this block. The historical numerical study is not being relabelled as a fresh experiment.\n\\par\\medskip\n\\pageheading{1. The actual odd-power generating function}{Lemma 4.1; manuscript page 10}\nWork in the rational formal power-series ring $\\Q[[t]]$. Define\n\\[\n S_n(p)=\\sum_{a=0}^{p-1}(-1)^a(2a+1)^n\\in\\Z,\n \\qquad F_p(t)=\\sum_{n\\ge0}S_n(p)\\frac{t^n}{n!}.\n\\]\nThe formal exponential is Mathlib's \\mcode{PowerSeries.exp}; scalar substitution is its actual \\mcode{rescale} homomorphism. We define\n\\[\n \\exp_a(t)=e^{at},\\qquad\n \\cosh_a(t)=\\frac{e^{at}+e^{-at}}2,\\qquad\n \\sech(t)=\\cosh_1(t)^{-1}.\n\\]\nThe constant term of $\\cosh_a$ is proved to be one, so the reciprocal used here is a genuine unit inverse.","r\\medskip\n\\pageheading{1. The actual odd-power generating function}{Lemma 4.1; manuscript page 10}\nWork in the rational formal power-series ring $\\Q[[t]]$. Define\n\\[\n S_n(p)=\\sum_{a=0}^{p-1}(-1)^a(2a+1)^n\\in\\Z,\n \\qquad F_p(t)=\\sum_{n\\ge0}S_n(p)\\frac{t^n}{n!}.\n\\]\nThe formal exponential is Mathlib's \\mcode{PowerSeries.exp}; scalar substitution is its actual \\mcode{rescale} homomorphism. We define\n\\[\n \\exp_a(t)=e^{at},\\qquad\n \\cosh_a(t)=\\frac{e^{at}+e^{-at}}2,\\qquad\n \\sech(t)=\\cosh_1(t)^{-1}.\n\\]\nThe constant term of $\\cosh_a$ is proved to be one, so the reciprocal used here is a genuine unit inverse.","formal power-series ring $\\Q[[t]]$. Define\n\\[\n S_n(p)=\\sum_{a=0}^{p-1}(-1)^a(2a+1)^n\\in\\Z,\n \\qquad F_p(t)=\\sum_{n\\ge0}S_n(p)\\frac{t^n}{n!}.\n\\]\nThe formal exponential is Mathlib's \\mcode{PowerSeries.exp}; scalar substitution is its actual \\mcode{rescale} homomorphism. We define\n\\[\n \\exp_a(t)=e^{at},\\qquad\n \\cosh_a(t)=\\frac{e^{at}+e^{-at}}2,\\qquad\n \\sech(t)=\\cosh_1(t)^{-1}.\n\\]\nThe constant term of $\\cosh_a$ is proved to be one, so the reciprocal used here is a genuine unit inverse.","\\subsection*{From finite sums to a series identity}\nCoefficient extraction proves\n\\[\n F_p(t)=\\sum_{a=0}^{p-1}(-1)^a e^{(2a+1)t}.\n\\]\nAn induction on the actual finite summation length establishes the telescoping identity\n\\[\n F_p(t)\\cosh t=\\frac{1-(-1)^p e^{2pt}}2.\n\\]\nFor odd $p$ this gives, without a geometric-series convergence assumption,\n\\[\n \\boxed{F_p(t)=\\frac{1+e^{2pt}}2\\,\\sech t.}\n\\]\nThis is an equality of complete formal series, not a fixed-degree polynomial check.","\\subsection*{Even projection is identified coefficientwise}\nFor any formal series $f$, let $\\operatorname{EvenPart}(f)=(f(t)+f(-t))/2$. Its $n$th coefficient is proved to be the coefficient of $f$ when $n$ is even and zero otherwise. Rescaling the inverse law proves $\\sech(-t)=\\sech(t)$. The exponential multiplication identity then gives\n\\[\n \\cosh^2(pt)=\\frac{2+e^{2pt}+e^{-2pt}}4,\n\\]\nand hence\n\\[\n \\boxed{\\sum_{r\\ge0}S_{2r}(p)\\frac{t^{2r}}{(2r)!}\n       =\\cosh^2(pt)\\,\\sech t.}\n\\]\nBoth the full and even EGF formulas are derived from the original integer sum. Neither is supplied as a hypothesis.","\\textbf{Principal targets.}\\par\n\\target{OddPower.source_as_exponentials}\\par\n\\target{OddPower.source_times_cosh}\\par\n\\target{OddPower.odd_source_egf, OddPower.odd_source_even_egf}\n\\par\\medskip\n\\pageheading{2. Integral Euler coefficients and the zigzag bridge}{Coefficient foundations for Theorem 4.2}\nThe first block already constructed $C_{N,j}(r)$ by the manuscript's triangular recurrence. At $(N,j)=(2,0)$, every diagonal binomial coefficient equals one. Its denominator-clearing product is therefore one, and $C_{2,0}(r)$ is an integer.","r\n\\target{OddPower.odd_source_egf, OddPower.odd_source_even_egf}\n\\par\\medskip\n\\pageheading{2. Integral Euler coefficients and the zigzag bridge}{Coefficient foundations for Theorem 4.2}\nThe first block already constructed $C_{N,j}(r)$ by the manuscript's triangular recurrence. At $(N,j)=(2,0)$, every diagonal binomial coefficient equals one. Its denominator-clearing product is therefore one, and $C_{2,0}(r)$ is an integer.","The source defines $E_{2r}$ as the integer numerator of that rational coefficient and proves that its rational cast equals $C_{2,0}(r)$. Set $E_{2r+1}=0$. The actual exponential-section identity from the first block proves\n\\[\n \\boxed{\\sum_{n\\ge0}E_n\\frac{t^n}{n!}=\\sech t.}\n\\]\nThus the integer sequence is connected to the intended signed Euler family, rather than merely assigned the name Euler.","\\subsection*{The inverse coefficients are integral by construction}\nDefine the integer binomial convolution\n\\[\n B_n=\\sum_{r=0}^{n}\\binom nr E_rE_{n-r},\\qquad b_m=B_{2m}.\n\\]\nThe product formula for actual EGFs gives\n\\[\n \\sum_{n\\ge0}B_n\\frac{t^n}{n!}=\\sech^2 t.\n\\]\nOdd coefficients vanish by parity. In particular, every $b_m$ is an integer; this is the integrality required by the later congruence.","\\subsection*{Connecting to the paper's $Z_n$ convention}\nThe paper defines $Z_n$ through $\\sec t+\\tan t$. The Lean construction uses Mathlib's actual formal sine and cosine series and defines\n\\[\n \\sec t=(\\cos t)^{-1},\\quad \\tan t=(\\sin t)\\sec t,\n \\quad Z_n=n![t^n](\\sec t+\\tan t).\n\\]\nIt proves the sine and cosine derivative identities, $\\sin^2t+\\cos^2t=1$, and\n\\[\n (\\tan t)'=\\sec^2t.\n\\]\nSign substitution in the squared variable carries the hyperbolic cosine kernel to the actual cosine kernel. The same homomorphism carries their inverses and their squares. Extracting the even coefficient in the derivative of tangent yields\n\\[\n \\boxed{b_m=(-1)^m Z_{2m+1}.}\n\\]\nThis formula is \\emph{proved}, not assumed as an input to reconstruction. The f","\\subsection*{Connecting to the paper's $Z_n$ convention}\nThe paper defines $Z_n$ through $\\sec t+\\tan t$. The Lean construction uses Mathlib's actual formal sine and cosine series and defines\n\\[\n \\sec t=(\\cos t)^{-1},\\quad \\tan t=(\\sin t)\\sec t,\n \\quad Z_n=n![t^n](\\sec t+\\tan t).\n\\]\nIt proves the sine and cosine derivative identities, $\\sin^2t+\\cos^2t=1$, and\n\\[\n (\\tan t)'=\\sec^2t.\n\\]\nSign substitution in the squared variable carries the hyperbolic cosine kernel to the actual cosine kernel. The same homomorphism carries their inverses and their squares. Extracting the even coefficient in the derivative of tangent yields\n\\[\n \\boxed{b_m=(-1)^m Z_{2m+1}.}\n\\]\nThis formula is \\emph{proved}, not assumed as an input to reconstruction. The formalization follows the generating-","\\subsection*{Connecting to the paper's $Z_n$ convention}\nThe paper defines $Z_n$ through $\\sec t+\\tan t$. The Lean construction uses Mathlib's actual formal sine and cosine series and defines\n\\[\n \\sec t=(\\cos t)^{-1},\\quad \\tan t=(\\sin t)\\sec t,\n \\quad Z_n=n![t^n](\\sec t+\\tan t).\n\\]\nIt proves the sine and cosine derivative identities, $\\sin^2t+\\cos^2t=1$, and\n\\[\n (\\tan t)'=\\sec^2t.\n\\]\nSign substitution in the squared variable carries the hyperbolic cosine kernel to the actual cosine kernel. The same homomorphism carries their inverses and their squares. Extracting the even coefficient in the derivative of tangent yields\n\\[\n \\boxed{b_m=(-1)^m Z_{2m+1}.}\n\\]\nThis formula is \\emph{proved}, not assumed as an input to reconstruction. The formalization follows the generating-function definition of z","\\subsection*{Connecting to the paper's $Z_n$ convention}\nThe paper defines $Z_n$ through $\\sec t+\\tan t$. The Lean construction uses Mathlib's actual formal sine and cosine series and defines\n\\[\n \\sec t=(\\cos t)^{-1},\\quad \\tan t=(\\sin t)\\sec t,\n \\quad Z_n=n![t^n](\\sec t+\\tan t).\n\\]\nIt proves the sine and cosine derivative identities, $\\sin^2t+\\cos^2t=1$, and\n\\[\n (\\tan t)'=\\sec^2t.\n\\]\nSign substitution in the squared variable carries the hyperbolic cosine kernel to the actual cosine kernel. The same homomorphism carries their inverses and their squares. Extracting the even coefficient in the derivative of tangent yields\n\\[\n \\boxed{b_m=(-1)^m Z_{2m+1}.}\n\\]\nThis formula is \\emph{proved}, not assumed as an input to reconstruction. The formalization follows the generating-function definition of zigzag numbers; it does not additionally formalize their interpretation as counts of alternating permutations.","(\\sin t)\\sec t,\n \\quad Z_n=n![t^n](\\sec t+\\tan t).\n\\]\nIt proves the sine and cosine derivative identities, $\\sin^2t+\\cos^2t=1$, and\n\\[\n (\\tan t)'=\\sec^2t.\n\\]\nSign substitution in the squared variable carries the hyperbolic cosine kernel to the actual cosine kernel. The same homomorphism carries their inverses and their squares. Extracting the even coefficient in the derivative of tangent yields\n\\[\n \\boxed{b_m=(-1)^m Z_{2m+1}.}\n\\]\nThis formula is \\emph{proved}, not assumed as an input to reconstruction. The formalization follows the generating-function definition of zigzag numbers; it does not additionally formalize their interpretation as counts of alternating permutations.","\\textbf{Principal targets.}\\par\n\\target{EulerCoefficients.euler_egf, EulerCoefficients.inverse_coeff_cast}\\par\n\\target{Zigzag.derivative_tan, Zigzag.inverse_is_signed_zigzag}\n\\par\\medskip\n\\pageheading{3. Exact reconstruction and the triangular matrix}{Theorem 4.2; manuscript page 11}\nFor odd $p$, rescaling the already proved inverse law gives\n\\[\n \\sech(pt)\\cosh(pt)=1.\n\\]\nMultiplying the even-source identity by $\\sech^2(pt)$ therefore proves\n\\[\n \\sech^2(pt)\\left(\\sum_{r\\ge0}S_{2r}(p)\\frac{t^{2r}}{(2r)!}\\right)=\\sech t.\n\\]\nActual EGF multiplication now gives the integer coefficient formula\n\\[\n \\boxed{E_{2r}=\\sum_{m=0}^{r}\\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).}\n\\]\nThe formal proof first extracts the sum over all indices and then proves that the odd summands vanish. A checked even-index summation identity supplies the displayed change of variables.","al targets.}\\par\n\\target{EulerCoefficients.euler_egf, EulerCoefficients.inverse_coeff_cast}\\par\n\\target{Zigzag.derivative_tan, Zigzag.inverse_is_signed_zigzag}\n\\par\\medskip\n\\pageheading{3. Exact reconstruction and the triangular matrix}{Theorem 4.2; manuscript page 11}\nFor odd $p$, rescaling the already proved inverse law gives\n\\[\n \\sech(pt)\\cosh(pt)=1.\n\\]\nMultiplying the even-source identity by $\\sech^2(pt)$ therefore proves\n\\[\n \\sech^2(pt)\\left(\\sum_{r\\ge0}S_{2r}(p)\\frac{t^{2r}}{(2r)!}\\right)=\\sech t.\n\\]\nActual EGF multiplication now gives the integer coefficient formula\n\\[\n \\boxed{E_{2r}=\\sum_{m=0}^{r}\\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).}\n\\]\nThe formal proof first extracts the sum over all indices and then proves that the odd summands vanish. A checked even-index summation identity supplies the displayed change of variables.","efficients.euler_egf, EulerCoefficients.inverse_coeff_cast}\\par\n\\target{Zigzag.derivative_tan, Zigzag.inverse_is_signed_zigzag}\n\\par\\medskip\n\\pageheading{3. Exact reconstruction and the triangular matrix}{Theorem 4.2; manuscript page 11}\nFor odd $p$, rescaling the already proved inverse law gives\n\\[\n \\sech(pt)\\cosh(pt)=1.\n\\]\nMultiplying the even-source identity by $\\sech^2(pt)$ therefore proves\n\\[\n \\sech^2(pt)\\left(\\sum_{r\\ge0}S_{2r}(p)\\frac{t^{2r}}{(2r)!}\\right)=\\sech t.\n\\]\nActual EGF multiplication now gives the integer coefficient formula\n\\[\n \\boxed{E_{2r}=\\sum_{m=0}^{r}\\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).}\n\\]\nThe formal proof first extracts the sum over all indices and then proves that the odd summands vanish. A checked even-index summation identity supplies the displayed change of variables.","ing{3. Exact reconstruction and the triangular matrix}{Theorem 4.2; manuscript page 11}\nFor odd $p$, rescaling the already proved inverse law gives\n\\[\n \\sech(pt)\\cosh(pt)=1.\n\\]\nMultiplying the even-source identity by $\\sech^2(pt)$ therefore proves\n\\[\n \\sech^2(pt)\\left(\\sum_{r\\ge0}S_{2r}(p)\\frac{t^{2r}}{(2r)!}\\right)=\\sech t.\n\\]\nActual EGF multiplication now gives the integer coefficient formula\n\\[\n \\boxed{E_{2r}=\\sum_{m=0}^{r}\\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).}\n\\]\nThe formal proof first extracts the sum over all indices and then proves that the odd summands vanish. A checked even-index summation identity supplies the displayed change of variables.","Substituting the independently proved signed-zigzag identity recovers the manuscript's notation. For every even degree $n$,\n\\[\n \\boxed{E_n=\\sum_{m=0}^{n/2}(-1)^m Z_{2m+1}\\binom n{2m}p^{2m}S_{n-2m}(p).}\n\\]\nThe Lean version uses $n=2r$ to avoid ambiguous natural-number division. Every even natural degree has that form. Primality is not used in this equality.","\\subsection*{The row action is connected to the matrix entries}\nReversing the finite sum and using binomial symmetry gives the actual source-index row action\n\\[\n E_{2r}=\\sum_{q=0}^{r}U_{r,q}(p)S_{2q}(p),\n\\]\nwhere\n\\[\n U_{r,q}(p)=\n \\begin{cases}\n \\displaystyle\\binom{2r}{2q}b_{r-q}p^{2(r-q)},&q\\le r,","\\subsection*{The row action is connected to the matrix entries}\nReversing the finite sum and using binomial symmetry gives the actual source-index row action\n\\[\n E_{2r}=\\sum_{q=0}^{r}U_{r,q}(p)S_{2q}(p),\n\\]\nwhere\n\\[\n U_{r,q}(p)=\n \\begin{cases}\n \\displaystyle\\binom{2r}{2q}b_{r-q}p^{2(r-q)},&q\\le r,\\\\\n 0,&q>r.\n \\end{cases}\n\\]\nThe row-action equality, the diagonal identity $U_{r,r}=1$, and vanishing above the diagonal are all checked. The arithmetic weight is also explicit:\n\\[\n p^{2(r-q)}\\mid U_{r,q}(p)\\quad\\text{in }\\Z,\\qquad 0\\le q\\le r.\n\\]\nNo inverse-coordinate identity, unit diagonal or weight bound is assumed for these rows.","0,&q>r.\n \\end{cases}\n\\]\nThe row-action equality, the diagonal identity $U_{r,r}=1$, and vanishing above the diagonal are all checked. The arithmetic weight is also explicit:\n\\[\n p^{2(r-q)}\\mid U_{r,q}(p)\\quad\\text{in }\\Z,\\qquad 0\\le q\\le r.\n\\]\nNo inverse-coordinate identity, unit diagonal or weight bound is assumed for these rows.","\\textbf{Principal targets.}\\par\n\\target{OddEuler.odd_reconstruction_series}\\par\n\\target{OddEuler.odd_euler_reconstruction}\\par\n\\target{OddEuler.odd_euler_reconstruction_zigzag}\\par\n\\target{OddEuler.row_action, OddEuler.row_diagonal, OddEuler.row_weight}\n\\par\\medskip\n\\pageheading{4. Uniform truncation and a sharp exact example}{Corollary 4.3 and Example 4.4}\nFor row $2r$ and depth $M$, the actual truncation is\n\\[\n T_M(p,2r)=\\sum_{m=0}^{\\min(M,r)}\\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).\n\\]\nEvery omitted term has $m\\ge M+1$. Its other factors are integers, so the first-block omitted-tail theorem, now applied to the \\emph{proved Euler identity}, gives\n\\[\n \\boxed{p^{2M+2}\\mid E_{2r}-T_M(p,2r)\\quad\\text{in }\\Z.}\n\\]\nThis is not divisibility in $\\Q$, where a nonzero divisor would make the assertion vacuous. The statement holds for every row and depth. For an odd prime it is precisely the claimed supercongruence. If $M\\ge r$, complete retention is proved exact.","al targets.}\\par\n\\target{OddEuler.odd_reconstruction_series}\\par\n\\target{OddEuler.odd_euler_reconstruction}\\par\n\\target{OddEuler.odd_euler_reconstruction_zigzag}\\par\n\\target{OddEuler.row_action, OddEuler.row_diagonal, OddEuler.row_weight}\n\\par\\medskip\n\\pageheading{4. Uniform truncation and a sharp exact example}{Corollary 4.3 and Example 4.4}\nFor row $2r$ and depth $M$, the actual truncation is\n\\[\n T_M(p,2r)=\\sum_{m=0}^{\\min(M,r)}\\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).\n\\]\nEvery omitted term has $m\\ge M+1$. Its other factors are integers, so the first-block omitted-tail theorem, now applied to the \\emph{proved Euler identity}, gives\n\\[\n \\boxed{p^{2M+2}\\mid E_{2r}-T_M(p,2r)\\quad\\text{in }\\Z.}\n\\]\nThis is not divisibility in $\\Q$, where a nonzero divisor would make the assertion vacuous. The statement holds for every row and depth. For an odd prime it is precisely the claimed supercongruence. If $M\\ge r$, complete retention is proved exact.","uction}\\par\n\\target{OddEuler.odd_euler_reconstruction_zigzag}\\par\n\\target{OddEuler.row_action, OddEuler.row_diagonal, OddEuler.row_weight}\n\\par\\medskip\n\\pageheading{4. Uniform truncation and a sharp exact example}{Corollary 4.3 and Example 4.4}\nFor row $2r$ and depth $M$, the actual truncation is\n\\[\n T_M(p,2r)=\\sum_{m=0}^{\\min(M,r)}\\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).\n\\]\nEvery omitted term has $m\\ge M+1$. Its other factors are integers, so the first-block omitted-tail theorem, now applied to the \\emph{proved Euler identity}, gives\n\\[\n \\boxed{p^{2M+2}\\mid E_{2r}-T_M(p,2r)\\quad\\text{in }\\Z.}\n\\]\nThis is not divisibility in $\\Q$, where a nonzero divisor would make the assertion vacuous. The statement holds for every row and depth. For an odd prime it is precisely the claimed supercongruence. If $M\\ge r$, complete retention is proved exact.","ar\\medskip\n\\pageheading{4. Uniform truncation and a sharp exact example}{Corollary 4.3 and Example 4.4}\nFor row $2r$ and depth $M$, the actual truncation is\n\\[\n T_M(p,2r)=\\sum_{m=0}^{\\min(M,r)}\\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).\n\\]\nEvery omitted term has $m\\ge M+1$. Its other factors are integers, so the first-block omitted-tail theorem, now applied to the \\emph{proved Euler identity}, gives\n\\[\n \\boxed{p^{2M+2}\\mid E_{2r}-T_M(p,2r)\\quad\\text{in }\\Z.}\n\\]\nThis is not divisibility in $\\Q$, where a nonzero divisor would make the assertion vacuous. The statement holds for every row and depth. For an odd prime it is precisely the claimed supercongruence. If $M\\ge r$, complete retention is proved exact.","Example 4.4}\nFor row $2r$ and depth $M$, the actual truncation is\n\\[\n T_M(p,2r)=\\sum_{m=0}^{\\min(M,r)}\\binom{2r}{2m}b_m p^{2m}S_{2(r-m)}(p).\n\\]\nEvery omitted term has $m\\ge M+1$. Its other factors are integers, so the first-block omitted-tail theorem, now applied to the \\emph{proved Euler identity}, gives\n\\[\n \\boxed{p^{2M+2}\\mid E_{2r}-T_M(p,2r)\\quad\\text{in }\\Z.}\n\\]\nThis is not divisibility in $\\Q$, where a nonzero divisor would make the assertion vacuous. The statement holds for every row and depth. For an odd prime it is precisely the claimed supercongruence. If $M\\ge r$, complete retention is proved exact.","M+1$. Its other factors are integers, so the first-block omitted-tail theorem, now applied to the \\emph{proved Euler identity}, gives\n\\[\n \\boxed{p^{2M+2}\\mid E_{2r}-T_M(p,2r)\\quad\\text{in }\\Z.}\n\\]\nThis is not divisibility in $\\Q$, where a nonzero divisor would make the assertion vacuous. The statement holds for every row and depth. For an odd prime it is precisely the claimed supercongruence. If $M\\ge r$, complete retention is proved exact.","\\subsection*{The displayed manuscript row is kernel-checked}\nThe scalar anchors are derived from the same general coefficient recurrence:\n\\[\n (E_0,E_2,E_4,E_6,E_8)=(1,-1,5,-61,1385),\n\\]\n\\[\n (b_0,b_1,b_2,b_3,b_4)=(1,-2,16,-272,7936).\n\\]\nThe literal source sums at $p=5$ give\n\\[\n (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985).\n\\]\nConsequently,\n\\[\n T_2(5,8)=2\\,731\\,001\\,385,\\qquad\n E_8-T_2(5,8)=-2\\,731\\,000\\,000.\n\\]\nThe source proves both\n\\[\n 5^6\\mid E_8-T_2(5,8),\\qquad 5^7\\nmid E_8-T_2(5,8).\n\\]\nThus the guaranteed power is attained in this example. No stronger uniform exponent is inferred.","kernel-checked}\nThe scalar anchors are derived from the same general coefficient recurrence:\n\\[\n (E_0,E_2,E_4,E_6,E_8)=(1,-1,5,-61,1385),\n\\]\n\\[\n (b_0,b_1,b_2,b_3,b_4)=(1,-2,16,-272,7936).\n\\]\nThe literal source sums at $p=5$ give\n\\[\n (S_0,S_2,S_4,S_6,S_8)=(1,49,4705,428689,37665985).\n\\]\nConsequently,\n\\[\n T_2(5,8)=2\\,731\\,001\\,385,\\qquad\n E_8-T_2(5,8)=-2\\,731\\,000\\,000.\n\\]\nThe source proves both\n\\[\n 5^6\\mid E_8-T_2(5,8),\\qquad 5^7\\nmid E_8-T_2(5,8).\n\\]\nThus the guaranteed power is attained in this example. No stronger uniform exponent is inferred.","\\subsection*{Oddness cannot be silently dropped}\nFor length two, $S_0(2)=1-1=0$, whereas $E_0=1$. The row-zero reconstruction already fails. The source proves this failure, and the release verifier separately attempts and rejects an invalid proof of the even-length identity.","\\subsubsection*{The completed mathematical chain}\nThe inverse coefficients are defined from a genuine unit power series and connected to $z^2/\\sinh^2z$. The source is the paper's explicit rational formula, not an abstract integral input. Its valuation calculation proves the stronger statement that \\emph{every even source coefficient is a $2$-adic unit}. Formal-series multiplication and coefficient extraction then yield the Euler identity with the correct falling factorial. The omitted-tail theorem gives divisibility in the actual completed integer ring $\\Z_2$.","rce is the paper's explicit rational formula, not an abstract integral input. Its valuation calculation proves the stronger statement that \\emph{every even source coefficient is a $2$-adic unit}. Formal-series multiplication and coefficient extraction then yield the Euler identity with the correct falling factorial. The omitted-tail theorem gives divisibility in the actual completed integer ring $\\Z_2$.","cients is not universally formalized here. Formal-series identities do not establish analytic convergence. The corrected global odd-prime finite-difference theorem remains pending. Historical numerical evidence is not presented as a fresh reproduction.\n\\par\\medskip\n\\pageheading{1. Integral coefficients and an exact unit source}{Lemmas 5.1 and 5.2; manuscript page 14}\nDefine the rational subring\n\\[\n \\mathcal O_2=\\{q\\in\\Q:|q|_2\\leq1\\}.\n\\]\nIts ring laws follow from the actual nonarchimedean norm. For every $q\\in\\mathcal O_2$, the code constructs its element in Mathlib's completed ring $\\Z_2$, and proves that the underlying value in $\\Q_2$ is the rational cast of $q$. No integrality claim is hidden in a coercion.","corrected global odd-prime finite-difference theorem remains pending. Historical numerical evidence is not presented as a fresh reproduction.\n\\par\\medskip\n\\pageheading{1. Integral coefficients and an exact unit source}{Lemmas 5.1 and 5.2; manuscript page 14}\nDefine the rational subring\n\\[\n \\mathcal O_2=\\{q\\in\\Q:|q|_2\\leq1\\}.\n\\]\nIts ring laws follow from the actual nonarchimedean norm. For every $q\\in\\mathcal O_2$, the code constructs its element in Mathlib's completed ring $\\Z_2$, and proves that the underlying value in $\\Q_2$ is the rational cast of $q$. No integrality claim is hidden in a coercion.","l numerical evidence is not presented as a fresh reproduction.\n\\par\\medskip\n\\pageheading{1. Integral coefficients and an exact unit source}{Lemmas 5.1 and 5.2; manuscript page 14}\nDefine the rational subring\n\\[\n \\mathcal O_2=\\{q\\in\\Q:|q|_2\\leq1\\}.\n\\]\nIts ring laws follow from the actual nonarchimedean norm. For every $q\\in\\mathcal O_2$, the code constructs its element in Mathlib's completed ring $\\Z_2$, and proves that the underlying value in $\\Q_2$ is the rational cast of $q$. No integrality claim is hidden in a coercion.","reproduction.\n\\par\\medskip\n\\pageheading{1. Integral coefficients and an exact unit source}{Lemmas 5.1 and 5.2; manuscript page 14}\nDefine the rational subring\n\\[\n \\mathcal O_2=\\{q\\in\\Q:|q|_2\\leq1\\}.\n\\]\nIts ring laws follow from the actual nonarchimedean norm. For every $q\\in\\mathcal O_2$, the code constructs its element in Mathlib's completed ring $\\Z_2$, and proves that the underlying value in $\\Q_2$ is the rational cast of $q$. No integrality claim is hidden in a coercion.","geheading{1. Integral coefficients and an exact unit source}{Lemmas 5.1 and 5.2; manuscript page 14}\nDefine the rational subring\n\\[\n \\mathcal O_2=\\{q\\in\\Q:|q|_2\\leq1\\}.\n\\]\nIts ring laws follow from the actual nonarchimedean norm. For every $q\\in\\mathcal O_2$, the code constructs its element in Mathlib's completed ring $\\Z_2$, and proves that the underlying value in $\\Q_2$ is the rational cast of $q$. No integrality claim is hidden in a coercion.","\\subsection*{The unit kernel and its reciprocal}\nUse a squared formal variable $u$ and define\n\\[\n k_m=\\frac{2^{2m+1}}{(2m+2)!},\\qquad\n K(u)=\\sum_{m\\geq0}k_mu^m,\\qquad\n \\beta_m=[u^m]K(u)^{-1}.\n\\]\nThe constant term is proved to be one. The factorial valuation bound gives\n\\[\n v_2(k_m)=2m+1-v_2((2m+2)!)\\geq0.\n\\]\nThus every $k_m$ belongs to $\\mathcal O_2$. A general coefficientwise subring theorem is proved by strong induction using the actual unit-inverse coefficient formula. Applied to $K$, it establishes\n\\[\n \\boxed{\\beta_m\\in\\mathcal O_2\\quad\\text{for every }m\\geq0.}\n\\]\nThe reciprocal's coefficients are therefore proved integral; the proof does not assume them integral in order to obtain the final congruence. Their connection to the intended hyperbolic function is proved o","\\subsection*{The unit kernel and its reciprocal}\nUse a squared formal variable $u$ and define\n\\[\n k_m=\\frac{2^{2m+1}}{(2m+2)!},\\qquad\n K(u)=\\sum_{m\\geq0}k_mu^m,\\qquad\n \\beta_m=[u^m]K(u)^{-1}.\n\\]\nThe constant term is proved to be one. The factorial valuation bound gives\n\\[\n v_2(k_m)=2m+1-v_2((2m+2)!)\\geq0.\n\\]\nThus every $k_m$ belongs to $\\mathcal O_2$. A general coefficientwise subring theorem is proved by strong induction using the actual unit-inverse coefficient formula. Applied to $K$, it establishes\n\\[\n \\boxed{\\beta_m\\in\\mathcal O_2\\quad\\text{for every }m\\geq0.}\n\\]\nThe reciprocal's coefficients are therefore proved integral; the proof does not assume them integral in order to obtain the final congruence. Their connection to the intended hyperbolic function is proved on the next page.","fine\n\\[\n k_m=\\frac{2^{2m+1}}{(2m+2)!},\\qquad\n K(u)=\\sum_{m\\geq0}k_mu^m,\\qquad\n \\beta_m=[u^m]K(u)^{-1}.\n\\]\nThe constant term is proved to be one. The factorial valuation bound gives\n\\[\n v_2(k_m)=2m+1-v_2((2m+2)!)\\geq0.\n\\]\nThus every $k_m$ belongs to $\\mathcal O_2$. A general coefficientwise subring theorem is proved by strong induction using the actual unit-inverse coefficient formula. Applied to $K$, it establishes\n\\[\n \\boxed{\\beta_m\\in\\mathcal O_2\\quad\\text{for every }m\\geq0.}\n\\]\nThe reciprocal's coefficients are therefore proved integral; the proof does not assume them integral in order to obtain the final congruence. Their connection to the intended hyperbolic function is proved on the next page.","{2m+1}}{(2m+2)!},\\qquad\n K(u)=\\sum_{m\\geq0}k_mu^m,\\qquad\n \\beta_m=[u^m]K(u)^{-1}.\n\\]\nThe constant term is proved to be one. The factorial valuation bound gives\n\\[\n v_2(k_m)=2m+1-v_2((2m+2)!)\\geq0.\n\\]\nThus every $k_m$ belongs to $\\mathcal O_2$. A general coefficientwise subring theorem is proved by strong induction using the actual unit-inverse coefficient formula. Applied to $K$, it establishes\n\\[\n \\boxed{\\beta_m\\in\\mathcal O_2\\quad\\text{for every }m\\geq0.}\n\\]\nThe reciprocal's coefficients are therefore proved integral; the proof does not assume them integral in order to obtain the final congruence. Their connection to the intended hyperbolic function is proved on the next page.","\\subsection*{The source has valuation exactly zero}\nAt even degree $2r$, define the literal rational source\n\\[\n R_{2r}=\\frac{3^{2r+2}-1}{4(2r+1)(2r+2)}.\n\\]\nBoth numerator and denominator are nonzero. The pinned Mathlib lifting-the-exponent theorem gives\n\\[\n v_2(3^{2r+2}-1)=2+v_2(2r+2).\n\\]\nBecause $2r+1$ is odd, the denominator has the same valuation. Consequently,\n\\[\n \\boxed{v_2(R_{2r})=0,\\qquad |R_{2r}|_2=1.}\n\\]\nThe code proves actual \\mcode{IsUnit} status for the constructed $\\Z_2$ element. This strengthens the manuscript's integrality conclusion without changing its source. The argument applies at $r=0$, where $R_0=1$, as well as at every later row.","literal rational source\n\\[\n R_{2r}=\\frac{3^{2r+2}-1}{4(2r+1)(2r+2)}.\n\\]\nBoth numerator and denominator are nonzero. The pinned Mathlib lifting-the-exponent theorem gives\n\\[\n v_2(3^{2r+2}-1)=2+v_2(2r+2).\n\\]\nBecause $2r+1$ is odd, the denominator has the same valuation. Consequently,\n\\[\n \\boxed{v_2(R_{2r})=0,\\qquad |R_{2r}|_2=1.}\n\\]\nThe code proves actual \\mcode{IsUnit} status for the constructed $\\Z_2$ element. This strengthens the manuscript's integrality conclusion without changing its source. The argument applies at $r=0$, where $R_0=1$, as well as at every later row.","d Mathlib lifting-the-exponent theorem gives\n\\[\n v_2(3^{2r+2}-1)=2+v_2(2r+2).\n\\]\nBecause $2r+1$ is odd, the denominator has the same valuation. Consequently,\n\\[\n \\boxed{v_2(R_{2r})=0,\\qquad |R_{2r}|_2=1.}\n\\]\nThe code proves actual \\mcode{IsUnit} status for the constructed $\\Z_2$ element. This strengthens the manuscript's integrality conclusion without changing its source. The argument applies at $r=0$, where $R_0=1$, as well as at every later row.","ting-the-exponent theorem gives\n\\[\n v_2(3^{2r+2}-1)=2+v_2(2r+2).\n\\]\nBecause $2r+1$ is odd, the denominator has the same valuation. Consequently,\n\\[\n \\boxed{v_2(R_{2r})=0,\\qquad |R_{2r}|_2=1.}\n\\]\nThe code proves actual \\mcode{IsUnit} status for the constructed $\\Z_2$ element. This strengthens the manuscript's integrality conclusion without changing its source. The argument applies at $r=0$, where $R_0=1$, as well as at every later row.","\\textbf{Principal targets.}\\par\n\\target{Dyadic.kernel_coefficient_integral, Dyadic.beta_integral}\\par\n\\target{Dyadic.source_valuation, Dyadic.sourcePadic_isUnit}\n\\par\\medskip\n\\pageheading{2. Connecting the definitions to actual formal series}{Generating-function foundations for Theorem 5.3}\nWork in $\\Q[[t]]$. The formal exponential and rescaling homomorphism are the same checked objects used in the second block. Define\n\\[\n \\sinh_a(t)=\\frac{e^{at}-e^{-at}}2,\\qquad\n \\cosh_a(t)=\\frac{e^{at}+e^{-at}}2.\n\\]\nLet $B(t)=K(t^2)^{-1}=\\sum_{m\\geq0}\\beta_mt^{2m}$ and let\n\\[\n F(t)=\\sum_{r\\geq0}R_{2r}\\frac{t^{2r}}{(2r)!}.\n\\]\nThese series are constructed independently from the definitions on the preceding page.",".}\\par\n\\target{Dyadic.kernel_coefficient_integral, Dyadic.beta_integral}\\par\n\\target{Dyadic.source_valuation, Dyadic.sourcePadic_isUnit}\n\\par\\medskip\n\\pageheading{2. Connecting the definitions to actual formal series}{Generating-function foundations for Theorem 5.3}\nWork in $\\Q[[t]]$. The formal exponential and rescaling homomorphism are the same checked objects used in the second block. Define\n\\[\n \\sinh_a(t)=\\frac{e^{at}-e^{-at}}2,\\qquad\n \\cosh_a(t)=\\frac{e^{at}+e^{-at}}2.\n\\]\nLet $B(t)=K(t^2)^{-1}=\\sum_{m\\geq0}\\beta_mt^{2m}$ and let\n\\[\n F(t)=\\sum_{r\\geq0}R_{2r}\\frac{t^{2r}}{(2r)!}.\n\\]\nThese series are constructed independently from the definitions on the preceding page.","adic.source_valuation, Dyadic.sourcePadic_isUnit}\n\\par\\medskip\n\\pageheading{2. Connecting the definitions to actual formal series}{Generating-function foundations for Theorem 5.3}\nWork in $\\Q[[t]]$. The formal exponential and rescaling homomorphism are the same checked objects used in the second block. Define\n\\[\n \\sinh_a(t)=\\frac{e^{at}-e^{-at}}2,\\qquad\n \\cosh_a(t)=\\frac{e^{at}+e^{-at}}2.\n\\]\nLet $B(t)=K(t^2)^{-1}=\\sum_{m\\geq0}\\beta_mt^{2m}$ and let\n\\[\n F(t)=\\sum_{r\\geq0}R_{2r}\\frac{t^{2r}}{(2r)!}.\n\\]\nThese series are constructed independently from the definitions on the preceding page.","medskip\n\\pageheading{2. Connecting the definitions to actual formal series}{Generating-function foundations for Theorem 5.3}\nWork in $\\Q[[t]]$. The formal exponential and rescaling homomorphism are the same checked objects used in the second block. Define\n\\[\n \\sinh_a(t)=\\frac{e^{at}-e^{-at}}2,\\qquad\n \\cosh_a(t)=\\frac{e^{at}+e^{-at}}2.\n\\]\nLet $B(t)=K(t^2)^{-1}=\\sum_{m\\geq0}\\beta_mt^{2m}$ and let\n\\[\n F(t)=\\sum_{r\\geq0}R_{2r}\\frac{t^{2r}}{(2r)!}.\n\\]\nThese series are constructed independently from the definitions on the preceding page.","\\subsection*{No inverse of a zero-constant series is assumed}\nCoefficient extraction proves the cleared identity\n\\[\n t^2K(t^2)=\\frac{\\cosh(2t)-1}{2}.\n\\]\nA separate identity from the formal exponential gives\n\\[\n \\sinh^2t=\\frac{\\cosh(2t)-1}{2}.\n\\]\nSince $K(t^2)$ has constant term one, its reciprocal exists as a unit. Multiplication then proves\n\\[\n \\boxed{B(t)\\sinh^2t=t^2.}\n\\]\nThis is the precise formal meaning of $B(t)=t^2/\\sinh^2t$. The forced quadratic zero is removed first; the proof does not claim that $\\sinh t$ itself is an invertible formal power series.","inverse of a zero-constant series is assumed}\nCoefficient extraction proves the cleared identity\n\\[\n t^2K(t^2)=\\frac{\\cosh(2t)-1}{2}.\n\\]\nA separate identity from the formal exponential gives\n\\[\n \\sinh^2t=\\frac{\\cosh(2t)-1}{2}.\n\\]\nSince $K(t^2)$ has constant term one, its reciprocal exists as a unit. Multiplication then proves\n\\[\n \\boxed{B(t)\\sinh^2t=t^2.}\n\\]\nThis is the precise formal meaning of $B(t)=t^2/\\sinh^2t$. The forced quadratic zero is removed first; the proof does not claim that $\\sinh t$ itself is an invertible formal power series.","y\n\\[\n t^2K(t^2)=\\frac{\\cosh(2t)-1}{2}.\n\\]\nA separate identity from the formal exponential gives\n\\[\n \\sinh^2t=\\frac{\\cosh(2t)-1}{2}.\n\\]\nSince $K(t^2)$ has constant term one, its reciprocal exists as a unit. Multiplication then proves\n\\[\n \\boxed{B(t)\\sinh^2t=t^2.}\n\\]\nThis is the precise formal meaning of $B(t)=t^2/\\sinh^2t$. The forced quadratic zero is removed first; the proof does not claim that $\\sinh t$ itself is an invertible formal power series.","\\subsection*{The explicit rational source supplies its own identity}\nFrom the formula for $R_{2r}$, coefficient arithmetic proves\n\\[\n \\boxed{t^2 F(t)=\\frac{\\cosh(3t)-\\cosh t}{4}.}\n\\]\nThe product relation for the formal hyperbolic cosines yields\n\\[\n F(t)\\cosh t=K(4t^2).\n\\]\nThis equality is proved after multiplication by $t^2$ and cancellation of that monomial in the formal-series ring. It is not an assumed source-transform relation.","Rescale the reciprocal by $t\\mapsto2t$. The unit inverse law and the preceding identity now give\n\\[\n \\boxed{B(2t)F(t)=\\sech t.}\n\\]\nThe right side is connected to the same integral signed Euler sequence constructed in the previous release. These are equalities of full formal series, not fixed-degree polynomial experiments.","\\textbf{Principal targets.}\\par\n\\target{DyadicSeries.beta_original_generating_function}\\par\n\\target{DyadicSeries.source_shift_identity, DyadicSeries.source_times_cosh}\\par\n\\target{DyadicSeries.dyadic_reconstruction_series}\n\\par\\medskip\n\\pageheading{3. Exact reconstruction and genuine dyadic congruence}{Theorem 5.3 and Corollary 5.4; manuscript pages 14--15}\nThe reciprocal coefficients $\\beta_m$ use an \\emph{ordinary} power expansion. In contrast, the source and Euler series use exponential coefficients. Extracting degree $2r$ from $B(2t)F(t)=\\sech t$ therefore gives\n\\[\n \\boxed{E_{2r}=\\sum_{m=0}^{r}\\beta_m2^{2m}\\fall{2r}{2m}R_{2(r-m)},}\n \\qquad\n \\fall{2r}{2m}=\\frac{(2r)!}{(2r-2m)!}.\n\\]\nThe formal proof uses the actual natural-number descending factorial and proves its factorial-quotient identity for $2m\\leq2r$. The odd coefficient terms vanish by a proved parity argument. The binomial EGF product then supplies the falling factorial, rather than treating it as an extra assumed factor.","entity, DyadicSeries.source_times_cosh}\\par\n\\target{DyadicSeries.dyadic_reconstruction_series}\n\\par\\medskip\n\\pageheading{3. Exact reconstruction and genuine dyadic congruence}{Theorem 5.3 and Corollary 5.4; manuscript pages 14--15}\nThe reciprocal coefficients $\\beta_m$ use an \\emph{ordinary} power expansion. In contrast, the source and Euler series use exponential coefficients. Extracting degree $2r$ from $B(2t)F(t)=\\sech t$ therefore gives\n\\[\n \\boxed{E_{2r}=\\sum_{m=0}^{r}\\beta_m2^{2m}\\fall{2r}{2m}R_{2(r-m)},}\n \\qquad\n \\fall{2r}{2m}=\\frac{(2r)!}{(2r-2m)!}.\n\\]\nThe formal proof uses the actual natural-number descending factorial and proves its factorial-quotient identity for $2m\\leq2r$. The odd coefficient terms vanish by a proved parity argument. The binomial EGF product then supplies the falling factorial, rather than treating it as an extra assumed factor.","adicSeries.source_times_cosh}\\par\n\\target{DyadicSeries.dyadic_reconstruction_series}\n\\par\\medskip\n\\pageheading{3. Exact reconstruction and genuine dyadic congruence}{Theorem 5.3 and Corollary 5.4; manuscript pages 14--15}\nThe reciprocal coefficients $\\beta_m$ use an \\emph{ordinary} power expansion. In contrast, the source and Euler series use exponential coefficients. Extracting degree $2r$ from $B(2t)F(t)=\\sech t$ therefore gives\n\\[\n \\boxed{E_{2r}=\\sum_{m=0}^{r}\\beta_m2^{2m}\\fall{2r}{2m}R_{2(r-m)},}\n \\qquad\n \\fall{2r}{2m}=\\frac{(2r)!}{(2r-2m)!}.\n\\]\nThe formal proof uses the actual natural-number descending factorial and proves its factorial-quotient identity for $2m\\leq2r$. The odd coefficient terms vanish by a proved parity argument. The binomial EGF product then supplies the falling factorial, rather than treating it as an extra assumed factor.","rget{DyadicSeries.dyadic_reconstruction_series}\n\\par\\medskip\n\\pageheading{3. Exact reconstruction and genuine dyadic congruence}{Theorem 5.3 and Corollary 5.4; manuscript pages 14--15}\nThe reciprocal coefficients $\\beta_m$ use an \\emph{ordinary} power expansion. In contrast, the source and Euler series use exponential coefficients. Extracting degree $2r$ from $B(2t)F(t)=\\sech t$ therefore gives\n\\[\n \\boxed{E_{2r}=\\sum_{m=0}^{r}\\beta_m2^{2m}\\fall{2r}{2m}R_{2(r-m)},}\n \\qquad\n \\fall{2r}{2m}=\\frac{(2r)!}{(2r-2m)!}.\n\\]\nThe formal proof uses the actual natural-number descending factorial and proves its factorial-quotient identity for $2m\\leq2r$. The odd coefficient terms vanish by a proved parity argument. The binomial EGF product then supplies the falling factorial, rather than treating it as an extra assumed factor.","ordinary} power expansion. In contrast, the source and Euler series use exponential coefficients. Extracting degree $2r$ from $B(2t)F(t)=\\sech t$ therefore gives\n\\[\n \\boxed{E_{2r}=\\sum_{m=0}^{r}\\beta_m2^{2m}\\fall{2r}{2m}R_{2(r-m)},}\n \\qquad\n \\fall{2r}{2m}=\\frac{(2r)!}{(2r-2m)!}.\n\\]\nThe formal proof uses the actual natural-number descending factorial and proves its factorial-quotient identity for $2m\\leq2r$. The odd coefficient terms vanish by a proved parity argument. The binomial EGF product then supplies the falling factorial, rather than treating it as an extra assumed factor.","Both row indexing by $r$ and the literal even-degree version with hypothesis $\\operatorname{Even}(n)$ are checked. All rows include degree zero. The exact identity is a rational equality with the Euler coefficient's integer-to-rational cast made explicit.","\\subsection*{The finite truncation}\nFor $M\\geq0$, define\n\\[\n T_M(2r)=\\sum_{m=0}^{\\min(M,r)}\\beta_m2^{2m}\\fall{2r}{2m}R_{2(r-m)}.\n\\]\nThe integral rational factors are embedded into $\\Z_2$ using their proved witnesses. The cast of this $\\Z_2$ truncation into $\\Q_2$ is proved to equal the cast of the literal rational truncation.","Each omitted term has $m\\geq M+1$. Its remaining factors lie in $\\Z_2$, so the checked weighted-tail theorem proves\n\\[\n \\boxed{2^{2M+2}\\mid E_{2r}-T_M(2r)\\quad\\text{in }\\Z_2.}\n\\]\nThis is not divisibility in $\\Q$ or $\\Q_2$, where division by any nonzero scalar would make the claim vacuous. A second theorem supplies the equivalent rational statement\n\\[\n \\boxed{\\frac{E_{2r}-T_M(2r)}{2^{2M+2}}\\in\\mathcal O_2.}\n\\]\nIt follows from actual $\\Z_2$ divisibility and the proved rational/completed-field bridge. No precision assertion is passed in as a hypothesis.","s remaining factors lie in $\\Z_2$, so the checked weighted-tail theorem proves\n\\[\n \\boxed{2^{2M+2}\\mid E_{2r}-T_M(2r)\\quad\\text{in }\\Z_2.}\n\\]\nThis is not divisibility in $\\Q$ or $\\Q_2$, where division by any nonzero scalar would make the claim vacuous. A second theorem supplies the equivalent rational statement\n\\[\n \\boxed{\\frac{E_{2r}-T_M(2r)}{2^{2M+2}}\\in\\mathcal O_2.}\n\\]\nIt follows from actual $\\Z_2$ divisibility and the proved rational/completed-field bridge. No precision assertion is passed in as a hypothesis.","\\subsection*{Endpoints and scope}\nWhen $M\\geq r$, the sum retains every term and $T_M(2r)=E_{2r}$ exactly, in both rational and dyadic coordinates. The zero-error case is included; no finite valuation of zero is used. This is a separate dyadic construction, not substitution of $p=2$ into the odd-length theorem.","\\textbf{Principal targets.}\\par\n\\target{DyadicReconstruction.dyadic_reconstruction}\\par\n\\target{DyadicReconstruction.dyadic_supercongruence}\\par\n\\target{DyadicReconstruction.normalized_error_integral}\n\\par\\medskip\n\\pageheading{4. Exact examples and adversarial checks}{Example 5.5 and controls against incorrect normalizations}\nThe actual reciprocal recurrence gives\n\\[\n (\\beta_0,\\beta_1,\\beta_2,\\beta_3,\\beta_4)\n =\\left(1,-\\frac13,\\frac1{15},-\\frac2{189},\\frac1{675}\\right).\n\\]\nThese values are obtained from the general formal-series definition. They are not substituted as independent input data for the reconstruction theorem. The source formula gives\n\\[\n (R_0,R_2,R_4,R_6,R_8)\n =\\left(1,\\frac53,\\frac{91}{15},\\frac{205}{7},\\frac{7381}{45}\\right).\n\\]\nUsing the retained Euler anchor $E_8=1385$, Lean checks the literal decomposition\n\\[\n E_8=\\frac{7381}{45}-\\frac{6560}{3}+\\frac{163072}{15}\n       -\\frac{204800}{9}+\\frac{229376}{15}=1385.\n\\]","al checks}{Example 5.5 and controls against incorrect normalizations}\nThe actual reciprocal recurrence gives\n\\[\n (\\beta_0,\\beta_1,\\beta_2,\\beta_3,\\beta_4)\n =\\left(1,-\\frac13,\\frac1{15},-\\frac2{189},\\frac1{675}\\right).\n\\]\nThese values are obtained from the general formal-series definition. They are not substituted as independent input data for the reconstruction theorem. The source formula gives\n\\[\n (R_0,R_2,R_4,R_6,R_8)\n =\\left(1,\\frac53,\\frac{91}{15},\\frac{205}{7},\\frac{7381}{45}\\right).\n\\]\nUsing the retained Euler anchor $E_8=1385$, Lean checks the literal decomposition\n\\[\n E_8=\\frac{7381}{45}-\\frac{6560}{3}+\\frac{163072}{15}\n       -\\frac{204800}{9}+\\frac{229376}{15}=1385.\n\\]","\\beta_2,\\beta_3,\\beta_4)\n =\\left(1,-\\frac13,\\frac1{15},-\\frac2{189},\\frac1{675}\\right).\n\\]\nThese values are obtained from the general formal-series definition. They are not substituted as independent input data for the reconstruction theorem. The source formula gives\n\\[\n (R_0,R_2,R_4,R_6,R_8)\n =\\left(1,\\frac53,\\frac{91}{15},\\frac{205}{7},\\frac{7381}{45}\\right).\n\\]\nUsing the retained Euler anchor $E_8=1385$, Lean checks the literal decomposition\n\\[\n E_8=\\frac{7381}{45}-\\frac{6560}{3}+\\frac{163072}{15}\n       -\\frac{204800}{9}+\\frac{229376}{15}=1385.\n\\]","15},-\\frac2{189},\\frac1{675}\\right).\n\\]\nThese values are obtained from the general formal-series definition. They are not substituted as independent input data for the reconstruction theorem. The source formula gives\n\\[\n (R_0,R_2,R_4,R_6,R_8)\n =\\left(1,\\frac53,\\frac{91}{15},\\frac{205}{7},\\frac{7381}{45}\\right).\n\\]\nUsing the retained Euler anchor $E_8=1385$, Lean checks the literal decomposition\n\\[\n E_8=\\frac{7381}{45}-\\frac{6560}{3}+\\frac{163072}{15}\n       -\\frac{204800}{9}+\\frac{229376}{15}=1385.\n\\]","\\subsection*{Truncations and actual valuations}\n\\begin{center}\n\\begin{tabular}{@{}crr@{}}\n\\toprule\nDepth $M$ & $T_M(8)$ & Guaranteed exponent $2M+2$","midrule\n0 & $7381/45$ & 2","[3pt]\n1 & $-91019/45$ & 4\\\\[3pt]\n2 & $398197/45$ & 6\\\\[3pt]\n3 & $-208601/15$ & 8\\\\[3pt]\n4 & $1385$ & 10\\\\\\bottomrule\n\\end{tabular}\n\\end{center}\nThe checked nonzero error valuations are\n\\[\n E_8-T_0(8)=\\frac{54944}{45},\\qquad v_2(E_8-T_0(8))=5,\n\\]\n\\[\n E_8-T_3(8)=\\frac{229376}{15},\\qquad v_2(E_8-T_3(8))=15.\n\\]\nThey exceed the uniform guarantees; these two examples do not establish a universally stronger exponent.","bottomrule\n\\end{tabular}\n\\end{center}\nThe checked nonzero error valuations are\n\\[\n E_8-T_0(8)=\\frac{54944}{45},\\qquad v_2(E_8-T_0(8))=5,\n\\]\n\\[\n E_8-T_3(8)=\\frac{229376}{15},\\qquad v_2(E_8-T_3(8))=15.\n\\]\nThey exceed the uniform guarantees; these two examples do not establish a universally stronger exponent.","\\subsection*{What the negative controls rule out}\nOmitting the rescaling $2^{2m}$ already breaks the degree-two identity: the altered sum is $1$, whereas $E_2=-1$. Confusing ordinary beta coefficients with EGF coefficients would replace $\\beta_1=-1/3$ by $-2/3$; a checked inequality excludes that normalization. A deliberately false claim that the depth-zero degree-eight error has valuation at least six is rejected.","\\subsubsection*{Two explicit specification boundaries}\nA finite sample through $N$ determines $\\Delta^s f(n)$ only when $n+s\\leq N$. For arbitrary orders, the finite-coefficient identity describes the constructed interpolant, not every unknown continuation of the sample. All three versions are separately checked.","The additive valuation is extended to $\\ord_p(0)=\\infty$. This avoids incorrectly applying Mathlib's finite-valued default at zero to an inequality that must also cover cancellation and identically zero differences.","d rejected 12 invalid-proof controls. All 74 infrastructure and release tests passed. Independent exact-rational mirrors are recorded separately from the universal proofs. The archive is also subjected to a fresh-extraction rebuild and integrity audit; its separate report records that outcome.\n\\par\\medskip\n\\pageheading{1. Reconstruction and finite-data scope}{Theorem 7.2; manuscript page 17}\nLet $R$ be any commutative ring. The implementation uses the actual forward-difference operator\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad c_r=\\Delta_1^r f(0).\n\\]\nFor a full sequence $f:\\N\\to R$, pinned Mathlib's Gregory--Newton formula gives\n\\[\n \\boxed{f(n)=\\sum_{r=0}^{n}c_r\\binom nr.}\n\\]\nThe source applies the theorem to these exact definitions. A separate explicit formula relates an $s$-th difference to its original sampled values, so the data dependence is inspectable:\n\\[\n \\Delta_h^s f(n)=\\sum_{j=0}^{s}(-1)^{s-j}\\binom sj f(n+jh).\n\\]\nApplying Newton reconstruction to $\\Delta_1^s f$ and composing the iterated operators proves\n\\[\n \\boxed{\\Delta_1^s f(n)=\\sum_{j=0}^{n}c_{s+j}\\binom nj.}\n\\]\nThe apparen…","dent exact-rational mirrors are recorded separately from the universal proofs. The archive is also subjected to a fresh-extraction rebuild and integrity audit; its separate report records that outcome.\n\\par\\medskip\n\\pageheading{1. Reconstruction and finite-data scope}{Theorem 7.2; manuscript page 17}\nLet $R$ be any commutative ring. The implementation uses the actual forward-difference operator\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad c_r=\\Delta_1^r f(0).\n\\]\nFor a full sequence $f:\\N\\to R$, pinned Mathlib's Gregory--Newton formula gives\n\\[\n \\boxed{f(n)=\\sum_{r=0}^{n}c_r\\binom nr.}\n\\]\nThe source applies the theorem to these exact definitions. A separate explicit formula relates an $s$-th difference to its original sampled values, so the data dependence is inspectable:\n\\[\n \\Delta_h^s f(n)=\\sum_{j=0}^{s}(-1)^{s-j}\\binom sj f(n+jh).\n\\]\nApplying Newton reconstruction to $\\Delta_1^s f$ and composing the iterated operators proves\n\\[\n \\boxed{\\Delta_1^s f(n)=\\sum_{j=0}^{n}c_{s+j}\\binom nj.}\n\\]\nThe apparently infinite coefficient sum is finite at each natural argument. Neither continuity nor conve…","he archive is also subjected to a fresh-extraction rebuild and integrity audit; its separate report records that outcome.\n\\par\\medskip\n\\pageheading{1. Reconstruction and finite-data scope}{Theorem 7.2; manuscript page 17}\nLet $R$ be any commutative ring. The implementation uses the actual forward-difference operator\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad c_r=\\Delta_1^r f(0).\n\\]\nFor a full sequence $f:\\N\\to R$, pinned Mathlib's Gregory--Newton formula gives\n\\[\n \\boxed{f(n)=\\sum_{r=0}^{n}c_r\\binom nr.}\n\\]\nThe source applies the theorem to these exact definitions. A separate explicit formula relates an $s$-th difference to its original sampled values, so the data dependence is inspectable:\n\\[\n \\Delta_h^s f(n)=\\sum_{j=0}^{s}(-1)^{s-j}\\binom sj f(n+jh).\n\\]\nApplying Newton reconstruction to $\\Delta_1^s f$ and composing the iterated operators proves\n\\[\n \\boxed{\\Delta_1^s f(n)=\\sum_{j=0}^{n}c_{s+j}\\binom nj.}\n\\]\nThe apparently infinite coefficient sum is finite at each natural argument. Neither continuity nor convergence is needed for these identities.","d integrity audit; its separate report records that outcome.\n\\par\\medskip\n\\pageheading{1. Reconstruction and finite-data scope}{Theorem 7.2; manuscript page 17}\nLet $R$ be any commutative ring. The implementation uses the actual forward-difference operator\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad c_r=\\Delta_1^r f(0).\n\\]\nFor a full sequence $f:\\N\\to R$, pinned Mathlib's Gregory--Newton formula gives\n\\[\n \\boxed{f(n)=\\sum_{r=0}^{n}c_r\\binom nr.}\n\\]\nThe source applies the theorem to these exact definitions. A separate explicit formula relates an $s$-th difference to its original sampled values, so the data dependence is inspectable:\n\\[\n \\Delta_h^s f(n)=\\sum_{j=0}^{s}(-1)^{s-j}\\binom sj f(n+jh).\n\\]\nApplying Newton reconstruction to $\\Delta_1^s f$ and composing the iterated operators proves\n\\[\n \\boxed{\\Delta_1^s f(n)=\\sum_{j=0}^{n}c_{s+j}\\binom nj.}\n\\]\nThe apparently infinite coefficient sum is finite at each natural argument. Neither continuity nor convergence is needed for these identities.","script page 17}\nLet $R$ be any commutative ring. The implementation uses the actual forward-difference operator\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad c_r=\\Delta_1^r f(0).\n\\]\nFor a full sequence $f:\\N\\to R$, pinned Mathlib's Gregory--Newton formula gives\n\\[\n \\boxed{f(n)=\\sum_{r=0}^{n}c_r\\binom nr.}\n\\]\nThe source applies the theorem to these exact definitions. A separate explicit formula relates an $s$-th difference to its original sampled values, so the data dependence is inspectable:\n\\[\n \\Delta_h^s f(n)=\\sum_{j=0}^{s}(-1)^{s-j}\\binom sj f(n+jh).\n\\]\nApplying Newton reconstruction to $\\Delta_1^s f$ and composing the iterated operators proves\n\\[\n \\boxed{\\Delta_1^s f(n)=\\sum_{j=0}^{n}c_{s+j}\\binom nj.}\n\\]\nThe apparently infinite coefficient sum is finite at each natural argument. Neither continuity nor convergence is needed for these identities.","d-difference operator\n\\[\n \\Delta_h f(n)=f(n+h)-f(n),\\qquad c_r=\\Delta_1^r f(0).\n\\]\nFor a full sequence $f:\\N\\to R$, pinned Mathlib's Gregory--Newton formula gives\n\\[\n \\boxed{f(n)=\\sum_{r=0}^{n}c_r\\binom nr.}\n\\]\nThe source applies the theorem to these exact definitions. A separate explicit formula relates an $s$-th difference to its original sampled values, so the data dependence is inspectable:\n\\[\n \\Delta_h^s f(n)=\\sum_{j=0}^{s}(-1)^{s-j}\\binom sj f(n+jh).\n\\]\nApplying Newton reconstruction to $\\Delta_1^s f$ and composing the iterated operators proves\n\\[\n \\boxed{\\Delta_1^s f(n)=\\sum_{j=0}^{n}c_{s+j}\\binom nj.}\n\\]\nThe apparently infinite coefficient sum is finite at each natural argument. Neither continuity nor convergence is needed for these identities.","nce $f:\\N\\to R$, pinned Mathlib's Gregory--Newton formula gives\n\\[\n \\boxed{f(n)=\\sum_{r=0}^{n}c_r\\binom nr.}\n\\]\nThe source applies the theorem to these exact definitions. A separate explicit formula relates an $s$-th difference to its original sampled values, so the data dependence is inspectable:\n\\[\n \\Delta_h^s f(n)=\\sum_{j=0}^{s}(-1)^{s-j}\\binom sj f(n+jh).\n\\]\nApplying Newton reconstruction to $\\Delta_1^s f$ and composing the iterated operators proves\n\\[\n \\boxed{\\Delta_1^s f(n)=\\sum_{j=0}^{n}c_{s+j}\\binom nj.}\n\\]\nThe apparently infinite coefficient sum is finite at each natural argument. Neither continuity nor convergence is needed for these identities.","ves\n\\[\n \\boxed{f(n)=\\sum_{r=0}^{n}c_r\\binom nr.}\n\\]\nThe source applies the theorem to these exact definitions. A separate explicit formula relates an $s$-th difference to its original sampled values, so the data dependence is inspectable:\n\\[\n \\Delta_h^s f(n)=\\sum_{j=0}^{s}(-1)^{s-j}\\binom sj f(n+jh).\n\\]\nApplying Newton reconstruction to $\\Delta_1^s f$ and composing the iterated operators proves\n\\[\n \\boxed{\\Delta_1^s f(n)=\\sum_{j=0}^{n}c_{s+j}\\binom nj.}\n\\]\nThe apparently infinite coefficient sum is finite at each natural argument. Neither continuity nor convergence is needed for these identities.","\\subsection*{An independently defined interpolant}\nFor a finite coefficient list, define\n\\[\n I_N(c)(n)=\\sum_{r=0}^{N}c_r\\binom nr.\n\\]\nThe source proves exact coefficient extraction:\n\\[\n \\Delta_1^s I_N(c)(0)=\n \\begin{cases}c_s,&s\\leq N,","\\subsection*{An independently defined interpolant}\nFor a finite coefficient list, define\n\\[\n I_N(c)(n)=\\sum_{r=0}^{N}c_r\\binom nr.\n\\]\nThe source proves exact coefficient extraction:\n\\[\n \\Delta_1^s I_N(c)(0)=\n \\begin{cases}c_s,&s\\leq N,\\\\0,&s>N.\\end{cases}\n\\]\nIt follows that $I_N(c)$ has the stated shifted-coefficient formula at every order, with zero coefficients beyond $N$. Every difference of order greater than $N$ vanishes. Its coefficients are uniquely determined by its values at $0,\\ldots,N$.","0,&s>N.\\end{cases}\n\\]\nIt follows that $I_N(c)$ has the stated shifted-coefficient formula at every order, with zero coefficients beyond $N$. Every difference of order greater than $N$ vanishes. Its coefficients are uniquely determined by its values at $0,\\ldots,N$.","\\subsection*{What the finite sample does and does not determine}\nFor $c_r=\\Delta_1^r f(0)$ with $r\\leq N$, the source proves\n\\[\n I_N(c)(n)=f(n)\\quad(n\\leq N),\\qquad\n \\Delta_1^s I_N(c)(n)=\\Delta_1^s f(n)\\quad(n+s\\leq N).\n\\]\nThe condition is essential. The sequences $f(n)=n$ and $g(n)=0$ agree on the sample $\\{0\\}$, but $\\Delta f(0)=1$ and $\\Delta g(0)=0$. This is a checked counterexample to unrestricted inference outside the observed stencil.","\\textbf{Principal targets.}\\par\n\\target{Mahler.finite_mahler_reconstruction, Mahler.coefficient_interpolant}\\par\n\\target{Mahler.difference_from_coefficients, Mahler.finite_sample_difference}\n\\par\\medskip\n\\pageheading{2. Weighted bounds in the actual ring $\\Z_p$}{Corollary 7.3; manuscript page 18}\nLet $p$ be prime. For $x\\in\\Z_p$, define\n\\[\n \\ord_p(x)=\\begin{cases}\\infty,&x=0,","\\textbf{Principal targets.}\\par\n\\target{Mahler.finite_mahler_reconstruction, Mahler.coefficient_interpolant}\\par\n\\target{Mahler.difference_from_coefficients, Mahler.finite_sample_difference}\n\\par\\medskip\n\\pageheading{2. Weighted bounds in the actual ring $\\Z_p$}{Corollary 7.3; manuscript page 18}\nLet $p$ be prime. For $x\\in\\Z_p$, define\n\\[\n \\ord_p(x)=\\begin{cases}\\infty,&x=0,\\\\v_p(x),&x\\neq0.\\end{cases}\n\\]\nThe nonzero branch uses Mathlib's actual $p$-adic valuation. The implementation proves its connection to genuine integral divisibility:\n\\[\n \\boxed{L\\leq\\ord_p(x)\\quad\\Longleftrightarrow\\quad p^L\\mid x\n \\ \\text{in }\\Z_p,\\qquad L\\in\\N.}\n\\]\nThis is not divisibility in $\\Q_p$, where a nonzero scalar can divide any field element. The zero case is proved separately, so zero does not spuriously violate a positive lower bound.","v_p(x),&x\\neq0.\\end{cases}\n\\]\nThe nonzero branch uses Mathlib's actual $p$-adic valuation. The implementation proves its connection to genuine integral divisibility:\n\\[\n \\boxed{L\\leq\\ord_p(x)\\quad\\Longleftrightarrow\\quad p^L\\mid x\n \\ \\text{in }\\Z_p,\\qquad L\\in\\N.}\n\\]\nThis is not divisibility in $\\Q_p$, where a nonzero scalar can divide any field element. The zero case is proved separately, so zero does not spuriously violate a positive lower bound.","\\subsection*{The binomial valuation is retained}\nLet $f:\\N\\to\\Z_p$, let $c_r=\\Delta_1^r f(0)$, and suppose\n\\[\n w(s+j)\\leq\\ord_p(c_{s+j})\\quad(0\\leq j\\leq n),\n \\qquad w:\\N\\to\\N\\cup\\{\\infty\\}.\n\\]\nSet\n\\[\n B_p(w;n,s)=\\min_{0\\leq j\\leq n}\n \\left\\{w(s+j)+v_p\\!\\binom nj\\right\\}.\n\\]\nEvery binomial in this range is nonzero. The code proves\n\\[\n \\boxed{B_p(w;n,s)\\leq\\ord_p(\\Delta_1^s f(n)).}\n\\]\nThis is the manuscript minimum after the substitution $r=s+j$. Coefficients outside this range have zero binomial multiplier. Weights equal to infinity are allowed, and a completely vanishing difference has order infinity.","ained}\nLet $f:\\N\\to\\Z_p$, let $c_r=\\Delta_1^r f(0)$, and suppose\n\\[\n w(s+j)\\leq\\ord_p(c_{s+j})\\quad(0\\leq j\\leq n),\n \\qquad w:\\N\\to\\N\\cup\\{\\infty\\}.\n\\]\nSet\n\\[\n B_p(w;n,s)=\\min_{0\\leq j\\leq n}\n \\left\\{w(s+j)+v_p\\!\\binom nj\\right\\}.\n\\]\nEvery binomial in this range is nonzero. The code proves\n\\[\n \\boxed{B_p(w;n,s)\\leq\\ord_p(\\Delta_1^s f(n)).}\n\\]\nThis is the manuscript minimum after the substitution $r=s+j$. Coefficients outside this range have zero binomial multiplier. Weights equal to infinity are allowed, and a completely vanishing difference has order infinity.","For a uniform coefficient hypothesis, the simpler conclusion follows:\n\\[\n \\boxed{\\inf_{j\\geq0} w(s+j)\\leq\\ord_p(\\Delta_1^s f(n)).}\n\\]\nFor each finite $L\\leq B_p(w;n,s)$, a separate theorem yields $p^L\\mid\\Delta_1^s f(n)$ inside $\\Z_p$. Another specialization handles arbitrary finite interpolants with infinite weights beyond their degree.","\\textbf{Principal targets.}\\par\n\\target{PadicBounds.nat_le_order_iff, PadicBounds.weighted_difference_bound}\\par\n\\target{PadicBounds.weighted_difference_divisibility}\\par\n\\target{PadicBounds.coefficient_tail_bound, PadicBounds.interpolant_weighted_bound}\n\\par\\medskip\n\\pageheading{3. Unit modes and residue-field reduction}{Theorem 7.4; manuscript page 18}\nFor $u$ in any commutative ring, and every natural step $h$, iteration proves\n\\[\n \\boxed{\\Delta_h^s(u^n)=u^n(u^h-1)^s.}\n\\]\nThe proof unfolds the actual forward-difference operator and uses the power multiplication law. No valuation premise or geometric-difference identity is assumed. More generally, a proved divisibility $p^k\\mid u^h-1$ implies\n\\[\n p^{ks}\\mid\\Delta_h^s(u^n).\n\\]","pal targets.}\\par\n\\target{PadicBounds.nat_le_order_iff, PadicBounds.weighted_difference_bound}\\par\n\\target{PadicBounds.weighted_difference_divisibility}\\par\n\\target{PadicBounds.coefficient_tail_bound, PadicBounds.interpolant_weighted_bound}\n\\par\\medskip\n\\pageheading{3. Unit modes and residue-field reduction}{Theorem 7.4; manuscript page 18}\nFor $u$ in any commutative ring, and every natural step $h$, iteration proves\n\\[\n \\boxed{\\Delta_h^s(u^n)=u^n(u^h-1)^s.}\n\\]\nThe proof unfolds the actual forward-difference operator and uses the power multiplication law. No valuation premise or geometric-difference identity is assumed. More generally, a proved divisibility $p^k\\mid u^h-1$ implies\n\\[\n p^{ks}\\mid\\Delta_h^s(u^n).\n\\]","et{PadicBounds.nat_le_order_iff, PadicBounds.weighted_difference_bound}\\par\n\\target{PadicBounds.weighted_difference_divisibility}\\par\n\\target{PadicBounds.coefficient_tail_bound, PadicBounds.interpolant_weighted_bound}\n\\par\\medskip\n\\pageheading{3. Unit modes and residue-field reduction}{Theorem 7.4; manuscript page 18}\nFor $u$ in any commutative ring, and every natural step $h$, iteration proves\n\\[\n \\boxed{\\Delta_h^s(u^n)=u^n(u^h-1)^s.}\n\\]\nThe proof unfolds the actual forward-difference operator and uses the power multiplication law. No valuation premise or geometric-difference identity is assumed. More generally, a proved divisibility $p^k\\mid u^h-1$ implies\n\\[\n p^{ks}\\mid\\Delta_h^s(u^n).\n\\]","3. Unit modes and residue-field reduction}{Theorem 7.4; manuscript page 18}\nFor $u$ in any commutative ring, and every natural step $h$, iteration proves\n\\[\n \\boxed{\\Delta_h^s(u^n)=u^n(u^h-1)^s.}\n\\]\nThe proof unfolds the actual forward-difference operator and uses the power multiplication law. No valuation premise or geometric-difference identity is assumed. More generally, a proved divisibility $p^k\\mid u^h-1$ implies\n\\[\n p^{ks}\\mid\\Delta_h^s(u^n).\n\\]","-field reduction}{Theorem 7.4; manuscript page 18}\nFor $u$ in any commutative ring, and every natural step $h$, iteration proves\n\\[\n \\boxed{\\Delta_h^s(u^n)=u^n(u^h-1)^s.}\n\\]\nThe proof unfolds the actual forward-difference operator and uses the power multiplication law. No valuation premise or geometric-difference identity is assumed. More generally, a proved divisibility $p^k\\mid u^h-1$ implies\n\\[\n p^{ks}\\mid\\Delta_h^s(u^n).\n\\]","\\subsection*{Why a $p$-adic unit supplies one factor of $p$}\nNow let $u\\in\\Z_p^\\times$. The actual ring homomorphism\n\\[\n \\pi:\\Z_p\\longrightarrow\\mathbb F_p\n\\]\n sends $u$ to a nonzero unit. Fermat's finite-field theorem therefore gives\n\\[\n \\pi(u)^{p-1}=1,\\qquad \\pi(u^{p-1}-1)=0.\n\\]\nUsing the proved kernel identity $\\ker\\pi=p\\Z_p$, the source derives\n\\[\n \\boxed{p\\mid u^{p-1}-1\\quad\\text{inside }\\Z_p.}\n\\]\nThe conclusion is not supplied as an abstract certificate. It follows from the unit hypothesis and the standard residue-field construction.","Combining it with the geometric identity establishes\n\\[\n \\boxed{p^s\\mid\\Delta_{p-1}^s(u^n),\\qquad\n s\\leq\\ord_p(\\Delta_{p-1}^s(u^n)).}\n\\]\nThese statements hold for every prime, including $p=2$ with step one. Zero differences and order zero are included. Every multiple of the Fermat step satisfies the same lower bound.","\\subsection*{Finite integral linear combinations}\nFor any finite index set $S$, coefficients $a_i\\in\\Z_p$ and units $u_i\\in\\Z_p^\\times$, linearity of the actual iterated operator gives\n\\[\n \\Delta_h^s\\!\\left(\\sum_{i\\in S}a_i u_i^n\\right)\n =\\sum_{i\\in S}a_i u_i^n(u_i^h-1)^s.\n\\]\nAt $h=p-1$, the result is divisible by $p^s$. The empty sum is permitted. This does not prove that the general congruential Euler packets of Theorem 7.5 admit such a representation.","rincipal targets.}\\par\n\\target{UnitModes.geometric_difference, UnitModes.unit_fermat_divisibility}\\par\n\\target{UnitModes.unit_mode_valuation_gain}\\par\n\\target{UnitModes.finite_unit_modes_valuation_gain}\n\\par\\medskip\n\\pageheading{4. Exact examples and adversarial checks}{The examples use the same definitions as the general results}\n\\subsection*{A polynomial anchor}\nThe actual difference table for $f(n)=n^2$ yields\n\\[\n (c_0,c_1,c_2,c_3)=(0,1,2,0).\n\\]\nThe finite interpolant therefore reconstructs $I_2(c)(7)=49$. This checks the sign and indexing conventions, not just the names of the defined objects.","par\n\\target{UnitModes.geometric_difference, UnitModes.unit_fermat_divisibility}\\par\n\\target{UnitModes.unit_mode_valuation_gain}\\par\n\\target{UnitModes.finite_unit_modes_valuation_gain}\n\\par\\medskip\n\\pageheading{4. Exact examples and adversarial checks}{The examples use the same definitions as the general results}\n\\subsection*{A polynomial anchor}\nThe actual difference table for $f(n)=n^2$ yields\n\\[\n (c_0,c_1,c_2,c_3)=(0,1,2,0).\n\\]\nThe finite interpolant therefore reconstructs $I_2(c)(7)=49$. This checks the sign and indexing conventions, not just the names of the defined objects.","ar\n\\target{UnitModes.unit_mode_valuation_gain}\\par\n\\target{UnitModes.finite_unit_modes_valuation_gain}\n\\par\\medskip\n\\pageheading{4. Exact examples and adversarial checks}{The examples use the same definitions as the general results}\n\\subsection*{A polynomial anchor}\nThe actual difference table for $f(n)=n^2$ yields\n\\[\n (c_0,c_1,c_2,c_3)=(0,1,2,0).\n\\]\nThe finite interpolant therefore reconstructs $I_2(c)(7)=49$. This checks the sign and indexing conventions, not just the names of the defined objects.","\\subsection*{A sharp binomial-weight improvement}\nLet\n\\[\n F(n)=81\\binom n1+3\\binom n2+3\\binom n3+81\\binom n4.\n\\]\nAt $n=3$ and $s=1$, the four coefficient valuations are $(4,1,1,4)$, while the binomial valuations are $(0,1,1,0)$. Thus the refined bound is\n\\[\n B_3(w;3,1)=\\min(4,2,2,4)=2,\n\\]\nstronger than the coefficient-only lower bound one. The source checks both the minimum and the actual result:\n\\[\n \\boxed{\\Delta F(3)=180,\\qquad\\ord_3(180)=2.}\n\\]\nThe valuation equality is proved in the actual completed ring $\\Z_3$, with an independent exact integer divisibility check as well.","ubsection*{A sharp binomial-weight improvement}\nLet\n\\[\n F(n)=81\\binom n1+3\\binom n2+3\\binom n3+81\\binom n4.\n\\]\nAt $n=3$ and $s=1$, the four coefficient valuations are $(4,1,1,4)$, while the binomial valuations are $(0,1,1,0)$. Thus the refined bound is\n\\[\n B_3(w;3,1)=\\min(4,2,2,4)=2,\n\\]\nstronger than the coefficient-only lower bound one. The source checks both the minimum and the actual result:\n\\[\n \\boxed{\\Delta F(3)=180,\\qquad\\ord_3(180)=2.}\n\\]\nThe valuation equality is proved in the actual completed ring $\\Z_3$, with an independent exact integer divisibility check as well.","$n=3$ and $s=1$, the four coefficient valuations are $(4,1,1,4)$, while the binomial valuations are $(0,1,1,0)$. Thus the refined bound is\n\\[\n B_3(w;3,1)=\\min(4,2,2,4)=2,\n\\]\nstronger than the coefficient-only lower bound one. The source checks both the minimum and the actual result:\n\\[\n \\boxed{\\Delta F(3)=180,\\qquad\\ord_3(180)=2.}\n\\]\nThe valuation equality is proved in the actual completed ring $\\Z_3$, with an independent exact integer divisibility check as well.","\\subsection*{A sharp unit-mode gain}\nFor $p=5$ and $u=2$,\n\\[\n \\Delta_4^2(u^n)\\big|_{n=0}=(2^4-1)^2=225.\n\\]\nThe source proves $5^2\\mid225$ and $5^3\\nmid225$. This example attains the guaranteed gain; no additional uniform power of $p$ is claimed.","\\subsection*{What fails when hypotheses or conventions are dropped}\nThe nonunit $u=3$ at $p=3$ gives $\\Delta_2(u^n)|_{n=0}=8$, with no factor of three. A constant unit has every positive-order difference zero, hence infinite extended valuation. The finite-sample counterexample on page 2 distinguishes a true interpolant identity from an unsupported claim about unknown continuation.","bottomrule\n\\end{tabular}\n\\end{center}\nThe 61 new theorems include supporting lemmas and examples, not 61 independent discoveries. Pinned Mathlib supplies finite-field, unit-group, ideal-power, $p$-adic residue and Chinese-remainder results. The new modules connect these to the actual packet coefficients, explicit finite lifts and existing selectors.","\\subsubsection*{The specification matters}\nRational coefficients are not reduced through an imaginary map $\\Q\\to\\F_p$. The proof first constructs their integral lifts in $\\Z_p$ and establishes equality in $\\Q_p$. Root representatives are actual units at the requested precision, not unchanged residues that need not satisfy the root equation modulo $p^\\ell$.","reports, and rejected all 17 invalid-proof controls. All 103 infrastructure and release tests passed. Independent finite regressions are recorded separately from the proofs. A separate release audit records the archive's fresh-extraction rebuild and integrity checks.\n\\par\\medskip\n\\pageheading{1. Integral reduction of the original coefficients}{Corollary 8.3; manuscript pages 19--20}\nFix a prime $p$, an integer $m\\geq1$ and a window address $0\\leq j<p$. The earlier project defines the rational packet coefficients by the actual recurrence\n\\[\n \\sum_{q=0}^{n}\\binom{mpn+j}{mpq}C_{mp,j}(q)=\\delta_{n0}.\n\\]\nThe previous Lucas theorem proves\n\\[\n \\binom{mpn+j}{mpq}\\equiv\\binom{mn}{mq}\\pmod p,\\qquad\n \\binom{mpn+j}{mpn}\\equiv1\\pmod p.\n\\]\nThe last congruence makes each diagonal entry a unit in the actual completed ring $\\Z_p$.","jected all 17 invalid-proof controls. All 103 infrastructure and release tests passed. Independent finite regressions are recorded separately from the proofs. A separate release audit records the archive's fresh-extraction rebuild and integrity checks.\n\\par\\medskip\n\\pageheading{1. Integral reduction of the original coefficients}{Corollary 8.3; manuscript pages 19--20}\nFix a prime $p$, an integer $m\\geq1$ and a window address $0\\leq j<p$. The earlier project defines the rational packet coefficients by the actual recurrence\n\\[\n \\sum_{q=0}^{n}\\binom{mpn+j}{mpq}C_{mp,j}(q)=\\delta_{n0}.\n\\]\nThe previous Lucas theorem proves\n\\[\n \\binom{mpn+j}{mpq}\\equiv\\binom{mn}{mq}\\pmod p,\\qquad\n \\binom{mpn+j}{mpn}\\equiv1\\pmod p.\n\\]\nThe last congruence makes each diagonal entry a unit in the actual completed ring $\\Z_p$.","ntrols. All 103 infrastructure and release tests passed. Independent finite regressions are recorded separately from the proofs. A separate release audit records the archive's fresh-extraction rebuild and integrity checks.\n\\par\\medskip\n\\pageheading{1. Integral reduction of the original coefficients}{Corollary 8.3; manuscript pages 19--20}\nFix a prime $p$, an integer $m\\geq1$ and a window address $0\\leq j<p$. The earlier project defines the rational packet coefficients by the actual recurrence\n\\[\n \\sum_{q=0}^{n}\\binom{mpn+j}{mpq}C_{mp,j}(q)=\\delta_{n0}.\n\\]\nThe previous Lucas theorem proves\n\\[\n \\binom{mpn+j}{mpq}\\equiv\\binom{mn}{mq}\\pmod p,\\qquad\n \\binom{mpn+j}{mpn}\\equiv1\\pmod p.\n\\]\nThe last congruence makes each diagonal entry a unit in the actual completed ring $\\Z_p$.","separately from the proofs. A separate release audit records the archive's fresh-extraction rebuild and integrity checks.\n\\par\\medskip\n\\pageheading{1. Integral reduction of the original coefficients}{Corollary 8.3; manuscript pages 19--20}\nFix a prime $p$, an integer $m\\geq1$ and a window address $0\\leq j<p$. The earlier project defines the rational packet coefficients by the actual recurrence\n\\[\n \\sum_{q=0}^{n}\\binom{mpn+j}{mpq}C_{mp,j}(q)=\\delta_{n0}.\n\\]\nThe previous Lucas theorem proves\n\\[\n \\binom{mpn+j}{mpq}\\equiv\\binom{mn}{mq}\\pmod p,\\qquad\n \\binom{mpn+j}{mpn}\\equiv1\\pmod p.\n\\]\nThe last congruence makes each diagonal entry a unit in the actual completed ring $\\Z_p$.","ld and integrity checks.\n\\par\\medskip\n\\pageheading{1. Integral reduction of the original coefficients}{Corollary 8.3; manuscript pages 19--20}\nFix a prime $p$, an integer $m\\geq1$ and a window address $0\\leq j<p$. The earlier project defines the rational packet coefficients by the actual recurrence\n\\[\n \\sum_{q=0}^{n}\\binom{mpn+j}{mpq}C_{mp,j}(q)=\\delta_{n0}.\n\\]\nThe previous Lucas theorem proves\n\\[\n \\binom{mpn+j}{mpq}\\equiv\\binom{mn}{mq}\\pmod p,\\qquad\n \\binom{mpn+j}{mpn}\\equiv1\\pmod p.\n\\]\nThe last congruence makes each diagonal entry a unit in the actual completed ring $\\Z_p$.","q1$ and a window address $0\\leq j<p$. The earlier project defines the rational packet coefficients by the actual recurrence\n\\[\n \\sum_{q=0}^{n}\\binom{mpn+j}{mpq}C_{mp,j}(q)=\\delta_{n0}.\n\\]\nThe previous Lucas theorem proves\n\\[\n \\binom{mpn+j}{mpq}\\equiv\\binom{mn}{mq}\\pmod p,\\qquad\n \\binom{mpn+j}{mpn}\\equiv1\\pmod p.\n\\]\nThe last congruence makes each diagonal entry a unit in the actual completed ring $\\Z_p$.","\\subsection*{First construct the integral solution}\nFor a ring $R$, a triangular row system $T$, right-hand side $b$, and proved unit pivots $d_n=T_{nn}$, define\n\\[\n s_n=d_n^{-1}\\left(b_n-\\sum_{q<n}T_{nq}s_q\\right).\n\\]\nThe new ring-level solver uses well-founded recursion. Its row identities and unique-solution property are proved. The source also proves compatibility of this solver with ring homomorphisms into fields.","Specializing to $R=\\Z_p$ constructs $c^{(p)}_{m,j}(n)$. Mapping to $\\Q_p$ and comparing with the original rational forward substitution gives\n\\[\n \\boxed{c^{(p)}_{m,j}(n)=C_{mp,j}(n)\\quad\\text{in }\\Q_p.}\n\\]\nThus every original standard-window coefficient is $p$-integral, for every row. Integrality is a conclusion, not an additional hypothesis on the coefficients.","\\subsection*{Then apply the residue map}\nIndependently define $e^{(m,p)}$ over $\\F_p$ by the unit-diagonal system\n\\[\n \\sum_{q=0}^{n}\\binom{mn}{mq}e^{(m,p)}_q=\\delta_{n0}.\n\\]\nIts recurrence and uniqueness are checked. Applying the actual homomorphism $\\pi:\\Z_p\\to\\F_p$ to the integral row, and then Lucas reduction, proves\n\\[\n \\boxed{\\pi(c^{(p)}_{m,j}(n))=e^{(m,p)}_n.}\n\\]\nFor $j,k<p$, a separate ideal statement gives\n\\[\n p\\mid c^{(p)}_{m,j}(n)-c^{(p)}_{m,k}(n)\\quad\\text{inside }\\Z_p.\n\\]\nThe formal solver also covers $m=0$; the manuscript uses $m\\geq1$. No global finite-difference theorem or period assertion enters this proof.","ply the residue map}\nIndependently define $e^{(m,p)}$ over $\\F_p$ by the unit-diagonal system\n\\[\n \\sum_{q=0}^{n}\\binom{mn}{mq}e^{(m,p)}_q=\\delta_{n0}.\n\\]\nIts recurrence and uniqueness are checked. Applying the actual homomorphism $\\pi:\\Z_p\\to\\F_p$ to the integral row, and then Lucas reduction, proves\n\\[\n \\boxed{\\pi(c^{(p)}_{m,j}(n))=e^{(m,p)}_n.}\n\\]\nFor $j,k<p$, a separate ideal statement gives\n\\[\n p\\mid c^{(p)}_{m,j}(n)-c^{(p)}_{m,k}(n)\\quad\\text{inside }\\Z_p.\n\\]\nThe formal solver also covers $m=0$; the manuscript uses $m\\geq1$. No global finite-difference theorem or period assertion enters this proof.","um_{q=0}^{n}\\binom{mn}{mq}e^{(m,p)}_q=\\delta_{n0}.\n\\]\nIts recurrence and uniqueness are checked. Applying the actual homomorphism $\\pi:\\Z_p\\to\\F_p$ to the integral row, and then Lucas reduction, proves\n\\[\n \\boxed{\\pi(c^{(p)}_{m,j}(n))=e^{(m,p)}_n.}\n\\]\nFor $j,k<p$, a separate ideal statement gives\n\\[\n p\\mid c^{(p)}_{m,j}(n)-c^{(p)}_{m,k}(n)\\quad\\text{inside }\\Z_p.\n\\]\nThe formal solver also covers $m=0$; the manuscript uses $m\\geq1$. No global finite-difference theorem or period assertion enters this proof.","^{(m,p)}_q=\\delta_{n0}.\n\\]\nIts recurrence and uniqueness are checked. Applying the actual homomorphism $\\pi:\\Z_p\\to\\F_p$ to the integral row, and then Lucas reduction, proves\n\\[\n \\boxed{\\pi(c^{(p)}_{m,j}(n))=e^{(m,p)}_n.}\n\\]\nFor $j,k<p$, a separate ideal statement gives\n\\[\n p\\mid c^{(p)}_{m,j}(n)-c^{(p)}_{m,k}(n)\\quad\\text{inside }\\Z_p.\n\\]\nThe formal solver also covers $m=0$; the manuscript uses $m\\geq1$. No global finite-difference theorem or period assertion enters this proof.","\\textbf{Principal targets.}\\par\n\\target{WindowIndependence.coeffPadic_coe, .standard_window_integral}\\par\n\\target{WindowIndependence.residue_independence, .reduced_unique}\\par\n\\target{WindowIndependence.two_window_addresses_congruent}\n\\par\\medskip\n\\pageheading{2. Explicit finite prime-power decomposition}{Proposition 6.2; odd-prime branch}\nLet $p$ be prime, $\\ell\\geq1$, and\n\\[\n a\\in\\bigl(\\Z/p^\\ell\\Z\\bigr)^\\times.\n\\]\nDefine actual units in the same ring by\n\\[\n \\boxed{\\omega_\\ell(a)=a^{p^{\\ell-1}},\\qquad\n \\langle a\\rangle_\\ell=a\\,\\omega_\\ell(a)^{-1}.}\n\\]\nThe Lean source writes $\\ell=L+1$, so zero precision is excluded without truncated natural-number subtraction.","ffPadic_coe, .standard_window_integral}\\par\n\\target{WindowIndependence.residue_independence, .reduced_unique}\\par\n\\target{WindowIndependence.two_window_addresses_congruent}\n\\par\\medskip\n\\pageheading{2. Explicit finite prime-power decomposition}{Proposition 6.2; odd-prime branch}\nLet $p$ be prime, $\\ell\\geq1$, and\n\\[\n a\\in\\bigl(\\Z/p^\\ell\\Z\\bigr)^\\times.\n\\]\nDefine actual units in the same ring by\n\\[\n \\boxed{\\omega_\\ell(a)=a^{p^{\\ell-1}},\\qquad\n \\langle a\\rangle_\\ell=a\\,\\omega_\\ell(a)^{-1}.}\n\\]\nThe Lean source writes $\\ell=L+1$, so zero precision is excluded without truncated natural-number subtraction.","{WindowIndependence.two_window_addresses_congruent}\n\\par\\medskip\n\\pageheading{2. Explicit finite prime-power decomposition}{Proposition 6.2; odd-prime branch}\nLet $p$ be prime, $\\ell\\geq1$, and\n\\[\n a\\in\\bigl(\\Z/p^\\ell\\Z\\bigr)^\\times.\n\\]\nDefine actual units in the same ring by\n\\[\n \\boxed{\\omega_\\ell(a)=a^{p^{\\ell-1}},\\qquad\n \\langle a\\rangle_\\ell=a\\,\\omega_\\ell(a)^{-1}.}\n\\]\nThe Lean source writes $\\ell=L+1$, so zero precision is excluded without truncated natural-number subtraction.","\\subsection*{The root equation and residue are derived}\nEuler's unit-group exponent gives\n\\[\n \\omega_\\ell(a)^{p-1}\n =a^{p^{\\ell-1}(p-1)}=a^{\\varphi(p^\\ell)}=1.\n\\]\nReduction to the actual field $\\F_p$ commutes with powers. Its Frobenius identity gives\n\\[\n \\overline{\\omega_\\ell(a)}\n =\\bar a^{p^{\\ell-1}}=\\bar a.\n\\]\nConsequently,\n\\[\n \\boxed{a=\\omega_\\ell(a)\\langle a\\rangle_\\ell,\\quad\n \\omega_\\ell(a)^{p-1}=1,\\quad\n \\overline{\\langle a\\rangle_\\ell}=1.}\n\\]\nThe manuscript's finite existence conclusion is obtained directly. The proof does not assume a Hensel lift or an already supplied root representative.","\\subsection*{Uniqueness requires a lifting argument}\nThe source proves the actual kernel statement\n\\[\n \\bar x=0\\quad\\Longleftrightarrow\\quad p\\mid x\n \\quad\\text{in }\\Z/p^\\ell\\Z.\n\\]\nFor $x\\equiv y\\pmod p$, the checked ideal-power lifting theorem gives\n\\[\n x^{p^{\\ell-1}}-y^{p^{\\ell-1}}\\in(p)^\\ell=(0).\n\\]\nIf $w^{p-1}=1$, then $w^p=w$ and hence $w^{p^k}=w$ for every $k$. Therefore two roots with the same residue are equal. The decomposition is unique whenever its second factor reduces to one. Multiplicativity of both factors is also proved.","For $p=3,\\ell=2,a=2$, the checked values are\n\\[\n \\omega_2(2)=8,\\qquad \\langle2\\rangle_2=7,\\qquad\n 8^2=1,\\quad8\\cdot7=2\\quad\\text{in }\\Z/9\\Z.\n\\]\nThe unlifted representative $2$ is not a square root of one modulo nine.","\\textbf{Principal targets.}\\par\n\\target{FiniteUnits.finite_unit_decomposition, .power_lift_congruence}\\par\n\\target{FiniteUnits.omega_unique, .decomposition_unique}\\par\n\\target{FiniteUnits.omega_mul, .principal_mul}\n\\par\\medskip\n\\pageheading{3. Precision compatibility and the dyadic sign}{Finite refinements and Proposition 6.2's dyadic branch}\nFor $1\\leq k\\leq\\ell$, let\n\\[\n \\rho_{\\ell,k}:\\Z/p^\\ell\\Z\\longrightarrow\\Z/p^k\\Z\n\\]\nbe the actual reduction homomorphism. It induces a map on units. The explicit construction satisfies\n\\[\n \\boxed{\\rho_{\\ell,k}(\\omega_\\ell(a))\n =\\omega_k(\\rho_{\\ell,k}(a)),}\n\\]\n\\[\n \\boxed{\\rho_{\\ell,k}(\\langle a\\rangle_\\ell)\n =\\langle\\rho_{\\ell,k}(a)\\rangle_k.}\n\\]\nThe proof uses the exponent factorization\n$p^{\\ell-1}=p^{k-1}p^{\\ell-k}$ and the already proved Frobenius invariance of a lifted root. Principal factors also satisfy\n\\[\n \\langle a\\rangle_\\ell^{\\,p^{\\ell-1}}=1.\n\\]\nAt precision one the principal factor is exactly one. These are coherent finite certificates; no infinite inverse limit or topological unit-group equivalence is","\\textbf{Principal targets.}\\par\n\\target{FiniteUnits.finite_unit_decomposition, .power_lift_congruence}\\par\n\\target{FiniteUnits.omega_unique, .decomposition_unique}\\par\n\\target{FiniteUnits.omega_mul, .principal_mul}\n\\par\\medskip\n\\pageheading{3. Precision compatibility and the dyadic sign}{Finite refinements and Proposition 6.2's dyadic branch}\nFor $1\\leq k\\leq\\ell$, let\n\\[\n \\rho_{\\ell,k}:\\Z/p^\\ell\\Z\\longrightarrow\\Z/p^k\\Z\n\\]\nbe the actual reduction homomorphism. It induces a map on units. The explicit construction satisfies\n\\[\n \\boxed{\\rho_{\\ell,k}(\\omega_\\ell(a))\n =\\omega_k(\\rho_{\\ell,k}(a)),}\n\\]\n\\[\n \\boxed{\\rho_{\\ell,k}(\\langle a\\rangle_\\ell)\n =\\langle\\rho_{\\ell,k}(a)\\rangle_k.}\n\\]\nThe proof uses the exponent factorization\n$p^{\\ell-1}=p^{k-1}p^{\\ell-k}$ and the already proved Frobenius invariance of a lifted root. Principal factors also satisfy\n\\[\n \\langle a\\rangle_\\ell^{\\,p^{\\ell-1}}=1.\n\\]\nAt precision one the principal factor is exactly one. These are coherent finite certificates; no infinite inverse limit or topological unit-group equivalence is constructed in this release.","a_unique, .decomposition_unique}\\par\n\\target{FiniteUnits.omega_mul, .principal_mul}\n\\par\\medskip\n\\pageheading{3. Precision compatibility and the dyadic sign}{Finite refinements and Proposition 6.2's dyadic branch}\nFor $1\\leq k\\leq\\ell$, let\n\\[\n \\rho_{\\ell,k}:\\Z/p^\\ell\\Z\\longrightarrow\\Z/p^k\\Z\n\\]\nbe the actual reduction homomorphism. It induces a map on units. The explicit construction satisfies\n\\[\n \\boxed{\\rho_{\\ell,k}(\\omega_\\ell(a))\n =\\omega_k(\\rho_{\\ell,k}(a)),}\n\\]\n\\[\n \\boxed{\\rho_{\\ell,k}(\\langle a\\rangle_\\ell)\n =\\langle\\rho_{\\ell,k}(a)\\rangle_k.}\n\\]\nThe proof uses the exponent factorization\n$p^{\\ell-1}=p^{k-1}p^{\\ell-k}$ and the already proved Frobenius invariance of a lifted root. Principal factors also satisfy\n\\[\n \\langle a\\rangle_\\ell^{\\,p^{\\ell-1}}=1.\n\\]\nAt precision one the principal factor is exactly one. These are coherent finite certificates; no infinite inverse limit or topological unit-group equivalence is constructed in this release.","pal_mul}\n\\par\\medskip\n\\pageheading{3. Precision compatibility and the dyadic sign}{Finite refinements and Proposition 6.2's dyadic branch}\nFor $1\\leq k\\leq\\ell$, let\n\\[\n \\rho_{\\ell,k}:\\Z/p^\\ell\\Z\\longrightarrow\\Z/p^k\\Z\n\\]\nbe the actual reduction homomorphism. It induces a map on units. The explicit construction satisfies\n\\[\n \\boxed{\\rho_{\\ell,k}(\\omega_\\ell(a))\n =\\omega_k(\\rho_{\\ell,k}(a)),}\n\\]\n\\[\n \\boxed{\\rho_{\\ell,k}(\\langle a\\rangle_\\ell)\n =\\langle\\rho_{\\ell,k}(a)\\rangle_k.}\n\\]\nThe proof uses the exponent factorization\n$p^{\\ell-1}=p^{k-1}p^{\\ell-k}$ and the already proved Frobenius invariance of a lifted root. Principal factors also satisfy\n\\[\n \\langle a\\rangle_\\ell^{\\,p^{\\ell-1}}=1.\n\\]\nAt precision one the principal factor is exactly one. These are coherent finite certificates; no infinite inverse limit or topological unit-group equivalence is constructed in this release.","c branch}\nFor $1\\leq k\\leq\\ell$, let\n\\[\n \\rho_{\\ell,k}:\\Z/p^\\ell\\Z\\longrightarrow\\Z/p^k\\Z\n\\]\nbe the actual reduction homomorphism. It induces a map on units. The explicit construction satisfies\n\\[\n \\boxed{\\rho_{\\ell,k}(\\omega_\\ell(a))\n =\\omega_k(\\rho_{\\ell,k}(a)),}\n\\]\n\\[\n \\boxed{\\rho_{\\ell,k}(\\langle a\\rangle_\\ell)\n =\\langle\\rho_{\\ell,k}(a)\\rangle_k.}\n\\]\nThe proof uses the exponent factorization\n$p^{\\ell-1}=p^{k-1}p^{\\ell-k}$ and the already proved Frobenius invariance of a lifted root. Principal factors also satisfy\n\\[\n \\langle a\\rangle_\\ell^{\\,p^{\\ell-1}}=1.\n\\]\nAt precision one the principal factor is exactly one. These are coherent finite certificates; no infinite inverse limit or topological unit-group equivalence is constructed in this release.","hism. It induces a map on units. The explicit construction satisfies\n\\[\n \\boxed{\\rho_{\\ell,k}(\\omega_\\ell(a))\n =\\omega_k(\\rho_{\\ell,k}(a)),}\n\\]\n\\[\n \\boxed{\\rho_{\\ell,k}(\\langle a\\rangle_\\ell)\n =\\langle\\rho_{\\ell,k}(a)\\rangle_k.}\n\\]\nThe proof uses the exponent factorization\n$p^{\\ell-1}=p^{k-1}p^{\\ell-k}$ and the already proved Frobenius invariance of a lifted root. Principal factors also satisfy\n\\[\n \\langle a\\rangle_\\ell^{\\,p^{\\ell-1}}=1.\n\\]\nAt precision one the principal factor is exactly one. These are coherent finite certificates; no infinite inverse limit or topological unit-group equivalence is constructed in this release.","\\subsection*{Why the mod-four branch is separate}\nNow let $\\ell\\geq2$ and $a\\in(\\Z/2^\\ell\\Z)^\\times$. The actual units of $\\Z/4\\Z$ are exhausted by $1$ and $-1$. Define\n\\[\n \\varepsilon_\\ell(a)=\n \\begin{cases}1,&a\\equiv1\\pmod4,","\\subsection*{Why the mod-four branch is separate}\nNow let $\\ell\\geq2$ and $a\\in(\\Z/2^\\ell\\Z)^\\times$. The actual units of $\\Z/4\\Z$ are exhausted by $1$ and $-1$. Define\n\\[\n \\varepsilon_\\ell(a)=\n \\begin{cases}1,&a\\equiv1\\pmod4,\\\\-1,&a\\equiv3\\pmod4,\\end{cases}\n \\qquad\n \\langle a\\rangle_{2,\\ell}=\\varepsilon_\\ell(a)a.\n\\]\nThe source uses $\\ell=L+2$. It proves\n\\[\n \\boxed{a=\\varepsilon_\\ell(a)\\langle a\\rangle_{2,\\ell},\\quad\n \\varepsilon_\\ell(a)\\in\\{1,-1\\},\\quad\n \\langle a\\rangle_{2,\\ell}\\equiv1\\pmod4.}\n\\]\nThe sign is uniquely determined by the residue modulo four, and cancellation gives uniqueness of the principal factor.","-1,&a\\equiv3\\pmod4,\\end{cases}\n \\qquad\n \\langle a\\rangle_{2,\\ell}=\\varepsilon_\\ell(a)a.\n\\]\nThe source uses $\\ell=L+2$. It proves\n\\[\n \\boxed{a=\\varepsilon_\\ell(a)\\langle a\\rangle_{2,\\ell},\\quad\n \\varepsilon_\\ell(a)\\in\\{1,-1\\},\\quad\n \\langle a\\rangle_{2,\\ell}\\equiv1\\pmod4.}\n\\]\nThe sign is uniquely determined by the residue modulo four, and cancellation gives uniqueness of the principal factor.","The checked example $a=3$ modulo eight has sign $7=-1$ and principal factor $5$:\n\\[\n 7\\cdot5=3\\pmod8,\\qquad5=1\\pmod4.\n\\]\nThe code separately proves $1=-1$ in $\\Z/2\\Z$ but $1\\neq-1$ in $\\Z/4\\Z$. Modulo two alone cannot identify the sign. The generic prime-power construction at $p=2$ only asks for a principal residue modulo two and is not silently substituted for this stronger mod-four factorization.","\\textbf{Principal targets.}\\par\n\\target{UnitRefinement.omega_compatible, .principal_compatible, .principal_power}\\par\n\\target{DyadicUnits.dyadic_unit_decomposition, .decomposition_unique}\n\\par\\medskip\n\\pageheading{4. CRT addresses and the actual packet selectors}{Theorem 10.1; manuscript page 22}\nLet $a\\geq0$, let $M$ be odd, and put $N=2^aM$. Coprimality is derived from oddness:\n\\[\n \\gcd(2^a,M)=1.\n\\]\nThe construction uses the actual Chinese-remainder ring equivalence\n\\[\n \\boxed{\\Z/N\\Z\\ \\simeq\\ \\Z/2^a\\Z\\times\\Z/M\\Z.}\n\\]\nIts bijectivity and component formula are checked. On a natural representative $r$, the components are precisely its reductions modulo the two factors, not arbitrary coordinate labels.","rincipal_compatible, .principal_power}\\par\n\\target{DyadicUnits.dyadic_unit_decomposition, .decomposition_unique}\n\\par\\medskip\n\\pageheading{4. CRT addresses and the actual packet selectors}{Theorem 10.1; manuscript page 22}\nLet $a\\geq0$, let $M$ be odd, and put $N=2^aM$. Coprimality is derived from oddness:\n\\[\n \\gcd(2^a,M)=1.\n\\]\nThe construction uses the actual Chinese-remainder ring equivalence\n\\[\n \\boxed{\\Z/N\\Z\\ \\simeq\\ \\Z/2^a\\Z\\times\\Z/M\\Z.}\n\\]\nIts bijectivity and component formula are checked. On a natural representative $r$, the components are precisely its reductions modulo the two factors, not arbitrary coordinate labels.","heading{4. CRT addresses and the actual packet selectors}{Theorem 10.1; manuscript page 22}\nLet $a\\geq0$, let $M$ be odd, and put $N=2^aM$. Coprimality is derived from oddness:\n\\[\n \\gcd(2^a,M)=1.\n\\]\nThe construction uses the actual Chinese-remainder ring equivalence\n\\[\n \\boxed{\\Z/N\\Z\\ \\simeq\\ \\Z/2^a\\Z\\times\\Z/M\\Z.}\n\\]\nIts bijectivity and component formula are checked. On a natural representative $r$, the components are precisely its reductions modulo the two factors, not arbitrary coordinate labels.","\\subsection*{The algebraic bridge to projectors}\nFor natural indices $n,r$, the source proves\n\\[\n n\\equiv r\\pmod{2^aM}\n \\quad\\Longleftrightarrow\\quad\n \\bigl(n\\equiv r\\pmod{2^a}\\text{ and }n\\equiv r\\pmod M\\bigr).\n\\]\nFor the original coefficient selector\n\\[\n (P_{N,r}f)(n)=\n \\begin{cases}f(n),&n\\bmod N=r,","\\subsection*{The algebraic bridge to projectors}\nFor natural indices $n,r$, the source proves\n\\[\n n\\equiv r\\pmod{2^aM}\n \\quad\\Longleftrightarrow\\quad\n \\bigl(n\\equiv r\\pmod{2^a}\\text{ and }n\\equiv r\\pmod M\\bigr).\n\\]\nFor the original coefficient selector\n\\[\n (P_{N,r}f)(n)=\n \\begin{cases}f(n),&n\\bmod N=r,\\\\0,&\\text{otherwise},\\end{cases}\n\\]\ntake the canonical label $0\\leq r<N$. The logical equivalence yields the actual equality of sequence operators\n\\[\n \\boxed{P_{N,r}f=\n P_{2^a,\\,r\\bmod2^a}\\bigl(P_{M,\\,r\\bmod M}f\\bigr).}\n\\]\nThis statement holds for coefficients in any type with zero. A separate theorem lifts the equality coefficientwise to the existing formal power-series selectors over any semiring.","0,&\\text{otherwise},\\end{cases}\n\\]\ntake the canonical label $0\\leq r<N$. The logical equivalence yields the actual equality of sequence operators\n\\[\n \\boxed{P_{N,r}f=\n P_{2^a,\\,r\\bmod2^a}\\bigl(P_{M,\\,r\\bmod M}f\\bigr).}\n\\]\nThis statement holds for coefficients in any type with zero. A separate theorem lifts the equality coefficientwise to the existing formal power-series selectors over any semiring.","The cases $a=0$ and $M=1$ are included. The corresponding one-element residue ring is not excluded. Oddness of $M$ already implies positivity.","Coprimality cannot be discarded. The distinct classes $0$ and $2$ modulo four both map to $(0,0)$ when both components are reduced modulo two. The source checks this counterexample. It also checks a composite-modulus failure of the Fermat exponent and a coefficient outside the standard window that is not $3$-adically integral.","_bijective, .address_components, .residue_equivalence}\\par\n\\target{CRTPackets.packet_factorization, .series_packet_factorization}\\par\n\\target{FiniteArithmeticExamples.outside_window_not_integral}\n\\par\\medskip\n\\pageheading{5. Verification, regressions and remaining work}{Cumulative release evidence}\nThe fresh build started with an empty output directory under Lean 4.34.0-rc2 and Mathlib revision\n\\[\n \\mcode{aeb72f1e9981b6d35b426b16dca7759be3e7ba06}.\n\\]\nAll nine dependency revisions matched the runtime manifest and repository heads.","\\subsubsection*{The main mathematical connection}\nA sampled state $u_k=v_{Nk+j}$ evolves by the actual jump matrix $C^N$. Its recurrence is derived from the characteristic polynomial of that matrix. No scalar recurrence or spectral decomposition is supplied as a substitute for the desired conclusion.","For Fibonacci numbers, the state is independently defined from Mathlib's actual sequence. The Lucas sequence is independently defined by its initial values and recurrence. The trace and determinant of $C^N$ are then proved to supply the sampled recurrence coefficients.","\\subsubsection*{Important qualification}\nA $d$-dimensional state system supplies a recurrence of \\textbf{order at most $d$}. Its minimal recurrence need not have order $d$. The original statement is read as the characteristic-polynomial recurrence, not a minimal-order claim. The formal equations also handle singular matrices, zero-dimensional spaces and zero step.","\\par\\medskip\n\\pageheading{1. Actual matrix orbits and arithmetic jumps}{Theorem 9.1; manuscript page 22}\nLet $R$ be a commutative ring, $I$ a finite coordinate type and $d=|I|$. For $C\\in M_I(R)$ and $v_0\\in R^I$, define\n\\[\n \\operatorname{orbit}(C,v_0,n)=C^nv_0,\n \\qquad u_k=\\operatorname{packet}(C,v_0,N,j,k)=C^{Nk+j}v_0.\n\\]\nThe implementation uses actual matrix powers and matrix--vector multiplication. A separately supplied sequence $v$ with\n\\[\n v_{n+1}=Cv_n\n\\]\nis proved by induction to equal this power orbit. Thus the source-sequence formulation is connected to the construction.","trix orbits and arithmetic jumps}{Theorem 9.1; manuscript page 22}\nLet $R$ be a commutative ring, $I$ a finite coordinate type and $d=|I|$. For $C\\in M_I(R)$ and $v_0\\in R^I$, define\n\\[\n \\operatorname{orbit}(C,v_0,n)=C^nv_0,\n \\qquad u_k=\\operatorname{packet}(C,v_0,N,j,k)=C^{Nk+j}v_0.\n\\]\nThe implementation uses actual matrix powers and matrix--vector multiplication. A separately supplied sequence $v$ with\n\\[\n v_{n+1}=Cv_n\n\\]\nis proved by induction to equal this power orbit. Thus the source-sequence formulation is connected to the construction.","s and arithmetic jumps}{Theorem 9.1; manuscript page 22}\nLet $R$ be a commutative ring, $I$ a finite coordinate type and $d=|I|$. For $C\\in M_I(R)$ and $v_0\\in R^I$, define\n\\[\n \\operatorname{orbit}(C,v_0,n)=C^nv_0,\n \\qquad u_k=\\operatorname{packet}(C,v_0,N,j,k)=C^{Nk+j}v_0.\n\\]\nThe implementation uses actual matrix powers and matrix--vector multiplication. A separately supplied sequence $v$ with\n\\[\n v_{n+1}=Cv_n\n\\]\nis proved by induction to equal this power orbit. Thus the source-sequence formulation is connected to the construction.","\\subsection*{The jump formula}\nAssociativity and the laws for powers of one matrix give\n\\[\n \\boxed{u_k=C^j(C^N)^kv_0=(C^N)^k(C^jv_0).}\n\\]\nOnly powers of the same matrix are interchanged. The argument does not assume that arbitrary matrices commute. In particular,\n\\[\n u_{k+1}=C^Nu_k,\\qquad u_{k+a}=(C^N)^a u_k.\n\\]\nThe step matrix is $C^N$, not $C$. A checked example with the Fibonacci matrix and $N=2$ rejects replacing the jump with a single original step.","\\subsection*{Refinement of an arithmetic address}\nTaking another subsequence with step $M$ and offset $s$ yields\n\\[\n u_{Mk+s}=v_{(NM)k+(Ns+j)}.\n\\]\nThe formal proof establishes this identity for the actual packet function by checking the index arithmetic. It is not inferred from a diagram or a naming convention.","\\subsection*{Endpoints and interpretation}\nThe algebraic identities allow $N=0$, in which case $u_k=v_j$ is constant. Positive-modulus residue packets are the special case $N>0$ and $0\\le j<N$. The formula also permits offsets beyond that standard range. No inverse matrix, diagonalization, eigenvalue, or characteristic-zero assumption is needed.","\\textbf{Principal targets.}\\par\n\\target{Companion.orbit_of_recurrence}\\par\n\\target{Companion.companion_jump, Companion.source_companion_jump}\\par\n\\target{Companion.packet_step, Companion.packet_advance}\\par\n\\target{Companion.packet_refinement, Companion.packet_zero_step}\n\\par\\medskip\n\\pageheading{2. From the characteristic polynomial to a recurrence}{Theorem 9.1, continued}\nPut $D=C^N$ and let\n\\[\n \\chi_D(X)=\\det(XI-D)=X^d+\\sum_{a=0}^{d-1}c_aX^a\n\\]\nbe its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands polynomial evaluation into a finite sum and proves\n\\[\n \\sum_{a=0}^{d}\\chi_D[a]D^a=0,\n \\qquad \\chi_D[d]=1.\n\\]\nHere $\\chi_D[a]$ denotes the coefficient at index $a$. The identities hold over any commutative coefficient ring; the zero-ring case is handled explicitly by the proof's subsingleton branch.","packet_refinement, Companion.packet_zero_step}\n\\par\\medskip\n\\pageheading{2. From the characteristic polynomial to a recurrence}{Theorem 9.1, continued}\nPut $D=C^N$ and let\n\\[\n \\chi_D(X)=\\det(XI-D)=X^d+\\sum_{a=0}^{d-1}c_aX^a\n\\]\nbe its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands polynomial evaluation into a finite sum and proves\n\\[\n \\sum_{a=0}^{d}\\chi_D[a]D^a=0,\n \\qquad \\chi_D[d]=1.\n\\]\nHere $\\chi_D[a]$ denotes the coefficient at index $a$. The identities hold over any commutative coefficient ring; the zero-ring case is handled explicitly by the proof's subsingleton branch.","ageheading{2. From the characteristic polynomial to a recurrence}{Theorem 9.1, continued}\nPut $D=C^N$ and let\n\\[\n \\chi_D(X)=\\det(XI-D)=X^d+\\sum_{a=0}^{d-1}c_aX^a\n\\]\nbe its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands polynomial evaluation into a finite sum and proves\n\\[\n \\sum_{a=0}^{d}\\chi_D[a]D^a=0,\n \\qquad \\chi_D[d]=1.\n\\]\nHere $\\chi_D[a]$ denotes the coefficient at index $a$. The identities hold over any commutative coefficient ring; the zero-ring case is handled explicitly by the proof's subsingleton branch.","ial to a recurrence}{Theorem 9.1, continued}\nPut $D=C^N$ and let\n\\[\n \\chi_D(X)=\\det(XI-D)=X^d+\\sum_{a=0}^{d-1}c_aX^a\n\\]\nbe its actual characteristic polynomial. The implementation reuses the pinned Mathlib Cayley--Hamilton theorem, expands polynomial evaluation into a finite sum and proves\n\\[\n \\sum_{a=0}^{d}\\chi_D[a]D^a=0,\n \\qquad \\chi_D[d]=1.\n\\]\nHere $\\chi_D[a]$ denotes the coefficient at index $a$. The identities hold over any commutative coefficient ring; the zero-ring case is handled explicitly by the proof's subsingleton branch.","\\subsection*{The complete vector recurrence}\nApply the matrix identity to the actual state $u_k$. The already proved advance identity identifies each $D^a u_k$ with $u_{k+a}$. Consequently,\n\\[\n \\sum_{a=0}^{d}\\chi_D[a]u_{k+a}=0,\n \\qquad\n \\boxed{u_{k+d}=-\\sum_{a=0}^{d-1}c_a u_{k+a}.}\n\\]\nThis is the desired recurrence, with its monic leading coefficient and all earlier coefficients supplied by $\\chi_D$. It is not an assumption on the packet sequence.","Every coordinate and every $R$-linear scalar observation $\\lambda:R^I\\to R$ satisfies the corresponding equation:\n\\[\n \\lambda(u_{k+d})=-\\sum_{a=0}^{d-1}c_a\\lambda(u_{k+a}).\n\\]\nA separate source-sequence theorem transports the recurrence back to the original $v_{Nk+j}$ notation.","\\subsection*{Order versus minimal order}\nThe characteristic polynomial gives a recurrence of order at most $d$. For example, $C=I_2$ gives a degree-two characteristic polynomial, but every orbit is constant and obeys $u_{k+1}=u_k$. Both statements are checked. No minimal-polynomial classification is claimed.","For $d=0$, the unique empty vector is zero and the formula remains valid. For singular matrices, Cayley--Hamilton still applies: invertibility is irrelevant. The nilpotent example $C=\\left(\\begin{smallmatrix}0&1","For $d=0$, the unique empty vector is zero and the formula remains valid. For singular matrices, Cayley--Hamilton still applies: invertibility is irrelevant. The nilpotent example $C=\\left(\\begin{smallmatrix}0&1\\\\0&0\\end{smallmatrix}\\right)$ has $C^2=0$ and $\\chi_C=X^2$.","0&0\\end{smallmatrix}\\right)$ has $C^2=0$ and $\\chi_C=X^2$.","s, Companion.charpoly_coeff_dimension}\\par\n\\target{Companion.packet_charpoly_annihilation}\\par\n\\target{Companion.packet_charpoly_recurrence}\\par\n\\target{Companion.packet_coordinate_recurrence}\\par\n\\target{Companion.packet_observable_recurrence}\n\\par\\medskip\n\\pageheading{3. Fibonacci and Lucas coefficients are derived}{Corollary 9.2; manuscript page 22}\nDefine the actual integer matrix and state\n\\[\n C=\\begin{pmatrix}1&1\\\\1&0\\end{pmatrix},\n \\qquad v_n=\\begin{pmatrix}F_{n+1}\\\\F_n\\end{pmatrix},\n \\qquad v_0=\\begin{pmatrix}1\\\\0\\end{pmatrix}.\n\\]\nThe values $F_n$ are Mathlib's \\mcode{Nat.fib}, cast into $\\Z$. Its basic recurrence proves $v_{n+1}=Cv_n$. The second coordinate of the packet state is therefore proved to be the intended $F_{Nk+j}$, not a newly assumed sequence.","0\\end{pmatrix}.\n\\]\nThe values $F_n$ are Mathlib's \\mcode{Nat.fib}, cast into $\\Z$. Its basic recurrence proves $v_{n+1}=Cv_n$. The second coordinate of the packet state is therefore proved to be the intended $F_{Nk+j}$, not a newly assumed sequence.","\\subsection*{An independent Lucas sequence}\nDefine\n\\[\n L_0=2,\\quad L_1=1,\\quad L_{n+2}=L_n+L_{n+1}.\n\\]\nDirect finite matrix arithmetic proves $C^2=C+I$. Multiplication by $C^n$ gives $C^{n+2}=C^n+C^{n+1}$. Taking traces shows that $\\operatorname{tr}(C^n)$ satisfies the same recurrence and initial values as $L_n$. Induction yields\n\\[\n \\boxed{\\operatorname{tr}(C^N)=L_N.}\n\\]\nThe implementation separately proves\n\\[\n \\det C=-1,\\qquad \\det(C^N)=(-1)^N,\n \\qquad L_N=2F_{N+1}-F_N.\n\\]\nNo Binet roots or diagonalizability argument is needed.","n independent Lucas sequence}\nDefine\n\\[\n L_0=2,\\quad L_1=1,\\quad L_{n+2}=L_n+L_{n+1}.\n\\]\nDirect finite matrix arithmetic proves $C^2=C+I$. Multiplication by $C^n$ gives $C^{n+2}=C^n+C^{n+1}$. Taking traces shows that $\\operatorname{tr}(C^n)$ satisfies the same recurrence and initial values as $L_n$. Induction yields\n\\[\n \\boxed{\\operatorname{tr}(C^N)=L_N.}\n\\]\nThe implementation separately proves\n\\[\n \\det C=-1,\\qquad \\det(C^N)=(-1)^N,\n \\qquad L_N=2F_{N+1}-F_N.\n\\]\nNo Binet roots or diagonalizability argument is needed.","\\subsection*{The sampled Fibonacci recurrence}\nThe actual two-by-two characteristic-polynomial formula gives\n\\[\n \\chi_{C^N}(X)=X^2-L_NX+(-1)^N.\n\\]\nInsert these coefficients into the general packet recurrence and take the second coordinate:\n\\[\n \\boxed{F_{N(k+2)+j}=L_NF_{N(k+1)+j}-(-1)^NF_{Nk+j}.}\n\\]\nThe full two-coordinate state satisfies the same equation. All subtraction is in $\\Z$, not truncated natural-number subtraction. The result holds for all natural $N,j,k$, including zero step and zero initial values.","y-two characteristic-polynomial formula gives\n\\[\n \\chi_{C^N}(X)=X^2-L_NX+(-1)^N.\n\\]\nInsert these coefficients into the general packet recurrence and take the second coordinate:\n\\[\n \\boxed{F_{N(k+2)+j}=L_NF_{N(k+1)+j}-(-1)^NF_{Nk+j}.}\n\\]\nThe full two-coordinate state satisfies the same equation. All subtraction is in $\\Z$, not truncated natural-number subtraction. The result holds for all natural $N,j,k$, including zero step and zero initial values.","s.state_is_orbit, FibonacciPackets.packet_is_fibonacci}\\par\n\\target{FibonacciPackets.trace_power_is_lucas, FibonacciPackets.det_power}\\par\n\\target{FibonacciPackets.jump_charpoly}\\par\n\\target{FibonacciPackets.fibonacci_packet_recurrence}\n\\par\\medskip\n\\pageheading{4. Exact examples and meaningful boundary checks}{Example 9.3 and supporting results}\nThe independently defined Lucas recurrence gives\n\\[\n (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11).\n\\]\nFor $N=3$, the general theorem therefore specializes to\n\\[\n u_{k+2}=4u_{k+1}+u_k.\n\\]\nThe implementation proves this for every offset $j$ and every $k$, not only the displayed sample. At $j=2$, direct evaluation of the actual Fibonacci sequence gives\n\\[\n (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).\n\\]\nThis is the exact manuscript example.","ci}\\par\n\\target{FibonacciPackets.trace_power_is_lucas, FibonacciPackets.det_power}\\par\n\\target{FibonacciPackets.jump_charpoly}\\par\n\\target{FibonacciPackets.fibonacci_packet_recurrence}\n\\par\\medskip\n\\pageheading{4. Exact examples and meaningful boundary checks}{Example 9.3 and supporting results}\nThe independently defined Lucas recurrence gives\n\\[\n (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11).\n\\]\nFor $N=3$, the general theorem therefore specializes to\n\\[\n u_{k+2}=4u_{k+1}+u_k.\n\\]\nThe implementation proves this for every offset $j$ and every $k$, not only the displayed sample. At $j=2$, direct evaluation of the actual Fibonacci sequence gives\n\\[\n (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).\n\\]\nThis is the exact manuscript example.",", FibonacciPackets.det_power}\\par\n\\target{FibonacciPackets.jump_charpoly}\\par\n\\target{FibonacciPackets.fibonacci_packet_recurrence}\n\\par\\medskip\n\\pageheading{4. Exact examples and meaningful boundary checks}{Example 9.3 and supporting results}\nThe independently defined Lucas recurrence gives\n\\[\n (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11).\n\\]\nFor $N=3$, the general theorem therefore specializes to\n\\[\n u_{k+2}=4u_{k+1}+u_k.\n\\]\nThe implementation proves this for every offset $j$ and every $k$, not only the displayed sample. At $j=2$, direct evaluation of the actual Fibonacci sequence gives\n\\[\n (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).\n\\]\nThis is the exact manuscript example.","r\n\\target{FibonacciPackets.fibonacci_packet_recurrence}\n\\par\\medskip\n\\pageheading{4. Exact examples and meaningful boundary checks}{Example 9.3 and supporting results}\nThe independently defined Lucas recurrence gives\n\\[\n (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11).\n\\]\nFor $N=3$, the general theorem therefore specializes to\n\\[\n u_{k+2}=4u_{k+1}+u_k.\n\\]\nThe implementation proves this for every offset $j$ and every $k$, not only the displayed sample. At $j=2$, direct evaluation of the actual Fibonacci sequence gives\n\\[\n (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).\n\\]\nThis is the exact manuscript example.","nacciPackets.fibonacci_packet_recurrence}\n\\par\\medskip\n\\pageheading{4. Exact examples and meaningful boundary checks}{Example 9.3 and supporting results}\nThe independently defined Lucas recurrence gives\n\\[\n (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11).\n\\]\nFor $N=3$, the general theorem therefore specializes to\n\\[\n u_{k+2}=4u_{k+1}+u_k.\n\\]\nThe implementation proves this for every offset $j$ and every $k$, not only the displayed sample. At $j=2$, direct evaluation of the actual Fibonacci sequence gives\n\\[\n (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).\n\\]\nThis is the exact manuscript example.","e}\n\\par\\medskip\n\\pageheading{4. Exact examples and meaningful boundary checks}{Example 9.3 and supporting results}\nThe independently defined Lucas recurrence gives\n\\[\n (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11).\n\\]\nFor $N=3$, the general theorem therefore specializes to\n\\[\n u_{k+2}=4u_{k+1}+u_k.\n\\]\nThe implementation proves this for every offset $j$ and every $k$, not only the displayed sample. At $j=2$, direct evaluation of the actual Fibonacci sequence gives\n\\[\n (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).\n\\]\nThis is the exact manuscript example.","ndary checks}{Example 9.3 and supporting results}\nThe independently defined Lucas recurrence gives\n\\[\n (L_0,L_1,L_2,L_3,L_4,L_5)=(2,1,3,4,7,11).\n\\]\nFor $N=3$, the general theorem therefore specializes to\n\\[\n u_{k+2}=4u_{k+1}+u_k.\n\\]\nThe implementation proves this for every offset $j$ and every $k$, not only the displayed sample. At $j=2$, direct evaluation of the actual Fibonacci sequence gives\n\\[\n (F_2,F_5,F_8,F_{11},F_{14},F_{17})=(1,5,21,89,377,1597).\n\\]\nThis is the exact manuscript example.","\\subsection*{The parity sign cannot be removed}\nFor $N=2$, the recurrence is\n\\[\n u_{k+2}=3u_{k+1}-u_k.\n\\]\nAt $j=1,k=0$, the correct equation is $5=3\\cdot2-1$. Replacing the minus sign with a plus would assert $5=7$. The source proves the inequality, and a separate invalid-proof control attempts and rejects the false equation.","At $N=1$ the ordinary Fibonacci recurrence is recovered. At $N=0$, $L_0=2$ and the packet is constant, so the same formula becomes $u=2u-u$.","\\subsection*{Uniqueness from the first two packet values}\nFor a fixed $N,j$, any integer sequence satisfying the proved sampled recurrence and starting at $F_j,F_{N+j}$ agrees with $F_{Nk+j}$ for all $k$. This is proved by two-step induction. It is a consequence of the established coefficients and recurrence, not a premise used to identify Fibonacci values initially.","\\subsubsection*{The independent route}\nIn characteristic $p$, both the first and $p$th coefficient shifts obey a Leibniz rule for Hurwitz convolution. A section supported at multiples of $m$, with $m\\mid p-1$, has equal first and $p$th shifts. Its constructed reciprocal inherits this equality. Identifying that reciprocal with the existing reduced Euler sequence gives its period on the positive-index tail.","\\subsubsection*{What remains}\nTheorem 7.5 asks for powers as high as $p^{\\min(k,mn)}$ and covers nonstandard residue classes. A period modulo $p$ does not establish that stronger result. This release proves a genuine standard-window mod-$p$ fragment and precise conditional interfaces for the remaining argument, while keeping item 7.5 pending.","\\par\\medskip\n\\pageheading{1. Coefficient products and reciprocal differentiation}{A characteristic-$p$ argument without factorial division}\nFor sequences $f,g:\\N\\to K$, where $K$ is a field, define\n\\[\n (f\\star g)(d)=\\sum_{i=0}^{d}\\binom di f(i)g(d-i),\n \\qquad (U^a f)(d)=f(d+a).\n\\]\nThese are Hurwitz coefficients. The notation does not divide by $d!$ in $\\F_p$. In particular, the implementation is not ordinary power-series multiplication with undefined inverse factorials in positive characteristic.","The identity sequence is $\\mathbf1(d)=\\delta_{d0}$. The new source proves the actual higher product rule\n\\[\n U^a(f\\star g)=\\sum_{i=0}^{a}\\binom ai\n       (U^i f)\\star(U^{a-i}g).\n\\]\nThe argument starts with Pascal's identity for the first shift, then inducts on $a$. All sums are finite at each coefficient.","\\subsection*{The prime-shift cancellation}\nFor prime $p$ and characteristic $p$, every interior binomial coefficient in row $p$ vanishes. Hence\n\\[\n U^p(f\\star g)=(U^pf)\\star g+f\\star(U^pg).\n\\]\nThe first-shift rule has exactly the same form. Both are checked for the literal coefficient product.","Suppose $g(0)=1$. The source constructs $f$ by the full triangular recurrence\n\\[\n f(d)=\\delta_{d0}-\\sum_{i<d}\\binom di f(i)g(d-i)\n\\]\nand proves $f\\star g=\\mathbf1$. It also proves cancellation: if $a\\star g=0$, then $a=0$, by induction on the first nonzero coefficient.","If $U^pg=Ug$, differentiate the inverse equation with both operators and subtract. The terms involving $f\\star(Ug)$ cancel, leaving\n\\[\n (U^pf-Uf)\\star g=0.\n\\]\nThe proved cancellation law yields\n\\[\n \\boxed{U^pf=Uf.}\n\\]\nThe section identity is an explicit premise of this general reciprocal lemma; the next page proves that premise for the specific section used by the Euler sequence.","\\par\\medskip\n\\pageheading{2. Identification with the reduced Euler sequence}{Corollary 8.5; manuscript page 21}\nWork over $\\F_p=\\Z/p\\Z$. For positive $m\\mid p-1$, take the literal section\n\\[\n g_m(d)=\\begin{cases}1,&m\\mid d,","\\par\\medskip\n\\pageheading{2. Identification with the reduced Euler sequence}{Corollary 8.5; manuscript page 21}\nWork over $\\F_p=\\Z/p\\Z$. For positive $m\\mid p-1$, take the literal section\n\\[\n g_m(d)=\\begin{cases}1,&m\\mid d,\\\\0,&m\\nmid d.\\end{cases}\n\\]\nIt satisfies $g_m(0)=1$. Since $p\\equiv1\\pmod m$, the conditions $m\\mid d+p$ and $m\\mid d+1$ are equivalent. The source therefore proves $U^pg_m=Ug_m$.","0,&m\\nmid d.\\end{cases}\n\\]\nIt satisfies $g_m(0)=1$. Since $p\\equiv1\\pmod m$, the conditions $m\\mid d+p$ and $m\\mid d+1$ are equivalent. The source therefore proves $U^pg_m=Ug_m$.","Let $f_m$ be the reciprocal constructed on the previous page. A separate strong-induction proof establishes its actual support:\n\\[\n m\\nmid d\\quad\\Longrightarrow\\quad f_m(d)=0.\n\\]\nThis support is not included as an assumption of the construction.","\\subsection*{The same residual sequence, not a renamed substitute}\nEvaluate $f_m\\star g_m=\\mathbf1$ at $d=mn$. Reindex the surviving summation indices by $i=mq$ using the proved support. The result is\n\\[\n \\sum_{q=0}^{n}\\binom{mn}{mq}f_m(mq)=\\delta_{n0}.\n\\]\nThe fifth block independently constructed the unique sequence $e_n^{(m,p)}$ satisfying this triangular system. Applying that uniqueness theorem gives\n\\[\n \\boxed{f_m(mn)=e_n^{(m,p)}.}\n\\]\nThus the reciprocal really is connected to the manuscript's Lucas-reduced inverse.","on*{The same residual sequence, not a renamed substitute}\nEvaluate $f_m\\star g_m=\\mathbf1$ at $d=mn$. Reindex the surviving summation indices by $i=mq$ using the proved support. The result is\n\\[\n \\sum_{q=0}^{n}\\binom{mn}{mq}f_m(mq)=\\delta_{n0}.\n\\]\nThe fifth block independently constructed the unique sequence $e_n^{(m,p)}$ satisfying this triangular system. Applying that uniqueness theorem gives\n\\[\n \\boxed{f_m(mn)=e_n^{(m,p)}.}\n\\]\nThus the reciprocal really is connected to the manuscript's Lucas-reduced inverse.","\\subsection*{The positive-tail period}\nWrite $h=(p-1)/m$. For $n\\ge1$, evaluate $U^pf_m=Uf_m$ at the nonnegative degree $mn-1$:\n\\[\n f_m(mn+p-1)=f_m(mn).\n\\]\nSince $mn+p-1=m(n+h)$, the identified sequence satisfies\n\\[\n \\boxed{e_{n+h}^{(m,p)}=e_n^{(m,p)}\\quad(n\\ge1).}\n\\]\nInduction gives $e_{n+uh}^{(m,p)}=e_n^{(m,p)}$ for every $u\\ge0$. This proves a period, not necessarily the least positive period. The proof also includes $p=2$, where $m=1$ and $h=1$.","-tail period}\nWrite $h=(p-1)/m$. For $n\\ge1$, evaluate $U^pf_m=Uf_m$ at the nonnegative degree $mn-1$:\n\\[\n f_m(mn+p-1)=f_m(mn).\n\\]\nSince $mn+p-1=m(n+h)$, the identified sequence satisfies\n\\[\n \\boxed{e_{n+h}^{(m,p)}=e_n^{(m,p)}\\quad(n\\ge1).}\n\\]\nInduction gives $e_{n+uh}^{(m,p)}=e_n^{(m,p)}$ for every $u\\ge0$. This proves a period, not necessarily the least positive period. The proof also includes $p=2$, where $m=1$ and $h=1$.","Write $h=(p-1)/m$. For $n\\ge1$, evaluate $U^pf_m=Uf_m$ at the nonnegative degree $mn-1$:\n\\[\n f_m(mn+p-1)=f_m(mn).\n\\]\nSince $mn+p-1=m(n+h)$, the identified sequence satisfies\n\\[\n \\boxed{e_{n+h}^{(m,p)}=e_n^{(m,p)}\\quad(n\\ge1).}\n\\]\nInduction gives $e_{n+uh}^{(m,p)}=e_n^{(m,p)}$ for every $u\\ge0$. This proves a period, not necessarily the least positive period. The proof also includes $p=2$, where $m=1$ and $h=1$.","(p-1)/m$. For $n\\ge1$, evaluate $U^pf_m=Uf_m$ at the nonnegative degree $mn-1$:\n\\[\n f_m(mn+p-1)=f_m(mn).\n\\]\nSince $mn+p-1=m(n+h)$, the identified sequence satisfies\n\\[\n \\boxed{e_{n+h}^{(m,p)}=e_n^{(m,p)}\\quad(n\\ge1).}\n\\]\nInduction gives $e_{n+uh}^{(m,p)}=e_n^{(m,p)}$ for every $u\\ge0$. This proves a period, not necessarily the least positive period. The proof also includes $p=2$, where $m=1$ and $h=1$.","\\par\\medskip\n\\pageheading{3. Actual packet congruences and sharp boundaries}{Consequences for the standard window; not the full Theorem 7.5}\nThe fifth block constructed integral lifts $\\widehat C_{mp,j}(n)\\in\\Z_p$ of the original rational coefficients when $0\\le j<p$, and proved that their residues equal $e_n^{(m,p)}$.","Applying the actual residue map and its kernel ideal to the new period gives\n\\[\n \\boxed{p\\mid\\widehat C_{mp,j}(n+uh)-\\widehat C_{mp,j}(n)\n       \\quad\\hbox{in }\\Z_p}\n\\]\nfor $n\\ge1$ and $u\\ge0$. Every higher positive difference is a difference of terms already divisible by $p$. Consequently,\n\\[\n p\\mid\\Delta_{uh}^{k}\\widehat C_{mp,j}(n)\\qquad(k\\ge1,\\ n\\ge1).\n\\]\nA ring-homomorphism lemma identifies the image of these differences in $\\Q_p$ with the differences of the original rational coefficient function. This is not vacuous field divisibility. The valuation version uses order infinity at zero.","d its kernel ideal to the new period gives\n\\[\n \\boxed{p\\mid\\widehat C_{mp,j}(n+uh)-\\widehat C_{mp,j}(n)\n       \\quad\\hbox{in }\\Z_p}\n\\]\nfor $n\\ge1$ and $u\\ge0$. Every higher positive difference is a difference of terms already divisible by $p$. Consequently,\n\\[\n p\\mid\\Delta_{uh}^{k}\\widehat C_{mp,j}(n)\\qquad(k\\ge1,\\ n\\ge1).\n\\]\nA ring-homomorphism lemma identifies the image of these differences in $\\Q_p$ with the differences of the original rational coefficient function. This is not vacuous field divisibility. The valuation version uses order infinity at zero.","\\subsection*{Three checked boundaries}\n\\textbf{The initial row.} For $(p,m)=(3,2)$, the same reduced recurrence gives $e_0=1$ and $e_1=-1=2$ in $\\F_3$. Its proposed period is one, so the positive-tail result cannot in general include row zero.","\\textbf{The divisor hypothesis.} At $(p,m)=(5,3)$, where $3\\nmid4$, the independently computed reduced row is\n\\[\n (e_0,e_1,e_2,e_3)=(1,4,4,2).\n\\]\nReplacing $h$ by the integer quotient $\\lfloor4/3\\rfloor=1$ does not give a period: $e_3\\ne e_2$.","\\textbf{No automatic extra power.} The original rational triangular coefficients satisfy\n\\[\n C_{6,0}(1)=-1,\\quad C_{6,0}(2)=923,\\quad\n v_3\\bigl(C_{6,0}(2)-C_{6,0}(1)\\bigr)=v_3(924)=1.\n\\]\nA period modulo $3$ therefore cannot simply be reported as a period modulo $9$. This is consistent with the global theorem's proposed first-difference bound; it is not a counterexample to Theorem 7.5.","\\par\\medskip\n\\pageheading{4. Exact interfaces for the remaining global argument}{Conditional results are not counted as completion of item 7.5}\nLet $p$ be prime and $m$ a positive divisor of $p-1$. For a sequence $e$ in a commutative coefficient ring, define its actual packed Hurwitz sequence\n\\[\n X(d)=\\begin{cases}e(d/N),&N\\mid d,","\\par\\medskip\n\\pageheading{4. Exact interfaces for the remaining global argument}{Conditional results are not counted as completion of item 7.5}\nLet $p$ be prime and $m$ a positive divisor of $p-1$. For a sequence $e$ in a commutative coefficient ring, define its actual packed Hurwitz sequence\n\\[\n X(d)=\\begin{cases}e(d/N),&N\\mid d,\\\\0,&N\\nmid d,\\end{cases}\n \\qquad N=mp.\n\\]\nLet $L=p(p-1)$ and $\\delta=U^{p+L}-U^p=U^{p^2}-U^p$. The source proves the operator dictionary\n\\[\n (\\delta^sX)(d)=\\Delta_L^s X(d+ps).\n\\]\nAt the valid nonnegative coefficient degree $d=mpn-ps$, with $s\\le mn$, this becomes\n\\[\n (\\delta^sX)(mpn-ps)=\\Delta_h^s e(n),\\qquad h=(p-1)/m.\n\\]\nThe packed sequence and difference operations are definitions. Their coefficient correspondence is proved before any divisibility assumption is introduced.","0,&N\\nmid d,\\end{cases}\n \\qquad N=mp.\n\\]\nLet $L=p(p-1)$ and $\\delta=U^{p+L}-U^p=U^{p^2}-U^p$. The source proves the operator dictionary\n\\[\n (\\delta^sX)(d)=\\Delta_L^s X(d+ps).\n\\]\nAt the valid nonnegative coefficient degree $d=mpn-ps$, with $s\\le mn$, this becomes\n\\[\n (\\delta^sX)(mpn-ps)=\\Delta_h^s e(n),\\qquad h=(p-1)/m.\n\\]\nThe packed sequence and difference operations are definitions. Their coefficient correspondence is proved before any divisibility assumption is introduced.","\\subsection*{A sufficient condition, explicitly conditional}\nDefine flatness by the actual coefficientwise divisibility assertion\n\\[\n \\delta^sX(d)\\in p^sR\\qquad(s,d\\ge0).\n\\]\nThe new theorem proves that this hypothesis implies\n\\[\n \\boxed{p^{\\min(k,mn)}\\mid\\Delta_h^k e(n).}\n\\]\nFirst, coefficient extraction proves the triangular region $k\\le mn$. A separate induction on difference order proves that this region is equivalent to the complete saturated estimate. This implication is useful over $\\Z_p$; interpreting it over a field would not supply the intended integral content.","fficientwise divisibility assertion\n\\[\n \\delta^sX(d)\\in p^sR\\qquad(s,d\\ge0).\n\\]\nThe new theorem proves that this hypothesis implies\n\\[\n \\boxed{p^{\\min(k,mn)}\\mid\\Delta_h^k e(n).}\n\\]\nFirst, coefficient extraction proves the triangular region $k\\le mn$. A separate induction on difference order proves that this region is equivalent to the complete saturated estimate. This implication is useful over $\\Z_p$; interpreting it over a field would not supply the intended integral content.","\\subsection*{Normalization is also an explicit implication}\nUsing the existing definition $E^{(N,j)}_{Nn}=j!C_{N,j}(n)$, the source proves that multiplication by $j!$ commutes with every iterated difference. If the unnormalized difference has precision $A$, the normalized difference has precision $A-v_p(j!)$. Precision is defined as\n\\[\n x=0\\quad\\hbox{or}\\quad v_p(x)\\ge A,\n\\]\nso zero and negative factorial-loss thresholds are handled correctly. For $j<p$, the loss is zero.","ation}\nUsing the existing definition $E^{(N,j)}_{Nn}=j!C_{N,j}(n)$, the source proves that multiplication by $j!$ commutes with every iterated difference. If the unnormalized difference has precision $A$, the normalized difference has precision $A-v_p(j!)$. Precision is defined as\n\\[\n x=0\\quad\\hbox{or}\\quad v_p(x)\\ge A,\n\\]\nso zero and negative factorial-loss thresholds are handled correctly. For $j<p$, the loss is zero.","\\subsection*{Independent exact finite checks}\nThe new rational and finite-field mirrors passed 4,848 reciprocal coefficient rows, 2,592 higher-Leibniz cases, 4,520 reciprocal prime-shift cases, 3,720 period comparisons, and 1,476 standard-window rational rows. They also checked 1,104 rational period differences, 620 higher differences modulo $p$, and 586 Dwork extraction interfaces. Eleven invalid input configurations were rejected.","omplete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and result}\nThe all-residue Theorem 7.5 remains pending. This release proves its \\textbf{full saturated higher-power estimate throughout the standard window}, rather than only a period modulo $p$. For every prime $p$, positive $m\\mid p-1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","etains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and result}\nThe all-residue Theorem 7.5 remains pending. This release proves its \\textbf{full saturated higher-power estimate throughout the standard window}, rather than only a period modulo $p$. For every prime $p$, positive $m\\mid p-1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","tus of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and result}\nThe all-residue Theorem 7.5 remains pending. This release proves its \\textbf{full saturated higher-power estimate throughout the standard window}, rather than only a period modulo $p$. For every prime $p$, positive $m\\mid p-1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","ted original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and result}\nThe all-residue Theorem 7.5 remains pending. This release proves its \\textbf{full saturated higher-power estimate throughout the standard window}, rather than only a period modulo $p$. For every prime $p$, positive $m\\mid p-1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and result}\nThe all-residue Theorem 7.5 remains pending. This release proves its \\textbf{full saturated higher-power estimate throughout the standard window}, rather than only a period modulo $p$. For every prime $p$, positive $m\\mid p-1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","ing or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and result}\nThe all-residue Theorem 7.5 remains pending. This release proves its \\textbf{full saturated higher-power estimate throughout the standard window}, rather than only a period modulo $p$. For every prime $p$, positive $m\\mid p-1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","rage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and result}\nThe all-residue Theorem 7.5 remains pending. This release proves its \\textbf{full saturated higher-power estimate throughout the standard window}, rather than only a period modulo $p$. For every prime $p$, positive $m\\mid p-1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.",". This release proves its \\textbf{full saturated higher-power estimate throughout the standard window}, rather than only a period modulo $p$. For every prime $p$, positive $m\\mid p-1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","every prime $p$, positive $m\\mid p-1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","\\mid p-1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","1$, $0\\le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","le j<p$, and natural $n,k,u$, put $h=(p-1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","1)/m$. Then\n\\[\n\\boxed{\\quad\n\\Delta_{uh}^{k}C_{mp,j}(n)=0\\quad\\text{or}\\quad\nv_p\\!\\left(\\Delta_{uh}^{k}C_{mp,j}(n)\\right)\\ge\\min\\{k,mn\\}.\n\\quad}\n\\]\nThe unnormalized family $j!C_{mp,j}(n)$ satisfies the same bound. Both statements concern the original rational coefficient function, identified with its actual integral $p$-adic lift. They include $p=2$, $k=0$, $n=0$, and the additional harmless case $u=0$.","\\textbf{Coverage remains 23 of 24 named manuscript-proved items.} Three classical inputs and two conjectures remain separate. The new result covers every permissible residue when $m=1$, but not the range $p\\le j<mp$ when $m>1$.","\\par\\medskip\n\\pageheading{1. The characteristic-zero product rule}{The normalized defect is not an ordinary derivation}\nLet $A$ be a commutative algebra over a characteristic-zero field $K$, and let $U$ be an actual $K$-derivation. The source defines\n\\[\n B=\\frac{U^{p^2}-U^p}{p},\\qquad\n M_q(f,g)=\\sum_{a=1}^{q-1}\\frac{1}{p}\\binom qa\n          (U^a f)(U^{q-a}g),\\quad q=p,p^2.\n\\]\nThe coefficients of each $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coe","\\par\\medskip\n\\pageheading{1. The characteristic-zero product rule}{The normalized defect is not an ordinary derivation}\nLet $A$ be a commutative algebra over a characteristic-zero field $K$, and let $U$ be an actual $K$-derivation. The source defines\n\\[\n B=\\frac{U^{p^2}-U^p}{p},\\qquad\n M_q(f,g)=\\sum_{a=1}^{q-1}\\frac{1}{p}\\binom qa\n          (U^a f)(U^{q-a}g),\\quad q=p,p^2.\n\\]\nThe coefficients of each $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is","\\par\\medskip\n\\pageheading{1. The characteristic-zero product rule}{The normalized defect is not an ordinary derivation}\nLet $A$ be a commutative algebra over a characteristic-zero field $K$, and let $U$ be an actual $K$-derivation. The source defines\n\\[\n B=\\frac{U^{p^2}-U^p}{p},\\qquad\n M_q(f,g)=\\sum_{a=1}^{q-1}\\frac{1}{p}\\binom qa\n          (U^a f)(U^{q-a}g),\\quad q=p,p^2.\n\\]\nThe coefficients of each $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This in","\\par\\medskip\n\\pageheading{1. The characteristic-zero product rule}{The normalized defect is not an ordinary derivation}\nLet $A$ be a commutative algebra over a characteristic-zero field $K$, and let $U$ be an actual $K$-derivation. The source defines\n\\[\n B=\\frac{U^{p^2}-U^p}{p},\\qquad\n M_q(f,g)=\\sum_{a=1}^{q-1}\\frac{1}{p}\\binom qa\n          (U^a f)(U^{q-a}g),\\quad q=p,p^2.\n\\]\nThe coefficients of each $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itsel","\\par\\medskip\n\\pageheading{1. The characteristic-zero product rule}{The normalized defect is not an ordinary derivation}\nLet $A$ be a commutative algebra over a characteristic-zero field $K$, and let $U$ be an actual $K$-derivation. The source defines\n\\[\n B=\\frac{U^{p^2}-U^p}{p},\\qquad\n M_q(f,g)=\\sum_{a=1}^{q-1}\\frac{1}{p}\\binom qa\n          (U^a f)(U^{q-a}g),\\quad q=p,p^2.\n\\]\nThe coefficients of each $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itself Lean-checked.","medskip\n\\pageheading{1. The characteristic-zero product rule}{The normalized defect is not an ordinary derivation}\nLet $A$ be a commutative algebra over a characteristic-zero field $K$, and let $U$ be an actual $K$-derivation. The source defines\n\\[\n B=\\frac{U^{p^2}-U^p}{p},\\qquad\n M_q(f,g)=\\sum_{a=1}^{q-1}\\frac{1}{p}\\binom qa\n          (U^a f)(U^{q-a}g),\\quad q=p,p^2.\n\\]\nThe coefficients of each $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itself Lean-checked.","ordinary derivation}\nLet $A$ be a commutative algebra over a characteristic-zero field $K$, and let $U$ be an actual $K$-derivation. The source defines\n\\[\n B=\\frac{U^{p^2}-U^p}{p},\\qquad\n M_q(f,g)=\\sum_{a=1}^{q-1}\\frac{1}{p}\\binom qa\n          (U^a f)(U^{q-a}g),\\quad q=p,p^2.\n\\]\nThe coefficients of each $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itself Lean-checked.","\\sum_{a=1}^{q-1}\\frac{1}{p}\\binom qa\n          (U^a f)(U^{q-a}g),\\quad q=p,p^2.\n\\]\nThe coefficients of each $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itself Lean-checked.","efficients of each $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itself Lean-checked.","nts of each $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itself Lean-checked.","h $M_q$ are proved integral: the middle entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itself Lean-checked.","le entries of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itself Lean-checked.","of a prime-power binomial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itself Lean-checked.","ial row are divisible by $p$. In Lean they are implemented as natural quotients, with exact division proved before the product identity is used. Induction on derivative order establishes the full Leibniz formula and hence\n\\[\n\\boxed{B(fg)=B(f)g+fB(g)+M_{p^2}(f,g)-M_p(f,g).}\n\\]\nThe middle terms are essential. At $p=3$, $f=t$ and $g=t^2$, the constant coefficient of $B(fg)$ is $-2$, whereas that of $B(f)g+fB(g)$ is zero. This inequality is itself Lean-checked.","\\subsection*{All orders, without a circular closure assumption}\nLet $I\\subset A$ be a subring stable under $U$. Define\n\\[\n \\operatorname{Bounded}_r(f)\\iff B^s f\\in I\\ (0\\le s\\le r),\n \\qquad \\operatorname{Flat}(f)\\iff B^s f\\in I\\ (s\\ge0).\n\\]\nThe definitions contain no product or inverse closure requirement. Since $B$ commutes with $U$, derivatives of a bounded element remain bounded. Induction on $r$ in the displayed product rule then proves closure under multiplication. The finite middle sums use the lower-order induction hypothesis, not the desired next-order conclusion.","For an actual inverse $fg=1$, applying the product rule and multiplying by $g$ gives\n\\[\n B(g)=-g^2B(f)-gM_{p^2}(f,g)+gM_p(f,g).\n\\]\nIf $g\\in I$, induction on $r$ proves boundedness of $g$ from boundedness of $f$. At each successor stage the preceding stage supplies boundedness of $g$ needed on the right. Passing to all finite orders proves reciprocal closure for an integral inverse.","\\par\\medskip\n\\pageheading{2. The actual integral Hurwitz lattice}{Connecting differential expressions to coefficient divisibility}\nThe specialization works inside $\\Q_p[[t]]$ and uses factorial-scaled coefficients\n\\[\n a_f(d)=d![t^d]f,\\qquad\n I_p=\\{f\\in\\Q_p[[t]]:a_f(d)\\in\\Z_p\\text{ for every }d\\}.\n\\]\nThe source constructs $I_p$ as a subring. Actual exponential-series multiplication gives\n\\[\n a_{fg}(d)=\\sum_{r=0}^{d}\\binom dr a_f(r)a_g(d-r),\n \\qquad a_{Uf}(d)=a_f(d+1).\n\\]\nThese identities prove multiplication and derivative stability of the lattice. Membership does not require ordinary power-series coefficients to lie in $\\Z_p$.","t^d]f,\\qquad\n I_p=\\{f\\in\\Q_p[[t]]:a_f(d)\\in\\Z_p\\text{ for every }d\\}.\n\\]\nThe source constructs $I_p$ as a subring. Actual exponential-series multiplication gives\n\\[\n a_{fg}(d)=\\sum_{r=0}^{d}\\binom dr a_f(r)a_g(d-r),\n \\qquad a_{Uf}(d)=a_f(d+1).\n\\]\nThese identities prove multiplication and derivative stability of the lattice. Membership does not require ordinary power-series coefficients to lie in $\\Z_p$.","\\subsection*{The reciprocal is genuinely integral}\nFor $f\\in I_p$ with constant coefficient one, the usual power-series reciprocal is used. Its inverse equation yields the exact coefficient recurrence\n\\[\n a_{f^{-1}}(d)=\\delta_{d0}-\\sum_{r=0}^{d-1}\n     \\binom dr a_{f^{-1}}(r)a_f(d-r).\n\\]\nStrong induction proves that all of these coefficients are integral. Combining this construction with the preceding all-order theorem proves flatness of the \\emph{actual reciprocal}. No flatness or integrality of that reciprocal is assumed in the normalized-series specialization.","\\subsection*{The earlier coefficient criterion is recovered exactly}\nDefine the literal coefficient defect\n\\[\n (\\delta a)(d)=a(d+p^2)-a(d+p).\n\\]\nThe new source proves, for every $s,d$,\n\\[\n a_{B^s f}(d)=p^{-s}(\\delta^s a_f)(d).\n\\]\nFor an integral coefficient $x$, it proves the bridge\n\\[\n\\boxed{p^{-s}x\\in\\Z_p\\quad\\Longleftrightarrow\\quad p^s\\mid x\\text{ in }\\Z_p.}\n\\]\nThus the differential definition is equivalent to the existing coefficient-level Dwork criterion. Division occurs in $\\Q_p$, while the assertion after division is membership in $\\Z_p$. It is not divisibility in the field. Zero coefficients satisfy every finite precision condition.","l coefficient defect\n\\[\n (\\delta a)(d)=a(d+p^2)-a(d+p).\n\\]\nThe new source proves, for every $s,d$,\n\\[\n a_{B^s f}(d)=p^{-s}(\\delta^s a_f)(d).\n\\]\nFor an integral coefficient $x$, it proves the bridge\n\\[\n\\boxed{p^{-s}x\\in\\Z_p\\quad\\Longleftrightarrow\\quad p^s\\mid x\\text{ in }\\Z_p.}\n\\]\nThus the differential definition is equivalent to the existing coefficient-level Dwork criterion. Division occurs in $\\Q_p$, while the assertion after division is membership in $\\Z_p$. It is not divisibility in the field. Zero coefficients satisfy every finite precision condition.","he new source proves, for every $s,d$,\n\\[\n a_{B^s f}(d)=p^{-s}(\\delta^s a_f)(d).\n\\]\nFor an integral coefficient $x$, it proves the bridge\n\\[\n\\boxed{p^{-s}x\\in\\Z_p\\quad\\Longleftrightarrow\\quad p^s\\mid x\\text{ in }\\Z_p.}\n\\]\nThus the differential definition is equivalent to the existing coefficient-level Dwork criterion. Division occurs in $\\Q_p$, while the assertion after division is membership in $\\Z_p$. It is not divisibility in the field. Zero coefficients satisfy every finite precision condition.","\\par\\medskip\n\\pageheading{3. A nonsingular Euler-operator argument}{The standard interval without an analytic power-series-in-the-index premise}\nOn Hurwitz coefficients, the actual Euler operator plus $c$ is\n\\[\n (\\Lambda_c a)(d)=(d+c)a(d).\n\\]\nDirect induction proves its commutator with every defect iterate:\n\\[\n \\delta^s\\Lambda_c a\n =\\Lambda_c\\delta^s a\n  +sp\\bigl(pU^{p^2}-U^p\\bigr)\\delta^{s-1}a,\n \\qquad s\\ge1.\n\\]\nSuppose $a$ is integral and supported at degrees divisible by $p$, and $p\\nmid c$. Every supported pivot $d+c$ is proved to be a unit of $\\Z_p$. The defect preserves this support. Induction in the commutator identity therefore proves\n\\[\n \\operatorname{Flat}(\\Lambda_c a)\\Longrightarrow\\operatorname{Flat}(a).\n\\]\nThis is a reflection theorem on the integral supported domain. It does not assume flatness of the proposed preimage, and does not assert invertibility at a singular pivot $p\\","\\par\\medskip\n\\pageheading{3. A nonsingular Euler-operator argument}{The standard interval without an analytic power-series-in-the-index premise}\nOn Hurwitz coefficients, the actual Euler operator plus $c$ is\n\\[\n (\\Lambda_c a)(d)=(d+c)a(d).\n\\]\nDirect induction proves its commutator with every defect iterate:\n\\[\n \\delta^s\\Lambda_c a\n =\\Lambda_c\\delta^s a\n  +sp\\bigl(pU^{p^2}-U^p\\bigr)\\delta^{s-1}a,\n \\qquad s\\ge1.\n\\]\nSuppose $a$ is integral and supported at degrees divisible by $p$, and $p\\nmid c$. Every supported pivot $d+c$ is proved to be a unit of $\\Z_p$. The defect preserves this support. Induction in the commutator identity therefore proves\n\\[\n \\operatorname{Flat}(\\Lambda_c a)\\Longrightarrow\\operatorname{Flat}(a).\n\\]\nThis is a reflection theorem on the integral supported domain. It does not assume flatness of the proposed preimage, and does not assert invertibility at a singular pivot $p\\mid c$.","3. A nonsingular Euler-operator argument}{The standard interval without an analytic power-series-in-the-index premise}\nOn Hurwitz coefficients, the actual Euler operator plus $c$ is\n\\[\n (\\Lambda_c a)(d)=(d+c)a(d).\n\\]\nDirect induction proves its commutator with every defect iterate:\n\\[\n \\delta^s\\Lambda_c a\n =\\Lambda_c\\delta^s a\n  +sp\\bigl(pU^{p^2}-U^p\\bigr)\\delta^{s-1}a,\n \\qquad s\\ge1.\n\\]\nSuppose $a$ is integral and supported at degrees divisible by $p$, and $p\\nmid c$. Every supported pivot $d+c$ is proved to be a unit of $\\Z_p$. The defect preserves this support. Induction in the commutator identity therefore proves\n\\[\n \\operatorname{Flat}(\\Lambda_c a)\\Longrightarrow\\operatorname{Flat}(a).\n\\]\nThis is a reflection theorem on the integral supported domain. It does not assume flatness of the proposed preimage, and does not assert invertibility at a singular pivot $p\\mid c$.","interval without an analytic power-series-in-the-index premise}\nOn Hurwitz coefficients, the actual Euler operator plus $c$ is\n\\[\n (\\Lambda_c a)(d)=(d+c)a(d).\n\\]\nDirect induction proves its commutator with every defect iterate:\n\\[\n \\delta^s\\Lambda_c a\n =\\Lambda_c\\delta^s a\n  +sp\\bigl(pU^{p^2}-U^p\\bigr)\\delta^{s-1}a,\n \\qquad s\\ge1.\n\\]\nSuppose $a$ is integral and supported at degrees divisible by $p$, and $p\\nmid c$. Every supported pivot $d+c$ is proved to be a unit of $\\Z_p$. The defect preserves this support. Induction in the commutator identity therefore proves\n\\[\n \\operatorname{Flat}(\\Lambda_c a)\\Longrightarrow\\operatorname{Flat}(a).\n\\]\nThis is a reflection theorem on the integral supported domain. It does not assume flatness of the proposed preimage, and does not assert invertibility at a singular pivot $p\\mid c$.","without an analytic power-series-in-the-index premise}\nOn Hurwitz coefficients, the actual Euler operator plus $c$ is\n\\[\n (\\Lambda_c a)(d)=(d+c)a(d).\n\\]\nDirect induction proves its commutator with every defect iterate:\n\\[\n \\delta^s\\Lambda_c a\n =\\Lambda_c\\delta^s a\n  +sp\\bigl(pU^{p^2}-U^p\\bigr)\\delta^{s-1}a,\n \\qquad s\\ge1.\n\\]\nSuppose $a$ is integral and supported at degrees divisible by $p$, and $p\\nmid c$. Every supported pivot $d+c$ is proved to be a unit of $\\Z_p$. The defect preserves this support. Induction in the commutator identity therefore proves\n\\[\n \\operatorname{Flat}(\\Lambda_c a)\\Longrightarrow\\operatorname{Flat}(a).\n\\]\nThis is a reflection theorem on the integral supported domain. It does not assume flatness of the proposed preimage, and does not assert invertibility at a singular pivot $p\\mid c$.","in-the-index premise}\nOn Hurwitz coefficients, the actual Euler operator plus $c$ is\n\\[\n (\\Lambda_c a)(d)=(d+c)a(d).\n\\]\nDirect induction proves its commutator with every defect iterate:\n\\[\n \\delta^s\\Lambda_c a\n =\\Lambda_c\\delta^s a\n  +sp\\bigl(pU^{p^2}-U^p\\bigr)\\delta^{s-1}a,\n \\qquad s\\ge1.\n\\]\nSuppose $a$ is integral and supported at degrees divisible by $p$, and $p\\nmid c$. Every supported pivot $d+c$ is proved to be a unit of $\\Z_p$. The defect preserves this support. Induction in the commutator identity therefore proves\n\\[\n \\operatorname{Flat}(\\Lambda_c a)\\Longrightarrow\\operatorname{Flat}(a).\n\\]\nThis is a reflection theorem on the integral supported domain. It does not assume flatness of the proposed preimage, and does not assert invertibility at a singular pivot $p\\mid c$.","coefficients, the actual Euler operator plus $c$ is\n\\[\n (\\Lambda_c a)(d)=(d+c)a(d).\n\\]\nDirect induction proves its commutator with every defect iterate:\n\\[\n \\delta^s\\Lambda_c a\n =\\Lambda_c\\delta^s a\n  +sp\\bigl(pU^{p^2}-U^p\\bigr)\\delta^{s-1}a,\n \\qquad s\\ge1.\n\\]\nSuppose $a$ is integral and supported at degrees divisible by $p$, and $p\\nmid c$. Every supported pivot $d+c$ is proved to be a unit of $\\Z_p$. The defect preserves this support. Induction in the commutator identity therefore proves\n\\[\n \\operatorname{Flat}(\\Lambda_c a)\\Longrightarrow\\operatorname{Flat}(a).\n\\]\nThis is a reflection theorem on the integral supported domain. It does not assume flatness of the proposed preimage, and does not assert invertibility at a singular pivot $p\\mid c$.","uction proves its commutator with every defect iterate:\n\\[\n \\delta^s\\Lambda_c a\n =\\Lambda_c\\delta^s a\n  +sp\\bigl(pU^{p^2}-U^p\\bigr)\\delta^{s-1}a,\n \\qquad s\\ge1.\n\\]\nSuppose $a$ is integral and supported at degrees divisible by $p$, and $p\\nmid c$. Every supported pivot $d+c$ is proved to be a unit of $\\Z_p$. The defect preserves this support. Induction in the commutator identity therefore proves\n\\[\n \\operatorname{Flat}(\\Lambda_c a)\\Longrightarrow\\operatorname{Flat}(a).\n\\]\nThis is a reflection theorem on the integral supported domain. It does not assume flatness of the proposed preimage, and does not assert invertibility at a singular pivot $p\\mid c$.","very supported pivot $d+c$ is proved to be a unit of $\\Z_p$. The defect preserves this support. Induction in the commutator identity therefore proves\n\\[\n \\operatorname{Flat}(\\Lambda_c a)\\Longrightarrow\\operatorname{Flat}(a).\n\\]\nThis is a reflection theorem on the integral supported domain. It does not assume flatness of the proposed preimage, and does not assert invertibility at a singular pivot $p\\mid c$.","\\subsection*{Apply it to the explicit normalized sections}\nSet $N=mp$ and, for $0\\le j<p$, define integral Hurwitz coefficients\n\\[\n g_j(d)=\n \\begin{cases}\n \\displaystyle\\binom{d+j}{d}^{-1},&N\\mid d,","\\subsection*{Apply it to the explicit normalized sections}\nSet $N=mp$ and, for $0\\le j<p$, define integral Hurwitz coefficients\n\\[\n g_j(d)=\n \\begin{cases}\n \\displaystyle\\binom{d+j}{d}^{-1},&N\\mid d,\\\\[1mm]\n 0,&N\\nmid d.\n \\end{cases}\n\\]\nThe inverse binomial is constructed using the actual unit pivots already proved by Lucas reduction. Its image in $\\Q_p$ is proved equal to the displayed reciprocal.","[1mm]\n 0,&N\\nmid d.\n \\end{cases}\n\\]\nThe inverse binomial is constructed using the actual unit pivots already proved by Lucas reduction. Its image in $\\Q_p$ is proved equal to the displayed reciprocal.","For $j=0$, $g_0$ is the literal $N$-divisibility indicator. Because $N\\mid p(p-1)$, the two shifts in $\\delta g_0$ agree and $\\delta g_0=0$. It is flat at every order. The actual binomial identity gives\n\\[\n \\Lambda_{j+1}g_{j+1}=(j+1)g_j.\n\\]\nAs $1\\le j+1<p$, the nonsingular reflection theorem inductively proves flatness of every $g_j$ in the standard window. There is no uniform-convergence assumption in this argument.","\\par\\medskip\n\\pageheading{4. From these sections to the original Euler packets}{Actual reciprocal, actual rational sequence, every order and step multiple}\nThe factorial-ratio identity is proved coefficient by coefficient:\n\\[\n \\sum_{d\\ge0}g_j(d)\\frac{t^d}{d!}\n =j!\\sum_{r\\ge0}\\frac{t^{Nr}}{(Nr+j)!}.\n\\]\nThe right side is precisely the normalized source kernel from the first block, not a replacement family. The earlier rational identity for its reciprocal is transported to $\\Q_p[[t]]$. A separate encoding theorem identifies the reciprocal's coefficients with the already constructed $\\Z_p$ lifts of the original $C_{N,j}(r)$.","ts}{Actual reciprocal, actual rational sequence, every order and step multiple}\nThe factorial-ratio identity is proved coefficient by coefficient:\n\\[\n \\sum_{d\\ge0}g_j(d)\\frac{t^d}{d!}\n =j!\\sum_{r\\ge0}\\frac{t^{Nr}}{(Nr+j)!}.\n\\]\nThe right side is precisely the normalized source kernel from the first block, not a replacement family. The earlier rational identity for its reciprocal is transported to $\\Q_p[[t]]$. A separate encoding theorem identifies the reciprocal's coefficients with the already constructed $\\Z_p$ lifts of the original $C_{N,j}(r)$.","proved coefficient by coefficient:\n\\[\n \\sum_{d\\ge0}g_j(d)\\frac{t^d}{d!}\n =j!\\sum_{r\\ge0}\\frac{t^{Nr}}{(Nr+j)!}.\n\\]\nThe right side is precisely the normalized source kernel from the first block, not a replacement family. The earlier rational identity for its reciprocal is transported to $\\Q_p[[t]]$. A separate encoding theorem identifies the reciprocal's coefficients with the already constructed $\\Z_p$ lifts of the original $C_{N,j}(r)$.","cient:\n\\[\n \\sum_{d\\ge0}g_j(d)\\frac{t^d}{d!}\n =j!\\sum_{r\\ge0}\\frac{t^{Nr}}{(Nr+j)!}.\n\\]\nThe right side is precisely the normalized source kernel from the first block, not a replacement family. The earlier rational identity for its reciprocal is transported to $\\Q_p[[t]]$. A separate encoding theorem identifies the reciprocal's coefficients with the already constructed $\\Z_p$ lifts of the original $C_{N,j}(r)$.","All-order reciprocal closure therefore gives flatness of the actual packed Euler coefficient sequence. The earlier coefficient-extraction theorem now has a \\emph{proved antecedent} for every $j<p$, and yields\n\\[\n p^{\\min(k,mn)}\\mid\\Delta_h^k C_{mp,j}(n)\\quad\\text{in }\\Z_p,\n \\qquad h=(p-1)/m.\n\\]\nA new finite telescoping argument proves the extension to arbitrary natural multiples of $h$. Its first-step identity is\n\\[\n f(n+uh)-f(n)=\\sum_{a=0}^{u-1}\\Delta_h f(n+ah).\n\\]\nCommutation of differences and induction on their order transfer any fixed ideal bound valid on a forward tail. The required precision $\\min(k,mn)$ is nondecreasing as the starting row advances. Hence\n\\[\n \\boxed{p^{\\min(k,mn)}\\mid\\Delta_{uh}^{k}C_{mp,j}(n)\\quad(0\\le j<p).}\n\\]","l closure therefore gives flatness of the actual packed Euler coefficient sequence. The earlier coefficient-extraction theorem now has a \\emph{proved antecedent} for every $j<p$, and yields\n\\[\n p^{\\min(k,mn)}\\mid\\Delta_h^k C_{mp,j}(n)\\quad\\text{in }\\Z_p,\n \\qquad h=(p-1)/m.\n\\]\nA new finite telescoping argument proves the extension to arbitrary natural multiples of $h$. Its first-step identity is\n\\[\n f(n+uh)-f(n)=\\sum_{a=0}^{u-1}\\Delta_h f(n+ah).\n\\]\nCommutation of differences and induction on their order transfer any fixed ideal bound valid on a forward tail. The required precision $\\min(k,mn)$ is nondecreasing as the starting row advances. Hence\n\\[\n \\boxed{p^{\\min(k,mn)}\\mid\\Delta_{uh}^{k}C_{mp,j}(n)\\quad(0\\le j<p).}\n\\]","s\n\\[\n p^{\\min(k,mn)}\\mid\\Delta_h^k C_{mp,j}(n)\\quad\\text{in }\\Z_p,\n \\qquad h=(p-1)/m.\n\\]\nA new finite telescoping argument proves the extension to arbitrary natural multiples of $h$. Its first-step identity is\n\\[\n f(n+uh)-f(n)=\\sum_{a=0}^{u-1}\\Delta_h f(n+ah).\n\\]\nCommutation of differences and induction on their order transfer any fixed ideal bound valid on a forward tail. The required precision $\\min(k,mn)$ is nondecreasing as the starting row advances. Hence\n\\[\n \\boxed{p^{\\min(k,mn)}\\mid\\Delta_{uh}^{k}C_{mp,j}(n)\\quad(0\\le j<p).}\n\\]","\\qquad h=(p-1)/m.\n\\]\nA new finite telescoping argument proves the extension to arbitrary natural multiples of $h$. Its first-step identity is\n\\[\n f(n+uh)-f(n)=\\sum_{a=0}^{u-1}\\Delta_h f(n+ah).\n\\]\nCommutation of differences and induction on their order transfer any fixed ideal bound valid on a forward tail. The required precision $\\min(k,mn)$ is nondecreasing as the starting row advances. Hence\n\\[\n \\boxed{p^{\\min(k,mn)}\\mid\\Delta_{uh}^{k}C_{mp,j}(n)\\quad(0\\le j<p).}\n\\]","The scalar-extension bridge identifies these differences with the original rational coefficient function. The rational theorem explicitly allows zero differences, avoiding the library's finite valuation convention at zero. Since $j!$ is a unit for $j<p$, both normalized and unnormalized versions have the same precision.","\\textbf{Scope.} The proof includes all primes, all rows and orders, and all natural step multipliers. For $m=1$, the standard window is the complete allowed residue range. For larger $m$, residues $p\\le j<mp$ are not covered by this theorem. The earlier one-power period is not the source of the new higher-power estimate.","\\par\\medskip\n\\pageheading{5. Examples, verification and the remaining gap}{The coverage count is unchanged; the unresolved theorem now has a stronger proved part}\nThe original rational recurrence gives\n\\[\n C_{6,2}(1)=-\\frac1{28},\\quad C_{6,2}(2)=\\frac{425}{364},\\quad\n C_{6,2}(3)=-\\frac{2039759}{2660}.\n\\]\nConsequently, the new source proves\n\\[\n \\Delta_1^2 C_{6,2}(1)=-\\frac{949959}{1235},\\qquad\n v_3\\!\\left(\\Delta_1^2 C_{6,2}(1)\\right)=2.\n\\]\nThis attains the bound $\\min(2,2\\cdot1)=2$. A claim of a third power is rejected. The separate cubic example proves that the normalized differential defect is not an ordinary derivation. Controls also reject an asserted unit pivot at $c=p$ and the prime-power binomial-divisibility formula at a composite base.","ng gap}{The coverage count is unchanged; the unresolved theorem now has a stronger proved part}\nThe original rational recurrence gives\n\\[\n C_{6,2}(1)=-\\frac1{28},\\quad C_{6,2}(2)=\\frac{425}{364},\\quad\n C_{6,2}(3)=-\\frac{2039759}{2660}.\n\\]\nConsequently, the new source proves\n\\[\n \\Delta_1^2 C_{6,2}(1)=-\\frac{949959}{1235},\\qquad\n v_3\\!\\left(\\Delta_1^2 C_{6,2}(1)\\right)=2.\n\\]\nThis attains the bound $\\min(2,2\\cdot1)=2$. A claim of a third power is rejected. The separate cubic example proves that the normalized differential defect is not an ordinary derivation. Controls also reject an asserted unit pivot at $c=p$ and the prime-power binomial-divisibility formula at a composite base.",")=-\\frac{2039759}{2660}.\n\\]\nConsequently, the new source proves\n\\[\n \\Delta_1^2 C_{6,2}(1)=-\\frac{949959}{1235},\\qquad\n v_3\\!\\left(\\Delta_1^2 C_{6,2}(1)\\right)=2.\n\\]\nThis attains the bound $\\min(2,2\\cdot1)=2$. A claim of a third power is rejected. The separate cubic example proves that the normalized differential defect is not an ordinary derivation. Controls also reject an asserted unit pivot at $c=p$ and the prime-power binomial-divisibility formula at a composite base.","\\subsection*{What is still unproved}\nThe all-residue Theorem 7.5 requires the additional range $p\\le j<mp$. Its supporting prose proof treats singular ratios at $c=p,2p,\\ldots,(m-1)p$ through weighted Amice estimates and a dual Mahler pairing, then propagates across the intervening indices. Those actual-family estimates and the full remaining propagation are not supplied by this release. A unit argument at $p\\nmid c$ cannot be used at $p\\mid c$.","*{What is still unproved}\nThe all-residue Theorem 7.5 requires the additional range $p\\le j<mp$. Its supporting prose proof treats singular ratios at $c=p,2p,\\ldots,(m-1)p$ through weighted Amice estimates and a dual Mahler pairing, then propagates across the intervening indices. Those actual-family estimates and the full remaining propagation are not supplied by this release. A unit argument at $p\\nmid c$ cannot be used at $p\\mid c$.","all-residue Theorem 7.5 requires the additional range $p\\le j<mp$. Its supporting prose proof treats singular ratios at $c=p,2p,\\ldots,(m-1)p$ through weighted Amice estimates and a dual Mahler pairing, then propagates across the intervening indices. Those actual-family estimates and the full remaining propagation are not supplied by this release. A unit argument at $p\\nmid c$ cannot be used at $p\\mid c$.","\\clearpage\\subsection{Singular Amice Supplement}\\label{app:sup9}\n\\noindent\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and result}\nLet $p$ be prime, let $m\\mid p-1$, and put $N=mp$. For $1\\le a<m$, put $c=ap$ and define the original exponential section and ratio by\n\\[\n H_{N,c}(t)=\\sum_{n\\ge0}\\frac{t^{Nn+c}}{(Nn+c)!},\\qquad\n R_c(t)=\\frac{tH_{N,c-1}(t)}{H_{N,c}(t)}.\n\\]\nThe quotient is interpreted after removing its forced zero. This block proves, for the actual rational coefficients $\\alpha_d$,\n\\[\n R_{ap}(\\log(1+T))=\\sum_{d\\ge0}\\alpha_dT^d,\n \\qquad\n \\boxed{\\alpha_d=0\\ \\text{or}\\ v_p(\\alpha_d)\\ge-\\left\\lfloor d/p\\right\\rfloor.}\n\\]\nEquivalently, every $p^{\\lfloor d/p\\rfloor}\\alpha_d$ belongs to the actual completed ring $\\Z_p$. The zero-safe alternative is explicit in the formal rational statement.","}\n\\noindent\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and result}\nLet $p$ be prime, let $m\\mid p-1$, and put $N=mp$. For $1\\le a<m$, put $c=ap$ and define the original exponential section and ratio by\n\\[\n H_{N,c}(t)=\\sum_{n\\ge0}\\frac{t^{Nn+c}}{(Nn+c)!},\\qquad\n R_c(t)=\\frac{tH_{N,c-1}(t)}{H_{N,c}(t)}.\n\\]\nThe quotient is interpreted after removing its forced zero. This block proves, for the actual rational coefficients $\\alpha_d$,\n\\[\n R_{ap}(\\log(1+T))=\\sum_{d\\ge0}\\alpha_dT^d,\n \\qquad\n \\boxed{\\alpha_d=0\\ \\text{or}\\ v_p(\\alpha_d)\\ge-\\left\\lfloor d/p\\right\\rfloor.}\n\\]\nEquivalently, every $p^{\\lfloor d/p\\rfloor}\\alpha_d$ belongs to the actual completed ring $\\Z_p$. The zero-safe alternative is explicit in the formal rational statement.","r\\medskip\n\\subsubsection*{Scope and result}\nLet $p$ be prime, let $m\\mid p-1$, and put $N=mp$. For $1\\le a<m$, put $c=ap$ and define the original exponential section and ratio by\n\\[\n H_{N,c}(t)=\\sum_{n\\ge0}\\frac{t^{Nn+c}}{(Nn+c)!},\\qquad\n R_c(t)=\\frac{tH_{N,c-1}(t)}{H_{N,c}(t)}.\n\\]\nThe quotient is interpreted after removing its forced zero. This block proves, for the actual rational coefficients $\\alpha_d$,\n\\[\n R_{ap}(\\log(1+T))=\\sum_{d\\ge0}\\alpha_dT^d,\n \\qquad\n \\boxed{\\alpha_d=0\\ \\text{or}\\ v_p(\\alpha_d)\\ge-\\left\\lfloor d/p\\right\\rfloor.}\n\\]\nEquivalently, every $p^{\\lfloor d/p\\rfloor}\\alpha_d$ belongs to the actual completed ring $\\Z_p$. The zero-safe alternative is explicit in the formal rational statement.","bsubsection*{Scope and result}\nLet $p$ be prime, let $m\\mid p-1$, and put $N=mp$. For $1\\le a<m$, put $c=ap$ and define the original exponential section and ratio by\n\\[\n H_{N,c}(t)=\\sum_{n\\ge0}\\frac{t^{Nn+c}}{(Nn+c)!},\\qquad\n R_c(t)=\\frac{tH_{N,c-1}(t)}{H_{N,c}(t)}.\n\\]\nThe quotient is interpreted after removing its forced zero. This block proves, for the actual rational coefficients $\\alpha_d$,\n\\[\n R_{ap}(\\log(1+T))=\\sum_{d\\ge0}\\alpha_dT^d,\n \\qquad\n \\boxed{\\alpha_d=0\\ \\text{or}\\ v_p(\\alpha_d)\\ge-\\left\\lfloor d/p\\right\\rfloor.}\n\\]\nEquivalently, every $p^{\\lfloor d/p\\rfloor}\\alpha_d$ belongs to the actual completed ring $\\Z_p$. The zero-safe alternative is explicit in the formal rational statement.","ratio by\n\\[\n H_{N,c}(t)=\\sum_{n\\ge0}\\frac{t^{Nn+c}}{(Nn+c)!},\\qquad\n R_c(t)=\\frac{tH_{N,c-1}(t)}{H_{N,c}(t)}.\n\\]\nThe quotient is interpreted after removing its forced zero. This block proves, for the actual rational coefficients $\\alpha_d$,\n\\[\n R_{ap}(\\log(1+T))=\\sum_{d\\ge0}\\alpha_dT^d,\n \\qquad\n \\boxed{\\alpha_d=0\\ \\text{or}\\ v_p(\\alpha_d)\\ge-\\left\\lfloor d/p\\right\\rfloor.}\n\\]\nEquivalently, every $p^{\\lfloor d/p\\rfloor}\\alpha_d$ belongs to the actual completed ring $\\Z_p$. The zero-safe alternative is explicit in the formal rational statement.","_c(t)=\\frac{tH_{N,c-1}(t)}{H_{N,c}(t)}.\n\\]\nThe quotient is interpreted after removing its forced zero. This block proves, for the actual rational coefficients $\\alpha_d$,\n\\[\n R_{ap}(\\log(1+T))=\\sum_{d\\ge0}\\alpha_dT^d,\n \\qquad\n \\boxed{\\alpha_d=0\\ \\text{or}\\ v_p(\\alpha_d)\\ge-\\left\\lfloor d/p\\right\\rfloor.}\n\\]\nEquivalently, every $p^{\\lfloor d/p\\rfloor}\\alpha_d$ belongs to the actual completed ring $\\Z_p$. The zero-safe alternative is explicit in the formal rational statement.","eorem assumes primality, the modulus divisibility and the specified residue bounds. It does not assume weighted section membership, reciprocal closure, the ratio identity, a Mahler pairing or the desired higher-difference bound. Pages 2--5 describe the constructed objects and proved bridges.\n\\par\\medskip\n\\subsubsection*{1. Integral cyclic coefficients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","sumes primality, the modulus divisibility and the specified residue bounds. It does not assume weighted section membership, reciprocal closure, the ratio identity, a Mahler pairing or the desired higher-difference bound. Pages 2--5 describe the constructed objects and proved bridges.\n\\par\\medskip\n\\subsubsection*{1. Integral cyclic coefficients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","y, the modulus divisibility and the specified residue bounds. It does not assume weighted section membership, reciprocal closure, the ratio identity, a Mahler pairing or the desired higher-difference bound. Pages 2--5 describe the constructed objects and proved bridges.\n\\par\\medskip\n\\subsubsection*{1. Integral cyclic coefficients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","ified residue bounds. It does not assume weighted section membership, reciprocal closure, the ratio identity, a Mahler pairing or the desired higher-difference bound. Pages 2--5 describe the constructed objects and proved bridges.\n\\par\\medskip\n\\subsubsection*{1. Integral cyclic coefficients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","he ratio identity, a Mahler pairing or the desired higher-difference bound. Pages 2--5 describe the constructed objects and proved bridges.\n\\par\\medskip\n\\subsubsection*{1. Integral cyclic coefficients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","entity, a Mahler pairing or the desired higher-difference bound. Pages 2--5 describe the constructed objects and proved bridges.\n\\par\\medskip\n\\subsubsection*{1. Integral cyclic coefficients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","higher-difference bound. Pages 2--5 describe the constructed objects and proved bridges.\n\\par\\medskip\n\\subsubsection*{1. Integral cyclic coefficients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","escribe the constructed objects and proved bridges.\n\\par\\medskip\n\\subsubsection*{1. Integral cyclic coefficients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","ted objects and proved bridges.\n\\par\\medskip\n\\subsubsection*{1. Integral cyclic coefficients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","section*{1. Integral cyclic coefficients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","ients}\nWork first in an arbitrary commutative ring $R$. If $z^p=1$, set\n\\[\n I=(p,z-1),\\qquad P=(p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","p).\n\\]\nExpanding $((z-1)+1)^p=1$ and using divisibility of every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","f every interior binomial coefficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","ficient in prime row $p$ gives\n\\[\n (z-1)^p\\in PI.\n\\]\nEvery other degree-$p$ product of the two generators of $I$ also belongs to $PI$. The formal ideal proof therefore establishes\n\\[\n \\boxed{I^p\\subseteq PI.}\n\\]\nFor $q=p\\lfloor q/p\\rfloor+(q\\bmod p)$, strong induction and the factorial valuation identity give\n\\[\n \\boxed{I^q\\subseteq P^{v_p(q!)}}.\n\\]\nIn detail, write $b=\\lfloor q/p\\rfloor$. Then\n\\[\n I^q\\subseteq P^b I^b\\subseteq P^{b+v_p(b!)}=P^{v_p(q!)}.\n\\]\nNo integral-domain hypothesis or division in a field is used here.","Now take the integral cyclic group ring\n\\[\n R_N=\\Z[\\Z/N\\Z],\\qquad x=[1],\\qquad x^N=1,\n\\]\nand assume $N\\mid p(p-1)$. Thus $(x^{p-1})^p=1$. Each genuine consecutive block\n\\[\n \\prod_{i=0}^{p-1}\\bigl(x-(pb+i)\\bigr)\n\\]\nbelongs to $(p,x^{p-1}-1)$. This follows from the actual polynomial identity\n\\[\n \\prod_{i\\in\\F_p}(X-i)=X^p-X\n\\]\nmodulo $p$, not from a hypothesized valuation of the product.","For the falling factorial $(x)_d=\\prod_{i=0}^{d-1}(x-i)$, grouping into consecutive blocks gives\n\\[\n (x)_d\\in I^{\\lfloor d/p\\rfloor}\n \\subseteq\\bigl(p^{v_p(\\lfloor d/p\\rfloor!)}\\bigr).\n\\]\nEvery integral residue coordinate therefore satisfies\n\\[\n \\boxed{p^{v_p(\\lfloor d/p\\rfloor!)}\\mid [(r)](x)_d\\quad\\text{in }\\Z.}\n\\]\nDividing a coordinate by $d!$ is done only afterward, using\n$v_p(d!)=\\lfloor d/p\\rfloor+v_p(\\lfloor d/p\\rfloor!)$. This establishes the weighted denominator estimate before any field-valued divisibility could make it vacuous.","\\textbf{Modules.} \\mcode{CyclotomicFiltration}, \\mcode{PrimeBlockPolynomial}, and \\mcode{CyclicAmiceCoefficients}.\n\\par\\medskip\n\\subsubsection*{2. Connection to the actual section}\nAn integral coordinate estimate becomes relevant only after connecting it to the manuscript's generating functions. For a residue $r\\in\\Z/N\\Z$, define\n\\[\n H_r(t)=\\sum_{d\\ge0}\\mathbf1_{d\\equiv r\\ (N)}\\frac{t^d}{d!},\n \\qquad F_r(T)=H_r(\\log(1+T)).\n\\]\nThe implementation uses the actual formal logarithm and substitution in $\\Q[[T]]$. It proves\n\\[\n H_r'=H_{r-1},\\qquad (1+T)F_r'=F_{r-1},\n\\]\nand hence the coefficient recurrence\n\\[\n (d+1)\\coef{T}{d+1}F_r\n =\\coef{T}{d}F_{r-1}-d\\coef{T}{d}F_r.\n\\]\nThe initial values are $F_r(0)=\\mathbf1_{r=0}$.","fficients}.\n\\par\\medskip\n\\subsubsection*{2. Connection to the actual section}\nAn integral coordinate estimate becomes relevant only after connecting it to the manuscript's generating functions. For a residue $r\\in\\Z/N\\Z$, define\n\\[\n H_r(t)=\\sum_{d\\ge0}\\mathbf1_{d\\equiv r\\ (N)}\\frac{t^d}{d!},\n \\qquad F_r(T)=H_r(\\log(1+T)).\n\\]\nThe implementation uses the actual formal logarithm and substitution in $\\Q[[T]]$. It proves\n\\[\n H_r'=H_{r-1},\\qquad (1+T)F_r'=F_{r-1},\n\\]\nand hence the coefficient recurrence\n\\[\n (d+1)\\coef{T}{d+1}F_r\n =\\coef{T}{d}F_{r-1}-d\\coef{T}{d}F_r.\n\\]\nThe initial values are $F_r(0)=\\mathbf1_{r=0}$.","ip\n\\subsubsection*{2. Connection to the actual section}\nAn integral coordinate estimate becomes relevant only after connecting it to the manuscript's generating functions. For a residue $r\\in\\Z/N\\Z$, define\n\\[\n H_r(t)=\\sum_{d\\ge0}\\mathbf1_{d\\equiv r\\ (N)}\\frac{t^d}{d!},\n \\qquad F_r(T)=H_r(\\log(1+T)).\n\\]\nThe implementation uses the actual formal logarithm and substitution in $\\Q[[T]]$. It proves\n\\[\n H_r'=H_{r-1},\\qquad (1+T)F_r'=F_{r-1},\n\\]\nand hence the coefficient recurrence\n\\[\n (d+1)\\coef{T}{d+1}F_r\n =\\coef{T}{d}F_{r-1}-d\\coef{T}{d}F_r.\n\\]\nThe initial values are $F_r(0)=\\mathbf1_{r=0}$.","mate becomes relevant only after connecting it to the manuscript's generating functions. For a residue $r\\in\\Z/N\\Z$, define\n\\[\n H_r(t)=\\sum_{d\\ge0}\\mathbf1_{d\\equiv r\\ (N)}\\frac{t^d}{d!},\n \\qquad F_r(T)=H_r(\\log(1+T)).\n\\]\nThe implementation uses the actual formal logarithm and substitution in $\\Q[[T]]$. It proves\n\\[\n H_r'=H_{r-1},\\qquad (1+T)F_r'=F_{r-1},\n\\]\nand hence the coefficient recurrence\n\\[\n (d+1)\\coef{T}{d+1}F_r\n =\\coef{T}{d}F_{r-1}-d\\coef{T}{d}F_r.\n\\]\nThe initial values are $F_r(0)=\\mathbf1_{r=0}$.","For a residue $r\\in\\Z/N\\Z$, define\n\\[\n H_r(t)=\\sum_{d\\ge0}\\mathbf1_{d\\equiv r\\ (N)}\\frac{t^d}{d!},\n \\qquad F_r(T)=H_r(\\log(1+T)).\n\\]\nThe implementation uses the actual formal logarithm and substitution in $\\Q[[T]]$. It proves\n\\[\n H_r'=H_{r-1},\\qquad (1+T)F_r'=F_{r-1},\n\\]\nand hence the coefficient recurrence\n\\[\n (d+1)\\coef{T}{d+1}F_r\n =\\coef{T}{d}F_{r-1}-d\\coef{T}{d}F_r.\n\\]\nThe initial values are $F_r(0)=\\mathbf1_{r=0}$.","Independently, let $b_{d,r}$ be the residue-$r$ coordinate of $(x)_d$ in $R_N$. Actual group-ring multiplication gives\n\\[\n b_{0,r}=\\mathbf1_{r=0},\\qquad\n b_{d+1,r}=b_{d,r-1}-d\\,b_{d,r}.\n\\]\nInduction comparing these two recurrences proves\n\\[\n \\boxed{d!\\coef{T}{d}H_r(\\log(1+T))=b_{d,r}.}\n\\]\nThus the cyclic coefficient is not an unrelated stand-in for a section coefficient. For the canonical label $0\\le c<N$, the formal section is also proved equal to the original \\mcode{ExponentialSection.}\\allowbreak\\mcode{originalSection} used in the existing Euler project.","\\subsection*{Weighted membership}\nThe class used in the formalization is\n\\[\n \\Wp=\\left\\{F\\in\\Q_p[[T]]:\n p^{\\lfloor d/p\\rfloor}\\coef{T}{d}F\\in\\Z_p\\ \\text{for every }d\\right\\}.\n\\]\nThe integral-coordinate estimate from page 2 now proves\n\\[\n \\boxed{H_r(\\log(1+T))\\in\\Wp}\n\\]\nfor every residue $r$, including the singular ones. Rational coefficients are mapped into $\\Q_p$ by the actual coefficient ring homomorphism.","The same source package retains the earlier independent formal identity\n\\[\n d!\\coef{T}{d}\\exp(x\\log(1+T))=(x)_d\n\\]\nover a commutative $\\Q$-algebra. The section bridge above does not rely on selecting or averaging roots in a cyclotomic extension.","\\textbf{Principal targets.}\\par \\mcode{factorial_coeff_section}; \\mcode{section_weighted_integral};\\par \\mcode{weighted_section}.\n\\par\\medskip\n\\subsubsection*{3. Weighted closure and the normalized denominator}\nLet $w(d)=\\lfloor d/p\\rfloor$. Since $w(i)+w(d-i)\\le w(d)$, the coefficient formula for a product proves closure of $\\Wp$ under multiplication. Additive closure is immediate. For $U\\in\\Wp$ with $U(0)=1$, strong induction in the actual reciprocal coefficient recurrence proves\n\\[\n \\boxed{U^{-1}\\in\\Wp.}\n\\]\nThis inverse property is a theorem, not an additional field in a structure encoding weighted series.","f{Principal targets.}\\par \\mcode{factorial_coeff_section}; \\mcode{section_weighted_integral};\\par \\mcode{weighted_section}.\n\\par\\medskip\n\\subsubsection*{3. Weighted closure and the normalized denominator}\nLet $w(d)=\\lfloor d/p\\rfloor$. Since $w(i)+w(d-i)\\le w(d)$, the coefficient formula for a product proves closure of $\\Wp$ under multiplication. Additive closure is immediate. For $U\\in\\Wp$ with $U(0)=1$, strong induction in the actual reciprocal coefficient recurrence proves\n\\[\n \\boxed{U^{-1}\\in\\Wp.}\n\\]\nThis inverse property is a theorem, not an additional field in a structure encoding weighted series.","integral};\\par \\mcode{weighted_section}.\n\\par\\medskip\n\\subsubsection*{3. Weighted closure and the normalized denominator}\nLet $w(d)=\\lfloor d/p\\rfloor$. Since $w(i)+w(d-i)\\le w(d)$, the coefficient formula for a product proves closure of $\\Wp$ under multiplication. Additive closure is immediate. For $U\\in\\Wp$ with $U(0)=1$, strong induction in the actual reciprocal coefficient recurrence proves\n\\[\n \\boxed{U^{-1}\\in\\Wp.}\n\\]\nThis inverse property is a theorem, not an additional field in a structure encoding weighted series.","\\subsection*{Differentiation and the Euler field}\nThe derivative coefficient is $(d+1)a_{d+1}$. If $p\\nmid d+1$, the adjacent weights agree. If $p\\mid d+1$, the integer factor $d+1$ compensates for the one-unit weight increase. Therefore ordinary differentiation preserves $\\Wp$.","For the actual logarithm, its coefficient denominator is $d$, and the proved inequality\n\\[\n v_p(d)\\le\\lfloor d/p\\rfloor\\qquad(d>0)\n\\]\nimplies $\\log(1+T)\\in\\Wp$. Consequently\n\\[\n L_TF=\\log(1+T)(1+T)F'\n\\]\nalso preserves $\\Wp$.","\\subsection*{Removing a singular forced zero}\nPut $c=ap<N$ and define the actual normalized kernel\n\\[\n K_c(t)=c!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+c)!},\\qquad K_c(0)=1.\n\\]\nLet $\\ell(T)=\\log(1+T)/T$, defined by shifting logarithm coefficients, not by inverting $T$. Set\n\\[\n U(T)=\\ell(T)^cK_c(\\log(1+T)).\n\\]\nThe formal proof establishes $U(0)=1$ and the exact factorization\n\\[\n H_{N,c}(\\log(1+T))=\\frac{T^c}{c!}U(T).\n\\]\nIt follows coefficientwise that\n\\[\n \\coef{T}{d}U=c!\\coef{T}{d+c}H_{N,c}(\\log(1+T)).\n\\]\nBecause $p^a\\mid(ap)!$ and $\\lfloor(d+ap)/p\\rfloor=a+\\lfloor d/p\\rfloor$, the already proved section estimate gives $U\\in\\Wp$. Thus $U^{-1}\\in\\Wp$ as well. The factorial divisibility is derived from the pinned Legendre valuation theorem.","ut $c=ap<N$ and define the actual normalized kernel\n\\[\n K_c(t)=c!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+c)!},\\qquad K_c(0)=1.\n\\]\nLet $\\ell(T)=\\log(1+T)/T$, defined by shifting logarithm coefficients, not by inverting $T$. Set\n\\[\n U(T)=\\ell(T)^cK_c(\\log(1+T)).\n\\]\nThe formal proof establishes $U(0)=1$ and the exact factorization\n\\[\n H_{N,c}(\\log(1+T))=\\frac{T^c}{c!}U(T).\n\\]\nIt follows coefficientwise that\n\\[\n \\coef{T}{d}U=c!\\coef{T}{d+c}H_{N,c}(\\log(1+T)).\n\\]\nBecause $p^a\\mid(ap)!$ and $\\lfloor(d+ap)/p\\rfloor=a+\\lfloor d/p\\rfloor$, the already proved section estimate gives $U\\in\\Wp$. Thus $U^{-1}\\in\\Wp$ as well. The factorial divisibility is derived from the pinned Legendre valuation theorem.","=c!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+c)!},\\qquad K_c(0)=1.\n\\]\nLet $\\ell(T)=\\log(1+T)/T$, defined by shifting logarithm coefficients, not by inverting $T$. Set\n\\[\n U(T)=\\ell(T)^cK_c(\\log(1+T)).\n\\]\nThe formal proof establishes $U(0)=1$ and the exact factorization\n\\[\n H_{N,c}(\\log(1+T))=\\frac{T^c}{c!}U(T).\n\\]\nIt follows coefficientwise that\n\\[\n \\coef{T}{d}U=c!\\coef{T}{d+c}H_{N,c}(\\log(1+T)).\n\\]\nBecause $p^a\\mid(ap)!$ and $\\lfloor(d+ap)/p\\rfloor=a+\\lfloor d/p\\rfloor$, the already proved section estimate gives $U\\in\\Wp$. Thus $U^{-1}\\in\\Wp$ as well. The factorial divisibility is derived from the pinned Legendre valuation theorem.","c{t^{Nn}}{(Nn+c)!},\\qquad K_c(0)=1.\n\\]\nLet $\\ell(T)=\\log(1+T)/T$, defined by shifting logarithm coefficients, not by inverting $T$. Set\n\\[\n U(T)=\\ell(T)^cK_c(\\log(1+T)).\n\\]\nThe formal proof establishes $U(0)=1$ and the exact factorization\n\\[\n H_{N,c}(\\log(1+T))=\\frac{T^c}{c!}U(T).\n\\]\nIt follows coefficientwise that\n\\[\n \\coef{T}{d}U=c!\\coef{T}{d+c}H_{N,c}(\\log(1+T)).\n\\]\nBecause $p^a\\mid(ap)!$ and $\\lfloor(d+ap)/p\\rfloor=a+\\lfloor d/p\\rfloor$, the already proved section estimate gives $U\\in\\Wp$. Thus $U^{-1}\\in\\Wp$ as well. The factorial divisibility is derived from the pinned Legendre valuation theorem.","arithm coefficients, not by inverting $T$. Set\n\\[\n U(T)=\\ell(T)^cK_c(\\log(1+T)).\n\\]\nThe formal proof establishes $U(0)=1$ and the exact factorization\n\\[\n H_{N,c}(\\log(1+T))=\\frac{T^c}{c!}U(T).\n\\]\nIt follows coefficientwise that\n\\[\n \\coef{T}{d}U=c!\\coef{T}{d+c}H_{N,c}(\\log(1+T)).\n\\]\nBecause $p^a\\mid(ap)!$ and $\\lfloor(d+ap)/p\\rfloor=a+\\lfloor d/p\\rfloor$, the already proved section estimate gives $U\\in\\Wp$. Thus $U^{-1}\\in\\Wp$ as well. The factorial divisibility is derived from the pinned Legendre valuation theorem.","nts, not by inverting $T$. Set\n\\[\n U(T)=\\ell(T)^cK_c(\\log(1+T)).\n\\]\nThe formal proof establishes $U(0)=1$ and the exact factorization\n\\[\n H_{N,c}(\\log(1+T))=\\frac{T^c}{c!}U(T).\n\\]\nIt follows coefficientwise that\n\\[\n \\coef{T}{d}U=c!\\coef{T}{d+c}H_{N,c}(\\log(1+T)).\n\\]\nBecause $p^a\\mid(ap)!$ and $\\lfloor(d+ap)/p\\rfloor=a+\\lfloor d/p\\rfloor$, the already proved section estimate gives $U\\in\\Wp$. Thus $U^{-1}\\in\\Wp$ as well. The factorial divisibility is derived from the pinned Legendre valuation theorem.","\\textbf{Boundary.} The original positive-address section has zero constant coefficient. It is not the series to which unit inverse closure is applied.\n\\par\\medskip\n\\subsubsection*{4. The singular ratio estimate}\nFor a positive address $c$, the rational ratio is independently constructed from the original kernels:\n\\[\n R_c(t)=c\\,K_{c-1}(t)K_c(t)^{-1}.\n\\]\nThe scalar factor $c$ is necessary. The source proves the cleared identity\n\\[\n \\boxed{R_c(t)H_{N,c}(t)=tH_{N,c-1}(t).}\n\\]\nThis connects the constructed object to the original unnormalized ratio $A_{c-1}/A_c$. It does not assign the desired transformed formula as its definition.","positive-address section has zero constant coefficient. It is not the series to which unit inverse closure is applied.\n\\par\\medskip\n\\subsubsection*{4. The singular ratio estimate}\nFor a positive address $c$, the rational ratio is independently constructed from the original kernels:\n\\[\n R_c(t)=c\\,K_{c-1}(t)K_c(t)^{-1}.\n\\]\nThe scalar factor $c$ is necessary. The source proves the cleared identity\n\\[\n \\boxed{R_c(t)H_{N,c}(t)=tH_{N,c-1}(t).}\n\\]\nThis connects the constructed object to the original unnormalized ratio $A_{c-1}/A_c$. It does not assign the desired transformed formula as its definition.","\\par\\medskip\n\\subsubsection*{4. The singular ratio estimate}\nFor a positive address $c$, the rational ratio is independently constructed from the original kernels:\n\\[\n R_c(t)=c\\,K_{c-1}(t)K_c(t)^{-1}.\n\\]\nThe scalar factor $c$ is necessary. The source proves the cleared identity\n\\[\n \\boxed{R_c(t)H_{N,c}(t)=tH_{N,c-1}(t).}\n\\]\nThis connects the constructed object to the original unnormalized ratio $A_{c-1}/A_c$. It does not assign the desired transformed formula as its definition.","After actual logarithmic substitution and the chain rule,\n\\[\n R_c(\\log(1+T))H_{N,c}(\\log(1+T))\n =L_TH_{N,c}(\\log(1+T)).\n\\]\nInsert the proved factorization $H_{N,c}(\\log(1+T))=T^cU/c!$. Leibniz differentiation and cancellation of that nonzero section give\n\\[\n \\boxed{R_c(\\log(1+T))\n =c(1+T)\\ell(T)+(L_TU)U^{-1}.}\n\\]\nAll differentiation, substitution, nonzeroness and cancellation obligations are proved in the source.","For $c=ap$, the second summand belongs to $\\Wp$ by the closure results on page 4. The first does too: shifting the logarithm coefficient loses at most one unit of the floor weight, so $p\\ell(T)\\in\\Wp$, and the integer $a$ supplies the remaining factor. Therefore\n\\[\n \\boxed{R_{ap}(\\log(1+T))\\in\\Wp.}\n\\]\nThe final rational coefficient theorem states the bound with an explicit zero alternative. No finite-valued default valuation at zero is used as the meaning of unlimited precision.","\\subsection*{Checks that prevent a normalization mistake}\nThe source proves that the actual ratio has constant coefficient $c$, not one. In the concrete case $N=6,c=3$, the universal theorem gives weighted membership and the constant coefficient is proved to be $3$.","It also derives $\\coef{T}{3}H_{3,0}(\\log(1+T))=1/6$ from the actual differential recurrence and checks its valuation $v_3(1/6)=-1$. Thus weighted membership does not mean ordinary coefficient integrality. This value attains the allowed floor weight at degree three.","bottomrule\n\\end{tabular}\\end{center}\nFor $p=3,N=6,c=3$, this independent computation obtains the degree-six ratio coefficient $1/10080$, with valuation $-2$, attaining the allowed weight. This particular coefficient calculation is recorded as a finite regression, not as a new Lean theorem. A violating modulus $p=3,N=4$ yields a section coefficient of valuation $-4$ at degree nine, showing that the modulus divisibility hypothesis cannot simply be discarded.","\\clearpage\\subsection{All Residue Completion Supplement}\\label{app:sup10}\n\\noindent\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and final statement}\nLet $p$ be a prime, let $m>0$ divide $p-1$, put $N=mp$ and $h=(p-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_","\\clearpage\\subsection{All Residue Completion Supplement}\\label{app:sup10}\n\\noindent\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and final statement}\nLet $p$ be a prime, let $m>0$ divide $p-1$, put $N=mp$ and $h=(p-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\big","\\clearpage\\subsection{All Residue Completion Supplement}\\label{app:sup10}\n\\noindent\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and final statement}\nLet $p$ be a prime, let $m>0$ divide $p-1$, put $N=mp$ and $h=(p-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,m","\\clearpage\\subsection{All Residue Completion Supplement}\\label{app:sup10}\n\\noindent\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and final statement}\nLet $p$ be a prime, let $m>0$ divide $p-1$, put $N=mp$ and $h=(p-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}","arpage\\subsection{All Residue Completion Supplement}\\label{app:sup10}\n\\noindent\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and final statement}\nLet $p$ be a prime, let $m>0$ divide $p-1$, put $N=mp$ and $h=(p-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}\n\\]\nThe…","Residue Completion Supplement}\\label{app:sup10}\n\\noindent\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and final statement}\nLet $p$ be a prime, let $m>0$ divide $p-1$, put $N=mp$ and $h=(p-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}\n\\]\nThe manuscript's odd-prim…","nt\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and final statement}\nLet $p$ be a prime, let $m>0$ divide $p-1$, put $N=mp$ and $h=(p-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}\n\\]\nThe manuscript's odd-prime, positive-$u$ theorem is a specialization. The formal…","ement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and final statement}\nLet $p$ be a prime, let $m>0$ divide $p-1$, put $N=mp$ and $h=(p-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}\n\\]\nThe manuscript's odd-prime, positive-$u$ theorem is a specialization. The formal statement also includes zero row, zero order, zero multiplier, and the $p=2,m=1$ subcase. This does …","are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{Scope and final statement}\nLet $p$ be a prime, let $m>0$ divide $p-1$, put $N=mp$ and $h=(p-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}\n\\]\nThe manuscript's odd-prime, positive-$u$ theorem is a specialization. The formal statement also includes zero row, zero order, zero multiplier, and the $p=2,m=1$ subcase. This does not assert a different or broader external dyadic …","Scope and final statement}\nLet $p$ be a prime, let $m>0$ divide $p-1$, put $N=mp$ and $h=(p-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}\n\\]\nThe manuscript's odd-prime, positive-$u$ theorem is a specialization. The formal statement also includes zero row, zero order, zero multiplier, and the $p=2,m=1$ subcase. This does not assert a different or broader external dyadic conjecture.","-1)/m$, and let $0\\le j<N$. The original rational packet is defined by\n\\[\n \\sum_{q=0}^{n}\\binom{Nn+j}{Nq}C_{N,j}(q)=\\delta_{n0},\n \\qquad E_j(n)=j!\\,C_{N,j}(n).\n\\]\nFor every $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}\n\\]\nThe manuscript's odd-prime, positive-$u$ theorem is a specialization. The formal statement also includes zero row, zero order, zero multiplier, and the $p=2,m=1$ subcase. This does not assert a different or broader external dyadic conjecture.","ery $n,k,u\\in\\N$, the new formal result establishes\n\\[\n \\boxed{E_j(n)\\in\\Z_p,\\qquad\n \\Delta_{uh}^{k}E_j(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nHere $\\Delta_h f(n)=f(n+h)-f(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}\n\\]\nThe manuscript's odd-prime, positive-$u$ theorem is a specialization. The formal statement also includes zero row, zero order, zero multiplier, and the $p=2,m=1$ subcase. This does not assert a different or broader external dyadic conjecture.","(n)$. The corresponding rational valuation statement is zero-safe:\n\\[\n \\Delta_{uh}^{k}E_j(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}\n\\]\nThe manuscript's odd-prime, positive-$u$ theorem is a specialization. The formal statement also includes zero row, zero order, zero multiplier, and the $p=2,m=1$ subcase. This does not assert a different or broader external dyadic conjecture.","v_p\\bigl(\\Delta_{uh}^{k}E_j(n)\\bigr)\\ge\\min(k,mn).\n\\]\nFor the normalized coefficients, the exact correction is retained:\n\\[\n \\boxed{\\Delta_{uh}^{k}C_{N,j}(n)=0\\quad\\text{or}\\quad\n v_p\\bigl(\\Delta_{uh}^{k}C_{N,j}(n)\\bigr)\\ge\\min(k,mn)-v_p(j!).}\n\\]\nThe manuscript's odd-prime, positive-$u$ theorem is a specialization. The formal statement also includes zero row, zero order, zero multiplier, and the $p=2,m=1$ subcase. This does not assert a different or broader external dyadic conjecture.","\\textbf{Principal declaration.} \\path{KomorebiPaperIX.Theorem75.global_finite_difference} states the integrality and both precision bounds together, for the original coefficient functions. Its hypotheses are primality, $m>0$, $m\\mid p-1$ and $j<mp$, not an assumed flatness or weightedness certificate.\n\\par\\medskip\n\\subsubsection*{1. Stability of the weighted differential ring}\nThe recovered checkpoint used the actual weighted subring\n\\[\n \\Wp=\\left\\{F(T)=\\sum_{d\\ge0}a_dT^d:\\\n p^{\\lfloor d/p\\rfloor}a_d\\in\\Z_p\\ \\text{for every }d\\right\\}.\n\\]\nIts coefficient condition is identified with the one in the previously checked singular-Amice development. The failed \\mcode{ThetaWeighted} draft is retained separately; the repaired active source checks the complete formal-series statement below.","morebiPaperIX.Theorem75.global_finite_difference} states the integrality and both precision bounds together, for the original coefficient functions. Its hypotheses are primality, $m>0$, $m\\mid p-1$ and $j<mp$, not an assumed flatness or weightedness certificate.\n\\par\\medskip\n\\subsubsection*{1. Stability of the weighted differential ring}\nThe recovered checkpoint used the actual weighted subring\n\\[\n \\Wp=\\left\\{F(T)=\\sum_{d\\ge0}a_dT^d:\\\n p^{\\lfloor d/p\\rfloor}a_d\\in\\Z_p\\ \\text{for every }d\\right\\}.\n\\]\nIts coefficient condition is identified with the one in the previously checked singular-Amice development. The failed \\mcode{ThetaWeighted} draft is retained separately; the repaired active source checks the complete formal-series statement below.","Put\n\\[\n \\Theta=(1+T)\\frac{d}{dT},\\qquad\n B_\\Theta=\\frac{\\Theta^{p^2}-\\Theta^p}{p}.\n\\]\nThe coefficient formula\n\\[\n [T^d]\\Theta F=(d+1)a_{d+1}+da_d\n\\]\nproves $\\Theta\\Wp\\subseteq\\Wp$. If the floor weight increases between $d$ and $d+1$, then $p\\mid d+1$, which compensates for that increase.","\\textbf{The normalized defect is not assumed to be a derivation.}\nIts proved product rule contains the two middle Leibniz sums:\n\\[\n B_\\Theta(FG)=(B_\\Theta F)G+F(B_\\Theta G)+\\mathcal B_p(F,G).\n\\]\nThe coefficients of $\\mathcal B_p$ are the actual integral quotients\n$\\binom{p^2}{a}/p$ and $\\binom pa/p$. Derivative stability and this correction formula show that the elements $F\\in\\Wp$ satisfying $B_\\Theta F\\in\\Wp$ are closed under the finite ring operations.","The scalar input is proved for every natural $b$, including nonunits:\n\\[\n \\boxed{p^2\\mid b^{p^2}-b^p\\quad\\text{in }\\Z.}\n\\]\nBecause $\\Theta(1+T)^b=b(1+T)^b$, the binomial expansion of $T^p=((1+T)-1)^p$ then proves stability of the divided generator $T^p/p$. For $d=pq+r$, $0\\le r<p$, every weighted monomial has the exact factorization\n\\[\n a_dT^d=(p^qa_d)\\left(\\frac{T^p}{p}\\right)^qT^r.\n\\]\nThe scalar $p^qa_d$ is integral by the actual weight condition. Thus all weighted monomials, and all finite weighted polynomials, satisfy the defect condition.","e scalar input is proved for every natural $b$, including nonunits:\n\\[\n \\boxed{p^2\\mid b^{p^2}-b^p\\quad\\text{in }\\Z.}\n\\]\nBecause $\\Theta(1+T)^b=b(1+T)^b$, the binomial expansion of $T^p=((1+T)-1)^p$ then proves stability of the divided generator $T^p/p$. For $d=pq+r$, $0\\le r<p$, every weighted monomial has the exact factorization\n\\[\n a_dT^d=(p^qa_d)\\left(\\frac{T^p}{p}\\right)^qT^r.\n\\]\nThe scalar $p^qa_d$ is integral by the actual weight condition. Thus all weighted monomials, and all finite weighted polynomials, satisfy the defect condition.","Finally, $[T^d]B_\\Theta F$ depends only on coefficients through $d+p^2$. Replacing $F$ by this finite prefix proves the result for the entire formal series, without an analytic convergence assumption:\n\\[\n \\boxed{B_\\Theta\\Wp\\subseteq\\Wp,\\qquad B_\\Theta^s\\Wp\\subseteq\\Wp\\ (s\\ge0).}\n\\]\n\\textbf{Modules.} \\mcode{WeightedAmiceRing}, \\mcode{FrobeniusSquare}, \\mcode{ThetaWeighted}.\n\\par\\medskip\n\\subsubsection*{2. Actual substitution and singular higher differences}\nLet\n\\[\n \\mathcal E(F)(t)=F(e^t-1),\\qquad D=\\frac{d}{dt},\\qquad\n B_D=\\frac{D^{p^2}-D^p}{p}.\n\\]\nAll substitutions use Mathlib's actual formal exponential, logarithm and substitution maps. The chain rule gives\n\\[\n D\\mathcal E(F)=\\mathcal E(\\Theta F),\\qquad\n B_D^s\\mat","Finally, $[T^d]B_\\Theta F$ depends only on coefficients through $d+p^2$. Replacing $F$ by this finite prefix proves the result for the entire formal series, without an analytic convergence assumption:\n\\[\n \\boxed{B_\\Theta\\Wp\\subseteq\\Wp,\\qquad B_\\Theta^s\\Wp\\subseteq\\Wp\\ (s\\ge0).}\n\\]\n\\textbf{Modules.} \\mcode{WeightedAmiceRing}, \\mcode{FrobeniusSquare}, \\mcode{ThetaWeighted}.\n\\par\\medskip\n\\subsubsection*{2. Actual substitution and singular higher differences}\nLet\n\\[\n \\mathcal E(F)(t)=F(e^t-1),\\qquad D=\\frac{d}{dt},\\qquad\n B_D=\\frac{D^{p^2}-D^p}{p}.\n\\]\nAll substitutions use Mathlib's actual formal exponential, logarithm and substitution maps. The chain rule gives\n\\[\n D\\mathcal E(F)=\\mathcal E(\\Theta F),\\qquad\n B_D^s\\mathcal E(F)=\\mathcal E(B_\\Theta^sF).\n\\]\nThe imp","Finally, $[T^d]B_\\Theta F$ depends only on coefficients through $d+p^2$. Replacing $F$ by this finite prefix proves the result for the entire formal series, without an analytic convergence assumption:\n\\[\n \\boxed{B_\\Theta\\Wp\\subseteq\\Wp,\\qquad B_\\Theta^s\\Wp\\subseteq\\Wp\\ (s\\ge0).}\n\\]\n\\textbf{Modules.} \\mcode{WeightedAmiceRing}, \\mcode{FrobeniusSquare}, \\mcode{ThetaWeighted}.\n\\par\\medskip\n\\subsubsection*{2. Actual substitution and singular higher differences}\nLet\n\\[\n \\mathcal E(F)(t)=F(e^t-1),\\qquad D=\\frac{d}{dt},\\qquad\n B_D=\\frac{D^{p^2}-D^p}{p}.\n\\]\nAll substitutions use Mathlib's actual formal exponential, logarithm and substitution maps. The chain rule gives\n\\[\n D\\mathcal E(F)=\\mathcal E(\\Theta F),\\qquad\n B_D^s\\mathcal E(F)=\\mathcal E(B_\\Theta^sF).\n\\]\nThe implementation als","Finally, $[T^d]B_\\Theta F$ depends only on coefficients through $d+p^2$. Replacing $F$ by this finite prefix proves the result for the entire formal series, without an analytic convergence assumption:\n\\[\n \\boxed{B_\\Theta\\Wp\\subseteq\\Wp,\\qquad B_\\Theta^s\\Wp\\subseteq\\Wp\\ (s\\ge0).}\n\\]\n\\textbf{Modules.} \\mcode{WeightedAmiceRing}, \\mcode{FrobeniusSquare}, \\mcode{ThetaWeighted}.\n\\par\\medskip\n\\subsubsection*{2. Actual substitution and singular higher differences}\nLet\n\\[\n \\mathcal E(F)(t)=F(e^t-1),\\qquad D=\\frac{d}{dt},\\qquad\n B_D=\\frac{D^{p^2}-D^p}{p}.\n\\]\nAll substitutions use Mathlib's actual formal exponential, logarithm and substitution maps. The chain rule gives\n\\[\n D\\mathcal E(F)=\\mathcal E(\\Theta F),\\qquad\n B_D^s\\mathcal E(F)=\\mathcal E(B_\\Theta^sF).\n\\]\nThe implementation also proves the inverse-substitution identity\n\\[\n \\mathcal E\\bigl(f(\\log(1+T))\\bigr)=f(t).\n\\]","p^2$. Replacing $F$ by this finite prefix proves the result for the entire formal series, without an analytic convergence assumption:\n\\[\n \\boxed{B_\\Theta\\Wp\\subseteq\\Wp,\\qquad B_\\Theta^s\\Wp\\subseteq\\Wp\\ (s\\ge0).}\n\\]\n\\textbf{Modules.} \\mcode{WeightedAmiceRing}, \\mcode{FrobeniusSquare}, \\mcode{ThetaWeighted}.\n\\par\\medskip\n\\subsubsection*{2. Actual substitution and singular higher differences}\nLet\n\\[\n \\mathcal E(F)(t)=F(e^t-1),\\qquad D=\\frac{d}{dt},\\qquad\n B_D=\\frac{D^{p^2}-D^p}{p}.\n\\]\nAll substitutions use Mathlib's actual formal exponential, logarithm and substitution maps. The chain rule gives\n\\[\n D\\mathcal E(F)=\\mathcal E(\\Theta F),\\qquad\n B_D^s\\mathcal E(F)=\\mathcal E(B_\\Theta^sF).\n\\]\nThe implementation also proves the inverse-substitution identity\n\\[\n \\mathcal E\\bigl(f(\\log(1+T))\\bigr)=f(t).\n\\]","(s\\ge0).}\n\\]\n\\textbf{Modules.} \\mcode{WeightedAmiceRing}, \\mcode{FrobeniusSquare}, \\mcode{ThetaWeighted}.\n\\par\\medskip\n\\subsubsection*{2. Actual substitution and singular higher differences}\nLet\n\\[\n \\mathcal E(F)(t)=F(e^t-1),\\qquad D=\\frac{d}{dt},\\qquad\n B_D=\\frac{D^{p^2}-D^p}{p}.\n\\]\nAll substitutions use Mathlib's actual formal exponential, logarithm and substitution maps. The chain rule gives\n\\[\n D\\mathcal E(F)=\\mathcal E(\\Theta F),\\qquad\n B_D^s\\mathcal E(F)=\\mathcal E(B_\\Theta^sF).\n\\]\nThe implementation also proves the inverse-substitution identity\n\\[\n \\mathcal E\\bigl(f(\\log(1+T))\\bigr)=f(t).\n\\]","d singular higher differences}\nLet\n\\[\n \\mathcal E(F)(t)=F(e^t-1),\\qquad D=\\frac{d}{dt},\\qquad\n B_D=\\frac{D^{p^2}-D^p}{p}.\n\\]\nAll substitutions use Mathlib's actual formal exponential, logarithm and substitution maps. The chain rule gives\n\\[\n D\\mathcal E(F)=\\mathcal E(\\Theta F),\\qquad\n B_D^s\\mathcal E(F)=\\mathcal E(B_\\Theta^sF).\n\\]\nThe implementation also proves the inverse-substitution identity\n\\[\n \\mathcal E\\bigl(f(\\log(1+T))\\bigr)=f(t).\n\\]","The relevant integral subring is the \\emph{Hurwitz} subring\n\\[\n \\Hp=\\{f\\in\\Q_p[[t]]:\\ n![t^n]f\\in\\Z_p\\ \\text{for every }n\\}.\n\\]\nIt is not coefficientwise integrality of ordinary power-series coefficients. The precise connection is\n\\[\n n![t^n]\\mathcal E(F)=[T^0]\\Theta^nF.\n\\]\nSince every iterate of $\\Theta$ preserves $\\Wp$, the right side is integral. Therefore $\\mathcal E(\\Wp)\\subseteq\\Hp$. Applying the same statement to every $B_\\Theta^sF$ proves\n\\[\n \\boxed{f(\\log(1+T))\\in\\Wp\\quad\\Longrightarrow\\quad\n B_D^sf\\in\\Hp\\text{ for every }s.}\n\\]\nThis is the proved transfer that was missing in the earlier attempt. No weighted Mahler pairing is assumed; the formal differential route supplies an alternative bridge.","s the \\emph{Hurwitz} subring\n\\[\n \\Hp=\\{f\\in\\Q_p[[t]]:\\ n![t^n]f\\in\\Z_p\\ \\text{for every }n\\}.\n\\]\nIt is not coefficientwise integrality of ordinary power-series coefficients. The precise connection is\n\\[\n n![t^n]\\mathcal E(F)=[T^0]\\Theta^nF.\n\\]\nSince every iterate of $\\Theta$ preserves $\\Wp$, the right side is integral. Therefore $\\mathcal E(\\Wp)\\subseteq\\Hp$. Applying the same statement to every $B_\\Theta^sF$ proves\n\\[\n \\boxed{f(\\log(1+T))\\in\\Wp\\quad\\Longrightarrow\\quad\n B_D^sf\\in\\Hp\\text{ for every }s.}\n\\]\nThis is the proved transfer that was missing in the earlier attempt. No weighted Mahler pairing is assumed; the formal differential route supplies an alternative bridge.","ubring\n\\[\n \\Hp=\\{f\\in\\Q_p[[t]]:\\ n![t^n]f\\in\\Z_p\\ \\text{for every }n\\}.\n\\]\nIt is not coefficientwise integrality of ordinary power-series coefficients. The precise connection is\n\\[\n n![t^n]\\mathcal E(F)=[T^0]\\Theta^nF.\n\\]\nSince every iterate of $\\Theta$ preserves $\\Wp$, the right side is integral. Therefore $\\mathcal E(\\Wp)\\subseteq\\Hp$. Applying the same statement to every $B_\\Theta^sF$ proves\n\\[\n \\boxed{f(\\log(1+T))\\in\\Wp\\quad\\Longrightarrow\\quad\n B_D^sf\\in\\Hp\\text{ for every }s.}\n\\]\nThis is the proved transfer that was missing in the earlier attempt. No weighted Mahler pairing is assumed; the formal differential route supplies an alternative bridge.","\\subsection*{Application to the actual singular ratios}\nThe preceding v0.9.0 source independently constructs\n\\[\n H_{N,c}(t)=\\sum_{n\\ge0}\\frac{t^{Nn+c}}{(Nn+c)!},\\quad\n A_c(t)=t^{-c}H_{N,c}(t),\\quad R_c=\\frac{A_{c-1}}{A_c}.\n\\]\nThe notation $t^{-c}H_{N,c}$ means removing its proved initial zero coefficients. For each $c=ap$, $1\\le a<m$, that source proves\n$R_{ap}(\\log(1+T))\\in\\Wp$. The new transfer consequently proves flatness of the \\emph{same} rational ratio after scalar extension to $\\Q_p$.","ndently constructs\n\\[\n H_{N,c}(t)=\\sum_{n\\ge0}\\frac{t^{Nn+c}}{(Nn+c)!},\\quad\n A_c(t)=t^{-c}H_{N,c}(t),\\quad R_c=\\frac{A_{c-1}}{A_c}.\n\\]\nThe notation $t^{-c}H_{N,c}$ means removing its proved initial zero coefficients. For each $c=ap$, $1\\le a<m$, that source proves\n$R_{ap}(\\log(1+T))\\in\\Wp$. The new transfer consequently proves flatness of the \\emph{same} rational ratio after scalar extension to $\\Q_p$.","The actual ratio is also proved to be an expansion in $t^N$, so its coefficients vanish outside multiples of $N$. Constructed $\\Z_p$ lifts and the established coefficient criterion now give, for its literal Hurwitz coefficients $r_a(n)$,\n\\[\n \\boxed{\\Delta_{uh}^{k}r_a(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nThe separate rational precision theorem is obtained through the coefficient-lift equality. Divisibility in the field $\\Q_p$ is never substituted for this assertion.","is also proved to be an expansion in $t^N$, so its coefficients vanish outside multiples of $N$. Constructed $\\Z_p$ lifts and the established coefficient criterion now give, for its literal Hurwitz coefficients $r_a(n)$,\n\\[\n \\boxed{\\Delta_{uh}^{k}r_a(n)\\in p^{\\min(k,mn)}\\Z_p.}\n\\]\nThe separate rational precision theorem is obtained through the coefficient-lift equality. Divisibility in the field $\\Q_p$ is never substituted for this assertion.","\\textbf{Modules.} \\mcode{WeightedDworkTransfer}, \\mcode{SeriesDifferenceBridge}, \\mcode{RatioSeriesSupport}.\n\\par\\medskip\n\\subsubsection*{3. Nonsingular propagation without an assumed limit}\nFor consecutive original sections, the already defined ratios satisfy\n\\[\n R_cH_{N,c}=tH_{N,c-1},\\qquad H_{N,c}'=H_{N,c-1}.\n\\]\nDifferentiating the first relation, using the relation for $R_{c-1}$, and cancelling a section whose leading coefficient is proved nonzero gives the actual Riccati equation\n\\[\n \\boxed{tR_c'=R_c(1+R_{c-1}-R_c),\\qquad 2\\le c<N.}\n\\]\nIt is not an equation imposed on a replacement ratio. Write\n\\[\n q=R_{c-1}-(c-1),\\qquad y=R_c-c.\n\\]\nBoth have zero constant coefficient. With $\\Lambda_c=tD+c$, the equation becomes\n\\[\n \\boxed{\\Lambda_c y=cq+yq-y^2.}\n\\]","oSeriesSupport}.\n\\par\\medskip\n\\subsubsection*{3. Nonsingular propagation without an assumed limit}\nFor consecutive original sections, the already defined ratios satisfy\n\\[\n R_cH_{N,c}=tH_{N,c-1},\\qquad H_{N,c}'=H_{N,c-1}.\n\\]\nDifferentiating the first relation, using the relation for $R_{c-1}$, and cancelling a section whose leading coefficient is proved nonzero gives the actual Riccati equation\n\\[\n \\boxed{tR_c'=R_c(1+R_{c-1}-R_c),\\qquad 2\\le c<N.}\n\\]\nIt is not an equation imposed on a replacement ratio. Write\n\\[\n q=R_{c-1}-(c-1),\\qquad y=R_c-c.\n\\]\nBoth have zero constant coefficient. With $\\Lambda_c=tD+c$, the equation becomes\n\\[\n \\boxed{\\Lambda_c y=cq+yq-y^2.}\n\\]","iginal sections, the already defined ratios satisfy\n\\[\n R_cH_{N,c}=tH_{N,c-1},\\qquad H_{N,c}'=H_{N,c-1}.\n\\]\nDifferentiating the first relation, using the relation for $R_{c-1}$, and cancelling a section whose leading coefficient is proved nonzero gives the actual Riccati equation\n\\[\n \\boxed{tR_c'=R_c(1+R_{c-1}-R_c),\\qquad 2\\le c<N.}\n\\]\nIt is not an equation imposed on a replacement ratio. Write\n\\[\n q=R_{c-1}-(c-1),\\qquad y=R_c-c.\n\\]\nBoth have zero constant coefficient. With $\\Lambda_c=tD+c$, the equation becomes\n\\[\n \\boxed{\\Lambda_c y=cq+yq-y^2.}\n\\]","quad H_{N,c}'=H_{N,c-1}.\n\\]\nDifferentiating the first relation, using the relation for $R_{c-1}$, and cancelling a section whose leading coefficient is proved nonzero gives the actual Riccati equation\n\\[\n \\boxed{tR_c'=R_c(1+R_{c-1}-R_c),\\qquad 2\\le c<N.}\n\\]\nIt is not an equation imposed on a replacement ratio. Write\n\\[\n q=R_{c-1}-(c-1),\\qquad y=R_c-c.\n\\]\nBoth have zero constant coefficient. With $\\Lambda_c=tD+c$, the equation becomes\n\\[\n \\boxed{\\Lambda_c y=cq+yq-y^2.}\n\\]","ing the first relation, using the relation for $R_{c-1}$, and cancelling a section whose leading coefficient is proved nonzero gives the actual Riccati equation\n\\[\n \\boxed{tR_c'=R_c(1+R_{c-1}-R_c),\\qquad 2\\le c<N.}\n\\]\nIt is not an equation imposed on a replacement ratio. Write\n\\[\n q=R_{c-1}-(c-1),\\qquad y=R_c-c.\n\\]\nBoth have zero constant coefficient. With $\\Lambda_c=tD+c$, the equation becomes\n\\[\n \\boxed{\\Lambda_c y=cq+yq-y^2.}\n\\]","When $p\\nmid c$, define the actual inverse coefficientwise:\n\\[\n [t^d]\\Lambda_c^{-1}f=\\frac{[t^d]f}{d+c}.\n\\]\nOn $p$-supported degrees, $d+c$ is a genuine $p$-adic unit. The proof establishes Hurwitz integrality of this solution, its support, both inverse identities, and preservation of full flatness using the earlier all-order Euler commutator theorem. The singular pivots $c=ap$ are never declared to be units.","Now define explicit iterations\n\\[\n y_0=0,\\qquad y_{r+1}=\\Lambda_c^{-1}(cq+y_rq-y_r^2).\n\\]\nIf $q$ is flat and $p$-supported, every iterate has those properties. The crucial link to the actual ratio is an exact finite-prefix theorem:\n\\[\n \\boxed{[t^d]y_r=[t^d]y\\qquad(d\\le r).}\n\\]\nIndeed,\n\\[\n (cq+zq-z^2)-(cq+yq-y^2)=(z-y)(q-z-y).\n\\]\nThe second factor has zero constant term. Multiplication therefore raises the established agreement prefix by one; coefficientwise division by $d+c$ preserves that prefix.","_rq-y_r^2).\n\\]\nIf $q$ is flat and $p$-supported, every iterate has those properties. The crucial link to the actual ratio is an exact finite-prefix theorem:\n\\[\n \\boxed{[t^d]y_r=[t^d]y\\qquad(d\\le r).}\n\\]\nIndeed,\n\\[\n (cq+zq-z^2)-(cq+yq-y^2)=(z-y)(q-z-y).\n\\]\nThe second factor has zero constant term. Multiplication therefore raises the established agreement prefix by one; coefficientwise division by $d+c$ preserves that prefix.","For fixed $s,d$, the coefficient $d![t^d]B_D^sy$ depends only on the coefficients of $y$ through $d+p^2s$. Choosing the matching finite iterate proves its integrality. This establishes flatness of the actual $y$, and hence of $R_c$, without postulating flatness of the solution or invoking an unproved analytic fixed-point theorem.","\\textbf{Modules.} \\mcode{EulerSeriesInverse}, \\mcode{RiccatiFlatPropagation}, \\mcode{RatioRiccati}.\n\\par\\medskip\n\\subsubsection*{4. All-residue assembly and original coefficients}\nThe standard-window source and inverse results supply $R_1$ and $F_{N,0}$. Induction over $1\\le c<N$ now has two proved cases. At $p\\mid c$, the singular weighted estimate and the new substitution theorem give flatness. At $p\\nmid c$, the preceding Riccati propagation applies. Thus every actual ratio $R_c$ is flat.","tbf{Modules.} \\mcode{EulerSeriesInverse}, \\mcode{RiccatiFlatPropagation}, \\mcode{RatioRiccati}.\n\\par\\medskip\n\\subsubsection*{4. All-residue assembly and original coefficients}\nThe standard-window source and inverse results supply $R_1$ and $F_{N,0}$. Induction over $1\\le c<N$ now has two proved cases. At $p\\mid c$, the singular weighted estimate and the new substitution theorem give flatness. At $p\\nmid c$, the preceding Riccati propagation applies. Thus every actual ratio $R_c$ is flat.","e{RatioRiccati}.\n\\par\\medskip\n\\subsubsection*{4. All-residue assembly and original coefficients}\nThe standard-window source and inverse results supply $R_1$ and $F_{N,0}$. Induction over $1\\le c<N$ now has two proved cases. At $p\\mid c$, the singular weighted estimate and the new substitution theorem give flatness. At $p\\nmid c$, the preceding Riccati propagation applies. Thus every actual ratio $R_c$ is flat.","Let $K_j(t)$ be the normalized section kernel with constant coefficient one. The constructed unnormalized Euler series is\n\\[\n F_{N,j}(t)=j!\\,K_j(t)^{-1}.\n\\]\nThe source/inverse identities and the factorial successor identity prove\n\\[\n F_{N,j+1}=F_{N,j}R_{j+1},\\qquad\n F_{N,j}=F_{N,0}\\prod_{c=1}^{j}R_c.\n\\]\nThe product formula is not an assembly hypothesis. Proved flatness of all factors and the original product-closure theorem establish flatness of the full family.","Coefficient extraction identifies, literally,\n\\[\n (Nn)![t^{Nn}]F_{N,j}=j!\\,C_{N,j}(n).\n\\]\nThe support outside multiples of $N$ is proved separately. The constructed integral lift therefore represents the original rational coefficient function. The existing saturation criterion and the proved step-multiple argument give the ideal bound at every order, row and multiplier. Multiplication by $j!$ commutes with finite differences, giving precisely the displayed normalization loss.","\\subsection*{Exact examples outside the standard window}\nFor $(p,m,j)=(3,2,3)$, the source derives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\quad C_{6,3}(2)=\\frac{703}{5460},\\quad\n C_{6,3}(3)=-\\frac{223199}{7980}.\n\\]\nConsequently,\n\\[\n \\Delta_1^2\\bigl(3!C_{6,3}\\bigr)(1)=-\\frac{1464768}{8645},\\qquad\n v_3\\!\\left(-\\frac{1464768}{8645}\\right)=2.\n\\]\nThis attains $\\min(2,2\\cdot1)=2$. At the regular address $j=4$ beyond the singular boundary, the source likewise proves\n\\[\n \\Delta_1^2\\bigl(4!C_{6,4}\\bigr)(1)=-\\frac{102745512}{2377375},\n\\]\nand applies the universal exponent-two bound to this actual coefficient function.","w}\nFor $(p,m,j)=(3,2,3)$, the source derives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\quad C_{6,3}(2)=\\frac{703}{5460},\\quad\n C_{6,3}(3)=-\\frac{223199}{7980}.\n\\]\nConsequently,\n\\[\n \\Delta_1^2\\bigl(3!C_{6,3}\\bigr)(1)=-\\frac{1464768}{8645},\\qquad\n v_3\\!\\left(-\\frac{1464768}{8645}\\right)=2.\n\\]\nThis attains $\\min(2,2\\cdot1)=2$. At the regular address $j=4$ beyond the singular boundary, the source likewise proves\n\\[\n \\Delta_1^2\\bigl(4!C_{6,4}\\bigr)(1)=-\\frac{102745512}{2377375},\n\\]\nand applies the universal exponent-two bound to this actual coefficient function.","The earlier no-loss counterexample is retained: $v_3(C_{6,3}(2)-C_{6,3}(1))=0$. Proving the corrected theorem does not make that false normalized claim true.","\\subsection*{Independent finite checks}\nThe standard-library Python mirror checks 328 complete residue families for primes $2,3,5,7,11$, including 233 higher-residue families. It checks 4,264 integral unnormalized coefficients, 22,137 normalized and 22,137 unnormalized difference bounds, and 1,640 comparisons with a separately computed ordinary reciprocal. Further checks cover 108 weighted differential iterations, 205 integer Frobenius-square cases and 180 exact Picard agreement prefixes.","\\clearpage\\subsection{Nishibuchi conjecture completion}\\label{app:sup11}\n\\noindent\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{1. The complete external statement}\nFor $N>0$ and $0\\le j<N$, use the original unnormalized convention\n\\[\n \\sum_{d\\ge0}\\EE_d^{(N,j)}\\frac{t^d}{d!}\n =\\left(\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!}\\right)^{-1}.\n\\]\nThe existing packet convention satisfies\n$\\EE_{Nn}^{(N,j)}=j!C_{N,j}(n)$. The formalization defines the degree-indexed coefficients through the actual factorial-scaled series coefficients, proves their packet identification, and checks the cleared generating identity with the original exponential section.","{Nishibuchi conjecture completion}\\label{app:sup11}\n\\noindent\\textbf{Checkpoint scope.} This complete source body retains the status of its dated original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{1. The complete external statement}\nFor $N>0$ and $0\\le j<N$, use the original unnormalized convention\n\\[\n \\sum_{d\\ge0}\\EE_d^{(N,j)}\\frac{t^d}{d!}\n =\\left(\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!}\\right)^{-1}.\n\\]\nThe existing packet convention satisfies\n$\\EE_{Nn}^{(N,j)}=j!C_{N,j}(n)$. The formalization defines the degree-indexed coefficients through the actual factorial-scaled series coefficients, proves their packet identification, and checks the cleared generating identity with the original exponential section.","ed original supplement. Earlier pending or partial-coverage labels are historical; the final core records their resolution.\\par\\medskip\n\\subsubsection*{1. The complete external statement}\nFor $N>0$ and $0\\le j<N$, use the original unnormalized convention\n\\[\n \\sum_{d\\ge0}\\EE_d^{(N,j)}\\frac{t^d}{d!}\n =\\left(\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!}\\right)^{-1}.\n\\]\nThe existing packet convention satisfies\n$\\EE_{Nn}^{(N,j)}=j!C_{N,j}(n)$. The formalization defines the degree-indexed coefficients through the actual factorial-scaled series coefficients, proves their packet identification, and checks the cleared generating identity with the original exponential section.","\\begin{theorem}\nLet $p$ be prime, $m>0$, and assume $m\\mid p-1$ or $m=2$. Put $N=mp$ and $q=\\operatorname{lcm}(2,p-1)$. Every $\\EE_{Nn}^{(N,j)}$ is $p$-adically integral. For every $r\\ge1$,\n\\[\n\\boxed{\\EE_{Nn+qp^r}^{(N,j)}\\equiv\\EE_{Nn}^{(N,j)}\\pmod{p^r}\n       \\qquad(0\\le j<N,\\ n\\ge r).}\n\\]\nIn particular, the positive sufficient threshold $n_0=r$ works uniformly in all allowed parameters.\n\\end{theorem}","This is the parameter range and displacement of Nishibuchi's Conjecture 3.9 in arXiv:2605.10677v2. The explicit threshold also establishes the suggested bound $n_0\\le r$. The original paper already proves the $p=2,m=2,j=0$ period as a special case, with all starting rows; this extension does not claim that special case as a new discovery.","The theorem type includes only the stated arithmetic hypotheses. There is no assumed congruence, weighted estimate, or flatness certificate. Precision is the zero-safe predicate $x=0$ or $v_p(x)\\ge r$. Its connection to genuine ideal divisibility in $\\Z_p$ is used in the proof.","\\par\\medskip\n\\subsubsection*{2. The missing dyadic branch}\nFor odd primes, the extra option $m=2$ already divides $p-1$. Therefore the only allowed family outside the preceding all-residue theorem is $p=2,m=2$, with $N=4$.","The new proof establishes a higher-difference estimate for every one of its four addresses:\n\\[\n\\boxed{v_2\\!\\left(\\Delta_u^k C_{4,j}(n)\\right)\\ge\\min(k,n)\n       \\qquad(0\\le j<4).}\n\\]\nAll natural $u,k,n$ are allowed, including zero step, zero order and zero differences. The same bound holds for $j!C_{4,j}(n)$. Both a genuine $\\Z_2$ divisibility theorem and a valuation theorem for the original rational sequence are included.","\\subsection*{An explicit source instead of an assumed regularity condition}\nLet $g_j(n)=\\binom{4n+j}{4n}^{-1}$. Its four formulas are\n\\[\n\\begin{aligned}\n g_0(n)&=1,","\\subsection*{An explicit source instead of an assumed regularity condition}\nLet $g_j(n)=\\binom{4n+j}{4n}^{-1}$. Its four formulas are\n\\[\n\\begin{aligned}\n g_0(n)&=1,\\\\\n g_1(n)&=(4n+1)^{-1},\\\\\n g_2(n)&=((4n+1)(2n+1))^{-1},\\\\\n g_3(n)&=3((4n+1)(2n+1)(4n+3))^{-1}.\n\\end{aligned}\n\\]\nEvery denominator is odd. The code constructs the actual unit inverses in $\\Z_2$ and proves these identities against the original binomial coefficients, including $n=0$.","g_3(n)&=3((4n+1)(2n+1)(4n+3))^{-1}.\n\\end{aligned}\n\\]\nEvery denominator is odd. The code constructs the actual unit inverses in $\\Z_2$ and proves these identities against the original binomial coefficients, including $n=0$.","Define a Kummer sequence over a commutative ring by\n\\[\n\\PP_a(f)\\quad\\Longleftrightarrow\\quad\n a^k\\mid\\Delta_1^k f(n)\\ \\text{for all }k,n.\n\\]\nThe exact discrete product rule\n\\[\n\\Delta_1^s(fg)(n)=\\sum_{i=0}^s\\binom si\n     \\Delta_1^i f(n)\\,\\Delta_1^{s-i}g(n+i)\n\\]\nis proved by induction. It proves closure under products. Applied to $fg=1$, the same formula isolates\n$\\Delta_1^s g(n)f(n+s)$; induction and multiplication by $g(n+s)$ prove closure under actual pointwise inverses.","e over a commutative ring by\n\\[\n\\PP_a(f)\\quad\\Longleftrightarrow\\quad\n a^k\\mid\\Delta_1^k f(n)\\ \\text{for all }k,n.\n\\]\nThe exact discrete product rule\n\\[\n\\Delta_1^s(fg)(n)=\\sum_{i=0}^s\\binom si\n     \\Delta_1^i f(n)\\,\\Delta_1^{s-i}g(n+i)\n\\]\nis proved by induction. It proves closure under products. Applied to $fg=1$, the same formula isolates\n$\\Delta_1^s g(n)f(n+s)$; induction and multiplication by $g(n+s)$ prove closure under actual pointwise inverses.","An affine sequence whose slope is divisible by $a$ satisfies $\\PP_a$. Consequently the four displayed source sequences satisfy $\\PP_2$. No numerical interpolation or finite set of samples is used in that deduction.","\\par\\medskip\n\\subsubsection*{3. The dyadic differential operator and reciprocity}\nThe operator required here is\n\\[\n\\boxed{B_4=\\frac{D^8-D^4}{2},\\qquad D=\\frac{d}{dt}.}\n\\]\nIt is not the preceding $p=2$ specialization $(D^4-D^2)/2$. It is also not an ordinary derivation. The new algebra module proves the actual product formula, including the middle derivative sums whose coefficients are $\\binom 8i/2$ and $\\binom 4i/2$.","dskip\n\\subsubsection*{3. The dyadic differential operator and reciprocity}\nThe operator required here is\n\\[\n\\boxed{B_4=\\frac{D^8-D^4}{2},\\qquad D=\\frac{d}{dt}.}\n\\]\nIt is not the preceding $p=2$ specialization $(D^4-D^2)/2$. It is also not an ordinary derivation. The new algebra module proves the actual product formula, including the middle derivative sums whose coefficients are $\\binom 8i/2$ and $\\binom 4i/2$.","Write $\\mathcal H_{\\mathrm{int}}$ for the subring of $\\Q_2[[t]]$ whose factorial-scaled coefficients belong to $\\Z_2$. Flatness means $B_4^sF\\in\\mathcal H_{\\mathrm{int}}$ for every $s$. The proof of multiplicative and reciprocal closure is induction on $s$, using the full product correction. It separately proves that the actual reciprocal has integral Hurwitz coefficients when $F(0)=1$.","Pack $g_j$ into degrees divisible by four and form its actual EGF. The formal coefficient identity is\n\\[\nG_j(t)=\\sum_{n\\ge0}g_j(n)\\frac{t^{4n}}{(4n)!}\n       =j!\\sum_{n\\ge0}\\frac{t^{4n}}{(4n+j)!}.\n\\]\nIts inverse is therefore the original normalized packet EGF. The source Kummer property gives flatness of $G_j$. The proved reciprocal closure gives flatness of $G_j^{-1}$.","For a packed coefficient sequence $e$, the exact extraction formula is\n\\[\n\\bigl(U^8-U^4\\bigr)^k\\operatorname{pack}_4(e)(4n-4k)\n =\\Delta_1^k e(n),\\qquad k\\le n.\n\\]\nHere $U$ is the coefficient shift. The nonnegative extraction index is why the first estimate covers $k\\le n$. The existing saturation lemma extends it to all $k$ with exponent $\\min(k,n)$, and the step-multiple lemma extends step one to every natural $u$.","fficient sequence $e$, the exact extraction formula is\n\\[\n\\bigl(U^8-U^4\\bigr)^k\\operatorname{pack}_4(e)(4n-4k)\n =\\Delta_1^k e(n),\\qquad k\\le n.\n\\]\nHere $U$ is the coefficient shift. The nonnegative extraction index is why the first estimate covers $k\\le n$. The existing saturation lemma extends it to all $k$ with exponent $\\min(k,n)$, and the step-multiple lemma extends step one to every natural $u$.","\\subsection*{Identity with the original rational coefficients}\nThe integral lift is constructed from the decoded coefficients of the actual reciprocal. The code proves\n\\[\n \\operatorname{coe}(e_j(n))=C_{4,j}(n)\\quad\\text{in }\\Q_2,\n\\]\nand proves that the coefficient map commutes with every difference stencil. Multiplying by $j!$ yields the unnormalized family. The valuation result is not divisibility in a field.","\\par\\medskip\n\\subsubsection*{4. From differences to the exact external period}\nLet $f$ take values in a commutative ring and suppose\n\\[\n p^{\\min(k,r)}\\mid\\Delta_h^k f(n)\\qquad\\text{for every }k.\n\\]\nThe proved finite Newton expansion is\n\\[\n f(n+p^{r-1}h)-f(n)=\n \\sum_{k=1}^{p^{r-1}}\\binom{p^{r-1}}k\\Delta_h^k f(n).\n\\]\nFor $k\\ge r$, the difference already supplies $p^r$. For $1\\le k<r$, the code proves the integer divisibility\n\\[\n p^r\\mid\\binom{p^{r-1}}k p^k.\n\\]\nIt follows from $k\\binom{p^{r-1}}k=p^{r-1}\\binom{p^{r-1}-1}{k-1}$ and $v_p(k)<k$. Thus every summand is divisible by $p^r$, including the endpoint $r=1$.","4. From differences to the exact external period}\nLet $f$ take values in a commutative ring and suppose\n\\[\n p^{\\min(k,r)}\\mid\\Delta_h^k f(n)\\qquad\\text{for every }k.\n\\]\nThe proved finite Newton expansion is\n\\[\n f(n+p^{r-1}h)-f(n)=\n \\sum_{k=1}^{p^{r-1}}\\binom{p^{r-1}}k\\Delta_h^k f(n).\n\\]\nFor $k\\ge r$, the difference already supplies $p^r$. For $1\\le k<r$, the code proves the integer divisibility\n\\[\n p^r\\mid\\binom{p^{r-1}}k p^k.\n\\]\nIt follows from $k\\binom{p^{r-1}}k=p^{r-1}\\binom{p^{r-1}-1}{k-1}$ and $v_p(k)<k$. Thus every summand is divisible by $p^r$, including the endpoint $r=1$.","et $f$ take values in a commutative ring and suppose\n\\[\n p^{\\min(k,r)}\\mid\\Delta_h^k f(n)\\qquad\\text{for every }k.\n\\]\nThe proved finite Newton expansion is\n\\[\n f(n+p^{r-1}h)-f(n)=\n \\sum_{k=1}^{p^{r-1}}\\binom{p^{r-1}}k\\Delta_h^k f(n).\n\\]\nFor $k\\ge r$, the difference already supplies $p^r$. For $1\\le k<r$, the code proves the integer divisibility\n\\[\n p^r\\mid\\binom{p^{r-1}}k p^k.\n\\]\nIt follows from $k\\binom{p^{r-1}}k=p^{r-1}\\binom{p^{r-1}-1}{k-1}$ and $v_p(k)<k$. Thus every summand is divisible by $p^r$, including the endpoint $r=1$.","[\n p^{\\min(k,r)}\\mid\\Delta_h^k f(n)\\qquad\\text{for every }k.\n\\]\nThe proved finite Newton expansion is\n\\[\n f(n+p^{r-1}h)-f(n)=\n \\sum_{k=1}^{p^{r-1}}\\binom{p^{r-1}}k\\Delta_h^k f(n).\n\\]\nFor $k\\ge r$, the difference already supplies $p^r$. For $1\\le k<r$, the code proves the integer divisibility\n\\[\n p^r\\mid\\binom{p^{r-1}}k p^k.\n\\]\nIt follows from $k\\binom{p^{r-1}}k=p^{r-1}\\binom{p^{r-1}-1}{k-1}$ and $v_p(k)<k$. Thus every summand is divisible by $p^r$, including the endpoint $r=1$.","for every }k.\n\\]\nThe proved finite Newton expansion is\n\\[\n f(n+p^{r-1}h)-f(n)=\n \\sum_{k=1}^{p^{r-1}}\\binom{p^{r-1}}k\\Delta_h^k f(n).\n\\]\nFor $k\\ge r$, the difference already supplies $p^r$. For $1\\le k<r$, the code proves the integer divisibility\n\\[\n p^r\\mid\\binom{p^{r-1}}k p^k.\n\\]\nIt follows from $k\\binom{p^{r-1}}k=p^{r-1}\\binom{p^{r-1}-1}{k-1}$ and $v_p(k)<k$. Thus every summand is divisible by $p^r$, including the endpoint $r=1$.","nite Newton expansion is\n\\[\n f(n+p^{r-1}h)-f(n)=\n \\sum_{k=1}^{p^{r-1}}\\binom{p^{r-1}}k\\Delta_h^k f(n).\n\\]\nFor $k\\ge r$, the difference already supplies $p^r$. For $1\\le k<r$, the code proves the integer divisibility\n\\[\n p^r\\mid\\binom{p^{r-1}}k p^k.\n\\]\nIt follows from $k\\binom{p^{r-1}}k=p^{r-1}\\binom{p^{r-1}-1}{k-1}$ and $v_p(k)<k$. Thus every summand is divisible by $p^r$, including the endpoint $r=1$.","\\subsection*{Divisor branch}\nFor $m\\mid p-1$, the preceding all-residue theorem provides the needed differences as soon as $r\\le mn$. With $q=\\operatorname{lcm}(2,p-1)$, the required packet step $q/m$ is proved to be an integer multiple of $(p-1)/m$. The degree conversion is checked exactly:\n\\[\n mp\\left(n+p^{r-1}\\frac qm\\right)=mpn+qp^r.\n\\]\nThis retains the stronger sufficient condition $r\\le mn$ on the divisor branch. The uniform choice $n\\ge r$ implies it for every positive $m$.","$, the preceding all-residue theorem provides the needed differences as soon as $r\\le mn$. With $q=\\operatorname{lcm}(2,p-1)$, the required packet step $q/m$ is proved to be an integer multiple of $(p-1)/m$. The degree conversion is checked exactly:\n\\[\n mp\\left(n+p^{r-1}\\frac qm\\right)=mpn+qp^r.\n\\]\nThis retains the stronger sufficient condition $r\\le mn$ on the divisor branch. The uniform choice $n\\ge r$ implies it for every positive $m$.","ovides the needed differences as soon as $r\\le mn$. With $q=\\operatorname{lcm}(2,p-1)$, the required packet step $q/m$ is proved to be an integer multiple of $(p-1)/m$. The degree conversion is checked exactly:\n\\[\n mp\\left(n+p^{r-1}\\frac qm\\right)=mpn+qp^r.\n\\]\nThis retains the stronger sufficient condition $r\\le mn$ on the divisor branch. The uniform choice $n\\ge r$ implies it for every positive $m$.","\\subsection*{Exceptional branch}\nFor $p=2,m=2$, $q=2$ and the packet displacement is $2^{r-1}$. The new dyadic bound supplies the hypothesis when $n\\ge r$, and\n\\[\n 4\\bigl(n+2^{r-1}\\bigr)=4n+2^{r+1}.\n\\]\nThe $p=2,m=1$ branch is also checked using $q=2$, not an incorrectly substituted $q=p-1=1$.","\\subsection*{An exact arithmetic check, not an optimality claim}\nThe independent rational recurrence gives\n$C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=121/45$. Their difference is $26/9$, of $2$-adic valuation one. This agrees with the new bound at $n=k=u=1$. The source also contains universal instantiations of the degree-indexed theorem at all four dyadic addresses and at an odd singular address.","\\subsection*{Reference}\nYuta Nishibuchi. \\emph{$p$-adic Congruences of Generalized Euler Numbers and Relations to Even Zeta Values}. arXiv:2605.10677v2, revised 1 June 2026. Definition 2.1, Conjecture 3.9 and Proposition 3.10. The version identifier is fixed for the statement comparison.","To distinguish ordinary signed Euler numbers from the generalized numbers\nof the main text, write\n\\[\n \\operatorname{sech}t=\\sum_{n\\ge0}\\mathcal E_n\\frac{t^n}{n!},\n \\qquad \\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\n \\qquad B_1=-\\frac12.\n\\]\nThroughout this appendix, $v_2(2)=1$ and $v_2(0)=\\infty$.\nLet $s_2(n)$ be the number of ones in the binary expansion of the\nnonnegative integer $n$. Define\n\\[\n A(N)=\\frac{3^{2N+2}-1}{4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\l","To distinguish ordinary signed Euler numbers from the generalized numbers\nof the main text, write\n\\[\n \\operatorname{sech}t=\\sum_{n\\ge0}\\mathcal E_n\\frac{t^n}{n!},\n \\qquad \\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\n \\qquad B_1=-\\frac12.\n\\]\nThroughout this appendix, $v_2(2)=1$ and $v_2(0)=\\infty$.\nLet $s_2(n)$ be the number of ones in the binary expansion of the\nnonnegative integer $n$. Define\n\\[\n A(N)=\\frac{3^{2N+2}-1}{4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nT","To distinguish ordinary signed Euler numbers from the generalized numbers\nof the main text, write\n\\[\n \\operatorname{sech}t=\\sum_{n\\ge0}\\mathcal E_n\\frac{t^n}{n!},\n \\qquad \\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\n \\qquad B_1=-\\frac12.\n\\]\nThroughout this appendix, $v_2(2)=1$ and $v_2(0)=\\infty$.\nLet $s_2(n)$ be the number of ones in the binary expansion of the\nnonnegative integer $n$. Define\n\\[\n A(N)=\\frac{3^{2N+2}-1}{4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\","To distinguish ordinary signed Euler numbers from the generalized numbers\nof the main text, write\n\\[\n \\operatorname{sech}t=\\sum_{n\\ge0}\\mathcal E_n\\frac{t^n}{n!},\n \\qquad \\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\n \\qquad B_1=-\\frac12.\n\\]\nThroughout this appendix, $v_2(2)=1$ and $v_2(0)=\\infty$.\nLet $s_2(n)$ be the number of ones in the binary expansion of the\nnonnegative integer $n$. Define\n\\[\n A(N)=\\frac{3^{2N+2}-1}{4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0","To distinguish ordinary signed Euler numbers from the generalized numbers\nof the main text, write\n\\[\n \\operatorname{sech}t=\\sum_{n\\ge0}\\mathcal E_n\\frac{t^n}{n!},\n \\qquad \\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\n \\qquad B_1=-\\frac12.\n\\]\nThroughout this appendix, $v_2(2)=1$ and $v_2(0)=\\infty$.\nLet $s_2(n)$ be the number of ones in the binary expansion of the\nnonnegative integer $n$. Define\n\\[\n A(N)=\\frac{3^{2N+2}-1}{4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after m","o distinguish ordinary signed Euler numbers from the generalized numbers\nof the main text, write\n\\[\n \\operatorname{sech}t=\\sum_{n\\ge0}\\mathcal E_n\\frac{t^n}{n!},\n \\qquad \\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\n \\qquad B_1=-\\frac12.\n\\]\nThroughout this appendix, $v_2(2)=1$ and $v_2(0)=\\infty$.\nLet $s_2(n)$ be the number of ones in the binary expansion of the\nnonnegative integer $n$. Define\n\\[\n A(N)=\\frac{3^{2N+2}-1}{4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplicatio…","rac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\n \\qquad B_1=-\\frac12.\n\\]\nThroughout this appendix, $v_2(2)=1$ and $v_2(0)=\\infty$.\nLet $s_2(n)$ be the number of ones in the binary expansion of the\nnonnegative integer $n$. Define\n\\[\n A(N)=\\frac{3^{2N+2}-1}{4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0…","B_n\\frac{t^n}{n!},\n \\qquad B_1=-\\frac12.\n\\]\nThroughout this appendix, $v_2(2)=1$ and $v_2(0)=\\infty$.\nLet $s_2(n)$ be the number of ones in the binary expansion of the\nnonnegative integer $n$. Define\n\\[\n A(N)=\\frac{3^{2N+2}-1}{4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)…","_1=-\\frac12.\n\\]\nThroughout this appendix, $v_2(2)=1$ and $v_2(0)=\\infty$.\nLet $s_2(n)$ be the number of ones in the binary expansion of the\nnonnegative integer $n$. Define\n\\[\n A(N)=\\frac{3^{2N+2}-1}{4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)=\\mathcal E_{2N}\n   -\\sum_{m…","appendix, $v_2(2)=1$ and $v_2(0)=\\infty$.\nLet $s_2(n)$ be the number of ones in the binary expansion of the\nnonnegative integer $n$. Define\n\\[\n A(N)=\\frac{3^{2N+2}-1}{4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)=\\mathcal E_{2N}\n   -\\sum_{m=0}^{\\min(M,N)}\\kappa_m\\binom{2N…","4(2N+1)(2N+2)},\\qquad\n \\left(\\frac z{\\sinh z}\\right)^2=\\sum_{m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)=\\mathcal E_{2N}\n   -\\sum_{m=0}^{\\min(M,N)}\\kappa_m\\binom{2N}{2m}A(N-m).\n\\]\nAn empty tail, $N\\le M$, is exactly zero.","m\\ge0}\\beta_m z^{2m},\n \\qquad \\kappa_m=2^{2m}(2m)!\\beta_m.\n\\]\nThe initial values are $\\beta_0=\\kappa_0=1$. For $m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)=\\mathcal E_{2N}\n   -\\sum_{m=0}^{\\min(M,N)}\\kappa_m\\binom{2N}{2m}A(N-m).\n\\]\nAn empty tail, $N\\le M$, is exactly zero.","m\\ge1$, differentiation\nof $z\\coth z$ gives\n\\begin{equation}\n \\beta_m=-\\frac{2^{2m}(2m-1)B_{2m}}{(2m)!},\\qquad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)=\\mathcal E_{2N}\n   -\\sum_{m=0}^{\\min(M,N)}\\kappa_m\\binom{2N}{2m}A(N-m).\n\\]\nAn empty tail, $N\\le M$, is exactly zero.","ad\n \\kappa_m=-2^{4m}(2m-1)B_{2m}.\n \\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)=\\mathcal E_{2N}\n   -\\sum_{m=0}^{\\min(M,N)}\\kappa_m\\binom{2N}{2m}A(N-m).\n\\]\nAn empty tail, $N\\le M$, is exactly zero.","\\label{eq:root:beta}\n\\end{equation}\nThe classical dyadic Bernoulli denominator law\n$v_2(B_{2m})=-1$ and Legendre's formula yield\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)=\\mathcal E_{2N}\n   -\\sum_{m=0}^{\\min(M,N)}\\kappa_m\\binom{2N}{2m}A(N-m).\n\\]\nAn empty tail, $N\\le M$, is exactly zero.","ld\n\\begin{equation}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)=\\mathcal E_{2N}\n   -\\sum_{m=0}^{\\min(M,N)}\\kappa_m\\binom{2N}{2m}A(N-m).\n\\]\nAn empty tail, $N\\le M$, is exactly zero.","tion}\n v_2(\\beta_m)=s_2(m)-1,\\qquad\n v_2(\\kappa_m)=4m-1.\\label{eq:root:weights}\n\\end{equation}\nThe source identity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)=\\mathcal E_{2N}\n   -\\sum_{m=0}^{\\min(M,N)}\\kappa_m\\binom{2N}{2m}A(N-m).\n\\]\nAn empty tail, $N\\le M$, is exactly zero.","ntity\n\\[\n \\sum_{N\\ge0}A(N)\\frac{t^{2N}}{(2N)!}\n =\\frac{\\sinh(2t)\\sinh t}{2t^2}\n\\]\nimplies, after multiplication by $4t^2/\\sinh^2(2t)$,\n\\begin{equation}\n \\mathcal E_{2N}=\\sum_{m=0}^N\n    \\kappa_m\\binom{2N}{2m}A(N-m).\n \\label{eq:root:packet}\n\\end{equation}\nFor a retained depth $M\\ge0$, put\n\\[\n \\mathcal T_M(N)=\\mathcal E_{2N}\n   -\\sum_{m=0}^{\\min(M,N)}\\kappa_m\\binom{2N}{2m}A(N-m).\n\\]\nAn empty tail, $N\\le M$, is exactly zero.","\\begin{proposition}[Sharp uniform classical truncation bound]\nFor every $M,N\\ge0$,\n\\[\n v_2\\bigl(\\mathcal T_M(N)\\bigr)\\ge4M+3.\n\\]\nFor each fixed $M$, equality holds at $N=M+1$.\n\\end{proposition}\n\\begin{proof}\nThe source $A(n)$ is a dyadic unit for every nonnegative integer $n$.\nEach nonempty summand has $m\\ge M+1$, so\n\\eqref{eq:root:weights} and integrality of the binomial coefficient give\nthe asserted bound. At $N=M+1$, only the first omitted summand remains;\nits binomial coefficient and $A(0)$ both equal one.\n\\end{proof}","on}[Sharp uniform classical truncation bound]\nFor every $M,N\\ge0$,\n\\[\n v_2\\bigl(\\mathcal T_M(N)\\bigr)\\ge4M+3.\n\\]\nFor each fixed $M$, equality holds at $N=M+1$.\n\\end{proposition}\n\\begin{proof}\nThe source $A(n)$ is a dyadic unit for every nonnegative integer $n$.\nEach nonempty summand has $m\\ge M+1$, so\n\\eqref{eq:root:weights} and integrality of the binomial coefficient give\nthe asserted bound. At $N=M+1$, only the first omitted summand remains;\nits binomial coefficient and $A(0)$ both equal one.\n\\end{proof}","l(\\mathcal T_M(N)\\bigr)\\ge4M+3.\n\\]\nFor each fixed $M$, equality holds at $N=M+1$.\n\\end{proposition}\n\\begin{proof}\nThe source $A(n)$ is a dyadic unit for every nonnegative integer $n$.\nEach nonempty summand has $m\\ge M+1$, so\n\\eqref{eq:root:weights} and integrality of the binomial coefficient give\nthe asserted bound. At $N=M+1$, only the first omitted summand remains;\nits binomial coefficient and $A(0)$ both equal one.\n\\end{proof}","Let $\\mathbb Z_2\\langle x\\rangle$ denote the restricted power series:\ntheir integral coefficients tend to zero dyadically. Extend $A$ by\n\\[\n \\mathscr A(x)=\\frac{9^{x+1}-1}{8(x+1)(2x+1)},\n \\qquad \\mathscr A(-1)=-\\frac{\\log9}{8}.\n\\]\nThe apparent singularity at $x=-1$ is removable. Since\n$v_2(\\log9)=3$, its convergent expansion is\n\\begin{equation}\n \\mathscr A(x)=\\frac{\\log9/8}{1+2x}\n   \\sum_{h\\ge0}\\frac{(\\log9)^h}{(h+1)!}(x+1)^h.\n \\label{eq:root:source}\n\\end{equation}\nIt follows coefficientwise that\n$\\mathscr A\\in\\mathbb Z_2\\langle x\\rangle^\\times$,\n$\\mathscr A\\equiv1\\pmod2$, and $\\mathscr A(N)=A(N)$.","to zero dyadically. Extend $A$ by\n\\[\n \\mathscr A(x)=\\frac{9^{x+1}-1}{8(x+1)(2x+1)},\n \\qquad \\mathscr A(-1)=-\\frac{\\log9}{8}.\n\\]\nThe apparent singularity at $x=-1$ is removable. Since\n$v_2(\\log9)=3$, its convergent expansion is\n\\begin{equation}\n \\mathscr A(x)=\\frac{\\log9/8}{1+2x}\n   \\sum_{h\\ge0}\\frac{(\\log9)^h}{(h+1)!}(x+1)^h.\n \\label{eq:root:source}\n\\end{equation}\nIt follows coefficientwise that\n$\\mathscr A\\in\\mathbb Z_2\\langle x\\rangle^\\times$,\n$\\mathscr A\\equiv1\\pmod2$, and $\\mathscr A(N)=A(N)$.","=\\frac{9^{x+1}-1}{8(x+1)(2x+1)},\n \\qquad \\mathscr A(-1)=-\\frac{\\log9}{8}.\n\\]\nThe apparent singularity at $x=-1$ is removable. Since\n$v_2(\\log9)=3$, its convergent expansion is\n\\begin{equation}\n \\mathscr A(x)=\\frac{\\log9/8}{1+2x}\n   \\sum_{h\\ge0}\\frac{(\\log9)^h}{(h+1)!}(x+1)^h.\n \\label{eq:root:source}\n\\end{equation}\nIt follows coefficientwise that\n$\\mathscr A\\in\\mathbb Z_2\\langle x\\rangle^\\times$,\n$\\mathscr A\\equiv1\\pmod2$, and $\\mathscr A(N)=A(N)$.",",\n \\qquad \\mathscr A(-1)=-\\frac{\\log9}{8}.\n\\]\nThe apparent singularity at $x=-1$ is removable. Since\n$v_2(\\log9)=3$, its convergent expansion is\n\\begin{equation}\n \\mathscr A(x)=\\frac{\\log9/8}{1+2x}\n   \\sum_{h\\ge0}\\frac{(\\log9)^h}{(h+1)!}(x+1)^h.\n \\label{eq:root:source}\n\\end{equation}\nIt follows coefficientwise that\n$\\mathscr A\\in\\mathbb Z_2\\langle x\\rangle^\\times$,\n$\\mathscr A\\equiv1\\pmod2$, and $\\mathscr A(N)=A(N)$.","Set $r=M+1\\ge1$ and define\n\\[\n P_q(x)=\\prod_{i=0}^{q-1}(x-i)(2x-(2i+1)),\\qquad P_0=1,\n\\]\n\\[\n H_r(x)=\\sum_{q\\ge0}8^q\\frac{\\beta_{r+q}}{\\beta_r}\n                    P_q(x)\\mathscr A(x-q).\n\\]\nThe coefficient valuations tend to infinity. At any nonnegative integer\n$x$, terms with $q>x$ vanish. Cancelling the factorials in\n\\eqref{eq:root:packet} therefore gives the exact identity\n\\begin{equation}\n \\mathcal T_{r-1}(N)=\\kappa_r\\binom{2N}{2r}H_r(N-r)\n \\qquad(N\\ge r).\\label{eq:root:factor}\n\\end{equation}\nThis is a finite rational identity at every original Euler row.","The scalar packet weights have exact valuations\n\\[\n \\lambda_r(q)=3q+s_2(r+q)-s_2(r).\n\\]\nPut\n\\[\n \\mu_r=\\min_{q\\ge0}\\lambda_r(q),\\qquad\n J_r=\\{q:\\lambda_r(q)=\\mu_r\\},\\qquad G_r=2^{-\\mu_r}H_r.\n\\]\nSince $\\lambda_r(0)=0$ and\n$\\lambda_r(q)\\ge3q+1-s_2(r)$, every minimizing address satisfies\n\\begin{equation}\n 0\\le q\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n \\label{eq:root:cutoff}\n\\end{equation}\nIn particular $\\mu_r\\le0$, and the minimum is found by a finite carry\ncalculation. Reduction of the whole series gives\n\\begin{equation}\n \\overline G_r(X)=\\sum_{q\\in J_r}X^{\\underline q}\n       \\quad\\text{in }\\mathbb F_2[X].\n \\label{eq:root:reduction}\n\\end{equation}\nIts highest degree is $\\max J_r$, with leading coefficient one.\nThus the normalization is the actual Gauss normalization.","packet weights have exact valuations\n\\[\n \\lambda_r(q)=3q+s_2(r+q)-s_2(r).\n\\]\nPut\n\\[\n \\mu_r=\\min_{q\\ge0}\\lambda_r(q),\\qquad\n J_r=\\{q:\\lambda_r(q)=\\mu_r\\},\\qquad G_r=2^{-\\mu_r}H_r.\n\\]\nSince $\\lambda_r(0)=0$ and\n$\\lambda_r(q)\\ge3q+1-s_2(r)$, every minimizing address satisfies\n\\begin{equation}\n 0\\le q\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n \\label{eq:root:cutoff}\n\\end{equation}\nIn particular $\\mu_r\\le0$, and the minimum is found by a finite carry\ncalculation. Reduction of the whole series gives\n\\begin{equation}\n \\overline G_r(X)=\\sum_{q\\in J_r}X^{\\underline q}\n       \\quad\\text{in }\\mathbb F_2[X].\n \\label{eq:root:reduction}\n\\end{equation}\nIts highest degree is $\\max J_r$, with leading coefficient one.\nThus the normalization is the actual Gauss normalization.","ad\n J_r=\\{q:\\lambda_r(q)=\\mu_r\\},\\qquad G_r=2^{-\\mu_r}H_r.\n\\]\nSince $\\lambda_r(0)=0$ and\n$\\lambda_r(q)\\ge3q+1-s_2(r)$, every minimizing address satisfies\n\\begin{equation}\n 0\\le q\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n \\label{eq:root:cutoff}\n\\end{equation}\nIn particular $\\mu_r\\le0$, and the minimum is found by a finite carry\ncalculation. Reduction of the whole series gives\n\\begin{equation}\n \\overline G_r(X)=\\sum_{q\\in J_r}X^{\\underline q}\n       \\quad\\text{in }\\mathbb F_2[X].\n \\label{eq:root:reduction}\n\\end{equation}\nIts highest degree is $\\max J_r$, with leading coefficient one.\nThus the normalization is the actual Gauss normalization.","\\begin{equation}\n 0\\le q\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n \\label{eq:root:cutoff}\n\\end{equation}\nIn particular $\\mu_r\\le0$, and the minimum is found by a finite carry\ncalculation. Reduction of the whole series gives\n\\begin{equation}\n \\overline G_r(X)=\\sum_{q\\in J_r}X^{\\underline q}\n       \\quad\\text{in }\\mathbb F_2[X].\n \\label{eq:root:reduction}\n\\end{equation}\nIts highest degree is $\\max J_r$, with leading coefficient one.\nThus the normalization is the actual Gauss normalization.","\\begin{theorem}[Exact classical tail valuation]\nFor every $r\\ge1$ and $N\\ge r$,\n\\begin{equation}\n v_2\\bigl(\\mathcal T_{r-1}(N)\\bigr)\n =4r-1+v_2\\binom Nr+\\mu_r+v_2G_r(N-r).\n \\label{eq:root:exactvaluation}\n\\end{equation}\nThere are unique factors\n\\[\n G_r(X)=W_r(X)U_r(X),\n\\]\nwhere $W_r\\in\\mathbb Z_2[X]$ is monic of degree $d_r=\\max J_r$ and\n$U_r\\in\\mathbb Z_2\\langle X\\rangle^\\times$ satisfies $U_r\\equiv1\\pmod2$.\nConsequently $v_2G_r(x)=v_2W_r(x)$ for every $x\\in\\mathbb Z_2$, and\n\\[\n \\deg W_r\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n\\]\n\\label{thm:root:finitepreparation}\n\\end{theorem}\n\\begin{proof}\nEquation~\\eqref{eq:root:factor}, the exact weight in\n\\eqref{eq:root:weights}, and the digit-sum identity\n$v_2\\binom{2N}{2r}=v_2\\binom Nr$ prove\n\\eqref{eq:root","\\begin{theorem}[Exact classical tail valuation]\nFor every $r\\ge1$ and $N\\ge r$,\n\\begin{equation}\n v_2\\bigl(\\mathcal T_{r-1}(N)\\bigr)\n =4r-1+v_2\\binom Nr+\\mu_r+v_2G_r(N-r).\n \\label{eq:root:exactvaluation}\n\\end{equation}\nThere are unique factors\n\\[\n G_r(X)=W_r(X)U_r(X),\n\\]\nwhere $W_r\\in\\mathbb Z_2[X]$ is monic of degree $d_r=\\max J_r$ and\n$U_r\\in\\mathbb Z_2\\langle X\\rangle^\\times$ satisfies $U_r\\equiv1\\pmod2$.\nConsequently $v_2G_r(x)=v_2W_r(x)$ for every $x\\in\\mathbb Z_2$, and\n\\[\n \\deg W_r\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n\\]\n\\label{thm:root:finitepreparation}\n\\end{theorem}\n\\begin{proof}\nEquation~\\eqref{eq:root:factor}, the exact weight in\n\\eqref{eq:root:weights}, and the digit-sum identity\n$v_2\\binom{2N}{2r}=v_2\\binom Nr$ prove\n\\eqref{eq:root:exactvaluati","\\begin{theorem}[Exact classical tail valuation]\nFor every $r\\ge1$ and $N\\ge r$,\n\\begin{equation}\n v_2\\bigl(\\mathcal T_{r-1}(N)\\bigr)\n =4r-1+v_2\\binom Nr+\\mu_r+v_2G_r(N-r).\n \\label{eq:root:exactvaluation}\n\\end{equation}\nThere are unique factors\n\\[\n G_r(X)=W_r(X)U_r(X),\n\\]\nwhere $W_r\\in\\mathbb Z_2[X]$ is monic of degree $d_r=\\max J_r$ and\n$U_r\\in\\mathbb Z_2\\langle X\\rangle^\\times$ satisfies $U_r\\equiv1\\pmod2$.\nConsequently $v_2G_r(x)=v_2W_r(x)$ for every $x\\in\\mathbb Z_2$, and\n\\[\n \\deg W_r\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n\\]\n\\label{thm:root:finitepreparation}\n\\end{theorem}\n\\begin{proof}\nEquation~\\eqref{eq:root:factor}, the exact weight in\n\\eqref{eq:root:weights}, and the digit-sum identity\n$v_2\\binom{2N}{2r}=v_2\\binom Nr$ prove\n\\eqref{eq:root:exactvaluation}. The binomial coefficients themselves\nneed not be equal.","sical tail valuation]\nFor every $r\\ge1$ and $N\\ge r$,\n\\begin{equation}\n v_2\\bigl(\\mathcal T_{r-1}(N)\\bigr)\n =4r-1+v_2\\binom Nr+\\mu_r+v_2G_r(N-r).\n \\label{eq:root:exactvaluation}\n\\end{equation}\nThere are unique factors\n\\[\n G_r(X)=W_r(X)U_r(X),\n\\]\nwhere $W_r\\in\\mathbb Z_2[X]$ is monic of degree $d_r=\\max J_r$ and\n$U_r\\in\\mathbb Z_2\\langle X\\rangle^\\times$ satisfies $U_r\\equiv1\\pmod2$.\nConsequently $v_2G_r(x)=v_2W_r(x)$ for every $x\\in\\mathbb Z_2$, and\n\\[\n \\deg W_r\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n\\]\n\\label{thm:root:finitepreparation}\n\\end{theorem}\n\\begin{proof}\nEquation~\\eqref{eq:root:factor}, the exact weight in\n\\eqref{eq:root:weights}, and the digit-sum identity\n$v_2\\binom{2N}{2r}=v_2\\binom Nr$ prove\n\\eqref{eq:root:exactvaluation}. The binomial coefficients themselves\nneed not be equal.","$r\\ge1$ and $N\\ge r$,\n\\begin{equation}\n v_2\\bigl(\\mathcal T_{r-1}(N)\\bigr)\n =4r-1+v_2\\binom Nr+\\mu_r+v_2G_r(N-r).\n \\label{eq:root:exactvaluation}\n\\end{equation}\nThere are unique factors\n\\[\n G_r(X)=W_r(X)U_r(X),\n\\]\nwhere $W_r\\in\\mathbb Z_2[X]$ is monic of degree $d_r=\\max J_r$ and\n$U_r\\in\\mathbb Z_2\\langle X\\rangle^\\times$ satisfies $U_r\\equiv1\\pmod2$.\nConsequently $v_2G_r(x)=v_2W_r(x)$ for every $x\\in\\mathbb Z_2$, and\n\\[\n \\deg W_r\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n\\]\n\\label{thm:root:finitepreparation}\n\\end{theorem}\n\\begin{proof}\nEquation~\\eqref{eq:root:factor}, the exact weight in\n\\eqref{eq:root:weights}, and the digit-sum identity\n$v_2\\binom{2N}{2r}=v_2\\binom Nr$ prove\n\\eqref{eq:root:exactvaluation}. The binomial coefficients themselves\nneed not be equal.","egin{equation}\n v_2\\bigl(\\mathcal T_{r-1}(N)\\bigr)\n =4r-1+v_2\\binom Nr+\\mu_r+v_2G_r(N-r).\n \\label{eq:root:exactvaluation}\n\\end{equation}\nThere are unique factors\n\\[\n G_r(X)=W_r(X)U_r(X),\n\\]\nwhere $W_r\\in\\mathbb Z_2[X]$ is monic of degree $d_r=\\max J_r$ and\n$U_r\\in\\mathbb Z_2\\langle X\\rangle^\\times$ satisfies $U_r\\equiv1\\pmod2$.\nConsequently $v_2G_r(x)=v_2W_r(x)$ for every $x\\in\\mathbb Z_2$, and\n\\[\n \\deg W_r\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n\\]\n\\label{thm:root:finitepreparation}\n\\end{theorem}\n\\begin{proof}\nEquation~\\eqref{eq:root:factor}, the exact weight in\n\\eqref{eq:root:weights}, and the digit-sum identity\n$v_2\\binom{2N}{2r}=v_2\\binom Nr$ prove\n\\eqref{eq:root:exactvaluation}. The binomial coefficients themselves\nneed not be equal.","degree $d_r=\\max J_r$ and\n$U_r\\in\\mathbb Z_2\\langle X\\rangle^\\times$ satisfies $U_r\\equiv1\\pmod2$.\nConsequently $v_2G_r(x)=v_2W_r(x)$ for every $x\\in\\mathbb Z_2$, and\n\\[\n \\deg W_r\\le\\left\\lfloor\\frac{s_2(r)-1}{3}\\right\\rfloor.\n\\]\n\\label{thm:root:finitepreparation}\n\\end{theorem}\n\\begin{proof}\nEquation~\\eqref{eq:root:factor}, the exact weight in\n\\eqref{eq:root:weights}, and the digit-sum identity\n$v_2\\binom{2N}{2r}=v_2\\binom Nr$ prove\n\\eqref{eq:root:exactvaluation}. The binomial coefficients themselves\nneed not be equal.","For preparation, start from the monic polynomial\n$g=\\overline G_r$ and its zero-one lift $W_1$, with $U_1=1$.\nSuppose $G_r-W_kU_k$ is coefficientwise divisible by $2^k$.\nIts divided reduction is a finite polynomial because $G_r$ is restricted.\nDivide it by $g$ in $\\mathbb F_2[X]$:\n\\[\n \\overline{(G_r-W_kU_k)/2^k}=gq+s,\\qquad\\deg s<d_r.\n\\]\nWith zero-one coefficient lifts, set\n$W_{k+1}=W_k+2^ks$ and $U_{k+1}=U_k+2^kq$.\nThe error is now divisible by $2^{k+1}$, the polynomial remains monic of\ndegree $d_r$, and the unit reduces to one. The sequences converge in the\nGauss norm. Uniqueness of quotient and remainder at successive binary\ndigits proves uniqueness of the factors. For $d_r=0$, take $W_r=1$.\nThe degree bound follows from \\eqref{eq:root:cutoff}.\n\\end{pr","For preparation, start from the monic polynomial\n$g=\\overline G_r$ and its zero-one lift $W_1$, with $U_1=1$.\nSuppose $G_r-W_kU_k$ is coefficientwise divisible by $2^k$.\nIts divided reduction is a finite polynomial because $G_r$ is restricted.\nDivide it by $g$ in $\\mathbb F_2[X]$:\n\\[\n \\overline{(G_r-W_kU_k)/2^k}=gq+s,\\qquad\\deg s<d_r.\n\\]\nWith zero-one coefficient lifts, set\n$W_{k+1}=W_k+2^ks$ and $U_{k+1}=U_k+2^kq$.\nThe error is now divisible by $2^{k+1}$, the polynomial remains monic of\ndegree $d_r$, and the unit reduces to one. The sequences converge in the\nGauss norm. Uniqueness of quotient and remainder at successive binary\ndigits proves uniqueness of the factors. For $d_r=0$, take $W_r=1$.\nThe degree bound follows from \\eqref{eq:root:cutoff}.\n\\end{proof}","or preparation, start from the monic polynomial\n$g=\\overline G_r$ and its zero-one lift $W_1$, with $U_1=1$.\nSuppose $G_r-W_kU_k$ is coefficientwise divisible by $2^k$.\nIts divided reduction is a finite polynomial because $G_r$ is restricted.\nDivide it by $g$ in $\\mathbb F_2[X]$:\n\\[\n \\overline{(G_r-W_kU_k)/2^k}=gq+s,\\qquad\\deg s<d_r.\n\\]\nWith zero-one coefficient lifts, set\n$W_{k+1}=W_k+2^ks$ and $U_{k+1}=U_k+2^kq$.\nThe error is now divisible by $2^{k+1}$, the polynomial remains monic of\ndegree $d_r$, and the unit reduces to one. The sequences converge in the\nGauss norm. Uniqueness of quotient and remainder at successive binary\ndigits proves uniqueness of the factors. For $d_r=0$, take $W_r=1$.\nThe degree bound follows from \\eqref{eq:root:cutoff}.\n\\end{proof}","=\\overline G_r$ and its zero-one lift $W_1$, with $U_1=1$.\nSuppose $G_r-W_kU_k$ is coefficientwise divisible by $2^k$.\nIts divided reduction is a finite polynomial because $G_r$ is restricted.\nDivide it by $g$ in $\\mathbb F_2[X]$:\n\\[\n \\overline{(G_r-W_kU_k)/2^k}=gq+s,\\qquad\\deg s<d_r.\n\\]\nWith zero-one coefficient lifts, set\n$W_{k+1}=W_k+2^ks$ and $U_{k+1}=U_k+2^kq$.\nThe error is now divisible by $2^{k+1}$, the polynomial remains monic of\ndegree $d_r$, and the unit reduces to one. The sequences converge in the\nGauss norm. Uniqueness of quotient and remainder at successive binary\ndigits proves uniqueness of the factors. For $d_r=0$, take $W_r=1$.\nThe degree bound follows from \\eqref{eq:root:cutoff}.\n\\end{proof}","$U_1=1$.\nSuppose $G_r-W_kU_k$ is coefficientwise divisible by $2^k$.\nIts divided reduction is a finite polynomial because $G_r$ is restricted.\nDivide it by $g$ in $\\mathbb F_2[X]$:\n\\[\n \\overline{(G_r-W_kU_k)/2^k}=gq+s,\\qquad\\deg s<d_r.\n\\]\nWith zero-one coefficient lifts, set\n$W_{k+1}=W_k+2^ks$ and $U_{k+1}=U_k+2^kq$.\nThe error is now divisible by $2^{k+1}$, the polynomial remains monic of\ndegree $d_r$, and the unit reduces to one. The sequences converge in the\nGauss norm. Uniqueness of quotient and remainder at successive binary\ndigits proves uniqueness of the factors. For $d_r=0$, take $W_r=1$.\nThe degree bound follows from \\eqref{eq:root:cutoff}.\n\\end{proof}","\\[\n \\overline{(G_r-W_kU_k)/2^k}=gq+s,\\qquad\\deg s<d_r.\n\\]\nWith zero-one coefficient lifts, set\n$W_{k+1}=W_k+2^ks$ and $U_{k+1}=U_k+2^kq$.\nThe error is now divisible by $2^{k+1}$, the polynomial remains monic of\ndegree $d_r$, and the unit reduces to one. The sequences converge in the\nGauss norm. Uniqueness of quotient and remainder at successive binary\ndigits proves uniqueness of the factors. For $d_r=0$, take $W_r=1$.\nThe degree bound follows from \\eqref{eq:root:cutoff}.\n\\end{proof}","(G_r-W_kU_k)/2^k}=gq+s,\\qquad\\deg s<d_r.\n\\]\nWith zero-one coefficient lifts, set\n$W_{k+1}=W_k+2^ks$ and $U_{k+1}=U_k+2^kq$.\nThe error is now divisible by $2^{k+1}$, the polynomial remains monic of\ndegree $d_r$, and the unit reduces to one. The sequences converge in the\nGauss norm. Uniqueness of quotient and remainder at successive binary\ndigits proves uniqueness of the factors. For $d_r=0$, take $W_r=1$.\nThe degree bound follows from \\eqref{eq:root:cutoff}.\n\\end{proof}","This theorem reduces every depth to a finite polynomial without discarding\nthe infinite analytic tail. It does not explicitly classify every\n$W_r$, decide every repeated root from a fixed precision, or prove that\nall its dyadic roots are simple.","For a wall label $j\\ge1$ and $0\\le q<j$, define\n\\[\n e_j(q)=3q+s_2(j-1)-s_2(j-q-1),\\qquad\n E_j=\\min_{q<j}e_j(q),\n\\]\n\\[\n \\Theta(j)=3j+1+s_2(j-1),\\qquad L_j=\\Theta(j)-E_j.\n\\]\nHere $E_j$ is a carry minimum, not an Euler number. A wall label is\n\\emph{regular} when $e_j(q)>0$ for $1\\le q<j$.\nFor $r=2^L-j$ with $L>L_j$, put $\\delta=L-L_j$.\nThe carry identities give\n\\[\n \\mu_r=E_j-\\delta,\\qquad J_r=\\{j\\},\\qquad\n G_r(x)=P_j(x)U(x-j)+2^\\delta V(x),\n\\]\nwhere $U\\equiv1\\pmod2$. For a regular label, also\n$E_j=0$ and $V\\equiv1\\pmod2$.","e_j(q)=3q+s_2(j-1)-s_2(j-q-1),\\qquad\n E_j=\\min_{q<j}e_j(q),\n\\]\n\\[\n \\Theta(j)=3j+1+s_2(j-1),\\qquad L_j=\\Theta(j)-E_j.\n\\]\nHere $E_j$ is a carry minimum, not an Euler number. A wall label is\n\\emph{regular} when $e_j(q)>0$ for $1\\le q<j$.\nFor $r=2^L-j$ with $L>L_j$, put $\\delta=L-L_j$.\nThe carry identities give\n\\[\n \\mu_r=E_j-\\delta,\\qquad J_r=\\{j\\},\\qquad\n G_r(x)=P_j(x)U(x-j)+2^\\delta V(x),\n\\]\nwhere $U\\equiv1\\pmod2$. For a regular label, also\n$E_j=0$ and $V\\equiv1\\pmod2$.","in_{q<j}e_j(q),\n\\]\n\\[\n \\Theta(j)=3j+1+s_2(j-1),\\qquad L_j=\\Theta(j)-E_j.\n\\]\nHere $E_j$ is a carry minimum, not an Euler number. A wall label is\n\\emph{regular} when $e_j(q)>0$ for $1\\le q<j$.\nFor $r=2^L-j$ with $L>L_j$, put $\\delta=L-L_j$.\nThe carry identities give\n\\[\n \\mu_r=E_j-\\delta,\\qquad J_r=\\{j\\},\\qquad\n G_r(x)=P_j(x)U(x-j)+2^\\delta V(x),\n\\]\nwhere $U\\equiv1\\pmod2$. For a regular label, also\n$E_j=0$ and $V\\equiv1\\pmod2$.","q),\n\\]\n\\[\n \\Theta(j)=3j+1+s_2(j-1),\\qquad L_j=\\Theta(j)-E_j.\n\\]\nHere $E_j$ is a carry minimum, not an Euler number. A wall label is\n\\emph{regular} when $e_j(q)>0$ for $1\\le q<j$.\nFor $r=2^L-j$ with $L>L_j$, put $\\delta=L-L_j$.\nThe carry identities give\n\\[\n \\mu_r=E_j-\\delta,\\qquad J_r=\\{j\\},\\qquad\n G_r(x)=P_j(x)U(x-j)+2^\\delta V(x),\n\\]\nwhere $U\\equiv1\\pmod2$. For a regular label, also\n$E_j=0$ and $V\\equiv1\\pmod2$.","For $0\\le i<j$, let\n\\[\n d_i=v_2(i!)+v_2((j-1-i)!),\\qquad\n k_i=1+\\left\\lfloor\\log_2\\max(i,j-1-i)\\right\\rfloor,\n\\]\n\\[\n B_i=i+2^{k_i}\\mathbb Z_2,\\qquad t_i=d_i+k_i.\n\\]\nAt $j=1$ use $d_0=k_0=t_0=0$. The balls $B_i$ partition\n$\\mathbb Z_2$. Indeed a residue represented twice among $0,\\ldots,j-1$\nsplits into its two children; every other represented residue remains\none ball. On each such ball,\n\\[\n P_j(i+2^{k_i}y)=2^{t_i}yQ_i(y),\\qquad\n Q_i\\in\\mathbb Z_2\\langle y\\rangle^\\times,\n \\qquad Q_i\\equiv1\\pmod2.\n\\]","\\begin{theorem}[Regular-wall root law]\nFor every regular label $j$ and every integer $\\delta\\ge1$,\n\\[\n v_2G_r(x)=\n \\begin{cases}\n  \\delta,&x\\in B_i,\\ \\delta<t_i,","\\begin{theorem}[Regular-wall root law]\nFor every regular label $j$ and every integer $\\delta\\ge1$,\n\\[\n v_2G_r(x)=\n \\begin{cases}\n  \\delta,&x\\in B_i,\\ \\delta<t_i,\\\\\n  d_i+v_2(x-\\alpha_i),&x\\in B_i,\\ \\delta\\ge t_i.\n \\end{cases}\n\\]\nIn the second case there is exactly one root $\\alpha_i$ in $B_i$,\nit is simple, and $v_2(\\alpha_i-i)=\\delta-d_i$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contribution","d_i+v_2(x-\\alpha_i),&x\\in B_i,\\ \\delta\\ge t_i.\n \\end{cases}\n\\]\nIn the second case there is exactly one root $\\alpha_i$ in $B_i$,\nit is simple, and $v_2(\\alpha_i-i)=\\delta-d_i$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exa","d_i+v_2(x-\\alpha_i),&x\\in B_i,\\ \\delta\\ge t_i.\n \\end{cases}\n\\]\nIn the second case there is exactly one root $\\alpha_i$ in $B_i$,\nit is simple, and $v_2(\\alpha_i-i)=\\delta-d_i$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distan","d_i+v_2(x-\\alpha_i),&x\\in B_i,\\ \\delta\\ge t_i.\n \\end{cases}\n\\]\nIn the second case there is exactly one root $\\alpha_i$ in $B_i$,\nit is simple, and $v_2(\\alpha_i-i)=\\delta-d_i$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated dis","d_i+v_2(x-\\alpha_i),&x\\in B_i,\\ \\delta\\ge t_i.\n \\end{cases}\n\\]\nIn the second case there is exactly one root $\\alpha_i$ in $B_i$,\nit is simple, and $v_2(\\alpha_i-i)=\\delta-d_i$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition","d_i+v_2(x-\\alpha_i),&x\\in B_i,\\ \\delta\\ge t_i.\n \\end{cases}\n\\]\nIn the second case there is exactly one root $\\alpha_i$ in $B_i$,\nit is simple, and $v_2(\\alpha_i-i)=\\delta-d_i$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition proves exhaustion of the roots.\n\\end{proof}","-\\alpha_i),&x\\in B_i,\\ \\delta\\ge t_i.\n \\end{cases}\n\\]\nIn the second case there is exactly one root $\\alpha_i$ in $B_i$,\nit is simple, and $v_2(\\alpha_i-i)=\\delta-d_i$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition proves exhaustion of the roots.\n\\end{proof}","i.\n \\end{cases}\n\\]\nIn the second case there is exactly one root $\\alpha_i$ in $B_i$,\nit is simple, and $v_2(\\alpha_i-i)=\\delta-d_i$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition proves exhaustion of the roots.\n\\end{proof}","\\alpha_i$ in $B_i$,\nit is simple, and $v_2(\\alpha_i-i)=\\delta-d_i$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition proves exhaustion of the roots.\n\\end{proof}","alpha_i-i)=\\delta-d_i$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition proves exhaustion of the roots.\n\\end{proof}","$.\nThus the number of dyadic roots is exactly\n\\[\n \\#\\{i:0\\le i<j,\\ t_i\\le\\delta\\}.\n\\]\n\\label{thm:root:regular}\n\\end{theorem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition proves exhaustion of the roots.\n\\end{proof}","rem}\n\\begin{proof}\nIf $\\delta<t_i$, the unit perturbation has smaller valuation than\nthe skeleton term throughout $B_i$, so the valuation is exactly\n$\\delta$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition proves exhaustion of the roots.\n\\end{proof}","a$. Otherwise, after the ball substitution and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition proves exhaustion of the roots.\n\\end{proof}","and division by\n$2^{t_i}$, the coefficientwise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition proves exhaustion of the roots.\n\\end{proof}","ise reduction is $y+c$ for a constant\n$c\\in\\mathbb F_2$. A restricted integral series with this reduction\nis a dyadic isometry: division of its difference by $y-z$ gives an\nodd linear coefficient plus even higher contributions. Binary lifting\ntherefore gives its unique root and exact distance law. Evaluating at\n$x=i$ gives the stated displacement, since\n$G_r(i)=2^\\delta V(i)$ has exact valuation $\\delta$.\nThe partition proves exhaustion of the roots.\n\\end{proof}","The largest emergence threshold is\n\\[\n \\max_i t_i=v_2((j-1)!)+\\lceil\\log_2 j\\rceil.\n\\]\nFor $j=2^a$, all thresholds equal $j-1$. There are no roots before\nthat distance and exactly $j$ simple roots at and after it, within the\npositive-distance range. The equality and pre-wall cases are also\nresolved: regular labels $j\\ge2$ have no roots there; $j=1$ has one\nsimple odd root at equality.","For example, $j=5$ has $(t_0,\\ldots,t_4)=(6,3,4,3,6)$.\nThe exact root counts at distances $1,2,3,4,5,\\ge6$ are\n$0,0,2,3,3,5$. The older sufficient stages $3,5,7$ are not the\nsharp emergence stages, which are $3,4,6$.\nThe first irregular label, $j=17$, has $V\\equiv1+X\\pmod2$;\nthe regular-wall theorem does not apply. Complete coefficient\ncertificates close distances $1$ through $19$, and the general separation\ntheorem gives all seventeen simple roots for $\\delta\\ge20$.\nThose two kinds of evidence together resolve that family.","exact root counts at distances $1,2,3,4,5,\\ge6$ are\n$0,0,2,3,3,5$. The older sufficient stages $3,5,7$ are not the\nsharp emergence stages, which are $3,4,6$.\nThe first irregular label, $j=17$, has $V\\equiv1+X\\pmod2$;\nthe regular-wall theorem does not apply. Complete coefficient\ncertificates close distances $1$ through $19$, and the general separation\ntheorem gives all seventeen simple roots for $\\delta\\ge20$.\nThose two kinds of evidence together resolve that family.","For a general depth $r$, call $j_r=\\max J_r$ its \\emph{true terminal}.\nThis definition depends on the actual carry minimum; rewriting the same\ninteger as another wall expression does not change $j_r$.\nThe universal simplicity target is\n\\begin{equation}\n \\text{for every integer }r\\ge1\\text{ and }x\\in\\mathbb Z_2,\n \\qquad G_r(x)=0\\ \\Longrightarrow\\ G_r'(x)\\ne0.\n \\label{eq:root:simplicitytarget}\n\\end{equation}\nThe following are completed restricted results of the later continuations.\nTheir family statements include a proof that the indicated label is the\ntrue terminal throughout the displayed ranges.","\\begin{center}\n\\begin{tabular}{lll}\n\\toprule\nDepth $r$ & Odd multiplier range & Outer scales","\\midrule\n$B2^L-129$ & $0<B<2^{89}$ & $L\\ge392$","$B2^L-161$ & $0<B<2^{932}$ & $L\\ge487$","\\bottomrule\n\\end{tabular}\n\\end{center}\nAll dyadic roots are simple in both infinite integer families.\nFor arbitrary larger admissible multipliers, a multiple root with true\nterminal $129$ can occur only at $L=515$ and in two necessary multiplier\nclasses modulo $2^{90}$. With true terminal $161$, it can occur only at\n$L=640$ and in one necessary multiplier class modulo $2^{935}$.\nThese classes are necessary conditions. The records do not assert\nexistence of a singular ordinary Euler index in them.","For terminal $161$, the neighboring phases are resolved uniformly:\n$L=639$ has exactly one simple root and $L=641$ exactly three simple\nroots, for every positive odd multiplier with that true terminal.\nThe middle phase is reduced, before the deeper certificate is applied,\nto\n\\[\n B\\equiv17659\\pmod{2^{18}},\\qquad\n x\\equiv10202\\pmod{2^{15}}.\n\\]\nThe $935$-bit multiplier condition composes this whole-phase reduction\nwith an inherited $917$-bit local parameter certificate. It is neither\na new $935$-bit reconstruction nor an all-precision interpolation of\nthe undepleted Euler family.","g phases are resolved uniformly:\n$L=639$ has exactly one simple root and $L=641$ exactly three simple\nroots, for every positive odd multiplier with that true terminal.\nThe middle phase is reduced, before the deeper certificate is applied,\nto\n\\[\n B\\equiv17659\\pmod{2^{18}},\\qquad\n x\\equiv10202\\pmod{2^{15}}.\n\\]\nThe $935$-bit multiplier condition composes this whole-phase reduction\nwith an inherited $917$-bit local parameter certificate. It is neither\na new $935$-bit reconstruction nor an all-precision interpolation of\nthe undepleted Euler family.","mple root and $L=641$ exactly three simple\nroots, for every positive odd multiplier with that true terminal.\nThe middle phase is reduced, before the deeper certificate is applied,\nto\n\\[\n B\\equiv17659\\pmod{2^{18}},\\qquad\n x\\equiv10202\\pmod{2^{15}}.\n\\]\nThe $935$-bit multiplier condition composes this whole-phase reduction\nwith an inherited $917$-bit local parameter certificate. It is neither\na new $935$-bit reconstruction nor an all-precision interpolation of\nthe undepleted Euler family.","The structural argument also extends beyond lower address one.\nFor a certified true-terminal wall, let $a$ be its last lower minimizing\naddress, $k=j_r-a$ the regular complement, and $t=v_2(k)$.\nUnder the regular lower-address hypotheses of the source theorem,\nall roots above emerged complement balls are simple. Every root is\nsimple at every wall distance in the band\n\\[\n \\Theta(a)\\le t<\\Theta(a)+v_2(a!).\n\\]\nEquality walls are simple even outside that band. This includes\ncertified terminal-$258$ walls with $a=2$, $k=256$, and $L\\ge777$.\nFor $a=3$, $k=4096$, only $L=16385$ remains a possible critical scale.\nOutside the band, the theorem gives necessary local valuation conditions;\nit does not exclude the remaining unit cancellation. Irregular lower\naddresses need their own prefix analysis.","ll, let $a$ be its last lower minimizing\naddress, $k=j_r-a$ the regular complement, and $t=v_2(k)$.\nUnder the regular lower-address hypotheses of the source theorem,\nall roots above emerged complement balls are simple. Every root is\nsimple at every wall distance in the band\n\\[\n \\Theta(a)\\le t<\\Theta(a)+v_2(a!).\n\\]\nEquality walls are simple even outside that band. This includes\ncertified terminal-$258$ walls with $a=2$, $k=256$, and $L\\ge777$.\nFor $a=3$, $k=4096$, only $L=16385$ remains a possible critical scale.\nOutside the band, the theorem gives necessary local valuation conditions;\nit does not exclude the remaining unit cancellation. Irregular lower\naddresses need their own prefix analysis.","s last lower minimizing\naddress, $k=j_r-a$ the regular complement, and $t=v_2(k)$.\nUnder the regular lower-address hypotheses of the source theorem,\nall roots above emerged complement balls are simple. Every root is\nsimple at every wall distance in the band\n\\[\n \\Theta(a)\\le t<\\Theta(a)+v_2(a!).\n\\]\nEquality walls are simple even outside that band. This includes\ncertified terminal-$258$ walls with $a=2$, $k=256$, and $L\\ge777$.\nFor $a=3$, $k=4096$, only $L=16385$ remains a possible critical scale.\nOutside the band, the theorem gives necessary local valuation conditions;\nit does not exclude the remaining unit cancellation. Irregular lower\naddresses need their own prefix analysis.","lower minimizing\naddress, $k=j_r-a$ the regular complement, and $t=v_2(k)$.\nUnder the regular lower-address hypotheses of the source theorem,\nall roots above emerged complement balls are simple. Every root is\nsimple at every wall distance in the band\n\\[\n \\Theta(a)\\le t<\\Theta(a)+v_2(a!).\n\\]\nEquality walls are simple even outside that band. This includes\ncertified terminal-$258$ walls with $a=2$, $k=256$, and $L\\ge777$.\nFor $a=3$, $k=4096$, only $L=16385$ remains a possible critical scale.\nOutside the band, the theorem gives necessary local valuation conditions;\nit does not exclude the remaining unit cancellation. Irregular lower\naddresses need their own prefix analysis.","ing\naddress, $k=j_r-a$ the regular complement, and $t=v_2(k)$.\nUnder the regular lower-address hypotheses of the source theorem,\nall roots above emerged complement balls are simple. Every root is\nsimple at every wall distance in the band\n\\[\n \\Theta(a)\\le t<\\Theta(a)+v_2(a!).\n\\]\nEquality walls are simple even outside that band. This includes\ncertified terminal-$258$ walls with $a=2$, $k=256$, and $L\\ge777$.\nFor $a=3$, $k=4096$, only $L=16385$ remains a possible critical scale.\nOutside the band, the theorem gives necessary local valuation conditions;\nit does not exclude the remaining unit cancellation. Irregular lower\naddresses need their own prefix analysis.","=j_r-a$ the regular complement, and $t=v_2(k)$.\nUnder the regular lower-address hypotheses of the source theorem,\nall roots above emerged complement balls are simple. Every root is\nsimple at every wall distance in the band\n\\[\n \\Theta(a)\\le t<\\Theta(a)+v_2(a!).\n\\]\nEquality walls are simple even outside that band. This includes\ncertified terminal-$258$ walls with $a=2$, $k=256$, and $L\\ge777$.\nFor $a=3$, $k=4096$, only $L=16385$ remains a possible critical scale.\nOutside the band, the theorem gives necessary local valuation conditions;\nit does not exclude the remaining unit cancellation. Irregular lower\naddresses need their own prefix analysis.","the regular complement, and $t=v_2(k)$.\nUnder the regular lower-address hypotheses of the source theorem,\nall roots above emerged complement balls are simple. Every root is\nsimple at every wall distance in the band\n\\[\n \\Theta(a)\\le t<\\Theta(a)+v_2(a!).\n\\]\nEquality walls are simple even outside that band. This includes\ncertified terminal-$258$ walls with $a=2$, $k=256$, and $L\\ge777$.\nFor $a=3$, $k=4096$, only $L=16385$ remains a possible critical scale.\nOutside the band, the theorem gives necessary local valuation conditions;\nit does not exclude the remaining unit cancellation. Irregular lower\naddresses need their own prefix analysis.","mplement, and $t=v_2(k)$.\nUnder the regular lower-address hypotheses of the source theorem,\nall roots above emerged complement balls are simple. Every root is\nsimple at every wall distance in the band\n\\[\n \\Theta(a)\\le t<\\Theta(a)+v_2(a!).\n\\]\nEquality walls are simple even outside that band. This includes\ncertified terminal-$258$ walls with $a=2$, $k=256$, and $L\\ge777$.\nFor $a=3$, $k=4096$, only $L=16385$ remains a possible critical scale.\nOutside the band, the theorem gives necessary local valuation conditions;\nit does not exclude the remaining unit cancellation. Irregular lower\naddresses need their own prefix analysis.","The current focused obstruction is the admitted ordinary-index family\n\\begin{equation}\n r=(17659+2^{18}T)2^{640}-161,\n \\qquad T\\in\\mathbb Z_{\\ge0},\\qquad j_r=161.\n \\label{eq:root:focusedfamily}\n\\end{equation}\nOn its surviving chart define the \\emph{actual}, undepleted function\n\\[\n K_r(Z)=2^{-178}G_r(10202+2^{15}Z).\n\\]\nThe complete coefficient theorem gives\n\\begin{equation}\n K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z),\n \\quad a_r,b_r\\in\\mathbb Z_2,\\quad\n u_r\\in\\mathbb Z_2^\\times,\\quad\n R_r\\in\\mathbb Z_2\\langle Z\\rangle.\n \\label{eq:root:quadraticclass}\n\\end{equation}\nIts half derivative is a dyadic isometry. It therefore has a unique\ncritical point $c_r\\in\\mathbb Z_2$. Set\n$\\Phi_r^{\\mathrm{act}}=K_r(c_r)$.\nThe earlier independent-undepletion notation for this same scalar is\n$\\Phi_r(2^{2r-25})$.\nExcluding a multiple root on the surviving chart is exactly the assertion\n$\\Phi_r^{\\mathrm{act}}\\","The current focused obstruction is the admitted ordinary-index family\n\\begin{equation}\n r=(17659+2^{18}T)2^{640}-161,\n \\qquad T\\in\\mathbb Z_{\\ge0},\\qquad j_r=161.\n \\label{eq:root:focusedfamily}\n\\end{equation}\nOn its surviving chart define the \\emph{actual}, undepleted function\n\\[\n K_r(Z)=2^{-178}G_r(10202+2^{15}Z).\n\\]\nThe complete coefficient theorem gives\n\\begin{equation}\n K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z),\n \\quad a_r,b_r\\in\\mathbb Z_2,\\quad\n u_r\\in\\mathbb Z_2^\\times,\\quad\n R_r\\in\\mathbb Z_2\\langle Z\\rangle.\n \\label{eq:root:quadraticclass}\n\\end{equation}\nIts half derivative is a dyadic isometry. It therefore has a unique\ncritical point $c_r\\in\\mathbb Z_2$. Set\n$\\Phi_r^{\\mathrm{act}}=K_r(c_r)$.\nThe earlier independent-undepletion notation for this same scalar is\n$\\Phi_r(2^{2r-25})$.\nExcluding a multiple root on the surviving chart is exactly the assertion\n$\\Phi_r^{\\mathrm{act}}\\ne0$.","actual}, undepleted function\n\\[\n K_r(Z)=2^{-178}G_r(10202+2^{15}Z).\n\\]\nThe complete coefficient theorem gives\n\\begin{equation}\n K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z),\n \\quad a_r,b_r\\in\\mathbb Z_2,\\quad\n u_r\\in\\mathbb Z_2^\\times,\\quad\n R_r\\in\\mathbb Z_2\\langle Z\\rangle.\n \\label{eq:root:quadraticclass}\n\\end{equation}\nIts half derivative is a dyadic isometry. It therefore has a unique\ncritical point $c_r\\in\\mathbb Z_2$. Set\n$\\Phi_r^{\\mathrm{act}}=K_r(c_r)$.\nThe earlier independent-undepletion notation for this same scalar is\n$\\Phi_r(2^{2r-25})$.\nExcluding a multiple root on the surviving chart is exactly the assertion\n$\\Phi_r^{\\mathrm{act}}\\ne0$.","ion\n\\[\n K_r(Z)=2^{-178}G_r(10202+2^{15}Z).\n\\]\nThe complete coefficient theorem gives\n\\begin{equation}\n K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z),\n \\quad a_r,b_r\\in\\mathbb Z_2,\\quad\n u_r\\in\\mathbb Z_2^\\times,\\quad\n R_r\\in\\mathbb Z_2\\langle Z\\rangle.\n \\label{eq:root:quadraticclass}\n\\end{equation}\nIts half derivative is a dyadic isometry. It therefore has a unique\ncritical point $c_r\\in\\mathbb Z_2$. Set\n$\\Phi_r^{\\mathrm{act}}=K_r(c_r)$.\nThe earlier independent-undepletion notation for this same scalar is\n$\\Phi_r(2^{2r-25})$.\nExcluding a multiple root on the surviving chart is exactly the assertion\n$\\Phi_r^{\\mathrm{act}}\\ne0$.","K_r(Z)=a_r+2b_rZ+u_rZ^2+256Z^3R_r(Z),\n \\quad a_r,b_r\\in\\mathbb Z_2,\\quad\n u_r\\in\\mathbb Z_2^\\times,\\quad\n R_r\\in\\mathbb Z_2\\langle Z\\rangle.\n \\label{eq:root:quadraticclass}\n\\end{equation}\nIts half derivative is a dyadic isometry. It therefore has a unique\ncritical point $c_r\\in\\mathbb Z_2$. Set\n$\\Phi_r^{\\mathrm{act}}=K_r(c_r)$.\nThe earlier independent-undepletion notation for this same scalar is\n$\\Phi_r(2^{2r-25})$.\nExcluding a multiple root on the surviving chart is exactly the assertion\n$\\Phi_r^{\\mathrm{act}}\\ne0$.","mes,\\quad\n R_r\\in\\mathbb Z_2\\langle Z\\rangle.\n \\label{eq:root:quadraticclass}\n\\end{equation}\nIts half derivative is a dyadic isometry. It therefore has a unique\ncritical point $c_r\\in\\mathbb Z_2$. Set\n$\\Phi_r^{\\mathrm{act}}=K_r(c_r)$.\nThe earlier independent-undepletion notation for this same scalar is\n$\\Phi_r(2^{2r-25})$.\nExcluding a multiple root on the surviving chart is exactly the assertion\n$\\Phi_r^{\\mathrm{act}}\\ne0$.","\\begin{proposition}[Exact quadratic compression]\nThere is a unique preparation\n\\[\n K_r(Z)=\\mathcal U_r(Z)\\mathcal W_r(Z),\\qquad\n \\mathcal W_r(Z)=Z^2+2\\beta_rZ+\\alpha_r,\\qquad\n \\mathcal U_r\\equiv u_r\\pmod{256}.\n\\]\nWith $\\delta_r=\\beta_r^2-\\alpha_r$ and\n$\\tau_r=\\mathcal U_r'(c_r)/\\mathcal U_r(c_r)$,\n\\begin{equation}\n \\delta_r=-\\frac{\\Phi_r^{\\mathrm{act}}}{\\mathcal U_r(c_r)}\n  \\left(1-\\frac{\\tau_r^2\\Phi_r^{\\mathrm{act}}}{4\\mathcal U_r(c_r)}\\right),\n \\qquad v_2(\\delta_r)=v_2(\\Phi_r^{\\mathrm{act}}).\n \\label{eq:root:discriminant}\n\\end{equation}\nThe equality includes the zero case.\n\\end{proposition}\n\\begin{proof}\nThe same successive monic polynomial division used in\nTheorem~\\ref{thm:root:finitepreparation}, started at precision eight,\npreserves the quadratic and unit reductions of\n\\eqref{eq:root:quadraticclass}. Write\n$W=Z^2+2\\beta_rZ+\\alpha_r$ and $w=W(c_r)$.\nStationarity gives $W'(c_r)=-\\tau_rw$, while","\\begin{proposition}[Exact quadratic compression]\nThere is a unique preparation\n\\[\n K_r(Z)=\\mathcal U_r(Z)\\mathcal W_r(Z),\\qquad\n \\mathcal W_r(Z)=Z^2+2\\beta_rZ+\\alpha_r,\\qquad\n \\mathcal U_r\\equiv u_r\\pmod{256}.\n\\]\nWith $\\delta_r=\\beta_r^2-\\alpha_r$ and\n$\\tau_r=\\mathcal U_r'(c_r)/\\mathcal U_r(c_r)$,\n\\begin{equation}\n \\delta_r=-\\frac{\\Phi_r^{\\mathrm{act}}}{\\mathcal U_r(c_r)}\n  \\left(1-\\frac{\\tau_r^2\\Phi_r^{\\mathrm{act}}}{4\\mathcal U_r(c_r)}\\right),\n \\qquad v_2(\\delta_r)=v_2(\\Phi_r^{\\mathrm{act}}).\n \\label{eq:root:discriminant}\n\\end{equation}\nThe equality includes the zero case.\n\\end{proposition}\n\\begin{proof}\nThe same successive monic polynomial division used in\nTheorem~\\ref{thm:root:finitepreparation}, started at precision eight,\npreserves the quadratic and unit reductions of\n\\eqref{eq:root:quadraticclass}. Write\n$W=Z^2+2\\beta_rZ+\\alpha_r$ and $w=W(c_r)$.\nStationarity gives $W'(c_r)=-\\tau_rw$, while\n$w=\\Phi_r^{\\mathrm{act}}/\\mathcal U","\\begin{proposition}[Exact quadratic compression]\nThere is a unique preparation\n\\[\n K_r(Z)=\\mathcal U_r(Z)\\mathcal W_r(Z),\\qquad\n \\mathcal W_r(Z)=Z^2+2\\beta_rZ+\\alpha_r,\\qquad\n \\mathcal U_r\\equiv u_r\\pmod{256}.\n\\]\nWith $\\delta_r=\\beta_r^2-\\alpha_r$ and\n$\\tau_r=\\mathcal U_r'(c_r)/\\mathcal U_r(c_r)$,\n\\begin{equation}\n \\delta_r=-\\frac{\\Phi_r^{\\mathrm{act}}}{\\mathcal U_r(c_r)}\n  \\left(1-\\frac{\\tau_r^2\\Phi_r^{\\mathrm{act}}}{4\\mathcal U_r(c_r)}\\right),\n \\qquad v_2(\\delta_r)=v_2(\\Phi_r^{\\mathrm{act}}).\n \\label{eq:root:discriminant}\n\\end{equation}\nThe equality includes the zero case.\n\\end{proposition}\n\\begin{proof}\nThe same successive monic polynomial division used in\nTheorem~\\ref{thm:root:finitepreparation}, started at precision eight,\npreserves the quadratic and unit reductions of\n\\eqref{eq:root:quadraticclass}. Write\n$W=Z^2+2\\beta_rZ+\\alpha_r$ and $w=W(c_r)$.\nStationarity gives $W'(c_r)=-\\tau_rw$, while\n$w=\\Phi_r^{\\mathrm{act}}/\\mathcal U_r(c_r)$.\nThe quadratic identity\n$\\beta_r^2-\\alpha","\\begin{proposition}[Exact quadratic compression]\nThere is a unique preparation\n\\[\n K_r(Z)=\\mathcal U_r(Z)\\mathcal W_r(Z),\\qquad\n \\mathcal W_r(Z)=Z^2+2\\beta_rZ+\\alpha_r,\\qquad\n \\mathcal U_r\\equiv u_r\\pmod{256}.\n\\]\nWith $\\delta_r=\\beta_r^2-\\alpha_r$ and\n$\\tau_r=\\mathcal U_r'(c_r)/\\mathcal U_r(c_r)$,\n\\begin{equation}\n \\delta_r=-\\frac{\\Phi_r^{\\mathrm{act}}}{\\mathcal U_r(c_r)}\n  \\left(1-\\frac{\\tau_r^2\\Phi_r^{\\mathrm{act}}}{4\\mathcal U_r(c_r)}\\right),\n \\qquad v_2(\\delta_r)=v_2(\\Phi_r^{\\mathrm{act}}).\n \\label{eq:root:discriminant}\n\\end{equation}\nThe equality includes the zero case.\n\\end{proposition}\n\\begin{proof}\nThe same successive monic polynomial division used in\nTheorem~\\ref{thm:root:finitepreparation}, started at precision eight,\npreserves the quadratic and unit reductions of\n\\eqref{eq:root:quadraticclass}. Write\n$W=Z^2+2\\beta_rZ+\\alpha_r$ and $w=W(c_r)$.\nStationarity gives $W'(c_r)=-\\tau_rw$, while\n$w=\\Phi_r^{\\mathrm{act}}/\\mathcal U_r(c_r)$.\nThe quadratic identity\n$\\beta_r^2-\\alpha_r=W'(c_r)^2/4-W(c_r)$ proves the formula.\nSince $\\tau_r\\in256\\mathbb Z_2$, its bracket is a unit.\n\\end{…","mathcal U_r(c_r)}\\right),\n \\qquad v_2(\\delta_r)=v_2(\\Phi_r^{\\mathrm{act}}).\n \\label{eq:root:discriminant}\n\\end{equation}\nThe equality includes the zero case.\n\\end{proposition}\n\\begin{proof}\nThe same successive monic polynomial division used in\nTheorem~\\ref{thm:root:finitepreparation}, started at precision eight,\npreserves the quadratic and unit reductions of\n\\eqref{eq:root:quadraticclass}. Write\n$W=Z^2+2\\beta_rZ+\\alpha_r$ and $w=W(c_r)$.\nStationarity gives $W'(c_r)=-\\tau_rw$, while\n$w=\\Phi_r^{\\mathrm{act}}/\\mathcal U_r(c_r)$.\nThe quadratic identity\n$\\beta_r^2-\\alpha_r=W'(c_r)^2/4-W(c_r)$ proves the formula.\nSince $\\tau_r\\in256\\mathbb Z_2$, its bracket is a unit.\n\\end{proof}","ad v_2(\\delta_r)=v_2(\\Phi_r^{\\mathrm{act}}).\n \\label{eq:root:discriminant}\n\\end{equation}\nThe equality includes the zero case.\n\\end{proposition}\n\\begin{proof}\nThe same successive monic polynomial division used in\nTheorem~\\ref{thm:root:finitepreparation}, started at precision eight,\npreserves the quadratic and unit reductions of\n\\eqref{eq:root:quadraticclass}. Write\n$W=Z^2+2\\beta_rZ+\\alpha_r$ and $w=W(c_r)$.\nStationarity gives $W'(c_r)=-\\tau_rw$, while\n$w=\\Phi_r^{\\mathrm{act}}/\\mathcal U_r(c_r)$.\nThe quadratic identity\n$\\beta_r^2-\\alpha_r=W'(c_r)^2/4-W(c_r)$ proves the formula.\nSince $\\tau_r\\in256\\mathbb Z_2$, its bracket is a unit.\n\\end{proof}","athrm{act}}).\n \\label{eq:root:discriminant}\n\\end{equation}\nThe equality includes the zero case.\n\\end{proposition}\n\\begin{proof}\nThe same successive monic polynomial division used in\nTheorem~\\ref{thm:root:finitepreparation}, started at precision eight,\npreserves the quadratic and unit reductions of\n\\eqref{eq:root:quadraticclass}. Write\n$W=Z^2+2\\beta_rZ+\\alpha_r$ and $w=W(c_r)$.\nStationarity gives $W'(c_r)=-\\tau_rw$, while\n$w=\\Phi_r^{\\mathrm{act}}/\\mathcal U_r(c_r)$.\nThe quadratic identity\n$\\beta_r^2-\\alpha_r=W'(c_r)^2/4-W(c_r)$ proves the formula.\nSince $\\tau_r\\in256\\mathbb Z_2$, its bracket is a unit.\n\\end{proof}","oot:discriminant}\n\\end{equation}\nThe equality includes the zero case.\n\\end{proposition}\n\\begin{proof}\nThe same successive monic polynomial division used in\nTheorem~\\ref{thm:root:finitepreparation}, started at precision eight,\npreserves the quadratic and unit reductions of\n\\eqref{eq:root:quadraticclass}. Write\n$W=Z^2+2\\beta_rZ+\\alpha_r$ and $w=W(c_r)$.\nStationarity gives $W'(c_r)=-\\tau_rw$, while\n$w=\\Phi_r^{\\mathrm{act}}/\\mathcal U_r(c_r)$.\nThe quadratic identity\n$\\beta_r^2-\\alpha_r=W'(c_r)^2/4-W(c_r)$ proves the formula.\nSince $\\tau_r\\in256\\mathbb Z_2$, its bracket is a unit.\n\\end{proof}",".\n\\end{proposition}\n\\begin{proof}\nThe same successive monic polynomial division used in\nTheorem~\\ref{thm:root:finitepreparation}, started at precision eight,\npreserves the quadratic and unit reductions of\n\\eqref{eq:root:quadraticclass}. Write\n$W=Z^2+2\\beta_rZ+\\alpha_r$ and $w=W(c_r)$.\nStationarity gives $W'(c_r)=-\\tau_rw$, while\n$w=\\Phi_r^{\\mathrm{act}}/\\mathcal U_r(c_r)$.\nThe quadratic identity\n$\\beta_r^2-\\alpha_r=W'(c_r)^2/4-W(c_r)$ proves the formula.\nSince $\\tau_r\\in256\\mathbb Z_2$, its bracket is a unit.\n\\end{proof}","mial division used in\nTheorem~\\ref{thm:root:finitepreparation}, started at precision eight,\npreserves the quadratic and unit reductions of\n\\eqref{eq:root:quadraticclass}. Write\n$W=Z^2+2\\beta_rZ+\\alpha_r$ and $w=W(c_r)$.\nStationarity gives $W'(c_r)=-\\tau_rw$, while\n$w=\\Phi_r^{\\mathrm{act}}/\\mathcal U_r(c_r)$.\nThe quadratic identity\n$\\beta_r^2-\\alpha_r=W'(c_r)^2/4-W(c_r)$ proves the formula.\nSince $\\tau_r\\in256\\mathbb Z_2$, its bracket is a unit.\n\\end{proof}","This compression retains the unit-cofactor derivative. It does not\nassert that $c_r$, $\\alpha_r$, or $\\beta_r$ is rational or algebraic\nover $\\mathbb Q$. The remaining arithmetic obligation is therefore\n\\begin{equation}\n \\delta_r\\ne0\\quad\\text{for every admitted ordinary index in\n \\eqref{eq:root:focusedfamily}.}\n \\label{eq:root:remaining}\n\\end{equation}\nEven this assertion would still leave the terminal-$129$ and higher-address\nobligations in the universal target \\eqref{eq:root:simplicitytarget}.","pression retains the unit-cofactor derivative. It does not\nassert that $c_r$, $\\alpha_r$, or $\\beta_r$ is rational or algebraic\nover $\\mathbb Q$. The remaining arithmetic obligation is therefore\n\\begin{equation}\n \\delta_r\\ne0\\quad\\text{for every admitted ordinary index in\n \\eqref{eq:root:focusedfamily}.}\n \\label{eq:root:remaining}\n\\end{equation}\nEven this assertion would still leave the terminal-$129$ and higher-address\nobligations in the universal target \\eqref{eq:root:simplicitytarget}.","The latest continuation retains the exact Bernoulli undepletion factor\n\\[\n e_{r,q}=\\frac{1-2^{2r-1}}{1-2^{2r-1}4^q},\\qquad q\\ge0.\n\\]\nIt is a dyadic unit, including $e_{r,0}=1$. Multiplication by this\nfactor preserves the all-$q$ error of the shifted Jacobi moment model.\nConsequently the actual model critical values satisfy, for every $m\\ge25$,\n\\begin{equation}\n v_2\\bigl(\\Phi_{r,m}^{\\mathrm{act}}-\\Phi_r^{\\mathrm{act}}\\bigr)\n \\ge L_m=8m-4s_2(m)-178,\n \\qquad L_m\\longrightarrow\\infty.\n \\label{eq:root:actualapprox}\n\\end{equation}\nThe estimate includes every packet and controls motion of the critical\npoint. There is no ceiling at the undepletion exponent $2r-25$.\nFor each fixed admitted $r$ it implies the exact equivalence\n\\begin{equation}\n \\Phi_r^{\\mathrm{act}}\\ne0\n \\quad\\Longleftrightarrow\\quad\n \\text{there exists }m\\ge25\\","The latest continuation retains the exact Bernoulli undepletion factor\n\\[\n e_{r,q}=\\frac{1-2^{2r-1}}{1-2^{2r-1}4^q},\\qquad q\\ge0.\n\\]\nIt is a dyadic unit, including $e_{r,0}=1$. Multiplication by this\nfactor preserves the all-$q$ error of the shifted Jacobi moment model.\nConsequently the actual model critical values satisfy, for every $m\\ge25$,\n\\begin{equation}\n v_2\\bigl(\\Phi_{r,m}^{\\mathrm{act}}-\\Phi_r^{\\mathrm{act}}\\bigr)\n \\ge L_m=8m-4s_2(m)-178,\n \\qquad L_m\\longrightarrow\\infty.\n \\label{eq:root:actualapprox}\n\\end{equation}\nThe estimate includes every packet and controls motion of the critical\npoint. There is no ceiling at the undepletion exponent $2r-25$.\nFor each fixed admitted $r$ it implies the exact equivalence\n\\begin{equation}\n \\Phi_r^{\\mathrm{act}}\\ne0\n \\quad\\Longleftrightarrow\\quad\n \\text{there exists }m\\ge25\\text{ such that }\n v_2(\\Phi_{r,m}^{\\mathrm{a","The latest continuation retains the exact Bernoulli undepletion factor\n\\[\n e_{r,q}=\\frac{1-2^{2r-1}}{1-2^{2r-1}4^q},\\qquad q\\ge0.\n\\]\nIt is a dyadic unit, including $e_{r,0}=1$. Multiplication by this\nfactor preserves the all-$q$ error of the shifted Jacobi moment model.\nConsequently the actual model critical values satisfy, for every $m\\ge25$,\n\\begin{equation}\n v_2\\bigl(\\Phi_{r,m}^{\\mathrm{act}}-\\Phi_r^{\\mathrm{act}}\\bigr)\n \\ge L_m=8m-4s_2(m)-178,\n \\qquad L_m\\longrightarrow\\infty.\n \\label{eq:root:actualapprox}\n\\end{equation}\nThe estimate includes every packet and controls motion of the critical\npoint. There is no ceiling at the undepletion exponent $2r-25$.\nFor each fixed admitted $r$ it implies the exact equivalence\n\\begin{equation}\n \\Phi_r^{\\mathrm{act}}\\ne0\n \\quad\\Longleftrightarrow\\quad\n \\text{there exists }m\\ge25\\text{ such that }\n v_2(\\Phi_{r,m}^{\\mathrm{act}})<L_m.\n \\label{eq:root:stricttest}\n\\end{equation}","The latest continuation retains the exact Bernoulli undepletion factor\n\\[\n e_{r,q}=\\frac{1-2^{2r-1}}{1-2^{2r-1}4^q},\\qquad q\\ge0.\n\\]\nIt is a dyadic unit, including $e_{r,0}=1$. Multiplication by this\nfactor preserves the all-$q$ error of the shifted Jacobi moment model.\nConsequently the actual model critical values satisfy, for every $m\\ge25$,\n\\begin{equation}\n v_2\\bigl(\\Phi_{r,m}^{\\mathrm{act}}-\\Phi_r^{\\mathrm{act}}\\bigr)\n \\ge L_m=8m-4s_2(m)-178,\n \\qquad L_m\\longrightarrow\\infty.\n \\label{eq:root:actualapprox}\n\\end{equation}\nThe estimate includes every packet and controls motion of the critical\npoint. There is no ceiling at the undepletion exponent $2r-25$.\nFor each fixed admitted $r$ it implies the exact equivalence\n\\begin{equation}\n \\Phi_r^{\\mathrm{act}}\\ne0\n \\quad\\Longleftrightarrow\\quad\n \\text{there exists }m\\ge25\\text{ such that }\n v_2(\\Phi_{r,m}^{\\mathrm{act}})<L_m.\n \\label{eq:root:stricttest}\n\\end{equation}\nIndeed a passing strict test has the same valuation as the limit; conversely\nany finite limit valuation is eventual","The latest continuation retains the exact Bernoulli undepletion factor\n\\[\n e_{r,q}=\\frac{1-2^{2r-1}}{1-2^{2r-1}4^q},\\qquad q\\ge0.\n\\]\nIt is a dyadic unit, including $e_{r,0}=1$. Multiplication by this\nfactor preserves the all-$q$ error of the shifted Jacobi moment model.\nConsequently the actual model critical values satisfy, for every $m\\ge25$,\n\\begin{equation}\n v_2\\bigl(\\Phi_{r,m}^{\\mathrm{act}}-\\Phi_r^{\\mathrm{act}}\\bigr)\n \\ge L_m=8m-4s_2(m)-178,\n \\qquad L_m\\longrightarrow\\infty.\n \\label{eq:root:actualapprox}\n\\end{equation}\nThe estimate includes every packet and controls motion of the critical\npoint. There is no ceiling at the undepletion exponent $2r-25$.\nFor each fixed admitted $r$ it implies the exact equivalence\n\\begin{equation}\n \\Phi_r^{\\mathrm{act}}\\ne0\n \\quad\\Longleftrightarrow\\quad\n \\text{there exists }m\\ge25\\text{ such that }\n v_2(\\Phi_{r,m}^{\\mathrm{act}})<L_m.\n \\label{eq:root:stricttest}\n\\end{equation}\nIndeed a passing strict test has the same valuation as the limit; conversely\nany finite limit valuation is eventually smaller than $L_m$.\nThe existence of a passing depth f…","oment model.\nConsequently the actual model critical values satisfy, for every $m\\ge25$,\n\\begin{equation}\n v_2\\bigl(\\Phi_{r,m}^{\\mathrm{act}}-\\Phi_r^{\\mathrm{act}}\\bigr)\n \\ge L_m=8m-4s_2(m)-178,\n \\qquad L_m\\longrightarrow\\infty.\n \\label{eq:root:actualapprox}\n\\end{equation}\nThe estimate includes every packet and controls motion of the critical\npoint. There is no ceiling at the undepletion exponent $2r-25$.\nFor each fixed admitted $r$ it implies the exact equivalence\n\\begin{equation}\n \\Phi_r^{\\mathrm{act}}\\ne0\n \\quad\\Longleftrightarrow\\quad\n \\text{there exists }m\\ge25\\text{ such that }\n v_2(\\Phi_{r,m}^{\\mathrm{act}})<L_m.\n \\label{eq:root:stricttest}\n\\end{equation}\nIndeed a passing strict test has the same valuation as the limit; conversely\nany finite limit valuation is eventually smaller than $L_m$.\nThe existence of a passing depth for every ordinary index is unproved.\nNonzero models at every depth are compatible with an exactly zero limit.","ual model critical values satisfy, for every $m\\ge25$,\n\\begin{equation}\n v_2\\bigl(\\Phi_{r,m}^{\\mathrm{act}}-\\Phi_r^{\\mathrm{act}}\\bigr)\n \\ge L_m=8m-4s_2(m)-178,\n \\qquad L_m\\longrightarrow\\infty.\n \\label{eq:root:actualapprox}\n\\end{equation}\nThe estimate includes every packet and controls motion of the critical\npoint. There is no ceiling at the undepletion exponent $2r-25$.\nFor each fixed admitted $r$ it implies the exact equivalence\n\\begin{equation}\n \\Phi_r^{\\mathrm{act}}\\ne0\n \\quad\\Longleftrightarrow\\quad\n \\text{there exists }m\\ge25\\text{ such that }\n v_2(\\Phi_{r,m}^{\\mathrm{act}})<L_m.\n \\label{eq:root:stricttest}\n\\end{equation}\nIndeed a passing strict test has the same valuation as the limit; conversely\nany finite limit valuation is eventually smaller than $L_m$.\nThe existence of a passing depth for every ordinary index is unproved.\nNonzero models at every depth are compatible with an exactly zero limit.","$,\n\\begin{equation}\n v_2\\bigl(\\Phi_{r,m}^{\\mathrm{act}}-\\Phi_r^{\\mathrm{act}}\\bigr)\n \\ge L_m=8m-4s_2(m)-178,\n \\qquad L_m\\longrightarrow\\infty.\n \\label{eq:root:actualapprox}\n\\end{equation}\nThe estimate includes every packet and controls motion of the critical\npoint. There is no ceiling at the undepletion exponent $2r-25$.\nFor each fixed admitted $r$ it implies the exact equivalence\n\\begin{equation}\n \\Phi_r^{\\mathrm{act}}\\ne0\n \\quad\\Longleftrightarrow\\quad\n \\text{there exists }m\\ge25\\text{ such that }\n v_2(\\Phi_{r,m}^{\\mathrm{act}})<L_m.\n \\label{eq:root:stricttest}\n\\end{equation}\nIndeed a passing strict test has the same valuation as the limit; conversely\nany finite limit valuation is eventually smaller than $L_m$.\nThe existence of a passing depth for every ordinary index is unproved.\nNonzero models at every depth are compatible with an exactly zero limit.","-25$.\nFor each fixed admitted $r$ it implies the exact equivalence\n\\begin{equation}\n \\Phi_r^{\\mathrm{act}}\\ne0\n \\quad\\Longleftrightarrow\\quad\n \\text{there exists }m\\ge25\\text{ such that }\n v_2(\\Phi_{r,m}^{\\mathrm{act}})<L_m.\n \\label{eq:root:stricttest}\n\\end{equation}\nIndeed a passing strict test has the same valuation as the limit; conversely\nany finite limit valuation is eventually smaller than $L_m$.\nThe existence of a passing depth for every ordinary index is unproved.\nNonzero models at every depth are compatible with an exactly zero limit.","The degree-two preparation also reduces the old representative-height\ncost to a linear cost, about $2P$ bits at precision $P$. This remains\nlarger than the $P$ bits needed for a strict nonzero residue certificate.\nThe fixed rational control $(Z-1/3)^2$ has zero critical value while\nits least-residue integer discriminant representatives are nonzero at\nevery even $P\\ge6$. Thus a nonzero integer representative alone cannot\nprove a nonzero dyadic discriminant.","One sufficient new arithmetic input would be an odd-denominator\nprojective quadratic\n\\[\n A Z^2+2BZ+C\\equiv A\\mathcal W_r(Z)\\pmod{2^P},\\qquad A\\text{ odd},\n \\qquad 0<|B^2-AC|<2^P,\n\\]\nat a proved precision for every admitted ordinary $r$.\nThen $(B^2-AC)/A^2\\equiv\\delta_r\\pmod{2^P}$ cannot vanish\nmodulo $2^P$. No depth-independent theorem supplying this inequality\nhas been established. The tested recurrence and exterior-contact\nidentities retain balancing terms and do not force a contradiction.\nFinite exponential models approximate the actual target at every depth,\nbut their order grows with precision, their clearing polynomial has\nzeros on the actual chart, and their proved construction-specific height\ncost does not give the required discriminant bound.","Sharp uniform ordinary Euler tail exponent $4M+3$\n & Proved, with earlier formal coverage","Universal simplicity of every $G_r$\n & Open","Throughout these sections, the normalized source and its reciprocal are\n\\begin{equation}\n G_{N,j}(t)=j!\\sum_{n\\geq0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n F_{N,j}(t)=G_{N,j}(t)^{-1}\n   =\\sum_{n\\geq0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!},\n \\label{eq:final-normalization}\n\\end{equation}\nwhere $N\\geq1$ and $0\\leq j<N$. In particular, $C_{N,j}(0)=1$.\nThe coefficients in the original convention satisfy\n\\begin{equation}\n E_{Nn}^{(N,j)}=j!C_{N,j}(n),\\qquad\n E_d^{(N,j)}=0\\quad(N\\nmid d).\n \\label{eq:final-original-normalization}\n\\end{equation}\nThus every valuation of a nonzero difference gains $v_p(j!)$ on returning\nto the original coefficients. This factor must be retained when stating\noptimal periods or the final congruence.","hroughout these sections, the normalized source and its reciprocal are\n\\begin{equation}\n G_{N,j}(t)=j!\\sum_{n\\geq0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n F_{N,j}(t)=G_{N,j}(t)^{-1}\n   =\\sum_{n\\geq0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!},\n \\label{eq:final-normalization}\n\\end{equation}\nwhere $N\\geq1$ and $0\\leq j<N$. In particular, $C_{N,j}(0)=1$.\nThe coefficients in the original convention satisfy\n\\begin{equation}\n E_{Nn}^{(N,j)}=j!C_{N,j}(n),\\qquad\n E_d^{(N,j)}=0\\quad(N\\nmid d).\n \\label{eq:final-original-normalization}\n\\end{equation}\nThus every valuation of a nonzero difference gains $v_p(j!)$ on returning\nto the original coefficients. This factor must be retained when stating\noptimal periods or the final congruence.","=\\sum_{n\\geq0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!},\n \\label{eq:final-normalization}\n\\end{equation}\nwhere $N\\geq1$ and $0\\leq j<N$. In particular, $C_{N,j}(0)=1$.\nThe coefficients in the original convention satisfy\n\\begin{equation}\n E_{Nn}^{(N,j)}=j!C_{N,j}(n),\\qquad\n E_d^{(N,j)}=0\\quad(N\\nmid d).\n \\label{eq:final-original-normalization}\n\\end{equation}\nThus every valuation of a nonzero difference gains $v_p(j!)$ on returning\nto the original coefficients. This factor must be retained when stating\noptimal periods or the final congruence.","Write $S f(n)=f(n+1)$ and $\\Delta_H=S^H-1$. A mixed difference\n$\\Delta_{H_1}\\cdots\\Delta_{H_k}$ acts on the row variable $n$, rather than\non the full Euler subscript. We use $v_p(0)=+\\infty$. Orders and starting\nrows are nonnegative integers; steps are positive integers unless stated\notherwise. An empty mixed difference is the original value, and a zero\nordinary step makes every nonempty product containing it zero.","\\begin{definition}[Hurwitz integral series]\nFor a prime $p$, let\n\\[\n \\mathcal H_p=\\left\\{\\sum_{d\\geq0}f_d\\frac{t^d}{d!}:\n                    f_d\\in\\mathbb Z_p\\right\\}\n       \\subset\\mathbb Q_p[[t]].\n\\]\nFor $N=p^a$ and $\\lambda\\in\\mathbb Z_p$, put\n\\[\n D=\\frac{d}{dt},\\qquad L_\\lambda=\\frac{D^N-\\lambda}{p}.\n\\]\nA series $A$ is $\\lambda$-flat if $L_\\lambda^k A\\in\\mathcal H_p$ for every\n$k\\geq0$.\n\\end{definition}","Products preserve $\\mathcal H_p$ because the coefficient at factorial\ndegree $d$ is\n\\[\n (AB)_d=\\sum_{i=0}^d\\binom di a_i b_{d-i}.\n\\]\nThe derivative $D$ acts by the coefficient shift $(DA)_d=a_{d+1}$.\nConsequently every nonnegative power of $D$ preserves the lattice.\nIf $A\\in\\mathcal H_p$ has unit constant term, its reciprocal belongs to\nthe same lattice: with $B=A^{-1}$,\n\\begin{equation}\n b_0=a_0^{-1},\\qquad\n b_d=-a_0^{-1}\\sum_{i=1}^d\\binom di a_i b_{d-i}.\n \\label{eq:hurwitz-reciprocal-recurrence}\n\\end{equation}\nIf $A$ is supported at degrees divisible by $N$, this recurrence also\nproves that $B$ has the same support.","_{d-i}.\n\\]\nThe derivative $D$ acts by the coefficient shift $(DA)_d=a_{d+1}$.\nConsequently every nonnegative power of $D$ preserves the lattice.\nIf $A\\in\\mathcal H_p$ has unit constant term, its reciprocal belongs to\nthe same lattice: with $B=A^{-1}$,\n\\begin{equation}\n b_0=a_0^{-1},\\qquad\n b_d=-a_0^{-1}\\sum_{i=1}^d\\binom di a_i b_{d-i}.\n \\label{eq:hurwitz-reciprocal-recurrence}\n\\end{equation}\nIf $A$ is supported at degrees divisible by $N$, this recurrence also\nproves that $B$ has the same support.","coefficient shift $(DA)_d=a_{d+1}$.\nConsequently every nonnegative power of $D$ preserves the lattice.\nIf $A\\in\\mathcal H_p$ has unit constant term, its reciprocal belongs to\nthe same lattice: with $B=A^{-1}$,\n\\begin{equation}\n b_0=a_0^{-1},\\qquad\n b_d=-a_0^{-1}\\sum_{i=1}^d\\binom di a_i b_{d-i}.\n \\label{eq:hurwitz-reciprocal-recurrence}\n\\end{equation}\nIf $A$ is supported at degrees divisible by $N$, this recurrence also\nproves that $B$ has the same support.","onsequently every nonnegative power of $D$ preserves the lattice.\nIf $A\\in\\mathcal H_p$ has unit constant term, its reciprocal belongs to\nthe same lattice: with $B=A^{-1}$,\n\\begin{equation}\n b_0=a_0^{-1},\\qquad\n b_d=-a_0^{-1}\\sum_{i=1}^d\\binom di a_i b_{d-i}.\n \\label{eq:hurwitz-reciprocal-recurrence}\n\\end{equation}\nIf $A$ is supported at degrees divisible by $N$, this recurrence also\nproves that $B$ has the same support.","\\begin{lemma}[Translated product formula]\n\\label{lem:translated-hurwitz-product}\nFor $N=p^a$, $a\\geq1$, and $\\lambda,\\mu\\in\\mathbb Z_p$,\n\\begin{equation}\n L_{\\lambda+\\mu}(AB)\n  =(L_\\lambda A)B+A(L_\\mu B)\n       +\\sum_{i=1}^{N-1}c_i(D^iA)(D^{N-i}B),\n \\qquad c_i=\\frac1p\\binom Ni\\in\\mathbb Z.\n \\label{eq:translated-product}\n\\end{equation}\nFlatness types add under multiplication.\n\\end{lemma}","\\begin{proof}\nThe identity $(1+X)^{p^a}=1+X^{p^a}$ over $\\mathbb F_p$ shows that every\ninterior binomial coefficient $\\binom Ni$ is divisible by $p$.\nSubtract $(\\lambda+\\mu)AB$ from the ordinary $N$th derivative product\nformula and divide by $p$ to obtain \\eqref{eq:translated-product}.","For completeness, the iterated formula is an integral multinomial\nformula, rather than a product rule with unspecified remainders. On the\ntensor of the two factors use the mutually commuting operators\n\\[\n L_\\lambda\\otimes1,\\quad 1\\otimes L_\\mu,\\quad\n c_iD^i\\otimes D^{N-i}\\quad(1\\leq i<N).\n\\]\nApplying multiplication after the $k$th power of their sum gives exactly\n$L_{\\lambda+\\mu}^k(AB)$. Expanding it yields terms indexed by\n$a,b,e_1,\\ldots,e_{N-1}\\geq0$, with $a+b+\\sum_i e_i=k$:\n\\begin{equation}\n \\frac{k!}{a!b!\\prod_i e_i!}\\prod_i c_i^{e_i}\n (D^\\alpha L_\\lambda^a A)(D^\\beta L_\\mu^b B),\n \\quad\n \\alpha=\\sum_i i e_i,\\quad\n \\beta=\\sum_i(N-i)e_i.\n \\label{eq:translated-multinomial}\n\\end{equation}\nEvery scalar is integral. If $A$ and $B$ have the stated flatness types,\nboth factors in every term belong to $\\mathcal H_p$, as does their product.\n\\end{proof}","d formula is an integral multinomial\nformula, rather than a product rule with unspecified remainders. On the\ntensor of the two factors use the mutually commuting operators\n\\[\n L_\\lambda\\otimes1,\\quad 1\\otimes L_\\mu,\\quad\n c_iD^i\\otimes D^{N-i}\\quad(1\\leq i<N).\n\\]\nApplying multiplication after the $k$th power of their sum gives exactly\n$L_{\\lambda+\\mu}^k(AB)$. Expanding it yields terms indexed by\n$a,b,e_1,\\ldots,e_{N-1}\\geq0$, with $a+b+\\sum_i e_i=k$:\n\\begin{equation}\n \\frac{k!}{a!b!\\prod_i e_i!}\\prod_i c_i^{e_i}\n (D^\\alpha L_\\lambda^a A)(D^\\beta L_\\mu^b B),\n \\quad\n \\alpha=\\sum_i i e_i,\\quad\n \\beta=\\sum_i(N-i)e_i.\n \\label{eq:translated-multinomial}\n\\end{equation}\nEvery scalar is integral. If $A$ and $B$ have the stated flatness types,\nboth factors in every term belong to $\\mathcal H_p$, as does their product.\n\\end{proof}","formula, rather than a product rule with unspecified remainders. On the\ntensor of the two factors use the mutually commuting operators\n\\[\n L_\\lambda\\otimes1,\\quad 1\\otimes L_\\mu,\\quad\n c_iD^i\\otimes D^{N-i}\\quad(1\\leq i<N).\n\\]\nApplying multiplication after the $k$th power of their sum gives exactly\n$L_{\\lambda+\\mu}^k(AB)$. Expanding it yields terms indexed by\n$a,b,e_1,\\ldots,e_{N-1}\\geq0$, with $a+b+\\sum_i e_i=k$:\n\\begin{equation}\n \\frac{k!}{a!b!\\prod_i e_i!}\\prod_i c_i^{e_i}\n (D^\\alpha L_\\lambda^a A)(D^\\beta L_\\mu^b B),\n \\quad\n \\alpha=\\sum_i i e_i,\\quad\n \\beta=\\sum_i(N-i)e_i.\n \\label{eq:translated-multinomial}\n\\end{equation}\nEvery scalar is integral. If $A$ and $B$ have the stated flatness types,\nboth factors in every term belong to $\\mathcal H_p$, as does their product.\n\\end{proof}","e with unspecified remainders. On the\ntensor of the two factors use the mutually commuting operators\n\\[\n L_\\lambda\\otimes1,\\quad 1\\otimes L_\\mu,\\quad\n c_iD^i\\otimes D^{N-i}\\quad(1\\leq i<N).\n\\]\nApplying multiplication after the $k$th power of their sum gives exactly\n$L_{\\lambda+\\mu}^k(AB)$. Expanding it yields terms indexed by\n$a,b,e_1,\\ldots,e_{N-1}\\geq0$, with $a+b+\\sum_i e_i=k$:\n\\begin{equation}\n \\frac{k!}{a!b!\\prod_i e_i!}\\prod_i c_i^{e_i}\n (D^\\alpha L_\\lambda^a A)(D^\\beta L_\\mu^b B),\n \\quad\n \\alpha=\\sum_i i e_i,\\quad\n \\beta=\\sum_i(N-i)e_i.\n \\label{eq:translated-multinomial}\n\\end{equation}\nEvery scalar is integral. If $A$ and $B$ have the stated flatness types,\nboth factors in every term belong to $\\mathcal H_p$, as does their product.\n\\end{proof}","th power of their sum gives exactly\n$L_{\\lambda+\\mu}^k(AB)$. Expanding it yields terms indexed by\n$a,b,e_1,\\ldots,e_{N-1}\\geq0$, with $a+b+\\sum_i e_i=k$:\n\\begin{equation}\n \\frac{k!}{a!b!\\prod_i e_i!}\\prod_i c_i^{e_i}\n (D^\\alpha L_\\lambda^a A)(D^\\beta L_\\mu^b B),\n \\quad\n \\alpha=\\sum_i i e_i,\\quad\n \\beta=\\sum_i(N-i)e_i.\n \\label{eq:translated-multinomial}\n\\end{equation}\nEvery scalar is integral. If $A$ and $B$ have the stated flatness types,\nboth factors in every term belong to $\\mathcal H_p$, as does their product.\n\\end{proof}","of their sum gives exactly\n$L_{\\lambda+\\mu}^k(AB)$. Expanding it yields terms indexed by\n$a,b,e_1,\\ldots,e_{N-1}\\geq0$, with $a+b+\\sum_i e_i=k$:\n\\begin{equation}\n \\frac{k!}{a!b!\\prod_i e_i!}\\prod_i c_i^{e_i}\n (D^\\alpha L_\\lambda^a A)(D^\\beta L_\\mu^b B),\n \\quad\n \\alpha=\\sum_i i e_i,\\quad\n \\beta=\\sum_i(N-i)e_i.\n \\label{eq:translated-multinomial}\n\\end{equation}\nEvery scalar is integral. If $A$ and $B$ have the stated flatness types,\nboth factors in every term belong to $\\mathcal H_p$, as does their product.\n\\end{proof}","terms indexed by\n$a,b,e_1,\\ldots,e_{N-1}\\geq0$, with $a+b+\\sum_i e_i=k$:\n\\begin{equation}\n \\frac{k!}{a!b!\\prod_i e_i!}\\prod_i c_i^{e_i}\n (D^\\alpha L_\\lambda^a A)(D^\\beta L_\\mu^b B),\n \\quad\n \\alpha=\\sum_i i e_i,\\quad\n \\beta=\\sum_i(N-i)e_i.\n \\label{eq:translated-multinomial}\n\\end{equation}\nEvery scalar is integral. If $A$ and $B$ have the stated flatness types,\nboth factors in every term belong to $\\mathcal H_p$, as does their product.\n\\end{proof}","\\begin{lemma}[Reciprocal transfer]\n\\label{lem:translated-hurwitz-inverse}\nIf $A$ is $\\lambda$-flat and has unit constant term, then $A^{-1}$ is\n$(-\\lambda)$-flat.\n\\end{lemma}","\\begin{proof}\nSet $B=A^{-1}$. Its order-zero integrality was established in\n\\eqref{eq:hurwitz-reciprocal-recurrence}. Assume by induction that\n$L_{-\\lambda}^bB\\in\\mathcal H_p$ for $b<k$. Since $AB=1$,\n\\[\n L_0^k(AB)=\\left(\\frac{D^N}{p}\\right)^k1=0\\qquad(k>0).\n\\]\nUse \\eqref{eq:translated-multinomial} with $\\mu=-\\lambda$.\nExactly one term has inverse order $b=k$: it has $a=0$ and all $e_i=0$,\nand is $A L_{-\\lambda}^k B$. In every other term $b<k$, so its second\nfactor is integral by induction and derivative-shift integrality. Its\nfirst factor is integral by the flatness of $A$, and its scalar is\nintegral. Hence $A L_{-\\lambda}^k B\\in\\mathcal H_p$. Multiplying by the\nalready integral $B$ proves $L_{-\\lambda}^k B\\in\\mathca","\\begin{proof}\nSet $B=A^{-1}$. Its order-zero integrality was established in\n\\eqref{eq:hurwitz-reciprocal-recurrence}. Assume by induction that\n$L_{-\\lambda}^bB\\in\\mathcal H_p$ for $b<k$. Since $AB=1$,\n\\[\n L_0^k(AB)=\\left(\\frac{D^N}{p}\\right)^k1=0\\qquad(k>0).\n\\]\nUse \\eqref{eq:translated-multinomial} with $\\mu=-\\lambda$.\nExactly one term has inverse order $b=k$: it has $a=0$ and all $e_i=0$,\nand is $A L_{-\\lambda}^k B$. In every other term $b<k$, so its second\nfactor is integral by induction and derivative-shift integrality. Its\nfirst factor is integral by the flatness of $A$, and its scalar is\nintegral. Hence $A L_{-\\lambda}^k B\\in\\mathcal H_p$. Multiplying by the\nalready integral $B$ proves $L_{-\\lambda}^k B\\in\\mathcal H_p$.\nThis induction takes place in the entire Hurwitz lattice. It places no\nrestriction on the coefficient degree or the starting row.\n\\end{proo","\\begin{proof}\nSet $B=A^{-1}$. Its order-zero integrality was established in\n\\eqref{eq:hurwitz-reciprocal-recurrence}. Assume by induction that\n$L_{-\\lambda}^bB\\in\\mathcal H_p$ for $b<k$. Since $AB=1$,\n\\[\n L_0^k(AB)=\\left(\\frac{D^N}{p}\\right)^k1=0\\qquad(k>0).\n\\]\nUse \\eqref{eq:translated-multinomial} with $\\mu=-\\lambda$.\nExactly one term has inverse order $b=k$: it has $a=0$ and all $e_i=0$,\nand is $A L_{-\\lambda}^k B$. In every other term $b<k$, so its second\nfactor is integral by induction and derivative-shift integrality. Its\nfirst factor is integral by the flatness of $A$, and its scalar is\nintegral. Hence $A L_{-\\lambda}^k B\\in\\mathcal H_p$. Multiplying by the\nalready integral $B$ proves $L_{-\\lambda}^k B\\in\\mathcal H_p$.\nThis induction takes place in the entire Hurwitz lattice. It places no\nrestriction on the coefficient degree or the starting row.\n\\end{proof}","begin{proof}\nSet $B=A^{-1}$. Its order-zero integrality was established in\n\\eqref{eq:hurwitz-reciprocal-recurrence}. Assume by induction that\n$L_{-\\lambda}^bB\\in\\mathcal H_p$ for $b<k$. Since $AB=1$,\n\\[\n L_0^k(AB)=\\left(\\frac{D^N}{p}\\right)^k1=0\\qquad(k>0).\n\\]\nUse \\eqref{eq:translated-multinomial} with $\\mu=-\\lambda$.\nExactly one term has inverse order $b=k$: it has $a=0$ and all $e_i=0$,\nand is $A L_{-\\lambda}^k B$. In every other term $b<k$, so its second\nfactor is integral by induction and derivative-shift integrality. Its\nfirst factor is integral by the flatness of $A$, and its scalar is\nintegral. Hence $A L_{-\\lambda}^k B\\in\\mathcal H_p$. Multiplying by the\nalready integral $B$ proves $L_{-\\lambda}^k B\\in\\mathcal H_p$.\nThis induction takes place in the entire Hurwitz lattice. It places no\nrestriction on the coefficient degree or the starting row.\n\\end{proof}","o integrality was established in\n\\eqref{eq:hurwitz-reciprocal-recurrence}. Assume by induction that\n$L_{-\\lambda}^bB\\in\\mathcal H_p$ for $b<k$. Since $AB=1$,\n\\[\n L_0^k(AB)=\\left(\\frac{D^N}{p}\\right)^k1=0\\qquad(k>0).\n\\]\nUse \\eqref{eq:translated-multinomial} with $\\mu=-\\lambda$.\nExactly one term has inverse order $b=k$: it has $a=0$ and all $e_i=0$,\nand is $A L_{-\\lambda}^k B$. In every other term $b<k$, so its second\nfactor is integral by induction and derivative-shift integrality. Its\nfirst factor is integral by the flatness of $A$, and its scalar is\nintegral. Hence $A L_{-\\lambda}^k B\\in\\mathcal H_p$. Multiplying by the\nalready integral $B$ proves $L_{-\\lambda}^k B\\in\\mathcal H_p$.\nThis induction takes place in the entire Hurwitz lattice. It places no\nrestriction on the coefficient degree or the starting row.\n\\end{proof}","r $b<k$. Since $AB=1$,\n\\[\n L_0^k(AB)=\\left(\\frac{D^N}{p}\\right)^k1=0\\qquad(k>0).\n\\]\nUse \\eqref{eq:translated-multinomial} with $\\mu=-\\lambda$.\nExactly one term has inverse order $b=k$: it has $a=0$ and all $e_i=0$,\nand is $A L_{-\\lambda}^k B$. In every other term $b<k$, so its second\nfactor is integral by induction and derivative-shift integrality. Its\nfirst factor is integral by the flatness of $A$, and its scalar is\nintegral. Hence $A L_{-\\lambda}^k B\\in\\mathcal H_p$. Multiplying by the\nalready integral $B$ proves $L_{-\\lambda}^k B\\in\\mathcal H_p$.\nThis induction takes place in the entire Hurwitz lattice. It places no\nrestriction on the coefficient degree or the starting row.\n\\end{proof}","eft(\\frac{D^N}{p}\\right)^k1=0\\qquad(k>0).\n\\]\nUse \\eqref{eq:translated-multinomial} with $\\mu=-\\lambda$.\nExactly one term has inverse order $b=k$: it has $a=0$ and all $e_i=0$,\nand is $A L_{-\\lambda}^k B$. In every other term $b<k$, so its second\nfactor is integral by induction and derivative-shift integrality. Its\nfirst factor is integral by the flatness of $A$, and its scalar is\nintegral. Hence $A L_{-\\lambda}^k B\\in\\mathcal H_p$. Multiplying by the\nalready integral $B$ proves $L_{-\\lambda}^k B\\in\\mathcal H_p$.\nThis induction takes place in the entire Hurwitz lattice. It places no\nrestriction on the coefficient degree or the starting row.\n\\end{proof}","tinomial} with $\\mu=-\\lambda$.\nExactly one term has inverse order $b=k$: it has $a=0$ and all $e_i=0$,\nand is $A L_{-\\lambda}^k B$. In every other term $b<k$, so its second\nfactor is integral by induction and derivative-shift integrality. Its\nfirst factor is integral by the flatness of $A$, and its scalar is\nintegral. Hence $A L_{-\\lambda}^k B\\in\\mathcal H_p$. Multiplying by the\nalready integral $B$ proves $L_{-\\lambda}^k B\\in\\mathcal H_p$.\nThis induction takes place in the entire Hurwitz lattice. It places no\nrestriction on the coefficient degree or the starting row.\n\\end{proof}","ith $\\mu=-\\lambda$.\nExactly one term has inverse order $b=k$: it has $a=0$ and all $e_i=0$,\nand is $A L_{-\\lambda}^k B$. In every other term $b<k$, so its second\nfactor is integral by induction and derivative-shift integrality. Its\nfirst factor is integral by the flatness of $A$, and its scalar is\nintegral. Hence $A L_{-\\lambda}^k B\\in\\mathcal H_p$. Multiplying by the\nalready integral $B$ proves $L_{-\\lambda}^k B\\in\\mathcal H_p$.\nThis induction takes place in the entire Hurwitz lattice. It places no\nrestriction on the coefficient degree or the starting row.\n\\end{proof}","The packed factorial coefficient of $G_{N,j}$ is\n\\begin{equation}\n g_{N,j}(n)=\\frac{j!(Nn)!}{(Nn+j)!}\n   =\\prod_{i=1}^j\\left(1+\\frac Ni n\\right)^{-1}.\n \\label{eq:actual-prime-power-kernel}\n\\end{equation}\nThe empty product is one. If $N=p^a$ and $1\\leq i<N$, then\n$v_p(i)<a$, so $N/i\\in p\\mathbb Z_p$. Every affine denominator in\n\\eqref{eq:actual-prime-power-kernel} is therefore a unit on $\\mathbb Z_p$.","For $s\\in p\\mathbb Z_p$, an exact finite-difference identity is\n\\begin{equation}\n \\Delta_1^k\\frac1{1+sn}\n  =\\frac{(-s)^k k!}{\\prod_{h=0}^k(1+s(n+h))}.\n \\label{eq:affine-reciprocal-exact}\n\\end{equation}\nIt follows by induction, subtracting adjacent rational functions; each\nsuccessive subtraction adds the factor $-(k+1)s$ in the numerator.\nIts denominator is a unit, so it belongs to $p^k\\mathbb Z_p$.\nThe discrete Leibniz formula\n\\begin{equation}\n \\Delta_1^k(fg)(n)\n =\\sum_{r=0}^k\\binom kr\n       (\\Delta_1^r f)(n)(\\Delta_1^{k-r}g)(n+r)\n \\label{eq:discrete-leibniz-final}\n\\end{equation}\nthen proves $\\Delta_1^k g_{N,j}(n)\\in p^k\\mathbb Z_p$ for every $k,n$.\nThe coefficient at degree $Nn$ of $L_1^kG_{N,j}$ is precisely\n$","For $s\\in p\\mathbb Z_p$, an exact finite-difference identity is\n\\begin{equation}\n \\Delta_1^k\\frac1{1+sn}\n  =\\frac{(-s)^k k!}{\\prod_{h=0}^k(1+s(n+h))}.\n \\label{eq:affine-reciprocal-exact}\n\\end{equation}\nIt follows by induction, subtracting adjacent rational functions; each\nsuccessive subtraction adds the factor $-(k+1)s$ in the numerator.\nIts denominator is a unit, so it belongs to $p^k\\mathbb Z_p$.\nThe discrete Leibniz formula\n\\begin{equation}\n \\Delta_1^k(fg)(n)\n =\\sum_{r=0}^k\\binom kr\n       (\\Delta_1^r f)(n)(\\Delta_1^{k-r}g)(n+r)\n \\label{eq:discrete-leibniz-final}\n\\end{equation}\nthen proves $\\Delta_1^k g_{N,j}(n)\\in p^k\\mathbb Z_p$ for every $k,n$.\nThe coefficient at degree $Nn$ of $L_1^kG_{N,j}$ is precisely\n$p^{-k}\\Delta_1^k g_{N,j}(n)$, and the other residue classes vanish.\nThus the actual source is $1$-flat.","ence identity is\n\\begin{equation}\n \\Delta_1^k\\frac1{1+sn}\n  =\\frac{(-s)^k k!}{\\prod_{h=0}^k(1+s(n+h))}.\n \\label{eq:affine-reciprocal-exact}\n\\end{equation}\nIt follows by induction, subtracting adjacent rational functions; each\nsuccessive subtraction adds the factor $-(k+1)s$ in the numerator.\nIts denominator is a unit, so it belongs to $p^k\\mathbb Z_p$.\nThe discrete Leibniz formula\n\\begin{equation}\n \\Delta_1^k(fg)(n)\n =\\sum_{r=0}^k\\binom kr\n       (\\Delta_1^r f)(n)(\\Delta_1^{k-r}g)(n+r)\n \\label{eq:discrete-leibniz-final}\n\\end{equation}\nthen proves $\\Delta_1^k g_{N,j}(n)\\in p^k\\mathbb Z_p$ for every $k,n$.\nThe coefficient at degree $Nn$ of $L_1^kG_{N,j}$ is precisely\n$p^{-k}\\Delta_1^k g_{N,j}(n)$, and the other residue classes vanish.\nThus the actual source is $1$-flat.","It follows by induction, subtracting adjacent rational functions; each\nsuccessive subtraction adds the factor $-(k+1)s$ in the numerator.\nIts denominator is a unit, so it belongs to $p^k\\mathbb Z_p$.\nThe discrete Leibniz formula\n\\begin{equation}\n \\Delta_1^k(fg)(n)\n =\\sum_{r=0}^k\\binom kr\n       (\\Delta_1^r f)(n)(\\Delta_1^{k-r}g)(n+r)\n \\label{eq:discrete-leibniz-final}\n\\end{equation}\nthen proves $\\Delta_1^k g_{N,j}(n)\\in p^k\\mathbb Z_p$ for every $k,n$.\nThe coefficient at degree $Nn$ of $L_1^kG_{N,j}$ is precisely\n$p^{-k}\\Delta_1^k g_{N,j}(n)$, and the other residue classes vanish.\nThus the actual source is $1$-flat.","nal functions; each\nsuccessive subtraction adds the factor $-(k+1)s$ in the numerator.\nIts denominator is a unit, so it belongs to $p^k\\mathbb Z_p$.\nThe discrete Leibniz formula\n\\begin{equation}\n \\Delta_1^k(fg)(n)\n =\\sum_{r=0}^k\\binom kr\n       (\\Delta_1^r f)(n)(\\Delta_1^{k-r}g)(n+r)\n \\label{eq:discrete-leibniz-final}\n\\end{equation}\nthen proves $\\Delta_1^k g_{N,j}(n)\\in p^k\\mathbb Z_p$ for every $k,n$.\nThe coefficient at degree $Nn$ of $L_1^kG_{N,j}$ is precisely\n$p^{-k}\\Delta_1^k g_{N,j}(n)$, and the other residue classes vanish.\nThus the actual source is $1$-flat.","btraction adds the factor $-(k+1)s$ in the numerator.\nIts denominator is a unit, so it belongs to $p^k\\mathbb Z_p$.\nThe discrete Leibniz formula\n\\begin{equation}\n \\Delta_1^k(fg)(n)\n =\\sum_{r=0}^k\\binom kr\n       (\\Delta_1^r f)(n)(\\Delta_1^{k-r}g)(n+r)\n \\label{eq:discrete-leibniz-final}\n\\end{equation}\nthen proves $\\Delta_1^k g_{N,j}(n)\\in p^k\\mathbb Z_p$ for every $k,n$.\nThe coefficient at degree $Nn$ of $L_1^kG_{N,j}$ is precisely\n$p^{-k}\\Delta_1^k g_{N,j}(n)$, and the other residue classes vanish.\nThus the actual source is $1$-flat.","n adds the factor $-(k+1)s$ in the numerator.\nIts denominator is a unit, so it belongs to $p^k\\mathbb Z_p$.\nThe discrete Leibniz formula\n\\begin{equation}\n \\Delta_1^k(fg)(n)\n =\\sum_{r=0}^k\\binom kr\n       (\\Delta_1^r f)(n)(\\Delta_1^{k-r}g)(n+r)\n \\label{eq:discrete-leibniz-final}\n\\end{equation}\nthen proves $\\Delta_1^k g_{N,j}(n)\\in p^k\\mathbb Z_p$ for every $k,n$.\nThe coefficient at degree $Nn$ of $L_1^kG_{N,j}$ is precisely\n$p^{-k}\\Delta_1^k g_{N,j}(n)$, and the other residue classes vanish.\nThus the actual source is $1$-flat.","in the numerator.\nIts denominator is a unit, so it belongs to $p^k\\mathbb Z_p$.\nThe discrete Leibniz formula\n\\begin{equation}\n \\Delta_1^k(fg)(n)\n =\\sum_{r=0}^k\\binom kr\n       (\\Delta_1^r f)(n)(\\Delta_1^{k-r}g)(n+r)\n \\label{eq:discrete-leibniz-final}\n\\end{equation}\nthen proves $\\Delta_1^k g_{N,j}(n)\\in p^k\\mathbb Z_p$ for every $k,n$.\nThe coefficient at degree $Nn$ of $L_1^kG_{N,j}$ is precisely\n$p^{-k}\\Delta_1^k g_{N,j}(n)$, and the other residue classes vanish.\nThus the actual source is $1$-flat.","ibniz formula\n\\begin{equation}\n \\Delta_1^k(fg)(n)\n =\\sum_{r=0}^k\\binom kr\n       (\\Delta_1^r f)(n)(\\Delta_1^{k-r}g)(n+r)\n \\label{eq:discrete-leibniz-final}\n\\end{equation}\nthen proves $\\Delta_1^k g_{N,j}(n)\\in p^k\\mathbb Z_p$ for every $k,n$.\nThe coefficient at degree $Nn$ of $L_1^kG_{N,j}$ is precisely\n$p^{-k}\\Delta_1^k g_{N,j}(n)$, and the other residue classes vanish.\nThus the actual source is $1$-flat.","\\begin{theorem}[All-row prime-power transfer]\n\\label{thm:all-prime-power-transfer}\nLet $p$ be prime, $a\\geq1$, $N=p^a$, and $0\\leq j<N$.\nThe coefficients $C_{N,j}(n)$ belong to $\\mathbb Z_p$ for all $n\\geq0$.\nSet $A_{N,j}(n)=(-1)^n C_{N,j}(n)$ and\n$T_H=S^H-(-1)^H$. For $k\\geq1$ and positive $H_1,\\ldots,H_k$,\n\\begin{align}\n v_p(\\Delta_{H_1}\\cdots\\Delta_{H_k}A_{N,j}(n))\n  &\\geq k+\\sum_{\\nu=1}^k v_p(H_\\nu),\n \\label{eq:prime-power-signed-bound}","\\begin{theorem}[All-row prime-power transfer]\n\\label{thm:all-prime-power-transfer}\nLet $p$ be prime, $a\\geq1$, $N=p^a$, and $0\\leq j<N$.\nThe coefficients $C_{N,j}(n)$ belong to $\\mathbb Z_p$ for all $n\\geq0$.\nSet $A_{N,j}(n)=(-1)^n C_{N,j}(n)$ and\n$T_H=S^H-(-1)^H$. For $k\\geq1$ and positive $H_1,\\ldots,H_k$,\n\\begin{align}\n v_p(\\Delta_{H_1}\\cdots\\Delta_{H_k}A_{N,j}(n))\n  &\\geq k+\\sum_{\\nu=1}^k v_p(H_\\nu),\n \\label{eq:prime-power-signed-bound}\\\\\n v_p(T_{H_1}\\cdots T_{H_k}E_{Nn}^{(N,j)})\n  &\\geq v_p(j!)+k+\\sum_{\\nu=1}^k v_p(H_\\nu).\n \\label{eq:prime-power-original-twist}\n\\end{align}\nAt $p=2$, the ordinary differences of $C_{N,j}$ and of the original\ncoefficients satisfy the same bounds. These dyadic statements also hold\nfor $N=1$.\n\\end{theorem}","v_p(T_{H_1}\\cdots T_{H_k}E_{Nn}^{(N,j)})\n  &\\geq v_p(j!)+k+\\sum_{\\nu=1}^k v_p(H_\\nu).\n \\label{eq:prime-power-original-twist}\n\\end{align}\nAt $p=2$, the ordinary differences of $C_{N,j}$ and of the original\ncoefficients satisfy the same bounds. These dyadic statements also hold\nfor $N=1$.\n\\end{theorem}","\\begin{proof}\nReciprocal transfer gives $L_{-1}^kF_{N,j}\\in\\mathcal H_p$.\nExtraction of the packed coefficient gives\n$(S+1)^k C_{N,j}(n)\\in p^k\\mathbb Z_p$. Moreover,\n\\[\n \\Delta_1^k A_{N,j}(n)=(-1)^{n+k}(S+1)^k C_{N,j}(n),\n\\]\nso $A_{N,j}$ has the ordinary $p^k$ difference property.\nAt $p=2$, $L_1=L_{-1}-1$; its integral binomial expansion on\n$F_{N,j}$ gives $\\Delta_1^k C_{N,j}(n)\\in2^k\\mathbb Z_2$ directly.\nFor $N=1$ the interior sum in the product formula is empty, so the same\nargument applies without change; equivalently, $F_{1,0}(t)=e^{-t}$.","ciprocal transfer gives $L_{-1}^kF_{N,j}\\in\\mathcal H_p$.\nExtraction of the packed coefficient gives\n$(S+1)^k C_{N,j}(n)\\in p^k\\mathbb Z_p$. Moreover,\n\\[\n \\Delta_1^k A_{N,j}(n)=(-1)^{n+k}(S+1)^k C_{N,j}(n),\n\\]\nso $A_{N,j}$ has the ordinary $p^k$ difference property.\nAt $p=2$, $L_1=L_{-1}-1$; its integral binomial expansion on\n$F_{N,j}$ gives $\\Delta_1^k C_{N,j}(n)\\in2^k\\mathbb Z_2$ directly.\nFor $N=1$ the interior sum in the product formula is empty, so the same\nargument applies without change; equivalently, $F_{1,0}(t)=e^{-t}$.","j}(n)\\in p^k\\mathbb Z_p$. Moreover,\n\\[\n \\Delta_1^k A_{N,j}(n)=(-1)^{n+k}(S+1)^k C_{N,j}(n),\n\\]\nso $A_{N,j}$ has the ordinary $p^k$ difference property.\nAt $p=2$, $L_1=L_{-1}-1$; its integral binomial expansion on\n$F_{N,j}$ gives $\\Delta_1^k C_{N,j}(n)\\in2^k\\mathbb Z_2$ directly.\nFor $N=1$ the interior sum in the product formula is empty, so the same\nargument applies without change; equivalently, $F_{1,0}(t)=e^{-t}$.","To retain the valuation of a step, let $B$ be any sequence satisfying\n$\\Delta_1^m B\\in p^m\\mathbb Z_p$ at every order and row. Use the finite\noperator identity\n\\[\n \\Delta_H=\\sum_{i=1}^H\\binom Hi\\Delta_1^i.\n\\]\nThe relation $i\\binom Hi=H\\binom{H-1}{i-1}$ implies\n\\[\n v_p\\left(p^i\\binom Hi\\right)\n \\geq i+v_p(H)-v_p(i)\\geq1+v_p(H).\n\\]\nMultiplying these identities for the $k$ steps, every term of total\norder $M$ has its scalar multiplied by $p^M$, and is divisible by\n$p^{k+\\sum_\\nu v_p(H_\\nu)}$. This proves the mixed-step bound.\nFinally,\n\\[\n \\Delta_H A_{N,j}(n)=(-1)^{n+H}T_H C_{N,j}(n).\n\\]\nIteration contributes only a global sign. Multiplying by $j!$ proves\n\\eqref{eq:prime-power-original-twist}.\n\\end{proof}","in the valuation of a step, let $B$ be any sequence satisfying\n$\\Delta_1^m B\\in p^m\\mathbb Z_p$ at every order and row. Use the finite\noperator identity\n\\[\n \\Delta_H=\\sum_{i=1}^H\\binom Hi\\Delta_1^i.\n\\]\nThe relation $i\\binom Hi=H\\binom{H-1}{i-1}$ implies\n\\[\n v_p\\left(p^i\\binom Hi\\right)\n \\geq i+v_p(H)-v_p(i)\\geq1+v_p(H).\n\\]\nMultiplying these identities for the $k$ steps, every term of total\norder $M$ has its scalar multiplied by $p^M$, and is divisible by\n$p^{k+\\sum_\\nu v_p(H_\\nu)}$. This proves the mixed-step bound.\nFinally,\n\\[\n \\Delta_H A_{N,j}(n)=(-1)^{n+H}T_H C_{N,j}(n).\n\\]\nIteration contributes only a global sign. Multiplying by $j!$ proves\n\\eqref{eq:prime-power-original-twist}.\n\\end{proof}","B$ be any sequence satisfying\n$\\Delta_1^m B\\in p^m\\mathbb Z_p$ at every order and row. Use the finite\noperator identity\n\\[\n \\Delta_H=\\sum_{i=1}^H\\binom Hi\\Delta_1^i.\n\\]\nThe relation $i\\binom Hi=H\\binom{H-1}{i-1}$ implies\n\\[\n v_p\\left(p^i\\binom Hi\\right)\n \\geq i+v_p(H)-v_p(i)\\geq1+v_p(H).\n\\]\nMultiplying these identities for the $k$ steps, every term of total\norder $M$ has its scalar multiplied by $p^M$, and is divisible by\n$p^{k+\\sum_\\nu v_p(H_\\nu)}$. This proves the mixed-step bound.\nFinally,\n\\[\n \\Delta_H A_{N,j}(n)=(-1)^{n+H}T_H C_{N,j}(n).\n\\]\nIteration contributes only a global sign. Multiplying by $j!$ proves\n\\eqref{eq:prime-power-original-twist}.\n\\end{proof}","ng\n$\\Delta_1^m B\\in p^m\\mathbb Z_p$ at every order and row. Use the finite\noperator identity\n\\[\n \\Delta_H=\\sum_{i=1}^H\\binom Hi\\Delta_1^i.\n\\]\nThe relation $i\\binom Hi=H\\binom{H-1}{i-1}$ implies\n\\[\n v_p\\left(p^i\\binom Hi\\right)\n \\geq i+v_p(H)-v_p(i)\\geq1+v_p(H).\n\\]\nMultiplying these identities for the $k$ steps, every term of total\norder $M$ has its scalar multiplied by $p^M$, and is divisible by\n$p^{k+\\sum_\\nu v_p(H_\\nu)}$. This proves the mixed-step bound.\nFinally,\n\\[\n \\Delta_H A_{N,j}(n)=(-1)^{n+H}T_H C_{N,j}(n).\n\\]\nIteration contributes only a global sign. Multiplying by $j!$ proves\n\\eqref{eq:prime-power-original-twist}.\n\\end{proof}","e\noperator identity\n\\[\n \\Delta_H=\\sum_{i=1}^H\\binom Hi\\Delta_1^i.\n\\]\nThe relation $i\\binom Hi=H\\binom{H-1}{i-1}$ implies\n\\[\n v_p\\left(p^i\\binom Hi\\right)\n \\geq i+v_p(H)-v_p(i)\\geq1+v_p(H).\n\\]\nMultiplying these identities for the $k$ steps, every term of total\norder $M$ has its scalar multiplied by $p^M$, and is divisible by\n$p^{k+\\sum_\\nu v_p(H_\\nu)}$. This proves the mixed-step bound.\nFinally,\n\\[\n \\Delta_H A_{N,j}(n)=(-1)^{n+H}T_H C_{N,j}(n).\n\\]\nIteration contributes only a global sign. Multiplying by $j!$ proves\n\\eqref{eq:prime-power-original-twist}.\n\\end{proof}","1}$ implies\n\\[\n v_p\\left(p^i\\binom Hi\\right)\n \\geq i+v_p(H)-v_p(i)\\geq1+v_p(H).\n\\]\nMultiplying these identities for the $k$ steps, every term of total\norder $M$ has its scalar multiplied by $p^M$, and is divisible by\n$p^{k+\\sum_\\nu v_p(H_\\nu)}$. This proves the mixed-step bound.\nFinally,\n\\[\n \\Delta_H A_{N,j}(n)=(-1)^{n+H}T_H C_{N,j}(n).\n\\]\nIteration contributes only a global sign. Multiplying by $j!$ proves\n\\eqref{eq:prime-power-original-twist}.\n\\end{proof}","\\begin{corollary}[Threshold-free dyadic congruence]\n\\label{cor:all-dyadic-final}\nFor $N=2^a$, $a\\geq0$, all permitted $j$, and $n\\geq0$,\n\\begin{equation}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}E_{Nn}^{(N,j)})\n  \\geq v_2(j!)+k+\\sum_{\\nu=1}^k v_2(H_\\nu).\n \\label{eq:all-dyadic-final-bound}\n\\end{equation}\nIn particular, $\\Delta_H^k$ gains $k(1+v_2(H))$ bits before normalization.\nFor $r>v_2(j!)$, the row displacement\n$H=2^{\\max(0,r-v_2(j!)-1)}$ is a sufficient period modulo $2^r$ from\nrow zero. If $r\\leq v_2(j!)$, all original coefficients vanish modulo\n$2^r$.\n\\end{corollary}","ary}[Threshold-free dyadic congruence]\n\\label{cor:all-dyadic-final}\nFor $N=2^a$, $a\\geq0$, all permitted $j$, and $n\\geq0$,\n\\begin{equation}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}E_{Nn}^{(N,j)})\n  \\geq v_2(j!)+k+\\sum_{\\nu=1}^k v_2(H_\\nu).\n \\label{eq:all-dyadic-final-bound}\n\\end{equation}\nIn particular, $\\Delta_H^k$ gains $k(1+v_2(H))$ bits before normalization.\nFor $r>v_2(j!)$, the row displacement\n$H=2^{\\max(0,r-v_2(j!)-1)}$ is a sufficient period modulo $2^r$ from\nrow zero. If $r\\leq v_2(j!)$, all original coefficients vanish modulo\n$2^r$.\n\\end{corollary}","ic-final}\nFor $N=2^a$, $a\\geq0$, all permitted $j$, and $n\\geq0$,\n\\begin{equation}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}E_{Nn}^{(N,j)})\n  \\geq v_2(j!)+k+\\sum_{\\nu=1}^k v_2(H_\\nu).\n \\label{eq:all-dyadic-final-bound}\n\\end{equation}\nIn particular, $\\Delta_H^k$ gains $k(1+v_2(H))$ bits before normalization.\nFor $r>v_2(j!)$, the row displacement\n$H=2^{\\max(0,r-v_2(j!)-1)}$ is a sufficient period modulo $2^r$ from\nrow zero. If $r\\leq v_2(j!)$, all original coefficients vanish modulo\n$2^r$.\n\\end{corollary}","a\\geq0$, all permitted $j$, and $n\\geq0$,\n\\begin{equation}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}E_{Nn}^{(N,j)})\n  \\geq v_2(j!)+k+\\sum_{\\nu=1}^k v_2(H_\\nu).\n \\label{eq:all-dyadic-final-bound}\n\\end{equation}\nIn particular, $\\Delta_H^k$ gains $k(1+v_2(H))$ bits before normalization.\nFor $r>v_2(j!)$, the row displacement\n$H=2^{\\max(0,r-v_2(j!)-1)}$ is a sufficient period modulo $2^r$ from\nrow zero. If $r\\leq v_2(j!)$, all original coefficients vanish modulo\n$2^r$.\n\\end{corollary}","}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}E_{Nn}^{(N,j)})\n  \\geq v_2(j!)+k+\\sum_{\\nu=1}^k v_2(H_\\nu).\n \\label{eq:all-dyadic-final-bound}\n\\end{equation}\nIn particular, $\\Delta_H^k$ gains $k(1+v_2(H))$ bits before normalization.\nFor $r>v_2(j!)$, the row displacement\n$H=2^{\\max(0,r-v_2(j!)-1)}$ is a sufficient period modulo $2^r$ from\nrow zero. If $r\\leq v_2(j!)$, all original coefficients vanish modulo\n$2^r$.\n\\end{corollary}","\\begin{remark}[Necessary boundaries]\nAt an odd prime the ordinary step-one difference cannot replace the\nsign-twisted operator in the general theorem. For $j=0$,\n$C_{N,0}(0)=1$ and $C_{N,0}(1)=-1$, so their ordinary difference is\n$-2$, a unit at every odd prime. The proof also does not extend\nunchanged to non-prime-power $N$, whose interior binomial coefficients\nneed not be divisible by the chosen prime. Neither observation is a\ncounterexample to Nishibuchi's original eventual congruence.\n\\end{remark}","Write $C_j=C_{4,j}$ and $F_j=F_{4,j}$. The all-row theorem removes the\nthreshold, but at $N=4$ it leaves a further factorial gain unused.\nThe exact algebra below proves that gain for all four addresses and\nidentifies several infinite families where it is sharp.","\\begin{theorem}[Four-address factorial bound]\n\\label{thm:four-factorial-bound}\nFor $j\\in\\{0,1,2,3\\}$, $n\\geq0$, $k\\geq1$, and positive mixed steps,\n\\begin{equation}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}C_j(n))\n \\geq k+v_2(k!)+\\sum_{\\nu=1}^k v_2(H_\\nu)+\\beta_j,\n \\quad (\\beta_0,\\beta_1,\\beta_2,\\beta_3)=(0,0,3,1).\n \\label{eq:factorial-four-lower}\n\\end{equation}\nAddress two also satisfies\n\\begin{equation}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}C_2(n))\n \\geq 2k+v_2(k!)+\\left\\lfloor\\frac{k}{2}\\right\\rfloor\n             +\\sum_{\\nu=1}^k v_2(H_\\nu).\n \\label{eq:factorial-four-j2-filtration}\n\\end{equation}\nFor the original coefficients, the offsets in\n\\eqref{eq:factorial-four-lower} are $(0,0,4,2)$.\nNo positive-order offset is asserted for the empty difference.\n\\end{theorem}","egin{theorem}[Four-address factorial bound]\n\\label{thm:four-factorial-bound}\nFor $j\\in\\{0,1,2,3\\}$, $n\\geq0$, $k\\geq1$, and positive mixed steps,\n\\begin{equation}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}C_j(n))\n \\geq k+v_2(k!)+\\sum_{\\nu=1}^k v_2(H_\\nu)+\\beta_j,\n \\quad (\\beta_0,\\beta_1,\\beta_2,\\beta_3)=(0,0,3,1).\n \\label{eq:factorial-four-lower}\n\\end{equation}\nAddress two also satisfies\n\\begin{equation}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}C_2(n))\n \\geq 2k+v_2(k!)+\\left\\lfloor\\frac{k}{2}\\right\\rfloor\n             +\\sum_{\\nu=1}^k v_2(H_\\nu).\n \\label{eq:factorial-four-j2-filtration}\n\\end{equation}\nFor the original coefficients, the offsets in\n\\eqref{eq:factorial-four-lower} are $(0,0,4,2)$.\nNo positive-order offset is asserted for the empty difference.\n\\end{theorem}","_1,\\beta_2,\\beta_3)=(0,0,3,1).\n \\label{eq:factorial-four-lower}\n\\end{equation}\nAddress two also satisfies\n\\begin{equation}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}C_2(n))\n \\geq 2k+v_2(k!)+\\left\\lfloor\\frac{k}{2}\\right\\rfloor\n             +\\sum_{\\nu=1}^k v_2(H_\\nu).\n \\label{eq:factorial-four-j2-filtration}\n\\end{equation}\nFor the original coefficients, the offsets in\n\\eqref{eq:factorial-four-lower} are $(0,0,4,2)$.\nNo positive-order offset is asserted for the empty difference.\n\\end{theorem}","The proof of the first two addresses uses one classical arithmetic\ninput: Frobenius's exact congruence for the ordinary Euler numbers in\nthe $\\operatorname{sech}$ convention, as stated and derived by\nGessel~\\cite{gessel}, Section~9. The address-two and address-three\nbounds below are finite Bernoulli calculations and do not require that\ninput. A divided-power argument is necessary: ordinary integral moments\nalone do not justify division by $k!$.","e classical arithmetic\ninput: Frobenius's exact congruence for the ordinary Euler numbers in\nthe $\\operatorname{sech}$ convention, as stated and derived by\nGessel~\\cite{gessel}, Section~9. The address-two and address-three\nbounds below are finite Bernoulli calculations and do not require that\ninput. A divided-power argument is necessary: ordinary integral moments\nalone do not justify division by $k!$.","Let $\\mathcal L$ be the ordinary Euler evaluation functional on a\nvariable $\\mathcal E$, defined by\n\\[\n \\mathcal L(\\mathcal E^d)=\\operatorname{Euler}_d,\n \\qquad \\sum_{d\\geq0}\\operatorname{Euler}_d\\frac{t^d}{d!}\n                  =\\operatorname{sech}t.\n\\]\nThe functional is linear; it is not a multiplicative evaluation.\nIndependent variables mean that moments of polynomials in separate\nvariables factor. Odd moments vanish. Frobenius's congruence is\n\\begin{equation}\n v_2\\left(\\sum_{r=0}^m(-1)^{m-r}\\binom mr\n                 \\operatorname{Euler}_{2n+2r}\\right)\n       =m+v_2(m!)\\qquad(n,m\\geq0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.","Let $\\mathcal L$ be the ordinary Euler evaluation functional on a\nvariable $\\mathcal E$, defined by\n\\[\n \\mathcal L(\\mathcal E^d)=\\operatorname{Euler}_d,\n \\qquad \\sum_{d\\geq0}\\operatorname{Euler}_d\\frac{t^d}{d!}\n                  =\\operatorname{sech}t.\n\\]\nThe functional is linear; it is not a multiplicative evaluation.\nIndependent variables mean that moments of polynomials in separate\nvariables factor. Odd moments vanish. Frobenius's congruence is\n\\begin{equation}\n v_2\\left(\\sum_{r=0}^m(-1)^{m-r}\\binom mr\n                 \\operatorname{Euler}_{2n+2r}\\right)\n       =m+v_2(m!)\\qquad(n,m\\geq0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.\nWith $P=(\\mathcal E^2-1)/2$ and\n$\\gamma_m(P)=P^m/m!$, the case $n=0$ s","Let $\\mathcal L$ be the ordinary Euler evaluation functional on a\nvariable $\\mathcal E$, defined by\n\\[\n \\mathcal L(\\mathcal E^d)=\\operatorname{Euler}_d,\n \\qquad \\sum_{d\\geq0}\\operatorname{Euler}_d\\frac{t^d}{d!}\n                  =\\operatorname{sech}t.\n\\]\nThe functional is linear; it is not a multiplicative evaluation.\nIndependent variables mean that moments of polynomials in separate\nvariables factor. Odd moments vanish. Frobenius's congruence is\n\\begin{equation}\n v_2\\left(\\sum_{r=0}^m(-1)^{m-r}\\binom mr\n                 \\operatorname{Euler}_{2n+2r}\\right)\n       =m+v_2(m!)\\qquad(n,m\\geq0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.\nWith $P=(\\mathcal E^2-1)/2$ and\n$\\gamma_m(P)=P^m/m!$, the case $n=0$ says\n\\begin{equation}\n \\mathcal L(\\gamma_m(P))\\in\\mathbb Z_2^{\\times}\n                 \\qquad(m\\geq0).\n \\label{eq:euler-pd-odd-moments}\n\\end{equation}\nIntroduce an ind","ariable $\\mathcal E$, defined by\n\\[\n \\mathcal L(\\mathcal E^d)=\\operatorname{Euler}_d,\n \\qquad \\sum_{d\\geq0}\\operatorname{Euler}_d\\frac{t^d}{d!}\n                  =\\operatorname{sech}t.\n\\]\nThe functional is linear; it is not a multiplicative evaluation.\nIndependent variables mean that moments of polynomials in separate\nvariables factor. Odd moments vanish. Frobenius's congruence is\n\\begin{equation}\n v_2\\left(\\sum_{r=0}^m(-1)^{m-r}\\binom mr\n                 \\operatorname{Euler}_{2n+2r}\\right)\n       =m+v_2(m!)\\qquad(n,m\\geq0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.\nWith $P=(\\mathcal E^2-1)/2$ and\n$\\gamma_m(P)=P^m/m!$, the case $n=0$ says\n\\begin{equation}\n \\mathcal L(\\gamma_m(P))\\in\\mathbb Z_2^{\\times}\n                 \\qquad(m\\geq0).\n \\label{eq:euler-pd-odd-moments}\n\\end{equation}\nIntroduce an independent Euler variable $\\mathcal F$ and\n$Q=(\\mathcal F^2-1)/2$.","he functional is linear; it is not a multiplicative evaluation.\nIndependent variables mean that moments of polynomials in separate\nvariables factor. Odd moments vanish. Frobenius's congruence is\n\\begin{equation}\n v_2\\left(\\sum_{r=0}^m(-1)^{m-r}\\binom mr\n                 \\operatorname{Euler}_{2n+2r}\\right)\n       =m+v_2(m!)\\qquad(n,m\\geq0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.\nWith $P=(\\mathcal E^2-1)/2$ and\n$\\gamma_m(P)=P^m/m!$, the case $n=0$ says\n\\begin{equation}\n \\mathcal L(\\gamma_m(P))\\in\\mathbb Z_2^{\\times}\n                 \\qquad(m\\geq0).\n \\label{eq:euler-pd-odd-moments}\n\\end{equation}\nIntroduce an independent Euler variable $\\mathcal F$ and\n$Q=(\\mathcal F^2-1)/2$.","dependent variables mean that moments of polynomials in separate\nvariables factor. Odd moments vanish. Frobenius's congruence is\n\\begin{equation}\n v_2\\left(\\sum_{r=0}^m(-1)^{m-r}\\binom mr\n                 \\operatorname{Euler}_{2n+2r}\\right)\n       =m+v_2(m!)\\qquad(n,m\\geq0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.\nWith $P=(\\mathcal E^2-1)/2$ and\n$\\gamma_m(P)=P^m/m!$, the case $n=0$ says\n\\begin{equation}\n \\mathcal L(\\gamma_m(P))\\in\\mathbb Z_2^{\\times}\n                 \\qquad(m\\geq0).\n \\label{eq:euler-pd-odd-moments}\n\\end{equation}\nIntroduce an independent Euler variable $\\mathcal F$ and\n$Q=(\\mathcal F^2-1)/2$.","at moments of polynomials in separate\nvariables factor. Odd moments vanish. Frobenius's congruence is\n\\begin{equation}\n v_2\\left(\\sum_{r=0}^m(-1)^{m-r}\\binom mr\n                 \\operatorname{Euler}_{2n+2r}\\right)\n       =m+v_2(m!)\\qquad(n,m\\geq0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.\nWith $P=(\\mathcal E^2-1)/2$ and\n$\\gamma_m(P)=P^m/m!$, the case $n=0$ says\n\\begin{equation}\n \\mathcal L(\\gamma_m(P))\\in\\mathbb Z_2^{\\times}\n                 \\qquad(m\\geq0).\n \\label{eq:euler-pd-odd-moments}\n\\end{equation}\nIntroduce an independent Euler variable $\\mathcal F$ and\n$Q=(\\mathcal F^2-1)/2$.","parate\nvariables factor. Odd moments vanish. Frobenius's congruence is\n\\begin{equation}\n v_2\\left(\\sum_{r=0}^m(-1)^{m-r}\\binom mr\n                 \\operatorname{Euler}_{2n+2r}\\right)\n       =m+v_2(m!)\\qquad(n,m\\geq0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.\nWith $P=(\\mathcal E^2-1)/2$ and\n$\\gamma_m(P)=P^m/m!$, the case $n=0$ says\n\\begin{equation}\n \\mathcal L(\\gamma_m(P))\\in\\mathbb Z_2^{\\times}\n                 \\qquad(m\\geq0).\n \\label{eq:euler-pd-odd-moments}\n\\end{equation}\nIntroduce an independent Euler variable $\\mathcal F$ and\n$Q=(\\mathcal F^2-1)/2$.","dd moments vanish. Frobenius's congruence is\n\\begin{equation}\n v_2\\left(\\sum_{r=0}^m(-1)^{m-r}\\binom mr\n                 \\operatorname{Euler}_{2n+2r}\\right)\n       =m+v_2(m!)\\qquad(n,m\\geq0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.\nWith $P=(\\mathcal E^2-1)/2$ and\n$\\gamma_m(P)=P^m/m!$, the case $n=0$ says\n\\begin{equation}\n \\mathcal L(\\gamma_m(P))\\in\\mathbb Z_2^{\\times}\n                 \\qquad(m\\geq0).\n \\label{eq:euler-pd-odd-moments}\n\\end{equation}\nIntroduce an independent Euler variable $\\mathcal F$ and\n$Q=(\\mathcal F^2-1)/2$.","m!)\\qquad(n,m\\geq0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.\nWith $P=(\\mathcal E^2-1)/2$ and\n$\\gamma_m(P)=P^m/m!$, the case $n=0$ says\n\\begin{equation}\n \\mathcal L(\\gamma_m(P))\\in\\mathbb Z_2^{\\times}\n                 \\qquad(m\\geq0).\n \\label{eq:euler-pd-odd-moments}\n\\end{equation}\nIntroduce an independent Euler variable $\\mathcal F$ and\n$Q=(\\mathcal F^2-1)/2$.","0).\n \\label{eq:classical-frobenius-exact}\n\\end{equation}\nFor $m=0$, the assertion is that every even Euler number is odd.\nWith $P=(\\mathcal E^2-1)/2$ and\n$\\gamma_m(P)=P^m/m!$, the case $n=0$ says\n\\begin{equation}\n \\mathcal L(\\gamma_m(P))\\in\\mathbb Z_2^{\\times}\n                 \\qquad(m\\geq0).\n \\label{eq:euler-pd-odd-moments}\n\\end{equation}\nIntroduce an independent Euler variable $\\mathcal F$ and\n$Q=(\\mathcal F^2-1)/2$.","Put $K=\\mathbb Q_2(i)$, $R=\\mathbb Z_2[i]$,\n$a=(1+i)/2$, and $b=(1-i)/2$. The exact identity\n\\begin{equation}\n F_0(t)=\\frac{2}{\\cosh t+\\cos t}\n       =\\operatorname{sech}(at)\\operatorname{sech}(bt)\n \\label{eq:four-euler-factorization}\n\\end{equation}\ngives, for $U=a\\mathcal E+b\\mathcal F$,\n$\\mathcal L(U^{4n})=C_0(n)$. Although $a$ and $b$ have negative\nvaluation, their denominators disappear from the following exact\npolynomial identities:\n\\begin{align}\n U^2&=\\mathcal E\\mathcal F+i(P-Q),\\nonumber",", and $b=(1-i)/2$. The exact identity\n\\begin{equation}\n F_0(t)=\\frac{2}{\\cosh t+\\cos t}\n       =\\operatorname{sech}(at)\\operatorname{sech}(bt)\n \\label{eq:four-euler-factorization}\n\\end{equation}\ngives, for $U=a\\mathcal E+b\\mathcal F$,\n$\\mathcal L(U^{4n})=C_0(n)$. Although $a$ and $b$ have negative\nvaluation, their denominators disappear from the following exact\npolynomial identities:\n\\begin{align}\n U^2&=\\mathcal E\\mathcal F+i(P-Q),\\nonumber\\\\\n H_0:=\\frac{U^4-1}{2}\n   &=P+Q-\\frac{P^2}{2}-\\frac{Q^2}{2}\n                 +3PQ+i\\mathcal E\\mathcal F(P-Q).\n \\label{eq:four-euler-H}\n\\end{align}","Let $\\Gamma_R(P,Q)$ be the $R$-subalgebra generated by all\n$\\gamma_m(P)$ and $\\gamma_r(Q)$. Adjoin $\\mathcal E,\\mathcal F$ with\n\\[\n \\mathcal E^2=1+2P,\\qquad \\mathcal F^2=1+2Q,\n\\]\nand call the resulting algebra $\\mathcal A_0$. Its elements are finite\nexpressions; no completion is being presumed. It is generated as a\n$\\Gamma_R(P,Q)$-module by $1,\\mathcal E,\\mathcal F,\\mathcal E\\mathcal F$.\nThe functional maps this module to $R$. For an even basis term this is\n\\eqref{eq:euler-pd-odd-moments} and independence. For a basis term with\nan odd Euler factor the moment is zero, since $P$ and $Q$ are even\npolynomials in their respective variables.","$R$-subalgebra generated by all\n$\\gamma_m(P)$ and $\\gamma_r(Q)$. Adjoin $\\mathcal E,\\mathcal F$ with\n\\[\n \\mathcal E^2=1+2P,\\qquad \\mathcal F^2=1+2Q,\n\\]\nand call the resulting algebra $\\mathcal A_0$. Its elements are finite\nexpressions; no completion is being presumed. It is generated as a\n$\\Gamma_R(P,Q)$-module by $1,\\mathcal E,\\mathcal F,\\mathcal E\\mathcal F$.\nThe functional maps this module to $R$. For an even basis term this is\n\\eqref{eq:euler-pd-odd-moments} and independence. For a basis term with\nan odd Euler factor the moment is zero, since $P$ and $Q$ are even\npolynomials in their respective variables.","d call the resulting algebra $\\mathcal A_0$. Its elements are finite\nexpressions; no completion is being presumed. It is generated as a\n$\\Gamma_R(P,Q)$-module by $1,\\mathcal E,\\mathcal F,\\mathcal E\\mathcal F$.\nThe functional maps this module to $R$. For an even basis term this is\n\\eqref{eq:euler-pd-odd-moments} and independence. For a basis term with\nan odd Euler factor the moment is zero, since $P$ and $Q$ are even\npolynomials in their respective variables.","he resulting algebra $\\mathcal A_0$. Its elements are finite\nexpressions; no completion is being presumed. It is generated as a\n$\\Gamma_R(P,Q)$-module by $1,\\mathcal E,\\mathcal F,\\mathcal E\\mathcal F$.\nThe functional maps this module to $R$. For an even basis term this is\n\\eqref{eq:euler-pd-odd-moments} and independence. For a basis term with\nan odd Euler factor the moment is zero, since $P$ and $Q$ are even\npolynomials in their respective variables.","Every displayed component of $H_0$ has all its divided powers in\n$\\mathcal A_0$. Besides the generators, the needed identities are\n\\begin{align}\n \\gamma_m(P^2/2)\n     &=\\frac{(2m)!}{2^m m!}\\gamma_{2m}(P),\\nonumber","Every displayed component of $H_0$ has all its divided powers in\n$\\mathcal A_0$. Besides the generators, the needed identities are\n\\begin{align}\n \\gamma_m(P^2/2)\n     &=\\frac{(2m)!}{2^m m!}\\gamma_{2m}(P),\\nonumber\\\\\n \\gamma_m(3PQ)&=3^m m!\\gamma_m(P)\\gamma_m(Q),\\nonumber\\\\\n \\gamma_m(P-Q)&=\\sum_{r=0}^m(-1)^r\n                   \\gamma_{m-r}(P)\\gamma_r(Q).\n \\label{eq:components-divided-powers}\n\\end{align}\nThe first scalar counts partitions into pairs and is an integer.\nFor the last term of $H_0$, multiply the third identity by\n$(i\\mathcal E\\mathcal F)^m$ and reduce the Euler powers using the\nintegral quadratic relations. Finally,\n\\[\n \\gamma_k(X+Y)=\\sum_{r=0}^k\\gamma_r(X)\\gamma_{k-r}(Y)\n\\]\nproves $\\gamma_k(H_0)\\in\\mathcal A_0$. Applying the functional and\nputting $W=U^4-1=2H_0$ gives\n\\begin{equation}\n \\mathcal L(W^m)\\in2^m m!R\\qquad(m\\geq0).\n \\label{eq:euler-W-factorial}\n\\end{equation}\nThese moments are rational. The identity\n$R\\cap\\mathbb Q_2=\\mathbb Z_2$, obtained from the linear independence\nof $1,i$ over $\\mathbb Q_2$, descends every such divisibility statement\nto $\\mathbb Z_2$","\\gamma_m(P-Q)&=\\sum_{r=0}^m(-1)^r\n                   \\gamma_{m-r}(P)\\gamma_r(Q).\n \\label{eq:components-divided-powers}\n\\end{align}\nThe first scalar counts partitions into pairs and is an integer.\nFor the last term of $H_0$, multiply the third identity by\n$(i\\mathcal E\\mathcal F)^m$ and reduce the Euler powers using the\nintegral quadratic relations. Finally,\n\\[\n \\gamma_k(X+Y)=\\sum_{r=0}^k\\gamma_r(X)\\gamma_{k-r}(Y)\n\\]\nproves $\\gamma_k(H_0)\\in\\mathcal A_0$. Applying the functional and\nputting $W=U^4-1=2H_0$ gives\n\\begin{equation}\n \\mathcal L(W^m)\\in2^m m!R\\qquad(m\\geq0).\n \\label{eq:euler-W-factorial}\n\\end{equation}\nThese moments are rational. The identity\n$R\\cap\\mathbb Q_2=\\mathbb Z_2$, obtained from the linear independence\nof $1,i$ over $\\mathbb Q_2$, descends every such divisibility statement\nto $\\mathbb Z_2$. No field property of $R/2R$ is needed.","-1)^r\n                   \\gamma_{m-r}(P)\\gamma_r(Q).\n \\label{eq:components-divided-powers}\n\\end{align}\nThe first scalar counts partitions into pairs and is an integer.\nFor the last term of $H_0$, multiply the third identity by\n$(i\\mathcal E\\mathcal F)^m$ and reduce the Euler powers using the\nintegral quadratic relations. Finally,\n\\[\n \\gamma_k(X+Y)=\\sum_{r=0}^k\\gamma_r(X)\\gamma_{k-r}(Y)\n\\]\nproves $\\gamma_k(H_0)\\in\\mathcal A_0$. Applying the functional and\nputting $W=U^4-1=2H_0$ gives\n\\begin{equation}\n \\mathcal L(W^m)\\in2^m m!R\\qquad(m\\geq0).\n \\label{eq:euler-W-factorial}\n\\end{equation}\nThese moments are rational. The identity\n$R\\cap\\mathbb Q_2=\\mathbb Z_2$, obtained from the linear independence\nof $1,i$ over $\\mathbb Q_2$, descends every such divisibility statement\nto $\\mathbb Z_2$. No field property of $R/2R$ is needed.","onents-divided-powers}\n\\end{align}\nThe first scalar counts partitions into pairs and is an integer.\nFor the last term of $H_0$, multiply the third identity by\n$(i\\mathcal E\\mathcal F)^m$ and reduce the Euler powers using the\nintegral quadratic relations. Finally,\n\\[\n \\gamma_k(X+Y)=\\sum_{r=0}^k\\gamma_r(X)\\gamma_{k-r}(Y)\n\\]\nproves $\\gamma_k(H_0)\\in\\mathcal A_0$. Applying the functional and\nputting $W=U^4-1=2H_0$ gives\n\\begin{equation}\n \\mathcal L(W^m)\\in2^m m!R\\qquad(m\\geq0).\n \\label{eq:euler-W-factorial}\n\\end{equation}\nThese moments are rational. The identity\n$R\\cap\\mathbb Q_2=\\mathbb Z_2$, obtained from the linear independence\nof $1,i$ over $\\mathbb Q_2$, descends every such divisibility statement\nto $\\mathbb Z_2$. No field property of $R/2R$ is needed.","he first scalar counts partitions into pairs and is an integer.\nFor the last term of $H_0$, multiply the third identity by\n$(i\\mathcal E\\mathcal F)^m$ and reduce the Euler powers using the\nintegral quadratic relations. Finally,\n\\[\n \\gamma_k(X+Y)=\\sum_{r=0}^k\\gamma_r(X)\\gamma_{k-r}(Y)\n\\]\nproves $\\gamma_k(H_0)\\in\\mathcal A_0$. Applying the functional and\nputting $W=U^4-1=2H_0$ gives\n\\begin{equation}\n \\mathcal L(W^m)\\in2^m m!R\\qquad(m\\geq0).\n \\label{eq:euler-W-factorial}\n\\end{equation}\nThese moments are rational. The identity\n$R\\cap\\mathbb Q_2=\\mathbb Z_2$, obtained from the linear independence\nof $1,i$ over $\\mathbb Q_2$, descends every such divisibility statement\nto $\\mathbb Z_2$. No field property of $R/2R$ is needed.","hcal E\\mathcal F)^m$ and reduce the Euler powers using the\nintegral quadratic relations. Finally,\n\\[\n \\gamma_k(X+Y)=\\sum_{r=0}^k\\gamma_r(X)\\gamma_{k-r}(Y)\n\\]\nproves $\\gamma_k(H_0)\\in\\mathcal A_0$. Applying the functional and\nputting $W=U^4-1=2H_0$ gives\n\\begin{equation}\n \\mathcal L(W^m)\\in2^m m!R\\qquad(m\\geq0).\n \\label{eq:euler-W-factorial}\n\\end{equation}\nThese moments are rational. The identity\n$R\\cap\\mathbb Q_2=\\mathbb Z_2$, obtained from the linear independence\nof $1,i$ over $\\mathbb Q_2$, descends every such divisibility statement\nto $\\mathbb Z_2$. No field property of $R/2R$ is needed.","atic relations. Finally,\n\\[\n \\gamma_k(X+Y)=\\sum_{r=0}^k\\gamma_r(X)\\gamma_{k-r}(Y)\n\\]\nproves $\\gamma_k(H_0)\\in\\mathcal A_0$. Applying the functional and\nputting $W=U^4-1=2H_0$ gives\n\\begin{equation}\n \\mathcal L(W^m)\\in2^m m!R\\qquad(m\\geq0).\n \\label{eq:euler-W-factorial}\n\\end{equation}\nThese moments are rational. The identity\n$R\\cap\\mathbb Q_2=\\mathbb Z_2$, obtained from the linear independence\nof $1,i$ over $\\mathbb Q_2$, descends every such divisibility statement\nto $\\mathbb Z_2$. No field property of $R/2R$ is needed.","ons. Finally,\n\\[\n \\gamma_k(X+Y)=\\sum_{r=0}^k\\gamma_r(X)\\gamma_{k-r}(Y)\n\\]\nproves $\\gamma_k(H_0)\\in\\mathcal A_0$. Applying the functional and\nputting $W=U^4-1=2H_0$ gives\n\\begin{equation}\n \\mathcal L(W^m)\\in2^m m!R\\qquad(m\\geq0).\n \\label{eq:euler-W-factorial}\n\\end{equation}\nThese moments are rational. The identity\n$R\\cap\\mathbb Q_2=\\mathbb Z_2$, obtained from the linear independence\nof $1,i$ over $\\mathbb Q_2$, descends every such divisibility statement\nto $\\mathbb Z_2$. No field property of $R/2R$ is needed.","_{k-r}(Y)\n\\]\nproves $\\gamma_k(H_0)\\in\\mathcal A_0$. Applying the functional and\nputting $W=U^4-1=2H_0$ gives\n\\begin{equation}\n \\mathcal L(W^m)\\in2^m m!R\\qquad(m\\geq0).\n \\label{eq:euler-W-factorial}\n\\end{equation}\nThese moments are rational. The identity\n$R\\cap\\mathbb Q_2=\\mathbb Z_2$, obtained from the linear independence\nof $1,i$ over $\\mathbb Q_2$, descends every such divisibility statement\nto $\\mathbb Z_2$. No field property of $R/2R$ is needed.","n\\mathcal A_0$. Applying the functional and\nputting $W=U^4-1=2H_0$ gives\n\\begin{equation}\n \\mathcal L(W^m)\\in2^m m!R\\qquad(m\\geq0).\n \\label{eq:euler-W-factorial}\n\\end{equation}\nThese moments are rational. The identity\n$R\\cap\\mathbb Q_2=\\mathbb Z_2$, obtained from the linear independence\nof $1,i$ over $\\mathbb Q_2$, descends every such divisibility statement\nto $\\mathbb Z_2$. No field property of $R/2R$ is needed.","For every starting row,\n\\begin{equation}\n \\Delta_1^k C_0(n)\n   =\\mathcal L(U^{4n}W^k)\n   =\\sum_{s=0}^n\\binom ns\\mathcal L(W^{k+s}).\n \\label{eq:base-row-finite-expansion}\n\\end{equation}\nEach summand lies in $2^k k!\\mathbb Z_2$, proving the factorial\nbound for address zero from row zero onward.","The conversion from unit steps to arbitrary mixed steps is used several\ntimes, so it is useful to make it explicit. Write $\\Delta=\\Delta_1$.\nFor $H>0$,\n\\begin{equation}\n \\Delta_H=\\sum_{m=1}^H\\binom Hm\\Delta^m.\n \\label{eq:finite-step-expansion}\n\\end{equation}\nA mixed term with local orders $m_1,\\ldots,m_k\\geq1$ and\n$M=\\sum_\\nu m_\\nu$ therefore has valuation at least\n\\[\n M+v_2(M!)+\\beta\n    +\\sum_\\nu\\bigl(v_2(H_\\nu)-v_2(m_\\nu)\\bigr)\n\\]\nif the unit-order estimate has offset $\\beta$ for every positive order.\nThe inequalities\n\\[\n m-1\\geq v_2(m),\\qquad M\\geq k,\\qquad v_2(M!)\\geq v_2(k!)\n\\]\nmake this at least $k+v_2(k!)+\\beta+\\sum_\\nu v_2(H_\\nu)$.\nEvery expansion is finite. This proves the desired mixed-step transfer\nwithout an exchange of limits.",", so it is useful to make it explicit. Write $\\Delta=\\Delta_1$.\nFor $H>0$,\n\\begin{equation}\n \\Delta_H=\\sum_{m=1}^H\\binom Hm\\Delta^m.\n \\label{eq:finite-step-expansion}\n\\end{equation}\nA mixed term with local orders $m_1,\\ldots,m_k\\geq1$ and\n$M=\\sum_\\nu m_\\nu$ therefore has valuation at least\n\\[\n M+v_2(M!)+\\beta\n    +\\sum_\\nu\\bigl(v_2(H_\\nu)-v_2(m_\\nu)\\bigr)\n\\]\nif the unit-order estimate has offset $\\beta$ for every positive order.\nThe inequalities\n\\[\n m-1\\geq v_2(m),\\qquad M\\geq k,\\qquad v_2(M!)\\geq v_2(k!)\n\\]\nmake this at least $k+v_2(k!)+\\beta+\\sum_\\nu v_2(H_\\nu)$.\nEvery expansion is finite. This proves the desired mixed-step transfer\nwithout an exchange of limits.","=\\Delta_1$.\nFor $H>0$,\n\\begin{equation}\n \\Delta_H=\\sum_{m=1}^H\\binom Hm\\Delta^m.\n \\label{eq:finite-step-expansion}\n\\end{equation}\nA mixed term with local orders $m_1,\\ldots,m_k\\geq1$ and\n$M=\\sum_\\nu m_\\nu$ therefore has valuation at least\n\\[\n M+v_2(M!)+\\beta\n    +\\sum_\\nu\\bigl(v_2(H_\\nu)-v_2(m_\\nu)\\bigr)\n\\]\nif the unit-order estimate has offset $\\beta$ for every positive order.\nThe inequalities\n\\[\n m-1\\geq v_2(m),\\qquad M\\geq k,\\qquad v_2(M!)\\geq v_2(k!)\n\\]\nmake this at least $k+v_2(k!)+\\beta+\\sum_\\nu v_2(H_\\nu)$.\nEvery expansion is finite. This proves the desired mixed-step transfer\nwithout an exchange of limits.","\\label{eq:finite-step-expansion}\n\\end{equation}\nA mixed term with local orders $m_1,\\ldots,m_k\\geq1$ and\n$M=\\sum_\\nu m_\\nu$ therefore has valuation at least\n\\[\n M+v_2(M!)+\\beta\n    +\\sum_\\nu\\bigl(v_2(H_\\nu)-v_2(m_\\nu)\\bigr)\n\\]\nif the unit-order estimate has offset $\\beta$ for every positive order.\nThe inequalities\n\\[\n m-1\\geq v_2(m),\\qquad M\\geq k,\\qquad v_2(M!)\\geq v_2(k!)\n\\]\nmake this at least $k+v_2(k!)+\\beta+\\sum_\\nu v_2(H_\\nu)$.\nEvery expansion is finite. This proves the desired mixed-step transfer\nwithout an exchange of limits.","Let $B_m$ denote the Bernoulli numbers with $B_1=-1/2$.\nThe identity obtained by substituting $u=e^t-1$ into\n$\\log(1+u)/u$ is\n\\begin{equation}\n B_m=\\sum_{r=0}^m(-1)^r\\frac{r!}{r+1}\n                                  \\left\\{\\begin{matrix}m","Let $B_m$ denote the Bernoulli numbers with $B_1=-1/2$.\nThe identity obtained by substituting $u=e^t-1$ into\n$\\log(1+u)/u$ is\n\\begin{equation}\n B_m=\\sum_{r=0}^m(-1)^r\\frac{r!}{r+1}\n                                  \\left\\{\\begin{matrix}m\\\\r\\end{matrix}\\right\\},\n \\label{eq:bernoulli-stirling-denominator}\n\\end{equation}\nwhere the braces are Stirling numbers of the second kind. Each\n$2r!/(r+1)$ is dyadically integral. Indeed, if $s=v_2(r+1)>0$, then\n$r\\geq2^s-1$ and\n\\[\n v_2(r!)\\geq v_2((2^s-1)!)=2^s-1-s\\geq s-1.\n\\]\nThe last inequality is $2^s\\geq2s$; the case $s=0$ is immediate.\nConsequently\n\\begin{equation}\n 2B_m\\in\\mathbb Z_2\\qquad(m\\geq0).\n \\label{eq:elementary-bernoulli-bound}\n\\end{equation}\nThus the transfer factors below require no appeal to the\nvon Staudt--Clausen theorem.","r\\end{matrix}\\right\\},\n \\label{eq:bernoulli-stirling-denominator}\n\\end{equation}\nwhere the braces are Stirling numbers of the second kind. Each\n$2r!/(r+1)$ is dyadically integral. Indeed, if $s=v_2(r+1)>0$, then\n$r\\geq2^s-1$ and\n\\[\n v_2(r!)\\geq v_2((2^s-1)!)=2^s-1-s\\geq s-1.\n\\]\nThe last inequality is $2^s\\geq2s$; the case $s=0$ is immediate.\nConsequently\n\\begin{equation}\n 2B_m\\in\\mathbb Z_2\\qquad(m\\geq0).\n \\label{eq:elementary-bernoulli-bound}\n\\end{equation}\nThus the transfer factors below require no appeal to the\nvon Staudt--Clausen theorem.","The address-one source is\n$G_{4,1}(t)=(\\sinh t+\\sin t)/(2t)$. The identities $a+b=1$,\n$a-b=i$, and $ab=1/2$ give\n\\begin{equation}\n D_1(t):=G_{4,1}(t)F_0(t)\n =\\frac12\\left(\\frac{\\tanh(at)}{at}\n                         +\\frac{\\tanh(bt)}{bt}\\right).\n \\label{eq:address-one-tangent-factor}\n\\end{equation}\nFor example, its numerator follows from\n\\[\n \\sinh t+\\sin t\n =2b\\sinh(at)\\cosh(bt)+2a\\cosh(at)\\sinh(bt),\n\\]\nwhile $\\cosh t+\\cos t=2\\cosh(at)\\cosh(bt)$.\nOnly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1","The address-one source is\n$G_{4,1}(t)=(\\sinh t+\\sin t)/(2t)$. The identities $a+b=1$,\n$a-b=i$, and $ab=1/2$ give\n\\begin{equation}\n D_1(t):=G_{4,1}(t)F_0(t)\n =\\frac12\\left(\\frac{\\tanh(at)}{at}\n                         +\\frac{\\tanh(bt)}{bt}\\right).\n \\label{eq:address-one-tangent-factor}\n\\end{equation}\nFor example, its numerator follows from\n\\[\n \\sinh t+\\sin t\n =2b\\sinh(at)\\cosh(bt)+2a\\cosh(at)\\sinh(bt),\n\\]\nwhile $\\cosh t+\\cos t=2\\cosh(at)\\cosh(bt)$.\nOnly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernou","The address-one source is\n$G_{4,1}(t)=(\\sinh t+\\sin t)/(2t)$. The identities $a+b=1$,\n$a-b=i$, and $ab=1/2$ give\n\\begin{equation}\n D_1(t):=G_{4,1}(t)F_0(t)\n =\\frac12\\left(\\frac{\\tanh(at)}{at}\n                         +\\frac{\\tanh(bt)}{bt}\\right).\n \\label{eq:address-one-tangent-factor}\n\\end{equation}\nFor example, its numerator follows from\n\\[\n \\sinh t+\\sin t\n =2b\\sinh(at)\\cosh(bt)+2a\\cosh(at)\\sinh(bt),\n\\]\nwhile $\\cosh t+\\cos t=2\\cosh(at)\\cosh(bt)$.\nOnly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound","The address-one source is\n$G_{4,1}(t)=(\\sinh t+\\sin t)/(2t)$. The identities $a+b=1$,\n$a-b=i$, and $ab=1/2$ give\n\\begin{equation}\n D_1(t):=G_{4,1}(t)F_0(t)\n =\\frac12\\left(\\frac{\\tanh(at)}{at}\n                         +\\frac{\\tanh(bt)}{bt}\\right).\n \\label{eq:address-one-tangent-factor}\n\\end{equation}\nFor example, its numerator follows from\n\\[\n \\sinh t+\\sin t\n =2b\\sinh(at)\\cosh(bt)+2a\\cosh(at)\\sinh(bt),\n\\]\nwhile $\\cosh t+\\cos t=2\\cosh(at)\\cosh(bt)$.\nOnly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_","The address-one source is\n$G_{4,1}(t)=(\\sinh t+\\sin t)/(2t)$. The identities $a+b=1$,\n$a-b=i$, and $ab=1/2$ give\n\\begin{equation}\n D_1(t):=G_{4,1}(t)F_0(t)\n =\\frac12\\left(\\frac{\\tanh(at)}{at}\n                         +\\frac{\\tanh(bt)}{bt}\\right).\n \\label{eq:address-one-tangent-factor}\n\\end{equation}\nFor example, its numerator follows from\n\\[\n \\sinh t+\\sin t\n =2b\\sinh(at)\\cosh(bt)+2a\\cosh(at)\\sinh(bt),\n\\]\nwhile $\\cosh t+\\cos t=2\\cosh(at)\\cosh(bt)$.\nOnly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse","The address-one source is\n$G_{4,1}(t)=(\\sinh t+\\sin t)/(2t)$. The identities $a+b=1$,\n$a-b=i$, and $ab=1/2$ give\n\\begin{equation}\n D_1(t):=G_{4,1}(t)F_0(t)\n =\\frac12\\left(\\frac{\\tanh(at)}{at}\n                         +\\frac{\\tanh(bt)}{bt}\\right).\n \\label{eq:address-one-tangent-factor}\n\\end{equation}\nFor example, its numerator follows from\n\\[\n \\sinh t+\\sin t\n =2b\\sinh(at)\\cosh(bt)+2a\\cosh(at)\\sinh(bt),\n\\]\nwhile $\\cosh t+\\cos t=2\\cosh(at)\\cosh(bt)$.\nOnly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multip…","source is\n$G_{4,1}(t)=(\\sinh t+\\sin t)/(2t)$. The identities $a+b=1$,\n$a-b=i$, and $ab=1/2$ give\n\\begin{equation}\n D_1(t):=G_{4,1}(t)F_0(t)\n =\\frac12\\left(\\frac{\\tanh(at)}{at}\n                         +\\frac{\\tanh(bt)}{bt}\\right).\n \\label{eq:address-one-tangent-factor}\n\\end{equation}\nFor example, its numerator follows from\n\\[\n \\sinh t+\\sin t\n =2b\\sinh(at)\\cosh(bt)+2a\\cosh(at)\\sinh(bt),\n\\]\nwhile $\\cosh t+\\cos t=2\\cosh(at)\\cosh(bt)$.\nOnly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral…","G_{4,1}(t)F_0(t)\n =\\frac12\\left(\\frac{\\tanh(at)}{at}\n                         +\\frac{\\tanh(bt)}{bt}\\right).\n \\label{eq:address-one-tangent-factor}\n\\end{equation}\nFor example, its numerator follows from\n\\[\n \\sinh t+\\sin t\n =2b\\sinh(at)\\cosh(bt)+2a\\cosh(at)\\sinh(bt),\n\\]\nwhile $\\cosh t+\\cos t=2\\cosh(at)\\cosh(bt)$.\nOnly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:addre…","nh t+\\sin t\n =2b\\sinh(at)\\cosh(bt)+2a\\cosh(at)\\sinh(bt),\n\\]\nwhile $\\cosh t+\\cos t=2\\cosh(at)\\cosh(bt)$.\nOnly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:address-one-exact-transfer}\n\\end{equation}\nAt a natural $n$ this is a finite sum, with $i\\leq n$.","bt),\n\\]\nwhile $\\cosh t+\\cos t=2\\cosh(at)\\cosh(bt)$.\nOnly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:address-one-exact-transfer}\n\\end{equation}\nAt a natural $n$ this is a finite sum, with $i\\leq n$.","nly degrees divisible by four survive. Its actual Hurwitz packets are\n\\begin{equation}\n D_1(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=\\left(-\\frac14\\right)^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:address-one-exact-transfer}\n\\end{equation}\nAt a natural $n$ this is a finite sum, with $i\\leq n$.",")^i\n \\frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:address-one-exact-transfer}\n\\end{equation}\nAt a natural $n$ this is a finite sum, with $i\\leq n$.","(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:address-one-exact-transfer}\n\\end{equation}\nAt a natural $n$ this is a finite sum, with $i\\leq n$.","\\label{eq:address-one-packets}\n\\end{equation}\nThe constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:address-one-exact-transfer}\n\\end{equation}\nAt a natural $n$ this is a finite sum, with $i\\leq n$.","stant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:address-one-exact-transfer}\n\\end{equation}\nAt a natural $n$ this is a finite sum, with $i\\leq n$.","the elementary Bernoulli bound\nshows $v_2(d_i)\\geq2i$. Hurwitz inversion therefore gives\n\\[\n D_1(t)^{-1}=\\sum_{i\\geq0}h_i\\frac{t^{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:address-one-exact-transfer}\n\\end{equation}\nAt a natural $n$ this is a finite sum, with $i\\leq n$.","{4i}}{(4i)!},\n \\qquad h_0=1,\\quad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:address-one-exact-transfer}\n\\end{equation}\nAt a natural $n$ this is a finite sum, with $i\\leq n$.","ad v_2(h_i)\\geq2i.\n\\]\nThis follows by induction from the inverse recurrence: for packet index\n$i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its\nbinomial multiplier is integral.\nBecause $F_1=F_0D_1^{-1}$,\n\\begin{equation}\n C_1(n)-C_0(n)\n     =\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_0(n-i).\n \\label{eq:address-one-exact-transfer}\n\\end{equation}\nAt a natural $n$ this is a finite sum, with $i\\leq n$.","Here is the polynomial estimate that supplies one additional bit.\nFor $i\\geq1$, put $P_i(x)=4^i\\binom{4x}{4i}$. Its constant is zero,\nand its degree-$d$ coefficient is\n\\begin{equation}\n [x^d]P_i(x)=\\frac{4^i4^d s(4i,d)}{(4i)!},\n \\label{eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nF","Here is the polynomial estimate that supplies one additional bit.\nFor $i\\geq1$, put $P_i(x)=4^i\\binom{4x}{4i}$. Its constant is zero,\nand its degree-$d$ coefficient is\n\\begin{equation}\n [x^d]P_i(x)=\\frac{4^i4^d s(4i,d)}{(4i)!},\n \\label{eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ va","Here is the polynomial estimate that supplies one additional bit.\nFor $i\\geq1$, put $P_i(x)=4^i\\binom{4x}{4i}$. Its constant is zero,\nand its degree-$d$ coefficient is\n\\begin{equation}\n [x^d]P_i(x)=\\frac{4^i4^d s(4i,d)}{(4i)!},\n \\label{eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polyno","Here is the polynomial estimate that supplies one additional bit.\nFor $i\\geq1$, put $P_i(x)=4^i\\binom{4x}{4i}$. Its constant is zero,\nand its degree-$d$ coefficient is\n\\begin{equation}\n [x^d]P_i(x)=\\frac{4^i4^d s(4i,d)}{(4i)!},\n \\label{eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","polynomial estimate that supplies one additional bit.\nFor $i\\geq1$, put $P_i(x)=4^i\\binom{4x}{4i}$. Its constant is zero,\nand its degree-$d$ coefficient is\n\\begin{equation}\n [x^d]P_i(x)=\\frac{4^i4^d s(4i,d)}{(4i)!},\n \\label{eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","t supplies one additional bit.\nFor $i\\geq1$, put $P_i(x)=4^i\\binom{4x}{4i}$. Its constant is zero,\nand its degree-$d$ coefficient is\n\\begin{equation}\n [x^d]P_i(x)=\\frac{4^i4^d s(4i,d)}{(4i)!},\n \\label{eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","geq1$, put $P_i(x)=4^i\\binom{4x}{4i}$. Its constant is zero,\nand its degree-$d$ coefficient is\n\\begin{equation}\n [x^d]P_i(x)=\\frac{4^i4^d s(4i,d)}{(4i)!},\n \\label{eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","onstant is zero,\nand its degree-$d$ coefficient is\n\\begin{equation}\n [x^d]P_i(x)=\\frac{4^i4^d s(4i,d)}{(4i)!},\n \\label{eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","quation}\n [x^d]P_i(x)=\\frac{4^i4^d s(4i,d)}{(4i)!},\n \\label{eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","P_i(x)=\\frac{4^i4^d s(4i,d)}{(4i)!},\n \\label{eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","eq:scaled-binomial-coefficients}\n\\end{equation}\nwhere $s(m,d)$ is a signed Stirling number of the first kind.\nIn each product contributing to $s(4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","4i,d)$, at most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","most $d-1$ of the\n$2i-1$ even integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","integers in $1,\\ldots,4i-1$ can be omitted. Thus\n\\[\n v_2(s(4i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","i,d))\\geq\\max(0,2i-d).\n\\]\nWriting $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d","g $s_2(i)$ for the number of ones in the binary expansion of $i$,\nand using $v_2((4i)!)=4i-s_2(i)$, gives\n\\begin{equation}\n v_2([x^d]P_i)\n \\geq2d-2i+s_2(i)+\\max(0,2i-d)\n \\geq d+s_2(i)\\geq d+1\\quad(d\\geq1).\n \\label{eq:scaled-binomial-degree-bound}\n\\end{equation}\nFor $k\\geq1$, $\\Delta_1^k x^d$ vanishes when $d<k$ and is an\ninteger polynomial with values divisible by $k!$. To see the latter,\nexpand $x^d=\\sum_r\\left\\{\\begin{smallmatrix}d\\\\r\\end{smallmatrix}\\right\\}\nr!\\binom xr$ and apply\n$\\Delta_1^k\\binom xr=\\binom{x}{r-k}$.\nIt follows that\n\\begin{equation}\n \\Delta_1^kP_i(x)\\in2^{k+1}k!\\mathbb Z_2\n                      \\qquad(k\\geq0,\\ x\\in\\mathbb Z_2).\n \\label{eq:scaled-binomial-difference-bound}\n\\end{equation}\nFor $k=0$ the bound follows directly from the coefficient estimate.","r\\end{smallmatrix}\\right\\}\nr!\\binom xr$ and apply\n$\\Delta_1^k\\binom xr=\\binom{x}{r-k}$.\nIt follows that\n\\begin{equation}\n \\Delta_1^kP_i(x)\\in2^{k+1}k!\\mathbb Z_2\n                      \\qquad(k\\geq0,\\ x\\in\\mathbb Z_2).\n \\label{eq:scaled-binomial-difference-bound}\n\\end{equation}\nFor $k=0$ the bound follows directly from the coefficient estimate.","To apply discrete products in \\eqref{eq:address-one-exact-transfer}\nwithout a moving truncation, first extend $C_0$ to $\\mathbb Z_2$ by its\nuniformly convergent Newton series. The all-row bound\n$v_2(\\Delta_1^m C_0(0))\\geq m$ proves this convergence. Its factorial\nbounds extend to every dyadic starting point by continuity and\napproximation with natural rows. The sum in\n\\eqref{eq:address-one-exact-transfer} is then uniformly convergent on\n$\\mathbb Z_2$: both the binomial polynomial and $C_0(x-i)$ are\nintegral-valued and $v_2(h_i)\\geq2i$.","dress-one-exact-transfer}\nwithout a moving truncation, first extend $C_0$ to $\\mathbb Z_2$ by its\nuniformly convergent Newton series. The all-row bound\n$v_2(\\Delta_1^m C_0(0))\\geq m$ proves this convergence. Its factorial\nbounds extend to every dyadic starting point by continuity and\napproximation with natural rows. The sum in\n\\eqref{eq:address-one-exact-transfer} is then uniformly convergent on\n$\\mathbb Z_2$: both the binomial polynomial and $C_0(x-i)$ are\nintegral-valued and $v_2(h_i)\\geq2i$.","ation, first extend $C_0$ to $\\mathbb Z_2$ by its\nuniformly convergent Newton series. The all-row bound\n$v_2(\\Delta_1^m C_0(0))\\geq m$ proves this convergence. Its factorial\nbounds extend to every dyadic starting point by continuity and\napproximation with natural rows. The sum in\n\\eqref{eq:address-one-exact-transfer} is then uniformly convergent on\n$\\mathbb Z_2$: both the binomial polynomial and $C_0(x-i)$ are\nintegral-valued and $v_2(h_i)\\geq2i$.","bb Z_2$ by its\nuniformly convergent Newton series. The all-row bound\n$v_2(\\Delta_1^m C_0(0))\\geq m$ proves this convergence. Its factorial\nbounds extend to every dyadic starting point by continuity and\napproximation with natural rows. The sum in\n\\eqref{eq:address-one-exact-transfer} is then uniformly convergent on\n$\\mathbb Z_2$: both the binomial polynomial and $C_0(x-i)$ are\nintegral-valued and $v_2(h_i)\\geq2i$.","For the pointwise product $P_i(x)C_0(x-i)$, the discrete Leibniz formula\nhas scalar divisibility\n\\[\n \\binom kr(2^{r+1}r!)(2^{k-r}(k-r)!)=2^{k+1}k!.\n\\]\nMultiplication by $h_i/4^i\\in\\mathbb Z_2$ and uniform summation now give\n\\begin{equation}\n v_2(\\Delta_1^k(C_1-C_0)(n))\n               \\geq k+v_2(k!)+1\\qquad(k,n\\geq0).\n \\label{eq:address-one-extra-bit}\n\\end{equation}\nThe same extra bit holds for positive mixed steps by\n\\eqref{eq:finite-step-expansion}. Combining this with the address-zero\nbound proves Theorem~\\ref{thm:four-factorial-bound} for address one.","\\begin{lemma}[Bernoulli divided-power moments]\n\\label{lem:bernoulli-central-pd}\nLet $Z$ be the ordinary Bernoulli variable,\n$\\mathcal L(Z^r)=B_r$, with exponential transform $t/(e^t-1)$.\nFor $P=Z(Z+1)$ and $\\mu_m=\\mathcal L(P^m/m!)$,\n\\begin{equation}\n \\mu_m\\in\\mathbb Z_2,\\qquad\n v_2(\\mu_m)\\geq s_2(m)-1\\quad(m\\geq1).\n \\label{eq:bernoulli-central-moment-bound}\n\\end{equation}\nMoreover, $\\mu_m$ is odd exactly when $m=0$ or $m$ is a positive power\nof two.\n\\end{lemma}","ernoulli divided-power moments]\n\\label{lem:bernoulli-central-pd}\nLet $Z$ be the ordinary Bernoulli variable,\n$\\mathcal L(Z^r)=B_r$, with exponential transform $t/(e^t-1)$.\nFor $P=Z(Z+1)$ and $\\mu_m=\\mathcal L(P^m/m!)$,\n\\begin{equation}\n \\mu_m\\in\\mathbb Z_2,\\qquad\n v_2(\\mu_m)\\geq s_2(m)-1\\quad(m\\geq1).\n \\label{eq:bernoulli-central-moment-bound}\n\\end{equation}\nMoreover, $\\mu_m$ is odd exactly when $m=0$ or $m$ is a positive power\nof two.\n\\end{lemma}","divided-power moments]\n\\label{lem:bernoulli-central-pd}\nLet $Z$ be the ordinary Bernoulli variable,\n$\\mathcal L(Z^r)=B_r$, with exponential transform $t/(e^t-1)$.\nFor $P=Z(Z+1)$ and $\\mu_m=\\mathcal L(P^m/m!)$,\n\\begin{equation}\n \\mu_m\\in\\mathbb Z_2,\\qquad\n v_2(\\mu_m)\\geq s_2(m)-1\\quad(m\\geq1).\n \\label{eq:bernoulli-central-moment-bound}\n\\end{equation}\nMoreover, $\\mu_m$ is odd exactly when $m=0$ or $m$ is a positive power\nof two.\n\\end{lemma}","\\begin{proof}\nDefine the central-factorial basis polynomials\n\\[\n P_r(X)=\\prod_{a=0}^{r-1}(X-a(a+1)).\n\\]\nAt $X=Z(Z+1)$, they have the exact factorization\n\\begin{equation}\n P_r(Z(Z+1))\n   =\\prod_{a=0}^{r-1}(Z-a)(Z+a+1)\n   =(2r)!\\binom{Z+r}{2r}.\n \\label{eq:central-factorial-substitution}\n\\end{equation}\nThe Bernoulli transform gives\n$\\mathcal L((1+T)^Z)=\\log(1+T)/T$, whence\n$\\mathcal L(\\binom Zj)=(-1)^j/(j+1)$. Finite Vandermonde expansion\ntherefore gives\n\\begin{align}\n \\mathcal L\\left(\\binom{Z+r}{2r}\\right)\n &=(-1)^r\\sum_{q=0}^r(-1)^q\\binom rq\\frac1{r+q+1}\\nonumber","X)=\\prod_{a=0}^{r-1}(X-a(a+1)).\n\\]\nAt $X=Z(Z+1)$, they have the exact factorization\n\\begin{equation}\n P_r(Z(Z+1))\n   =\\prod_{a=0}^{r-1}(Z-a)(Z+a+1)\n   =(2r)!\\binom{Z+r}{2r}.\n \\label{eq:central-factorial-substitution}\n\\end{equation}\nThe Bernoulli transform gives\n$\\mathcal L((1+T)^Z)=\\log(1+T)/T$, whence\n$\\mathcal L(\\binom Zj)=(-1)^j/(j+1)$. Finite Vandermonde expansion\ntherefore gives\n\\begin{align}\n \\mathcal L\\left(\\binom{Z+r}{2r}\\right)\n &=(-1)^r\\sum_{q=0}^r(-1)^q\\binom rq\\frac1{r+q+1}\\nonumber\\\\\n &=(-1)^r\\frac{(r!)^2}{(2r+1)!}.\n \\label{eq:central-factorial-evaluation}\n\\end{align}\nThe last equality is the rational identity obtained by expanding\n$\\int_0^1t^r(1-t)^r\\,dt$. This use of an ordinary finite integral makes\nno assertion about a positive or bounded $p$-adic measure. In\nparticular,\n\\[\n \\mathcal L(P_r(P))=(-1)^r\\frac{(r!)^2}{2r+1}.\n\\]","&=(-1)^r\\frac{(r!)^2}{(2r+1)!}.\n \\label{eq:central-factorial-evaluation}\n\\end{align}\nThe last equality is the rational identity obtained by expanding\n$\\int_0^1t^r(1-t)^r\\,dt$. This use of an ordinary finite integral makes\nno assertion about a positive or bounded $p$-adic measure. In\nparticular,\n\\[\n \\mathcal L(P_r(P))=(-1)^r\\frac{(r!)^2}{2r+1}.\n\\]","Expand $X^m=\\sum_{r=0}^mT(m,r)P_r(X)$. Since\n$XP_r(X)=P_{r+1}(X)+r(r+1)P_r(X)$, these integer coefficients satisfy\n\\begin{equation}\n T(m+1,r)=T(m,r-1)+r(r+1)T(m,r),\n \\quad T(0,0)=1.\n \\label{eq:central-factorial-recurrence}\n\\end{equation}\nEvery $r(r+1)$ is even. Induction, including the boundary indices,\nproves $2^{m-r}\\mid T(m,r)$ for $0\\leq r\\leq m$.\nFor $m>0$ the $r=0$ term vanishes. Each remaining summand of\n$\\mathcal L(P^m/m!)$ has valuation at least\n\\begin{equation}\n m-r+2v_2(r!)-v_2(m!)\n   =r+s_2(m)-2s_2(r)\\geq s_2(m)-1\\geq0.\n \\label{eq:central-factorial-term-bound}\n\\end{equation}\nHere $r\\geq2s_2(r)-1$ follows from\n$r\\geq2^{s_2(r)}-1$ and $2^u\\geq2u$ for $u\\geq1$.\nThis proves integrality and the stronger bound.","_{r=0}^mT(m,r)P_r(X)$. Since\n$XP_r(X)=P_{r+1}(X)+r(r+1)P_r(X)$, these integer coefficients satisfy\n\\begin{equation}\n T(m+1,r)=T(m,r-1)+r(r+1)T(m,r),\n \\quad T(0,0)=1.\n \\label{eq:central-factorial-recurrence}\n\\end{equation}\nEvery $r(r+1)$ is even. Induction, including the boundary indices,\nproves $2^{m-r}\\mid T(m,r)$ for $0\\leq r\\leq m$.\nFor $m>0$ the $r=0$ term vanishes. Each remaining summand of\n$\\mathcal L(P^m/m!)$ has valuation at least\n\\begin{equation}\n m-r+2v_2(r!)-v_2(m!)\n   =r+s_2(m)-2s_2(r)\\geq s_2(m)-1\\geq0.\n \\label{eq:central-factorial-term-bound}\n\\end{equation}\nHere $r\\geq2s_2(r)-1$ follows from\n$r\\geq2^{s_2(r)}-1$ and $2^u\\geq2u$ for $u\\geq1$.\nThis proves integrality and the stronger bound.","X)$, these integer coefficients satisfy\n\\begin{equation}\n T(m+1,r)=T(m,r-1)+r(r+1)T(m,r),\n \\quad T(0,0)=1.\n \\label{eq:central-factorial-recurrence}\n\\end{equation}\nEvery $r(r+1)$ is even. Induction, including the boundary indices,\nproves $2^{m-r}\\mid T(m,r)$ for $0\\leq r\\leq m$.\nFor $m>0$ the $r=0$ term vanishes. Each remaining summand of\n$\\mathcal L(P^m/m!)$ has valuation at least\n\\begin{equation}\n m-r+2v_2(r!)-v_2(m!)\n   =r+s_2(m)-2s_2(r)\\geq s_2(m)-1\\geq0.\n \\label{eq:central-factorial-term-bound}\n\\end{equation}\nHere $r\\geq2s_2(r)-1$ follows from\n$r\\geq2^{s_2(r)}-1$ and $2^u\\geq2u$ for $u\\geq1$.\nThis proves integrality and the stronger bound.","-factorial-recurrence}\n\\end{equation}\nEvery $r(r+1)$ is even. Induction, including the boundary indices,\nproves $2^{m-r}\\mid T(m,r)$ for $0\\leq r\\leq m$.\nFor $m>0$ the $r=0$ term vanishes. Each remaining summand of\n$\\mathcal L(P^m/m!)$ has valuation at least\n\\begin{equation}\n m-r+2v_2(r!)-v_2(m!)\n   =r+s_2(m)-2s_2(r)\\geq s_2(m)-1\\geq0.\n \\label{eq:central-factorial-term-bound}\n\\end{equation}\nHere $r\\geq2s_2(r)-1$ follows from\n$r\\geq2^{s_2(r)}-1$ and $2^u\\geq2u$ for $u\\geq1$.\nThis proves integrality and the stronger bound.","uation}\nEvery $r(r+1)$ is even. Induction, including the boundary indices,\nproves $2^{m-r}\\mid T(m,r)$ for $0\\leq r\\leq m$.\nFor $m>0$ the $r=0$ term vanishes. Each remaining summand of\n$\\mathcal L(P^m/m!)$ has valuation at least\n\\begin{equation}\n m-r+2v_2(r!)-v_2(m!)\n   =r+s_2(m)-2s_2(r)\\geq s_2(m)-1\\geq0.\n \\label{eq:central-factorial-term-bound}\n\\end{equation}\nHere $r\\geq2s_2(r)-1$ follows from\n$r\\geq2^{s_2(r)}-1$ and $2^u\\geq2u$ for $u\\geq1$.\nThis proves integrality and the stronger bound.","s even. Induction, including the boundary indices,\nproves $2^{m-r}\\mid T(m,r)$ for $0\\leq r\\leq m$.\nFor $m>0$ the $r=0$ term vanishes. Each remaining summand of\n$\\mathcal L(P^m/m!)$ has valuation at least\n\\begin{equation}\n m-r+2v_2(r!)-v_2(m!)\n   =r+s_2(m)-2s_2(r)\\geq s_2(m)-1\\geq0.\n \\label{eq:central-factorial-term-bound}\n\\end{equation}\nHere $r\\geq2s_2(r)-1$ follows from\n$r\\geq2^{s_2(r)}-1$ and $2^u\\geq2u$ for $u\\geq1$.\nThis proves integrality and the stronger bound.","n, including the boundary indices,\nproves $2^{m-r}\\mid T(m,r)$ for $0\\leq r\\leq m$.\nFor $m>0$ the $r=0$ term vanishes. Each remaining summand of\n$\\mathcal L(P^m/m!)$ has valuation at least\n\\begin{equation}\n m-r+2v_2(r!)-v_2(m!)\n   =r+s_2(m)-2s_2(r)\\geq s_2(m)-1\\geq0.\n \\label{eq:central-factorial-term-bound}\n\\end{equation}\nHere $r\\geq2s_2(r)-1$ follows from\n$r\\geq2^{s_2(r)}-1$ and $2^u\\geq2u$ for $u\\geq1$.\nThis proves integrality and the stronger bound.","If $m$ is not a power of two, $s_2(m)\\geq2$ makes $\\mu_m$ even.\nIf $m$ is a positive power of two, the only central indices which can\ncontribute an odd residue are $r=1$ and $r=3$, since\n$r-2s_2(r)=-1$ only at these two positive integers. The $r=1$ term is\nodd: $T(m,1)=2^{m-1}$ and $v_2(m!)=m-1$. For $m\\geq4$ a power of\ntwo, the $r=3$ term is even. More explicitly,\n\\[\n \\frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),\n \\qquad\n \\sum_{d\\geq0}h_d(1,3,6)t^d\n       =\\frac1{(1-t)(1-3t)(1-6t)}.\n\\]\nReduction modulo two gives $(1-t)^{-2}$, whose degree $m-3$\ncoefficient has parity $m-2=0$. At $m=1,2$ the $r=3$ term is absent,\nand the $r=1$ term is again odd. Finally $\\mu_0=1$.\n\\end{proof}","al indices which can\ncontribute an odd residue are $r=1$ and $r=3$, since\n$r-2s_2(r)=-1$ only at these two positive integers. The $r=1$ term is\nodd: $T(m,1)=2^{m-1}$ and $v_2(m!)=m-1$. For $m\\geq4$ a power of\ntwo, the $r=3$ term is even. More explicitly,\n\\[\n \\frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),\n \\qquad\n \\sum_{d\\geq0}h_d(1,3,6)t^d\n       =\\frac1{(1-t)(1-3t)(1-6t)}.\n\\]\nReduction modulo two gives $(1-t)^{-2}$, whose degree $m-3$\ncoefficient has parity $m-2=0$. At $m=1,2$ the $r=3$ term is absent,\nand the $r=1$ term is again odd. Finally $\\mu_0=1$.\n\\end{proof}","bute an odd residue are $r=1$ and $r=3$, since\n$r-2s_2(r)=-1$ only at these two positive integers. The $r=1$ term is\nodd: $T(m,1)=2^{m-1}$ and $v_2(m!)=m-1$. For $m\\geq4$ a power of\ntwo, the $r=3$ term is even. More explicitly,\n\\[\n \\frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),\n \\qquad\n \\sum_{d\\geq0}h_d(1,3,6)t^d\n       =\\frac1{(1-t)(1-3t)(1-6t)}.\n\\]\nReduction modulo two gives $(1-t)^{-2}$, whose degree $m-3$\ncoefficient has parity $m-2=0$. At $m=1,2$ the $r=3$ term is absent,\nand the $r=1$ term is again odd. Finally $\\mu_0=1$.\n\\end{proof}","and $r=3$, since\n$r-2s_2(r)=-1$ only at these two positive integers. The $r=1$ term is\nodd: $T(m,1)=2^{m-1}$ and $v_2(m!)=m-1$. For $m\\geq4$ a power of\ntwo, the $r=3$ term is even. More explicitly,\n\\[\n \\frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),\n \\qquad\n \\sum_{d\\geq0}h_d(1,3,6)t^d\n       =\\frac1{(1-t)(1-3t)(1-6t)}.\n\\]\nReduction modulo two gives $(1-t)^{-2}$, whose degree $m-3$\ncoefficient has parity $m-2=0$. At $m=1,2$ the $r=3$ term is absent,\nand the $r=1$ term is again odd. Finally $\\mu_0=1$.\n\\end{proof}","since\n$r-2s_2(r)=-1$ only at these two positive integers. The $r=1$ term is\nodd: $T(m,1)=2^{m-1}$ and $v_2(m!)=m-1$. For $m\\geq4$ a power of\ntwo, the $r=3$ term is even. More explicitly,\n\\[\n \\frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),\n \\qquad\n \\sum_{d\\geq0}h_d(1,3,6)t^d\n       =\\frac1{(1-t)(1-3t)(1-6t)}.\n\\]\nReduction modulo two gives $(1-t)^{-2}$, whose degree $m-3$\ncoefficient has parity $m-2=0$. At $m=1,2$ the $r=3$ term is absent,\nand the $r=1$ term is again odd. Finally $\\mu_0=1$.\n\\end{proof}","2(r)=-1$ only at these two positive integers. The $r=1$ term is\nodd: $T(m,1)=2^{m-1}$ and $v_2(m!)=m-1$. For $m\\geq4$ a power of\ntwo, the $r=3$ term is even. More explicitly,\n\\[\n \\frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),\n \\qquad\n \\sum_{d\\geq0}h_d(1,3,6)t^d\n       =\\frac1{(1-t)(1-3t)(1-6t)}.\n\\]\nReduction modulo two gives $(1-t)^{-2}$, whose degree $m-3$\ncoefficient has parity $m-2=0$. At $m=1,2$ the $r=3$ term is absent,\nand the $r=1$ term is again odd. Finally $\\mu_0=1$.\n\\end{proof}","itive integers. The $r=1$ term is\nodd: $T(m,1)=2^{m-1}$ and $v_2(m!)=m-1$. For $m\\geq4$ a power of\ntwo, the $r=3$ term is even. More explicitly,\n\\[\n \\frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),\n \\qquad\n \\sum_{d\\geq0}h_d(1,3,6)t^d\n       =\\frac1{(1-t)(1-3t)(1-6t)}.\n\\]\nReduction modulo two gives $(1-t)^{-2}$, whose degree $m-3$\ncoefficient has parity $m-2=0$. At $m=1,2$ the $r=3$ term is absent,\nand the $r=1$ term is again odd. Finally $\\mu_0=1$.\n\\end{proof}","odd: $T(m,1)=2^{m-1}$ and $v_2(m!)=m-1$. For $m\\geq4$ a power of\ntwo, the $r=3$ term is even. More explicitly,\n\\[\n \\frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),\n \\qquad\n \\sum_{d\\geq0}h_d(1,3,6)t^d\n       =\\frac1{(1-t)(1-3t)(1-6t)}.\n\\]\nReduction modulo two gives $(1-t)^{-2}$, whose degree $m-3$\ncoefficient has parity $m-2=0$. At $m=1,2$ the $r=3$ term is absent,\nand the $r=1$ term is again odd. Finally $\\mu_0=1$.\n\\end{proof}","The variable $\\mathcal B=2Z+1$ has transform\n\\[\n e^t\\frac{2t}{e^{2t}-1}=\\frac{t}{\\sinh t}.\n\\]\nIts odd moments vanish, and\n$P=(\\mathcal B^2-1)/4=Z(Z+1)$ has the divided-power moments of\nLemma~\\ref{lem:bernoulli-central-pd}. Introduce an independent copy\n$\\mathcal D$ and $Q=(\\mathcal D^2-1)/4$. The exact factorization\n\\begin{equation}\n F_2(t)=\\frac{t^2}{\\cosh t-\\cos t}\n       =\\frac{at}{\\sinh(at)}\\frac{bt}{\\sinh(bt)}\n \\label{eq:address-two-factorization}\n\\end{equation}\nshows that $U=a\\mathcal B+b\\mathcal D$ satisfies\n$\\mathcal L(U^{4n})=C_2(n)$. Direct expansion gives\n\\begin{equation}\n H_2:=\\frac{U^4-1}{4}\n     =P+Q-P^2-Q^2+6PQ+i\\mathcal B\\mathcal D(P-Q).\n \\label{eq:address-two-H}\n\\end{equation}","nh t}.\n\\]\nIts odd moments vanish, and\n$P=(\\mathcal B^2-1)/4=Z(Z+1)$ has the divided-power moments of\nLemma~\\ref{lem:bernoulli-central-pd}. Introduce an independent copy\n$\\mathcal D$ and $Q=(\\mathcal D^2-1)/4$. The exact factorization\n\\begin{equation}\n F_2(t)=\\frac{t^2}{\\cosh t-\\cos t}\n       =\\frac{at}{\\sinh(at)}\\frac{bt}{\\sinh(bt)}\n \\label{eq:address-two-factorization}\n\\end{equation}\nshows that $U=a\\mathcal B+b\\mathcal D$ satisfies\n$\\mathcal L(U^{4n})=C_2(n)$. Direct expansion gives\n\\begin{equation}\n H_2:=\\frac{U^4-1}{4}\n     =P+Q-P^2-Q^2+6PQ+i\\mathcal B\\mathcal D(P-Q).\n \\label{eq:address-two-H}\n\\end{equation}","$P=(\\mathcal B^2-1)/4=Z(Z+1)$ has the divided-power moments of\nLemma~\\ref{lem:bernoulli-central-pd}. Introduce an independent copy\n$\\mathcal D$ and $Q=(\\mathcal D^2-1)/4$. The exact factorization\n\\begin{equation}\n F_2(t)=\\frac{t^2}{\\cosh t-\\cos t}\n       =\\frac{at}{\\sinh(at)}\\frac{bt}{\\sinh(bt)}\n \\label{eq:address-two-factorization}\n\\end{equation}\nshows that $U=a\\mathcal B+b\\mathcal D$ satisfies\n$\\mathcal L(U^{4n})=C_2(n)$. Direct expansion gives\n\\begin{equation}\n H_2:=\\frac{U^4-1}{4}\n     =P+Q-P^2-Q^2+6PQ+i\\mathcal B\\mathcal D(P-Q).\n \\label{eq:address-two-H}\n\\end{equation}","a~\\ref{lem:bernoulli-central-pd}. Introduce an independent copy\n$\\mathcal D$ and $Q=(\\mathcal D^2-1)/4$. The exact factorization\n\\begin{equation}\n F_2(t)=\\frac{t^2}{\\cosh t-\\cos t}\n       =\\frac{at}{\\sinh(at)}\\frac{bt}{\\sinh(bt)}\n \\label{eq:address-two-factorization}\n\\end{equation}\nshows that $U=a\\mathcal B+b\\mathcal D$ satisfies\n$\\mathcal L(U^{4n})=C_2(n)$. Direct expansion gives\n\\begin{equation}\n H_2:=\\frac{U^4-1}{4}\n     =P+Q-P^2-Q^2+6PQ+i\\mathcal B\\mathcal D(P-Q).\n \\label{eq:address-two-H}\n\\end{equation}","Let $\\mathcal A_2$ be the divided-power algebra over $R$ generated by\n$P,Q$, adjoining $\\mathcal B,\\mathcal D$ with\n$\\mathcal B^2=1+4P$, $\\mathcal D^2=1+4Q$.\nThe functional is integral on its module basis\n$1,\\mathcal B,\\mathcal D,\\mathcal B\\mathcal D$ by the Bernoulli lemma,\nindependence, and vanishing of odd moments. Every component in\n\\eqref{eq:address-two-H} has integral divided powers, by the same\nfinite identities as before; $P^2$ is even easier than $P^2/2$.\nConsequently\n\\begin{equation}\n \\mathcal L((U^4-1)^m)\\in4^m m!\\mathbb Z_2,\n \\qquad\n v_2(\\Delta_1^kC_2(n))\\geq2k+v_2(k!).\n \\label{eq:address-two-basic-factorial}\n\\end{equation}\nThe starting-row conclusion follows from the finite base expansion\n\\eqref","Let $\\mathcal A_2$ be the divided-power algebra over $R$ generated by\n$P,Q$, adjoining $\\mathcal B,\\mathcal D$ with\n$\\mathcal B^2=1+4P$, $\\mathcal D^2=1+4Q$.\nThe functional is integral on its module basis\n$1,\\mathcal B,\\mathcal D,\\mathcal B\\mathcal D$ by the Bernoulli lemma,\nindependence, and vanishing of odd moments. Every component in\n\\eqref{eq:address-two-H} has integral divided powers, by the same\nfinite identities as before; $P^2$ is even easier than $P^2/2$.\nConsequently\n\\begin{equation}\n \\mathcal L((U^4-1)^m)\\in4^m m!\\mathbb Z_2,\n \\qquad\n v_2(\\Delta_1^kC_2(n))\\geq2k+v_2(k!).\n \\label{eq:address-two-basic-factorial}\n\\end{equation}\nThe starting-row conclusion follows from the finite base expansion\n\\eqref{eq:base-row-finite-expansion}, now wi","Let $\\mathcal A_2$ be the divided-power algebra over $R$ generated by\n$P,Q$, adjoining $\\mathcal B,\\mathcal D$ with\n$\\mathcal B^2=1+4P$, $\\mathcal D^2=1+4Q$.\nThe functional is integral on its module basis\n$1,\\mathcal B,\\mathcal D,\\mathcal B\\mathcal D$ by the Bernoulli lemma,\nindependence, and vanishing of odd moments. Every component in\n\\eqref{eq:address-two-H} has integral divided powers, by the same\nfinite identities as before; $P^2$ is even easier than $P^2/2$.\nConsequently\n\\begin{equation}\n \\mathcal L((U^4-1)^m)\\in4^m m!\\mathbb Z_2,\n \\qquad\n v_2(\\Delta_1^kC_2(n))\\geq2k+v_2(k!).\n \\label{eq:address-two-basic-factorial}\n\\end{equation}\nThe starting-row conclusion follows from the finite base expansion\n\\eqref{eq:base-row-finite-expansion}, now with $W=4H_2$.","ower algebra over $R$ generated by\n$P,Q$, adjoining $\\mathcal B,\\mathcal D$ with\n$\\mathcal B^2=1+4P$, $\\mathcal D^2=1+4Q$.\nThe functional is integral on its module basis\n$1,\\mathcal B,\\mathcal D,\\mathcal B\\mathcal D$ by the Bernoulli lemma,\nindependence, and vanishing of odd moments. Every component in\n\\eqref{eq:address-two-H} has integral divided powers, by the same\nfinite identities as before; $P^2$ is even easier than $P^2/2$.\nConsequently\n\\begin{equation}\n \\mathcal L((U^4-1)^m)\\in4^m m!\\mathbb Z_2,\n \\qquad\n v_2(\\Delta_1^kC_2(n))\\geq2k+v_2(k!).\n \\label{eq:address-two-basic-factorial}\n\\end{equation}\nThe starting-row conclusion follows from the finite base expansion\n\\eqref{eq:base-row-finite-expansion}, now with $W=4H_2$.","rated by\n$P,Q$, adjoining $\\mathcal B,\\mathcal D$ with\n$\\mathcal B^2=1+4P$, $\\mathcal D^2=1+4Q$.\nThe functional is integral on its module basis\n$1,\\mathcal B,\\mathcal D,\\mathcal B\\mathcal D$ by the Bernoulli lemma,\nindependence, and vanishing of odd moments. Every component in\n\\eqref{eq:address-two-H} has integral divided powers, by the same\nfinite identities as before; $P^2$ is even easier than $P^2/2$.\nConsequently\n\\begin{equation}\n \\mathcal L((U^4-1)^m)\\in4^m m!\\mathbb Z_2,\n \\qquad\n v_2(\\Delta_1^kC_2(n))\\geq2k+v_2(k!).\n \\label{eq:address-two-basic-factorial}\n\\end{equation}\nThe starting-row conclusion follows from the finite base expansion\n\\eqref{eq:base-row-finite-expansion}, now with $W=4H_2$.",",\\mathcal D$ with\n$\\mathcal B^2=1+4P$, $\\mathcal D^2=1+4Q$.\nThe functional is integral on its module basis\n$1,\\mathcal B,\\mathcal D,\\mathcal B\\mathcal D$ by the Bernoulli lemma,\nindependence, and vanishing of odd moments. Every component in\n\\eqref{eq:address-two-H} has integral divided powers, by the same\nfinite identities as before; $P^2$ is even easier than $P^2/2$.\nConsequently\n\\begin{equation}\n \\mathcal L((U^4-1)^m)\\in4^m m!\\mathbb Z_2,\n \\qquad\n v_2(\\Delta_1^kC_2(n))\\geq2k+v_2(k!).\n \\label{eq:address-two-basic-factorial}\n\\end{equation}\nThe starting-row conclusion follows from the finite base expansion\n\\eqref{eq:base-row-finite-expansion}, now with $W=4H_2$.","o-H} has integral divided powers, by the same\nfinite identities as before; $P^2$ is even easier than $P^2/2$.\nConsequently\n\\begin{equation}\n \\mathcal L((U^4-1)^m)\\in4^m m!\\mathbb Z_2,\n \\qquad\n v_2(\\Delta_1^kC_2(n))\\geq2k+v_2(k!).\n \\label{eq:address-two-basic-factorial}\n\\end{equation}\nThe starting-row conclusion follows from the finite base expansion\n\\eqref{eq:base-row-finite-expansion}, now with $W=4H_2$.","The low Newton coefficients of the actual reciprocal are\n\\begin{equation}\n C_2(0)=1,\\quad C_2(1)=-\\frac1{15},\\quad C_2(2)=\\frac{13}{45},\n \\qquad a_1=-\\frac{16}{15},\\quad a_2=\\frac{64}{45},\n \\label{eq:address-two-initial-newton}\n\\end{equation}\nwhere $a_m=\\Delta_1^m C_2(0)$. Thus $v_2(a_1)=4$ and $v_2(a_2)=6$.\nFor $m\\geq3$, \\eqref{eq:address-two-basic-factorial} bounds $a_m$ by\n$2m+v_2(m!)$. The finite shifted Newton formula\n\\begin{equation}\n \\Delta_1^kC_2(n)=\\sum_{m=k}^{n+k}a_m\\binom n{m-k}\n \\label{eq:shifted-newton-finite}\n\\end{equation}\ntherefore gives\n$v_2(\\Delta_1^kC_2(n))\\geq k+v_2(k!)+3$ for every $k\\geq1$.\nFor $k\\geq3$ this is $2k\\geq k+3$; for $k=1,2$ it uses the exact\ninitial coefficients and the higher tail bound. The finite step\nexpansion proves the offset three in\n\\eqref{eq:factorial-four-lower}.","eciprocal are\n\\begin{equation}\n C_2(0)=1,\\quad C_2(1)=-\\frac1{15},\\quad C_2(2)=\\frac{13}{45},\n \\qquad a_1=-\\frac{16}{15},\\quad a_2=\\frac{64}{45},\n \\label{eq:address-two-initial-newton}\n\\end{equation}\nwhere $a_m=\\Delta_1^m C_2(0)$. Thus $v_2(a_1)=4$ and $v_2(a_2)=6$.\nFor $m\\geq3$, \\eqref{eq:address-two-basic-factorial} bounds $a_m$ by\n$2m+v_2(m!)$. The finite shifted Newton formula\n\\begin{equation}\n \\Delta_1^kC_2(n)=\\sum_{m=k}^{n+k}a_m\\binom n{m-k}\n \\label{eq:shifted-newton-finite}\n\\end{equation}\ntherefore gives\n$v_2(\\Delta_1^kC_2(n))\\geq k+v_2(k!)+3$ for every $k\\geq1$.\nFor $k\\geq3$ this is $2k\\geq k+3$; for $k=1,2$ it uses the exact\ninitial coefficients and the higher tail bound. The finite step\nexpansion proves the offset three in\n\\eqref{eq:factorial-four-lower}.","5},\\quad a_2=\\frac{64}{45},\n \\label{eq:address-two-initial-newton}\n\\end{equation}\nwhere $a_m=\\Delta_1^m C_2(0)$. Thus $v_2(a_1)=4$ and $v_2(a_2)=6$.\nFor $m\\geq3$, \\eqref{eq:address-two-basic-factorial} bounds $a_m$ by\n$2m+v_2(m!)$. The finite shifted Newton formula\n\\begin{equation}\n \\Delta_1^kC_2(n)=\\sum_{m=k}^{n+k}a_m\\binom n{m-k}\n \\label{eq:shifted-newton-finite}\n\\end{equation}\ntherefore gives\n$v_2(\\Delta_1^kC_2(n))\\geq k+v_2(k!)+3$ for every $k\\geq1$.\nFor $k\\geq3$ this is $2k\\geq k+3$; for $k=1,2$ it uses the exact\ninitial coefficients and the higher tail bound. The finite step\nexpansion proves the offset three in\n\\eqref{eq:factorial-four-lower}.","-initial-newton}\n\\end{equation}\nwhere $a_m=\\Delta_1^m C_2(0)$. Thus $v_2(a_1)=4$ and $v_2(a_2)=6$.\nFor $m\\geq3$, \\eqref{eq:address-two-basic-factorial} bounds $a_m$ by\n$2m+v_2(m!)$. The finite shifted Newton formula\n\\begin{equation}\n \\Delta_1^kC_2(n)=\\sum_{m=k}^{n+k}a_m\\binom n{m-k}\n \\label{eq:shifted-newton-finite}\n\\end{equation}\ntherefore gives\n$v_2(\\Delta_1^kC_2(n))\\geq k+v_2(k!)+3$ for every $k\\geq1$.\nFor $k\\geq3$ this is $2k\\geq k+3$; for $k=1,2$ it uses the exact\ninitial coefficients and the higher tail bound. The finite step\nexpansion proves the offset three in\n\\eqref{eq:factorial-four-lower}.","n}\n\\end{equation}\nwhere $a_m=\\Delta_1^m C_2(0)$. Thus $v_2(a_1)=4$ and $v_2(a_2)=6$.\nFor $m\\geq3$, \\eqref{eq:address-two-basic-factorial} bounds $a_m$ by\n$2m+v_2(m!)$. The finite shifted Newton formula\n\\begin{equation}\n \\Delta_1^kC_2(n)=\\sum_{m=k}^{n+k}a_m\\binom n{m-k}\n \\label{eq:shifted-newton-finite}\n\\end{equation}\ntherefore gives\n$v_2(\\Delta_1^kC_2(n))\\geq k+v_2(k!)+3$ for every $k\\geq1$.\nFor $k\\geq3$ this is $2k\\geq k+3$; for $k=1,2$ it uses the exact\ninitial coefficients and the higher tail bound. The finite step\nexpansion proves the offset three in\n\\eqref{eq:factorial-four-lower}.","where $a_m=\\Delta_1^m C_2(0)$. Thus $v_2(a_1)=4$ and $v_2(a_2)=6$.\nFor $m\\geq3$, \\eqref{eq:address-two-basic-factorial} bounds $a_m$ by\n$2m+v_2(m!)$. The finite shifted Newton formula\n\\begin{equation}\n \\Delta_1^kC_2(n)=\\sum_{m=k}^{n+k}a_m\\binom n{m-k}\n \\label{eq:shifted-newton-finite}\n\\end{equation}\ntherefore gives\n$v_2(\\Delta_1^kC_2(n))\\geq k+v_2(k!)+3$ for every $k\\geq1$.\nFor $k\\geq3$ this is $2k\\geq k+3$; for $k=1,2$ it uses the exact\ninitial coefficients and the higher tail bound. The finite step\nexpansion proves the offset three in\n\\eqref{eq:factorial-four-lower}.","_1^m C_2(0)$. Thus $v_2(a_1)=4$ and $v_2(a_2)=6$.\nFor $m\\geq3$, \\eqref{eq:address-two-basic-factorial} bounds $a_m$ by\n$2m+v_2(m!)$. The finite shifted Newton formula\n\\begin{equation}\n \\Delta_1^kC_2(n)=\\sum_{m=k}^{n+k}a_m\\binom n{m-k}\n \\label{eq:shifted-newton-finite}\n\\end{equation}\ntherefore gives\n$v_2(\\Delta_1^kC_2(n))\\geq k+v_2(k!)+3$ for every $k\\geq1$.\nFor $k\\geq3$ this is $2k\\geq k+3$; for $k=1,2$ it uses the exact\ninitial coefficients and the higher tail bound. The finite step\nexpansion proves the offset three in\n\\eqref{eq:factorial-four-lower}.","There is a still stronger filtration. Put\n\\[\n X=P+Q,\\qquad \\tau=i\\mathcal B\\mathcal D,\\qquad \\epsilon=1+\\tau.\n\\]\nThen the exact identities are\n\\begin{equation}\n H_2=\\epsilon X+2G,\\quad\n G=-\\gamma_2(P)-\\gamma_2(Q)+3PQ-\\tau Q,\\quad\n \\epsilon^2=2V,\\quad V=\\tau-2X-8PQ.\n \\label{eq:address-two-filtration-algebra}\n\\end{equation}\nEvery divided power of $G$ is integral component by component.\nThe finite addition identity gives\n\\begin{equation}\n \\gamma_k(H_2)=\\sum_{r=0}^k2^r\\gamma_r(G)\n                  \\epsilon^{k-r}\\gamma_{k-r}(X)\n          \\in2^{\\lfloor k/2\\rfloor}\\mathcal A_2,\n \\label{eq:address-two-filtration-pd}\n\\end{equation}\nbecause the $r$th term belongs to\n$2^{r+\\lfloor(k-r)/2\\rfloor}\\mathcal A_2$.\nApplying the functional, rational descent, and the base-index expansion\nproves the unit-step version o","There is a still stronger filtration. Put\n\\[\n X=P+Q,\\qquad \\tau=i\\mathcal B\\mathcal D,\\qquad \\epsilon=1+\\tau.\n\\]\nThen the exact identities are\n\\begin{equation}\n H_2=\\epsilon X+2G,\\quad\n G=-\\gamma_2(P)-\\gamma_2(Q)+3PQ-\\tau Q,\\quad\n \\epsilon^2=2V,\\quad V=\\tau-2X-8PQ.\n \\label{eq:address-two-filtration-algebra}\n\\end{equation}\nEvery divided power of $G$ is integral component by component.\nThe finite addition identity gives\n\\begin{equation}\n \\gamma_k(H_2)=\\sum_{r=0}^k2^r\\gamma_r(G)\n                  \\epsilon^{k-r}\\gamma_{k-r}(X)\n          \\in2^{\\lfloor k/2\\rfloor}\\mathcal A_2,\n \\label{eq:address-two-filtration-pd}\n\\end{equation}\nbecause the $r$th term belongs to\n$2^{r+\\lfloor(k-r)/2\\rfloor}\\mathcal A_2$.\nApplying the functional, rational descent, and the base-index expansion\nproves the unit-step version of\n\\eqref{eq:factorial-four-j2-filtration}.\nFor a mixed term of total order $M$, its lower bound exceeds the target\nat order $k$ by at least\n\\[\n 2(M-k)+\\left\\lfloor\\frac M2\\right\\rfloor\n          -\\le","There is a still stronger filtration. Put\n\\[\n X=P+Q,\\qquad \\tau=i\\mathcal B\\mathcal D,\\qquad \\epsilon=1+\\tau.\n\\]\nThen the exact identities are\n\\begin{equation}\n H_2=\\epsilon X+2G,\\quad\n G=-\\gamma_2(P)-\\gamma_2(Q)+3PQ-\\tau Q,\\quad\n \\epsilon^2=2V,\\quad V=\\tau-2X-8PQ.\n \\label{eq:address-two-filtration-algebra}\n\\end{equation}\nEvery divided power of $G$ is integral component by component.\nThe finite addition identity gives\n\\begin{equation}\n \\gamma_k(H_2)=\\sum_{r=0}^k2^r\\gamma_r(G)\n                  \\epsilon^{k-r}\\gamma_{k-r}(X)\n          \\in2^{\\lfloor k/2\\rfloor}\\mathcal A_2,\n \\label{eq:address-two-filtration-pd}\n\\end{equation}\nbecause the $r$th term belongs to\n$2^{r+\\lfloor(k-r)/2\\rfloor}\\mathcal A_2$.\nApplying the functional, rational descent, and the base-index expansion\nproves the unit-step version of\n\\eqref{eq:factorial-four-j2-filtration}.\nFor a mixed term of total order $M$, its lower bound exceeds the target\nat order $k$ by at least\n\\[\n 2(M-k)+\\left\\lfloor\\frac M2\\right\\rfloor\n          -\\left\\lfloor\\frac k2\\right\\rfloor\n       +v","Put\n\\[\n X=P+Q,\\qquad \\tau=i\\mathcal B\\mathcal D,\\qquad \\epsilon=1+\\tau.\n\\]\nThen the exact identities are\n\\begin{equation}\n H_2=\\epsilon X+2G,\\quad\n G=-\\gamma_2(P)-\\gamma_2(Q)+3PQ-\\tau Q,\\quad\n \\epsilon^2=2V,\\quad V=\\tau-2X-8PQ.\n \\label{eq:address-two-filtration-algebra}\n\\end{equation}\nEvery divided power of $G$ is integral component by component.\nThe finite addition identity gives\n\\begin{equation}\n \\gamma_k(H_2)=\\sum_{r=0}^k2^r\\gamma_r(G)\n                  \\epsilon^{k-r}\\gamma_{k-r}(X)\n          \\in2^{\\lfloor k/2\\rfloor}\\mathcal A_2,\n \\label{eq:address-two-filtration-pd}\n\\end{equation}\nbecause the $r$th term belongs to\n$2^{r+\\lfloor(k-r)/2\\rfloor}\\mathcal A_2$.\nApplying the functional, rational descent, and the base-index expansion\nproves the unit-step version of\n\\eqref{eq:factorial-four-j2-filtration}.\nFor a mixed term of total order $M$, its lower bound exceeds the target\nat order $k$ by at least\n\\[\n 2(M-k)+\\left\\lfloor\\frac M2\\right\\rfloor\n          -\\left\\lfloor\\frac k2\\right\\rfloor\n       +v_2(M!)-v_2(k!)-\\sum_\\nu v_2(m_\\nu)\\geq0.\n\\]\nIndeed $\\sum_\\nu v_2(m_\\nu)\\leq M-k$. This pr…",",\\quad V=\\tau-2X-8PQ.\n \\label{eq:address-two-filtration-algebra}\n\\end{equation}\nEvery divided power of $G$ is integral component by component.\nThe finite addition identity gives\n\\begin{equation}\n \\gamma_k(H_2)=\\sum_{r=0}^k2^r\\gamma_r(G)\n                  \\epsilon^{k-r}\\gamma_{k-r}(X)\n          \\in2^{\\lfloor k/2\\rfloor}\\mathcal A_2,\n \\label{eq:address-two-filtration-pd}\n\\end{equation}\nbecause the $r$th term belongs to\n$2^{r+\\lfloor(k-r)/2\\rfloor}\\mathcal A_2$.\nApplying the functional, rational descent, and the base-index expansion\nproves the unit-step version of\n\\eqref{eq:factorial-four-j2-filtration}.\nFor a mixed term of total order $M$, its lower bound exceeds the target\nat order $k$ by at least\n\\[\n 2(M-k)+\\left\\lfloor\\frac M2\\right\\rfloor\n          -\\left\\lfloor\\frac k2\\right\\rfloor\n       +v_2(M!)-v_2(k!)-\\sum_\\nu v_2(m_\\nu)\\geq0.\n\\]\nIndeed $\\sum_\\nu v_2(m_\\nu)\\leq M-k$. This proves the stronger mixed\nbound. The earlier offset-three bound is retained, since it is stronger\nat the lowest orders.","\\label{eq:address-two-filtration-algebra}\n\\end{equation}\nEvery divided power of $G$ is integral component by component.\nThe finite addition identity gives\n\\begin{equation}\n \\gamma_k(H_2)=\\sum_{r=0}^k2^r\\gamma_r(G)\n                  \\epsilon^{k-r}\\gamma_{k-r}(X)\n          \\in2^{\\lfloor k/2\\rfloor}\\mathcal A_2,\n \\label{eq:address-two-filtration-pd}\n\\end{equation}\nbecause the $r$th term belongs to\n$2^{r+\\lfloor(k-r)/2\\rfloor}\\mathcal A_2$.\nApplying the functional, rational descent, and the base-index expansion\nproves the unit-step version of\n\\eqref{eq:factorial-four-j2-filtration}.\nFor a mixed term of total order $M$, its lower bound exceeds the target\nat order $k$ by at least\n\\[\n 2(M-k)+\\left\\lfloor\\frac M2\\right\\rfloor\n          -\\left\\lfloor\\frac k2\\right\\rfloor\n       +v_2(M!)-v_2(k!)-\\sum_\\nu v_2(m_\\nu)\\geq0.\n\\]\nIndeed $\\sum_\\nu v_2(m_\\nu)\\leq M-k$. This proves the stronger mixed\nbound. The earlier offset-three bound is retained, since it is stronger\nat the lowest orders.","\\epsilon^{k-r}\\gamma_{k-r}(X)\n          \\in2^{\\lfloor k/2\\rfloor}\\mathcal A_2,\n \\label{eq:address-two-filtration-pd}\n\\end{equation}\nbecause the $r$th term belongs to\n$2^{r+\\lfloor(k-r)/2\\rfloor}\\mathcal A_2$.\nApplying the functional, rational descent, and the base-index expansion\nproves the unit-step version of\n\\eqref{eq:factorial-four-j2-filtration}.\nFor a mixed term of total order $M$, its lower bound exceeds the target\nat order $k$ by at least\n\\[\n 2(M-k)+\\left\\lfloor\\frac M2\\right\\rfloor\n          -\\left\\lfloor\\frac k2\\right\\rfloor\n       +v_2(M!)-v_2(k!)-\\sum_\\nu v_2(m_\\nu)\\geq0.\n\\]\nIndeed $\\sum_\\nu v_2(m_\\nu)\\leq M-k$. This proves the stronger mixed\nbound. The earlier offset-three bound is retained, since it is stronger\nat the lowest orders.","\\in2^{\\lfloor k/2\\rfloor}\\mathcal A_2,\n \\label{eq:address-two-filtration-pd}\n\\end{equation}\nbecause the $r$th term belongs to\n$2^{r+\\lfloor(k-r)/2\\rfloor}\\mathcal A_2$.\nApplying the functional, rational descent, and the base-index expansion\nproves the unit-step version of\n\\eqref{eq:factorial-four-j2-filtration}.\nFor a mixed term of total order $M$, its lower bound exceeds the target\nat order $k$ by at least\n\\[\n 2(M-k)+\\left\\lfloor\\frac M2\\right\\rfloor\n          -\\left\\lfloor\\frac k2\\right\\rfloor\n       +v_2(M!)-v_2(k!)-\\sum_\\nu v_2(m_\\nu)\\geq0.\n\\]\nIndeed $\\sum_\\nu v_2(m_\\nu)\\leq M-k$. This proves the stronger mixed\nbound. The earlier offset-three bound is retained, since it is stronger\nat the lowest orders.","k/2\\rfloor}\\mathcal A_2,\n \\label{eq:address-two-filtration-pd}\n\\end{equation}\nbecause the $r$th term belongs to\n$2^{r+\\lfloor(k-r)/2\\rfloor}\\mathcal A_2$.\nApplying the functional, rational descent, and the base-index expansion\nproves the unit-step version of\n\\eqref{eq:factorial-four-j2-filtration}.\nFor a mixed term of total order $M$, its lower bound exceeds the target\nat order $k$ by at least\n\\[\n 2(M-k)+\\left\\lfloor\\frac M2\\right\\rfloor\n          -\\left\\lfloor\\frac k2\\right\\rfloor\n       +v_2(M!)-v_2(k!)-\\sum_\\nu v_2(m_\\nu)\\geq0.\n\\]\nIndeed $\\sum_\\nu v_2(m_\\nu)\\leq M-k$. This proves the stronger mixed\nbound. The earlier offset-three bound is retained, since it is stronger\nat the lowest orders.","A_2$.\nApplying the functional, rational descent, and the base-index expansion\nproves the unit-step version of\n\\eqref{eq:factorial-four-j2-filtration}.\nFor a mixed term of total order $M$, its lower bound exceeds the target\nat order $k$ by at least\n\\[\n 2(M-k)+\\left\\lfloor\\frac M2\\right\\rfloor\n          -\\left\\lfloor\\frac k2\\right\\rfloor\n       +v_2(M!)-v_2(k!)-\\sum_\\nu v_2(m_\\nu)\\geq0.\n\\]\nIndeed $\\sum_\\nu v_2(m_\\nu)\\leq M-k$. This proves the stronger mixed\nbound. The earlier offset-three bound is retained, since it is stronger\nat the lowest orders.","The address-three source is\n$G_{4,3}(t)=3(\\sinh t-\\sin t)/t^3$. Addition identities give\n\\[\n \\sinh t-\\sin t\n =2a\\sinh(at)\\cosh(bt)+2b\\cosh(at)\\sinh(bt),\n \\qquad \\cosh t-\\cos t=2\\sinh(at)\\sinh(bt).\n\\]\nConsequently\n\\begin{equation}\n D_3(t):=G_{4,3}(t)F_2(t)\n =\\frac32\\left(\\frac{\\coth(at)}{at}\n                            +\\frac{\\coth(bt)}{bt}\\right).\n \\label{eq:address-three-cotangent-factor}\n\\end{equation}\nThis expression is a power series: its two polar terms cancel because\n$a^{-2}+b^{-2}=0$. Its constant is one, only degrees divisible by four\nsurvive, and the Bernoulli expansion gives\n\\begin{equation}\n D_3(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=3\\left(-\\frac14\\right)^i\n          \\frac{2^{4i+2}B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-th","The address-three source is\n$G_{4,3}(t)=3(\\sinh t-\\sin t)/t^3$. Addition identities give\n\\[\n \\sinh t-\\sin t\n =2a\\sinh(at)\\cosh(bt)+2b\\cosh(at)\\sinh(bt),\n \\qquad \\cosh t-\\cos t=2\\sinh(at)\\sinh(bt).\n\\]\nConsequently\n\\begin{equation}\n D_3(t):=G_{4,3}(t)F_2(t)\n =\\frac32\\left(\\frac{\\coth(at)}{at}\n                            +\\frac{\\coth(bt)}{bt}\\right).\n \\label{eq:address-three-cotangent-factor}\n\\end{equation}\nThis expression is a power series: its two polar terms cancel because\n$a^{-2}+b^{-2}=0$. Its constant is one, only degrees divisible by four\nsurvive, and the Bernoulli expansion gives\n\\begin{equation}\n D_3(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=3\\left(-\\frac14\\right)^i\n          \\frac{2^{4i+2}B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-three-packets}\n\\end{equation}\nAgain $v_2(d_i)\\geq2i$, and the inverse coefficients $h_i$ have the same\nbound. Since $F_3=F_2D_3^{-1}$,\n\\be","The address-three source is\n$G_{4,3}(t)=3(\\sinh t-\\sin t)/t^3$. Addition identities give\n\\[\n \\sinh t-\\sin t\n =2a\\sinh(at)\\cosh(bt)+2b\\cosh(at)\\sinh(bt),\n \\qquad \\cosh t-\\cos t=2\\sinh(at)\\sinh(bt).\n\\]\nConsequently\n\\begin{equation}\n D_3(t):=G_{4,3}(t)F_2(t)\n =\\frac32\\left(\\frac{\\coth(at)}{at}\n                            +\\frac{\\coth(bt)}{bt}\\right).\n \\label{eq:address-three-cotangent-factor}\n\\end{equation}\nThis expression is a power series: its two polar terms cancel because\n$a^{-2}+b^{-2}=0$. Its constant is one, only degrees divisible by four\nsurvive, and the Bernoulli expansion gives\n\\begin{equation}\n D_3(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=3\\left(-\\frac14\\right)^i\n          \\frac{2^{4i+2}B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-three-packets}\n\\end{equation}\nAgain $v_2(d_i)\\geq2i$, and the inverse coefficients $h_i$ have the same\nbound. Since $F_3=F_2D_3^{-1}$,\n\\begin{equation}\n C_3(n)-C_2(n)=\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_2(n-i).\n \\label{eq:address-three-exact-transfer}\n\\end{equation}\nThe scaled-binomial argument\n\\eqref{eq:scaled-binomial-difference-bound","ties give\n\\[\n \\sinh t-\\sin t\n =2a\\sinh(at)\\cosh(bt)+2b\\cosh(at)\\sinh(bt),\n \\qquad \\cosh t-\\cos t=2\\sinh(at)\\sinh(bt).\n\\]\nConsequently\n\\begin{equation}\n D_3(t):=G_{4,3}(t)F_2(t)\n =\\frac32\\left(\\frac{\\coth(at)}{at}\n                            +\\frac{\\coth(bt)}{bt}\\right).\n \\label{eq:address-three-cotangent-factor}\n\\end{equation}\nThis expression is a power series: its two polar terms cancel because\n$a^{-2}+b^{-2}=0$. Its constant is one, only degrees divisible by four\nsurvive, and the Bernoulli expansion gives\n\\begin{equation}\n D_3(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=3\\left(-\\frac14\\right)^i\n          \\frac{2^{4i+2}B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-three-packets}\n\\end{equation}\nAgain $v_2(d_i)\\geq2i$, and the inverse coefficients $h_i$ have the same\nbound. Since $F_3=F_2D_3^{-1}$,\n\\begin{equation}\n C_3(n)-C_2(n)=\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_2(n-i).\n \\label{eq:address-three-exact-transfer}\n\\end{equation}\nThe scaled-binomial argument\n\\eqref{eq:scaled-binomial-difference-bound} applies unchanged.\nThe basic bound for $C_2$ includes order zero; its continuous ext…","ntly\n\\begin{equation}\n D_3(t):=G_{4,3}(t)F_2(t)\n =\\frac32\\left(\\frac{\\coth(at)}{at}\n                            +\\frac{\\coth(bt)}{bt}\\right).\n \\label{eq:address-three-cotangent-factor}\n\\end{equation}\nThis expression is a power series: its two polar terms cancel because\n$a^{-2}+b^{-2}=0$. Its constant is one, only degrees divisible by four\nsurvive, and the Bernoulli expansion gives\n\\begin{equation}\n D_3(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=3\\left(-\\frac14\\right)^i\n          \\frac{2^{4i+2}B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-three-packets}\n\\end{equation}\nAgain $v_2(d_i)\\geq2i$, and the inverse coefficients $h_i$ have the same\nbound. Since $F_3=F_2D_3^{-1}$,\n\\begin{equation}\n C_3(n)-C_2(n)=\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_2(n-i).\n \\label{eq:address-three-exact-transfer}\n\\end{equation}\nThe scaled-binomial argument\n\\eqref{eq:scaled-binomial-difference-bound} applies unchanged.\nThe basic bound for $C_2$ includes order zero; its continuous extension\nis available from the threshold-free theorem. Thus the shifted sums\nconverge uniformly on $\\mathbb Z_2$, and the pointwise…","d{equation}\nThis expression is a power series: its two polar terms cancel because\n$a^{-2}+b^{-2}=0$. Its constant is one, only degrees divisible by four\nsurvive, and the Bernoulli expansion gives\n\\begin{equation}\n D_3(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=3\\left(-\\frac14\\right)^i\n          \\frac{2^{4i+2}B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-three-packets}\n\\end{equation}\nAgain $v_2(d_i)\\geq2i$, and the inverse coefficients $h_i$ have the same\nbound. Since $F_3=F_2D_3^{-1}$,\n\\begin{equation}\n C_3(n)-C_2(n)=\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_2(n-i).\n \\label{eq:address-three-exact-transfer}\n\\end{equation}\nThe scaled-binomial argument\n\\eqref{eq:scaled-binomial-difference-bound} applies unchanged.\nThe basic bound for $C_2$ includes order zero; its continuous extension\nis available from the threshold-free theorem. Thus the shifted sums\nconverge uniformly on $\\mathbb Z_2$, and the pointwise product rule\nproves\n\\begin{equation}\n v_2(\\Delta_1^k(C_3-C_2)(n))\\geq k+v_2(k!)+1\n                                      \\qquad(k,n\\geq0).\n \\label{eq:address-three-extra-bit}\n\\end{equation…","its two polar terms cancel because\n$a^{-2}+b^{-2}=0$. Its constant is one, only degrees divisible by four\nsurvive, and the Bernoulli expansion gives\n\\begin{equation}\n D_3(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=3\\left(-\\frac14\\right)^i\n          \\frac{2^{4i+2}B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-three-packets}\n\\end{equation}\nAgain $v_2(d_i)\\geq2i$, and the inverse coefficients $h_i$ have the same\nbound. Since $F_3=F_2D_3^{-1}$,\n\\begin{equation}\n C_3(n)-C_2(n)=\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_2(n-i).\n \\label{eq:address-three-exact-transfer}\n\\end{equation}\nThe scaled-binomial argument\n\\eqref{eq:scaled-binomial-difference-bound} applies unchanged.\nThe basic bound for $C_2$ includes order zero; its continuous extension\nis available from the threshold-free theorem. Thus the shifted sums\nconverge uniformly on $\\mathbb Z_2$, and the pointwise product rule\nproves\n\\begin{equation}\n v_2(\\Delta_1^k(C_3-C_2)(n))\\geq k+v_2(k!)+1\n                                      \\qquad(k,n\\geq0).\n \\label{eq:address-three-extra-bit}\n\\end{equation}\nAt every positive order $C_2$ has three extra…","e\n$a^{-2}+b^{-2}=0$. Its constant is one, only degrees divisible by four\nsurvive, and the Bernoulli expansion gives\n\\begin{equation}\n D_3(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=3\\left(-\\frac14\\right)^i\n          \\frac{2^{4i+2}B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-three-packets}\n\\end{equation}\nAgain $v_2(d_i)\\geq2i$, and the inverse coefficients $h_i$ have the same\nbound. Since $F_3=F_2D_3^{-1}$,\n\\begin{equation}\n C_3(n)-C_2(n)=\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_2(n-i).\n \\label{eq:address-three-exact-transfer}\n\\end{equation}\nThe scaled-binomial argument\n\\eqref{eq:scaled-binomial-difference-bound} applies unchanged.\nThe basic bound for $C_2$ includes order zero; its continuous extension\nis available from the threshold-free theorem. Thus the shifted sums\nconverge uniformly on $\\mathbb Z_2$, and the pointwise product rule\nproves\n\\begin{equation}\n v_2(\\Delta_1^k(C_3-C_2)(n))\\geq k+v_2(k!)+1\n                                      \\qquad(k,n\\geq0).\n \\label{eq:address-three-extra-bit}\n\\end{equation}\nAt every positive order $C_2$ has three extra bits, so this gives the\none-bit …","is one, only degrees divisible by four\nsurvive, and the Bernoulli expansion gives\n\\begin{equation}\n D_3(t)=\\sum_{i\\geq0}d_i\\frac{t^{4i}}{(4i)!},\\qquad\n d_i=3\\left(-\\frac14\\right)^i\n          \\frac{2^{4i+2}B_{4i+2}}{(4i+2)(4i+1)}.\n \\label{eq:address-three-packets}\n\\end{equation}\nAgain $v_2(d_i)\\geq2i$, and the inverse coefficients $h_i$ have the same\nbound. Since $F_3=F_2D_3^{-1}$,\n\\begin{equation}\n C_3(n)-C_2(n)=\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_2(n-i).\n \\label{eq:address-three-exact-transfer}\n\\end{equation}\nThe scaled-binomial argument\n\\eqref{eq:scaled-binomial-difference-bound} applies unchanged.\nThe basic bound for $C_2$ includes order zero; its continuous extension\nis available from the threshold-free theorem. Thus the shifted sums\nconverge uniformly on $\\mathbb Z_2$, and the pointwise product rule\nproves\n\\begin{equation}\n v_2(\\Delta_1^k(C_3-C_2)(n))\\geq k+v_2(k!)+1\n                                      \\qquad(k,n\\geq0).\n \\label{eq:address-three-extra-bit}\n\\end{equation}\nAt every positive order $C_2$ has three extra bits, so this gives the\none-bit offset for $C_3$. Finite step expa…","1)}.\n \\label{eq:address-three-packets}\n\\end{equation}\nAgain $v_2(d_i)\\geq2i$, and the inverse coefficients $h_i$ have the same\nbound. Since $F_3=F_2D_3^{-1}$,\n\\begin{equation}\n C_3(n)-C_2(n)=\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_2(n-i).\n \\label{eq:address-three-exact-transfer}\n\\end{equation}\nThe scaled-binomial argument\n\\eqref{eq:scaled-binomial-difference-bound} applies unchanged.\nThe basic bound for $C_2$ includes order zero; its continuous extension\nis available from the threshold-free theorem. Thus the shifted sums\nconverge uniformly on $\\mathbb Z_2$, and the pointwise product rule\nproves\n\\begin{equation}\n v_2(\\Delta_1^k(C_3-C_2)(n))\\geq k+v_2(k!)+1\n                                      \\qquad(k,n\\geq0).\n \\label{eq:address-three-extra-bit}\n\\end{equation}\nAt every positive order $C_2$ has three extra bits, so this gives the\none-bit offset for $C_3$. Finite step expansion finishes the proof of\nTheorem~\\ref{thm:four-factorial-bound}.","_3=F_2D_3^{-1}$,\n\\begin{equation}\n C_3(n)-C_2(n)=\\sum_{i\\geq1}h_i\\binom{4n}{4i}C_2(n-i).\n \\label{eq:address-three-exact-transfer}\n\\end{equation}\nThe scaled-binomial argument\n\\eqref{eq:scaled-binomial-difference-bound} applies unchanged.\nThe basic bound for $C_2$ includes order zero; its continuous extension\nis available from the threshold-free theorem. Thus the shifted sums\nconverge uniformly on $\\mathbb Z_2$, and the pointwise product rule\nproves\n\\begin{equation}\n v_2(\\Delta_1^k(C_3-C_2)(n))\\geq k+v_2(k!)+1\n                                      \\qquad(k,n\\geq0).\n \\label{eq:address-three-extra-bit}\n\\end{equation}\nAt every positive order $C_2$ has three extra bits, so this gives the\none-bit offset for $C_3$. Finite step expansion finishes the proof of\nTheorem~\\ref{thm:four-factorial-bound}.","nom{4n}{4i}C_2(n-i).\n \\label{eq:address-three-exact-transfer}\n\\end{equation}\nThe scaled-binomial argument\n\\eqref{eq:scaled-binomial-difference-bound} applies unchanged.\nThe basic bound for $C_2$ includes order zero; its continuous extension\nis available from the threshold-free theorem. Thus the shifted sums\nconverge uniformly on $\\mathbb Z_2$, and the pointwise product rule\nproves\n\\begin{equation}\n v_2(\\Delta_1^k(C_3-C_2)(n))\\geq k+v_2(k!)+1\n                                      \\qquad(k,n\\geq0).\n \\label{eq:address-three-extra-bit}\n\\end{equation}\nAt every positive order $C_2$ has three extra bits, so this gives the\none-bit offset for $C_3$. Finite step expansion finishes the proof of\nTheorem~\\ref{thm:four-factorial-bound}.","bound for $C_2$ includes order zero; its continuous extension\nis available from the threshold-free theorem. Thus the shifted sums\nconverge uniformly on $\\mathbb Z_2$, and the pointwise product rule\nproves\n\\begin{equation}\n v_2(\\Delta_1^k(C_3-C_2)(n))\\geq k+v_2(k!)+1\n                                      \\qquad(k,n\\geq0).\n \\label{eq:address-three-extra-bit}\n\\end{equation}\nAt every positive order $C_2$ has three extra bits, so this gives the\none-bit offset for $C_3$. Finite step expansion finishes the proof of\nTheorem~\\ref{thm:four-factorial-bound}.","available from the threshold-free theorem. Thus the shifted sums\nconverge uniformly on $\\mathbb Z_2$, and the pointwise product rule\nproves\n\\begin{equation}\n v_2(\\Delta_1^k(C_3-C_2)(n))\\geq k+v_2(k!)+1\n                                      \\qquad(k,n\\geq0).\n \\label{eq:address-three-extra-bit}\n\\end{equation}\nAt every positive order $C_2$ has three extra bits, so this gives the\none-bit offset for $C_3$. Finite step expansion finishes the proof of\nTheorem~\\ref{thm:four-factorial-bound}.","\\begin{theorem}[Exact factorial precision at addresses zero and one]\n\\label{thm:zero-one-exact-mixed}\nLet $j\\in\\{0,1\\}$, $k\\geq1$, $n\\geq0$, and $H_1,\\ldots,H_k>0$.\nPut\n\\[\n L=k+v_2(k!)+\\sum_{\\nu=1}^k v_2(H_\\nu),\\qquad\n e=\\#\\{\\nu:2\\mid H_\\nu\\}.\n\\]\nThe mixed difference has valuation exactly $L$ if $k$ is odd or $e$ is\neven. If $k$ is even and $e$ is odd, it has valuation at least $L+1$.\nIn particular, for every homogeneous positive step,\n\\begin{equation}\n v_2(\\Delta_H^kC_j(n))=k+v_2(k!)+k v_2(H).\n \\label{eq:zero-one-homogeneous-exact}\n\\end{equation}\n\\end{theorem}","ero and one]\n\\label{thm:zero-one-exact-mixed}\nLet $j\\in\\{0,1\\}$, $k\\geq1$, $n\\geq0$, and $H_1,\\ldots,H_k>0$.\nPut\n\\[\n L=k+v_2(k!)+\\sum_{\\nu=1}^k v_2(H_\\nu),\\qquad\n e=\\#\\{\\nu:2\\mid H_\\nu\\}.\n\\]\nThe mixed difference has valuation exactly $L$ if $k$ is odd or $e$ is\neven. If $k$ is even and $e$ is odd, it has valuation at least $L+1$.\nIn particular, for every homogeneous positive step,\n\\begin{equation}\n v_2(\\Delta_H^kC_j(n))=k+v_2(k!)+k v_2(H).\n \\label{eq:zero-one-homogeneous-exact}\n\\end{equation}\n\\end{theorem}","\\begin{proof}\nReturn to the Euler variables of address zero and put\n$X=P+Q$, $\\tau=i\\mathcal E\\mathcal F$, $\\epsilon=1+\\tau$.\nDirect rearrangement of \\eqref{eq:four-euler-H} gives\n\\begin{equation}\n H_0=\\gamma_2(X)+\\epsilon X+2G,\\qquad\n G=-\\gamma_2(P)-\\gamma_2(Q)+PQ-\\tau Q,\n \\qquad \\epsilon^2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equati","\\begin{proof}\nReturn to the Euler variables of address zero and put\n$X=P+Q$, $\\tau=i\\mathcal E\\mathcal F$, $\\epsilon=1+\\tau$.\nDirect rearrangement of \\eqref{eq:four-euler-H} gives\n\\begin{equation}\n H_0=\\gamma_2(X)+\\epsilon X+2G,\\qquad\n G=-\\gamma_2(P)-\\gamma_2(Q)+PQ-\\tau Q,\n \\qquad \\epsilon^2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv","\\begin{proof}\nReturn to the Euler variables of address zero and put\n$X=P+Q$, $\\tau=i\\mathcal E\\mathcal F$, $\\epsilon=1+\\tau$.\nDirect rearrangement of \\eqref{eq:four-euler-H} gives\n\\begin{equation}\n H_0=\\gamma_2(X)+\\epsilon X+2G,\\qquad\n G=-\\gamma_2(P)-\\gamma_2(Q)+PQ-\\tau Q,\n \\qquad \\epsilon^2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\e","\\begin{proof}\nReturn to the Euler variables of address zero and put\n$X=P+Q$, $\\tau=i\\mathcal E\\mathcal F$, $\\epsilon=1+\\tau$.\nDirect rearrangement of \\eqref{eq:four-euler-H} gives\n\\begin{equation}\n H_0=\\gamma_2(X)+\\epsilon X+2G,\\qquad\n G=-\\gamma_2(P)-\\gamma_2(Q)+PQ-\\tau Q,\n \\qquad \\epsilon^2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref","\\begin{proof}\nReturn to the Euler variables of address zero and put\n$X=P+Q$, $\\tau=i\\mathcal E\\mathcal F$, $\\epsilon=1+\\tau$.\nDirect rearrangement of \\eqref{eq:four-euler-H} gives\n\\begin{equation}\n H_0=\\gamma_2(X)+\\epsilon X+2G,\\qquad\n G=-\\gamma_2(P)-\\gamma_2(Q)+PQ-\\tau Q,\n \\qquad \\epsilon^2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independe","o and put\n$X=P+Q$, $\\tau=i\\mathcal E\\mathcal F$, $\\epsilon=1+\\tau$.\nDirect rearrangement of \\eqref{eq:four-euler-H} gives\n\\begin{equation}\n H_0=\\gamma_2(X)+\\epsilon X+2G,\\qquad\n G=-\\gamma_2(P)-\\gamma_2(Q)+PQ-\\tau Q,\n \\qquad \\epsilon^2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weigh…",", $\\tau=i\\mathcal E\\mathcal F$, $\\epsilon=1+\\tau$.\nDirect rearrangement of \\eqref{eq:four-euler-H} gives\n\\begin{equation}\n H_0=\\gamma_2(X)+\\epsilon X+2G,\\qquad\n G=-\\gamma_2(P)-\\gamma_2(Q)+PQ-\\tau Q,\n \\qquad \\epsilon^2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanish…","\\tau$.\nDirect rearrangement of \\eqref{eq:four-euler-H} gives\n\\begin{equation}\n H_0=\\gamma_2(X)+\\epsilon X+2G,\\qquad\n G=-\\gamma_2(P)-\\gamma_2(Q)+PQ-\\tau Q,\n \\qquad \\epsilon^2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanishes because it contains an odd power of\neach …",":four-euler-H} gives\n\\begin{equation}\n H_0=\\gamma_2(X)+\\epsilon X+2G,\\qquad\n G=-\\gamma_2(P)-\\gamma_2(Q)+PQ-\\tau Q,\n \\qquad \\epsilon^2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanishes because it contains an odd power of\neach Euler variable. Therefore the first term…","\\qquad \\epsilon^2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanishes because it contains an odd power of\neach Euler variable. Therefore the first term of\n\\eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the\nsecond to zero. Hence $\\mathcal L(\\gamma_k…","2=2(\\tau-P-Q-2PQ).\n \\label{eq:zero-parity-decomposition}\n\\end{equation}\nThe divided powers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanishes because it contains an odd power of\neach Euler variable. Therefore the first term of\n\\eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the\nsecond to zero. Hence $\\mathcal L(\\gamma_k(H_0))$ is odd a…","wers of $G$ are integral by the same component formulas\nused above. Thus modulo $2\\mathcal A_0$, $\\gamma_k(H_0)$ equals the\ndivided power of $\\gamma_2(X)+\\epsilon X$. The identity\n\\[\n \\gamma_r(\\gamma_2(X))\n   =\\frac{(2r)!}{2^r r!}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanishes because it contains an odd power of\neach Euler variable. Therefore the first term of\n\\eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the\nsecond to zero. Hence $\\mathcal L(\\gamma_k(H_0))$ is odd at every order,\nincluding order zero. This proves\n\\[\n v_2(\\Delta_1^k C_0(0))=k+v_2(k!).…","}\\gamma_{2r}(X)\n\\]\nhas an odd integer scalar. Every term involving $\\epsilon^j$ for\n$j\\geq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanishes because it contains an odd power of\neach Euler variable. Therefore the first term of\n\\eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the\nsecond to zero. Hence $\\mathcal L(\\gamma_k(H_0))$ is odd at every order,\nincluding order zero. This proves\n\\[\n v_2(\\Delta_1^k C_0(0))=k+v_2(k!).\n\\]\nIn the base-row expansion \\eqref{eq:base-row-finite-expansion}, every\n$s>0$ term has at least one extra bit after division by $2^k k!$.\nThus the same exact valuation holds at every starting row.\nThe extra-bit transfer \\eqref{…","eq2$ vanishes modulo two. The surviving zero and first powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanishes because it contains an odd power of\neach Euler variable. Therefore the first term of\n\\eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the\nsecond to zero. Hence $\\mathcal L(\\gamma_k(H_0))$ is odd at every order,\nincluding order zero. This proves\n\\[\n v_2(\\Delta_1^k C_0(0))=k+v_2(k!).\n\\]\nIn the base-row expansion \\eqref{eq:base-row-finite-expansion}, every\n$s>0$ term has at least one extra bit after division by $2^k k!$.\nThus the same exact valuation holds at every starting row.\nThe extra-bit transfer \\eqref{eq:address-one-extra-bit} proves it for\naddress one as well.","t powers give\n\\begin{equation}\n \\gamma_k(H_0)\\equiv\n       \\gamma_{2k}(X)+\\epsilon\\gamma_{2k-1}(X)\n                                  \\pmod{2\\mathcal A_0}.\n \\label{eq:zero-parity-leading-pd}\n\\end{equation}\nThe multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanishes because it contains an odd power of\neach Euler variable. Therefore the first term of\n\\eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the\nsecond to zero. Hence $\\mathcal L(\\gamma_k(H_0))$ is odd at every order,\nincluding order zero. This proves\n\\[\n v_2(\\Delta_1^k C_0(0))=k+v_2(k!).\n\\]\nIn the base-row expansion \\eqref{eq:base-row-finite-expansion}, every\n$s>0$ term has at least one extra bit after division by $2^k k!$.\nThus the same exact valuation holds at every starting row.\nThe extra-bit transfer \\eqref{eq:address-one-extra-bit} proves it for\naddress one as well.","multiplication coefficient in the second term is $2k-1$, also odd.\nBy the odd moments \\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanishes because it contains an odd power of\neach Euler variable. Therefore the first term of\n\\eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the\nsecond to zero. Hence $\\mathcal L(\\gamma_k(H_0))$ is odd at every order,\nincluding order zero. This proves\n\\[\n v_2(\\Delta_1^k C_0(0))=k+v_2(k!).\n\\]\nIn the base-row expansion \\eqref{eq:base-row-finite-expansion}, every\n$s>0$ term has at least one extra bit after division by $2^k k!$.\nThus the same exact valuation holds at every starting row.\nThe extra-bit transfer \\eqref{eq:address-one-extra-bit} proves it for\naddress one as well.","\\eqref{eq:euler-pd-odd-moments} and independence,\n\\[\n \\mathcal L(\\gamma_m(P+Q))\\equiv m+1\\pmod2.\n\\]\nThe $\\tau$-weighted moment vanishes because it contains an odd power of\neach Euler variable. Therefore the first term of\n\\eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the\nsecond to zero. Hence $\\mathcal L(\\gamma_k(H_0))$ is odd at every order,\nincluding order zero. This proves\n\\[\n v_2(\\Delta_1^k C_0(0))=k+v_2(k!).\n\\]\nIn the base-row expansion \\eqref{eq:base-row-finite-expansion}, every\n$s>0$ term has at least one extra bit after division by $2^k k!$.\nThus the same exact valuation holds at every starting row.\nThe extra-bit transfer \\eqref{eq:address-one-extra-bit} proves it for\naddress one as well.","-weighted moment vanishes because it contains an odd power of\neach Euler variable. Therefore the first term of\n\\eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the\nsecond to zero. Hence $\\mathcal L(\\gamma_k(H_0))$ is odd at every order,\nincluding order zero. This proves\n\\[\n v_2(\\Delta_1^k C_0(0))=k+v_2(k!).\n\\]\nIn the base-row expansion \\eqref{eq:base-row-finite-expansion}, every\n$s>0$ term has at least one extra bit after division by $2^k k!$.\nThus the same exact valuation holds at every starting row.\nThe extra-bit transfer \\eqref{eq:address-one-extra-bit} proves it for\naddress one as well.","er of\neach Euler variable. Therefore the first term of\n\\eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the\nsecond to zero. Hence $\\mathcal L(\\gamma_k(H_0))$ is odd at every order,\nincluding order zero. This proves\n\\[\n v_2(\\Delta_1^k C_0(0))=k+v_2(k!).\n\\]\nIn the base-row expansion \\eqref{eq:base-row-finite-expansion}, every\n$s>0$ term has at least one extra bit after division by $2^k k!$.\nThus the same exact valuation holds at every starting row.\nThe extra-bit transfer \\eqref{eq:address-one-extra-bit} proves it for\naddress one as well.","It remains to track possible ties in the mixed-step expansion.\nFor local orders $m_\\nu\\geq1$ and total $M$, the excess over $L$ is\nat least\n\\begin{equation}\n \\left(M-k-\\sum_\\nu v_2(m_\\nu)\\right)\n                       +\\left(v_2(M!)-v_2(k!)\\right).\n \\label{eq:zero-one-mixed-excess}\n\\end{equation}\nThe first bracket is nonnegative, and equality requires every\n$m_\\nu$ to be one or two. Two or more occurrences of two make the\nsecond bracket positive, since the intervening factorial interval then\ncontains an even number. Exactly one occurrence of two can tie only\nwhen $k$ is even; its binomial coefficient can have the minimally\nallowed valuation only if its associated step is even.","m_\\nu v_2(m_\\nu)\\right)\n                       +\\left(v_2(M!)-v_2(k!)\\right).\n \\label{eq:zero-one-mixed-excess}\n\\end{equation}\nThe first bracket is nonnegative, and equality requires every\n$m_\\nu$ to be one or two. Two or more occurrences of two make the\nsecond bracket positive, since the intervening factorial interval then\ncontains an even number. Exactly one occurrence of two can tie only\nwhen $k$ is even; its binomial coefficient can have the minimally\nallowed valuation only if its associated step is even.","The all-one multi-index contributes one after normalization by $2^L$.\nWhen $k$ is even, each even step supplies exactly one further tied term,\nagain with normalized residue one. For example,\n$\\binom H2=H(H-1)/2$ attains $v_2(H)-1$ exactly when $H$ is even;\nall remaining factors and the exact normalized order-$k+1$ difference\nare odd. Thus the total normalized parity is one for odd $k$, and\n$1+e$ for even $k$. This proves the classification.\nIn the homogeneous case, $e$ is zero or $k$, so it is even whenever\n$k$ is even, proving \\eqref{eq:zero-one-homogeneous-exact}.\n\\end{proof}","ne multi-index contributes one after normalization by $2^L$.\nWhen $k$ is even, each even step supplies exactly one further tied term,\nagain with normalized residue one. For example,\n$\\binom H2=H(H-1)/2$ attains $v_2(H)-1$ exactly when $H$ is even;\nall remaining factors and the exact normalized order-$k+1$ difference\nare odd. Thus the total normalized parity is one for odd $k$, and\n$1+e$ for even $k$. This proves the classification.\nIn the homogeneous case, $e$ is zero or $k$, so it is even whenever\n$k$ is even, proving \\eqref{eq:zero-one-homogeneous-exact}.\n\\end{proof}","hen $k$ is even, each even step supplies exactly one further tied term,\nagain with normalized residue one. For example,\n$\\binom H2=H(H-1)/2$ attains $v_2(H)-1$ exactly when $H$ is even;\nall remaining factors and the exact normalized order-$k+1$ difference\nare odd. Thus the total normalized parity is one for odd $k$, and\n$1+e$ for even $k$. This proves the classification.\nIn the homogeneous case, $e$ is zero or $k$, so it is even whenever\n$k$ is even, proving \\eqref{eq:zero-one-homogeneous-exact}.\n\\end{proof}",", each even step supplies exactly one further tied term,\nagain with normalized residue one. For example,\n$\\binom H2=H(H-1)/2$ attains $v_2(H)-1$ exactly when $H$ is even;\nall remaining factors and the exact normalized order-$k+1$ difference\nare odd. Thus the total normalized parity is one for odd $k$, and\n$1+e$ for even $k$. This proves the classification.\nIn the homogeneous case, $e$ is zero or $k$, so it is even whenever\n$k$ is even, proving \\eqref{eq:zero-one-homogeneous-exact}.\n\\end{proof}","actly one further tied term,\nagain with normalized residue one. For example,\n$\\binom H2=H(H-1)/2$ attains $v_2(H)-1$ exactly when $H$ is even;\nall remaining factors and the exact normalized order-$k+1$ difference\nare odd. Thus the total normalized parity is one for odd $k$, and\n$1+e$ for even $k$. This proves the classification.\nIn the homogeneous case, $e$ is zero or $k$, so it is even whenever\n$k$ is even, proving \\eqref{eq:zero-one-homogeneous-exact}.\n\\end{proof}","\\begin{remark}[A mixed cancellation]\nThe exceptional class can gain more than one bit. The exact values\n\\[\n \\Delta_1\\Delta_2C_0(0)=-33728,\\qquad\n \\Delta_1\\Delta_2C_1(0)=-\\frac{150208}{585}\n\\]\nboth have valuation six, whereas their baseline $L$ is four.\nThe theorem classifies attainment of the baseline; it does not give a\ncomplete deeper valuation in the cancellation class.\n\\end{remark}","\\begin{theorem}[Exact address-two power orders]\n\\label{thm:two-exact-power-orders}\nFor every starting row $n\\geq0$ and all positive mixed steps,\n\\begin{equation}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}C_2(n))\n       =V_2(k)+\\sum_{\\nu=1}^k v_2(H_\\nu)\n \\label{eq:two-power-orders-exact}\n\\end{equation}\nwhenever $k$ is a positive power of two, where\n\\[\n V_2(1)=4,\\qquad V_2(2)=6,\\qquad\n V_2(k)=2k+v_2(k!)+\\frac{k}{2}\\quad(k\\geq4).\n\\]\nFor the original coefficients, the right side gains one bit.\n\\end{theorem}","\\begin{proof}\nFor $k=2^r\\geq4$, put $q=k/2$, which is even. Normalize the exact\nexpansion \\eqref{eq:address-two-filtration-pd} by $2^q$ before reducing\nmodulo two. Its $u$th auxiliary term has excess\n$u+\\lfloor(k-u)/2\\rfloor-q=\\lfloor u/2\\rfloor$.\nOnly auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\","\\begin{proof}\nFor $k=2^r\\geq4$, put $q=k/2$, which is even. Normalize the exact\nexpansion \\eqref{eq:address-two-filtration-pd} by $2^q$ before reducing\nmodulo two. Its $u$th auxiliary term has excess\n$u+\\lfloor(k-u)/2\\rfloor-q=\\lfloor u/2\\rfloor$.\nOnly auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter","\\begin{proof}\nFor $k=2^r\\geq4$, put $q=k/2$, which is even. Normalize the exact\nexpansion \\eqref{eq:address-two-filtration-pd} by $2^q$ before reducing\nmodulo two. Its $u$th auxiliary term has excess\n$u+\\lfloor(k-u)/2\\rfloor-q=\\lfloor u/2\\rfloor$.\nOnly auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to","\\begin{proof}\nFor $k=2^r\\geq4$, put $q=k/2$, which is even. Normalize the exact\nexpansion \\eqref{eq:address-two-filtration-pd} by $2^q$ before reducing\nmodulo two. Its $u$th auxiliary term has excess\n$u+\\lfloor(k-u)/2\\rfloor-q=\\lfloor u/2\\rfloor$.\nOnly auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)","\\begin{proof}\nFor $k=2^r\\geq4$, put $q=k/2$, which is even. Normalize the exact\nexpansion \\eqref{eq:address-two-filtration-pd} by $2^q$ before reducing\nmodulo two. Its $u$th auxiliary term has excess\n$u+\\lfloor(k-u)/2\\rfloor-q=\\lfloor u/2\\rfloor$.\nOnly auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}","\\begin{proof}\nFor $k=2^r\\geq4$, put $q=k/2$, which is even. Normalize the exact\nexpansion \\eqref{eq:address-two-filtration-pd} by $2^q$ before reducing\nmodulo two. Its $u$th auxiliary term has excess\n$u+\\lfloor(k-u)/2\\rfloor-q=\\lfloor u/2\\rfloor$.\nOnly auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\m…","2^q$ before reducing\nmodulo two. Its $u$th auxiliary term has excess\n$u+\\lfloor(k-u)/2\\rfloor-q=\\lfloor u/2\\rfloor$.\nOnly auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. …","ulo two. Its $u$th auxiliary term has excess\n$u+\\lfloor(k-u)/2\\rfloor-q=\\lfloor u/2\\rfloor$.\nOnly auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernou…","term has excess\n$u+\\lfloor(k-u)/2\\rfloor-q=\\lfloor u/2\\rfloor$.\nOnly auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this require…","$u+\\lfloor(k-u)/2\\rfloor-q=\\lfloor u/2\\rfloor$.\nOnly auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen …","auxiliary indices zero and one can therefore survive:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. He…","ve:\n\\begin{equation}\n \\frac{\\gamma_k(H_2)}{2^q}\n \\equiv V^q\\gamma_k(X)\n       +\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","+\\epsilon V^{q-1}G\\gamma_{k-1}(X)\n                                \\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","\\pmod{2\\mathcal A_2}.\n \\label{eq:two-power-leading-normalized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","malized}\n\\end{equation}\nHere $V\\equiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","quiv\\tau\\pmod2$ and $\\tau^2\\equiv1\\pmod2$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","$.\nSince $q$ is even and $\\epsilon\\tau\\equiv\\epsilon\\pmod2$, this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","this reduces\nto $\\gamma_k(X)+\\epsilon G\\gamma_{k-1}(X)$.\nMoreover, $G\\equiv\\gamma_2(X)+\\tau Q\\pmod2$.\nAfter evaluation, every term with a single $\\tau$ vanishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","nishes, so the\nsecond contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","d contribution reduces to\n\\[\n \\mathcal L\\bigl((\\gamma_2(X)+Q)\\gamma_{k-1}(X)\\bigr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","igr).\n\\]\nThe $\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","\\gamma_2(X)$ product has scalar $\\binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","binom{k+1}{2}$, which is even\nbecause $4\\mid k$. The other contribution expands as\n\\begin{equation}\n \\mathcal L(Q\\gamma_{k-1}(P+Q))\n    =\\sum_{a=0}^{k-1}(k-a)\\mu_a\\mu_{k-a}.\n \\label{eq:two-power-Q-evaluation}\n\\end{equation}\nAn odd summand would require $k-a$ to be odd and $\\mu_{k-a}$ to be\nodd. By Lemma~\\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.\nThen $a=k-1$ is not a power of two and $\\mu_a$ is even. Hence the\nsecond contribution is zero modulo two.","For the first contribution, independence gives\n$\\mathcal L(\\gamma_k(P+Q))=\\sum_{a=0}^k\\mu_a\\mu_{k-a}$.\nThe parity generating series of these convolution moments is\n\\[\n (1+t+t^2+t^4+t^8+\\cdots)^2\n        =1+t^2+t^4+t^8+\\cdots\\quad\\text{in }\\mathbb F_2[[t]].\n\\]\nIts coefficient is one at every $k=2^r\\geq4$. Thus the normalized\nmoment in \\eqref{eq:two-power-leading-normalized} is odd and\n\\[\n v_2(\\Delta_1^kC_2(0))=2k+v_2(k!)+k/2.\n\\]\nInserting $U^{4n}=(1+4H_2)^n$ introduces only corrections with strictly\nhigher valuation, by the filtered bound at orders $k+s$ for $s>0$.\nThe equality consequently holds at every starting row.","ves\n$\\mathcal L(\\gamma_k(P+Q))=\\sum_{a=0}^k\\mu_a\\mu_{k-a}$.\nThe parity generating series of these convolution moments is\n\\[\n (1+t+t^2+t^4+t^8+\\cdots)^2\n        =1+t^2+t^4+t^8+\\cdots\\quad\\text{in }\\mathbb F_2[[t]].\n\\]\nIts coefficient is one at every $k=2^r\\geq4$. Thus the normalized\nmoment in \\eqref{eq:two-power-leading-normalized} is odd and\n\\[\n v_2(\\Delta_1^kC_2(0))=2k+v_2(k!)+k/2.\n\\]\nInserting $U^{4n}=(1+4H_2)^n$ introduces only corrections with strictly\nhigher valuation, by the filtered bound at orders $k+s$ for $s>0$.\nThe equality consequently holds at every starting row.","ments is\n\\[\n (1+t+t^2+t^4+t^8+\\cdots)^2\n        =1+t^2+t^4+t^8+\\cdots\\quad\\text{in }\\mathbb F_2[[t]].\n\\]\nIts coefficient is one at every $k=2^r\\geq4$. Thus the normalized\nmoment in \\eqref{eq:two-power-leading-normalized} is odd and\n\\[\n v_2(\\Delta_1^kC_2(0))=2k+v_2(k!)+k/2.\n\\]\nInserting $U^{4n}=(1+4H_2)^n$ introduces only corrections with strictly\nhigher valuation, by the filtered bound at orders $k+s$ for $s>0$.\nThe equality consequently holds at every starting row.","[\n (1+t+t^2+t^4+t^8+\\cdots)^2\n        =1+t^2+t^4+t^8+\\cdots\\quad\\text{in }\\mathbb F_2[[t]].\n\\]\nIts coefficient is one at every $k=2^r\\geq4$. Thus the normalized\nmoment in \\eqref{eq:two-power-leading-normalized} is odd and\n\\[\n v_2(\\Delta_1^kC_2(0))=2k+v_2(k!)+k/2.\n\\]\nInserting $U^{4n}=(1+4H_2)^n$ introduces only corrections with strictly\nhigher valuation, by the filtered bound at orders $k+s$ for $s>0$.\nThe equality consequently holds at every starting row.","For mixed steps the all-one local multi-index gives the claimed leading\nterm. A higher term of total order $M>k$ has excess at least\n\\[\n 2(M-k)+\\left\\lfloor\\frac M2\\right\\rfloor-\\frac k2\n       +v_2(M!)-v_2(k!)-\\sum_\\nu v_2(m_\\nu)\n       \\geq M-k\\geq1.\n\\]\nHence the leading term cannot cancel.","At order one, $a_1$ in \\eqref{eq:address-two-initial-newton} has\nvaluation four and the Newton tail starts at valuation six; its shifted\nvalue remains exact. In the step expansion a local order two can lose\nat most one bit through its binomial coefficient, so every correction\nis still above $4+v_2(H_1)$. At order two, $a_2$ has valuation six and\nthe next coefficient has lower bound eight from the stronger\nfiltration. A total-order-three correction loses at most one bit,\nso it stays above the leading valuation six, while later orders gain\nstill more. The same argument proves exactness for both mixed steps.\n\\end{proof}","\\begin{lemma}[A first-difference criterion]\n\\label{lem:newton-exact-isometry-criterion}\nLet $C:\\mathbb Z_{\\geq0}\\to\\mathbb Z_2$ have Newton coefficients\n$a_k=\\Delta_1^kC(0)$. Suppose for an integer $b\\geq0$ that\n\\begin{equation}\n v_2(a_1)=b,\\qquad\n v_2(a_k)\\geq b+1+\\lfloor\\log_2k\\rfloor\\quad(k\\geq2).\n \\label{eq:newton-isometry-criterion}\n\\end{equation}\nThen for every $n\\geq0$ and $H>0$,\n\\begin{equation}\n v_2(C(n+H)-C(n))=b+v_2(H).\n \\label{eq:newton-isometry-conclusion}\n\\end{equation}\nThe Newton interpolation extends continuously to $\\mathbb Z_2$ and\nsatisfies the same identity for any two distinct dyadic arguments.\n\\end{lemma}","{lemma}[A first-difference criterion]\n\\label{lem:newton-exact-isometry-criterion}\nLet $C:\\mathbb Z_{\\geq0}\\to\\mathbb Z_2$ have Newton coefficients\n$a_k=\\Delta_1^kC(0)$. Suppose for an integer $b\\geq0$ that\n\\begin{equation}\n v_2(a_1)=b,\\qquad\n v_2(a_k)\\geq b+1+\\lfloor\\log_2k\\rfloor\\quad(k\\geq2).\n \\label{eq:newton-isometry-criterion}\n\\end{equation}\nThen for every $n\\geq0$ and $H>0$,\n\\begin{equation}\n v_2(C(n+H)-C(n))=b+v_2(H).\n \\label{eq:newton-isometry-conclusion}\n\\end{equation}\nThe Newton interpolation extends continuously to $\\mathbb Z_2$ and\nsatisfies the same identity for any two distinct dyadic arguments.\n\\end{lemma}","coefficients\n$a_k=\\Delta_1^kC(0)$. Suppose for an integer $b\\geq0$ that\n\\begin{equation}\n v_2(a_1)=b,\\qquad\n v_2(a_k)\\geq b+1+\\lfloor\\log_2k\\rfloor\\quad(k\\geq2).\n \\label{eq:newton-isometry-criterion}\n\\end{equation}\nThen for every $n\\geq0$ and $H>0$,\n\\begin{equation}\n v_2(C(n+H)-C(n))=b+v_2(H).\n \\label{eq:newton-isometry-conclusion}\n\\end{equation}\nThe Newton interpolation extends continuously to $\\mathbb Z_2$ and\nsatisfies the same identity for any two distinct dyadic arguments.\n\\end{lemma}","\\begin{proof}\nAt a natural argument the exact Newton expansion is finite.\nVandermonde gives\n\\[\n \\binom{n+H}{k}-\\binom nk\n   =\\sum_{i=1}^k\\binom Hi\\binom n{k-i}.\n\\]\nFor $1\\leq i\\leq H$,\n$v_2(\\binom Hi)\\geq v_2(H)-v_2(i)\n \\geq v_2(H)-\\lfloor\\log_2k\\rfloor$; terms with $i>H$ vanish.\nThus each Newton summand with $k\\geq2$ has valuation at least\n$b+v_2(H)+1$. The $k=1$ summand is $Ha_1$, of exact valuation\n$b+v_2(H)$, proving \\eqref{eq:newton-isometry-conclusion}.","{proof}\nAt a natural argument the exact Newton expansion is finite.\nVandermonde gives\n\\[\n \\binom{n+H}{k}-\\binom nk\n   =\\sum_{i=1}^k\\binom Hi\\binom n{k-i}.\n\\]\nFor $1\\leq i\\leq H$,\n$v_2(\\binom Hi)\\geq v_2(H)-v_2(i)\n \\geq v_2(H)-\\lfloor\\log_2k\\rfloor$; terms with $i>H$ vanish.\nThus each Newton summand with $k\\geq2$ has valuation at least\n$b+v_2(H)+1$. The $k=1$ summand is $Ha_1$, of exact valuation\n$b+v_2(H)$, proving \\eqref{eq:newton-isometry-conclusion}.","The coefficient hypothesis implies $a_k\\to0$ dyadically, so the\nNewton series converges uniformly on $\\mathbb Z_2$, where\n$\\binom xk$ is integral-valued. For distinct $x,y\\in\\mathbb Z_2$,\napproximate the pair by natural integer pairs with their difference\nvaluation eventually equal to $v_2(x-y)$. The corresponding function\ndifferences have the fixed exact valuation $b+v_2(x-y)$.\nA limit of elements with that fixed valuation has the same valuation:\nthe normalized residues are units, and $\\mathbb Z_2^\\times$ is closed.\n\\end{proof}","ormly on $\\mathbb Z_2$, where\n$\\binom xk$ is integral-valued. For distinct $x,y\\in\\mathbb Z_2$,\napproximate the pair by natural integer pairs with their difference\nvaluation eventually equal to $v_2(x-y)$. The corresponding function\ndifferences have the fixed exact valuation $b+v_2(x-y)$.\nA limit of elements with that fixed valuation has the same valuation:\nthe normalized residues are units, and $\\mathbb Z_2^\\times$ is closed.\n\\end{proof}","\\begin{theorem}[Exact four-address first differences]\n\\label{thm:rank-four-exact-periods}\nFor every $n\\geq0$, $H>0$, and $j\\in\\{0,1,2,3\\}$,\n\\begin{align}\n v_2(C_j(n+H)-C_j(n))&=v_2(H)+b_j,\n      &(b_0,b_1,b_2,b_3)&=(1,1,4,2),\\label{eq:rank-four-normalized-isometry}","\\begin{theorem}[Exact four-address first differences]\n\\label{thm:rank-four-exact-periods}\nFor every $n\\geq0$, $H>0$, and $j\\in\\{0,1,2,3\\}$,\n\\begin{align}\n v_2(C_j(n+H)-C_j(n))&=v_2(H)+b_j,\n      &(b_0,b_1,b_2,b_3)&=(1,1,4,2),\\label{eq:rank-four-normalized-isometry}\\\\\n v_2(E_{4(n+H)}^{(4,j)}-E_{4n}^{(4,j)})&=v_2(H)+B_j,\n      &(B_0,B_1,B_2,B_3)&=(1,1,5,3).\n \\label{eq:rank-four-original-isometry}\n\\end{align}\nConsequently, the least positive row period modulo $2^r$ is exactly\n\\begin{equation}\n P_j(r)=2^{\\max(0,r-B_j)}\\qquad(r\\geq1).\n \\label{eq:rank-four-least-period-final}\n\\end{equation}\nThe corresponding displacement in the original Euler subscript is\n$4P_j(r)$.\n\\end{theorem}","v_2(E_{4(n+H)}^{(4,j)}-E_{4n}^{(4,j)})&=v_2(H)+B_j,\n      &(B_0,B_1,B_2,B_3)&=(1,1,5,3).\n \\label{eq:rank-four-original-isometry}\n\\end{align}\nConsequently, the least positive row period modulo $2^r$ is exactly\n\\begin{equation}\n P_j(r)=2^{\\max(0,r-B_j)}\\qquad(r\\geq1).\n \\label{eq:rank-four-least-period-final}\n\\end{equation}\nThe corresponding displacement in the original Euler subscript is\n$4P_j(r)$.\n\\end{theorem}","\\begin{proof}\nThe actual reciprocal recurrence is\n\\begin{equation}\n C_{N,j}(n)=-\\sum_{q=1}^n\\binom{Nn}{Nq}\n           g_{N,j}(q)C_{N,j}(n-q),\\qquad C_{N,j}(0)=1.\n \\label{eq:exact-reciprocal-row-recurrence}\n\\end{equation}\nFor $N=4$, the four kernels are\n\\[\n 1,\\qquad\\frac1{4q+1},\\qquad\n \\frac1{(4q+1)(2q+1)},\\qquad\n \\frac3{(4q+1)(2q+1)(4q+3)}.\n\\]\nThe threshold-free theorem gives $v_2(a_k)\\geq k$ in every address.\nThe following exact values cover all orders where that bound does not\nalready imply \\eqref{eq:newton-isometry-criterion}:\n\\begin{center}\n\\begin{tabular}{c r l r}\n$j$&$k$&$a_k$&$v_2(a_k)$","t-reciprocal-row-recurrence}\n\\end{equation}\nFor $N=4$, the four kernels are\n\\[\n 1,\\qquad\\frac1{4q+1},\\qquad\n \\frac1{(4q+1)(2q+1)},\\qquad\n \\frac3{(4q+1)(2q+1)(4q+3)}.\n\\]\nThe threshold-free theorem gives $v_2(a_k)\\geq k$ in every address.\nThe following exact values cover all orders where that bound does not\nalready imply \\eqref{eq:newton-isometry-criterion}:\n\\begin{center}\n\\begin{tabular}{c r l r}\n$j$&$k$&$a_k$&$v_2(a_k)$","ciprocal-row-recurrence}\n\\end{equation}\nFor $N=4$, the four kernels are\n\\[\n 1,\\qquad\\frac1{4q+1},\\qquad\n \\frac1{(4q+1)(2q+1)},\\qquad\n \\frac3{(4q+1)(2q+1)(4q+3)}.\n\\]\nThe threshold-free theorem gives $v_2(a_k)\\geq k$ in every address.\nThe following exact values cover all orders where that bound does not\nalready imply \\eqref{eq:newton-isometry-criterion}:\n\\begin{center}\n\\begin{tabular}{c r l r}\n$j$&$k$&$a_k$&$v_2(a_k)$","ocal-row-recurrence}\n\\end{equation}\nFor $N=4$, the four kernels are\n\\[\n 1,\\qquad\\frac1{4q+1},\\qquad\n \\frac1{(4q+1)(2q+1)},\\qquad\n \\frac3{(4q+1)(2q+1)(4q+3)}.\n\\]\nThe threshold-free theorem gives $v_2(a_k)\\geq k$ in every address.\nThe following exact values cover all orders where that bound does not\nalready imply \\eqref{eq:newton-isometry-criterion}:\n\\begin{center}\n\\begin{tabular}{c r l r}\n$j$&$k$&$a_k$&$v_2(a_k)$","ow-recurrence}\n\\end{equation}\nFor $N=4$, the four kernels are\n\\[\n 1,\\qquad\\frac1{4q+1},\\qquad\n \\frac1{(4q+1)(2q+1)},\\qquad\n \\frac3{(4q+1)(2q+1)(4q+3)}.\n\\]\nThe threshold-free theorem gives $v_2(a_k)\\geq k$ in every address.\nThe following exact values cover all orders where that bound does not\nalready imply \\eqref{eq:newton-isometry-criterion}:\n\\begin{center}\n\\begin{tabular}{c r l r}\n$j$&$k$&$a_k$&$v_2(a_k)$","hline\n0&1&$-2$&1","0&2&$72$&3","1&1&$-6/5$&1","1&2&$184/45$&3","2&1&$-16/15$&4","2&2&$64/45$&6","2&3&$-14848/1365$&9","2&4&$71426048/69615$&13","2&5&$-25509691392/85085$&16","2&6&$149113198870528/765765$&22","3&1&$-36/35$&2","3&2&$256/231$&8","3&3&$-65792/35035$&8","3&4&$161644544/4849845$&15\n\\end{tabular}\n\\end{center}\nThey are finite rational evaluations of\n\\eqref{eq:exact-reciprocal-row-recurrence}, followed by\n$a_k=\\sum_{h=0}^k(-1)^{k-h}\\binom kh C_j(h)$.\nFor addresses zero and one, $k\\geq3$ is covered by the generic bound;\nfor address two, $k\\geq7$ is covered; for address three, $k\\geq5$ is\ncovered. In each range the elementary inequality is\n$k-\\lfloor\\log_2k\\rfloor\\geq b_j+1$.\nLemma~\\ref{lem:newton-exact-isometry-criterion} proves the normalized\nidentity. Multiplication by $j!$ proves the original identity.","w-recurrence}, followed by\n$a_k=\\sum_{h=0}^k(-1)^{k-h}\\binom kh C_j(h)$.\nFor addresses zero and one, $k\\geq3$ is covered by the generic bound;\nfor address two, $k\\geq7$ is covered; for address three, $k\\geq5$ is\ncovered. In each range the elementary inequality is\n$k-\\lfloor\\log_2k\\rfloor\\geq b_j+1$.\nLemma~\\ref{lem:newton-exact-isometry-criterion} proves the normalized\nidentity. Multiplication by $j!$ proves the original identity.","A row displacement $H$ preserves the original coefficient modulo\n$2^r$ at even one row exactly when $v_2(H)+B_j\\geq r$.\nThe least positive such integer is\n$2^{\\max(0,r-B_j)}$, and the same displacement works at every row.\nThis proves both optimality and the absence of a starting threshold.\n\\end{proof}","\\begin{corollary}[Exact image and residue enumeration]\n\\label{cor:four-exact-images}\nLet $\\widetilde C_j$ denote the dyadic Newton interpolation.\nThe map\n\\[\n T_j(x)=\\frac{\\widetilde C_j(x)-1}{2^{b_j}}\n\\]\nis a bijective isometry $\\mathbb Z_2\\to\\mathbb Z_2$, and\n\\begin{equation}\n \\widetilde C_j(\\mathbb Z_2)=1+2^{b_j}\\mathbb Z_2,\n \\qquad j!\\widetilde C_j(\\mathbb Z_2)=j!+2^{B_j}\\mathbb Z_2.\n \\label{eq:four-interpolation-images-final}\n\\end{equation}\nFor $r>B_j$, one least period modulo $2^r$ visits every residue in\n$j!+2^{B_j}\\mathbb Z/2^r\\mathbb Z$ exactly once. For $r\\leq B_j$,\nthere is one residue and the least row period is one.\n\\end{corollary}","\\label{cor:four-exact-images}\nLet $\\widetilde C_j$ denote the dyadic Newton interpolation.\nThe map\n\\[\n T_j(x)=\\frac{\\widetilde C_j(x)-1}{2^{b_j}}\n\\]\nis a bijective isometry $\\mathbb Z_2\\to\\mathbb Z_2$, and\n\\begin{equation}\n \\widetilde C_j(\\mathbb Z_2)=1+2^{b_j}\\mathbb Z_2,\n \\qquad j!\\widetilde C_j(\\mathbb Z_2)=j!+2^{B_j}\\mathbb Z_2.\n \\label{eq:four-interpolation-images-final}\n\\end{equation}\nFor $r>B_j$, one least period modulo $2^r$ visits every residue in\n$j!+2^{B_j}\\mathbb Z/2^r\\mathbb Z$ exactly once. For $r\\leq B_j$,\nthere is one residue and the least row period is one.\n\\end{corollary}","t $\\widetilde C_j$ denote the dyadic Newton interpolation.\nThe map\n\\[\n T_j(x)=\\frac{\\widetilde C_j(x)-1}{2^{b_j}}\n\\]\nis a bijective isometry $\\mathbb Z_2\\to\\mathbb Z_2$, and\n\\begin{equation}\n \\widetilde C_j(\\mathbb Z_2)=1+2^{b_j}\\mathbb Z_2,\n \\qquad j!\\widetilde C_j(\\mathbb Z_2)=j!+2^{B_j}\\mathbb Z_2.\n \\label{eq:four-interpolation-images-final}\n\\end{equation}\nFor $r>B_j$, one least period modulo $2^r$ visits every residue in\n$j!+2^{B_j}\\mathbb Z/2^r\\mathbb Z$ exactly once. For $r\\leq B_j$,\nthere is one residue and the least row period is one.\n\\end{corollary}","dyadic Newton interpolation.\nThe map\n\\[\n T_j(x)=\\frac{\\widetilde C_j(x)-1}{2^{b_j}}\n\\]\nis a bijective isometry $\\mathbb Z_2\\to\\mathbb Z_2$, and\n\\begin{equation}\n \\widetilde C_j(\\mathbb Z_2)=1+2^{b_j}\\mathbb Z_2,\n \\qquad j!\\widetilde C_j(\\mathbb Z_2)=j!+2^{B_j}\\mathbb Z_2.\n \\label{eq:four-interpolation-images-final}\n\\end{equation}\nFor $r>B_j$, one least period modulo $2^r$ visits every residue in\n$j!+2^{B_j}\\mathbb Z/2^r\\mathbb Z$ exactly once. For $r\\leq B_j$,\nthere is one residue and the least row period is one.\n\\end{corollary}","{\\widetilde C_j(x)-1}{2^{b_j}}\n\\]\nis a bijective isometry $\\mathbb Z_2\\to\\mathbb Z_2$, and\n\\begin{equation}\n \\widetilde C_j(\\mathbb Z_2)=1+2^{b_j}\\mathbb Z_2,\n \\qquad j!\\widetilde C_j(\\mathbb Z_2)=j!+2^{B_j}\\mathbb Z_2.\n \\label{eq:four-interpolation-images-final}\n\\end{equation}\nFor $r>B_j$, one least period modulo $2^r$ visits every residue in\n$j!+2^{B_j}\\mathbb Z/2^r\\mathbb Z$ exactly once. For $r\\leq B_j$,\nthere is one residue and the least row period is one.\n\\end{corollary}","\\begin{proof}\nThe exact difference identity gives\n$v_2(T_j(x)-T_j(y))=v_2(x-y)$ and $T_j(0)=0$.\nFor every $s\\geq1$, the map induces an injection of\n$\\mathbb Z_2/2^s\\mathbb Z_2$ into itself. It is a bijection because\nboth sets have $2^s$ elements. These quotient bijections are compatible\nwith reduction. Inverting them gives a compatible preimage system for\nevery dyadic integer, proving surjectivity. Multiplication by $j!$\nchanges the image ideal by exactly $v_2(j!)$.\nThe finite quotient bijection is also precisely the residue enumeration\nin a least row period.\n\\end{proof}","ct difference identity gives\n$v_2(T_j(x)-T_j(y))=v_2(x-y)$ and $T_j(0)=0$.\nFor every $s\\geq1$, the map induces an injection of\n$\\mathbb Z_2/2^s\\mathbb Z_2$ into itself. It is a bijection because\nboth sets have $2^s$ elements. These quotient bijections are compatible\nwith reduction. Inverting them gives a compatible preimage system for\nevery dyadic integer, proving surjectivity. Multiplication by $j!$\nchanges the image ideal by exactly $v_2(j!)$.\nThe finite quotient bijection is also precisely the residue enumeration\nin a least row period.\n\\end{proof}","-T_j(y))=v_2(x-y)$ and $T_j(0)=0$.\nFor every $s\\geq1$, the map induces an injection of\n$\\mathbb Z_2/2^s\\mathbb Z_2$ into itself. It is a bijection because\nboth sets have $2^s$ elements. These quotient bijections are compatible\nwith reduction. Inverting them gives a compatible preimage system for\nevery dyadic integer, proving surjectivity. Multiplication by $j!$\nchanges the image ideal by exactly $v_2(j!)$.\nThe finite quotient bijection is also precisely the residue enumeration\nin a least row period.\n\\end{proof}","\\begin{theorem}[Lower-half exact laws]\n\\label{thm:lower-half-all-dyadic-isometry}\nLet $N=2^a\\geq4$ and $0\\leq j<N/2$. Then for all $n\\geq0$ and $H>0$,\n\\begin{equation}\n v_2(C_{N,j}(n+H)-C_{N,j}(n))=1+v_2(H).\n \\label{eq:lower-half-all-dyadic-exact}\n\\end{equation}\nIts normalized interpolation is a bijection onto $1+2\\mathbb Z_2$.\nFor the original coefficients, the least row period modulo $2^r$ is\n$2^{\\max(0,r-1-v_2(j!))}$.\n\\end{theorem}","\\begin{proof}\nEvery slope $N/i$ in the source has valuation at least two, since\n$i\\leq j<N/2$. Hence $g(1)\\equiv1\\pmod4$ and\n$g(2)\\equiv1\\pmod8$. The first two Newton coefficients from the\nreciprocal recurrence are\n\\begin{equation}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\e","\\begin{proof}\nEvery slope $N/i$ in the source has valuation at least two, since\n$i\\leq j<N/2$. Hence $g(1)\\equiv1\\pmod4$ and\n$g(2)\\equiv1\\pmod8$. The first two Newton coefficients from the\nreciprocal recurrence are\n\\begin{equation}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\ge","\\begin{proof}\nEvery slope $N/i$ in the source has valuation at least two, since\n$i\\leq j<N/2$. Hence $g(1)\\equiv1\\pmod4$ and\n$g(2)\\equiv1\\pmod8$. The first two Newton coefficients from the\nreciprocal recurrence are\n\\begin{equation}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criteri","\\begin{proof}\nEvery slope $N/i$ in the source has valuation at least two, since\n$i\\leq j<N/2$. Hence $g(1)\\equiv1\\pmod4$ and\n$g(2)\\equiv1\\pmod8$. The first two Newton coefficients from the\nreciprocal recurrence are\n\\begin{equation}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The perio","\\begin{proof}\nEvery slope $N/i$ in the source has valuation at least two, since\n$i\\leq j<N/2$. Hence $g(1)\\equiv1\\pmod4$ and\n$g(2)\\equiv1\\pmod8$. The first two Newton coefficients from the\nreciprocal recurrence are\n\\begin{equation}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","$N/i$ in the source has valuation at least two, since\n$i\\leq j<N/2$. Hence $g(1)\\equiv1\\pmod4$ and\n$g(2)\\equiv1\\pmod8$. The first two Newton coefficients from the\nreciprocal recurrence are\n\\begin{equation}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","at least two, since\n$i\\leq j<N/2$. Hence $g(1)\\equiv1\\pmod4$ and\n$g(2)\\equiv1\\pmod8$. The first two Newton coefficients from the\nreciprocal recurrence are\n\\begin{equation}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","o Newton coefficients from the\nreciprocal recurrence are\n\\begin{equation}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","ents from the\nreciprocal recurrence are\n\\begin{equation}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","rocal recurrence are\n\\begin{equation}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","}\n a_1=-g(1)-1,\\qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","qquad\n a_2=\\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","2-g(2)+2g(1)+1.\n \\label{eq:lower-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","r-half-initial-newton}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","on}\n\\end{equation}\nThe first has valuation exactly one. Also\n$\\binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","binom{2N}{N}\\equiv6\\pmod8$ for every $N=2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","2^a\\geq2$.\nTo see this, write\n\\[\n B_a=\\binom{2^{a+1}}{2^a}\n       =2\\prod_{h=1}^{2^a-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","-1}\\left(1+\\frac{2^a}{h}\\right).\n\\]\nThe even-$h$ factors give $B_{a-1}$; the odd-$h$ factors are one\nmodulo four for $a\\geq2$. Since $B_{a-1}$ has valuation one,\n$B_a\\equiv B_{a-1}\\pmod8$, and $B_1=6$.\nTherefore $a_2\\equiv6-1+2+1=0\\pmod8$.\nThe global Mahler bound covers all $k\\geq3$, completing the criterion\nwith $b=1$. The period and image deductions are the same as above.\nThe proof also covers $N=2,j=0$, whose kernel is identically one.\n\\end{proof}","\\begin{proposition}[Small-rank complete isometry table]\n\\label{prop:two-four-eight-isometry-table}\nEvery address at $N\\in\\{2,4,8\\}$ satisfies\n\\[\n v_2(C_{N,j}(n+H)-C_{N,j}(n))=b_{N,j}+v_2(H).\n\\]\nThe following entries are ordered by $j=0,\\ldots,N-1$:\n\\begin{center}\n\\begin{tabular}{c l l}\n$N$&normalized shifts $b_{N,j}$&original shifts $B_{N,j}=b_{N,j}+v_2(j!)$","hline\n2&$1,2$&$1,2$","4&$1,1,4,2$&$1,1,5,3$","8&$1,1,1,1,4,3,2,2$&$1,1,2,2,7,6,6,6$\n\\end{tabular}\n\\end{center}\nFor each entry, the least original row period is\n$2^{\\max(0,r-B_{N,j})}$, its Euler subscript displacement is $N$ times\nthat period, and the exact-image and residue-enumeration conclusions\nhold with the corresponding shifts.\n\\end{proposition}","\\begin{proof}\nRank four was proved above. At ranks two and eight the finite\ncriterion is verified by the actual rational recurrence\n\\eqref{eq:exact-reciprocal-row-recurrence}. The following valuation\nlists record every initial Newton coefficient needed in addition to\nthe generic tail $v_2(a_k)\\geq k$:\n\\begin{center}\n\\begin{tabular}{c c c l}\n$N$&$j$&$b_{N,j}$&$v_2(a_1),v_2(a_2),\\ldots$ through the finite cutoff","hline\n2&0&1&$1,3$","2&1&2&$2,5,9,11$","8&0&1&$1,3$","8&1&1&$1,3$","8&2&1&$1,3$","8&3&1&$1,3$","8&4&4&$4,6,9,13,16,20$","8&5&3&$3,6,10,13,17$","8&6&2&$2,6,8,13$","8&7&2&$2,11,8,15$\n\\end{tabular}\n\\end{center}\nThe first entries can also be read directly from\n$a_1=-1-\\binom{N+j}{j}^{-1}$. At rank eight they are respectively\n\\[\n -2,\\ -\\frac{10}{9},\\ -\\frac{46}{45},\\ -\\frac{166}{165},\\\n -\\frac{496}{495},\\ -\\frac{1288}{1287},\\\n -\\frac{3004}{3003},\\ -\\frac{6436}{6435}.\n\\]\nAll later entries in the table are finite rational arithmetic, with\nfull numerator and denominator records retained in the accompanying\nexact certificate \\texttt{all\\_rank\\_extensions\\_verified.json}.\nEvery entry with $k\\geq2$ meets\n$b_{N,j}+1+\\lfloor\\log_2k\\rfloor$. Beyond the displayed cutoff, the\ninequality $k\\geq b_{N,j}+1+\\lfloor\\log_2k\\rfloor$ holds at every\norder. Thus the finite table and the prove","8&7&2&$2,11,8,15$\n\\end{tabular}\n\\end{center}\nThe first entries can also be read directly from\n$a_1=-1-\\binom{N+j}{j}^{-1}$. At rank eight they are respectively\n\\[\n -2,\\ -\\frac{10}{9},\\ -\\frac{46}{45},\\ -\\frac{166}{165},\\\n -\\frac{496}{495},\\ -\\frac{1288}{1287},\\\n -\\frac{3004}{3003},\\ -\\frac{6436}{6435}.\n\\]\nAll later entries in the table are finite rational arithmetic, with\nfull numerator and denominator records retained in the accompanying\nexact certificate \\texttt{all\\_rank\\_extensions\\_verified.json}.\nEvery entry with $k\\geq2$ meets\n$b_{N,j}+1+\\lfloor\\log_2k\\rfloor$. Beyond the displayed cutoff, the\ninequality $k\\geq b_{N,j}+1+\\lfloor\\log_2k\\rfloor$ holds at every\norder. Thus the finite table and the proved infinite tail establish\nthe full hypothesis of Lemma~\\ref{lem:newton-exact-isometry-criterion}.\nThis is","8&7&2&$2,11,8,15$\n\\end{tabular}\n\\end{center}\nThe first entries can also be read directly from\n$a_1=-1-\\binom{N+j}{j}^{-1}$. At rank eight they are respectively\n\\[\n -2,\\ -\\frac{10}{9},\\ -\\frac{46}{45},\\ -\\frac{166}{165},\\\n -\\frac{496}{495},\\ -\\frac{1288}{1287},\\\n -\\frac{3004}{3003},\\ -\\frac{6436}{6435}.\n\\]\nAll later entries in the table are finite rational arithmetic, with\nfull numerator and denominator records retained in the accompanying\nexact certificate \\texttt{all\\_rank\\_extensions\\_verified.json}.\nEvery entry with $k\\geq2$ meets\n$b_{N,j}+1+\\lfloor\\log_2k\\rfloor$. Beyond the displayed cutoff, the\ninequality $k\\geq b_{N,j}+1+\\lfloor\\log_2k\\rfloor$ holds at every\norder. Thus the finite table and the proved infinite tail establish\nthe full hypothesis of Lemma~\\ref{lem:newton-exact-isometry-criterion}.\nThis is a finite coefficient proof followed by a universal tail\nargument, rather than extrapolation from sampled rows or steps.\n\\end{proof}","rank eight they are respectively\n\\[\n -2,\\ -\\frac{10}{9},\\ -\\frac{46}{45},\\ -\\frac{166}{165},\\\n -\\frac{496}{495},\\ -\\frac{1288}{1287},\\\n -\\frac{3004}{3003},\\ -\\frac{6436}{6435}.\n\\]\nAll later entries in the table are finite rational arithmetic, with\nfull numerator and denominator records retained in the accompanying\nexact certificate \\texttt{all\\_rank\\_extensions\\_verified.json}.\nEvery entry with $k\\geq2$ meets\n$b_{N,j}+1+\\lfloor\\log_2k\\rfloor$. Beyond the displayed cutoff, the\ninequality $k\\geq b_{N,j}+1+\\lfloor\\log_2k\\rfloor$ holds at every\norder. Thus the finite table and the proved infinite tail establish\nthe full hypothesis of Lemma~\\ref{lem:newton-exact-isometry-criterion}.\nThis is a finite coefficient proof followed by a universal tail\nargument, rather than extrapolation from sampled rows or steps.\n\\end{proof}","y are respectively\n\\[\n -2,\\ -\\frac{10}{9},\\ -\\frac{46}{45},\\ -\\frac{166}{165},\\\n -\\frac{496}{495},\\ -\\frac{1288}{1287},\\\n -\\frac{3004}{3003},\\ -\\frac{6436}{6435}.\n\\]\nAll later entries in the table are finite rational arithmetic, with\nfull numerator and denominator records retained in the accompanying\nexact certificate \\texttt{all\\_rank\\_extensions\\_verified.json}.\nEvery entry with $k\\geq2$ meets\n$b_{N,j}+1+\\lfloor\\log_2k\\rfloor$. Beyond the displayed cutoff, the\ninequality $k\\geq b_{N,j}+1+\\lfloor\\log_2k\\rfloor$ holds at every\norder. Thus the finite table and the proved infinite tail establish\nthe full hypothesis of Lemma~\\ref{lem:newton-exact-isometry-criterion}.\nThis is a finite coefficient proof followed by a universal tail\nargument, rather than extrapolation from sampled rows or steps.\n\\end{proof}","-\\frac{496}{495},\\ -\\frac{1288}{1287},\\\n -\\frac{3004}{3003},\\ -\\frac{6436}{6435}.\n\\]\nAll later entries in the table are finite rational arithmetic, with\nfull numerator and denominator records retained in the accompanying\nexact certificate \\texttt{all\\_rank\\_extensions\\_verified.json}.\nEvery entry with $k\\geq2$ meets\n$b_{N,j}+1+\\lfloor\\log_2k\\rfloor$. Beyond the displayed cutoff, the\ninequality $k\\geq b_{N,j}+1+\\lfloor\\log_2k\\rfloor$ holds at every\norder. Thus the finite table and the proved infinite tail establish\nthe full hypothesis of Lemma~\\ref{lem:newton-exact-isometry-criterion}.\nThis is a finite coefficient proof followed by a universal tail\nargument, rather than extrapolation from sampled rows or steps.\n\\end{proof}","\\begin{remark}[Failure of a universal upper-half exact law]\nThe all-rank lower bound remains valid in every address. The exact\nfirst-difference law does not automatically extend to all upper-half\naddresses. Exact rational recurrence calculations give\n\\begin{center}\n\\begin{tabular}{c c c c}\n$(N,j)$&$v_2(C(1)-C(0))$&$v_2(C(2)-C(0))$&naive $b+v_2(2)$","hline\n$(16,9)$&5&7&6","$(32,18)$&6&6&7","$(32,19)$&5&7&6\n\\end{tabular}\n\\end{center}\nFor example, at $(N,j)=(16,9)$ the first difference is\n$-2042976/2042975$, whereas the two-row difference is\n$-2215603929728/2215923048625$. These exact counterexamples refute\nthe additional universal isometry proposal. They preserve the\nthreshold-free lower bound and the original conjecture.\n\\end{remark}","\\begin{theorem}[Taylor coefficient bounds and radii]\n\\label{thm:four-taylor-radii}\nFor $j=0,1,2,3$, the dyadic interpolation has an ordinary power series\n\\[\n \\widetilde C_j(x)=1+\\sum_{\\ell\\geq1}d_{j,\\ell}x^\\ell,\n \\qquad v_2(d_{j,\\ell})\\geq\\ell+\\beta_j.\n\\]\nIts Taylor series converges for $|x|_2<2$.\nAt addresses zero and one,\n\\begin{equation}\n v_2(d_{j,\\ell})=\\ell\\quad(\\ell\\geq1),\n \\qquad R_0=R_1=2.\n \\label{eq:zero-one-exact-taylor-radius}\n\\end{equation}\nAt address two,\n\\begin{equation}\n v_2(d_{2,\\ell})\\geq\n       \\max\\left(2\\ell+\\left\\lfloor\\frac\\ell2\\right\\rfloor,\n                          \\ell+3\\right),\n \\qquad R_2=2^{5/2}.\n \\label{eq:two-exact-taylor-radius}\n\\end{equation}\nThe radius at address three is at least two; its exact radius is not\nasserted here. The stated exact radii at add","\\begin{theorem}[Taylor coefficient bounds and radii]\n\\label{thm:four-taylor-radii}\nFor $j=0,1,2,3$, the dyadic interpolation has an ordinary power series\n\\[\n \\widetilde C_j(x)=1+\\sum_{\\ell\\geq1}d_{j,\\ell}x^\\ell,\n \\qquad v_2(d_{j,\\ell})\\geq\\ell+\\beta_j.\n\\]\nIts Taylor series converges for $|x|_2<2$.\nAt addresses zero and one,\n\\begin{equation}\n v_2(d_{j,\\ell})=\\ell\\quad(\\ell\\geq1),\n \\qquad R_0=R_1=2.\n \\label{eq:zero-one-exact-taylor-radius}\n\\end{equation}\nAt address two,\n\\begin{equation}\n v_2(d_{2,\\ell})\\geq\n       \\max\\left(2\\ell+\\left\\lfloor\\frac\\ell2\\right\\rfloor,\n                          \\ell+3\\right),\n \\qquad R_2=2^{5/2}.\n \\label{eq:two-exact-taylor-radius}\n\\end{equation}\nThe radius at address three is at least two; its exact radius is not\nasserted here. The stated exact radii at addresses zero, one, and two\nhold at every center in $\\mathbb Z_2$.\n\\end{theorem}","=0,1,2,3$, the dyadic interpolation has an ordinary power series\n\\[\n \\widetilde C_j(x)=1+\\sum_{\\ell\\geq1}d_{j,\\ell}x^\\ell,\n \\qquad v_2(d_{j,\\ell})\\geq\\ell+\\beta_j.\n\\]\nIts Taylor series converges for $|x|_2<2$.\nAt addresses zero and one,\n\\begin{equation}\n v_2(d_{j,\\ell})=\\ell\\quad(\\ell\\geq1),\n \\qquad R_0=R_1=2.\n \\label{eq:zero-one-exact-taylor-radius}\n\\end{equation}\nAt address two,\n\\begin{equation}\n v_2(d_{2,\\ell})\\geq\n       \\max\\left(2\\ell+\\left\\lfloor\\frac\\ell2\\right\\rfloor,\n                          \\ell+3\\right),\n \\qquad R_2=2^{5/2}.\n \\label{eq:two-exact-taylor-radius}\n\\end{equation}\nThe radius at address three is at least two; its exact radius is not\nasserted here. The stated exact radii at addresses zero, one, and two\nhold at every center in $\\mathbb Z_2$.\n\\end{theorem}","quation}\nAt address two,\n\\begin{equation}\n v_2(d_{2,\\ell})\\geq\n       \\max\\left(2\\ell+\\left\\lfloor\\frac\\ell2\\right\\rfloor,\n                          \\ell+3\\right),\n \\qquad R_2=2^{5/2}.\n \\label{eq:two-exact-taylor-radius}\n\\end{equation}\nThe radius at address three is at least two; its exact radius is not\nasserted here. The stated exact radii at addresses zero, one, and two\nhold at every center in $\\mathbb Z_2$.\n\\end{theorem}","\\begin{proof}\nPut $a_m=\\Delta_1^mC_j(0)$ and expand the falling factorial in the\nNewton series. The ordinary coefficients are\n\\begin{equation}\n d_{j,\\ell}=\\sum_{m\\geq\\ell}\\frac{a_m}{m!}s(m,\\ell).\n \\label{eq:newton-taylor-stirling}\n\\end{equation}\nSince $s(m,\\ell)$ is integral and\n$v_2(a_m/m!)\\geq m+\\beta_j$, the coefficient sum converges and has\nvaluation at least $\\ell+\\beta_j$. The sum exchange is justified on\n$\\mathbb Z_2$ by these uniform geometric coefficient bounds.\nThe ordinary power series agrees there with the Newton interpolation\nand converges whenever $|x|_2<2$.","t $a_m=\\Delta_1^mC_j(0)$ and expand the falling factorial in the\nNewton series. The ordinary coefficients are\n\\begin{equation}\n d_{j,\\ell}=\\sum_{m\\geq\\ell}\\frac{a_m}{m!}s(m,\\ell).\n \\label{eq:newton-taylor-stirling}\n\\end{equation}\nSince $s(m,\\ell)$ is integral and\n$v_2(a_m/m!)\\geq m+\\beta_j$, the coefficient sum converges and has\nvaluation at least $\\ell+\\beta_j$. The sum exchange is justified on\n$\\mathbb Z_2$ by these uniform geometric coefficient bounds.\nThe ordinary power series agrees there with the Newton interpolation\nand converges whenever $|x|_2<2$.","l}\\frac{a_m}{m!}s(m,\\ell).\n \\label{eq:newton-taylor-stirling}\n\\end{equation}\nSince $s(m,\\ell)$ is integral and\n$v_2(a_m/m!)\\geq m+\\beta_j$, the coefficient sum converges and has\nvaluation at least $\\ell+\\beta_j$. The sum exchange is justified on\n$\\mathbb Z_2$ by these uniform geometric coefficient bounds.\nThe ordinary power series agrees there with the Newton interpolation\nand converges whenever $|x|_2<2$.","For $j=0,1$, the term $m=\\ell$ in\n\\eqref{eq:newton-taylor-stirling} has exact valuation $\\ell$, since\n$s(\\ell,\\ell)=1$ and the Newton valuation is exact. Every later term\nhas valuation at least $\\ell+1$. Thus\n$v_2(d_{j,\\ell})=\\ell$, and the coefficient radius is exactly two.\nAt a center $x_0\\in\\mathbb Z_2$, the coefficient of $y^\\ell$ in the\ntranslated series is\n\\begin{equation}\n \\sum_{m\\geq\\ell}d_{j,m}\\binom m\\ell x_0^{m-\\ell}.\n \\label{eq:taylor-translation-coefficient}\n\\end{equation}\nIts initial term again has exact valuation $\\ell$, and all later\nterms have at least one more bit. The exact radius is therefore two\nat every integral center.","aluation is exact. Every later term\nhas valuation at least $\\ell+1$. Thus\n$v_2(d_{j,\\ell})=\\ell$, and the coefficient radius is exactly two.\nAt a center $x_0\\in\\mathbb Z_2$, the coefficient of $y^\\ell$ in the\ntranslated series is\n\\begin{equation}\n \\sum_{m\\geq\\ell}d_{j,m}\\binom m\\ell x_0^{m-\\ell}.\n \\label{eq:taylor-translation-coefficient}\n\\end{equation}\nIts initial term again has exact valuation $\\ell$, and all later\nterms have at least one more bit. The exact radius is therefore two\nat every integral center.","For address two, the stronger factorial filtration gives the lower\nbound in \\eqref{eq:two-exact-taylor-radius}. At each\n$\\ell=2^r\\geq4$, Theorem~\\ref{thm:two-exact-power-orders} makes the\n$m=\\ell$ term in \\eqref{eq:newton-taylor-stirling} have valuation\n$2\\ell+\\ell/2=5\\ell/2$. Every $m>\\ell$ term has strictly greater\nlower bound $2m+\\lfloor m/2\\rfloor$. Hence\n\\[\n v_2(d_{2,\\ell})=5\\ell/2\\qquad(\\ell=2^r\\geq4).\n\\]\nThe universal lower bound and this unbounded exact subsequence prove\nthat the coefficient radius is exactly $2^{5/2}$. Translation by an\nintegral center preserves the same unique leading coefficients at\nthese degrees in \\eqref{eq:taylor-translation-coefficient}, proving\nthe exact radius at every such center.\n\\end{proof}","l=2^r\\geq4$, Theorem~\\ref{thm:two-exact-power-orders} makes the\n$m=\\ell$ term in \\eqref{eq:newton-taylor-stirling} have valuation\n$2\\ell+\\ell/2=5\\ell/2$. Every $m>\\ell$ term has strictly greater\nlower bound $2m+\\lfloor m/2\\rfloor$. Hence\n\\[\n v_2(d_{2,\\ell})=5\\ell/2\\qquad(\\ell=2^r\\geq4).\n\\]\nThe universal lower bound and this unbounded exact subsequence prove\nthat the coefficient radius is exactly $2^{5/2}$. Translation by an\nintegral center preserves the same unique leading coefficients at\nthese degrees in \\eqref{eq:taylor-translation-coefficient}, proving\nthe exact radius at every such center.\n\\end{proof}","\\begin{remark}[Boundary interpretation]\nFor addresses zero and one, at an argument of absolute value two every\npositive-degree Taylor term has valuation zero, so that ordinary Taylor\nseries diverges. For address two, over a complete extension admitting\narguments of absolute value $2^{5/2}$, the power-of-two-degree terms\nhave valuation zero and fail to tend to zero. These statements concern\nthe Taylor expansions just constructed; they do not prohibit a\ndifferent analytic continuation. On $\\mathbb Q_2$-valued arguments,\nopen-disc notation must be read with its discrete value group. The\ncoefficient radius is the precise invariant in the theorem.\n\\end{remark}","e value two every\npositive-degree Taylor term has valuation zero, so that ordinary Taylor\nseries diverges. For address two, over a complete extension admitting\narguments of absolute value $2^{5/2}$, the power-of-two-degree terms\nhave valuation zero and fail to tend to zero. These statements concern\nthe Taylor expansions just constructed; they do not prohibit a\ndifferent analytic continuation. On $\\mathbb Q_2$-valued arguments,\nopen-disc notation must be read with its discrete value group. The\ncoefficient radius is the precise invariant in the theorem.\n\\end{remark}","\\begin{corollary}[Final common-step congruence]\n\\label{cor:final-euler-congruence}\nLet $j\\in\\{0,1,2,3\\}$, $n\\geq0$, and $k,H\\geq1$. Define\n\\[\n A_0(k)=A_1(k)=k,\\qquad\n A_2(k)=\\max\\left(k+4,\\ 2k+\\left\\lfloor\\frac k2\\right\\rfloor+1\\right),\n \\qquad A_3(k)=k+2.\n\\]\nThen\n\\begin{equation}\n \\boxed{\\displaystyle\n \\sum_{i=0}^k(-1)^{k-i}\\binom ki E_{4(n+iH)}^{(4,j)}\n       \\equiv0\\pmod{2^{\\,v_2(k!)+k v_2(H)+A_j(k)}}.}\n \\label{eq:boxed-final-euler-congruence}\n\\end{equation}\nFor $j=0,1$, this displayed exponent is the exact valuation for every\n$n,k,H$ in the stated range. For $j=2$, the exact positive\npower-of-two-order valuations are given by\nTheorem~\\ref{thm:two-exact-power-orders}, with one added bit.\n\\end{corollary}","congruence}\nLet $j\\in\\{0,1,2,3\\}$, $n\\geq0$, and $k,H\\geq1$. Define\n\\[\n A_0(k)=A_1(k)=k,\\qquad\n A_2(k)=\\max\\left(k+4,\\ 2k+\\left\\lfloor\\frac k2\\right\\rfloor+1\\right),\n \\qquad A_3(k)=k+2.\n\\]\nThen\n\\begin{equation}\n \\boxed{\\displaystyle\n \\sum_{i=0}^k(-1)^{k-i}\\binom ki E_{4(n+iH)}^{(4,j)}\n       \\equiv0\\pmod{2^{\\,v_2(k!)+k v_2(H)+A_j(k)}}.}\n \\label{eq:boxed-final-euler-congruence}\n\\end{equation}\nFor $j=0,1$, this displayed exponent is the exact valuation for every\n$n,k,H$ in the stated range. For $j=2$, the exact positive\npower-of-two-order valuations are given by\nTheorem~\\ref{thm:two-exact-power-orders}, with one added bit.\n\\end{corollary}","e\n\\[\n A_0(k)=A_1(k)=k,\\qquad\n A_2(k)=\\max\\left(k+4,\\ 2k+\\left\\lfloor\\frac k2\\right\\rfloor+1\\right),\n \\qquad A_3(k)=k+2.\n\\]\nThen\n\\begin{equation}\n \\boxed{\\displaystyle\n \\sum_{i=0}^k(-1)^{k-i}\\binom ki E_{4(n+iH)}^{(4,j)}\n       \\equiv0\\pmod{2^{\\,v_2(k!)+k v_2(H)+A_j(k)}}.}\n \\label{eq:boxed-final-euler-congruence}\n\\end{equation}\nFor $j=0,1$, this displayed exponent is the exact valuation for every\n$n,k,H$ in the stated range. For $j=2$, the exact positive\npower-of-two-order valuations are given by\nTheorem~\\ref{thm:two-exact-power-orders}, with one added bit.\n\\end{corollary}","(k)=\\max\\left(k+4,\\ 2k+\\left\\lfloor\\frac k2\\right\\rfloor+1\\right),\n \\qquad A_3(k)=k+2.\n\\]\nThen\n\\begin{equation}\n \\boxed{\\displaystyle\n \\sum_{i=0}^k(-1)^{k-i}\\binom ki E_{4(n+iH)}^{(4,j)}\n       \\equiv0\\pmod{2^{\\,v_2(k!)+k v_2(H)+A_j(k)}}.}\n \\label{eq:boxed-final-euler-congruence}\n\\end{equation}\nFor $j=0,1$, this displayed exponent is the exact valuation for every\n$n,k,H$ in the stated range. For $j=2$, the exact positive\npower-of-two-order valuations are given by\nTheorem~\\ref{thm:two-exact-power-orders}, with one added bit.\n\\end{corollary}","\\begin{proof}\nThe left side is $j!\\Delta_H^kC_j(n)$.\nAt addresses zero and one, $v_2(j!)=0$ and the exact homogeneous\nformula gives the exponent. At address two, $v_2(2!)=1$; the maximum\nof \\eqref{eq:factorial-four-lower} and\n\\eqref{eq:factorial-four-j2-filtration} yields\n$\\max(k+4,2k+\\lfloor k/2\\rfloor+1)$ after removing the common\n$v_2(k!)+k v_2(H)$. At address three, the one-bit normalized offset\nand $v_2(3!)=1$ give $k+2$.\n\\end{proof}","roof}\nThe left side is $j!\\Delta_H^kC_j(n)$.\nAt addresses zero and one, $v_2(j!)=0$ and the exact homogeneous\nformula gives the exponent. At address two, $v_2(2!)=1$; the maximum\nof \\eqref{eq:factorial-four-lower} and\n\\eqref{eq:factorial-four-j2-filtration} yields\n$\\max(k+4,2k+\\lfloor k/2\\rfloor+1)$ after removing the common\n$v_2(k!)+k v_2(H)$. At address three, the one-bit normalized offset\nand $v_2(3!)=1$ give $k+2$.\n\\end{proof}","de is $j!\\Delta_H^kC_j(n)$.\nAt addresses zero and one, $v_2(j!)=0$ and the exact homogeneous\nformula gives the exponent. At address two, $v_2(2!)=1$; the maximum\nof \\eqref{eq:factorial-four-lower} and\n\\eqref{eq:factorial-four-j2-filtration} yields\n$\\max(k+4,2k+\\lfloor k/2\\rfloor+1)$ after removing the common\n$v_2(k!)+k v_2(H)$. At address three, the one-bit normalized offset\nand $v_2(3!)=1$ give $k+2$.\n\\end{proof}","For mixed steps the same corollary holds with $k v_2(H)$ replaced by\n$\\sum_\\nu v_2(H_\\nu)$ and the binomial stencil replaced by\n$j!\\Delta_{H_1}\\cdots\\Delta_{H_k}C_j(n)$.\nAt addresses zero and one, exactness is then governed by the parity\nclassification in Theorem~\\ref{thm:zero-one-exact-mixed}.\nAt first order these bounds are exact in every address, with original\nvaluation offsets $(1,1,5,3)$; the least periods are those in\n\\eqref{eq:rank-four-least-period-final}.","There are two coefficient conventions, and they must stay distinct:\n\\[\n C_{N,j}(0)=1,\n \\qquad E_0^{(N,j)}=j!,\n \\qquad\n v_2(E_{Nn}^{(N,j)})=v_2(C_{N,j}(n))+v_2(j!).\n\\]\nThe last equality uses $v_2(0)=+\\infty$. Congruences of dyadic rational numbers\nmean membership in the corresponding ideal of $\\mathbb Z_2$; a displayed\nrational denominator is invertible modulo $2^r$ when it is odd.","\\midrule\n$n$ & packed row index & Euler degree $4n$","$H$ & positive row step & Euler degree step $4H$","$k$ & number of differences & order $k$, independent of $n$","$j$ & address within a section & $0,1,2,3$","For example, $C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=121/45$.\nTheir difference is $26/9$, of valuation one. The row difference joins Euler\ndegrees $4$ and $8$.","Nishibuchi's Conjecture 3.9 concerns a prime $p$, a positive integer $m$ with\n$m\\mid p-1$ or $m=2$, and every address $0\\le j<mp$. Put\n\\[\n N=mp, \\qquad q=\\operatorname{lcm}(2,p-1).\n\\]\nIt asserts $p$-adic integrality of the actual generalized Euler coefficients and\neventual periodicity at every precision:\n\\[\n \\text{for every }r\\ge1,\\quad\n E_{mpn+qp^r}^{(mp,j)}\\equiv E_{mpn}^{(mp,j)}\\pmod {p^r}\n \\quad\\text{for all sufficiently large }n.\n\\]\nThe completed theorem gives the explicit sufficient threshold $n\\ge r$ in\nevery permitted branch. Its formal capstone is the theorem named\n\\texttt{NishibuchiCompletion.conjecture\\_3\\_9}. No conjectural flatness premise\nis supplied to that theorem.","very address $0\\le j<mp$. Put\n\\[\n N=mp, \\qquad q=\\operatorname{lcm}(2,p-1).\n\\]\nIt asserts $p$-adic integrality of the actual generalized Euler coefficients and\neventual periodicity at every precision:\n\\[\n \\text{for every }r\\ge1,\\quad\n E_{mpn+qp^r}^{(mp,j)}\\equiv E_{mpn}^{(mp,j)}\\pmod {p^r}\n \\quad\\text{for all sufficiently large }n.\n\\]\nThe completed theorem gives the explicit sufficient threshold $n\\ge r$ in\nevery permitted branch. Its formal capstone is the theorem named\n\\texttt{NishibuchiCompletion.conjecture\\_3\\_9}. No conjectural flatness premise\nis supplied to that theorem.","\\begin{center}\n\\begin{tabular}{llll}\n\\toprule\nPrime branch & Allowed $m$ & $q$ & Row step for the original shift","\\midrule\n$p=2$ & $1$ & $2$ & $2^r$","$p=2$ & $2$ & $2$ & $2^{r-1}$","$p$ odd & $m\\mid p-1$ & $p-1$ & $\\dfrac{p-1}{m}p^{r-1}$","For odd primes, $m=2$ is already in the divisor branch. The conversion is exact:\n\\[\n N\\left(\\frac qm p^{r-1}\\right)=qp^r.\n\\]\nAt $p=2,m=2$, the original conjectured Euler-degree shift $2^{r+1}$ is the row\nstep $2^{r-1}$ for the rank-four sequence.","The conjecture's primary source is Yuta Nishibuchi, \\emph{$p$-adic Congruences of\nGeneralized Euler Numbers and Relations to Even Zeta Values},\narXiv:2605.10677v2, revised 1 June 2026. The statement above uses that version's\nnormalization and actual Euler degrees.","At rank four the earlier completed estimate was\n$v_2(\\Delta_H^k C_j(n))\\ge\\min(k,n)$. The strengthened theorem reaches every\nstarting row, including $n=0$, and every order, including $k>n$.","\\begin{figure}[ht]\n\\centering\n\\includegraphics[width=\\linewidth]{../figures/threshold.pdf}\n\\caption{Unit-step lower bounds on a finite display window. Each colored cell\nis obtained from a proved formula, rather than from sampled coefficients.\nThe dashed line indicates $k=n$.}\n\\end{figure}","For \\emph{every} dyadic section size $N=2^a$, $a\\ge0$, every permitted address,\nand all positive steps,\n\\[\n v_2\\!\\left(\\Delta_{H_1}\\cdots\\Delta_{H_k}C_{N,j}(n)\\right)\n \\ge k+\\sum_{\\nu=1}^k v_2(H_\\nu),\n \\qquad n\\ge0.\n\\]\nThe corresponding statement for $E_{Nn}^{(N,j)}$ adds $v_2(j!)$.","The proof acts on the full factorial-scaled power series before it extracts a\npacked row. This avoids the old extraction constraint $k\\le n$. Reciprocal\ninduction controls the unique term with the highest inverse order in the\ncommuting product expansion; all other terms are covered by the preceding\ninduction stages.","Write $Sf(n)=f(n+1)$ and $\\Delta_H=S^H-1$. Mixed differences commute. Their\nexact finite expression is\n\\[\n \\Delta_{H_1}\\cdots\\Delta_{H_k}f(n)\n =\\sum_{A\\subseteq\\{1,\\ldots,k\\}}\n (-1)^{k-|A|}\n f\\!\\left(n+\\sum_{i\\in A}H_i\\right).\n\\]\nRepeated subset offsets are combined with their signed multiplicities.","\\begin{figure}[ht]\n\\centering\n\\includegraphics[width=\\linewidth]{../figures/mixed_stencil.pdf}\n\\caption{For steps $(1,2,4)$ all eight offsets are distinct. This is an exact\nsigned stencil, rather than an approximate derivative.}\n\\end{figure}","For this stencil, the general dyadic bound is\n\\[\n 3+v_2(1)+v_2(2)+v_2(4)=6.\n\\]\nAt rank four the factorial theorem improves it to seven for addresses zero\nand one. Exact rational evaluation gives\n$v_2(\\Delta_1\\Delta_2\\Delta_4 C_0(0))=7$.\nThe order is odd, so the mixed-step equality criterion also proves this\nvaluation at \\emph{every} starting row.","There is a prime-power extension. For $N=p^a$, $a\\ge1$, the sign-corrected\noperator is\n\\[\n T_H=S^H-(-1)^H,\n \\qquad\n v_p\\!\\left(T_{H_1}\\cdots T_{H_k}E^{(N,j)}(n)\\right)\n \\ge v_p(j!)+\\sum_{i=1}^k\\bigl(1+v_p(H_i)\\bigr).\n\\]\nHere $E^{(N,j)}(n)=E_{Nn}^{(N,j)}$. For an odd prime, an ordinary first\ndifference need not have positive valuation: at address zero its first two\nnormalized values are $1$ and $-1$, with difference $-2$. The sign correction\nis therefore essential to this prime-power theorem.","er extension. For $N=p^a$, $a\\ge1$, the sign-corrected\noperator is\n\\[\n T_H=S^H-(-1)^H,\n \\qquad\n v_p\\!\\left(T_{H_1}\\cdots T_{H_k}E^{(N,j)}(n)\\right)\n \\ge v_p(j!)+\\sum_{i=1}^k\\bigl(1+v_p(H_i)\\bigr).\n\\]\nHere $E^{(N,j)}(n)=E_{Nn}^{(N,j)}$. For an odd prime, an ordinary first\ndifference need not have positive valuation: at address zero its first two\nnormalized values are $1$ and $-1$, with difference $-2$. The sign correction\nis therefore essential to this prime-power theorem.","For $n\\ge0$, $k\\ge1$, and $H\\ge1$, the actual Euler coefficients satisfy\n\\[\n \\sum_{i=0}^k(-1)^{k-i}\\binom ki\n E_{4(n+iH)}^{(4,j)}\n \\equiv0\\pmod {2^{\\,v_2(k!)+k v_2(H)+A_j(k)}},\n\\]\nwhere\n\\[\n\\begin{aligned}\n A_0(k)=A_1(k)&=k,","For $n\\ge0$, $k\\ge1$, and $H\\ge1$, the actual Euler coefficients satisfy\n\\[\n \\sum_{i=0}^k(-1)^{k-i}\\binom ki\n E_{4(n+iH)}^{(4,j)}\n \\equiv0\\pmod {2^{\\,v_2(k!)+k v_2(H)+A_j(k)}},\n\\]\nwhere\n\\[\n\\begin{aligned}\n A_0(k)=A_1(k)&=k,\\\\\n A_2(k)&=\\max\\{k+4,\\ 2k+\\lfloor k/2\\rfloor+1\\},\\\\\n A_3(k)&=k+2.\n\\end{aligned}\n\\]","At a common step with $v_2(H)=s$, add $ks$ to every displayed exponent. Thus a\ndoubling of the step gains one additional bit \\emph{per difference}, rather\nthan one bit for the whole expression.","The normalized address-two sequence obeys both\n\\[\n\\begin{aligned}\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}C_2(n))\n &\\ge k+v_2(k!)+3+\\sum_i v_2(H_i),\\\\\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}C_2(n))\n &\\ge2k+v_2(k!)+\\lfloor k/2\\rfloor+\\sum_i v_2(H_i).\n\\end{aligned}\n\\]\nTheir maximum supplies the useful envelope.","For every base row and every positive step list, equality is proved when\n$k$ is a power of two:\n\\[\n v_2(\\Delta_{H_1}\\cdots\\Delta_{H_k}C_2(n))\n =V_2(k)+\\sum_i v_2(H_i),\n\\]\nwith $V_2(1)=4$, $V_2(2)=6$, and\n$V_2(k)=2k+v_2(k!)+k/2$ for $k=2^r\\ge4$.\nThe infinite marked family establishes sharpness at unbounded orders.","For original Euler coefficients add one, since $v_2(2!)=1$.\nThe extra $\\lfloor k/2\\rfloor$ comes from a paired divided-power filtration in\nthe Bernoulli transformation. It does not arise by fitting the plotted points.","For addresses $j=0,1$, put\n\\[\n V=k+v_2(k!)+\\sum_i v_2(H_i),\n \\qquad q=\\#\\{i:H_i\\text{ is even}\\}.\n\\]\nThe valuation equals $V$ exactly when $k$ is odd or $q$ is even.","\\midrule\n$k$ odd & either parity & exact valuation $V$","$k$ even & $q$ even & exact valuation $V$","$k$ even & $q$ odd & valuation at least $V+1$","For instance,\n\\[\n \\Delta_1\\Delta_2 C_0(0)=-33728,\n \\qquad v_2(-33728)=6,\n \\qquad V=4.\n\\]\nCancellation removes the leading term, and here it removes two bits beyond\nthe baseline. The parity theorem proves strictness in this case; it does not\ngive a complete formula for the further depth.","With a common step, $q$ is either zero or $k$. Consequently the exceptional\nparity never occurs, and the homogeneous address-zero/one formula is always\nexact.","All four rank-four sequences satisfy the sharp first-difference law\n\\[\n v_2(C_j(n+H)-C_j(n))=v_2(H)+b_j,\n \\qquad b=(1,1,4,2).\n\\]\nFor the actual Euler coefficients this becomes\n\\[\n v_2(E_{4(n+H)}^{(4,j)}-E_{4n}^{(4,j)})=v_2(H)+B_j,\n \\qquad B=(1,1,5,3).\n\\]\nBoth laws hold at every starting row and every positive integer step.","\\begin{figure}[ht]\n\\centering\n\\includegraphics[width=\\linewidth]{../figures/metric.pdf}\n\\caption{The proved Euler metric laws, with exact modular sample points at\nbase zero and steps $H=2^s$. Addresses zero and one have the same scale;\ntheir circle and square marks overlap.}\n\\end{figure}","In dyadic distance, $|x|_2=2^{-v_2(x)}$. After continuous interpolation to\n$\\mathbb Z_2$, the law reads\n\\[\n |E^{(4,j)}(x)-E^{(4,j)}(y)|_2\n =2^{-B_j}|x-y|_2.\n\\]\nIt says exactly how much the map shrinks distances. It also prevents any\nhidden collision at finer precision: distinct inputs remain distinct after\nthe known scale is removed.","The scale of address two is $2^{-5}$ in the original normalization and\n$2^{-4}$ in the normalized sequence. Omitting its factor $2!$ would change\nthe period claimed for the original coefficients.","At precision $2^r$ the least positive \\emph{row} period is\n\\[\n P_j(r)=2^{\\max(0,r-B_j)},\\qquad B=(1,1,5,3).\n\\]\nThe exact valuation law proves both periodicity and minimality. A step $H$\nis a period precisely when $v_2(H)+B_j\\ge r$.","Modulo $256=2^8$, the least periods are\n\\[\n (P_0(8),P_1(8),P_2(8),P_3(8))=(128,128,8,32).\n\\]\nIn actual Euler degree, multiply by four:\n\\[\n (512,512,32,128).\n\\]\nFor address two, precision at most $2^5$ already has row period one. Its\nEuler coefficients all lie in $2+32\\mathbb Z_2$; further precision resolves\nadditional dyadic digits within that residue class.","The original rank-four conjectural shift has row step $2^{r-1}$. It remains\na valid period for every address, but addresses two and three usually admit\nshorter periods. The sharp theorem identifies the shortest one.","For each address the exact interpolation image is\n\\[\n E^{(4,j)}(\\mathbb Z_2)=j!+2^{B_j}\\mathbb Z_2.\n\\]\nWhen $r>B_j$, a least period lists every residue in this coset modulo $2^r$\nexactly once. The following figure checks this assertion by exact arithmetic\nat precision $256$.","\\begin{figure}[ht]\n\\centering\n\\includegraphics[width=\\linewidth]{../figures/residue_permutations.pdf}\n\\caption{Exact residue permutations modulo $256$. The vertical coordinate is\n$y_n=(E_{4n}^{(4,j)}-j!)/2^{B_j}$ modulo $P_j(8)$.\nEach panel contains one full least period; every vertical residue occurs once.}\n\\end{figure}","No decimal approximation is used to form these residues. The generating\nseries is inverted modulo $2^{256}$ using odd denominators, then the exact\nresult is reduced modulo $256$. An independent rational inversion checks the\ninitial coefficients.","The map\n\\[\n x\\longmapsto \\frac{E^{(4,j)}(x)-j!}{2^{B_j}}\n\\]\nis a bijective isometry of $\\mathbb Z_2$. Its finite reductions preserve the\nnumber of common low binary digits.","\\begin{figure}[ht]\n\\centering\n\\includegraphics[width=\\linewidth]{../figures/isometry_matrices.pdf}\n\\caption{Address two modulo $256$ gives an eight-element quotient. The\ntwo exact matrices coincide: every pair of distinct rows has the same dyadic\nvaluation before and after the scale is removed. The diagonal represents\n$v_2(0)=+\\infty$ and is left blank.}\n\\end{figure}","If $n-m$ is odd, the rows disagree in their first binary digit and the matrix\nentry is zero. If it is divisible by two but not four, they share exactly one\nlow digit. An entry two means agreement through two low digits.","The finite matrix is an illustration of that proof. Its eight rows alone\nwould not establish a statement about the whole of $\\mathbb Z_2$.","Sharp first-difference laws hold for every address at ranks $2$, $4$, and $8$.\nThe arrays below give their normalized scales $b_{N,j}$ and original Euler\nscales $B_{N,j}=b_{N,j}+v_2(j!)$.","There is also an infinite family: at every $N=2^a\\ge4$, all lower-half\naddresses $j<N/2$ have normalized scale one. Their Euler scale is\n$1+v_2(j!)$.","The same constant-scale proposal fails at some upper-half addresses:\n\\begin{center}\n\\begin{tabular}{rrrr}\n\\toprule\n$(N,j)$ & $v_2(C(1)-C(0))$ & Step-two prediction & Actual step-two valuation","\\midrule\n$(16,9)$ & $5$ & $6$ & $7$","$(32,18)$ & $6$ & $7$ & $6$","$(32,19)$ & $5$ & $6$ & $7$","At every center $x_0\\in\\mathbb Z_2$, the interpolations admit ordinary Taylor\nseries $C_j(x_0+z)=\\sum_{d\\ge0}a_{j,d}(x_0)z^d$. For positive degrees,\n\\[\n\\begin{aligned}\n v_2(a_{j,d}(x_0))&=d &&(j=0,1),","At every center $x_0\\in\\mathbb Z_2$, the interpolations admit ordinary Taylor\nseries $C_j(x_0+z)=\\sum_{d\\ge0}a_{j,d}(x_0)z^d$. For positive degrees,\n\\[\n\\begin{aligned}\n v_2(a_{j,d}(x_0))&=d &&(j=0,1),\\\\\n v_2(a_{2,d}(x_0))&\\ge\\max\\{2d+\\lfloor d/2\\rfloor,\\ d+3\\},\\\\\n v_2(a_{3,d}(x_0))&\\ge d.\n\\end{aligned}\n\\]\nAt address two equality with $5d/2$ holds along every power-of-two degree\n$d\\ge4$.","v_2(a_{3,d}(x_0))&\\ge d.\n\\end{aligned}\n\\]\nAt address two equality with $5d/2$ holds along every power-of-two degree\n$d\\ge4$.","The radius is $2^{\\liminf_d v_2(a_d)/d}$ in the normalized dyadic absolute\nvalue. Hence\n\\[\n R_0=R_1=2,\\qquad R_2=2^{5/2},\\qquad R_3\\ge2.\n\\]\nThe exact radius of address three is not asserted. For address two, the\nlower bound gives convergence inside the claimed radius and the unbounded\nequality family prevents a larger one.","The broad exact regression through row $512$ contains more than $1.8$ million\nbounded difference assertions. Smaller later runs test the strengthened\naddress-two filtration and the exact mixed parity law. Their manifests pin\nprecision, coefficient definitions, input versions, and executed ranges.","For a common step, the final congruence to use is\n\\[\n \\Delta_H^k E^{(4,j)}(n)\n \\equiv0\\pmod {2^{v_2(k!)+k v_2(H)+A_j(k)}},\n \\quad n\\ge0,\\ k,H\\ge1,\n\\]\nwith\n\\[\n (A_0,A_1,A_2,A_3)\n =\\bigl(k,k,\\max\\{k+4,2k+\\lfloor k/2\\rfloor+1\\},k+2\\bigr).\n\\]\nFor different steps, replace $k v_2(H)$ by $\\sum_i v_2(H_i)$ in the lower\nbound. The address-zero/one equality rule then depends on order parity and the\nnumber of even steps.","All common-step exact valuations, $j=0,1$ & proved at all rows and orders","is checked by the accompanying exact script.\n\\fi\n\\newcommand{\\AtlasHead}[2]{\\ifdefined\\AtlasStarted\\clearpage\\fi\\ifdefined\\IXMainAtlas\\subsection{#1}\\else\\section{#1}\\fi\\label{#2}\\def\\AtlasStarted{1}}\n\\newcommand{\\AtlasImage}[1]{\\begin{center}\\includegraphics[width=.91\\linewidth]{\\AtlasFigDir/#1.pdf}\\end{center}}\n\\AtlasHead{Compute the reciprocal before estimating it}{atlas:reciprocal}\nFor width $N$ and address $j$, the normalized source and its reciprocal are\n\\[\n G_{N,j}(t)=j!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n G_{N,j}(t)^{-1}=\\sum_{n\\ge0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!}.\n\\]\nMultiplying the two series and clearing the factorials gives an exact triangular recurrence:\n\\[\n C_{N,j}(0)=1,\\qquad\n C_{N,j}(n)=-\\frac{\\displaystyle\\sum_{r=0}^{n-1}\n \\binom{Nn+j}{Nr}C_{N,j}(r)}{\\binom{Nn+j}{Nn}}.\n\\]\n\\AtlasImage{reciprocal_rows}\nThe table is for $N=4$. The first two address-zero rows illustrate the computation without any valuation estimate:\n\\[\n C_{4,0}(1)=-1,\\qquad\n C_{4,0}(2)=-\\bigl(1+\\binom84(-1)\\bigr)=69.\n\\]\nAt address one, the same recurrence yields $C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=12…","e accompanying exact script.\n\\fi\n\\newcommand{\\AtlasHead}[2]{\\ifdefined\\AtlasStarted\\clearpage\\fi\\ifdefined\\IXMainAtlas\\subsection{#1}\\else\\section{#1}\\fi\\label{#2}\\def\\AtlasStarted{1}}\n\\newcommand{\\AtlasImage}[1]{\\begin{center}\\includegraphics[width=.91\\linewidth]{\\AtlasFigDir/#1.pdf}\\end{center}}\n\\AtlasHead{Compute the reciprocal before estimating it}{atlas:reciprocal}\nFor width $N$ and address $j$, the normalized source and its reciprocal are\n\\[\n G_{N,j}(t)=j!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n G_{N,j}(t)^{-1}=\\sum_{n\\ge0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!}.\n\\]\nMultiplying the two series and clearing the factorials gives an exact triangular recurrence:\n\\[\n C_{N,j}(0)=1,\\qquad\n C_{N,j}(n)=-\\frac{\\displaystyle\\sum_{r=0}^{n-1}\n \\binom{Nn+j}{Nr}C_{N,j}(r)}{\\binom{Nn+j}{Nn}}.\n\\]\n\\AtlasImage{reciprocal_rows}\nThe table is for $N=4$. The first two address-zero rows illustrate the computation without any valuation estimate:\n\\[\n C_{4,0}(1)=-1,\\qquad\n C_{4,0}(2)=-\\bigl(1+\\binom84(-1)\\bigr)=69.\n\\]\nAt address one, the same recurrence yields $C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=121/45$. These are","Head}[2]{\\ifdefined\\AtlasStarted\\clearpage\\fi\\ifdefined\\IXMainAtlas\\subsection{#1}\\else\\section{#1}\\fi\\label{#2}\\def\\AtlasStarted{1}}\n\\newcommand{\\AtlasImage}[1]{\\begin{center}\\includegraphics[width=.91\\linewidth]{\\AtlasFigDir/#1.pdf}\\end{center}}\n\\AtlasHead{Compute the reciprocal before estimating it}{atlas:reciprocal}\nFor width $N$ and address $j$, the normalized source and its reciprocal are\n\\[\n G_{N,j}(t)=j!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n G_{N,j}(t)^{-1}=\\sum_{n\\ge0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!}.\n\\]\nMultiplying the two series and clearing the factorials gives an exact triangular recurrence:\n\\[\n C_{N,j}(0)=1,\\qquad\n C_{N,j}(n)=-\\frac{\\displaystyle\\sum_{r=0}^{n-1}\n \\binom{Nn+j}{Nr}C_{N,j}(r)}{\\binom{Nn+j}{Nn}}.\n\\]\n\\AtlasImage{reciprocal_rows}\nThe table is for $N=4$. The first two address-zero rows illustrate the computation without any valuation estimate:\n\\[\n C_{4,0}(1)=-1,\\qquad\n C_{4,0}(2)=-\\bigl(1+\\binom84(-1)\\bigr)=69.\n\\]\nAt address one, the same recurrence yields $C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=121/45$. These are exact rational coefficients, not rounded approxima…","tlasFigDir/#1.pdf}\\end{center}}\n\\AtlasHead{Compute the reciprocal before estimating it}{atlas:reciprocal}\nFor width $N$ and address $j$, the normalized source and its reciprocal are\n\\[\n G_{N,j}(t)=j!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n G_{N,j}(t)^{-1}=\\sum_{n\\ge0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!}.\n\\]\nMultiplying the two series and clearing the factorials gives an exact triangular recurrence:\n\\[\n C_{N,j}(0)=1,\\qquad\n C_{N,j}(n)=-\\frac{\\displaystyle\\sum_{r=0}^{n-1}\n \\binom{Nn+j}{Nr}C_{N,j}(r)}{\\binom{Nn+j}{Nn}}.\n\\]\n\\AtlasImage{reciprocal_rows}\nThe table is for $N=4$. The first two address-zero rows illustrate the computation without any valuation estimate:\n\\[\n C_{4,0}(1)=-1,\\qquad\n C_{4,0}(2)=-\\bigl(1+\\binom84(-1)\\bigr)=69.\n\\]\nAt address one, the same recurrence yields $C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=121/45$. These are exact rational coefficients, not rounded approximations. The second implementation in the supplied script inverts the ordinary power series first, then multiplies by $(Nn)!$. Agreement between the two constructions checks the factorial placement.","ciprocal are\n\\[\n G_{N,j}(t)=j!\\sum_{n\\ge0}\\frac{t^{Nn}}{(Nn+j)!},\\qquad\n G_{N,j}(t)^{-1}=\\sum_{n\\ge0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!}.\n\\]\nMultiplying the two series and clearing the factorials gives an exact triangular recurrence:\n\\[\n C_{N,j}(0)=1,\\qquad\n C_{N,j}(n)=-\\frac{\\displaystyle\\sum_{r=0}^{n-1}\n \\binom{Nn+j}{Nr}C_{N,j}(r)}{\\binom{Nn+j}{Nn}}.\n\\]\n\\AtlasImage{reciprocal_rows}\nThe table is for $N=4$. The first two address-zero rows illustrate the computation without any valuation estimate:\n\\[\n C_{4,0}(1)=-1,\\qquad\n C_{4,0}(2)=-\\bigl(1+\\binom84(-1)\\bigr)=69.\n\\]\nAt address one, the same recurrence yields $C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=121/45$. These are exact rational coefficients, not rounded approximations. The second implementation in the supplied script inverts the ordinary power series first, then multiplies by $(Nn)!$. Agreement between the two constructions checks the factorial placement.","0}C_{N,j}(n)\\frac{t^{Nn}}{(Nn)!}.\n\\]\nMultiplying the two series and clearing the factorials gives an exact triangular recurrence:\n\\[\n C_{N,j}(0)=1,\\qquad\n C_{N,j}(n)=-\\frac{\\displaystyle\\sum_{r=0}^{n-1}\n \\binom{Nn+j}{Nr}C_{N,j}(r)}{\\binom{Nn+j}{Nn}}.\n\\]\n\\AtlasImage{reciprocal_rows}\nThe table is for $N=4$. The first two address-zero rows illustrate the computation without any valuation estimate:\n\\[\n C_{4,0}(1)=-1,\\qquad\n C_{4,0}(2)=-\\bigl(1+\\binom84(-1)\\bigr)=69.\n\\]\nAt address one, the same recurrence yields $C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=121/45$. These are exact rational coefficients, not rounded approximations. The second implementation in the supplied script inverts the ordinary power series first, then multiplies by $(Nn)!$. Agreement between the two constructions checks the factorial placement.","lar recurrence:\n\\[\n C_{N,j}(0)=1,\\qquad\n C_{N,j}(n)=-\\frac{\\displaystyle\\sum_{r=0}^{n-1}\n \\binom{Nn+j}{Nr}C_{N,j}(r)}{\\binom{Nn+j}{Nn}}.\n\\]\n\\AtlasImage{reciprocal_rows}\nThe table is for $N=4$. The first two address-zero rows illustrate the computation without any valuation estimate:\n\\[\n C_{4,0}(1)=-1,\\qquad\n C_{4,0}(2)=-\\bigl(1+\\binom84(-1)\\bigr)=69.\n\\]\nAt address one, the same recurrence yields $C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=121/45$. These are exact rational coefficients, not rounded approximations. The second implementation in the supplied script inverts the ordinary power series first, then multiplies by $(Nn)!$. Agreement between the two constructions checks the factorial placement.","{N,j}(0)=1,\\qquad\n C_{N,j}(n)=-\\frac{\\displaystyle\\sum_{r=0}^{n-1}\n \\binom{Nn+j}{Nr}C_{N,j}(r)}{\\binom{Nn+j}{Nn}}.\n\\]\n\\AtlasImage{reciprocal_rows}\nThe table is for $N=4$. The first two address-zero rows illustrate the computation without any valuation estimate:\n\\[\n C_{4,0}(1)=-1,\\qquad\n C_{4,0}(2)=-\\bigl(1+\\binom84(-1)\\bigr)=69.\n\\]\nAt address one, the same recurrence yields $C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=121/45$. These are exact rational coefficients, not rounded approximations. The second implementation in the supplied script inverts the ordinary power series first, then multiplies by $(Nn)!$. Agreement between the two constructions checks the factorial placement.","rows illustrate the computation without any valuation estimate:\n\\[\n C_{4,0}(1)=-1,\\qquad\n C_{4,0}(2)=-\\bigl(1+\\binom84(-1)\\bigr)=69.\n\\]\nAt address one, the same recurrence yields $C_{4,1}(1)=-1/5$ and $C_{4,1}(2)=121/45$. These are exact rational coefficients, not rounded approximations. The second implementation in the supplied script inverts the ordinary power series first, then multiplies by $(Nn)!$. Agreement between the two constructions checks the factorial placement.","\\AtlasHead{The normalization loss is a real boundary}{atlas:normalization}\nThe original Euler coefficient is $E_{Nn}^{(N,j)}=j!C_{N,j}(n)$. Multiplication by $j!$ changes the available $p$-adic precision. That change cannot be discarded outside the unit-factorial window.","Take $p=3$, $N=6$, $j=3$, and compare rows one and two. Exact inversion gives\n\\[\n C_{6,3}(1)=-\\frac1{84},\\quad\n C_{6,3}(2)=\\frac{703}{5460},\\quad\n C_{6,3}(2)-C_{6,3}(1)=\\frac{64}{455}.\n\\]\n\\AtlasImage{normalization_loss}\nNeither $64$ nor $455$ is divisible by three, so the normalized difference has valuation zero. The original coefficient difference is $3!\\cdot64/455=384/455$, of valuation one.","This is the counterexample behind the corrected factorial-loss statement. A bound for the original Euler normalization must lose $v_p(j!)$ when translated to $C_{N,j}$. For $0\\le j<p$, that loss is zero. The distinction concerns which sequence the theorem measures, not a numerical weakness of the calculation.","At $N=p=3$ and $j=0$, put $A(n)=(-1)^nC_{3,0}(n)$. The first three rows are\n\\AtlasImage{odd_prime_sign}\nThe ordinary first difference at row zero is $-1-1=-2$. It is a $3$-adic unit, so an asserted lower bound of one for that difference would be false. The twisted first difference is zero, whose mathematical valuation is $+\\infty$. At the next row, the twisted difference is $19-1=18$, of valuation two.","Equivalently, the untwisted sequence is acted on by $T_H=S^H-(-1)^H$, where $S$ advances one row. At odd step one this is $S+1$, not $S-1$. The zero in the example is handled by a zero-safe divisibility statement rather than a finite valuation equality. The explicit sign and the zero convention are both part of the theorem's interface.","\\AtlasHead{Newton coordinates turn precision into a finite computation}{atlas:newton}\nLet $b_k=\\Delta_1^kC_{4,0}(0)$. The first coefficients obtained from the difference triangle are\n\\input{generated/worked_newton_values.tex}\nThe exact Newton identity recovers each nonnegative integer row:\n\\[\n C_{4,0}(n)=\\sum_{k=0}^{n}b_k\\binom nk.\n\\]\n\\AtlasImage{newton_cutoff}\nFor this address the proved law is $v_2(b_k)=2k-s_2(k)$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ throug","\\AtlasHead{Newton coordinates turn precision into a finite computation}{atlas:newton}\nLet $b_k=\\Delta_1^kC_{4,0}(0)$. The first coefficients obtained from the difference triangle are\n\\input{generated/worked_newton_values.tex}\nThe exact Newton identity recovers each nonnegative integer row:\n\\[\n C_{4,0}(n)=\\sum_{k=0}^{n}b_k\\binom nk.\n\\]\n\\AtlasImage{newton_cutoff}\nFor this address the proved law is $v_2(b_k)=2k-s_2(k)$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ through $32$. Those finite checks validate the worked arithmetic without replacing the theorem.","ates turn precision into a finite computation}{atlas:newton}\nLet $b_k=\\Delta_1^kC_{4,0}(0)$. The first coefficients obtained from the difference triangle are\n\\input{generated/worked_newton_values.tex}\nThe exact Newton identity recovers each nonnegative integer row:\n\\[\n C_{4,0}(n)=\\sum_{k=0}^{n}b_k\\binom nk.\n\\]\n\\AtlasImage{newton_cutoff}\nFor this address the proved law is $v_2(b_k)=2k-s_2(k)$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ through $32$. Those finite checks validate the worked arithmetic without replacing the theorem.","sion into a finite computation}{atlas:newton}\nLet $b_k=\\Delta_1^kC_{4,0}(0)$. The first coefficients obtained from the difference triangle are\n\\input{generated/worked_newton_values.tex}\nThe exact Newton identity recovers each nonnegative integer row:\n\\[\n C_{4,0}(n)=\\sum_{k=0}^{n}b_k\\binom nk.\n\\]\n\\AtlasImage{newton_cutoff}\nFor this address the proved law is $v_2(b_k)=2k-s_2(k)$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ through $32$. Those finite checks validate the worked arithmetic without replacing the theorem.","putation}{atlas:newton}\nLet $b_k=\\Delta_1^kC_{4,0}(0)$. The first coefficients obtained from the difference triangle are\n\\input{generated/worked_newton_values.tex}\nThe exact Newton identity recovers each nonnegative integer row:\n\\[\n C_{4,0}(n)=\\sum_{k=0}^{n}b_k\\binom nk.\n\\]\n\\AtlasImage{newton_cutoff}\nFor this address the proved law is $v_2(b_k)=2k-s_2(k)$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ through $32$. Those finite checks validate the worked arithmetic without replacing the theorem.","}\nLet $b_k=\\Delta_1^kC_{4,0}(0)$. The first coefficients obtained from the difference triangle are\n\\input{generated/worked_newton_values.tex}\nThe exact Newton identity recovers each nonnegative integer row:\n\\[\n C_{4,0}(n)=\\sum_{k=0}^{n}b_k\\binom nk.\n\\]\n\\AtlasImage{newton_cutoff}\nFor this address the proved law is $v_2(b_k)=2k-s_2(k)$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ through $32$. Those finite checks validate the worked arithmetic without replacing the theorem.","ients obtained from the difference triangle are\n\\input{generated/worked_newton_values.tex}\nThe exact Newton identity recovers each nonnegative integer row:\n\\[\n C_{4,0}(n)=\\sum_{k=0}^{n}b_k\\binom nk.\n\\]\n\\AtlasImage{newton_cutoff}\nFor this address the proved law is $v_2(b_k)=2k-s_2(k)$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ through $32$. Those finite checks validate the worked arithmetic without replacing the theorem.","difference triangle are\n\\input{generated/worked_newton_values.tex}\nThe exact Newton identity recovers each nonnegative integer row:\n\\[\n C_{4,0}(n)=\\sum_{k=0}^{n}b_k\\binom nk.\n\\]\n\\AtlasImage{newton_cutoff}\nFor this address the proved law is $v_2(b_k)=2k-s_2(k)$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ through $32$. Those finite checks validate the worked arithmetic without replacing the theorem.","e triangle are\n\\input{generated/worked_newton_values.tex}\nThe exact Newton identity recovers each nonnegative integer row:\n\\[\n C_{4,0}(n)=\\sum_{k=0}^{n}b_k\\binom nk.\n\\]\n\\AtlasImage{newton_cutoff}\nFor this address the proved law is $v_2(b_k)=2k-s_2(k)$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ through $32$. Those finite checks validate the worked arithmetic without replacing the theorem.","k)=2k-s_2(k)$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ through $32$. Those finite checks validate the worked arithmetic without replacing the theorem.","$ for $k\\ge1$. Every $k\\ge5$ therefore has $v_2(b_k)\\ge8$. Since $\\binom nk$ is integral, those terms vanish modulo $256$. Thus, for every $n\\ge0$,\n\\[\n C_{4,0}(n)\\equiv1+254\\binom n1+72\\binom n2+\n 176\\binom n3+128\\binom n4\\pmod{256}.\n\\]\nThe uniform truncation follows from the coefficient theorem. The script separately checks the exact Newton reconstruction and this reduction for rows $0$ through $32$. Those finite checks validate the worked arithmetic without replacing the theorem.","\\AtlasHead{A compatible root is built one digit at a time}{atlas:smooth}\nTake $F(x)=x^2-7$ over the $3$-adic integers and start with $x_1=1$ modulo three. If $F(x_L)=3^Lr_L$, the next representative has the form $x_{L+1}=x_L+3^Lt_L$. Its correcting digit satisfies\n\\[\n r_L+2x_Lt_L\\equiv0\\pmod3.\n\\]\n\\AtlasImage{smooth_tower}\nFor example, $F(1)/3=-2$ and $2x_1=2$, so the next digit is one and $x_2=4$. Then $F(4)/9=1$, again giving digit one and $x_3=13$. At the next stage the residual is divisible by three, so digit zero is correct. A zero digit is not a stalled construction.","ead{A compatible root is built one digit at a time}{atlas:smooth}\nTake $F(x)=x^2-7$ over the $3$-adic integers and start with $x_1=1$ modulo three. If $F(x_L)=3^Lr_L$, the next representative has the form $x_{L+1}=x_L+3^Lt_L$. Its correcting digit satisfies\n\\[\n r_L+2x_Lt_L\\equiv0\\pmod3.\n\\]\n\\AtlasImage{smooth_tower}\nFor example, $F(1)/3=-2$ and $2x_1=2$, so the next digit is one and $x_2=4$. Then $F(4)/9=1$, again giving digit one and $x_3=13$. At the next stage the residual is divisible by three, so digit zero is correct. A zero digit is not a stalled construction.","t at a time}{atlas:smooth}\nTake $F(x)=x^2-7$ over the $3$-adic integers and start with $x_1=1$ modulo three. If $F(x_L)=3^Lr_L$, the next representative has the form $x_{L+1}=x_L+3^Lt_L$. Its correcting digit satisfies\n\\[\n r_L+2x_Lt_L\\equiv0\\pmod3.\n\\]\n\\AtlasImage{smooth_tower}\nFor example, $F(1)/3=-2$ and $2x_1=2$, so the next digit is one and $x_2=4$. Then $F(4)/9=1$, again giving digit one and $x_3=13$. At the next stage the residual is divisible by three, so digit zero is correct. A zero digit is not a stalled construction.","Every center remains congruent to one modulo three. Hence $2x_L$ stays a unit, and the correction equation has a unique digit at every stage. The representatives are compatible and their increments are divisible by $3^L$. Completeness gives a limit, and continuity of the polynomial gives a root of $F$. The finite table exhibits the construction. The unit derivative is what supports indefinite continuation.","\\AtlasHead{A singular residual can stop a lift}{atlas:blocked}\nNow take $F(x)=x^2-3$ at the residue $x=0$ modulo three. The residual is $r=F(0)/3=-1$ and the Jacobian is zero modulo three. The next-digit equation would require\n\\[\n -1+0\\cdot t\\equiv0\\pmod3,\n\\]\nwhich has no solution.\n\\AtlasImage{singular_block}\nThe three representatives of the old root class modulo nine all leave residual six modulo nine. Thus the root class exists at precision three but does not lift to precision nine. This is an exact obstruction, not failure of an iterative search.","A singular Jacobian alone does not imply failure. For $F(x)=x^2$ at the same center, the residual is zero and all three candidate digits lift to roots modulo nine. The obstruction is the residual class in the cokernel of the actual Jacobian. It is this class, rather than the rank by itself, that decides whether a next lift exists.","\\AtlasHead{Separate corrections need not be one shared correction}{atlas:shared}\nConsider the two-component system $F(x)=(x,x-3)$ at $x=0$, with $q=3$ and $L=1$. Both components vanish modulo three. After writing a candidate as $y=3t$, its residual vector modulo three is\n\\[\n \\overline r+J\\overline t=(0,2)+(t,t)=(t,t-1).\n\\]\n\\AtlasImage{shared_obstruction}\nThe first coordinate can be corrected by $t=0$. The second can be corrected by $t=1$. No digit corrects both. The available residuals are $(0,2)$, $(1,0)$, and $(2,1)$, none of which is $(0,0)$.","not be one shared correction}{atlas:shared}\nConsider the two-component system $F(x)=(x,x-3)$ at $x=0$, with $q=3$ and $L=1$. Both components vanish modulo three. After writing a candidate as $y=3t$, its residual vector modulo three is\n\\[\n \\overline r+J\\overline t=(0,2)+(t,t)=(t,t-1).\n\\]\n\\AtlasImage{shared_obstruction}\nThe first coordinate can be corrected by $t=0$. The second can be corrected by $t=1$. No digit corrects both. The available residuals are $(0,2)$, $(1,0)$, and $(2,1)$, none of which is $(0,0)$.","o-component system $F(x)=(x,x-3)$ at $x=0$, with $q=3$ and $L=1$. Both components vanish modulo three. After writing a candidate as $y=3t$, its residual vector modulo three is\n\\[\n \\overline r+J\\overline t=(0,2)+(t,t)=(t,t-1).\n\\]\n\\AtlasImage{shared_obstruction}\nThe first coordinate can be corrected by $t=0$. The second can be corrected by $t=1$. No digit corrects both. The available residuals are $(0,2)$, $(1,0)$, and $(2,1)$, none of which is $(0,0)$.","nt system $F(x)=(x,x-3)$ at $x=0$, with $q=3$ and $L=1$. Both components vanish modulo three. After writing a candidate as $y=3t$, its residual vector modulo three is\n\\[\n \\overline r+J\\overline t=(0,2)+(t,t)=(t,t-1).\n\\]\n\\AtlasImage{shared_obstruction}\nThe first coordinate can be corrected by $t=0$. The second can be corrected by $t=1$. No digit corrects both. The available residuals are $(0,2)$, $(1,0)$, and $(2,1)$, none of which is $(0,0)$.","x)=(x,x-3)$ at $x=0$, with $q=3$ and $L=1$. Both components vanish modulo three. After writing a candidate as $y=3t$, its residual vector modulo three is\n\\[\n \\overline r+J\\overline t=(0,2)+(t,t)=(t,t-1).\n\\]\n\\AtlasImage{shared_obstruction}\nThe first coordinate can be corrected by $t=0$. The second can be corrected by $t=1$. No digit corrects both. The available residuals are $(0,2)$, $(1,0)$, and $(2,1)$, none of which is $(0,0)$.","$q=3$ and $L=1$. Both components vanish modulo three. After writing a candidate as $y=3t$, its residual vector modulo three is\n\\[\n \\overline r+J\\overline t=(0,2)+(t,t)=(t,t-1).\n\\]\n\\AtlasImage{shared_obstruction}\nThe first coordinate can be corrected by $t=0$. The second can be corrected by $t=1$. No digit corrects both. The available residuals are $(0,2)$, $(1,0)$, and $(2,1)$, none of which is $(0,0)$.","The combined Jacobian maps $t$ to the diagonal vector $(t,t)$. Its image is smaller than the product of its coordinate images. Checking the two projected obstructions independently therefore loses the compatibility condition. The common-correction theorem retains the full residual vector and the shared digit space, which is exactly the information needed to detect this failure.","\\AtlasHead{A derivative weight can cost one digit of root precision}{atlas:weighted}\nCompare the linear equations $F(x)=3x=0$ and $G_A(x)=3x+3^A=0$, where $A\\ge2$. Their coefficient difference has $3$-adic valuation $A$, but their roots differ by\n\\[\n x_A-0=-3^{A-1},\\qquad v_3(x_A)=A-1.\n\\]\n\\AtlasImage{weighted_motion}\nThe derivative has valuation one. Solving the correction equation divides by three, so one digit of coefficient precision is lost in root displacement. The calculation is exact for every displayed $A$ and also follows immediately from the two linear equations.","x)=3x=0$ and $G_A(x)=3x+3^A=0$, where $A\\ge2$. Their coefficient difference has $3$-adic valuation $A$, but their roots differ by\n\\[\n x_A-0=-3^{A-1},\\qquad v_3(x_A)=A-1.\n\\]\n\\AtlasImage{weighted_motion}\nThe derivative has valuation one. Solving the correction equation divides by three, so one digit of coefficient precision is lost in root displacement. The calculation is exact for every displayed $A$ and also follows immediately from the two linear equations.","This example explains why weighted lifting and deformation estimates track a Jacobian loss rather than silently treating every invertible derivative over $\\mathbb Q_p$ as a unit over $\\mathbb Z_p$. A valuation-zero derivative would not incur this particular loss. Nonlinear systems additionally require their stated neighborhood, residual, and right-inverse hypotheses."]}