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normalized rational sequences start with\n\\begin{equation}\n (u_0,u_1)=(1,7/4),\\qquad(v_0,v_1)=(0,13/8).\n \\label{eq:app-initial}\n\\end{equation}\nFor $n\\geq1$, set\n\\begin{align}\n p(n)&=20n^2-8n+1,\\notag","The normalized rational sequences start with\n\\begin{equation}\n (u_0,u_1)=(1,7/4),\\qquad(v_0,v_1)=(0,13/8).\n \\label{eq:app-initial}\n\\end{equation}\nFor $n\\geq1$, set\n\\begin{align}\n p(n)&=20n^2-8n+1,\\notag\\\\\n q(n)&=3520n^6+5632n^5+2064n^4\\notag\\\\\n &\\hspace{1.6em}-384n^3-156n^2+16n+7,\\notag\\\\\n a_n&=(2n+1)^2(2n+2)^2p(n),\\notag\\\\\n c_n&=(2n-1)^2(2n)^2p(n+1).\n \\label{eq:app-recurrence-coeffs}\n\\end{align}\nEach sequence $w=u$ or $w=v$ obeys\n\\begin{equation}\n a_nw_{n+1}=q(n)w_n+c_nw_{n-1}.\n \\label{eq:app-recurrence}\n\\end{equation}\nThe forward recurrence is used only at $n\\geq1$. Its coefficients $a_n$ are positive there. The first additional terms are\n\\begin{equation}\n u_2=\\frac{649}{64},\\quad u_3=\\frac{19471}{256},\\qquad\n v_2=\\frac{10699}{1152},\\quad v_3=\\frac{8025653}{115200}.\n \\label{eq:app-small-sequences}\n\\end{equation}","c_n&=(2n-1)^2(2n)^2p(n+1).\n \\label{eq:app-recurrence-coeffs}\n\\end{align}\nEach sequence $w=u$ or $w=v$ obeys\n\\begin{equation}\n a_nw_{n+1}=q(n)w_n+c_nw_{n-1}.\n \\label{eq:app-recurrence}\n\\end{equation}\nThe forward recurrence is used only at $n\\geq1$. Its coefficients $a_n$ are positive there. The first additional terms are\n\\begin{equation}\n u_2=\\frac{649}{64},\\quad u_3=\\frac{19471}{256},\\qquad\n v_2=\\frac{10699}{1152},\\quad v_3=\\frac{8025653}{115200}.\n \\label{eq:app-small-sequences}\n\\end{equation}","The Wronskian in this edition uses the orientation\n\\begin{equation}\n W_n=u_{n+1}v_n-u_nv_{n+1}\n =\\frac{(-1)^{n+1}p(n+1)}{2(2n+1)^2(2n+2)^2}.\n \\label{eq:app-Wronskian}\n\\end{equation}\nReversing its two terms reverses the sign. The formula applies to $n\\geq0$, with the base case supplied by the initial pair and the induction step supplied by \\eqref{eq:app-recurrence}.","Set $T_n=16^nu_n$. Its forward recurrence is\n\\begin{equation}\n a_nT_{n+1}=16q(n)T_n+256c_nT_{n-1},\n \\qquad T_0=1,\\quad T_1=28.\n \\label{eq:app-T-recurrence}\n\\end{equation}\nThe initial terms are $1,28,2596,311536$. The universally integral expression is\n\\begin{equation}\n u_n=\\sum_{j=0}^n\\binom nj\n \\binom{n-\\tfrac12}{j}\\binom{n+j-\\tfrac12}{j},\n \\label{eq:app-half-binomial}\n\\end{equation}\nwith the external coefficient identity and the written half-binomial denominator argument kept distinct.","For $n,m\\geq0$, the integer and rational packets are\n\\begin{equation}\n S_{n,m}=\\sum_{i=0}^m\\binom miT_{n+i},\\qquad\n H_{n,m}=\\sum_{i=0}^m\\binom mi16^{n+i}v_{n+i}.\n \\label{eq:app-packets}\n\\end{equation}\nTheir raw linear form and integral filter satisfy\n\\begin{align}\n F_{n,m}&=8(-1)^n(S_{n,m}G-H_{n,m}),\\notag","For $n,m\\geq0$, the integer and rational packets are\n\\begin{equation}\n S_{n,m}=\\sum_{i=0}^m\\binom miT_{n+i},\\qquad\n H_{n,m}=\\sum_{i=0}^m\\binom mi16^{n+i}v_{n+i}.\n \\label{eq:app-packets}\n\\end{equation}\nTheir raw linear form and integral filter satisfy\n\\begin{align}\n F_{n,m}&=8(-1)^n(S_{n,m}G-H_{n,m}),\\notag\\\\\n \\widetilde J_{n,m}\n &=\\sum_{i=0}^m(-16)^i\\binom miJ_{n+i}\n =16^{-n}F_{n,m}.\n \\label{eq:app-filter}\n\\end{align}\nThe corrected unfiltered integral is\n\\begin{equation}\n J_n=\\int_0^1\\!\\int_0^1\n \\frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}}\n {(1-xy)^{n+1}}\\,\\dd x\\,\\dd y\n =8(-1)^n(u_nG-v_n).\n \\label{eq:app-J}\n\\end{equation}\nThese are fixed normalization identities. Their factors should be checked together when translating a differently normalized source.","\\widetilde J_{n,m}\n &=\\sum_{i=0}^m(-16)^i\\binom miJ_{n+i}\n =16^{-n}F_{n,m}.\n \\label{eq:app-filter}\n\\end{align}\nThe corrected unfiltered integral is\n\\begin{equation}\n J_n=\\int_0^1\\!\\int_0^1\n \\frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}}\n {(1-xy)^{n+1}}\\,\\dd x\\,\\dd y\n =8(-1)^n(u_nG-v_n).\n \\label{eq:app-J}\n\\end{equation}\nThese are fixed normalization identities. Their factors should be checked together when translating a differently normalized source.","For a fixed pair, put $U=S_{n,m}$ and write $V=H_{n,m}=N/D$ in reduced form. The exact dictionary is\n\\begin{align}\n Q&=8L_{2(n+m)}^2,\\notag","For a fixed pair, put $U=S_{n,m}$ and write $V=H_{n,m}=N/D$ in reduced form. The exact dictionary is\n\\begin{align}\n Q&=8L_{2(n+m)}^2,\\notag\\\\\n A&=(-1)^nQU,\\qquad B=(-1)^nQV,\\notag\\\\\n h&=\\gcd(DU,N)=\\gcd(U,N),\\notag\\\\\n g&=\\gcd(|A|,|B|)=\\frac QD\\,h,\\notag\\\\\n q_{\\rm red}&=\\frac{DU}{h},\\qquad p_{\\rm red}=\\frac Nh,\\notag\\\\\n \\Lambda_{n,m}&=\\frac AgG-\\frac Bg\n =\\frac{16^nL_{2(n+m)}^2}{g}\\widetilde J_{n,m}.\n \\label{eq:app-reduction-dictionary}\n\\end{align}\nThe symbol $q_{\\rm red}$ denotes a reduced denominator. It is unrelated to the polynomial $q(n)$ in the recurrence. The rational coefficients of Eskandari's integral normalization are $\\widetilde U_{n,m}=8(-1)^n16^{-n}S_{n,m}$ and $\\widetilde V_{n,m}=8(-1)^n16^{-n}H_{n,m}$. They must not be mistaken for the fixed-pair aliases $U,V$ above.","\\Lambda_{n,m}&=\\frac AgG-\\frac Bg\n =\\frac{16^nL_{2(n+m)}^2}{g}\\widetilde J_{n,m}.\n \\label{eq:app-reduction-dictionary}\n\\end{align}\nThe symbol $q_{\\rm red}$ denotes a reduced denominator. It is unrelated to the polynomial $q(n)$ in the recurrence. The rational coefficients of Eskandari's integral normalization are $\\widetilde U_{n,m}=8(-1)^n16^{-n}S_{n,m}$ and $\\widetilde V_{n,m}=8(-1)^n16^{-n}H_{n,m}$. They must not be mistaken for the fixed-pair aliases $U,V$ above.","g}\\widetilde J_{n,m}.\n \\label{eq:app-reduction-dictionary}\n\\end{align}\nThe symbol $q_{\\rm red}$ denotes a reduced denominator. It is unrelated to the polynomial $q(n)$ in the recurrence. The rational coefficients of Eskandari's integral normalization are $\\widetilde U_{n,m}=8(-1)^n16^{-n}S_{n,m}$ and $\\widetilde V_{n,m}=8(-1)^n16^{-n}H_{n,m}$. They must not be mistaken for the fixed-pair aliases $U,V$ above.","\\midrule\n\\endhead\n\\bottomrule\n\\endfoot\nLC01 & Zudilin, 2003 & Recurrence, initial pairs, and identification of the limit with $G$. The recurrence is solved forwards only for $n\\geq1$.","[5pt]\nLC03 & Zudilin, 2003 & Integral identity with an explicitly audited normalization discrepancy. This edition uses the corrected factor $1/8$.","[5pt]\nLC04 & Zudilin, 2002 & Asymptotic denominator bounds. They do not alone prove the exact integrality of every $T_n$.","[5pt]\nLC06 & Eskandari, 2026 & The filter and its coefficient dictionary. The base measure has $1-xy$ outside the square root.","[5pt]\nLC07 & Eskandari, 2026 & The bound $M(\\alpha)$ and approximation exponent. It is a preprint input below the irrationality threshold.","[5pt]\nLC09 & Zudilin, 2019 & Background disjunction: at least one of six even beta values is irrational. It does not isolate $\\beta(2)=G$.","[5pt]\nLC10 & Zagier, 1997 & Prime number theorem input and the conversion to $\\log L_k=k+o(k)$.","At $n=1$, direct substitution in \\eqref{eq:app-recurrence-coeffs} gives\n\\begin{equation*}\n p(1)=13,\\quad p(2)=65,\\quad q(1)=10699,\n \\quad a_1=1872,\\quad c_1=260.\n\\end{equation*}\nThe recurrence therefore yields\n\\begin{align*}\n u_2&=\\frac{10699\\cdot(7/4)+260}{1872}\n =\\frac{649}{64},","At $n=1$, direct substitution in \\eqref{eq:app-recurrence-coeffs} gives\n\\begin{equation*}\n p(1)=13,\\quad p(2)=65,\\quad q(1)=10699,\n \\quad a_1=1872,\\quad c_1=260.\n\\end{equation*}\nThe recurrence therefore yields\n\\begin{align*}\n u_2&=\\frac{10699\\cdot(7/4)+260}{1872}\n =\\frac{649}{64},\\\\\n v_2&=\\frac{10699\\cdot(13/8)}{1872}\n =\\frac{10699}{1152}.\n\\end{align*}\nMultiplying $u_2$ by $16^2$ gives $T_2=2596$. Alternatively, the integer recurrence numerator is\n\\begin{equation*}\n 16\\cdot10699\\cdot28+256\\cdot260\\cdot1\n =4\\,859\\,712=1872\\cdot2596.\n\\end{equation*}\nThe exact division has no remainder. This calculation illustrates the program's division check. The all-index integrality proof uses the separate half-binomial argument.","v_2&=\\frac{10699\\cdot(13/8)}{1872}\n =\\frac{10699}{1152}.\n\\end{align*}\nMultiplying $u_2$ by $16^2$ gives $T_2=2596$. Alternatively, the integer recurrence numerator is\n\\begin{equation*}\n 16\\cdot10699\\cdot28+256\\cdot260\\cdot1\n =4\\,859\\,712=1872\\cdot2596.\n\\end{equation*}\nThe exact division has no remainder. This calculation illustrates the program's division check. The all-index integrality proof uses the separate half-binomial argument.","The same small case checks the Wronskian orientation:\n\\begin{align*}\n W_0&=\\frac74\\cdot0-1\\cdot\\frac{13}{8}\n =-\\frac{13}{8},\\\\\n W_1&=\\frac{649}{64}\\cdot\\frac{13}{8}\n -\\frac74\\cdot\\frac{10699}{1152}\n =\\frac{65}{288}.\n\\end{align*}\nBoth agree with \\eqref{eq:app-Wronskian}. They show why the sign is $(-1)^{n+1}$ for this orientation.","W_1&=\\frac{649}{64}\\cdot\\frac{13}{8}\n -\\frac74\\cdot\\frac{10699}{1152}\n =\\frac{65}{288}.\n\\end{align*}\nBoth agree with \\eqref{eq:app-Wronskian}. They show why the sign is $(-1)^{n+1}$ for this orientation.","For $(n,m)=(1,1)$,\n\\begin{equation*}\n S_{1,1}=T_1+T_2=28+2596=2624=2^6\\cdot41.\n\\end{equation*}\nIts exact valuation is $v_2(S_{1,1})=6$. Pascal evolution gives the same result from the initial row: $S_{1,1}=S_{1,0}+S_{2,0}$.","If this calculation is carried out modulo $2^4$, its residue is zero. The residue therefore certifies only $v_2(S_{1,1})\\geq4$. It does not identify the exact valuation as $4$. At modulus $2^8$, the residue is $64$, whose valuation is $6<8$, so the modular valuation criterion certifies the exact value $6$. This concrete example explains both the unresolved branch and the sufficient precision branch in the scan.","The associated rational packet is\n\\begin{equation*}\n H_{1,1}=16v_1+16^2v_2\n =26+\\frac{21398}{9}\n =\\frac{21632}{9}.\n\\end{equation*}\nTherefore the raw form is\n\\begin{equation*}\n F_{1,1}=-8\\left(2624G-\\frac{21632}{9}\\right)\n =\\frac{173056}{9}-20992G.\n\\end{equation*}\nA certified interval for $G$ makes this value positive. This is the smallest counterexample in the finite audit to the proposed sign rule $\\operatorname{sign}F_{n,m}=(-1)^m$.","associated rational packet is\n\\begin{equation*}\n H_{1,1}=16v_1+16^2v_2\n =26+\\frac{21398}{9}\n =\\frac{21632}{9}.\n\\end{equation*}\nTherefore the raw form is\n\\begin{equation*}\n F_{1,1}=-8\\left(2624G-\\frac{21632}{9}\\right)\n =\\frac{173056}{9}-20992G.\n\\end{equation*}\nA certified interval for $G$ makes this value positive. This is the smallest counterexample in the finite audit to the proposed sign rule $\\operatorname{sign}F_{n,m}=(-1)^m$.","Continue with $(n,m)=(1,1)$. The reduced rational packet has $N=21632$ and $D=9$. The minimal pair is\n\\begin{equation*}\n (DS_{1,1},N)=(23616,21632).\n\\end{equation*}\nIts gcd is $h=64$. In fact $23616=64\\cdot369$ and $21632=64\\cdot338$, and $369$ and $338$ are coprime. Thus the reduced approximation is\n\\begin{equation*}\n \\frac{H_{1,1}}{S_{1,1}}=\\frac{338}{369}.\n\\end{equation*}","The universal factor uses $L_4=12$, hence $Q=8L_4^2=1152$. Its redundant scale is $Q/D=128$. Consequently\n\\begin{equation*}\n g=128\\cdot64=8192.\n\\end{equation*}\nThe signed cleared pair is $(-3\\,022\\,848,-2\\,768\\,896)$. Dividing by $8192$ gives $(-369,-338)$, so the primitive form is\n\\begin{equation}\n \\Lambda_{1,1}=338-369G.\n \\label{eq:app-small-primitive}\n\\end{equation}\nThe integral identity gives the same expression. Since $\\widetilde J_{1,1}=F_{1,1}/16$ and $16L_4^2/g=9/32$,\n\\begin{equation*}\n \\frac{16L_4^2}{g}\\widetilde J_{1,1}\n =\\frac9{512}F_{1,1}=338-369G.\n\\end{equation*}\nThis checks the factor $8$, the sign $(-1)^n$, and the denominator index in one exact example.","_4=12$, hence $Q=8L_4^2=1152$. Its redundant scale is $Q/D=128$. Consequently\n\\begin{equation*}\n g=128\\cdot64=8192.\n\\end{equation*}\nThe signed cleared pair is $(-3\\,022\\,848,-2\\,768\\,896)$. Dividing by $8192$ gives $(-369,-338)$, so the primitive form is\n\\begin{equation}\n \\Lambda_{1,1}=338-369G.\n \\label{eq:app-small-primitive}\n\\end{equation}\nThe integral identity gives the same expression. Since $\\widetilde J_{1,1}=F_{1,1}/16$ and $16L_4^2/g=9/32$,\n\\begin{equation*}\n \\frac{16L_4^2}{g}\\widetilde J_{1,1}\n =\\frac9{512}F_{1,1}=338-369G.\n\\end{equation*}\nThis checks the factor $8$, the sign $(-1)^n$, and the denominator index in one exact example.","edundant scale is $Q/D=128$. Consequently\n\\begin{equation*}\n g=128\\cdot64=8192.\n\\end{equation*}\nThe signed cleared pair is $(-3\\,022\\,848,-2\\,768\\,896)$. Dividing by $8192$ gives $(-369,-338)$, so the primitive form is\n\\begin{equation}\n \\Lambda_{1,1}=338-369G.\n \\label{eq:app-small-primitive}\n\\end{equation}\nThe integral identity gives the same expression. Since $\\widetilde J_{1,1}=F_{1,1}/16$ and $16L_4^2/g=9/32$,\n\\begin{equation*}\n \\frac{16L_4^2}{g}\\widetilde J_{1,1}\n =\\frac9{512}F_{1,1}=338-369G.\n\\end{equation*}\nThis checks the factor $8$, the sign $(-1)^n$, and the denominator index in one exact example.","quently\n\\begin{equation*}\n g=128\\cdot64=8192.\n\\end{equation*}\nThe signed cleared pair is $(-3\\,022\\,848,-2\\,768\\,896)$. Dividing by $8192$ gives $(-369,-338)$, so the primitive form is\n\\begin{equation}\n \\Lambda_{1,1}=338-369G.\n \\label{eq:app-small-primitive}\n\\end{equation}\nThe integral identity gives the same expression. Since $\\widetilde J_{1,1}=F_{1,1}/16$ and $16L_4^2/g=9/32$,\n\\begin{equation*}\n \\frac{16L_4^2}{g}\\widetilde J_{1,1}\n =\\frac9{512}F_{1,1}=338-369G.\n\\end{equation*}\nThis checks the factor $8$, the sign $(-1)^n$, and the denominator index in one exact example.","68\\,896)$. Dividing by $8192$ gives $(-369,-338)$, so the primitive form is\n\\begin{equation}\n \\Lambda_{1,1}=338-369G.\n \\label{eq:app-small-primitive}\n\\end{equation}\nThe integral identity gives the same expression. Since $\\widetilde J_{1,1}=F_{1,1}/16$ and $16L_4^2/g=9/32$,\n\\begin{equation*}\n \\frac{16L_4^2}{g}\\widetilde J_{1,1}\n =\\frac9{512}F_{1,1}=338-369G.\n\\end{equation*}\nThis checks the factor $8$, the sign $(-1)^n$, and the denominator index in one exact example.","iding by $8192$ gives $(-369,-338)$, so the primitive form is\n\\begin{equation}\n \\Lambda_{1,1}=338-369G.\n \\label{eq:app-small-primitive}\n\\end{equation}\nThe integral identity gives the same expression. Since $\\widetilde J_{1,1}=F_{1,1}/16$ and $16L_4^2/g=9/32$,\n\\begin{equation*}\n \\frac{16L_4^2}{g}\\widetilde J_{1,1}\n =\\frac9{512}F_{1,1}=338-369G.\n\\end{equation*}\nThis checks the factor $8$, the sign $(-1)^n$, and the denominator index in one exact example.","For comparison, $(n,m)=(1,0)$ has $S=28,H=26,D=1,h=2,Q=32,g=64$. Its primitive form is $13-14G=J_1$. At $(0,0)$, the convention $H=0,D=1$ gives $S=1,Q=8,g=8$ and primitive form $G=J_0/8$. These boundary cases require no valuation of the integer zero.","\\midrule\n\\endhead\n\\bottomrule\n\\endfoot\nWronskian run & $0\\leq n\\leq200$ & Exact rational equalities at all recorded indices.","[5pt]\nInteger $T$ run & $0\\leq n\\leq10000$ & Every forward exact division has zero remainder.","[5pt]\nLarge Pascal scan & $1\\leq n\\leq2000$, $0\\leq m\\leq4n$, modulus $2^{128}$ & All $8\\,006\\,000$ valuations are exact, maximum $31$.","[5pt]\nDirect grid & $1\\leq n\\leq100$, $0\\leq m\\leq4n$ & All $20\\,300$ direct sums agree with the modular scan.","[5pt]\nIndependent recurrence & $0\\leq n\\leq500$ & Rational sequence agreement and stated clearing checks.","[5pt]\nHalf-binomial check & $0\\leq n\\leq50$, $0\\leq j\\leq n$ & All $1\\,326$ summands and sequence sums agree.","[5pt]\nCertified $G$ interval & Euler terms $j=0,\\ldots,1000$ & Exact rational endpoints with a tail of at most $2^{-1001}$.","[5pt]\nRaw filter signs & $1\\leq n\\leq20$, $0\\leq m\\leq4n$ & All $860$ forms are nonzero, and $138$ disagree with the sign guess.","[5pt]\nPrecision negative control & $1\\leq n\\leq3$, $0\\leq m\\leq4n$, modulus $2^4$ & Eight zero residues are unresolved, and the direct maximum is $6$.","The Euler interval uses $a_k=(2k+1)^{-2}$ and the positive difference convention $\\Delta a_k=a_k-a_{k+1}$. The identity\n\\begin{equation*}\n \\Delta^ja_0=\\int_0^1(1-x^2)^j(-\\log x)\\,\\dd x\n\\end{equation*}\ngives $0\\leq\\Delta^ja_0\\leq1$. The positive Euler series for $G$ is $\\sum_{j\\geq0}2^{-j-1}\\Delta^ja_0$. Keeping the first $1001$ terms therefore gives a lower endpoint and an upper endpoint obtained by adding $2^{-1001}$. All retained terms are rational. The analytic tail bound is a written argument. The computed interval is its finite arithmetic application.","nterval uses $a_k=(2k+1)^{-2}$ and the positive difference convention $\\Delta a_k=a_k-a_{k+1}$. The identity\n\\begin{equation*}\n \\Delta^ja_0=\\int_0^1(1-x^2)^j(-\\log x)\\,\\dd x\n\\end{equation*}\ngives $0\\leq\\Delta^ja_0\\leq1$. The positive Euler series for $G$ is $\\sum_{j\\geq0}2^{-j-1}\\Delta^ja_0$. Keeping the first $1001$ terms therefore gives a lower endpoint and an upper endpoint obtained by adding $2^{-1001}$. All retained terms are rational. The analytic tail bound is a written argument. The computed interval is its finite arithmetic application.","\\bibitem{Eskandari2026}\nPayman Eskandari.\n\\emph{Improved rational approximations to Catalan's constant}.\nPreprint \\href{https://arxiv.org/abs/2609.26354v1}{2609.26354v1}, 22 September 2026.\nIntegral filter and a guaranteed approximation exponent at least $0.62$, below the irrationality threshold.","The elementary series gives rational approximations to $G$ term by term.\nZudilin's recurrence gives a different family whose errors decrease much\nfaster \\cite{Zudilin2003}. Its coefficients and initial values are rational,\nso every finite calculation can be performed exactly. The recurrence is\nalso a useful structural object: a two dimensional transfer matrix carries\neach pair of adjacent values to the next, and a determinant tracks the\nindependence of its two distinguished solutions.","For integer $n\\geq0$, set\n\\begin{align}\n p(n)&=20n^2-8n+1,\\label{cb_eq_p_polynomial}","For integer $n\\geq0$, set\n\\begin{align}\n p(n)&=20n^2-8n+1,\\label{cb_eq_p_polynomial}\\\\\n q(n)&=3520n^6+5632n^5+2064n^4-384n^3-156n^2+16n+7.\n \\label{cb_eq_q_polynomial}\n\\end{align}\nFor $n\\geq1$, abbreviate the leading and trailing coefficients by\n\\begin{equation}\\label{cb_eq_a_c_def}\n a_n=(2n+1)^2(2n+2)^2p(n),\\qquad\n c_n=(2n-1)^2(2n)^2p(n+1).\n\\end{equation}\nThe recurrence is\n\\begin{equation}\\label{cb_eq_zudilin_recurrence}\n a_nw_{n+1}-q(n)w_n-c_nw_{n-1}=0,\\qquad n\\geq1.\n\\end{equation}\nThe signs of the last two terms are part of the normalization. Solving\nforward gives $w_{n+1}=(q(n)w_n+c_nw_{n-1})/a_n$.","q(n)&=3520n^6+5632n^5+2064n^4-384n^3-156n^2+16n+7.\n \\label{cb_eq_q_polynomial}\n\\end{align}\nFor $n\\geq1$, abbreviate the leading and trailing coefficients by\n\\begin{equation}\\label{cb_eq_a_c_def}\n a_n=(2n+1)^2(2n+2)^2p(n),\\qquad\n c_n=(2n-1)^2(2n)^2p(n+1).\n\\end{equation}\nThe recurrence is\n\\begin{equation}\\label{cb_eq_zudilin_recurrence}\n a_nw_{n+1}-q(n)w_n-c_nw_{n-1}=0,\\qquad n\\geq1.\n\\end{equation}\nThe signs of the last two terms are part of the normalization. Solving\nforward gives $w_{n+1}=(q(n)w_n+c_nw_{n-1})/a_n$.","ign}\nFor $n\\geq1$, abbreviate the leading and trailing coefficients by\n\\begin{equation}\\label{cb_eq_a_c_def}\n a_n=(2n+1)^2(2n+2)^2p(n),\\qquad\n c_n=(2n-1)^2(2n)^2p(n+1).\n\\end{equation}\nThe recurrence is\n\\begin{equation}\\label{cb_eq_zudilin_recurrence}\n a_nw_{n+1}-q(n)w_n-c_nw_{n-1}=0,\\qquad n\\geq1.\n\\end{equation}\nThe signs of the last two terms are part of the normalization. Solving\nforward gives $w_{n+1}=(q(n)w_n+c_nw_{n-1})/a_n$.","\\begin{definition}[The distinguished rational sequences]\\label{cb_def_uv}\nLet $(u_n)_{n\\geq0}$ and $(v_n)_{n\\geq0}$ satisfy\n\\eqref{cb_eq_zudilin_recurrence}, with\n\\begin{equation}\\label{cb_eq_uv_initial}\n (u_0,u_1)=\\left(1,\\frac74\\right),\\qquad\n (v_0,v_1)=\\left(0,\\frac{13}{8}\\right).\n\\end{equation}\nTheir rational approximants are\n\\begin{equation}\\label{cb_eq_r_def}\n r_n=\\frac{v_n}{u_n}.\n\\end{equation}\n\\end{definition}","The notation $a_n,c_n$ in this chapter and the next denotes recurrence\ncoefficients. The later source notation $a_n,b_n$ is translated explicitly\nwhen it occurs, and the cleared packet coefficients $A_{n,m},B_{n,m}$ have\ntwo indices. These are different objects.","The forward equation starts at $n=1$. Substituting $n=0$ into its polynomial\nexpression would remove the trailing term and impose $4w_1=7w_0$. Although\nthe initial values of $u$ happen to satisfy that extra equation, those of\n$v$ do not. It is therefore not a valid additional recurrence condition\nfor the pair. At $n=1$, both $w_0$ and $w_1$ are already specified and\n$w_2$ is the first value to be computed.","\\begin{proposition}[Well defined rational solutions]\\label{cb_prop_forward}\nEvery initial pair $(w_0,w_1)\\in\\Q^2$ determines exactly one rational\nsequence satisfying \\eqref{cb_eq_zudilin_recurrence}. In particular the\nsequences in Definition~\\ref{cb_def_uv} exist uniquely. Moreover,\n\\[\n u_n>0\\quad(n\\geq0),\\qquad v_n>0\\quad(n\\geq1),\\qquad v_0=0.\n\\]\n\\end{proposition}","\\begin{proof}\nCompleting the square gives\n\\[\n p(n)=20\\left(n-\\frac15\\right)^2+\\frac15>0.\n\\]\nThus $a_n,c_n>0$ for $n\\geq1$. Also, for such $n$,\n\\[\n 2064n^4-384n^3-156n^2\\geq1524n^4>0,\n\\]\nbecause $n^3\\leq n^4$ and $n^2\\leq n^4$. All the remaining terms of $q(n)$\nare positive, so $q(n)>0$. Division by $a_n$ therefore makes forward\niteration possible. It preserves rationality and determines the next term\nuniquely. Induction proves existence and uniqueness at every index.","The initial $u$ values are positive, and each subsequent value is a positive\nlinear combination of its two predecessors. For $v$, the initial $v_0$ is\nzero and $v_1>0$. The first forward step gives $v_2>0$, and the same\ninduction applies thereafter. Consequently the quotient $r_n$ is defined\nat every index.\n\\end{proof}","The solution space also has a concrete basis. For any rational solution\n$w$, put\n\\[\n x=w_0,\\qquad y=\\frac8{13}\\left(w_1-\\frac74w_0\\right).\n\\]\nThe sequence $xu_n+yv_n$ has the same initial pair as $w_n$. By uniqueness,\n\\begin{equation}\\label{cb_eq_solution_basis}\n w_n=xu_n+yv_n\\qquad(n\\geq0).\n\\end{equation}\nThus the entire space of rational solutions is two dimensional. The factor\n$13/8$ in the initial data is nonzero, so there is no degeneracy in this\nbasis calculation. The same argument works for real initial values,\nyielding the two dimensional real solution space.","\\begin{figure}[tbp]\n\\centering\\includegraphics[width=0.93\\linewidth]{../visual/figures/09_recurrence_structure.pdf}\n\\caption{Two preceding values determine the next rational value. The\nforward rule begins at $n=1$ and uses the stated initial pair. Here and in\nsubsequent recurrence diagrams, $A_n=a_n$, $Q_n=q(n)$ and $C_n=c_n$.}\n\\label{cb_fig_recurrence}\n\\end{figure}","The first steps expose all constants and signs. At $n=1$,\n\\[\n p(1)=13,\\quad p(2)=65,\\quad q(1)=10699,\\qquad\n a_1=1872,\\quad c_1=260.\n\\]\nFor $u$, the recurrence gives\n\\[\n u_2=\\frac{10699(7/4)+260}{1872}\n     =\\frac{75933/4}{1872}=\\frac{649}{64}.\n\\]\nFor $v$, its zero initial term removes the trailing contribution:\n\\[\n v_2=\\frac{10699(13/8)+260\\cdot0}{1872}\n     =\\frac{139087/8}{1872}=\\frac{10699}{1152}.\n\\]\nTherefore the second rational approximant is\n\\[\n r_2=\\frac{10699}{11682}.\n\\]","$,\n\\[\n p(1)=13,\\quad p(2)=65,\\quad q(1)=10699,\\qquad\n a_1=1872,\\quad c_1=260.\n\\]\nFor $u$, the recurrence gives\n\\[\n u_2=\\frac{10699(7/4)+260}{1872}\n     =\\frac{75933/4}{1872}=\\frac{649}{64}.\n\\]\nFor $v$, its zero initial term removes the trailing contribution:\n\\[\n v_2=\\frac{10699(13/8)+260\\cdot0}{1872}\n     =\\frac{139087/8}{1872}=\\frac{10699}{1152}.\n\\]\nTherefore the second rational approximant is\n\\[\n r_2=\\frac{10699}{11682}.\n\\]","At $n=2$ the coefficients become\n\\[\n p(2)=65,\\quad p(3)=157,\\quad q(2)=434871,\\qquad\n a_2=58500,\\quad c_2=22608.\n\\]\nUsing the fractions just obtained,\n\\begin{align*}\n u_3\n &=\\frac{434871(649/64)+22608(7/4)}{58500}","At $n=2$ the coefficients become\n\\[\n p(2)=65,\\quad p(3)=157,\\quad q(2)=434871,\\qquad\n a_2=58500,\\quad c_2=22608.\n\\]\nUsing the fractions just obtained,\n\\begin{align*}\n u_3\n &=\\frac{434871(649/64)+22608(7/4)}{58500}\\\\\n &=\\frac{284763375/64}{58500}=\\frac{19471}{256},\\\\[3pt]\n v_3\n &=\\frac{434871(10699/1152)+22608(13/8)}{58500}\\\\\n &=\\frac{521667445/128}{58500}=\\frac{8025653}{115200}.\n\\end{align*}\nIt follows that $r_3=8025653/8761950$. All these values were also checked\nagainst the recovered exact rational script. A floating point calculation\nis unnecessary for the equality at any stage.","&=\\frac{521667445/128}{58500}=\\frac{8025653}{115200}.\n\\end{align*}\nIt follows that $r_3=8025653/8761950$. All these values were also checked\nagainst the recovered exact rational script. A floating point calculation\nis unnecessary for the equality at any stage.","The first ratios are $r_0=0$ and $r_1=13/14$. The subsequent values lie\nnear the decimal for $G$, but proximity is only a first observation.\nChapter~\\ref{cb_ch_wronskian} proves exact nested brackets and an error\nsign at every index. The external identification below then makes those\nbrackets enclosures of Catalan's constant itself.","Normalize the recurrence by putting\n\\[\n \\alpha_n=\\frac{q(n)}{a_n},\\qquad\n \\beta_n=\\frac{c_n}{a_n}\\qquad(n\\geq1).\n\\]\nBoth quantities are positive rationals. For the adjacent state vector\n$X_n(w)=(w_n,w_{n-1})^{\\mathsf T}$, define\n\\begin{equation}\\label{cb_eq_transfer}\n M_n=\\begin{pmatrix}\\alpha_n&\\beta_n","Normalize the recurrence by putting\n\\[\n \\alpha_n=\\frac{q(n)}{a_n},\\qquad\n \\beta_n=\\frac{c_n}{a_n}\\qquad(n\\geq1).\n\\]\nBoth quantities are positive rationals. For the adjacent state vector\n$X_n(w)=(w_n,w_{n-1})^{\\mathsf T}$, define\n\\begin{equation}\\label{cb_eq_transfer}\n M_n=\\begin{pmatrix}\\alpha_n&\\beta_n\\\\1&0\\end{pmatrix}.\n \\qquad X_{n+1}(w)=M_nX_n(w).\n\\end{equation}\nHence\n\\[\n X_{N+1}(w)=M_NM_{N-1}\\cdots M_1X_1(w)\\qquad(N\\geq1).\n\\]\nThe order of multiplication follows the temporal order: $M_1$ acts first.\nThis is a product of changing matrices, not a power of one fixed matrix.","1&0\\end{pmatrix}.\n \\qquad X_{n+1}(w)=M_nX_n(w).\n\\end{equation}\nHence\n\\[\n X_{N+1}(w)=M_NM_{N-1}\\cdots M_1X_1(w)\\qquad(N\\geq1).\n\\]\nThe order of multiplication follows the temporal order: $M_1$ acts first.\nThis is a product of changing matrices, not a power of one fixed matrix.","The determinant is\n\\[\n \\det M_n=-\\beta_n=-\\frac{c_n}{a_n}<0.\n\\]\nEvery transfer is invertible, with\n\\[\n M_n^{-1}=\\begin{pmatrix}0&1","The determinant is\n\\[\n \\det M_n=-\\beta_n=-\\frac{c_n}{a_n}<0.\n\\]\nEvery transfer is invertible, with\n\\[\n M_n^{-1}=\\begin{pmatrix}0&1\\\\1/\\beta_n&-\\alpha_n/\\beta_n\\end{pmatrix}.\n\\]\nAn adjacent pair at index $n+1$ therefore determines the preceding pair at\nindex $n$, within the stated recurrence domain. This does not extend the\nequation through $n=0$, where the trailing coefficient would vanish.","1/\\beta_n&-\\alpha_n/\\beta_n\\end{pmatrix}.\n\\]\nAn adjacent pair at index $n+1$ therefore determines the preceding pair at\nindex $n$, within the stated recurrence domain. This does not extend the\nequation through $n=0$, where the trailing coefficient would vanish.","The two distinguished solutions can be collected into the fundamental\nmatrix\n\\[\n \\mathcal F_n=\n \\begin{pmatrix}u_n&v_n\\\\u_{n-1}&v_{n-1}\\end{pmatrix},\\qquad n\\geq1.\n\\]\nIt satisfies $\\mathcal F_{n+1}=M_n\\mathcal F_n$. Its initial determinant is\n\\[\n \\det\\mathcal F_1=\n \\det\\begin{pmatrix}7/4&13/8","u_{n-1}&v_{n-1}\\end{pmatrix},\\qquad n\\geq1.\n\\]\nIt satisfies $\\mathcal F_{n+1}=M_n\\mathcal F_n$. Its initial determinant is\n\\[\n \\det\\mathcal F_1=\n \\det\\begin{pmatrix}7/4&13/8","u_{n-1}&v_{n-1}\\end{pmatrix},\\qquad n\\geq1.\n\\]\nIt satisfies $\\mathcal F_{n+1}=M_n\\mathcal F_n$. Its initial determinant is\n\\[\n \\det\\mathcal F_1=\n \\det\\begin{pmatrix}7/4&13/8\\\\1&0\\end{pmatrix}=-\\frac{13}{8}.\n\\]\nMultiplicativity of determinants now gives\n\\[\n \\det\\mathcal F_{N+1}\n =-\\frac{13}{8}\\prod_{n=1}^N(-\\beta_n)\\ne0.\n\\]\nThus adjacent values of the two solutions remain independent at every\nindex. Each transfer reverses the determinant sign. The closed evaluation\nof this product is the Wronskian theorem in\nChapter~\\ref{cb_ch_wronskian}, where the factors telescope and yield exact\nquotient increments. Here the matrix description identifies the mechanism\nwithout duplicating that full proof.","1&0\\end{pmatrix}=-\\frac{13}{8}.\n\\]\nMultiplicativity of determinants now gives\n\\[\n \\det\\mathcal F_{N+1}\n =-\\frac{13}{8}\\prod_{n=1}^N(-\\beta_n)\\ne0.\n\\]\nThus adjacent values of the two solutions remain independent at every\nindex. Each transfer reverses the determinant sign. The closed evaluation\nof this product is the Wronskian theorem in\nChapter~\\ref{cb_ch_wronskian}, where the factors telescope and yield exact\nquotient increments. Here the matrix description identifies the mechanism\nwithout duplicating that full proof.","The transfer product can also be written using continuants. For integers\n$1\\leq r\\leq s$, let $K_{r,s}$ be the determinant of the tridiagonal matrix\nwhose diagonal is $\\alpha_r,\\ldots,\\alpha_s$, lower adjacent diagonal is\none, and upper adjacent diagonal is\n$-\\beta_{r+1},\\ldots,-\\beta_s$. For instance,\n\\[\n K_{r,r}=\\alpha_r,\\qquad\n K_{r,r+1}=\\det\n \\begin{pmatrix}\\alpha_r&-\\beta_{r+1}","The transfer product can also be written using continuants. For integers\n$1\\leq r\\leq s$, let $K_{r,s}$ be the determinant of the tridiagonal matrix\nwhose diagonal is $\\alpha_r,\\ldots,\\alpha_s$, lower adjacent diagonal is\none, and upper adjacent diagonal is\n$-\\beta_{r+1},\\ldots,-\\beta_s$. For instance,\n\\[\n K_{r,r}=\\alpha_r,\\qquad\n K_{r,r+1}=\\det\n \\begin{pmatrix}\\alpha_r&-\\beta_{r+1}\\\\1&\\alpha_{r+1}\\end{pmatrix}\n =\\alpha_r\\alpha_{r+1}+\\beta_{r+1}.\n\\]\nSet $K_{r,r-1}=1$ for the empty determinant. Expansion at the last row gives\n\\begin{equation}\\label{cb_eq_continuant}\n K_{r,s}=\\alpha_sK_{r,s-1}+\\beta_sK_{r,s-2}\n \\qquad(s\\geq r+1).\n\\end{equation}\nThe plus sign comes from the minus sign on the upper adjacent diagonal.\nThis convention matches the positive trailing term in the recurrence.\nAll these continuants are positive rationals.","1&\\alpha_{r+1}\\end{pmatrix}\n =\\alpha_r\\alpha_{r+1}+\\beta_{r+1}.\n\\]\nSet $K_{r,r-1}=1$ for the empty determinant. Expansion at the last row gives\n\\begin{equation}\\label{cb_eq_continuant}\n K_{r,s}=\\alpha_sK_{r,s-1}+\\beta_sK_{r,s-2}\n \\qquad(s\\geq r+1).\n\\end{equation}\nThe plus sign comes from the minus sign on the upper adjacent diagonal.\nThis convention matches the positive trailing term in the recurrence.\nAll these continuants are positive rationals.","\\begin{proposition}[Initial data through continuants]\\label{cb_prop_continuant}\nFor any solution $w$ and every $n\\geq2$,\n\\begin{equation}\\label{cb_eq_continuant_solution}\n w_n=K_{1,n-1}w_1+\\beta_1K_{2,n-1}w_0.\n\\end{equation}\n\\end{proposition}","\\begin{proof}\nAt $n=2$, the formula is\n$w_2=\\alpha_1w_1+\\beta_1w_0$, using the empty determinant for $K_{2,1}$.\nAt $n=3$, direct iteration gives\n\\[\n w_3=(\\alpha_2\\alpha_1+\\beta_2)w_1+\\alpha_2\\beta_1w_0,\n\\]\nwhich agrees with the formula. For later indices, both coefficients in\n\\eqref{cb_eq_continuant_solution} satisfy the same recurrence by\n\\eqref{cb_eq_continuant}. Induction finishes the proof.\n\\end{proof}","For a positive solution, ratios of consecutive values give another finite\ndescription. If $x_n=w_n/w_{n-1}$ is defined, then\n\\[\n x_{n+1}=\\alpha_n+\\frac{\\beta_n}{x_n}.\n\\]\nFor $u$, every such ratio is defined. The first two iterations are\n\\[\n \\frac{u_2}{u_1}=\\alpha_1+\\frac{\\beta_1}{7/4},\\qquad\n \\frac{u_3}{u_2}=\n \\alpha_2+\\frac{\\beta_2}{\\alpha_1+\\beta_1/(7/4)}.\n\\]\nRepeated substitution yields a finite generalized continued fraction.\nFor $v$, one cannot begin with $v_1/v_0$, because $v_0=0$. Its ratio\niteration instead begins at $v_2/v_1=\\alpha_1$. These formulas are exact\nfinite identities. Establishing a limit for an infinite continued fraction\nrequires an additional argument and is not implied merely by writing the\nfinite nested fractions.","ecutive values give another finite\ndescription. If $x_n=w_n/w_{n-1}$ is defined, then\n\\[\n x_{n+1}=\\alpha_n+\\frac{\\beta_n}{x_n}.\n\\]\nFor $u$, every such ratio is defined. The first two iterations are\n\\[\n \\frac{u_2}{u_1}=\\alpha_1+\\frac{\\beta_1}{7/4},\\qquad\n \\frac{u_3}{u_2}=\n \\alpha_2+\\frac{\\beta_2}{\\alpha_1+\\beta_1/(7/4)}.\n\\]\nRepeated substitution yields a finite generalized continued fraction.\nFor $v$, one cannot begin with $v_1/v_0$, because $v_0=0$. Its ratio\niteration instead begins at $v_2/v_1=\\alpha_1$. These formulas are exact\nfinite identities. Establishing a limit for an infinite continued fraction\nrequires an additional argument and is not implied merely by writing the\nfinite nested fractions.","finite\ndescription. If $x_n=w_n/w_{n-1}$ is defined, then\n\\[\n x_{n+1}=\\alpha_n+\\frac{\\beta_n}{x_n}.\n\\]\nFor $u$, every such ratio is defined. The first two iterations are\n\\[\n \\frac{u_2}{u_1}=\\alpha_1+\\frac{\\beta_1}{7/4},\\qquad\n \\frac{u_3}{u_2}=\n \\alpha_2+\\frac{\\beta_2}{\\alpha_1+\\beta_1/(7/4)}.\n\\]\nRepeated substitution yields a finite generalized continued fraction.\nFor $v$, one cannot begin with $v_1/v_0$, because $v_0=0$. Its ratio\niteration instead begins at $v_2/v_1=\\alpha_1$. These formulas are exact\nfinite identities. Establishing a limit for an infinite continued fraction\nrequires an additional argument and is not implied merely by writing the\nfinite nested fractions.",". If $x_n=w_n/w_{n-1}$ is defined, then\n\\[\n x_{n+1}=\\alpha_n+\\frac{\\beta_n}{x_n}.\n\\]\nFor $u$, every such ratio is defined. The first two iterations are\n\\[\n \\frac{u_2}{u_1}=\\alpha_1+\\frac{\\beta_1}{7/4},\\qquad\n \\frac{u_3}{u_2}=\n \\alpha_2+\\frac{\\beta_2}{\\alpha_1+\\beta_1/(7/4)}.\n\\]\nRepeated substitution yields a finite generalized continued fraction.\nFor $v$, one cannot begin with $v_1/v_0$, because $v_0=0$. Its ratio\niteration instead begins at $v_2/v_1=\\alpha_1$. These formulas are exact\nfinite identities. Establishing a limit for an infinite continued fraction\nrequires an additional argument and is not implied merely by writing the\nfinite nested fractions.","n+1}=\\alpha_n+\\frac{\\beta_n}{x_n}.\n\\]\nFor $u$, every such ratio is defined. The first two iterations are\n\\[\n \\frac{u_2}{u_1}=\\alpha_1+\\frac{\\beta_1}{7/4},\\qquad\n \\frac{u_3}{u_2}=\n \\alpha_2+\\frac{\\beta_2}{\\alpha_1+\\beta_1/(7/4)}.\n\\]\nRepeated substitution yields a finite generalized continued fraction.\nFor $v$, one cannot begin with $v_1/v_0$, because $v_0=0$. Its ratio\niteration instead begins at $v_2/v_1=\\alpha_1$. These formulas are exact\nfinite identities. Establishing a limit for an infinite continued fraction\nrequires an additional argument and is not implied merely by writing the\nfinite nested fractions.","The leading terms of the coefficient polynomials are\n\\[\n a_n=320n^6+O(n^5),\\qquad\n q(n)=3520n^6+O(n^5),\\qquad\n c_n=320n^6+O(n^5).\n\\]\nConsequently $\\alpha_n\\to11$, $\\beta_n\\to1$, and\n\\[\n M_n\\longrightarrow M_\\infty=\\begin{pmatrix}11&1","The leading terms of the coefficient polynomials are\n\\[\n a_n=320n^6+O(n^5),\\qquad\n q(n)=3520n^6+O(n^5),\\qquad\n c_n=320n^6+O(n^5).\n\\]\nConsequently $\\alpha_n\\to11$, $\\beta_n\\to1$, and\n\\[\n M_n\\longrightarrow M_\\infty=\\begin{pmatrix}11&1\\\\1&0\\end{pmatrix}.\n\\]\nThe characteristic polynomial of this limiting matrix is\n\\begin{equation}\\label{cb_eq_characteristic}\n \\lambda^2-11\\lambda-1.\n\\end{equation}\nIts two roots are\n\\begin{equation}\\label{cb_eq_characteristic_roots}\n \\lambda_+=\\frac{11+5\\sqrt5}{2}=\\phi^5,\\qquad\n \\lambda_-=\\frac{11-5\\sqrt5}{2}=-\\phi^{-5}.\n\\end{equation}\nHere $\\phi=(1+\\sqrt5)/2$. The identity $\\phi^5=5\\phi+3$ follows by\nrepeatedly using $\\phi^2=\\phi+1$. The roots have product $-1$, as required\nby the limiting determinant, and very different magnitudes.","1&0\\end{pmatrix}.\n\\]\nThe characteristic polynomial of this limiting matrix is\n\\begin{equation}\\label{cb_eq_characteristic}\n \\lambda^2-11\\lambda-1.\n\\end{equation}\nIts two roots are\n\\begin{equation}\\label{cb_eq_characteristic_roots}\n \\lambda_+=\\frac{11+5\\sqrt5}{2}=\\phi^5,\\qquad\n \\lambda_-=\\frac{11-5\\sqrt5}{2}=-\\phi^{-5}.\n\\end{equation}\nHere $\\phi=(1+\\sqrt5)/2$. The identity $\\phi^5=5\\phi+3$ follows by\nrepeatedly using $\\phi^2=\\phi+1$. The roots have product $-1$, as required\nby the limiting determinant, and very different magnitudes.","The characteristic polynomial of this limiting matrix is\n\\begin{equation}\\label{cb_eq_characteristic}\n \\lambda^2-11\\lambda-1.\n\\end{equation}\nIts two roots are\n\\begin{equation}\\label{cb_eq_characteristic_roots}\n \\lambda_+=\\frac{11+5\\sqrt5}{2}=\\phi^5,\\qquad\n \\lambda_-=\\frac{11-5\\sqrt5}{2}=-\\phi^{-5}.\n\\end{equation}\nHere $\\phi=(1+\\sqrt5)/2$. The identity $\\phi^5=5\\phi+3$ follows by\nrepeatedly using $\\phi^2=\\phi+1$. The roots have product $-1$, as required\nby the limiting determinant, and very different magnitudes.","miting matrix is\n\\begin{equation}\\label{cb_eq_characteristic}\n \\lambda^2-11\\lambda-1.\n\\end{equation}\nIts two roots are\n\\begin{equation}\\label{cb_eq_characteristic_roots}\n \\lambda_+=\\frac{11+5\\sqrt5}{2}=\\phi^5,\\qquad\n \\lambda_-=\\frac{11-5\\sqrt5}{2}=-\\phi^{-5}.\n\\end{equation}\nHere $\\phi=(1+\\sqrt5)/2$. The identity $\\phi^5=5\\phi+3$ follows by\nrepeatedly using $\\phi^2=\\phi+1$. The roots have product $-1$, as required\nby the limiting determinant, and very different magnitudes.","The limiting polynomial explains the scales that appear in the analytic\ntheorem, but calculating its roots is not a substitute for that theorem.\nThe original transfer matrices vary with $n$. One must also identify which\nsolution or linear combination occupies which asymptotic scale. In\nparticular, numerical agreement between the ratio and $G$ cannot supply\nthe identification of the decaying combination $u_nG-v_n$.","iting polynomial explains the scales that appear in the analytic\ntheorem, but calculating its roots is not a substitute for that theorem.\nThe original transfer matrices vary with $n$. One must also identify which\nsolution or linear combination occupies which asymptotic scale. In\nparticular, numerical agreement between the ratio and $G$ cannot supply\nthe identification of the decaying combination $u_nG-v_n$.","\\begin{theorem}[External recurrence identification and rates]\\label{cb_thm_source_rates}\nFor the exact normalization \\eqref{cb_eq_uv_initial}, Zudilin's theorem and\nits stated asymptotic rates give\n\\begin{align}\n \\lim_{n\\to\\infty}\\frac{v_n}{u_n}&=G,\\label{cb_eq_source_limit}\\\\\n \\lim_{n\\to\\infty}u_n^{1/n}&=\\phi^5,\\label{cb_eq_source_u_rate}\\\\\n \\lim_{n\\to\\infty}|u_nG-v_n|^{1/n}&=\\phi^{-5}.\n \\label{cb_eq_source_error_rate}\n\\end{align}\nThese are external mathematical inputs, recorded precisely in source\ncontracts LC01 and LC02 \\cite{Zudilin2003}.\n\\end{theorem}","Chapter~\\ref{cb_ch_wronskian} gives a recurrence only proof that the ratios\nhave a limit, with strict alternating brackets. It then invokes\n\\eqref{cb_eq_source_limit} to name that limit $G$. This division of labor\nkeeps the exact recurrence reasoning visible while acknowledging the\nexternal analytic identification.","An immediate consequence of the two source rates is\n\\begin{equation}\\label{cb_eq_ratio_error_rate}\n \\lim_{n\\to\\infty}|G-r_n|^{1/n}\n =\\lim_{n\\to\\infty}\\frac{|u_nG-v_n|^{1/n}}{u_n^{1/n}}\n =\\phi^{-10}.\n\\end{equation}\nThus the ratio error has a smaller exponential scale than the unnormalized\nlinear form. The distinction comes from division by the growing\ncoefficient $u_n$. Also, since $r_n\\to G>0$, one has\n$r_n^{1/n}\\to1$ and consequently $v_n^{1/n}\\to\\phi^5$. Both individual\nsequences grow, while their specified linear combination decays.","diate consequence of the two source rates is\n\\begin{equation}\\label{cb_eq_ratio_error_rate}\n \\lim_{n\\to\\infty}|G-r_n|^{1/n}\n =\\lim_{n\\to\\infty}\\frac{|u_nG-v_n|^{1/n}}{u_n^{1/n}}\n =\\phi^{-10}.\n\\end{equation}\nThus the ratio error has a smaller exponential scale than the unnormalized\nlinear form. The distinction comes from division by the growing\ncoefficient $u_n$. Also, since $r_n\\to G>0$, one has\n$r_n^{1/n}\\to1$ and consequently $v_n^{1/n}\\to\\phi^5$. Both individual\nsequences grow, while their specified linear combination decays.","ce rates is\n\\begin{equation}\\label{cb_eq_ratio_error_rate}\n \\lim_{n\\to\\infty}|G-r_n|^{1/n}\n =\\lim_{n\\to\\infty}\\frac{|u_nG-v_n|^{1/n}}{u_n^{1/n}}\n =\\phi^{-10}.\n\\end{equation}\nThus the ratio error has a smaller exponential scale than the unnormalized\nlinear form. The distinction comes from division by the growing\ncoefficient $u_n$. Also, since $r_n\\to G>0$, one has\n$r_n^{1/n}\\to1$ and consequently $v_n^{1/n}\\to\\phi^5$. Both individual\nsequences grow, while their specified linear combination decays.","The scaled coefficient used in the packet construction has rate\n\\[\n \\lim_{n\\to\\infty}(16^nu_n)^{1/n}\n =16\\phi^5=88+40\\sqrt5.\n\\]\nChapter~\\ref{cb_ch_integrality} proves that this scaled coefficient is an\ninteger at every index using a separate half binomial identity. Its growth\nrate and its integrality are distinct mathematical facts.","The approximants therefore do much more than the original alternating\npartial sums in numerical accuracy. Their rapid convergence still does\nnot classify the limit as irrational. The fractions $u_n$ and $v_n$ carry\ndenominators, and the denominator of their ratio may grow rapidly too.\nTo apply the integer lattice criterion, one must turn the error into a\nnonzero integer linear form and show that its size tends to zero after\nall necessary clearing. The recurrence supplies the coefficients and the\nanalytic theorem supplies the error scale. The subsequent determinant,\nintegrality and packet chapters examine the additional arithmetic needed\nto connect these facts to that criterion.","\\begin{definition}[Catalan's constant]\n\\label{def:c01-G}\nThe real number studied in this book is\n\\begin{equation}\nG=\\sum_{k=0}^{\\infty}\\frac{(-1)^k}{(2k+1)^2}.\n\\label{eq:c01-definition}\n\\end{equation}\nThe series converges absolutely because\n\\[\n\\sum_{k=0}^{\\infty}\\frac1{(2k+1)^2}\n\\leq\\sum_{j=1}^{\\infty}\\frac1{j^2}<\\infty.\n\\]\n\\end{definition}","The familiar decimal display begins\n\\[\nG\\approx0.91596559417721901505460351493238411077.\n\\]\nThe approximation sign matters. A decimal is a coordinate for seeing the number, while a rational enclosure is evidence for a particular numerical assertion. A proof of irrationality requires a different assertion again.","\\begin{conjecture}[The irrationality question]\n\\label{conj:c01-irrationality}\nCatalan's constant does not belong to $\\Q$.\n\\end{conjecture}","Irrationality would mean that no integers $a,b$ with $b\\ne0$ satisfy $G=a/b$. Transcendence would mean that no nonzero polynomial with integer coefficients vanishes at $G$. These are different questions. The second implies the first, because a rational number $a/b$ is a zero of the polynomial $bx-a$. Our target throughout is the first assertion.","For real $s>1$, the Dirichlet beta value is\n\\begin{equation}\n\\beta(s)=\\sum_{k=0}^{\\infty}\\frac{(-1)^k}{(2k+1)^s}.\n\\label{eq:c01-beta}\n\\end{equation}\nAbsolute convergence follows from comparison with $\\sum j^{-s}$. Thus $G=\\beta(2)$. The alternating series also defines $\\beta(s)$ for $s>0$, though absolute convergence is no longer available throughout that wider range. At $s=1$, integration of the finite geometric expansion of $(1+x^2)^{-1}$ gives $\\beta(1)=\\pi/4$. The integrated remainder is bounded by $\\int_0^1x^{2N}\\dd x=(2N+1)^{-1}$ and tends to zero, so this passage does not require absolute convergence of the beta series at $1$.","eta value is\n\\begin{equation}\n\\beta(s)=\\sum_{k=0}^{\\infty}\\frac{(-1)^k}{(2k+1)^s}.\n\\label{eq:c01-beta}\n\\end{equation}\nAbsolute convergence follows from comparison with $\\sum j^{-s}$. Thus $G=\\beta(2)$. The alternating series also defines $\\beta(s)$ for $s>0$, though absolute convergence is no longer available throughout that wider range. At $s=1$, integration of the finite geometric expansion of $(1+x^2)^{-1}$ gives $\\beta(1)=\\pi/4$. The integrated remainder is bounded by $\\int_0^1x^{2N}\\dd x=(2N+1)^{-1}$ and tends to zero, so this passage does not require absolute convergence of the beta series at $1$.","quation}\n\\beta(s)=\\sum_{k=0}^{\\infty}\\frac{(-1)^k}{(2k+1)^s}.\n\\label{eq:c01-beta}\n\\end{equation}\nAbsolute convergence follows from comparison with $\\sum j^{-s}$. Thus $G=\\beta(2)$. The alternating series also defines $\\beta(s)$ for $s>0$, though absolute convergence is no longer available throughout that wider range. At $s=1$, integration of the finite geometric expansion of $(1+x^2)^{-1}$ gives $\\beta(1)=\\pi/4$. The integrated remainder is bounded by $\\int_0^1x^{2N}\\dd x=(2N+1)^{-1}$ and tends to zero, so this passage does not require absolute convergence of the beta series at $1$.","^s}.\n\\label{eq:c01-beta}\n\\end{equation}\nAbsolute convergence follows from comparison with $\\sum j^{-s}$. Thus $G=\\beta(2)$. The alternating series also defines $\\beta(s)$ for $s>0$, though absolute convergence is no longer available throughout that wider range. At $s=1$, integration of the finite geometric expansion of $(1+x^2)^{-1}$ gives $\\beta(1)=\\pi/4$. The integrated remainder is bounded by $\\int_0^1x^{2N}\\dd x=(2N+1)^{-1}$ and tends to zero, so this passage does not require absolute convergence of the beta series at $1$.","lternating series also defines $\\beta(s)$ for $s>0$, though absolute convergence is no longer available throughout that wider range. At $s=1$, integration of the finite geometric expansion of $(1+x^2)^{-1}$ gives $\\beta(1)=\\pi/4$. The integrated remainder is bounded by $\\int_0^1x^{2N}\\dd x=(2N+1)^{-1}$ and tends to zero, so this passage does not require absolute convergence of the beta series at $1$.","The family gives useful context without collapsing its separate arithmetic questions. Zudilin proved that at least one of\n\\[\n\\beta(2),\\ \\beta(4),\\ \\beta(6),\\ \\beta(8),\\ \\beta(10),\\ \\beta(12)\n\\]\nis irrational \\cite{Zudilin2019}. This is a disjunction. It permits the irrational member to be a value other than $\\beta(2)$. A proof about the collection therefore cannot be inserted as a proof about $G$. The precise source contract is LC09 in the evidence bundle.","Write\n\\begin{equation}\ns_N=\\sum_{k=0}^{N-1}\\frac{(-1)^k}{(2k+1)^2},\\qquad N\\geq0,\n\\label{eq:c01-partial-sum}\n\\end{equation}\nwhere $s_0=0$ is the empty sum. Thus $N$ counts retained terms, as it does in the visual guide. Each $s_N$ is an exact rational number. The terms decrease strictly in magnitude to zero. Grouping successive terms shows that the even partial sums increase and the odd partial sums decrease. These two sequences approach the same limit.","\\begin{proposition}[Alternating enclosure]\n\\label{prop:c01-alternating}\nFor every integer $r\\geq0$,\n\\begin{equation}\ns_{2r}<G<s_{2r+1}.\n\\label{eq:c01-parity-enclosure}\n\\end{equation}\nFor every integer $N\\geq0$,\n\\begin{equation}\n0<(-1)^N(G-s_N)<\\frac1{(2N+1)^2}.\n\\label{eq:c01-alternating-error}\n\\end{equation}\n\\end{proposition}","\\begin{proof}\nThe difference between two consecutive even sums is\n\\[\ns_{2r+2}-s_{2r}\n=\\frac1{(4r+1)^2}-\\frac1{(4r+3)^2}>0.\n\\]\nSimilarly,\n\\[\ns_{2r+3}-s_{2r+1}\n=-\\frac1{(4r+3)^2}+\\frac1{(4r+5)^2}<0.\n\\]\nConsecutive sums differ by a term tending to zero, so the increasing even sums and decreasing odd sums have the same limit $G$. Strictness follows because the next sum of the same parity lies strictly between the current endpoint and the limit on the other side.","To obtain the error bound, apply this enclosure to $s_N$ and $s_{N+1}$. The limit lies strictly between them. Their distance is $(2N+1)^{-2}$, and the direction from $s_N$ to $s_{N+1}$ has sign $(-1)^N$.\n\\end{proof}","For example,\n\\[\ns_0=0,\\qquad s_1=1,\\qquad s_2=\\frac89,\\qquad\ns_3=\\frac{209}{225},\\qquad\ns_4=\\frac{10016}{11025}.\n\\]\nHence\n\\begin{equation}\n\\frac{10016}{11025}<G<\\frac{209}{225}.\n\\label{eq:c01-small-bracket}\n\\end{equation}\nThese statements use only rational arithmetic. The strict inequality also proves $0<G<1$, which will simplify later tail estimates.","The error bound is transparent, but its decay is slow. To make the next omitted term smaller than $10^{-d}$, one needs $2N+1>10^{d/2}$. Increasing numerical accuracy this way rapidly becomes expensive. The Euler transform below produces a much faster certified enclosure. Before using it, we establish the positive integral that controls its terms.","\\begin{lemma}[A basic moment]\n\\label{lem:c01-moment}\nFor every real $a>-1$,\n\\begin{equation}\n\\int_0^1x^a(-\\log x)\\,\\dd x=\\frac1{(a+1)^2}.\n\\label{eq:c01-moment}\n\\end{equation}\nThe integral is understood as an improper integral at $0$.\n\\end{lemma}","\\begin{proof}\nSet $x=e^{-t}$. Then $-\\log x=t$, $\\dd x=-e^{-t}\\dd t$, and the integral becomes\n\\[\n\\int_0^{\\infty}t e^{-(a+1)t}\\,\\dd t.\n\\]\nIntegration by parts gives $(a+1)^{-2}$. The boundary term at infinity is zero because an exponential with positive decay rate dominates $t$. The boundary term at zero is also zero.\n\\end{proof}","\\begin{proposition}[Single integral]\n\\label{prop:c01-log-integral}\nCatalan's constant satisfies\n\\begin{equation}\nG=\\int_0^1\\frac{-\\log x}{1+x^2}\\,\\dd x.\n\\label{eq:c01-log-integral}\n\\end{equation}\n\\end{proposition}","\\begin{proof}\nFor $0\\leq x<1$, the geometric expansion is\n\\[\n\\frac1{1+x^2}=\\sum_{k=0}^{\\infty}(-1)^kx^{2k}.\n\\]\nThe integrated absolute values of the proposed terms have finite sum:\n\\[\n\\sum_{k=0}^{\\infty}\\int_0^1x^{2k}(-\\log x)\\,\\dd x\n=\\sum_{k=0}^{\\infty}\\frac1{(2k+1)^2}<\\infty.\n\\]\nTherefore the series of integrable functions can be summed under the integral. The moment identity gives the defining series for $G$. The point $x=1$ has measure zero and does not affect the integral.\n\\end{proof}","}\nFor $0\\leq x<1$, the geometric expansion is\n\\[\n\\frac1{1+x^2}=\\sum_{k=0}^{\\infty}(-1)^kx^{2k}.\n\\]\nThe integrated absolute values of the proposed terms have finite sum:\n\\[\n\\sum_{k=0}^{\\infty}\\int_0^1x^{2k}(-\\log x)\\,\\dd x\n=\\sum_{k=0}^{\\infty}\\frac1{(2k+1)^2}<\\infty.\n\\]\nTherefore the series of integrable functions can be summed under the integral. The moment identity gives the defining series for $G$. The point $x=1$ has measure zero and does not affect the integral.\n\\end{proof}","1$, the geometric expansion is\n\\[\n\\frac1{1+x^2}=\\sum_{k=0}^{\\infty}(-1)^kx^{2k}.\n\\]\nThe integrated absolute values of the proposed terms have finite sum:\n\\[\n\\sum_{k=0}^{\\infty}\\int_0^1x^{2k}(-\\log x)\\,\\dd x\n=\\sum_{k=0}^{\\infty}\\frac1{(2k+1)^2}<\\infty.\n\\]\nTherefore the series of integrable functions can be summed under the integral. The moment identity gives the defining series for $G$. The point $x=1$ has measure zero and does not affect the integral.\n\\end{proof}","The singularity in $-\\log x$ is integrable. This is a useful example of why an unbounded integrand is not the same as an infinite integral. In fact, $\\int_0^1-\\log x\\,\\dd x=1$. The factor $(1+x^2)^{-1}$ is bounded between $1/2$ and $1$, so the integral representation independently places $G$ between $1/2$ and $1$.","\\begin{proposition}[Arctangent and tangent forms]\n\\label{prop:c01-tangent}\nThe following equalities hold:\n\\begin{equation}\nG=\\int_0^1\\frac{\\arctan x}{x}\\,\\dd x\n=-\\int_0^{\\pi/4}\\log(\\tan\\theta)\\,\\dd\\theta.\n\\label{eq:c01-tangent-forms}\n\\end{equation}\nThe integrand $\\arctan x/x$ is assigned its continuous value $1$ at $x=0$.\n\\end{proposition}","\\begin{proof}\nFor $0<\\delta<1$, integration by parts gives\n\\[\n\\int_\\delta^1\\frac{\\arctan x}{x}\\,\\dd x\n=-\\arctan(\\delta)\\log\\delta\n-\\int_\\delta^1\\frac{\\log x}{1+x^2}\\,\\dd x.\n\\]\nThe boundary expression tends to zero as $\\delta\\downarrow0$, since $0\\leq\\arctan\\delta\\leq\\delta$ and $\\delta\\abs{\\log\\delta}\\to0$. The first equality follows from Proposition~\\ref{prop:c01-log-integral}.","For the second equality, substitute $x=\\tan\\theta$ into the logarithmic integral. On $0<\\theta\\leq\\pi/4$,\n\\[\n\\frac{\\dd x}{1+x^2}=\\dd\\theta.\n\\]\nThe substitution is valid first on a truncated interval. Its limit exists by the integrability already established for $-\\log x$.\n\\end{proof}","\\begin{proposition}[Elementary double integral]\n\\label{prop:c01-double}\nCatalan's constant also satisfies\n\\begin{equation}\nG=\\int_0^1\\int_0^1\\frac{\\dd x\\,\\dd y}{1+x^2y^2}.\n\\label{eq:c01-double}\n\\end{equation}\n\\end{proposition}","\\begin{proof}\nOne proof integrates with respect to $y$. For $x>0$,\n\\[\n\\int_0^1\\frac{\\dd y}{1+x^2y^2}=\\frac{\\arctan x}{x}.\n\\]\nThe value at $x=0$ is $1$. Proposition~\\ref{prop:c01-tangent} now gives the result.","There is also a direct series justification. Expanding $(1+x^2y^2)^{-1}$ on the square except at the corner $(1,1)$ gives geometric terms $(-1)^kx^{2k}y^{2k}$. Their integrated absolute values sum to\n\\[\n\\sum_{k=0}^{\\infty}\\left(\\int_0^1x^{2k}\\,\\dd x\\right)^2\n=\\sum_{k=0}^{\\infty}\\frac1{(2k+1)^2}<\\infty.\n\\]\nThus summation and integration may be interchanged, giving the defining series again.\n\\end{proof}","Let\n\\[\na_k=\\frac1{(2k+1)^2},\\qquad k\\geq0,\n\\]\nand define the positive difference convention\n\\begin{equation}\n\\Delta a_k=a_k-a_{k+1}.\n\\label{eq:c01-difference-convention}\n\\end{equation}\nThe sign convention is fixed here because the other common convention would change every odd difference. Iteration gives\n\\begin{equation}\n\\Delta^j a_0=\\sum_{k=0}^{j}(-1)^k\\binom jk\\frac1{(2k+1)^2}.\n\\label{eq:c01-finite-difference}\n\\end{equation}\nThis is an exact rational number for every $j$.","rence convention\n\\begin{equation}\n\\Delta a_k=a_k-a_{k+1}.\n\\label{eq:c01-difference-convention}\n\\end{equation}\nThe sign convention is fixed here because the other common convention would change every odd difference. Iteration gives\n\\begin{equation}\n\\Delta^j a_0=\\sum_{k=0}^{j}(-1)^k\\binom jk\\frac1{(2k+1)^2}.\n\\label{eq:c01-finite-difference}\n\\end{equation}\nThis is an exact rational number for every $j$.","\\begin{lemma}[Positive differences]\n\\label{lem:c01-positive-differences}\nFor every integer $j\\geq0$,\n\\begin{equation}\n\\Delta^j a_0\n=\\int_0^1(1-x^2)^j(-\\log x)\\,\\dd x,\n\\qquad 0<\\Delta^j a_0\\leq1.\n\\label{eq:c01-positive-differences}\n\\end{equation}\n\\end{lemma}","\\begin{proof}\nInsert the moment formula into the finite sum in \\eqref{eq:c01-finite-difference}. A finite sum can be interchanged with an integral without any convergence issue. The binomial theorem gives\n\\[\n\\sum_{k=0}^{j}(-1)^k\\binom jkx^{2k}=(1-x^2)^j.\n\\]\nThe integrand is positive for $0<x<1$. It is at most $-\\log x$, whose integral is $1$.\n\\end{proof}","\\begin{theorem}[Certified Euler enclosure]\n\\label{thm:c01-euler}\nDefine\n\\begin{equation}\nE_N=\\sum_{j=0}^{N}\\frac{\\Delta^j a_0}{2^{j+1}},\\qquad N\\geq0.\n\\label{eq:c01-E}\n\\end{equation}\nThen\n\\begin{equation}\nG=\\sum_{j=0}^{\\infty}\\frac{\\Delta^j a_0}{2^{j+1}},\n\\qquad\nE_N<G<E_N+2^{-N-1}.\n\\label{eq:c01-euler-enclosure}\n\\end{equation}\nMore precisely,\n\\begin{equation}\nG-E_N\n=\\int_0^1\\frac{-\\log x}{1+x^2}\n\\left(\\frac{1-x^2}{2}\\right)^{N+1}\\,\\dd x.\n\\label{eq:c01-euler-remainder}\n\\end{equation}\n\\end{theorem}","\\begin{proof}\nBy the preceding lemma all terms of the series are nonnegative. Monotone convergence permits the sum to pass under the integral. For $0\\leq x\\leq1$,\n\\[\n\\sum_{j=0}^{\\infty}\\frac{(1-x^2)^j}{2^{j+1}}\n=\\frac1{1+x^2}.\n\\]\nThe resulting integral is $G$ by Proposition~\\ref{prop:c01-log-integral}.","The finite geometric identity is\n\\[\n\\sum_{j=0}^{N}\\frac{(1-x^2)^j}{2^{j+1}}\n=\\frac{1-((1-x^2)/2)^{N+1}}{1+x^2}.\n\\]\nIntegration gives the exact remainder formula. Its integrand is positive on $(0,1)$, so $G-E_N>0$. Since $(1-x^2)/2\\leq1/2$,\n\\[\nG-E_N\\leq2^{-N-1}G<2^{-N-1}.\n\\]\nThis gives the stated enclosure. An alternative bound sums the omitted terms using $\\Delta^ja_0\\leq1$, with the same geometric tail $2^{-N-1}$.\n\\end{proof}","te geometric identity is\n\\[\n\\sum_{j=0}^{N}\\frac{(1-x^2)^j}{2^{j+1}}\n=\\frac{1-((1-x^2)/2)^{N+1}}{1+x^2}.\n\\]\nIntegration gives the exact remainder formula. Its integrand is positive on $(0,1)$, so $G-E_N>0$. Since $(1-x^2)/2\\leq1/2$,\n\\[\nG-E_N\\leq2^{-N-1}G<2^{-N-1}.\n\\]\nThis gives the stated enclosure. An alternative bound sums the omitted terms using $\\Delta^ja_0\\leq1$, with the same geometric tail $2^{-N-1}$.\n\\end{proof}","The stronger intermediate inequality also gives\n\\begin{equation}\nG\\leq\\frac{E_N}{1-2^{-N-1}}.\n\\label{eq:c01-euler-sharper}\n\\end{equation}\nThe denominator is positive for every $N\\geq0$. The release computations use the simpler interval $[E_N,E_N+2^{-N-1}]$, because that interval has an immediate rational width and is already sufficient for the tested signs.","\\begin{example}[Small Euler intervals]\nThe first difference values are\n\\[\n\\Delta^0a_0=1,\\qquad \\Delta a_0=\\frac89,\\qquad\n\\Delta^2a_0=\\frac{184}{225}.\n\\]\nThey give\n\\[\nE_0=\\frac12,\\qquad E_1=\\frac{13}{18},\\qquad\nE_2=\\frac{371}{450}.\n\\]\nThus, for example,\n\\[\n\\frac{371}{450}<G<\\frac{1709}{1800}.\n\\]\nThese short transformed sums are useful for checking conventions. Their principal advantage appears as $N$ grows, because the certified width decreases geometrically.\n\\end{example}","The independent audit retains $j=0,\\ldots,1000$. Its exact rational endpoints differ by $2^{-1001}$. The finite differences and endpoint numerators are recorded in the evidence bundle. The preceding theorem proves the tail bound for every $N$. The program computes the finite rational portion at one chosen $N$ and checks its internal arithmetic. These are complementary forms of evidence with different scopes.","Suppose $l\\leq G\\leq h$ is a proven rational enclosure, and suppose $a,b\\in\\Q$. Then the interval for $aG-b$ is\n\\begin{equation}\n\\begin{cases}\n[al-b,ah-b],&a\\geq0,","Suppose $l\\leq G\\leq h$ is a proven rational enclosure, and suppose $a,b\\in\\Q$. Then the interval for $aG-b$ is\n\\begin{equation}\n\\begin{cases}\n[al-b,ah-b],&a\\geq0,\\\\\n[ah-b,al-b],&a<0.\n\\end{cases}\n\\label{eq:c01-linear-interval}\n\\end{equation}\nAll endpoints are rational. If the entire interval is positive, the form is positive. If the entire interval is negative, the form is negative. If it contains zero, the enclosure has not decided the sign. That last outcome is a limitation of the chosen certificate, not evidence that the form actually vanishes.","The real approximation problem asks for rational numbers near $G$. The arithmetic problem asks for nonzero integer linear forms in $G$ whose absolute values tend to zero. These problems are related, but the passage between them has a cost. A coefficient denominator can absorb the analytic gain. A common divisor can recover some of that gain. Nonvanishing has to survive the whole construction.","\\begin{lemma}[Rational lattice gap]\n\\label{lem:c02-gap}\nSuppose $\\alpha=a/b$ with $a\\in\\Z$ and $b\\in\\Z_{>0}$. For every pair $P,Q\\in\\Z$, either $Q\\alpha-P=0$ or\n\\begin{equation}\n\\abs{Q\\alpha-P}\\geq\\frac1b.\n\\label{eq:c02-gap}\n\\end{equation}\n\\end{lemma}","\\begin{proof}\nMultiplication by $b$ gives\n\\[\nb(Q\\alpha-P)=Qa-Pb\\in\\Z.\n\\]\nA nonzero integer has absolute value at least $1$. Division by the positive integer $b$ gives the result. No coprimality hypothesis is needed.\n\\end{proof}","For a fixed rational $\\alpha$, every integer form $Q\\alpha-P$ lies in the discrete set $b^{-1}\\Z$. Its nonzero values cannot enter the open interval $(-1/b,1/b)$. The denominator $b$ need not be known to use this obstruction. If a constructed infinite family has nonzero values tending to zero, it eventually violates the gap for any proposed $b$.","\\begin{theorem}[Integer form criterion]\n\\label{thm:c02-criterion}\nLet $\\alpha\\in\\R$. Suppose there are integer pairs $(P_j,Q_j)$ for $j\\geq1$ such that\n\\begin{equation}\n0<\\abs{Q_j\\alpha-P_j}\\longrightarrow0.\n\\label{eq:c02-form-criterion}\n\\end{equation}\nThen $\\alpha$ is irrational. The same conclusion holds if these conditions are verified along any infinite subsequence.\n\\end{theorem}","\\begin{proof}\nIf $\\alpha=a/b$ were rational with $b>0$, Lemma~\\ref{lem:c02-gap} would bound every nonzero form below by $1/b$. This contradicts convergence to zero. Restricting to a subsequence does not change the argument.\n\\end{proof}","\\begin{figure}[tbp]\n\\centering\n\\includegraphics[width=0.94\\linewidth]{../visual/figures/06_rationality_lattice.pdf}\n\\caption{Under the assumption $G=a/b$, integer linear forms lie on a lattice with spacing $1/b$. A sequence of nonzero forms approaching zero would contradict that spacing. The spacing is conceptual here because the hypothetical denominator is unknown.}\n\\label{fig:c02-lattice}\n\\end{figure}","For instance, setting $P_j=Q_j=0$ gives integer coefficients and zero error at every index, but proves nothing. Taking a rational approximation with a tiny error gives a small real form, but its coefficients may have large rational denominators. Clearing those denominators may keep the resulting integer form far from zero. These examples explain why the hypotheses are kept distinct throughout this book.","\\begin{proposition}[Elementary approximation by integer forms]\n\\label{prop:c02-pigeonhole}\nIf $\\alpha$ is irrational, then for every integer $H\\geq1$ there are integers $P,Q$ with $1\\leq Q\\leq H$ satisfying\n\\begin{equation}\n0<\\abs{Q\\alpha-P}<\\frac1H.\n\\label{eq:c02-pigeonhole}\n\\end{equation}\nIn particular, an infinite family as in Theorem~\\ref{thm:c02-criterion} exists.\n\\end{proposition}","\\begin{proof}\nConsider the $H+1$ fractional parts of $0,\\alpha,2\\alpha,\\ldots,H\\alpha$. Partition $[0,1)$ into $H$ intervals of length $1/H$, using half open intervals. Two fractional parts lie in one interval. Their difference has absolute value strictly smaller than $1/H$. Subtracting the corresponding multiples of $\\alpha$ gives $Q\\alpha-P$ with $1\\leq Q\\leq H$ after choosing the order of the two indices.","The form is nonzero because $\\alpha$ is irrational and $Q\\ne0$. Choosing such a pair for every $H$ gives nonzero forms tending to zero. The denominators $Q$ cannot remain bounded. For any fixed finite set of positive $Q$, the distances from $Q\\alpha$ to the nearest integers have a positive minimum. That minimum would contradict the bound as $H\\to\\infty$.\n\\end{proof}","This proof assumes irrationality in order to ensure nonvanishing. It cannot be used to prove irrationality of $G$ by inserting $\\alpha=G$ before that fact is known. The lattice criterion runs in the other direction: first establish small nonzero forms by independent mathematics, then deduce irrationality.","Let $P_j/Q_j$ be rational approximants written in lowest terms, with $Q_j>0$. Their ordinary errors and their integer form errors are related by\n\\begin{equation}\n\\abs{Q_j\\alpha-P_j}\n=Q_j\\abs{\\alpha-\\frac{P_j}{Q_j}}.\n\\label{eq:c02-error-conversion}\n\\end{equation}\nThus a plot of the ordinary error omits one factor from the arithmetic test. A rapid decrease in that plot can coexist with a denominator that grows just as rapidly.","\\begin{corollary}[Reduced rational approximation criterion]\n\\label{cor:c02-reduced}\nSuppose $P_j/Q_j$ is in lowest terms, $Q_j>0$, and $P_j/Q_j\\ne\\alpha$ at all indices used. If\n\\begin{equation}\nQ_j\\abs{\\alpha-\\frac{P_j}{Q_j}}\\longrightarrow0,\n\\label{eq:c02-reduced-condition}\n\\end{equation}\nthen $\\alpha$ is irrational.\n\\end{corollary}","\\begin{proof}\nThe left side equals the absolute value of the nonzero integer form $Q_j\\alpha-P_j$. Apply Theorem~\\ref{thm:c02-criterion}.\n\\end{proof}","Reduction to lowest terms is not necessary for the theorem, but it gives the smallest denominator belonging to that rational number. If a common factor is left in $P_j,Q_j$, the form is multiplied by that factor. Keeping a redundant factor can conceal an available decay margin. Removing it cannot create nonvanishing if the original form was zero.","\\begin{proposition}[Separation from a rational target]\n\\label{prop:c02-rational-distance}\nIf $\\alpha=a/b$ with $b>0$ and $P/Q\\ne\\alpha$ with $P\\in\\Z$, $Q\\in\\Z_{>0}$, then\n\\begin{equation}\n\\abs{\\alpha-\\frac PQ}\\geq\\frac1{bQ}.\n\\label{eq:c02-rational-distance}\n\\end{equation}\n\\end{proposition}","\\begin{proof}\nThe numerator of the difference is the nonzero integer $aQ-bP$:\n\\[\n\\abs{\\frac ab-\\frac PQ}=\\frac{\\abs{aQ-bP}}{bQ}\\geq\\frac1{bQ}.\n\\]\n\\end{proof}","This lower bound explains a familiar sufficient exponent. If distinct approximants have $Q_j\\to\\infty$ and\n\\begin{equation}\n\\abs{\\alpha-\\frac{P_j}{Q_j}}\\leq C Q_j^{-\\theta}\n\\label{eq:c02-exponent-bound}\n\\end{equation}\nfor a fixed constant $C>0$ and an exponent $\\theta>1$, then multiplying by $Q_j$ gives a quantity tending to zero. The criterion applies. An upper bound with $\\theta\\leq1$ does not supply that conclusion on its own. It may still describe useful approximation, and another argument may still prove irrationality, but this particular inequality does not cross the required threshold.","\\begin{example}[A convergent rational sequence with a fixed form error]\n\\label{ex:c02-linear-rational}\nLet $\\alpha=1/2$ and take\n\\[\n\\frac{P_n}{Q_n}=\\frac{n}{2n+1},\\qquad n\\geq1.\n\\]\nThe numerator and denominator are coprime because any common divisor of $n$ and $2n+1$ divides $1$. The ordinary error is\n\\[\n\\frac12-\\frac{n}{2n+1}=\\frac1{2(2n+1)}\\longrightarrow0.\n\\]\nNevertheless,\n\\[\nQ_n\\alpha-P_n=\\frac12\n\\]\nfor every $n$. The approximants are distinct from the target, all coefficients are integers, and the approximations converge. The integer form does not tend to zero.\n\\end{example}","convergent rational sequence with a fixed form error]\n\\label{ex:c02-linear-rational}\nLet $\\alpha=1/2$ and take\n\\[\n\\frac{P_n}{Q_n}=\\frac{n}{2n+1},\\qquad n\\geq1.\n\\]\nThe numerator and denominator are coprime because any common divisor of $n$ and $2n+1$ divides $1$. The ordinary error is\n\\[\n\\frac12-\\frac{n}{2n+1}=\\frac1{2(2n+1)}\\longrightarrow0.\n\\]\nNevertheless,\n\\[\nQ_n\\alpha-P_n=\\frac12\n\\]\nfor every $n$. The approximants are distinct from the target, all coefficients are integers, and the approximations converge. The integer form does not tend to zero.\n\\end{example}","\\begin{example}[Exponential approximation to a rational number]\n\\label{ex:c02-exponential-rational}\nAgain let $\\alpha=1/2$. For $n\\geq2$, set\n\\[\nQ_n=2^n,\\qquad P_n=2^{n-1}+1.\n\\]\nThe numerator is odd, so $P_n/Q_n$ is reduced. We have\n\\begin{equation}\n\\frac{P_n}{Q_n}=\\frac12+2^{-n},\\qquad\n\\abs{\\alpha-\\frac{P_n}{Q_n}}=2^{-n}.\n\\label{eq:c02-exponential-rational}\n\\end{equation}\nThe real error decreases exponentially. Yet\n\\[\nQ_n\\alpha-P_n=-1\n\\]\nat every index. Exponential convergence of the ordinary error is therefore insufficient without the denominator comparison.\n\\end{example}","ntial approximation to a rational number]\n\\label{ex:c02-exponential-rational}\nAgain let $\\alpha=1/2$. For $n\\geq2$, set\n\\[\nQ_n=2^n,\\qquad P_n=2^{n-1}+1.\n\\]\nThe numerator is odd, so $P_n/Q_n$ is reduced. We have\n\\begin{equation}\n\\frac{P_n}{Q_n}=\\frac12+2^{-n},\\qquad\n\\abs{\\alpha-\\frac{P_n}{Q_n}}=2^{-n}.\n\\label{eq:c02-exponential-rational}\n\\end{equation}\nThe real error decreases exponentially. Yet\n\\[\nQ_n\\alpha-P_n=-1\n\\]\nat every index. Exponential convergence of the ordinary error is therefore insufficient without the denominator comparison.\n\\end{example}","The second example can also be written as a failure of denominator clearing. The rational coefficient form\n\\[\n\\ell_n=\\alpha-\\left(\\frac12+2^{-n}\\right)=-2^{-n}\n\\]\nis nonzero and tends to zero. Its second coefficient is not an integer. Multiplication by $2^n$ makes both coefficients integral and gives the constant form $-1$. The analytic gain has been exactly consumed by the arithmetic scale.","In the later chapters the starting form is\n\\begin{equation}\n\\ell_j=u_j\\alpha-v_j,\\qquad u_j,v_j\\in\\Q.\n\\label{eq:c02-rational-form}\n\\end{equation}\nChoose a positive integer $D_j$ for which $D_ju_j$ and $D_jv_j$ are integers. Then\n\\begin{equation}\nL_j=D_j\\ell_j=(D_ju_j)\\alpha-D_jv_j\n\\label{eq:c02-cleared-form}\n\\end{equation}\nhas integer coefficients. Its size is $D_j\\abs{\\ell_j}$, so the clearance factor must be included in any decay estimate.","Such a $D_j$ always exists for one finite rational pair. Write $u_j=a_j/b_j$ and $v_j=c_j/d_j$ in lowest terms with positive denominators. The smallest positive common clearing factor is\n\\begin{equation}\nD_j^{\\min}=\\operatorname{lcm}(b_j,d_j).\n\\label{eq:c02-minimal-clearance}\n\\end{equation}\nIndeed, integrality of $D_ja_j/b_j$ forces $b_j\\mid D_j$ by coprimality, and integrality of $D_jc_j/d_j$ forces $d_j\\mid D_j$. Conversely, their least common multiple clears both denominators.","a $D_j$ always exists for one finite rational pair. Write $u_j=a_j/b_j$ and $v_j=c_j/d_j$ in lowest terms with positive denominators. The smallest positive common clearing factor is\n\\begin{equation}\nD_j^{\\min}=\\operatorname{lcm}(b_j,d_j).\n\\label{eq:c02-minimal-clearance}\n\\end{equation}\nIndeed, integrality of $D_ja_j/b_j$ forces $b_j\\mid D_j$ by coprimality, and integrality of $D_jc_j/d_j$ forces $d_j\\mid D_j$. Conversely, their least common multiple clears both denominators.","Suppose a cleared form is $A_j\\alpha-B_j$ with $A_j,B_j\\in\\Z$ and $(A_j,B_j)\\ne(0,0)$. Define its common content by\n\\begin{equation}\nh_j=\\gcd(\\abs{A_j},\\abs{B_j})>0.\n\\label{eq:c02-content}\n\\end{equation}\nThe primitive pair is\n\\[\n\\left(\\frac{A_j}{h_j},\\frac{B_j}{h_j}\\right),\n\\]\nand the associated form has size\n\\begin{equation}\n\\frac{\\abs{A_j\\alpha-B_j}}{h_j}.\n\\label{eq:c02-primitive-form}\n\\end{equation}\nBoth coefficients remain integers. Thus a proved lower bound on common content may improve an irrationality argument. The word ``proved'' is essential when the improvement is used at infinitely many indices.","If $A_j\\ne0$, reducing the rational approximant $B_j/A_j$ gives denominator $\\abs{A_j}/h_j$. This denominator controls the ordinary approximation criterion. The common content controls the primitive linear form. These are two descriptions of the same cancellation, not two independent gains that can be counted twice.","There is also a useful invariance. Multiplying a cleared pair by a nonzero integer $c$ multiplies its common content by $\\abs c$. After primitive reduction, the resulting pair is unchanged up to a common sign. A large gcd produced solely by an unnecessary clearance multiplier cannot improve the primitive form. The later gcd decomposition makes this observation exact for the historical filter data.","\\begin{theorem}[Exponential clearance criterion]\n\\label{thm:c02-rate}\nLet $\\alpha\\in\\R$, and let $u_j,v_j\\in\\Q$ for $j\\geq1$. Let $D_j$ be positive integers such that $D_ju_j,D_jv_j\\in\\Z$. Assume $u_j\\alpha-v_j\\ne0$ for all sufficiently large $j$. Suppose\n\\begin{equation}\n\\limsup_{j\\to\\infty}\\abs{u_j\\alpha-v_j}^{1/j}\\leq\\lambda,\n\\qquad\n\\limsup_{j\\to\\infty}D_j^{1/j}\\leq\\delta,\n\\label{eq:c02-root-bounds}\n\\end{equation}\nwhere $\\lambda\\geq0$, $\\delta<\\infty$ and $\\delta\\lambda<1$. Then $\\alpha$ is irrational.\n\\end{theorem}","clearance criterion]\n\\label{thm:c02-rate}\nLet $\\alpha\\in\\R$, and let $u_j,v_j\\in\\Q$ for $j\\geq1$. Let $D_j$ be positive integers such that $D_ju_j,D_jv_j\\in\\Z$. Assume $u_j\\alpha-v_j\\ne0$ for all sufficiently large $j$. Suppose\n\\begin{equation}\n\\limsup_{j\\to\\infty}\\abs{u_j\\alpha-v_j}^{1/j}\\leq\\lambda,\n\\qquad\n\\limsup_{j\\to\\infty}D_j^{1/j}\\leq\\delta,\n\\label{eq:c02-root-bounds}\n\\end{equation}\nwhere $\\lambda\\geq0$, $\\delta<\\infty$ and $\\delta\\lambda<1$. Then $\\alpha$ is irrational.\n\\end{theorem}","ion]\n\\label{thm:c02-rate}\nLet $\\alpha\\in\\R$, and let $u_j,v_j\\in\\Q$ for $j\\geq1$. Let $D_j$ be positive integers such that $D_ju_j,D_jv_j\\in\\Z$. Assume $u_j\\alpha-v_j\\ne0$ for all sufficiently large $j$. Suppose\n\\begin{equation}\n\\limsup_{j\\to\\infty}\\abs{u_j\\alpha-v_j}^{1/j}\\leq\\lambda,\n\\qquad\n\\limsup_{j\\to\\infty}D_j^{1/j}\\leq\\delta,\n\\label{eq:c02-root-bounds}\n\\end{equation}\nwhere $\\lambda\\geq0$, $\\delta<\\infty$ and $\\delta\\lambda<1$. Then $\\alpha$ is irrational.\n\\end{theorem}","rate}\nLet $\\alpha\\in\\R$, and let $u_j,v_j\\in\\Q$ for $j\\geq1$. Let $D_j$ be positive integers such that $D_ju_j,D_jv_j\\in\\Z$. Assume $u_j\\alpha-v_j\\ne0$ for all sufficiently large $j$. Suppose\n\\begin{equation}\n\\limsup_{j\\to\\infty}\\abs{u_j\\alpha-v_j}^{1/j}\\leq\\lambda,\n\\qquad\n\\limsup_{j\\to\\infty}D_j^{1/j}\\leq\\delta,\n\\label{eq:c02-root-bounds}\n\\end{equation}\nwhere $\\lambda\\geq0$, $\\delta<\\infty$ and $\\delta\\lambda<1$. Then $\\alpha$ is irrational.\n\\end{theorem}","and let $u_j,v_j\\in\\Q$ for $j\\geq1$. Let $D_j$ be positive integers such that $D_ju_j,D_jv_j\\in\\Z$. Assume $u_j\\alpha-v_j\\ne0$ for all sufficiently large $j$. Suppose\n\\begin{equation}\n\\limsup_{j\\to\\infty}\\abs{u_j\\alpha-v_j}^{1/j}\\leq\\lambda,\n\\qquad\n\\limsup_{j\\to\\infty}D_j^{1/j}\\leq\\delta,\n\\label{eq:c02-root-bounds}\n\\end{equation}\nwhere $\\lambda\\geq0$, $\\delta<\\infty$ and $\\delta\\lambda<1$. Then $\\alpha$ is irrational.\n\\end{theorem}","\\begin{proof}\nSince $D_j\\geq1$, the second hypothesis implies $\\delta\\geq1$. Choose positive numbers $\\lambda'>\\lambda$ and $\\delta'>\\delta$ with $\\lambda'\\delta'<1$. Such a choice is possible by the strict product inequality, including when $\\lambda=0$.","By the definitions of the two upper limits, for all sufficiently large $j$,\n\\[\n\\abs{u_j\\alpha-v_j}\\leq(\\lambda')^j,\n\\qquad D_j\\leq(\\delta')^j.\n\\]\nConsequently,\n\\[\n0<\\abs{D_j(u_j\\alpha-v_j)}\n\\leq(\\delta'\\lambda')^j\\longrightarrow0.\n\\]\nThe form has integer coefficients by hypothesis. Apply Theorem~\\ref{thm:c02-criterion}.\n\\end{proof}","\\begin{corollary}[Logarithmic rate accounting]\n\\label{cor:c02-log-rate}\nUnder the integrality and eventual nonvanishing hypotheses above, suppose\n\\begin{equation}\n\\limsup_{j\\to\\infty}\\frac{\\log D_j}{j}\\leq\\sigma,\n\\qquad\n\\limsup_{j\\to\\infty}\\frac{\\log\\abs{u_j\\alpha-v_j}}{j}\\leq-\\tau,\n\\label{eq:c02-log-bounds}\n\\end{equation}\nwith finite $\\sigma$ and $\\tau>\\sigma$. Then $\\alpha$ is irrational.\n\\end{corollary}","\\begin{proof}\nChoose $\\eps>0$ with $\\sigma-\\tau+2\\eps<0$. Eventually $D_j\\leq e^{(\\sigma+\\eps)j}$ and $\\abs{u_j\\alpha-v_j}\\leq e^{(-\\tau+\\eps)j}$. Their product tends to zero. Apply the integer form criterion.\n\\end{proof}","\\begin{proposition}[Content can recover a proved rate]\n\\label{prop:c02-content-rate}\nAssume the setup of Corollary~\\ref{cor:c02-log-rate}, except that $\\tau>\\sigma$ is not required. Discard the finite initial set of indices before nonvanishing holds. For the remaining sufficiently large $j$, define\n\\[\nh_j=\\gcd(\\abs{D_ju_j},\\abs{D_jv_j}),\n\\]\nwhich is positive because the form is nonzero at those indices. If\n\\begin{equation}\n\\liminf_{j\\to\\infty}\\frac{\\log h_j}{j}\\geq\\kappa\n\\quad\\hbox{and}\\quad \\tau+\\kappa>\\sigma,\n\\label{eq:c02-content-rate}\n\\end{equation}\nthen $\\alpha$ is irrational.\n\\end{proposition}","t can recover a proved rate]\n\\label{prop:c02-content-rate}\nAssume the setup of Corollary~\\ref{cor:c02-log-rate}, except that $\\tau>\\sigma$ is not required. Discard the finite initial set of indices before nonvanishing holds. For the remaining sufficiently large $j$, define\n\\[\nh_j=\\gcd(\\abs{D_ju_j},\\abs{D_jv_j}),\n\\]\nwhich is positive because the form is nonzero at those indices. If\n\\begin{equation}\n\\liminf_{j\\to\\infty}\\frac{\\log h_j}{j}\\geq\\kappa\n\\quad\\hbox{and}\\quad \\tau+\\kappa>\\sigma,\n\\label{eq:c02-content-rate}\n\\end{equation}\nthen $\\alpha$ is irrational.\n\\end{proposition}","s not required. Discard the finite initial set of indices before nonvanishing holds. For the remaining sufficiently large $j$, define\n\\[\nh_j=\\gcd(\\abs{D_ju_j},\\abs{D_jv_j}),\n\\]\nwhich is positive because the form is nonzero at those indices. If\n\\begin{equation}\n\\liminf_{j\\to\\infty}\\frac{\\log h_j}{j}\\geq\\kappa\n\\quad\\hbox{and}\\quad \\tau+\\kappa>\\sigma,\n\\label{eq:c02-content-rate}\n\\end{equation}\nthen $\\alpha$ is irrational.\n\\end{proposition}","\\begin{proof}\nThe primitive form $D_j(u_j\\alpha-v_j)/h_j$ has integer coefficients and remains nonzero. Choose $\\eps>0$ so that $\\sigma-\\tau-\\kappa+3\\eps<0$. The upper limits for $D_j$ and the error, and the lower limit for $h_j$, give eventually\n\\[\n\\frac{D_j\\abs{u_j\\alpha-v_j}}{h_j}\n\\leq\\exp\\bigl((\\sigma-\\tau-\\kappa+3\\eps)j\\bigr).\n\\]\nThe right side tends to zero. Apply Theorem~\\ref{thm:c02-criterion} to the primitive pair.\n\\end{proof}","The strictness of the rate comparison is not cosmetic. In Example~\\ref{ex:c02-exponential-rational}, the rational coefficient form has root error rate $\\lambda=1/2$ and clearance rate $\\delta=2$. The product is exactly $1$. The cleared form is constantly $-1$, and the target is rational. Thus replacing $\\delta\\lambda<1$ by $\\delta\\lambda\\leq1$ would make Theorem~\\ref{thm:c02-rate} false.","Equality in a root or logarithmic rate can still leave room for a more delicate argument. For example, an additional factor tending to zero can matter even when its logarithm divided by $j$ tends to zero. But such a factor needs to be proved. Rate equality alone leaves the behavior undetermined. A positive upper exponent likewise says that the displayed bound does not force decay. It does not prove that the actual forms grow.","For Catalan's constant, the procedure has a definite order. First construct rational coefficient pairs by a recurrence or integral identity. Then justify a universal integer clearance. Then establish nonvanishing of the chosen forms without assuming $G$ irrational. Finally compare the analytic decay with the full arithmetic scale, including any rigorously controlled common content.","An exact integral of fixed sign is one possible source of nonvanishing. A parity bracket around a limit is another, once that limit has independently been identified with $G$. An interval computation can certify a particular finite form, but it cannot supply the universal step by simply repeating the test many times.","Write the recurrence in the form\n\\begin{equation}\\label{cb_eq_rec_short}\n a_n w_{n+1}=q(n)w_n+c_nw_{n-1},\\qquad n\\geq1,\n\\end{equation}\nwhere, in this chapter only,\n\\[\n a_n=(2n+1)^2(2n+2)^2p(n),\\qquad\n c_n=(2n-1)^2(2n)^2p(n+1).\n\\]\nThese are recurrence coefficients, not the integer coefficients of a cleared\nlinear form used later. In particular, both are positive. The initial data are\n\\[\n u_0=1,\\quad u_1=\\frac74,\\qquad\n v_0=0,\\quad v_1=\\frac{13}{8}.\n\\]\nThe polynomial $p(n)=20n^2-8n+1$ is positive for integer $n\\geq0$.\nDefine the discrete Wronskian\n\\begin{equation}\\label{cb_eq_W_def}\n W_n=u_{n+1}v_n-u_nv_{n+1},\\qquad n\\geq0.\n\\end{equation}\nThis convention fixes its sign. Reversing the two terms reverses every sign\nin the subsequent formulas.","}\n a_n w_{n+1}=q(n)w_n+c_nw_{n-1},\\qquad n\\geq1,\n\\end{equation}\nwhere, in this chapter only,\n\\[\n a_n=(2n+1)^2(2n+2)^2p(n),\\qquad\n c_n=(2n-1)^2(2n)^2p(n+1).\n\\]\nThese are recurrence coefficients, not the integer coefficients of a cleared\nlinear form used later. In particular, both are positive. The initial data are\n\\[\n u_0=1,\\quad u_1=\\frac74,\\qquad\n v_0=0,\\quad v_1=\\frac{13}{8}.\n\\]\nThe polynomial $p(n)=20n^2-8n+1$ is positive for integer $n\\geq0$.\nDefine the discrete Wronskian\n\\begin{equation}\\label{cb_eq_W_def}\n W_n=u_{n+1}v_n-u_nv_{n+1},\\qquad n\\geq0.\n\\end{equation}\nThis convention fixes its sign. Reversing the two terms reverses every sign\nin the subsequent formulas.","\\geq1,\n\\end{equation}\nwhere, in this chapter only,\n\\[\n a_n=(2n+1)^2(2n+2)^2p(n),\\qquad\n c_n=(2n-1)^2(2n)^2p(n+1).\n\\]\nThese are recurrence coefficients, not the integer coefficients of a cleared\nlinear form used later. In particular, both are positive. The initial data are\n\\[\n u_0=1,\\quad u_1=\\frac74,\\qquad\n v_0=0,\\quad v_1=\\frac{13}{8}.\n\\]\nThe polynomial $p(n)=20n^2-8n+1$ is positive for integer $n\\geq0$.\nDefine the discrete Wronskian\n\\begin{equation}\\label{cb_eq_W_def}\n W_n=u_{n+1}v_n-u_nv_{n+1},\\qquad n\\geq0.\n\\end{equation}\nThis convention fixes its sign. Reversing the two terms reverses every sign\nin the subsequent formulas.","a_n=(2n+1)^2(2n+2)^2p(n),\\qquad\n c_n=(2n-1)^2(2n)^2p(n+1).\n\\]\nThese are recurrence coefficients, not the integer coefficients of a cleared\nlinear form used later. In particular, both are positive. The initial data are\n\\[\n u_0=1,\\quad u_1=\\frac74,\\qquad\n v_0=0,\\quad v_1=\\frac{13}{8}.\n\\]\nThe polynomial $p(n)=20n^2-8n+1$ is positive for integer $n\\geq0$.\nDefine the discrete Wronskian\n\\begin{equation}\\label{cb_eq_W_def}\n W_n=u_{n+1}v_n-u_nv_{n+1},\\qquad n\\geq0.\n\\end{equation}\nThis convention fixes its sign. Reversing the two terms reverses every sign\nin the subsequent formulas.","\\begin{theorem}[Exact Wronskian]\\label{cb_thm_W}\nFor every integer $n\\geq0$,\n\\begin{equation}\\label{cb_eq_W_closed}\n W_n=\\frac{(-1)^{n+1}p(n+1)}{2(2n+1)^2(2n+2)^2}.\n\\end{equation}\nIn particular, $W_n\\ne0$.\n\\end{theorem}","\\begin{proof}\nAt $n=0$ the determinant is $-13/8$, which agrees with the displayed\nformula because $p(1)=13$. For $n\\geq1$, insert\n\\eqref{cb_eq_rec_short} into both terms of \\eqref{cb_eq_W_def}. The terms\ncontaining $q(n)u_nv_n$ cancel, leaving\n\\[\n W_n=\\frac{c_n}{a_n}(u_{n-1}v_n-u_nv_{n-1})\n     =-\\frac{c_n}{a_n}W_{n-1}.\n\\]\nIterating this equality gives\n\\[\n W_n=(-1)^nW_0\n \\prod_{j=1}^n\n \\frac{(2j-1)^2(2j)^2p(j+1)}{(2j+1)^2(2j+2)^2p(j)}.\n\\]\nThere are two independent telescopes. The polynomial factors leave\n$p(n+1)/p(1)$. The integer factors leave\n$4/((2n+1)^2(2n+2)^2)$. Substituting $W_0=-13/8$ and $p(1)=13$\nproves the identity. The empty product gives the same calculation at $n=0$.\n\\end{proof}","r $n\\geq1$, insert\n\\eqref{cb_eq_rec_short} into both terms of \\eqref{cb_eq_W_def}. The terms\ncontaining $q(n)u_nv_n$ cancel, leaving\n\\[\n W_n=\\frac{c_n}{a_n}(u_{n-1}v_n-u_nv_{n-1})\n     =-\\frac{c_n}{a_n}W_{n-1}.\n\\]\nIterating this equality gives\n\\[\n W_n=(-1)^nW_0\n \\prod_{j=1}^n\n \\frac{(2j-1)^2(2j)^2p(j+1)}{(2j+1)^2(2j+2)^2p(j)}.\n\\]\nThere are two independent telescopes. The polynomial factors leave\n$p(n+1)/p(1)$. The integer factors leave\n$4/((2n+1)^2(2n+2)^2)$. Substituting $W_0=-13/8$ and $p(1)=13$\nproves the identity. The empty product gives the same calculation at $n=0$.\n\\end{proof}","} into both terms of \\eqref{cb_eq_W_def}. The terms\ncontaining $q(n)u_nv_n$ cancel, leaving\n\\[\n W_n=\\frac{c_n}{a_n}(u_{n-1}v_n-u_nv_{n-1})\n     =-\\frac{c_n}{a_n}W_{n-1}.\n\\]\nIterating this equality gives\n\\[\n W_n=(-1)^nW_0\n \\prod_{j=1}^n\n \\frac{(2j-1)^2(2j)^2p(j+1)}{(2j+1)^2(2j+2)^2p(j)}.\n\\]\nThere are two independent telescopes. The polynomial factors leave\n$p(n+1)/p(1)$. The integer factors leave\n$4/((2n+1)^2(2n+2)^2)$. Substituting $W_0=-13/8$ and $p(1)=13$\nproves the identity. The empty product gives the same calculation at $n=0$.\n\\end{proof}","ef}. The terms\ncontaining $q(n)u_nv_n$ cancel, leaving\n\\[\n W_n=\\frac{c_n}{a_n}(u_{n-1}v_n-u_nv_{n-1})\n     =-\\frac{c_n}{a_n}W_{n-1}.\n\\]\nIterating this equality gives\n\\[\n W_n=(-1)^nW_0\n \\prod_{j=1}^n\n \\frac{(2j-1)^2(2j)^2p(j+1)}{(2j+1)^2(2j+2)^2p(j)}.\n\\]\nThere are two independent telescopes. The polynomial factors leave\n$p(n+1)/p(1)$. The integer factors leave\n$4/((2n+1)^2(2n+2)^2)$. Substituting $W_0=-13/8$ and $p(1)=13$\nproves the identity. The empty product gives the same calculation at $n=0$.\n\\end{proof}","ontaining $q(n)u_nv_n$ cancel, leaving\n\\[\n W_n=\\frac{c_n}{a_n}(u_{n-1}v_n-u_nv_{n-1})\n     =-\\frac{c_n}{a_n}W_{n-1}.\n\\]\nIterating this equality gives\n\\[\n W_n=(-1)^nW_0\n \\prod_{j=1}^n\n \\frac{(2j-1)^2(2j)^2p(j+1)}{(2j+1)^2(2j+2)^2p(j)}.\n\\]\nThere are two independent telescopes. The polynomial factors leave\n$p(n+1)/p(1)$. The integer factors leave\n$4/((2n+1)^2(2n+2)^2)$. Substituting $W_0=-13/8$ and $p(1)=13$\nproves the identity. The empty product gives the same calculation at $n=0$.\n\\end{proof}","{n-1})\n     =-\\frac{c_n}{a_n}W_{n-1}.\n\\]\nIterating this equality gives\n\\[\n W_n=(-1)^nW_0\n \\prod_{j=1}^n\n \\frac{(2j-1)^2(2j)^2p(j+1)}{(2j+1)^2(2j+2)^2p(j)}.\n\\]\nThere are two independent telescopes. The polynomial factors leave\n$p(n+1)/p(1)$. The integer factors leave\n$4/((2n+1)^2(2n+2)^2)$. Substituting $W_0=-13/8$ and $p(1)=13$\nproves the identity. The empty product gives the same calculation at $n=0$.\n\\end{proof}","\\begin{center}\n\\begin{tabular}{@{}rrrrr@{}}\n\\toprule\n$n$ & $0$ & $1$ & $2$ & $3$","\\midrule\n$W_n$ & $-13/8$ & $65/288$ & $-157/1800$ & $289/6272$","The next argument needs the positivity of $u_n$. It also needs a modest lower\nbound on it, rather than a full asymptotic theorem. Both follow from the\nrecurrence. To see the relevant sign without relying on a plot, put $t=n-1$.\nDirect polynomial expansion gives\n\\begin{align*}\n q(n)-a_n={}&3200t^6+24000t^5+73392t^4+117072t^3","The next argument needs the positivity of $u_n$. It also needs a modest lower\nbound on it, rather than a full asymptotic theorem. Both follow from the\nrecurrence. To see the relevant sign without relying on a plot, put $t=n-1$.\nDirect polynomial expansion gives\n\\begin{align*}\n q(n)-a_n={}&3200t^6+24000t^5+73392t^4+117072t^3\\\\\n            &+102768t^2+47112t+8827.\n\\end{align*}\nThus $q(n)>a_n>0$ for every $n\\geq1$. Starting from $u_1>u_0>0$, the\nrecurrence implies\n\\[\n u_{n+1}>\\frac{q(n)}{a_n}u_n>u_n.\n\\]\nIn particular $u_n\\geq1$. This elementary bound is enough for the convergence\nargument below.","&+102768t^2+47112t+8827.\n\\end{align*}\nThus $q(n)>a_n>0$ for every $n\\geq1$. Starting from $u_1>u_0>0$, the\nrecurrence implies\n\\[\n u_{n+1}>\\frac{q(n)}{a_n}u_n>u_n.\n\\]\nIn particular $u_n\\geq1$. This elementary bound is enough for the convergence\nargument below.","Set\n\\begin{equation}\\label{cb_eq_ratio_steps}\n r_n=\\frac{v_n}{u_n},\\qquad\n d_n=\\frac{p(n+1)}{2(2n+1)^2(2n+2)^2u_nu_{n+1}}.\n\\end{equation}\nThe symbol $d_n$ denotes a positive rational step size. It is unrelated to\nthe least common multiples $L_N$ used in denominator clearing. Dividing the\nWronskian identity by $u_nu_{n+1}$ yields\n\\begin{equation}\\label{cb_eq_alternating_steps}\n r_{n+1}-r_n=(-1)^n d_n.\n\\end{equation}\nThe movement of the approximants therefore alternates exactly.","\\begin{lemma}[Strict decrease of the step sizes]\\label{cb_lem_d_decrease}\nThe rational numbers $d_n$ are strictly positive, decrease strictly, and tend\nto zero.\n\\end{lemma}","\\begin{proof}\nPositivity is immediate. For $n\\geq1$, the determinant recurrence gives\n\\[\n \\frac{d_n}{d_{n-1}}\n =\\frac{c_n}{a_n}\\frac{u_{n-1}}{u_{n+1}}<1.\n\\]\nThe strict inequality follows from\n$a_nu_{n+1}=q(n)u_n+c_nu_{n-1}>c_nu_{n-1}$.\nFinally $u_nu_{n+1}\\geq1$, so\n\\[\n 0<d_n\\leq\n \\frac{p(n+1)}{2(2n+1)^2(2n+2)^2}\\longrightarrow0.\n\\]\nNo limiting characteristic root is needed for this conclusion.\n\\end{proof}","\\begin{theorem}[Alternating brackets]\\label{cb_thm_brackets}\nThere is a real number $L$ such that $r_n\\to L$. For every $k\\geq0$,\n\\begin{equation}\\label{cb_eq_bracket}\n r_{2k}<L<r_{2k+1}.\n\\end{equation}\nThe even subsequence increases strictly and the odd subsequence decreases\nstrictly. For every $n\\geq0$,\n\\begin{equation}\\label{cb_eq_error_bounds}\n d_n-d_{n+1}<|L-r_n|<d_n,\n \\qquad \\operatorname{sign}(L-r_n)=(-1)^n.\n\\end{equation}\n\\end{theorem}","\\begin{proof}\nEquation \\eqref{cb_eq_alternating_steps} and the preceding lemma identify\n$r_n$ with partial sums of a convergent alternating series:\n\\[\n r_n=\\sum_{j=0}^{n-1}(-1)^j d_j,\n \\qquad L=\\sum_{j=0}^{\\infty}(-1)^j d_j.\n\\]\nFor the even subsequence,\n$r_{2k+2}-r_{2k}=d_{2k}-d_{2k+1}>0$.\nFor the odd subsequence,\n$r_{2k+3}-r_{2k+1}=-d_{2k+1}+d_{2k+2}<0$.\nTheir adjacent separation $d_{2k}$ tends to zero, so they have the same limit.\nStrictness of both subsequences places that limit strictly between every\ndisplayed pair. The alternating tail beginning with $d_n$ gives the sign\nand upper bound in \\eqref{cb_eq_error_bounds}. Pairing its first two terms\nand then successive pairs gives the strict lower bound. All omitted pairs\nare positive after multiplication by $(-1)^n$.\n\\end{proof}","=0}^{n-1}(-1)^j d_j,\n \\qquad L=\\sum_{j=0}^{\\infty}(-1)^j d_j.\n\\]\nFor the even subsequence,\n$r_{2k+2}-r_{2k}=d_{2k}-d_{2k+1}>0$.\nFor the odd subsequence,\n$r_{2k+3}-r_{2k+1}=-d_{2k+1}+d_{2k+2}<0$.\nTheir adjacent separation $d_{2k}$ tends to zero, so they have the same limit.\nStrictness of both subsequences places that limit strictly between every\ndisplayed pair. The alternating tail beginning with $d_n$ gives the sign\nand upper bound in \\eqref{cb_eq_error_bounds}. Pairing its first two terms\nand then successive pairs gives the strict lower bound. All omitted pairs\nare positive after multiplication by $(-1)^n$.\n\\end{proof}","nt separation $d_{2k}$ tends to zero, so they have the same limit.\nStrictness of both subsequences places that limit strictly between every\ndisplayed pair. The alternating tail beginning with $d_n$ gives the sign\nand upper bound in \\eqref{cb_eq_error_bounds}. Pairing its first two terms\nand then successive pairs gives the strict lower bound. All omitted pairs\nare positive after multiplication by $(-1)^n$.\n\\end{proof}","The endpoints in \\eqref{cb_eq_bracket} are exact rationals. For example,\nthe first few approximants are\n\\begin{center}\n\\begin{tabular}{@{}rl@{}}\n\\toprule\n$n$ & $r_n$","\\midrule\n$0$ & $0$","$1$ & $13/14$","$2$ & $10699/11682$","$3$ & $8025653/8761950$","The recurrence and its initial values establish a limit $L$. The statement\nthat this limit is $G$ is a separate mathematical identification. We use\nZudilin's Theorem 1 for that identification, with the normalization recorded\nin source contract LC01 \\cite{Zudilin2003}. After this external input,\n\\eqref{cb_eq_bracket} and \\eqref{cb_eq_error_bounds} become exact bounds for\nCatalan's constant and its rational approximants. In particular,\n\\begin{equation}\\label{cb_eq_linear_error_sign}\n (-1)^n(u_nG-v_n)>0\\qquad(n\\geq0).\n\\end{equation}","t $L$. The statement\nthat this limit is $G$ is a separate mathematical identification. We use\nZudilin's Theorem 1 for that identification, with the normalization recorded\nin source contract LC01 \\cite{Zudilin2003}. After this external input,\n\\eqref{cb_eq_bracket} and \\eqref{cb_eq_error_bounds} become exact bounds for\nCatalan's constant and its rational approximants. In particular,\n\\begin{equation}\\label{cb_eq_linear_error_sign}\n (-1)^n(u_nG-v_n)>0\\qquad(n\\geq0).\n\\end{equation}","This order of reasoning prevents a circular shortcut. Decimal agreement\nbetween $r_n$ and a familiar constant does not identify a limit. Nor does\nthe nonzero Wronskian alone exclude equality between one particular\napproximant and an arbitrary real number. The strict brackets supply\nnonvanishing for the recurrence limit, and the source theorem then identifies\nthat limit with $G$.","Krattenthaler and Rivoal use coefficients $a_n,b_n$ for a linear form in\nCatalan's constant \\cite{KR2008}. These letters belong to their source and\ndo not denote the recurrence coefficients of the previous chapter. In the\nnormalization of this book their coefficients satisfy\n\\begin{equation}\\label{cb_eq_KR_dictionary}\n a_n=-4u_n,\\qquad b_n=-4v_n.\n\\end{equation}\nThe notation bridge is recorded in source contract LC05. After shifting the\nsource summation variable by one, their hypergeometric series is $-1/2$\ntimes Zudilin's series. The source partial fraction expansion identifies\nthe two rational coefficients separately, giving\n\\eqref{cb_eq_KR_dictionary}.","These letters belong to their source and\ndo not denote the recurrence coefficients of the previous chapter. In the\nnormalization of this book their coefficients satisfy\n\\begin{equation}\\label{cb_eq_KR_dictionary}\n a_n=-4u_n,\\qquad b_n=-4v_n.\n\\end{equation}\nThe notation bridge is recorded in source contract LC05. After shifting the\nsource summation variable by one, their hypergeometric series is $-1/2$\ntimes Zudilin's series. The source partial fraction expansion identifies\nthe two rational coefficients separately, giving\n\\eqref{cb_eq_KR_dictionary}.","It would be invalid to obtain this dictionary merely by equating two real\nexpressions in $G$ and then treating $1$ and $G$ as linearly independent over\n$\\Q$. That independence is exactly the irrationality under investigation.\nInstead, take a formal variable $X$. The partial fraction coefficient\ncalculation identifies the polynomial $a_nX-b_n$ with\n$-4(u_nX-v_n)$ in $\\Q[X]$. Equality of coefficients in a polynomial ring\ndoes not depend on the arithmetic nature of any later substitution for\n$X$. This distinction also applies to the packet coefficients below.","d then treating $1$ and $G$ as linearly independent over\n$\\Q$. That independence is exactly the irrationality under investigation.\nInstead, take a formal variable $X$. The partial fraction coefficient\ncalculation identifies the polynomial $a_nX-b_n$ with\n$-4(u_nX-v_n)$ in $\\Q[X]$. Equality of coefficients in a polynomial ring\ndoes not depend on the arithmetic nature of any later substitution for\n$X$. This distinction also applies to the packet coefficients below.","The same source gives the explicit finite sum\n\\begin{equation}\\label{cb_eq_halfbinomial}\n u_n=\\sum_{j=0}^n\n \\binom nj\\binom{n-\\tfrac12}{j}\\binom{n+j-\\tfrac12}{j},\\qquad n\\geq0.\n\\end{equation}\nHere\n\\[\n \\binom{x}{0}=1,\\qquad\n \\binom{x}{j}=\\frac{x(x-1)\\cdots(x-j+1)}{j!}\\quad(j\\geq1)\n\\]\ndefines the generalized binomial coefficient. Formula\n\\eqref{cb_eq_halfbinomial} is Lemma 3, equation (3.1), after the coefficient\nnormalization above. Its identification with the recurrence is an external\nsource input. The denominator argument that follows is elementary and is\nwritten in full.","Define\n\\begin{equation}\\label{cb_eq_T_def}\n T_n=16^nu_n.\n\\end{equation}\nThe importance of the power $16^n$ is that it clears each summand of\n\\eqref{cb_eq_halfbinomial}, rather than relying on cancellation between\ndifferent summands.","\\begin{lemma}[Half binomial denominators]\\label{cb_lem_half_denom}\nFor integers $h,j\\geq0$, the number\n$4^j\\binom{h-\\tfrac12}{j}$ is an integer.\n\\end{lemma}","\\begin{proof}\nWork with formal power series, so no convergence issue arises. The central\nbinomial expansion gives\n\\[\n (1+4z)^{-1/2}=\\sum_{k\\geq0}(-1)^k\\binom{2k}{k}z^k.\n\\]\nEvery coefficient is an integer. Multiplying by the integer polynomial\n$(1+4z)^h$ preserves this property. The coefficient of $z^j$ in the product\nis $4^j\\binom{h-\\tfrac12}{j}$. More explicitly it is\n\\[\n \\sum_{\\ell=0}^{\\min(h,j)}\n \\binom h\\ell 4^\\ell(-1)^{j-\\ell}\n \\binom{2(j-\\ell)}{j-\\ell},\n\\]\nan integer sum.\n\\end{proof}","\\begin{theorem}[Universal integrality of $T_n$]\\label{cb_thm_T_integer}\nFor every integer $n\\geq0$, $T_n$ is a positive integer. In fact, after\nmultiplication by $16^n$, each summand in\n\\eqref{cb_eq_halfbinomial} is an integer.\n\\end{theorem}","\\begin{proof}\nFor $0\\leq j\\leq n$, apply the preceding lemma with $h=n$ and $h=n+j$.\nThe scaled summand is\n\\[\n 16^{n-j}\\binom nj\n \\left(4^j\\binom{n-\\tfrac12}{j}\\right)\n \\left(4^j\\binom{n+j-\\tfrac12}{j}\\right).\n\\]\nAll four factors are integers. Summing proves integrality. Every original\nsummand is positive: the generalized binomial factors have only positive\nfactors in their defining products when $j\\leq n$. The $j=0$ term is one,\nincluding at $n=0$. Thus the sum is positive.\n\\end{proof}","begin{proof}\nFor $0\\leq j\\leq n$, apply the preceding lemma with $h=n$ and $h=n+j$.\nThe scaled summand is\n\\[\n 16^{n-j}\\binom nj\n \\left(4^j\\binom{n-\\tfrac12}{j}\\right)\n \\left(4^j\\binom{n+j-\\tfrac12}{j}\\right).\n\\]\nAll four factors are integers. Summing proves integrality. Every original\nsummand is positive: the generalized binomial factors have only positive\nfactors in their defining products when $j\\leq n$. The $j=0$ term is one,\nincluding at $n=0$. Thus the sum is positive.\n\\end{proof}","r $0\\leq j\\leq n$, apply the preceding lemma with $h=n$ and $h=n+j$.\nThe scaled summand is\n\\[\n 16^{n-j}\\binom nj\n \\left(4^j\\binom{n-\\tfrac12}{j}\\right)\n \\left(4^j\\binom{n+j-\\tfrac12}{j}\\right).\n\\]\nAll four factors are integers. Summing proves integrality. Every original\nsummand is positive: the generalized binomial factors have only positive\nfactors in their defining products when $j\\leq n$. The $j=0$ term is one,\nincluding at $n=0$. Thus the sum is positive.\n\\end{proof}","ceding lemma with $h=n$ and $h=n+j$.\nThe scaled summand is\n\\[\n 16^{n-j}\\binom nj\n \\left(4^j\\binom{n-\\tfrac12}{j}\\right)\n \\left(4^j\\binom{n+j-\\tfrac12}{j}\\right).\n\\]\nAll four factors are integers. Summing proves integrality. Every original\nsummand is positive: the generalized binomial factors have only positive\nfactors in their defining products when $j\\leq n$. The $j=0$ term is one,\nincluding at $n=0$. Thus the sum is positive.\n\\end{proof}","For example, the $n=2$ sum is\n\\[\n u_2=1+\\frac{15}{2}+\\frac{105}{64}=\\frac{649}{64}.\n\\]\nMultiplication by $16^2$ gives the three integers $256$, $1920$, and $420$,\nwhose sum is $T_2=2596$. This illustrates the proof at the level of individual\nterms. The first sequence values are\n\\[\n T_0=1,\\quad T_1=28,\\quad T_2=2596,\\quad\n T_3=311536,\\quad T_4=41759524.\n\\]","The source theorem $16^na_n\\in\\Z$, combined with\n\\eqref{cb_eq_KR_dictionary}, would establish only $4T_n\\in\\Z$ directly.\nDivision by four is not an integrality argument. The half binomial proof\nabove establishes the stronger statement actually needed by the integer\npacket construction. It removes a possible hidden dependence on a finite\nexact division experiment.","The recurrence for $T_n$ is obtained by multiplying the original recurrence\nby $16^{n+1}$:\n\\begin{equation}\\label{cb_eq_T_recurrence}\n a_nT_{n+1}=16q(n)T_n+256c_nT_{n-1}\\qquad(n\\geq1).\n\\end{equation}\nA program can therefore generate $T_n$ using integer numerators and exact\ndivision by $a_n$. A nonzero remainder is a failed implementation check and\nmust terminate the computation. The written theorem establishes why exact\ndivision succeeds universally for the correctly normalized recurrence.","\\begin{figure}[tbp]\n\\centering\\includegraphics[width=0.93\\linewidth]{../visual/figures/15_scaled_coefficients.pdf}\n\\caption{The recurrence coefficients become an integer sequence after\nscaling by $16^n$. The finite values shown here accompany a universal\nhalf binomial denominator proof.}\\label{cb_fig_T}\n\\end{figure}","Put\n\\[\n L_N=\\operatorname{lcm}(1,2,\\ldots,N),\\qquad L_0=1.\n\\]\nThe source's least common multiple notation is translated to $L_N$ here.\nThe letter $D$ will later denote an actual reduced denominator, which can\nbe much smaller than this universal clearing bound.\nKrattenthaler and Rivoal's Theorem 2 gives\n$16^nL_{2n}^2b_n\\in\\Z$ for positive integer $n$\n\\cite{KR2008}. In the present normalization it implies\n\\begin{equation}\\label{cb_eq_v_sharp_clear}\n 4\\cdot16^nL_{2n}^2v_n\\in\\Z\\qquad(n\\geq0).\n\\end{equation}\nThe case $n=0$ is immediate from $v_0=0$. In particular the historical\nscript's weaker clearing factor, with $8$ in place of $4$, is valid at every\nindex. This is a source theorem about the rational coefficient. The\nhalf binomial argument for $u_n$ d","Put\n\\[\n L_N=\\operatorname{lcm}(1,2,\\ldots,N),\\qquad L_0=1.\n\\]\nThe source's least common multiple notation is translated to $L_N$ here.\nThe letter $D$ will later denote an actual reduced denominator, which can\nbe much smaller than this universal clearing bound.\nKrattenthaler and Rivoal's Theorem 2 gives\n$16^nL_{2n}^2b_n\\in\\Z$ for positive integer $n$\n\\cite{KR2008}. In the present normalization it implies\n\\begin{equation}\\label{cb_eq_v_sharp_clear}\n 4\\cdot16^nL_{2n}^2v_n\\in\\Z\\qquad(n\\geq0).\n\\end{equation}\nThe case $n=0$ is immediate from $v_0=0$. In particular the historical\nscript's weaker clearing factor, with $8$ in place of $4$, is valid at every\nindex. This is a source theorem about the rational coefficient. The\nhalf binomial argument for $u_n$ does not automatically prove it.","me{lcm}(1,2,\\ldots,N),\\qquad L_0=1.\n\\]\nThe source's least common multiple notation is translated to $L_N$ here.\nThe letter $D$ will later denote an actual reduced denominator, which can\nbe much smaller than this universal clearing bound.\nKrattenthaler and Rivoal's Theorem 2 gives\n$16^nL_{2n}^2b_n\\in\\Z$ for positive integer $n$\n\\cite{KR2008}. In the present normalization it implies\n\\begin{equation}\\label{cb_eq_v_sharp_clear}\n 4\\cdot16^nL_{2n}^2v_n\\in\\Z\\qquad(n\\geq0).\n\\end{equation}\nThe case $n=0$ is immediate from $v_0=0$. In particular the historical\nscript's weaker clearing factor, with $8$ in place of $4$, is valid at every\nindex. This is a source theorem about the rational coefficient. The\nhalf binomial argument for $u_n$ does not automatically prove it.","to $L_N$ here.\nThe letter $D$ will later denote an actual reduced denominator, which can\nbe much smaller than this universal clearing bound.\nKrattenthaler and Rivoal's Theorem 2 gives\n$16^nL_{2n}^2b_n\\in\\Z$ for positive integer $n$\n\\cite{KR2008}. In the present normalization it implies\n\\begin{equation}\\label{cb_eq_v_sharp_clear}\n 4\\cdot16^nL_{2n}^2v_n\\in\\Z\\qquad(n\\geq0).\n\\end{equation}\nThe case $n=0$ is immediate from $v_0=0$. In particular the historical\nscript's weaker clearing factor, with $8$ in place of $4$, is valid at every\nindex. This is a source theorem about the rational coefficient. The\nhalf binomial argument for $u_n$ does not automatically prove it.","er $D$ will later denote an actual reduced denominator, which can\nbe much smaller than this universal clearing bound.\nKrattenthaler and Rivoal's Theorem 2 gives\n$16^nL_{2n}^2b_n\\in\\Z$ for positive integer $n$\n\\cite{KR2008}. In the present normalization it implies\n\\begin{equation}\\label{cb_eq_v_sharp_clear}\n 4\\cdot16^nL_{2n}^2v_n\\in\\Z\\qquad(n\\geq0).\n\\end{equation}\nThe case $n=0$ is immediate from $v_0=0$. In particular the historical\nscript's weaker clearing factor, with $8$ in place of $4$, is valid at every\nindex. This is a source theorem about the rational coefficient. The\nhalf binomial argument for $u_n$ does not automatically prove it.","aller than this universal clearing bound.\nKrattenthaler and Rivoal's Theorem 2 gives\n$16^nL_{2n}^2b_n\\in\\Z$ for positive integer $n$\n\\cite{KR2008}. In the present normalization it implies\n\\begin{equation}\\label{cb_eq_v_sharp_clear}\n 4\\cdot16^nL_{2n}^2v_n\\in\\Z\\qquad(n\\geq0).\n\\end{equation}\nThe case $n=0$ is immediate from $v_0=0$. In particular the historical\nscript's weaker clearing factor, with $8$ in place of $4$, is valid at every\nindex. This is a source theorem about the rational coefficient. The\nhalf binomial argument for $u_n$ does not automatically prove it.","universal clearing bound.\nKrattenthaler and Rivoal's Theorem 2 gives\n$16^nL_{2n}^2b_n\\in\\Z$ for positive integer $n$\n\\cite{KR2008}. In the present normalization it implies\n\\begin{equation}\\label{cb_eq_v_sharp_clear}\n 4\\cdot16^nL_{2n}^2v_n\\in\\Z\\qquad(n\\geq0).\n\\end{equation}\nThe case $n=0$ is immediate from $v_0=0$. In particular the historical\nscript's weaker clearing factor, with $8$ in place of $4$, is valid at every\nindex. This is a source theorem about the rational coefficient. The\nhalf binomial argument for $u_n$ does not automatically prove it.","cite{KR2008}. In the present normalization it implies\n\\begin{equation}\\label{cb_eq_v_sharp_clear}\n 4\\cdot16^nL_{2n}^2v_n\\in\\Z\\qquad(n\\geq0).\n\\end{equation}\nThe case $n=0$ is immediate from $v_0=0$. In particular the historical\nscript's weaker clearing factor, with $8$ in place of $4$, is valid at every\nindex. This is a source theorem about the rational coefficient. The\nhalf binomial argument for $u_n$ does not automatically prove it.","Earlier asymptotic denominator bounds in Zudilin's remarks establish the\nappropriate exponential scale with a logarithmic correction in the exponent\n\\cite{Zudilin2002}. They serve a different purpose from exact clearing at\nevery index. An asymptotic expression involving $o(n)$ should not be read\nas a precise power of two that can be inserted into a finite computation.\nFor the packet arithmetic we use \\eqref{cb_eq_v_sharp_clear} and the exact\ninteger theorem for $T_n$.","h the\nappropriate exponential scale with a logarithmic correction in the exponent\n\\cite{Zudilin2002}. They serve a different purpose from exact clearing at\nevery index. An asymptotic expression involving $o(n)$ should not be read\nas a precise power of two that can be inserted into a finite computation.\nFor the packet arithmetic we use \\eqref{cb_eq_v_sharp_clear} and the exact\ninteger theorem for $T_n$.","The finite direct product check in this release compares\n\\eqref{cb_eq_halfbinomial} with a separately indexed rational recurrence for\nevery $n=0,\\ldots,50$. It checks 1326 summands, including the two half\nbinomial denominator statements and scaled integrality of each term. The\nindependent recurrence check extends through $n=500$ and verifies the weaker\nfactor $8$ clearing there. These are useful finite corroborations. The\nexternal identity and the two universal arithmetic arguments have their\nown evidence and remain distinct from those checks.","Consider the positive double integral\n\\begin{equation}\\label{cb_eq_J_def}\n J_n=\\int_0^1\\!\\int_0^1\n \\frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}}\n {(1-xy)^{n+1}}\\,\\dd x\\,\\dd y,\n \\qquad n\\geq0.\n\\end{equation}\nZudilin's Theorem 3 prints $u_nG-v_n=(-1)^nJ_n/4$ in both the authenticated\narXiv version and the published journal PDF \\cite{Zudilin2003}. With\n$u_0=1$ and $v_0=0$, this would force $J_0=4G$.\nThe integral itself gives $J_0=8G$. Thus that printed factor cannot be used\nwith these initial conditions. This edition uses\n\\begin{equation}\\label{cb_eq_J_correct}\n J_n=8(-1)^n(u_nG-v_n).\n\\end{equation}\nNo erratum was recovered in the bounded source search. The correction is\nsupported by the explicit derivation below, rather than an assertion that\na published correction exists.","double integral\n\\begin{equation}\\label{cb_eq_J_def}\n J_n=\\int_0^1\\!\\int_0^1\n \\frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}}\n {(1-xy)^{n+1}}\\,\\dd x\\,\\dd y,\n \\qquad n\\geq0.\n\\end{equation}\nZudilin's Theorem 3 prints $u_nG-v_n=(-1)^nJ_n/4$ in both the authenticated\narXiv version and the published journal PDF \\cite{Zudilin2003}. With\n$u_0=1$ and $v_0=0$, this would force $J_0=4G$.\nThe integral itself gives $J_0=8G$. Thus that printed factor cannot be used\nwith these initial conditions. This edition uses\n\\begin{equation}\\label{cb_eq_J_correct}\n J_n=8(-1)^n(u_nG-v_n).\n\\end{equation}\nNo erratum was recovered in the bounded source search. The correction is\nsupported by the explicit derivation below, rather than an assertion that\na published correction exists.","2}}\n {(1-xy)^{n+1}}\\,\\dd x\\,\\dd y,\n \\qquad n\\geq0.\n\\end{equation}\nZudilin's Theorem 3 prints $u_nG-v_n=(-1)^nJ_n/4$ in both the authenticated\narXiv version and the published journal PDF \\cite{Zudilin2003}. With\n$u_0=1$ and $v_0=0$, this would force $J_0=4G$.\nThe integral itself gives $J_0=8G$. Thus that printed factor cannot be used\nwith these initial conditions. This edition uses\n\\begin{equation}\\label{cb_eq_J_correct}\n J_n=8(-1)^n(u_nG-v_n).\n\\end{equation}\nNo erratum was recovered in the bounded source search. The correction is\nsupported by the explicit derivation below, rather than an assertion that\na published correction exists.","The initial integral supplies a short diagnostic. Substitute $x=t^2$ and\n$y=1-s^2$. Then\n\\begin{align*}\n J_0\n &=4\\int_0^1\\!\\int_0^1\\frac{\\dd s\\,\\dd t}{1-t^2+t^2s^2}","The initial integral supplies a short diagnostic. Substitute $x=t^2$ and\n$y=1-s^2$. Then\n\\begin{align*}\n J_0\n &=4\\int_0^1\\!\\int_0^1\\frac{\\dd s\\,\\dd t}{1-t^2+t^2s^2}\\\\\n &=4\\int_0^1\\frac{\\arcsin t}{t\\sqrt{1-t^2}}\\,\\dd t\n  =4\\int_0^{\\pi/2}\\theta\\csc\\theta\\,\\dd\\theta.\n\\end{align*}\nSince an antiderivative of $\\csc\\theta$ is $\\log\\tan(\\theta/2)$, integration\nby parts yields\n\\[\n J_0=-4\\int_0^{\\pi/2}\\log\\tan(\\theta/2)\\,\\dd\\theta\n     =-8\\int_0^{\\pi/4}\\log\\tan z\\,\\dd z=8G.\n\\]\nBoth boundary terms vanish. At zero this uses\n$\\theta\\log\\tan(\\theta/2)\\to0$, and at $\\pi/2$ the logarithm is zero.","&=4\\int_0^1\\frac{\\arcsin t}{t\\sqrt{1-t^2}}\\,\\dd t\n  =4\\int_0^{\\pi/2}\\theta\\csc\\theta\\,\\dd\\theta.\n\\end{align*}\nSince an antiderivative of $\\csc\\theta$ is $\\log\\tan(\\theta/2)$, integration\nby parts yields\n\\[\n J_0=-4\\int_0^{\\pi/2}\\log\\tan(\\theta/2)\\,\\dd\\theta\n     =-8\\int_0^{\\pi/4}\\log\\tan z\\,\\dd z=8G.\n\\]\nBoth boundary terms vanish. At zero this uses\n$\\theta\\log\\tan(\\theta/2)\\to0$, and at $\\pi/2$ the logarithm is zero.","Checking one index detects the wrong normalization. It does not by itself\nprove \\eqref{cb_eq_J_correct} for every $n$. The next calculation explains\nthe all index extension and identifies its external transformation input.","Expanding $(1-xy)^{-n-1}$ as its nonnegative binomial series and applying\nTonelli's theorem gives\n\\begin{equation}\\label{cb_eq_beta_H}\n J_n=C_nH_n,\n\\end{equation}\nwhere\n\\[\n C_n=\\frac{\\Gamma(n+1)^2\\Gamma(n+\\tfrac12)^2}\n {\\Gamma(2n+\\tfrac32)^2},\n\\]\nand\n\\[\n H_n={}_3F_2\\!\\left(\n \\begin{matrix}n+1,n+\\tfrac12,n+1","Expanding $(1-xy)^{-n-1}$ as its nonnegative binomial series and applying\nTonelli's theorem gives\n\\begin{equation}\\label{cb_eq_beta_H}\n J_n=C_nH_n,\n\\end{equation}\nwhere\n\\[\n C_n=\\frac{\\Gamma(n+1)^2\\Gamma(n+\\tfrac12)^2}\n {\\Gamma(2n+\\tfrac32)^2},\n\\]\nand\n\\[\n H_n={}_3F_2\\!\\left(\n \\begin{matrix}n+1,n+\\tfrac12,n+1\\\\\n 2n+\\tfrac32,2n+\\tfrac32\\end{matrix};1\\right).\n\\]\nIndeed, the two beta integrals in the term with index $k$ are\n$B(n+\\tfrac12+k,n+1)$ and $B(n+1+k,n+\\tfrac12)$.\nTheir ratios to the corresponding $k=0$ integrals give exactly the displayed\nhypergeometric parameters. The excess of the lower parameters over the\nupper parameters is $n+1/2>0$, so the series converges for every $n\\geq0$.","2n+\\tfrac32,2n+\\tfrac32\\end{matrix};1\\right).\n\\]\nIndeed, the two beta integrals in the term with index $k$ are\n$B(n+\\tfrac12+k,n+1)$ and $B(n+1+k,n+\\tfrac12)$.\nTheir ratios to the corresponding $k=0$ integrals give exactly the displayed\nhypergeometric parameters. The excess of the lower parameters over the\nupper parameters is $n+1/2>0$, so the series converges for every $n\\geq0$.","Let $E_n$ denote the well poised ${}_6F_5$ series at $-1$ in Zudilin's\ndisplayed Whipple transformation, with\n$a=3n+1$, $b=c=d=n+1/2$, and $e=n+1$. The transformation states\n\\[\n H_n=\\frac{\\Gamma(3n+2)\\Gamma(n+\\tfrac12)}\n {\\Gamma(2n+\\tfrac32)\\Gamma(2n+1)}E_n.\n\\]\nIts condition $\\operatorname{Re}(1+a-d-e)>0$ becomes $n+1/2>0$ here.\nThe displayed prefactor of the source hypergeometric form $F_n^{\\rm src}$\nis\n\\[\n F_n^{\\rm src}=(-1)^n\n \\frac{\\Gamma(n+1)^2\\Gamma(3n+2)\\Gamma(n+\\tfrac12)^3}\n {\\Gamma(2n+\\tfrac32)^3\\Gamma(2n+1)}E_n.\n\\]\nSubstitution in \\eqref{cb_eq_beta_H} therefore gives\n$F_n^{\\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma\ncancellation. The formal partial fraction coeffici","Let $E_n$ denote the well poised ${}_6F_5$ series at $-1$ in Zudilin's\ndisplayed Whipple transformation, with\n$a=3n+1$, $b=c=d=n+1/2$, and $e=n+1$. The transformation states\n\\[\n H_n=\\frac{\\Gamma(3n+2)\\Gamma(n+\\tfrac12)}\n {\\Gamma(2n+\\tfrac32)\\Gamma(2n+1)}E_n.\n\\]\nIts condition $\\operatorname{Re}(1+a-d-e)>0$ becomes $n+1/2>0$ here.\nThe displayed prefactor of the source hypergeometric form $F_n^{\\rm src}$\nis\n\\[\n F_n^{\\rm src}=(-1)^n\n \\frac{\\Gamma(n+1)^2\\Gamma(3n+2)\\Gamma(n+\\tfrac12)^3}\n {\\Gamma(2n+\\tfrac32)^3\\Gamma(2n+1)}E_n.\n\\]\nSubstitution in \\eqref{cb_eq_beta_H} therefore gives\n$F_n^{\\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma\ncancellation. The formal partial fraction coefficient identification in\nthe same so","Let $E_n$ denote the well poised ${}_6F_5$ series at $-1$ in Zudilin's\ndisplayed Whipple transformation, with\n$a=3n+1$, $b=c=d=n+1/2$, and $e=n+1$. The transformation states\n\\[\n H_n=\\frac{\\Gamma(3n+2)\\Gamma(n+\\tfrac12)}\n {\\Gamma(2n+\\tfrac32)\\Gamma(2n+1)}E_n.\n\\]\nIts condition $\\operatorname{Re}(1+a-d-e)>0$ becomes $n+1/2>0$ here.\nThe displayed prefactor of the source hypergeometric form $F_n^{\\rm src}$\nis\n\\[\n F_n^{\\rm src}=(-1)^n\n \\frac{\\Gamma(n+1)^2\\Gamma(3n+2)\\Gamma(n+\\tfrac12)^3}\n {\\Gamma(2n+\\tfrac32)^3\\Gamma(2n+1)}E_n.\n\\]\nSubstitution in \\eqref{cb_eq_beta_H} therefore gives\n$F_n^{\\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma\ncancellation. The formal partial fraction coefficient identification in\nthe same source gives $F_n","Let $E_n$ denote the well poised ${}_6F_5$ series at $-1$ in Zudilin's\ndisplayed Whipple transformation, with\n$a=3n+1$, $b=c=d=n+1/2$, and $e=n+1$. The transformation states\n\\[\n H_n=\\frac{\\Gamma(3n+2)\\Gamma(n+\\tfrac12)}\n {\\Gamma(2n+\\tfrac32)\\Gamma(2n+1)}E_n.\n\\]\nIts condition $\\operatorname{Re}(1+a-d-e)>0$ becomes $n+1/2>0$ here.\nThe displayed prefactor of the source hypergeometric form $F_n^{\\rm src}$\nis\n\\[\n F_n^{\\rm src}=(-1)^n\n \\frac{\\Gamma(n+1)^2\\Gamma(3n+2)\\Gamma(n+\\tfrac12)^3}\n {\\Gamma(2n+\\tfrac32)^3\\Gamma(2n+1)}E_n.\n\\]\nSubstitution in \\eqref{cb_eq_beta_H} therefore gives\n$F_n^{\\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma\ncancellation. The formal partial fraction coefficient identification in\nthe same source gives $F_n^{\\rm src}=8(u_nG-v_n)$, with the recurrence\ninitial data use","Let $E_n$ denote the well poised ${}_6F_5$ series at $-1$ in Zudilin's\ndisplayed Whipple transformation, with\n$a=3n+1$, $b=c=d=n+1/2$, and $e=n+1$. The transformation states\n\\[\n H_n=\\frac{\\Gamma(3n+2)\\Gamma(n+\\tfrac12)}\n {\\Gamma(2n+\\tfrac32)\\Gamma(2n+1)}E_n.\n\\]\nIts condition $\\operatorname{Re}(1+a-d-e)>0$ becomes $n+1/2>0$ here.\nThe displayed prefactor of the source hypergeometric form $F_n^{\\rm src}$\nis\n\\[\n F_n^{\\rm src}=(-1)^n\n \\frac{\\Gamma(n+1)^2\\Gamma(3n+2)\\Gamma(n+\\tfrac12)^3}\n {\\Gamma(2n+\\tfrac32)^3\\Gamma(2n+1)}E_n.\n\\]\nSubstitution in \\eqref{cb_eq_beta_H} therefore gives\n$F_n^{\\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma\ncancellation. The formal partial fraction coefficient identification in\nthe same source gives $F_n^{\\rm src}=8(u_nG-v_n)$, with the recurrence\ninitial data used here. Combini","Let $E_n$ denote the well poised ${}_6F_5$ series at $-1$ in Zudilin's\ndisplayed Whipple transformation, with\n$a=3n+1$, $b=c=d=n+1/2$, and $e=n+1$. The transformation states\n\\[\n H_n=\\frac{\\Gamma(3n+2)\\Gamma(n+\\tfrac12)}\n {\\Gamma(2n+\\tfrac32)\\Gamma(2n+1)}E_n.\n\\]\nIts condition $\\operatorname{Re}(1+a-d-e)>0$ becomes $n+1/2>0$ here.\nThe displayed prefactor of the source hypergeometric form $F_n^{\\rm src}$\nis\n\\[\n F_n^{\\rm src}=(-1)^n\n \\frac{\\Gamma(n+1)^2\\Gamma(3n+2)\\Gamma(n+\\tfrac12)^3}\n {\\Gamma(2n+\\tfrac32)^3\\Gamma(2n+1)}E_n.\n\\]\nSubstitution in \\eqref{cb_eq_beta_H} therefore gives\n$F_n^{\\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma\ncancellation. The formal partial fraction coefficient identification in\nthe same source gives $F_n^{\\rm src}=8(u_nG-v_n)$, with the recurrence\ninitial data used here. Combining these two","Let $E_n$ denote the well poised ${}_6F_5$ series at $-1$ in Zudilin's\ndisplayed Whipple transformation, with\n$a=3n+1$, $b=c=d=n+1/2$, and $e=n+1$. The transformation states\n\\[\n H_n=\\frac{\\Gamma(3n+2)\\Gamma(n+\\tfrac12)}\n {\\Gamma(2n+\\tfrac32)\\Gamma(2n+1)}E_n.\n\\]\nIts condition $\\operatorname{Re}(1+a-d-e)>0$ becomes $n+1/2>0$ here.\nThe displayed prefactor of the source hypergeometric form $F_n^{\\rm src}$\nis\n\\[\n F_n^{\\rm src}=(-1)^n\n \\frac{\\Gamma(n+1)^2\\Gamma(3n+2)\\Gamma(n+\\tfrac12)^3}\n {\\Gamma(2n+\\tfrac32)^3\\Gamma(2n+1)}E_n.\n\\]\nSubstitution in \\eqref{cb_eq_beta_H} therefore gives\n$F_n^{\\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma\ncancellation. The formal partial fraction coefficient identification in\nthe same source gives $F_n^{\\rm src}=8(u_nG-v_n)$, with the recurrence\ninitial data used here. Combining these two identities proves\n\\eqref{cb_eq_J_correct} at every index.","denote the well poised ${}_6F_5$ series at $-1$ in Zudilin's\ndisplayed Whipple transformation, with\n$a=3n+1$, $b=c=d=n+1/2$, and $e=n+1$. The transformation states\n\\[\n H_n=\\frac{\\Gamma(3n+2)\\Gamma(n+\\tfrac12)}\n {\\Gamma(2n+\\tfrac32)\\Gamma(2n+1)}E_n.\n\\]\nIts condition $\\operatorname{Re}(1+a-d-e)>0$ becomes $n+1/2>0$ here.\nThe displayed prefactor of the source hypergeometric form $F_n^{\\rm src}$\nis\n\\[\n F_n^{\\rm src}=(-1)^n\n \\frac{\\Gamma(n+1)^2\\Gamma(3n+2)\\Gamma(n+\\tfrac12)^3}\n {\\Gamma(2n+\\tfrac32)^3\\Gamma(2n+1)}E_n.\n\\]\nSubstitution in \\eqref{cb_eq_beta_H} therefore gives\n$F_n^{\\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma\ncancellation. The formal partial fraction coefficient identification in\nthe same source gives $F_n^{\\rm src}=8(u_nG-v_n)$, with the recurrence\ninitial data used here. Combining these two identities proves\n\\eqref{cb_eq_J_correct} at every index.","ac{\\Gamma(3n+2)\\Gamma(n+\\tfrac12)}\n {\\Gamma(2n+\\tfrac32)\\Gamma(2n+1)}E_n.\n\\]\nIts condition $\\operatorname{Re}(1+a-d-e)>0$ becomes $n+1/2>0$ here.\nThe displayed prefactor of the source hypergeometric form $F_n^{\\rm src}$\nis\n\\[\n F_n^{\\rm src}=(-1)^n\n \\frac{\\Gamma(n+1)^2\\Gamma(3n+2)\\Gamma(n+\\tfrac12)^3}\n {\\Gamma(2n+\\tfrac32)^3\\Gamma(2n+1)}E_n.\n\\]\nSubstitution in \\eqref{cb_eq_beta_H} therefore gives\n$F_n^{\\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma\ncancellation. The formal partial fraction coefficient identification in\nthe same source gives $F_n^{\\rm src}=8(u_nG-v_n)$, with the recurrence\ninitial data used here. Combining these two identities proves\n\\eqref{cb_eq_J_correct} at every index.","r of the source hypergeometric form $F_n^{\\rm src}$\nis\n\\[\n F_n^{\\rm src}=(-1)^n\n \\frac{\\Gamma(n+1)^2\\Gamma(3n+2)\\Gamma(n+\\tfrac12)^3}\n {\\Gamma(2n+\\tfrac32)^3\\Gamma(2n+1)}E_n.\n\\]\nSubstitution in \\eqref{cb_eq_beta_H} therefore gives\n$F_n^{\\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma\ncancellation. The formal partial fraction coefficient identification in\nthe same source gives $F_n^{\\rm src}=8(u_nG-v_n)$, with the recurrence\ninitial data used here. Combining these two identities proves\n\\eqref{cb_eq_J_correct} at every index.","al expansion and prefactor cancellation are written\ncalculations. The Whipple transformation and the partial fraction\nidentification are external mathematical inputs from the authenticated\npaper. Their precise roles are recorded in LC03 and the accompanying\nnormalization audit. The independent initial integral agrees with the\nresult, and Eskandari's coefficient normalization uses the same factor $8$\n\\cite{Eskandari2026}.","The recurrence permits efficient exact computation. The half binomial\nformula proves that $16^nu_n$ is always an integer. The integral describes\nthe error and makes its sign transparent after the corrected normalization.\nThese descriptions agree, but each answers a different question. Keeping\ntheir roles distinct prevents a finite arithmetic check from being mistaken\nfor an analytic theorem, or a positive integral from being mistaken for\ndenominator control. The packet construction will use all three.","For integers $n,m\\geq0$, define\n\\begin{align}\n S_{n,m}&=\\sum_{i=0}^m\\binom miT_{n+i},\\label{cb_eq_S_def}","For integers $n,m\\geq0$, define\n\\begin{align}\n S_{n,m}&=\\sum_{i=0}^m\\binom miT_{n+i},\\label{cb_eq_S_def}\\\\\n H_{n,m}&=\\sum_{i=0}^m\\binom mi16^{n+i}v_{n+i}.\n \\label{cb_eq_H_packet_def}\n\\end{align}\nThe two index symbol $H_{n,m}$ denotes a rational packet coefficient. It is\ndistinct from the hypergeometric function $H_n$ used in the integral\nnormalization derivation. The theorem on $T_n$ proves\n$S_{n,m}\\in\\Z_{>0}$ at every pair. The coefficient $H_{n,m}$ is rational\nand requires separate denominator clearing.","H_{n,m}&=\\sum_{i=0}^m\\binom mi16^{n+i}v_{n+i}.\n \\label{cb_eq_H_packet_def}\n\\end{align}\nThe two index symbol $H_{n,m}$ denotes a rational packet coefficient. It is\ndistinct from the hypergeometric function $H_n$ used in the integral\nnormalization derivation. The theorem on $T_n$ proves\n$S_{n,m}\\in\\Z_{>0}$ at every pair. The coefficient $H_{n,m}$ is rational\nand requires separate denominator clearing.","\\begin{proposition}[Pascal evolution]\\label{cb_prop_Pascal}\nFor $n,m\\geq0$,\n\\begin{equation}\\label{cb_eq_Pascal}\n S_{n,m+1}=S_{n,m}+S_{n+1,m}.\n\\end{equation}\nThe same identity holds with $H$ in place of $S$.\n\\end{proposition}","\\begin{proof}\nUse $\\binom{m+1}{i}=\\binom mi+\\binom m{i-1}$, with out of range binomial\ncoefficients interpreted as zero. The first contribution is $S_{n,m}$.\nChanging $i$ to $i+1$ in the second contribution gives $S_{n+1,m}$. The\nargument uses only the finite weights, so it applies to either seed\nsequence.\n\\end{proof}","At $m=0$, the packet is just $S_{n,0}=T_n$. The first two nontrivial rows\nare\n\\[\n S_{n,1}=T_n+T_{n+1},\\qquad\n S_{n,2}=T_n+2T_{n+1}+T_{n+2}.\n\\]\nFor example $S_{1,1}=28+2596=2624$. The Pascal identity produces exactly\nthe same integer as direct evaluation of its defining sum. This agreement\nbetween an additive row evolution and a direct weighted sum gives two\ndifferent implementations of the finite transform.","If $E$ denotes the shift operator $(Ef)_n=f_{n+1}$, the packet is\n$(1+E)^mT$ at index $n$. This is an identity about a finite polynomial in\none shift. It does not introduce a further assumption about recurrence\nsolutions or convergence. Reduction modulo a power of two commutes with\nthis polynomial, which is the basis of the valuation computation in the\nnext chapter.","Eskandari's preprint applies the filter\n\\begin{equation}\\label{cb_eq_Jtilde}\n \\widetilde J_{n,m}=\\sum_{i=0}^m(-16)^i\\binom miJ_{n+i}\n \\qquad(n,m\\geq0)\n\\end{equation}\nto the integral family \\cite{Eskandari2026}. The filter and its approximation\nanalysis are external work. This book examines its coefficient arithmetic\nand the finite evidence recovered from the project's script.","Substitute the corrected normalization\n$J_k=8(-1)^k(u_kG-v_k)$ into \\eqref{cb_eq_Jtilde}. The two factors\n$(-1)^i$ cancel. Thus\n\\begin{equation}\\label{cb_eq_Jtilde_coeffs}\n \\widetilde J_{n,m}=8(-1)^n16^{-n}(S_{n,m}G-H_{n,m}).\n\\end{equation}\nDefine the raw packet form\n\\begin{equation}\\label{cb_eq_raw_F}\n F_{n,m}=16^n\\widetilde J_{n,m}\n        =8(-1)^n(S_{n,m}G-H_{n,m}).\n\\end{equation}\nThe adjective raw means that the common denominator of $H_{n,m}$ has not\nyet been cleared. It does not mean that $F_{n,m}$ is an integer linear form.\nThe coefficient of $G$ is an integer, but the other coefficient may be\nrational.","=8(-1)^k(u_kG-v_k)$ into \\eqref{cb_eq_Jtilde}. The two factors\n$(-1)^i$ cancel. Thus\n\\begin{equation}\\label{cb_eq_Jtilde_coeffs}\n \\widetilde J_{n,m}=8(-1)^n16^{-n}(S_{n,m}G-H_{n,m}).\n\\end{equation}\nDefine the raw packet form\n\\begin{equation}\\label{cb_eq_raw_F}\n F_{n,m}=16^n\\widetilde J_{n,m}\n        =8(-1)^n(S_{n,m}G-H_{n,m}).\n\\end{equation}\nThe adjective raw means that the common denominator of $H_{n,m}$ has not\nyet been cleared. It does not mean that $F_{n,m}$ is an integer linear form.\nThe coefficient of $G$ is an integer, but the other coefficient may be\nrational.","tors\n$(-1)^i$ cancel. Thus\n\\begin{equation}\\label{cb_eq_Jtilde_coeffs}\n \\widetilde J_{n,m}=8(-1)^n16^{-n}(S_{n,m}G-H_{n,m}).\n\\end{equation}\nDefine the raw packet form\n\\begin{equation}\\label{cb_eq_raw_F}\n F_{n,m}=16^n\\widetilde J_{n,m}\n        =8(-1)^n(S_{n,m}G-H_{n,m}).\n\\end{equation}\nThe adjective raw means that the common denominator of $H_{n,m}$ has not\nyet been cleared. It does not mean that $F_{n,m}$ is an integer linear form.\nThe coefficient of $G$ is an integer, but the other coefficient may be\nrational.","Jtilde_coeffs}\n \\widetilde J_{n,m}=8(-1)^n16^{-n}(S_{n,m}G-H_{n,m}).\n\\end{equation}\nDefine the raw packet form\n\\begin{equation}\\label{cb_eq_raw_F}\n F_{n,m}=16^n\\widetilde J_{n,m}\n        =8(-1)^n(S_{n,m}G-H_{n,m}).\n\\end{equation}\nThe adjective raw means that the common denominator of $H_{n,m}$ has not\nyet been cleared. It does not mean that $F_{n,m}$ is an integer linear form.\nThe coefficient of $G$ is an integer, but the other coefficient may be\nrational.","This normalization is important for comparison with historical data.\nMultiplying or dividing a linear form by a growing power changes its\ndisplayed decay rate. The integral filter is $\\widetilde J_{n,m}$, while\nthe recorded arithmetic form $F_{n,m}$ is $16^n$ times it. Their signs and\nzeros agree because $16^n>0$. Their logarithmic size per unit $n$ differs by\n$\\log16$.","On the open unit square, set\n\\begin{equation}\\label{cb_eq_psi_mu}\n \\psi(x,y)=\\frac{xy(1-x)(1-y)}{1-xy},\\qquad\n \\dd\\mu=\\frac{\\dd x\\,\\dd y}{\\sqrt{x(1-y)}(1-xy)}.\n\\end{equation}\nThe factor $1-xy$ is outside the square root. Moving that factor inside, as in\n\\[\n \\frac{\\dd x\\,\\dd y}{\\sqrt{x(1-y)(1-xy)}},\n\\]\nwould represent a different integral. With the measure in \\eqref{cb_eq_psi_mu},\n\\[\n J_n=\\int_{(0,1)^2}\\psi^n\\,\\dd\\mu,\n \\qquad\n \\widetilde J_{n,m}=\\int_{(0,1)^2}\\psi^n(1-16\\psi)^m\\,\\dd\\mu.\n\\]\nThe second identity follows by the ordinary binomial theorem inside a\nfinite sum. The total measure is $J_0=8G<\\infty$.","tside the square root. Moving that factor inside, as in\n\\[\n \\frac{\\dd x\\,\\dd y}{\\sqrt{x(1-y)(1-xy)}},\n\\]\nwould represent a different integral. With the measure in \\eqref{cb_eq_psi_mu},\n\\[\n J_n=\\int_{(0,1)^2}\\psi^n\\,\\dd\\mu,\n \\qquad\n \\widetilde J_{n,m}=\\int_{(0,1)^2}\\psi^n(1-16\\psi)^m\\,\\dd\\mu.\n\\]\nThe second identity follows by the ordinary binomial theorem inside a\nfinite sum. The total measure is $J_0=8G<\\infty$.","The kernel has a simple maximum that also explains why the filter changes\nsign. Put $r=\\sqrt{xy}$. Since $x+y\\geq2r$,\n\\[\n \\psi(x,y)\\leq\\frac{r^2(1-r)^2}{1-r^2}\n              =\\frac{r^2(1-r)}{1+r},\\qquad0<r<1.\n\\]\nEquality is achieved when $x=y=r$. The derivative of the last expression\nis\n\\[\n \\frac{2r(1-r-r^2)}{(1+r)^2}.\n\\]\nIt has its interior maximum at $r=\\phi^{-1}$, where\n$\\phi=(1+\\sqrt5)/2$. Since $1-r=r^2$ there, the maximum is\n\\begin{equation}\\label{cb_eq_psi_max}\n \\max\\psi=\\phi^{-5}\\approx0.09017.\n\\end{equation}\nThis is greater than $1/16$. One can already see a negative filter value\nat $x=y=1/2$, where $\\psi=1/12$ and $1-16\\psi=-1/3$.\nNear the sides of the square $\\psi$ is small and the filter is positive.","kernel has a simple maximum that also explains why the filter changes\nsign. Put $r=\\sqrt{xy}$. Since $x+y\\geq2r$,\n\\[\n \\psi(x,y)\\leq\\frac{r^2(1-r)^2}{1-r^2}\n              =\\frac{r^2(1-r)}{1+r},\\qquad0<r<1.\n\\]\nEquality is achieved when $x=y=r$. The derivative of the last expression\nis\n\\[\n \\frac{2r(1-r-r^2)}{(1+r)^2}.\n\\]\nIt has its interior maximum at $r=\\phi^{-1}$, where\n$\\phi=(1+\\sqrt5)/2$. Since $1-r=r^2$ there, the maximum is\n\\begin{equation}\\label{cb_eq_psi_max}\n \\max\\psi=\\phi^{-5}\\approx0.09017.\n\\end{equation}\nThis is greater than $1/16$. One can already see a negative filter value\nat $x=y=1/2$, where $\\psi=1/12$ and $1-16\\psi=-1/3$.\nNear the sides of the square $\\psi$ is small and the filter is positive.","why the filter changes\nsign. Put $r=\\sqrt{xy}$. Since $x+y\\geq2r$,\n\\[\n \\psi(x,y)\\leq\\frac{r^2(1-r)^2}{1-r^2}\n              =\\frac{r^2(1-r)}{1+r},\\qquad0<r<1.\n\\]\nEquality is achieved when $x=y=r$. The derivative of the last expression\nis\n\\[\n \\frac{2r(1-r-r^2)}{(1+r)^2}.\n\\]\nIt has its interior maximum at $r=\\phi^{-1}$, where\n$\\phi=(1+\\sqrt5)/2$. Since $1-r=r^2$ there, the maximum is\n\\begin{equation}\\label{cb_eq_psi_max}\n \\max\\psi=\\phi^{-5}\\approx0.09017.\n\\end{equation}\nThis is greater than $1/16$. One can already see a negative filter value\nat $x=y=1/2$, where $\\psi=1/12$ and $1-16\\psi=-1/3$.\nNear the sides of the square $\\psi$ is small and the filter is positive.","1-r)^2}{1-r^2}\n              =\\frac{r^2(1-r)}{1+r},\\qquad0<r<1.\n\\]\nEquality is achieved when $x=y=r$. The derivative of the last expression\nis\n\\[\n \\frac{2r(1-r-r^2)}{(1+r)^2}.\n\\]\nIt has its interior maximum at $r=\\phi^{-1}$, where\n$\\phi=(1+\\sqrt5)/2$. Since $1-r=r^2$ there, the maximum is\n\\begin{equation}\\label{cb_eq_psi_max}\n \\max\\psi=\\phi^{-5}\\approx0.09017.\n\\end{equation}\nThis is greater than $1/16$. One can already see a negative filter value\nat $x=y=1/2$, where $\\psi=1/12$ and $1-16\\psi=-1/3$.\nNear the sides of the square $\\psi$ is small and the filter is positive.","ad0<r<1.\n\\]\nEquality is achieved when $x=y=r$. The derivative of the last expression\nis\n\\[\n \\frac{2r(1-r-r^2)}{(1+r)^2}.\n\\]\nIt has its interior maximum at $r=\\phi^{-1}$, where\n$\\phi=(1+\\sqrt5)/2$. Since $1-r=r^2$ there, the maximum is\n\\begin{equation}\\label{cb_eq_psi_max}\n \\max\\psi=\\phi^{-5}\\approx0.09017.\n\\end{equation}\nThis is greater than $1/16$. One can already see a negative filter value\nat $x=y=1/2$, where $\\psi=1/12$ and $1-16\\psi=-1/3$.\nNear the sides of the square $\\psi$ is small and the filter is positive.","ity is achieved when $x=y=r$. The derivative of the last expression\nis\n\\[\n \\frac{2r(1-r-r^2)}{(1+r)^2}.\n\\]\nIt has its interior maximum at $r=\\phi^{-1}$, where\n$\\phi=(1+\\sqrt5)/2$. Since $1-r=r^2$ there, the maximum is\n\\begin{equation}\\label{cb_eq_psi_max}\n \\max\\psi=\\phi^{-5}\\approx0.09017.\n\\end{equation}\nThis is greater than $1/16$. One can already see a negative filter value\nat $x=y=1/2$, where $\\psi=1/12$ and $1-16\\psi=-1/3$.\nNear the sides of the square $\\psi$ is small and the filter is positive.","when $x=y=r$. The derivative of the last expression\nis\n\\[\n \\frac{2r(1-r-r^2)}{(1+r)^2}.\n\\]\nIt has its interior maximum at $r=\\phi^{-1}$, where\n$\\phi=(1+\\sqrt5)/2$. Since $1-r=r^2$ there, the maximum is\n\\begin{equation}\\label{cb_eq_psi_max}\n \\max\\psi=\\phi^{-5}\\approx0.09017.\n\\end{equation}\nThis is greater than $1/16$. One can already see a negative filter value\nat $x=y=1/2$, where $\\psi=1/12$ and $1-16\\psi=-1/3$.\nNear the sides of the square $\\psi$ is small and the filter is positive.","ssion\nis\n\\[\n \\frac{2r(1-r-r^2)}{(1+r)^2}.\n\\]\nIt has its interior maximum at $r=\\phi^{-1}$, where\n$\\phi=(1+\\sqrt5)/2$. Since $1-r=r^2$ there, the maximum is\n\\begin{equation}\\label{cb_eq_psi_max}\n \\max\\psi=\\phi^{-5}\\approx0.09017.\n\\end{equation}\nThis is greater than $1/16$. One can already see a negative filter value\nat $x=y=1/2$, where $\\psi=1/12$ and $1-16\\psi=-1/3$.\nNear the sides of the square $\\psi$ is small and the filter is positive.","\\begin{theorem}[Nonvanishing for even filter order]\\label{cb_thm_even_positive}\nFor every $n\\geq0$ and every even integer $m\\geq0$,\n$\\widetilde J_{n,m}>0$ and $F_{n,m}>0$.\n\\end{theorem}","\\begin{proof}\nFor even $m$, the integrand $\\psi^n(1-16\\psi)^m$ is nonnegative. It is\nstrictly positive on an open subset of the square with $0<\\psi<1/16$.\nThe measure has positive density there. Hence its integral is strictly\npositive. It is finite because $\\psi$ and $1-16\\psi$ are bounded and\n$\\mu$ has finite total mass. Multiplication by $16^n$ preserves positivity.\n\\end{proof}","This theorem includes $m=0$ and $n=0$. It does not require that $G$ be\nirrational. Indeed, positivity is one of the ingredients that an eventual\nirrationality criterion would combine with sufficiently small cleared\ninteger forms. The even order restriction is easy to impose along an\nasymptotic slope by choosing $m(n)$ to be the nearest suitable even integer.\nThe change in $m/n$ then tends to zero.","For odd $m$, the polynomial factor changes sign, so the preceding positivity\nproof does not apply. A proposed shortcut was the sign pattern\n$\\operatorname{sign}F_{n,m}=(-1)^m$. The finite audit falsified it. The first\ncounterexample is the pair $(n,m)=(1,1)$.","Using $u_1=7/4$, $v_1=13/8$, $u_2=649/64$ and $v_2=10699/1152$, one obtains\n\\[\n S_{1,1}=2624,\\qquad H_{1,1}=\\frac{21632}{9}.\n\\]\nConsequently\n\\begin{equation}\\label{cb_eq_odd_counterexample}\n F_{1,1}=\\frac{173056}{9}-20992G>0.\n\\end{equation}\nThe inequality is certified by the exact rational enclosure of $G$ recorded\nin the independent computation. The threshold for it is $G<338/369$.\nThe enclosure lies strictly below that rational number. Thus this form is\npositive although $m$ is odd, contradicting the proposed pattern.\nThe sign is decided by rational interval endpoints, rather than floating\npoint subtraction between nearly equal numbers.","S_{1,1}=2624,\\qquad H_{1,1}=\\frac{21632}{9}.\n\\]\nConsequently\n\\begin{equation}\\label{cb_eq_odd_counterexample}\n F_{1,1}=\\frac{173056}{9}-20992G>0.\n\\end{equation}\nThe inequality is certified by the exact rational enclosure of $G$ recorded\nin the independent computation. The threshold for it is $G<338/369$.\nThe enclosure lies strictly below that rational number. Thus this form is\npositive although $m$ is odd, contradicting the proposed pattern.\nThe sign is decided by rational interval endpoints, rather than floating\npoint subtraction between nearly equal numbers.","The audit checks 860 pairs with $1\\leq n\\leq20$ and $0\\leq m\\leq4n$.\nEvery tested form is certified nonzero, but a number of odd pairs violate\nthe proposed pattern. Both positive and negative odd order forms occur in\nthat finite domain. For example the recorded signs at $(1,1)$ and $(2,1)$\nare opposite. These observations do not establish nonvanishing for every\nodd pair. This edition retains the failed hypothesis and its diagnostic\nrecord, and uses the universal even order theorem when nonvanishing is\nneeded in a global implication.","Let $N=n+m$ and put\n\\begin{equation}\\label{cb_eq_C_clear}\n C_{n,m}=L_{2N}^2.\n\\end{equation}\nThe pair produced by the historical script is\n\\begin{align}\n A_{n,m}&=8(-1)^nC_{n,m}S_{n,m},\\label{cb_eq_historical_A}","Let $N=n+m$ and put\n\\begin{equation}\\label{cb_eq_C_clear}\n C_{n,m}=L_{2N}^2.\n\\end{equation}\nThe pair produced by the historical script is\n\\begin{align}\n A_{n,m}&=8(-1)^nC_{n,m}S_{n,m},\\label{cb_eq_historical_A}\\\\\n B_{n,m}&=8(-1)^nC_{n,m}H_{n,m}.\n \\label{cb_eq_historical_B}\n\\end{align}\nBoth coefficients are integers. Integrality of $A$ follows from that of\n$S$. For $B$, each index $k=n+i$ satisfies $k\\leq N$, so $L_{2k}$ divides\n$L_{2N}$. The sharp source clearing\n$4\\cdot16^kL_{2k}^2v_k\\in\\Z$ therefore proves integrality term by term\nafter multiplication by $8L_{2N}^2$. The binomial weights introduce no\nfurther denominators.","B_{n,m}&=8(-1)^nC_{n,m}H_{n,m}.\n \\label{cb_eq_historical_B}\n\\end{align}\nBoth coefficients are integers. Integrality of $A$ follows from that of\n$S$. For $B$, each index $k=n+i$ satisfies $k\\leq N$, so $L_{2k}$ divides\n$L_{2N}$. The sharp source clearing\n$4\\cdot16^kL_{2k}^2v_k\\in\\Z$ therefore proves integrality term by term\nafter multiplication by $8L_{2N}^2$. The binomial weights introduce no\nfurther denominators.","The cleared linear form is exactly\n\\begin{equation}\\label{cb_eq_cleared_integral}\n A_{n,m}G-B_{n,m}\n =C_{n,m}F_{n,m}\n =16^nL_{2N}^2\\widetilde J_{n,m}.\n\\end{equation}\nEvery factor in this identity has a fixed meaning. In particular, the\nhistorical pair includes the constant factor $8$ and uses the possibly\nredundant common clearing $L_{2N}^2$. It is not automatically a primitive\ninteger pair. Dividing out its common content is a separate arithmetic\noperation examined later.","At $m=0$ the construction reduces to the cleared original recurrence form.\nAt $n=0$ all identities remain valid with $T_0=1$, $v_0=0$, and $L_0=1$.\nThe large valuation scan starts at $n=1$ for its stated triangular domain,\nbut that computational choice does not restrict the algebraic identities\nproved in this chapter.","For a nonzero integer $z$, let $v_2(z)$ be the largest nonnegative integer\n$t$ such that $2^t$ divides $z$. Thus $v_2(12)=2$ and $v_2(1)=0$.\nThe value $v_2(0)$ is infinite rather than a finite integer. Our packet\n$S_{n,m}$ is strictly positive, so its valuation is always finite, even when\na finite precision residue is zero.","\\begin{lemma}[Exactness below the precision ceiling]\\label{cb_lem_mod_v2}\nLet $K\\geq1$, let $z\\ne0$ be an integer, and let $\\rho$ be its least\nnonnegative residue modulo $2^K$. If $\\rho\\ne0$, then\n\\[\n v_2(z)=v_2(\\rho)<K.\n\\]\nIf $\\rho=0$, the residue gives only $v_2(z)\\geq K$.\n\\end{lemma}","\\begin{proof}\nWrite $z=\\rho+2^Kh$. If $\\rho=2^ta$ with $a$ odd and $t<K$, then\n$z=2^t(a+2^{K-t}h)$ and the parenthesis is odd. Hence the valuation is\nexactly $t$. If $\\rho=0$, the congruence says that $2^K$ divides $z$, but\ncontains no information about further powers of two.\n\\end{proof}","For example, the residue of $S_{1,1}=2624$ modulo $2^{128}$ is nonzero and\nhas valuation six. Its residue modulo $16$ is zero, which says only that\nits valuation is at least four. The integer itself did not change. The\nprecision changed the information available about it.","A program that returns a number $K$ for a zero residue must attach a lower\nbound interpretation to that number. It must not use it as an exact\nvaluation or as an exact maximum of the domain. The historical script\ncounts unresolved residues and includes an explicit exactness flag. The\nrelease tests that flag deliberately.","\\begin{figure}[tbp]\n\\centering\\includegraphics[width=0.91\\linewidth]{../visual/figures/18_modular_exactness.pdf}\n\\caption{A nonzero residue modulo $2^K$ determines the exact valuation.\nA zero residue reaches the precision ceiling and supplies only a lower\nbound.}\\label{cb_fig_mod_exactness}\n\\end{figure}","For a positive integer $N$, the scanned domain is\n\\begin{equation}\\label{cb_eq_scan_domain}\n \\mathcal D_N=\\{(n,m):1\\leq n\\leq N,\\ 0\\leq m\\leq4n\\}.\n\\end{equation}\nThe number of scanned pairs is\n\\begin{equation}\\label{cb_eq_scan_count}\n \\sum_{n=1}^N(4n+1)=2N^2+3N.\n\\end{equation}\nThe upper slope $4$ is part of the experiment's definition. The\nscan supplies no information about pairs above this boundary or with\n$n>N$.","ve integer $N$, the scanned domain is\n\\begin{equation}\\label{cb_eq_scan_domain}\n \\mathcal D_N=\\{(n,m):1\\leq n\\leq N,\\ 0\\leq m\\leq4n\\}.\n\\end{equation}\nThe number of scanned pairs is\n\\begin{equation}\\label{cb_eq_scan_count}\n \\sum_{n=1}^N(4n+1)=2N^2+3N.\n\\end{equation}\nThe upper slope $4$ is part of the experiment's definition. The\nscan supplies no information about pairs above this boundary or with\n$n>N$.","The computation first constructs $T_0,\\ldots,T_{5N}$ through the exact\ninteger recurrence. It then reduces these integers modulo $2^K$. The\nresulting array is the packet row with $m=0$. At each subsequent row,\nPascal evolution replaces adjacent entries by their sum modulo $2^K$.\nInduction from \\eqref{cb_eq_Pascal} proves that every entry read by the\nscan is the exact residue of its packet sum.","For a fixed row $m$, the allowed indices are\n\\[\n \\max(1,\\lceil m/4\\rceil)\\leq n\\leq N.\n\\]\nRows are needed through $m=4N$. In row $m$, the seed array is sufficient\nfor every $n\\leq5N-m$. At the extreme row $m=4N$ this bound is exactly\n$n\\leq N$. Thus no scanned entry needs an unconstructed seed term.\nThe algorithm evaluates additional row entries to support later rows.\nThose additions do not enlarge the declared set of scanned pairs.","The historical run takes $N=2000$ and $K=128$. The release reran the\nunchanged historical algorithm and compared its result with the recovered\noutput. The finite arithmetic results agree. In the domain\n$\\mathcal D_{2000}$ there are exactly $8{,}006{,}000$ pairs. Every observed\nresidue is nonzero, so the residue lemma applies at every pair. The exact\nmaximum valuation in this finite domain is\n\\begin{equation}\\label{cb_eq_finite_max}\n \\max_{(n,m)\\in\\mathcal D_{2000}}v_2(S_{n,m})=31.\n\\end{equation}\nThe recorded maximizing pair is $(1466,2113)$. The record reports the\nfirst maximum encountered in its scan order and makes no assertion that\nthis maximizing pair is unique.","cal run takes $N=2000$ and $K=128$. The release reran the\nunchanged historical algorithm and compared its result with the recovered\noutput. The finite arithmetic results agree. In the domain\n$\\mathcal D_{2000}$ there are exactly $8{,}006{,}000$ pairs. Every observed\nresidue is nonzero, so the residue lemma applies at every pair. The exact\nmaximum valuation in this finite domain is\n\\begin{equation}\\label{cb_eq_finite_max}\n \\max_{(n,m)\\in\\mathcal D_{2000}}v_2(S_{n,m})=31.\n\\end{equation}\nThe recorded maximizing pair is $(1466,2113)$. The record reports the\nfirst maximum encountered in its scan order and makes no assertion that\nthis maximizing pair is unique.","ecovered\noutput. The finite arithmetic results agree. In the domain\n$\\mathcal D_{2000}$ there are exactly $8{,}006{,}000$ pairs. Every observed\nresidue is nonzero, so the residue lemma applies at every pair. The exact\nmaximum valuation in this finite domain is\n\\begin{equation}\\label{cb_eq_finite_max}\n \\max_{(n,m)\\in\\mathcal D_{2000}}v_2(S_{n,m})=31.\n\\end{equation}\nThe recorded maximizing pair is $(1466,2113)$. The record reports the\nfirst maximum encountered in its scan order and makes no assertion that\nthis maximizing pair is unique.","There is a generous precision margin: the largest observed valuation is\n$31<128$, and there are zero unresolved residues. Consequently the finite\nmaximum is exact, rather than a maximum of truncated lower bounds. No\nstatistical confidence interval is involved. The essential qualification\nis the finite domain in \\eqref{cb_eq_finite_max}.","The historical smaller crosscheck reports the same maximum from direct\nsums through $N=100$. Matching only two summary maxima would be a weak\ncomparison: different entry values can share the same maximum. The release\ntherefore compares every pair in $\\mathcal D_{100}$.","For each of its $20{,}300$ pairs, a separate implementation evaluates\n\\[\n \\sum_{i=0}^m\\binom miT_{n+i}\n\\]\nas a full positive integer and compares both its residue modulo $2^{128}$\nand its exact valuation with the Pascal row entry. All pairs agree. There\nare no zero residues. The finite maximum is $20$, recorded first at\n$(52,6)$ in the direct comparison's order.","The direct summation and Pascal evolution use different operations on the\nsame seed data. This catches errors in the packet transform that agreement\nbetween two copies of the same row update could miss. A separately indexed\nrational recurrence also checks the seed construction through $n=500$.\nThese comparisons still share the intended recurrence formulas and require\nthe literature normalization contract. Their independence is operational,\nnot a claim that all common source risk has disappeared.","The pair $(1466,2113)$ was additionally evaluated as a full integer, along\nwith four neighboring pairs. The maximizing witness has bit length\n$26{,}743$ and exact valuation $31$. Its complete hexadecimal integer is\nincluded in the evidence bundle with a hash. The neighboring valuations\nare much smaller:\n\\begin{center}\n\\begin{tabular}{@{}rrr@{}}\n\\toprule\n$n$ & $m$ & $v_2(S_{n,m})$","\\midrule\n$1465$ & $2113$ & $12$","$1466$ & $2112$ & $13$","$1466$ & $2113$ & $31$","$1466$ & $2114$ & $11$","$1467$ & $2113$ & $11$","The release includes a diagnostic with $N=3$ and just $K=4$ bits of\nprecision. There are 27 pairs. Eight residues are zero, so those eight\nvaluations are unresolved at this precision. The historical scan records\na ceiling value of four and sets its exactness flag to false. Full integer\nevaluation shows that the actual finite maximum is six.","The largest valuation seen so far is small compared with the largest\nindex. That observation motivates a question such as\n\\[\n v_2(S_{n,m})=o(n+m)\n\\]\nalong a specified family of packet indices. It does not prove the answer.\nA universal statement needs quantified control beyond the scanned domain,\nwith the family of $m$ values and the meaning of uniformity stated\nexplicitly. A maximum found on a bounded grid cannot rule out later spikes\nof a different scale.","Even a proved estimate for $v_2(S_{n,m})$ would address only one prime and\none coefficient. The common content of the full integer pair involves both\ncoefficients, their clearing factor, and odd prime divisibility. The scan\ntherefore supplies a precise finite arithmetic fact and a well defined\nresearch question. It is not, by itself, the missing denominator theorem\nor an irrationality proof.","Fix integers $n,m\\geq0$, and write\n\\begin{equation}\n U=\\sum_{i=0}^{m}\\binom miT_{n+i},\\qquad\n V=\\sum_{i=0}^{m}\\binom mi16^{n+i}v_{n+i}.\n \\label{eq:primitive-UV}\n\\end{equation}\nThese are fixed-pair aliases $U=S_{n,m}$ and $V=H_{n,m}$ for the packets defined earlier, not the integral coefficients of a different source normalization. Here $T_k=16^ku_k$ is a positive integer. Consequently $U$ is a positive integer. Write $V=N/D$ in lowest terms, with $D>0$ and $\\gcd(N,D)=1$. In the exceptional case $V=0$, the same convention sets $N=0,D=1$. Set\n\\begin{equation}\n \\ell=2(n+m),\\qquad L_\\ell=\\operatorname{lcm}(1,\\ldots,\\ell),\n \\qquad Q=8L_\\ell^2,\n \\label{eq:Q-clearance}\n\\end{equation}\nwith $L_0=1$.","$, and write\n\\begin{equation}\n U=\\sum_{i=0}^{m}\\binom miT_{n+i},\\qquad\n V=\\sum_{i=0}^{m}\\binom mi16^{n+i}v_{n+i}.\n \\label{eq:primitive-UV}\n\\end{equation}\nThese are fixed-pair aliases $U=S_{n,m}$ and $V=H_{n,m}$ for the packets defined earlier, not the integral coefficients of a different source normalization. Here $T_k=16^ku_k$ is a positive integer. Consequently $U$ is a positive integer. Write $V=N/D$ in lowest terms, with $D>0$ and $\\gcd(N,D)=1$. In the exceptional case $V=0$, the same convention sets $N=0,D=1$. Set\n\\begin{equation}\n \\ell=2(n+m),\\qquad L_\\ell=\\operatorname{lcm}(1,\\ldots,\\ell),\n \\qquad Q=8L_\\ell^2,\n \\label{eq:Q-clearance}\n\\end{equation}\nwith $L_0=1$.","U=\\sum_{i=0}^{m}\\binom miT_{n+i},\\qquad\n V=\\sum_{i=0}^{m}\\binom mi16^{n+i}v_{n+i}.\n \\label{eq:primitive-UV}\n\\end{equation}\nThese are fixed-pair aliases $U=S_{n,m}$ and $V=H_{n,m}$ for the packets defined earlier, not the integral coefficients of a different source normalization. Here $T_k=16^ku_k$ is a positive integer. Consequently $U$ is a positive integer. Write $V=N/D$ in lowest terms, with $D>0$ and $\\gcd(N,D)=1$. In the exceptional case $V=0$, the same convention sets $N=0,D=1$. Set\n\\begin{equation}\n \\ell=2(n+m),\\qquad L_\\ell=\\operatorname{lcm}(1,\\ldots,\\ell),\n \\qquad Q=8L_\\ell^2,\n \\label{eq:Q-clearance}\n\\end{equation}\nwith $L_0=1$.","=0}^{m}\\binom miT_{n+i},\\qquad\n V=\\sum_{i=0}^{m}\\binom mi16^{n+i}v_{n+i}.\n \\label{eq:primitive-UV}\n\\end{equation}\nThese are fixed-pair aliases $U=S_{n,m}$ and $V=H_{n,m}$ for the packets defined earlier, not the integral coefficients of a different source normalization. Here $T_k=16^ku_k$ is a positive integer. Consequently $U$ is a positive integer. Write $V=N/D$ in lowest terms, with $D>0$ and $\\gcd(N,D)=1$. In the exceptional case $V=0$, the same convention sets $N=0,D=1$. Set\n\\begin{equation}\n \\ell=2(n+m),\\qquad L_\\ell=\\operatorname{lcm}(1,\\ldots,\\ell),\n \\qquad Q=8L_\\ell^2,\n \\label{eq:Q-clearance}\n\\end{equation}\nwith $L_0=1$.","{i=0}^{m}\\binom mi16^{n+i}v_{n+i}.\n \\label{eq:primitive-UV}\n\\end{equation}\nThese are fixed-pair aliases $U=S_{n,m}$ and $V=H_{n,m}$ for the packets defined earlier, not the integral coefficients of a different source normalization. Here $T_k=16^ku_k$ is a positive integer. Consequently $U$ is a positive integer. Write $V=N/D$ in lowest terms, with $D>0$ and $\\gcd(N,D)=1$. In the exceptional case $V=0$, the same convention sets $N=0,D=1$. Set\n\\begin{equation}\n \\ell=2(n+m),\\qquad L_\\ell=\\operatorname{lcm}(1,\\ldots,\\ell),\n \\qquad Q=8L_\\ell^2,\n \\label{eq:Q-clearance}\n\\end{equation}\nwith $L_0=1$.","}.\n \\label{eq:primitive-UV}\n\\end{equation}\nThese are fixed-pair aliases $U=S_{n,m}$ and $V=H_{n,m}$ for the packets defined earlier, not the integral coefficients of a different source normalization. Here $T_k=16^ku_k$ is a positive integer. Consequently $U$ is a positive integer. Write $V=N/D$ in lowest terms, with $D>0$ and $\\gcd(N,D)=1$. In the exceptional case $V=0$, the same convention sets $N=0,D=1$. Set\n\\begin{equation}\n \\ell=2(n+m),\\qquad L_\\ell=\\operatorname{lcm}(1,\\ldots,\\ell),\n \\qquad Q=8L_\\ell^2,\n \\label{eq:Q-clearance}\n\\end{equation}\nwith $L_0=1$.","nd{equation}\nThese are fixed-pair aliases $U=S_{n,m}$ and $V=H_{n,m}$ for the packets defined earlier, not the integral coefficients of a different source normalization. Here $T_k=16^ku_k$ is a positive integer. Consequently $U$ is a positive integer. Write $V=N/D$ in lowest terms, with $D>0$ and $\\gcd(N,D)=1$. In the exceptional case $V=0$, the same convention sets $N=0,D=1$. Set\n\\begin{equation}\n \\ell=2(n+m),\\qquad L_\\ell=\\operatorname{lcm}(1,\\ldots,\\ell),\n \\qquad Q=8L_\\ell^2,\n \\label{eq:Q-clearance}\n\\end{equation}\nwith $L_0=1$.","nts of a different source normalization. Here $T_k=16^ku_k$ is a positive integer. Consequently $U$ is a positive integer. Write $V=N/D$ in lowest terms, with $D>0$ and $\\gcd(N,D)=1$. In the exceptional case $V=0$, the same convention sets $N=0,D=1$. Set\n\\begin{equation}\n \\ell=2(n+m),\\qquad L_\\ell=\\operatorname{lcm}(1,\\ldots,\\ell),\n \\qquad Q=8L_\\ell^2,\n \\label{eq:Q-clearance}\n\\end{equation}\nwith $L_0=1$.","The source arithmetic ensures $QV\\in\\Z$. Indeed, the strong coefficient theorem gives $4\\cdot16^kL_{2k}^2v_k\\in\\Z$, and $L_{2k}\\mid L_\\ell$ for $k\\leq n+m$. The weaker factor $8$ in $Q$ therefore clears every summand in \\eqref{eq:primitive-UV}. Since $V=N/D$ is reduced, this proves $D\\mid Q$. The factor $Q$ is a convenient universal choice. The factor $D$ is the actual least denominator of this particular rational packet.","The historical cleared coefficients are\n\\begin{equation}\n A=(-1)^nQU,\\qquad B=(-1)^nQV,\n \\qquad g=\\gcd(|A|,|B|).\n \\label{eq:cleared-pair}\n\\end{equation}\nThe sign changes neither the gcd nor the approximation $B/A=V/U$. Finally put\n\\begin{equation}\n h=\\gcd(DU,N).\n \\label{eq:minimal-content}\n\\end{equation}\nThis is the content of the minimally cleared pair. It must not be confused with $g$. The three quantities $Q$, $D$, and $h$ record, respectively, the chosen clearing factor, the actual packet denominator, and the common content left after minimal clearing.","l cleared coefficients are\n\\begin{equation}\n A=(-1)^nQU,\\qquad B=(-1)^nQV,\n \\qquad g=\\gcd(|A|,|B|).\n \\label{eq:cleared-pair}\n\\end{equation}\nThe sign changes neither the gcd nor the approximation $B/A=V/U$. Finally put\n\\begin{equation}\n h=\\gcd(DU,N).\n \\label{eq:minimal-content}\n\\end{equation}\nThis is the content of the minimally cleared pair. It must not be confused with $g$. The three quantities $Q$, $D$, and $h$ record, respectively, the chosen clearing factor, the actual packet denominator, and the common content left after minimal clearing.","ared coefficients are\n\\begin{equation}\n A=(-1)^nQU,\\qquad B=(-1)^nQV,\n \\qquad g=\\gcd(|A|,|B|).\n \\label{eq:cleared-pair}\n\\end{equation}\nThe sign changes neither the gcd nor the approximation $B/A=V/U$. Finally put\n\\begin{equation}\n h=\\gcd(DU,N).\n \\label{eq:minimal-content}\n\\end{equation}\nThis is the content of the minimally cleared pair. It must not be confused with $g$. The three quantities $Q$, $D$, and $h$ record, respectively, the chosen clearing factor, the actual packet denominator, and the common content left after minimal clearing.","ficients are\n\\begin{equation}\n A=(-1)^nQU,\\qquad B=(-1)^nQV,\n \\qquad g=\\gcd(|A|,|B|).\n \\label{eq:cleared-pair}\n\\end{equation}\nThe sign changes neither the gcd nor the approximation $B/A=V/U$. Finally put\n\\begin{equation}\n h=\\gcd(DU,N).\n \\label{eq:minimal-content}\n\\end{equation}\nThis is the content of the minimally cleared pair. It must not be confused with $g$. The three quantities $Q$, $D$, and $h$ record, respectively, the chosen clearing factor, the actual packet denominator, and the common content left after minimal clearing.","\\begin{proposition}[Redundant clearance and primitive content]\n\\label{prop:content-decomposition}\nFor the coefficients above,\n\\begin{equation}\n h=\\gcd(U,N),\\qquad\n g=\\frac QD\\,h.\n \\label{eq:gcd-decomposition}\n\\end{equation}\nThe reduced rational approximation and primitive coefficients are\n\\begin{equation}\n \\frac VU=\\frac{N/h}{DU/h},\\qquad\n \\frac Ag=(-1)^n\\frac{DU}{h},\\qquad\n \\frac Bg=(-1)^n\\frac Nh.\n \\label{eq:primitive-pair}\n\\end{equation}\nIn particular the positive reduced denominator is $q=DU/h$.\n\\end{proposition}","}\nFor the coefficients above,\n\\begin{equation}\n h=\\gcd(U,N),\\qquad\n g=\\frac QD\\,h.\n \\label{eq:gcd-decomposition}\n\\end{equation}\nThe reduced rational approximation and primitive coefficients are\n\\begin{equation}\n \\frac VU=\\frac{N/h}{DU/h},\\qquad\n \\frac Ag=(-1)^n\\frac{DU}{h},\\qquad\n \\frac Bg=(-1)^n\\frac Nh.\n \\label{eq:primitive-pair}\n\\end{equation}\nIn particular the positive reduced denominator is $q=DU/h$.\n\\end{proposition}","\\begin{proof}\nIf $N=0$, the reduced convention gives $D=1$. Since $U>0$, one has $h=U$, $g=QU$, and the unsigned primitive pair is $(1,0)$. All the stated identities follow directly in this case. Assume now $N\\ne0$. Since $\\gcd(D,N)=1$, multiplication of $U$ by $D$ adds no prime factor that can divide $N$. Prime by prime, $\\min\\{v_p(DU),v_p(|N|)\\}=\\min\\{v_p(U),v_p(|N|)\\}$. This proves the first equality. Next write $Q=cD$, where $c$ is a positive integer. The unsigned cleared pair is $(cDU,cN)$, so its gcd is $c\\gcd(DU,N)=ch$. Dividing both coefficients by this gcd proves the remaining identities. The pair $(DU/h,N/h)$ is coprime by the definition of $h$, which proves the claim about the reduced denominator.\n\\en","\\begin{proof}\nIf $N=0$, the reduced convention gives $D=1$. Since $U>0$, one has $h=U$, $g=QU$, and the unsigned primitive pair is $(1,0)$. All the stated identities follow directly in this case. Assume now $N\\ne0$. Since $\\gcd(D,N)=1$, multiplication of $U$ by $D$ adds no prime factor that can divide $N$. Prime by prime, $\\min\\{v_p(DU),v_p(|N|)\\}=\\min\\{v_p(U),v_p(|N|)\\}$. This proves the first equality. Next write $Q=cD$, where $c$ is a positive integer. The unsigned cleared pair is $(cDU,cN)$, so its gcd is $c\\gcd(DU,N)=ch$. Dividing both coefficients by this gcd proves the remaining identities. The pair $(DU/h,N/h)$ is coprime by the definition of $h$, which proves the claim about the reduced denominator.\n\\end{proof}","0$, the reduced convention gives $D=1$. Since $U>0$, one has $h=U$, $g=QU$, and the unsigned primitive pair is $(1,0)$. All the stated identities follow directly in this case. Assume now $N\\ne0$. Since $\\gcd(D,N)=1$, multiplication of $U$ by $D$ adds no prime factor that can divide $N$. Prime by prime, $\\min\\{v_p(DU),v_p(|N|)\\}=\\min\\{v_p(U),v_p(|N|)\\}$. This proves the first equality. Next write $Q=cD$, where $c$ is a positive integer. The unsigned cleared pair is $(cDU,cN)$, so its gcd is $c\\gcd(DU,N)=ch$. Dividing both coefficients by this gcd proves the remaining identities. The pair $(DU/h,N/h)$ is coprime by the definition of $h$, which proves the claim about the reduced denominator.\n\\end{proof}","d convention gives $D=1$. Since $U>0$, one has $h=U$, $g=QU$, and the unsigned primitive pair is $(1,0)$. All the stated identities follow directly in this case. Assume now $N\\ne0$. Since $\\gcd(D,N)=1$, multiplication of $U$ by $D$ adds no prime factor that can divide $N$. Prime by prime, $\\min\\{v_p(DU),v_p(|N|)\\}=\\min\\{v_p(U),v_p(|N|)\\}$. This proves the first equality. Next write $Q=cD$, where $c$ is a positive integer. The unsigned cleared pair is $(cDU,cN)$, so its gcd is $c\\gcd(DU,N)=ch$. Dividing both coefficients by this gcd proves the remaining identities. The pair $(DU/h,N/h)$ is coprime by the definition of $h$, which proves the claim about the reduced denominator.\n\\end{proof}","=QU$, and the unsigned primitive pair is $(1,0)$. All the stated identities follow directly in this case. Assume now $N\\ne0$. Since $\\gcd(D,N)=1$, multiplication of $U$ by $D$ adds no prime factor that can divide $N$. Prime by prime, $\\min\\{v_p(DU),v_p(|N|)\\}=\\min\\{v_p(U),v_p(|N|)\\}$. This proves the first equality. Next write $Q=cD$, where $c$ is a positive integer. The unsigned cleared pair is $(cDU,cN)$, so its gcd is $c\\gcd(DU,N)=ch$. Dividing both coefficients by this gcd proves the remaining identities. The pair $(DU/h,N/h)$ is coprime by the definition of $h$, which proves the claim about the reduced denominator.\n\\end{proof}","mitive pair is $(1,0)$. All the stated identities follow directly in this case. Assume now $N\\ne0$. Since $\\gcd(D,N)=1$, multiplication of $U$ by $D$ adds no prime factor that can divide $N$. Prime by prime, $\\min\\{v_p(DU),v_p(|N|)\\}=\\min\\{v_p(U),v_p(|N|)\\}$. This proves the first equality. Next write $Q=cD$, where $c$ is a positive integer. The unsigned cleared pair is $(cDU,cN)$, so its gcd is $c\\gcd(DU,N)=ch$. Dividing both coefficients by this gcd proves the remaining identities. The pair $(DU/h,N/h)$ is coprime by the definition of $h$, which proves the claim about the reduced denominator.\n\\end{proof}","Since $\\gcd(D,N)=1$, multiplication of $U$ by $D$ adds no prime factor that can divide $N$. Prime by prime, $\\min\\{v_p(DU),v_p(|N|)\\}=\\min\\{v_p(U),v_p(|N|)\\}$. This proves the first equality. Next write $Q=cD$, where $c$ is a positive integer. The unsigned cleared pair is $(cDU,cN)$, so its gcd is $c\\gcd(DU,N)=ch$. Dividing both coefficients by this gcd proves the remaining identities. The pair $(DU/h,N/h)$ is coprime by the definition of $h$, which proves the claim about the reduced denominator.\n\\end{proof}","y $D$ adds no prime factor that can divide $N$. Prime by prime, $\\min\\{v_p(DU),v_p(|N|)\\}=\\min\\{v_p(U),v_p(|N|)\\}$. This proves the first equality. Next write $Q=cD$, where $c$ is a positive integer. The unsigned cleared pair is $(cDU,cN)$, so its gcd is $c\\gcd(DU,N)=ch$. Dividing both coefficients by this gcd proves the remaining identities. The pair $(DU/h,N/h)$ is coprime by the definition of $h$, which proves the claim about the reduced denominator.\n\\end{proof}","The decomposition contains no asymptotic assumption. It holds at each individual pair $(n,m)$. Its usefulness is that it assigns every common factor to a precise source. A large value of $g$ can be caused by a large value of $Q/D$, even when $h$ is small. Such a factor reflects an unnecessarily generous initial clearing choice. It is not additional content in the minimally cleared pair.","For the historical normalization of the integral filter,\n\\begin{equation}\n \\widetilde J_{n,m}\n =8(-1)^n16^{-n}(UG-V).\n \\label{eq:primitive-filter-normalization}\n\\end{equation}\nThus the primitive integer linear form is\n\\begin{align}\n \\Lambda_{n,m}\n &=\\frac AgG-\\frac Bg\n =(-1)^n\\frac D h(UG-V)\\notag","For the historical normalization of the integral filter,\n\\begin{equation}\n \\widetilde J_{n,m}\n =8(-1)^n16^{-n}(UG-V).\n \\label{eq:primitive-filter-normalization}\n\\end{equation}\nThus the primitive integer linear form is\n\\begin{align}\n \\Lambda_{n,m}\n &=\\frac AgG-\\frac Bg\n =(-1)^n\\frac D h(UG-V)\\notag\\\\\n &=\\frac{16^nL_\\ell^2}{g}\\,\\widetilde J_{n,m}.\n \\label{eq:primitive-error}\n\\end{align}\nEvery normalization factor in this formula matters. In particular, the $8$ in $Q$ cancels the $8$ in \\eqref{eq:primitive-filter-normalization}. The right side is a real expression for a linear form with integer coefficients. Its absolute value is independent of whether the signs are carried in $A,B$ or in the filter.","&=\\frac{16^nL_\\ell^2}{g}\\,\\widetilde J_{n,m}.\n \\label{eq:primitive-error}\n\\end{align}\nEvery normalization factor in this formula matters. In particular, the $8$ in $Q$ cancels the $8$ in \\eqref{eq:primitive-filter-normalization}. The right side is a real expression for a linear form with integer coefficients. Its absolute value is independent of whether the signs are carried in $A,B$ or in the filter.","\\begin{proposition}[Invariance under additional clearing]\n\\label{prop:scale-invariance}\nIf an integer pair $(A,B)$ is replaced by $(cA,cB)$ for an integer $c>0$, then its primitive pair and primitive linear form remain unchanged. More generally, any integer clearing factor $Q'$ divisible by $D$ gives the same unsigned primitive pair $(DU/h,N/h)$.\n\\end{proposition}","\\begin{proof}\nThe equality $\\gcd(cA,cB)=c\\gcd(A,B)$ proves the first statement. For the second, apply Proposition~\\ref{prop:content-decomposition} with $Q'$ in place of $Q$. The factor $Q'/D$ cancels exactly on division by the new gcd.\n\\end{proof}","This observation blocks a tempting but invalid arithmetic argument. One can make the cleared gcd arbitrarily large by multiplying both coefficients by a large integer. Doing so also enlarges the error by precisely the same factor. After reduction there is no change. Any proposed gain must be measured in the primitive form \\eqref{eq:primitive-error}, or equivalently in the reduced denominator $q=DU/h$.","An elementary example makes the cancellation visible. Take $U=12$ and $V=10/9$. The minimally cleared pair is $(108,10)$ and has gcd $2$. Its primitive pair is $(54,5)$. Clearing instead with $Q=90$ gives $(1080,100)$ and gcd $20$. The apparent tenfold increase in gcd is exactly the tenfold excess in the clearing factor. Both pairs reduce to $(54,5)$ and both represent the approximation $5/54$.","There is still a legitimate role for a redundant clearing factor. It can make a universal theorem easy to state and a program easy to reproduce. The correction is accounting, not a ban on convenient denominators. A calculation with $Q$ is valid when the reduction by $g$ is kept in the same formula. A calculation that counts $g$ as a gain while ignoring the corresponding size of $Q$ is not valid.","The historical script recorded five pairs near the slope used in Eskandari's filter. The fresh exact run reproduces their gcds and factorizations. An independent rational calculation also extracts $D$, $Q/D$, and $h$. The resulting decomposition is as follows.","\\begin{table}[htbp]\n\\centering\\small\n\\begin{tabular}{@{}rrrrr@{}}\n\\toprule\n$(n,m)$ & $Q/D$ & $h$ & $g$ & $v_2(g)$","\\midrule\n$(20,66)$ & $1\\,703\\,936$ & $8$ & $13\\,631\\,488$ & $20$","$(30,100)$ & $129\\,499\\,136$ & $4$ & $517\\,996\\,544$ & $21$","$(50,166)$ & $2\\,779\\,250\\,688$ & $8$ & $22\\,234\\,005\\,504$ & $22$","$(70,232)$ & $5\\,257\\,560\\,064$ & $64$ & $336\\,483\\,844\\,096$ & $27$","$(100,332)$ & $1\\,084\\,227\\,584$ & $8$ & $8\\,673\\,820\\,672$ & $24$","The factorizations of the cleared gcds are\n\\begin{align*}\n13\\,631\\,488&=2^{20}\\cdot13,\\\\\n517\\,996\\,544&=2^{21}\\cdot13\\cdot19,\\\\\n22\\,234\\,005\\,504&=2^{22}\\cdot3^2\\cdot19\\cdot31,\\\\\n336\\,483\\,844\\,096&=2^{27}\\cdot23\\cdot109,\\\\\n8\\,673\\,820\\,672&=2^{24}\\cdot11\\cdot47.\n\\end{align*}\nFor these rows the minimally cleared gcd has only the powers of $2$ displayed in $h$. All odd factors in the listed cleared gcds belong to $Q/D$. That conclusion is specific to these five pairs. It does not establish that odd primitive content is always absent.","8\\,673\\,820\\,672&=2^{24}\\cdot11\\cdot47.\n\\end{align*}\nFor these rows the minimally cleared gcd has only the powers of $2$ displayed in $h$. All odd factors in the listed cleared gcds belong to $Q/D$. That conclusion is specific to these five pairs. It does not establish that odd primitive content is always absent.","The first row illustrates the exact check without printing its very large numerator and denominator. The independently reduced rational packet has denominator\n\\begin{equation*}\n D=\\operatorname{den}\\left(\\sum_{i=0}^{66}\\binom{66}{i}16^{20+i}v_{20+i}\\right).\n\\end{equation*}\nIts recorded quotient $Q/D$ is $1\\,703\\,936$, and the minimal pair has gcd $8$. Multiplication gives $1\\,703\\,936\\cdot8=13\\,631\\,488$, exactly the historical value. The complete decimal integers $U,N,D,Q$ are retained in the diagnostic JSON. They are inputs to an exact equality check, rather than values inferred from rounded logarithms.","rge numerator and denominator. The independently reduced rational packet has denominator\n\\begin{equation*}\n D=\\operatorname{den}\\left(\\sum_{i=0}^{66}\\binom{66}{i}16^{20+i}v_{20+i}\\right).\n\\end{equation*}\nIts recorded quotient $Q/D$ is $1\\,703\\,936$, and the minimal pair has gcd $8$. Multiplication gives $1\\,703\\,936\\cdot8=13\\,631\\,488$, exactly the historical value. The complete decimal integers $U,N,D,Q$ are retained in the diagnostic JSON. They are inputs to an exact equality check, rather than values inferred from rounded logarithms.","The distinction also affects the dyadic interpretation. At $(20,66)$, $v_2(h)=3$, while $v_2(g)=20$. The remaining seventeen powers of $2$ occur in $Q/D$. At $(70,232)$, the corresponding split is $6+21=27$. A plotted valuation of the cleared gcd does not by itself measure the dyadic content of the packet.","The historical display divides $\\log g$ by $n+m$. Its five values decrease from approximately $0.19102$ to $0.05297$. These are useful diagnostics of the sampled pairs. They do not prove $\\log g=o(n+m)$, a monotonicity theorem, or a bound at every subsequent pair. Even the sampling rule matters. A result along one sequence $m=m(n)$ is weaker than a bound uniform over a two dimensional range of indices.","The exact decomposition yields a more informative identity for rate analysis:\n\\begin{equation}\n \\log g=\\log Q-\\log D+\\log h.\n \\label{eq:log-content-split}\n\\end{equation}\nSuppose a future proof supplies limits for the three terms on the right along a chosen family. Then \\eqref{eq:log-content-split} determines the cleared content rate. At present, the five rows supply values, not those limits. The identity separates the questions that a proof would need to answer. How much smaller is the true packet denominator than the universal denominator? How large is primitive common content? Are the bounds valid along every index in the chosen family?","There is a further one-sided relation to the binomial valuation scan. Because $h\\mid U$, one always has $v_2(h)\\leq v_2(U)$. If a universal estimate $v_2(U)=o(n+m)$ were proved, it would force the dyadic part of $h$ to be subexponential. It would not control odd factors of $h$, nor the factor $Q/D$. Conversely, a large cleared dyadic gcd can coexist with small $v_2(U)$ because most of its dyadic factor may come from redundant clearance.","An irrationality argument needs an infinite family, not simply many successful finite values. For the present filter, a convenient family chooses an even integer $m=m(n)$ with $m/n\\to\\alpha\\geq0$. Evenness supplies nonvanishing by a written integral argument. The slope supplies a definite analytic scale.","Write\n\\begin{equation}\n \\phi=\\frac{1+\\sqrt5}{2},\\qquad t=\\phi^{-5},\n \\qquad R=16\\phi^5=88+40\\sqrt5.\n \\label{eq:asymptotic-constants}\n\\end{equation}\nThe unfiltered integral has the representation\n\\begin{equation}\n J_k=\\int_{[0,1]^2}\\psi(x,y)^k\\,\\dd\\mu(x,y),\n \\quad\n \\psi=\\frac{xy(1-x)(1-y)}{1-xy},\n \\quad\n \\dd\\mu=\\frac{\\dd x\\,\\dd y}{\\sqrt{x(1-y)}(1-xy)}.\n \\label{eq:bottleneck-measure}\n\\end{equation}\nThe measure has finite total mass $J_0=8G$. The factor $1-xy$ in its denominator lies outside the square root. Moving it inside changes the integral and invalidates the coefficient identification.","1+\\sqrt5}{2},\\qquad t=\\phi^{-5},\n \\qquad R=16\\phi^5=88+40\\sqrt5.\n \\label{eq:asymptotic-constants}\n\\end{equation}\nThe unfiltered integral has the representation\n\\begin{equation}\n J_k=\\int_{[0,1]^2}\\psi(x,y)^k\\,\\dd\\mu(x,y),\n \\quad\n \\psi=\\frac{xy(1-x)(1-y)}{1-xy},\n \\quad\n \\dd\\mu=\\frac{\\dd x\\,\\dd y}{\\sqrt{x(1-y)}(1-xy)}.\n \\label{eq:bottleneck-measure}\n\\end{equation}\nThe measure has finite total mass $J_0=8G$. The factor $1-xy$ in its denominator lies outside the square root. Moving it inside changes the integral and invalidates the coefficient identification.","=\\phi^{-5},\n \\qquad R=16\\phi^5=88+40\\sqrt5.\n \\label{eq:asymptotic-constants}\n\\end{equation}\nThe unfiltered integral has the representation\n\\begin{equation}\n J_k=\\int_{[0,1]^2}\\psi(x,y)^k\\,\\dd\\mu(x,y),\n \\quad\n \\psi=\\frac{xy(1-x)(1-y)}{1-xy},\n \\quad\n \\dd\\mu=\\frac{\\dd x\\,\\dd y}{\\sqrt{x(1-y)}(1-xy)}.\n \\label{eq:bottleneck-measure}\n\\end{equation}\nThe measure has finite total mass $J_0=8G$. The factor $1-xy$ in its denominator lies outside the square root. Moving it inside changes the integral and invalidates the coefficient identification.","Expanding the filter gives\n\\begin{equation}\n \\widetilde J_{n,m}\n =\\int_{[0,1]^2}\\psi^n(1-16\\psi)^m\\,\\dd\\mu.\n \\label{eq:bottleneck-filter}\n\\end{equation}\nThe range of $\\psi$ is contained in $[0,t]$ and reaches its positive maximum $t$. In particular $t>1/16$. The filter factor therefore changes sign within the integration region. When $m$ is even, the integrand is nonnegative. It is strictly positive on an open subset of the interior on which $\\psi>0$ and $\\psi\\ne1/16$. The density of $\\mu$ is positive there. Hence\n\\begin{equation}\n \\widetilde J_{n,m}>0\\qquad(m\\ \\text{even}).\n \\label{eq:even-filter-positive}\n\\end{equation}\nThis is a universal nonvanishing result. It uses neither a decimal value of $G$ nor a finite sign scan. For odd $m$, the same proof does not apply.","idetilde J_{n,m}\n =\\int_{[0,1]^2}\\psi^n(1-16\\psi)^m\\,\\dd\\mu.\n \\label{eq:bottleneck-filter}\n\\end{equation}\nThe range of $\\psi$ is contained in $[0,t]$ and reaches its positive maximum $t$. In particular $t>1/16$. The filter factor therefore changes sign within the integration region. When $m$ is even, the integrand is nonnegative. It is strictly positive on an open subset of the interior on which $\\psi>0$ and $\\psi\\ne1/16$. The density of $\\mu$ is positive there. Hence\n\\begin{equation}\n \\widetilde J_{n,m}>0\\qquad(m\\ \\text{even}).\n \\label{eq:even-filter-positive}\n\\end{equation}\nThis is a universal nonvanishing result. It uses neither a decimal value of $G$ nor a finite sign scan. For odd $m$, the same proof does not apply.",",m}\n =\\int_{[0,1]^2}\\psi^n(1-16\\psi)^m\\,\\dd\\mu.\n \\label{eq:bottleneck-filter}\n\\end{equation}\nThe range of $\\psi$ is contained in $[0,t]$ and reaches its positive maximum $t$. In particular $t>1/16$. The filter factor therefore changes sign within the integration region. When $m$ is even, the integrand is nonnegative. It is strictly positive on an open subset of the interior on which $\\psi>0$ and $\\psi\\ne1/16$. The density of $\\mu$ is positive there. Hence\n\\begin{equation}\n \\widetilde J_{n,m}>0\\qquad(m\\ \\text{even}).\n \\label{eq:even-filter-positive}\n\\end{equation}\nThis is a universal nonvanishing result. It uses neither a decimal value of $G$ nor a finite sign scan. For odd $m$, the same proof does not apply.","6\\psi)^m\\,\\dd\\mu.\n \\label{eq:bottleneck-filter}\n\\end{equation}\nThe range of $\\psi$ is contained in $[0,t]$ and reaches its positive maximum $t$. In particular $t>1/16$. The filter factor therefore changes sign within the integration region. When $m$ is even, the integrand is nonnegative. It is strictly positive on an open subset of the interior on which $\\psi>0$ and $\\psi\\ne1/16$. The density of $\\mu$ is positive there. Hence\n\\begin{equation}\n \\widetilde J_{n,m}>0\\qquad(m\\ \\text{even}).\n \\label{eq:even-filter-positive}\n\\end{equation}\nThis is a universal nonvanishing result. It uses neither a decimal value of $G$ nor a finite sign scan. For odd $m$, the same proof does not apply.","}\n\\end{equation}\nThe range of $\\psi$ is contained in $[0,t]$ and reaches its positive maximum $t$. In particular $t>1/16$. The filter factor therefore changes sign within the integration region. When $m$ is even, the integrand is nonnegative. It is strictly positive on an open subset of the interior on which $\\psi>0$ and $\\psi\\ne1/16$. The density of $\\mu$ is positive there. Hence\n\\begin{equation}\n \\widetilde J_{n,m}>0\\qquad(m\\ \\text{even}).\n \\label{eq:even-filter-positive}\n\\end{equation}\nThis is a universal nonvanishing result. It uses neither a decimal value of $G$ nor a finite sign scan. For odd $m$, the same proof does not apply.","the integration region. When $m$ is even, the integrand is nonnegative. It is strictly positive on an open subset of the interior on which $\\psi>0$ and $\\psi\\ne1/16$. The density of $\\mu$ is positive there. Hence\n\\begin{equation}\n \\widetilde J_{n,m}>0\\qquad(m\\ \\text{even}).\n \\label{eq:even-filter-positive}\n\\end{equation}\nThis is a universal nonvanishing result. It uses neither a decimal value of $G$ nor a finite sign scan. For odd $m$, the same proof does not apply.","even, the integrand is nonnegative. It is strictly positive on an open subset of the interior on which $\\psi>0$ and $\\psi\\ne1/16$. The density of $\\mu$ is positive there. Hence\n\\begin{equation}\n \\widetilde J_{n,m}>0\\qquad(m\\ \\text{even}).\n \\label{eq:even-filter-positive}\n\\end{equation}\nThis is a universal nonvanishing result. It uses neither a decimal value of $G$ nor a finite sign scan. For odd $m$, the same proof does not apply.","\\section{Where the two branches of $M$ come from}","For $\\alpha\\geq0$, define\n\\begin{equation}\n M(\\alpha)=\\max\\left\\{\n \\frac{1}{16(1+\\alpha)}\n \\left(\\frac\\alpha{1+\\alpha}\\right)^\\alpha,\n \\ t(16t-1)^\\alpha\n \\right\\}.\n \\label{eq:M-alpha}\n\\end{equation}\nAt $\\alpha=0$ the first expression is assigned its continuous value $1/16$. This avoids treating the displayed $0^0$ as an algebraic input. The formula is Eskandari's source bound, with the recurrence normalization translated explicitly in LC06 and LC07~\\cite{Eskandari2026}.","It can also be understood directly from the filter. On $0\\leq s\\leq1/16$, maximize $s(1-16s)^\\alpha$. For $\\alpha>0$, logarithmic differentiation gives\n\\begin{equation*}\n \\frac1s-\\frac{16\\alpha}{1-16s}=0,\n \\qquad s=\\frac{1}{16(1+\\alpha)}.\n\\end{equation*}\nSubstituting this point gives the first branch in \\eqref{eq:M-alpha}. The endpoints give zero for $\\alpha>0$, so this critical point is the maximum. At $\\alpha=0$, the maximum on the interval is its right endpoint $1/16$.","lso be understood directly from the filter. On $0\\leq s\\leq1/16$, maximize $s(1-16s)^\\alpha$. For $\\alpha>0$, logarithmic differentiation gives\n\\begin{equation*}\n \\frac1s-\\frac{16\\alpha}{1-16s}=0,\n \\qquad s=\\frac{1}{16(1+\\alpha)}.\n\\end{equation*}\nSubstituting this point gives the first branch in \\eqref{eq:M-alpha}. The endpoints give zero for $\\alpha>0$, so this critical point is the maximum. At $\\alpha=0$, the maximum on the interval is its right endpoint $1/16$.","6$, maximize $s(1-16s)^\\alpha$. For $\\alpha>0$, logarithmic differentiation gives\n\\begin{equation*}\n \\frac1s-\\frac{16\\alpha}{1-16s}=0,\n \\qquad s=\\frac{1}{16(1+\\alpha)}.\n\\end{equation*}\nSubstituting this point gives the first branch in \\eqref{eq:M-alpha}. The endpoints give zero for $\\alpha>0$, so this critical point is the maximum. At $\\alpha=0$, the maximum on the interval is its right endpoint $1/16$.","On $1/16\\leq s\\leq t$, maximize $s(16s-1)^\\alpha$. For $\\alpha>0$ its logarithmic derivative is\n\\begin{equation*}\n \\frac1s+\\frac{16\\alpha}{16s-1}>0.\n\\end{equation*}\nIts maximum is therefore $t(16t-1)^\\alpha$. The same conclusion holds at $\\alpha=0$ because the function is then $s$. These two regions explain both branches and the need to retain their maximum.","Let $\\alpha_n=m(n)/n$. The finite measure in \\eqref{eq:bottleneck-filter} gives\n\\begin{equation}\n |\\widetilde J_{n,m(n)}|\n \\leq 8G\\,M(\\alpha_n)^n.\n \\label{eq:actual-M-bound}\n\\end{equation}\nContinuity of the two branches, including their stated extension at zero, then proves\n\\begin{equation}\n \\limsup_{n\\to\\infty}\n |\\widetilde J_{n,m(n)}|^{1/n}\\leq M(\\alpha).\n \\label{eq:M-limsup}\n\\end{equation}\nThis is an upper bound. An equality for the actual integral would require a further argument. In particular, an upper bound does not measure the effect of sign cancellation in an odd filter.","The prime number theorem supplies the standard asymptotic~\\cite{Zagier1997}\n\\begin{equation}\n \\log L_k=k+o(k).\n \\label{eq:PNT-LCM}\n\\end{equation}\nIndeed $\\log L_k=\\sum_{p^a\\leq k}\\log p$ is the Chebyshev function $\\psi_{\\rm Ch}(k)$, and its ratio to $k$ tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write $\\log L_k=\\sum_{a\\geq1}\\vartheta(k^{1/a})$, where $\\vartheta(x)=\\sum_{p\\leq x}\\log p$. The first term is $k+o(k)$. The remaining terms are $O(\\sqrt{k}\\log k)=o(k)$, using $\\vartheta(x)=O(x)$ and at most $\\log_2 k$ nonzero terms. The symbol $\\psi_{\\rm Ch}$ here denotes the arithmetic function, not the two variable kernel in \\eqref{eq:bottleneck-measure}. The prime number theorem is an external theorem. The finite lcm calculations in this publi","The prime number theorem supplies the standard asymptotic~\\cite{Zagier1997}\n\\begin{equation}\n \\log L_k=k+o(k).\n \\label{eq:PNT-LCM}\n\\end{equation}\nIndeed $\\log L_k=\\sum_{p^a\\leq k}\\log p$ is the Chebyshev function $\\psi_{\\rm Ch}(k)$, and its ratio to $k$ tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write $\\log L_k=\\sum_{a\\geq1}\\vartheta(k^{1/a})$, where $\\vartheta(x)=\\sum_{p\\leq x}\\log p$. The first term is $k+o(k)$. The remaining terms are $O(\\sqrt{k}\\log k)=o(k)$, using $\\vartheta(x)=O(x)$ and at most $\\log_2 k$ nonzero terms. The symbol $\\psi_{\\rm Ch}$ here denotes the arithmetic function, not the two variable kernel in \\eqref{eq:bottleneck-measure}. The prime number theorem is an external theorem. The finite lcm calculations in this publication do not reprove it.","e standard asymptotic~\\cite{Zagier1997}\n\\begin{equation}\n \\log L_k=k+o(k).\n \\label{eq:PNT-LCM}\n\\end{equation}\nIndeed $\\log L_k=\\sum_{p^a\\leq k}\\log p$ is the Chebyshev function $\\psi_{\\rm Ch}(k)$, and its ratio to $k$ tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write $\\log L_k=\\sum_{a\\geq1}\\vartheta(k^{1/a})$, where $\\vartheta(x)=\\sum_{p\\leq x}\\log p$. The first term is $k+o(k)$. The remaining terms are $O(\\sqrt{k}\\log k)=o(k)$, using $\\vartheta(x)=O(x)$ and at most $\\log_2 k$ nonzero terms. The symbol $\\psi_{\\rm Ch}$ here denotes the arithmetic function, not the two variable kernel in \\eqref{eq:bottleneck-measure}. The prime number theorem is an external theorem. The finite lcm calculations in this publication do not reprove it.","}\n \\log L_k=k+o(k).\n \\label{eq:PNT-LCM}\n\\end{equation}\nIndeed $\\log L_k=\\sum_{p^a\\leq k}\\log p$ is the Chebyshev function $\\psi_{\\rm Ch}(k)$, and its ratio to $k$ tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write $\\log L_k=\\sum_{a\\geq1}\\vartheta(k^{1/a})$, where $\\vartheta(x)=\\sum_{p\\leq x}\\log p$. The first term is $k+o(k)$. The remaining terms are $O(\\sqrt{k}\\log k)=o(k)$, using $\\vartheta(x)=O(x)$ and at most $\\log_2 k$ nonzero terms. The symbol $\\psi_{\\rm Ch}$ here denotes the arithmetic function, not the two variable kernel in \\eqref{eq:bottleneck-measure}. The prime number theorem is an external theorem. The finite lcm calculations in this publication do not reprove it.","-LCM}\n\\end{equation}\nIndeed $\\log L_k=\\sum_{p^a\\leq k}\\log p$ is the Chebyshev function $\\psi_{\\rm Ch}(k)$, and its ratio to $k$ tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write $\\log L_k=\\sum_{a\\geq1}\\vartheta(k^{1/a})$, where $\\vartheta(x)=\\sum_{p\\leq x}\\log p$. The first term is $k+o(k)$. The remaining terms are $O(\\sqrt{k}\\log k)=o(k)$, using $\\vartheta(x)=O(x)$ and at most $\\log_2 k$ nonzero terms. The symbol $\\psi_{\\rm Ch}$ here denotes the arithmetic function, not the two variable kernel in \\eqref{eq:bottleneck-measure}. The prime number theorem is an external theorem. The finite lcm calculations in this publication do not reprove it.","g L_k=\\sum_{p^a\\leq k}\\log p$ is the Chebyshev function $\\psi_{\\rm Ch}(k)$, and its ratio to $k$ tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write $\\log L_k=\\sum_{a\\geq1}\\vartheta(k^{1/a})$, where $\\vartheta(x)=\\sum_{p\\leq x}\\log p$. The first term is $k+o(k)$. The remaining terms are $O(\\sqrt{k}\\log k)=o(k)$, using $\\vartheta(x)=O(x)$ and at most $\\log_2 k$ nonzero terms. The symbol $\\psi_{\\rm Ch}$ here denotes the arithmetic function, not the two variable kernel in \\eqref{eq:bottleneck-measure}. The prime number theorem is an external theorem. The finite lcm calculations in this publication do not reprove it.","the Chebyshev function $\\psi_{\\rm Ch}(k)$, and its ratio to $k$ tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write $\\log L_k=\\sum_{a\\geq1}\\vartheta(k^{1/a})$, where $\\vartheta(x)=\\sum_{p\\leq x}\\log p$. The first term is $k+o(k)$. The remaining terms are $O(\\sqrt{k}\\log k)=o(k)$, using $\\vartheta(x)=O(x)$ and at most $\\log_2 k$ nonzero terms. The symbol $\\psi_{\\rm Ch}$ here denotes the arithmetic function, not the two variable kernel in \\eqref{eq:bottleneck-measure}. The prime number theorem is an external theorem. The finite lcm calculations in this publication do not reprove it.","}(k)$, and its ratio to $k$ tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write $\\log L_k=\\sum_{a\\geq1}\\vartheta(k^{1/a})$, where $\\vartheta(x)=\\sum_{p\\leq x}\\log p$. The first term is $k+o(k)$. The remaining terms are $O(\\sqrt{k}\\log k)=o(k)$, using $\\vartheta(x)=O(x)$ and at most $\\log_2 k$ nonzero terms. The symbol $\\psi_{\\rm Ch}$ here denotes the arithmetic function, not the two variable kernel in \\eqref{eq:bottleneck-measure}. The prime number theorem is an external theorem. The finite lcm calculations in this publication do not reprove it.","Set\n\\begin{equation}\n K_n=16^nL_{2(n+m(n))}^2.\n \\label{eq:K-asymptotic}\n\\end{equation}\nFor $m(n)/n\\to\\alpha$, \\eqref{eq:PNT-LCM} gives\n\\begin{equation}\n \\frac{\\log K_n}{n}\\longrightarrow\n \\kappa_0(\\alpha)=\\log16+4(1+\\alpha).\n \\label{eq:kappa-alpha}\n\\end{equation}\nThe factor $4$ has a precise origin. The lcm index is $2(n+m)$, and the lcm is squared. Missing either factor would create an incorrect apparent decay margin. The subscript in $\\kappa_0$ emphasizes that this is the clearing cost alone. Define the combined bound exponent by $E(\\alpha)=\\kappa_0(\\alpha)+\\log M(\\alpha)$.",")}^2.\n \\label{eq:K-asymptotic}\n\\end{equation}\nFor $m(n)/n\\to\\alpha$, \\eqref{eq:PNT-LCM} gives\n\\begin{equation}\n \\frac{\\log K_n}{n}\\longrightarrow\n \\kappa_0(\\alpha)=\\log16+4(1+\\alpha).\n \\label{eq:kappa-alpha}\n\\end{equation}\nThe factor $4$ has a precise origin. The lcm index is $2(n+m)$, and the lcm is squared. Missing either factor would create an incorrect apparent decay margin. The subscript in $\\kappa_0$ emphasizes that this is the clearing cost alone. Define the combined bound exponent by $E(\\alpha)=\\kappa_0(\\alpha)+\\log M(\\alpha)$.","in{equation}\n \\frac{\\log K_n}{n}\\longrightarrow\n \\kappa_0(\\alpha)=\\log16+4(1+\\alpha).\n \\label{eq:kappa-alpha}\n\\end{equation}\nThe factor $4$ has a precise origin. The lcm index is $2(n+m)$, and the lcm is squared. Missing either factor would create an incorrect apparent decay margin. The subscript in $\\kappa_0$ emphasizes that this is the clearing cost alone. Define the combined bound exponent by $E(\\alpha)=\\kappa_0(\\alpha)+\\log M(\\alpha)$.","With the cleared gcd $g_n=g_{n,m(n)}$, the primitive form is exactly\n\\begin{equation}\n \\Lambda_n=\\frac{K_n}{g_n}\\widetilde J_{n,m(n)}.\n \\label{eq:Lambda-bottleneck}\n\\end{equation}\nSuppose a future arithmetic argument establishes\n\\begin{equation}\n \\delta=\\liminf_{n\\to\\infty}\\frac{\\log g_n}{n}.\n \\label{eq:delta-content}\n\\end{equation}\nFor finite $\\delta$, the present analytic estimate yields\n\\begin{equation}\n \\limsup_{n\\to\\infty}\\frac{\\log|\\Lambda_n|}{n}\n \\leq\\kappa_0(\\alpha)+\\log M(\\alpha)-\\delta.\n \\label{eq:margin-inequality}\n\\end{equation}\nThe same interpretation applies when $\\delta=+\\infty$: any sufficiently large finite lower bound for the content rate would suffice. All logarithms of nonzero forms are legitimate for the e","With the cleared gcd $g_n=g_{n,m(n)}$, the primitive form is exactly\n\\begin{equation}\n \\Lambda_n=\\frac{K_n}{g_n}\\widetilde J_{n,m(n)}.\n \\label{eq:Lambda-bottleneck}\n\\end{equation}\nSuppose a future arithmetic argument establishes\n\\begin{equation}\n \\delta=\\liminf_{n\\to\\infty}\\frac{\\log g_n}{n}.\n \\label{eq:delta-content}\n\\end{equation}\nFor finite $\\delta$, the present analytic estimate yields\n\\begin{equation}\n \\limsup_{n\\to\\infty}\\frac{\\log|\\Lambda_n|}{n}\n \\leq\\kappa_0(\\alpha)+\\log M(\\alpha)-\\delta.\n \\label{eq:margin-inequality}\n\\end{equation}\nThe same interpretation applies when $\\delta=+\\infty$: any sufficiently large finite lower bound for the content rate would suffice. All logarithms of nonzero forms are legitimate for the even family by \\eqref{eq:even-filter-positive}. The strict negativity of the right side is the missing sufficient margin. A zero m","With the cleared gcd $g_n=g_{n,m(n)}$, the primitive form is exactly\n\\begin{equation}\n \\Lambda_n=\\frac{K_n}{g_n}\\widetilde J_{n,m(n)}.\n \\label{eq:Lambda-bottleneck}\n\\end{equation}\nSuppose a future arithmetic argument establishes\n\\begin{equation}\n \\delta=\\liminf_{n\\to\\infty}\\frac{\\log g_n}{n}.\n \\label{eq:delta-content}\n\\end{equation}\nFor finite $\\delta$, the present analytic estimate yields\n\\begin{equation}\n \\limsup_{n\\to\\infty}\\frac{\\log|\\Lambda_n|}{n}\n \\leq\\kappa_0(\\alpha)+\\log M(\\alpha)-\\delta.\n \\label{eq:margin-inequality}\n\\end{equation}\nThe same interpretation applies when $\\delta=+\\infty$: any sufficiently large finite lower bound for the content rate would suffice. All logarithms of nonzero forms are legitimate for the even family by \\eqref{eq:even-filter-positive}. The strict negativity of the right side is the missing sufficient margin. A zero margin would not suffice, because lower order terms could still prevent convergence to zero.","cleared gcd $g_n=g_{n,m(n)}$, the primitive form is exactly\n\\begin{equation}\n \\Lambda_n=\\frac{K_n}{g_n}\\widetilde J_{n,m(n)}.\n \\label{eq:Lambda-bottleneck}\n\\end{equation}\nSuppose a future arithmetic argument establishes\n\\begin{equation}\n \\delta=\\liminf_{n\\to\\infty}\\frac{\\log g_n}{n}.\n \\label{eq:delta-content}\n\\end{equation}\nFor finite $\\delta$, the present analytic estimate yields\n\\begin{equation}\n \\limsup_{n\\to\\infty}\\frac{\\log|\\Lambda_n|}{n}\n \\leq\\kappa_0(\\alpha)+\\log M(\\alpha)-\\delta.\n \\label{eq:margin-inequality}\n\\end{equation}\nThe same interpretation applies when $\\delta=+\\infty$: any sufficiently large finite lower bound for the content rate would suffice. All logarithms of nonzero forms are legitimate for the even family by \\eqref{eq:even-filter-positive}. The strict negativity of the right side is the missing sufficient margin. A zero margin would not suffice, because lower order terms could still prevent convergence to zero.","se a future arithmetic argument establishes\n\\begin{equation}\n \\delta=\\liminf_{n\\to\\infty}\\frac{\\log g_n}{n}.\n \\label{eq:delta-content}\n\\end{equation}\nFor finite $\\delta$, the present analytic estimate yields\n\\begin{equation}\n \\limsup_{n\\to\\infty}\\frac{\\log|\\Lambda_n|}{n}\n \\leq\\kappa_0(\\alpha)+\\log M(\\alpha)-\\delta.\n \\label{eq:margin-inequality}\n\\end{equation}\nThe same interpretation applies when $\\delta=+\\infty$: any sufficiently large finite lower bound for the content rate would suffice. All logarithms of nonzero forms are legitimate for the even family by \\eqref{eq:even-filter-positive}. The strict negativity of the right side is the missing sufficient margin. A zero margin would not suffice, because lower order terms could still prevent convergence to zero.","\\begin{proposition}[The displayed upper bound has no cleared decay margin]\n\\label{prop:positive-cleared-upper}\nFor every $\\alpha\\geq0$,\n\\begin{equation}\n \\kappa_0(\\alpha)+\\log M(\\alpha)>0.\n \\label{eq:positive-bound-rate}\n\\end{equation}\nThis statement concerns the exponent of the available upper bound. It is not a lower bound on the actual primitive forms.\n\\end{proposition}","\\begin{proof}\nPut $c=16t-1$. The second branch in \\eqref{eq:M-alpha} gives\n\\begin{equation*}\n \\kappa_0(\\alpha)+\\log M(\\alpha)\n \\geq 4+\\log(16t)+\\alpha(4+\\log c).\n\\end{equation*}\nNow $t=(5\\sqrt5-11)/2>5/64$. For example, this inequality follows by squaring the positive rational bound $\\sqrt5>357/160$, since $5\\cdot160^2>357^2$. Thus $16t>5/4>1$ and $c>1/4$. Consequently the last display is greater than $4+\\alpha(4-\\log4)>0$.\n\\end{proof}","{proof}\nPut $c=16t-1$. The second branch in \\eqref{eq:M-alpha} gives\n\\begin{equation*}\n \\kappa_0(\\alpha)+\\log M(\\alpha)\n \\geq 4+\\log(16t)+\\alpha(4+\\log c).\n\\end{equation*}\nNow $t=(5\\sqrt5-11)/2>5/64$. For example, this inequality follows by squaring the positive rational bound $\\sqrt5>357/160$, since $5\\cdot160^2>357^2$. Thus $16t>5/4>1$ and $c>1/4$. Consequently the last display is greater than $4+\\alpha(4-\\log4)>0$.\n\\end{proof}","The proposition says that the direct combination of $M$ with the chosen denominator bound does not prove shrinking integer forms without a quantified content gain. It does not say that the actual forms diverge. A growing upper bound is compatible with a quantity that decays rapidly. Stronger analytic estimates or smaller actual denominators can change the conclusion, but each requires its own proof.","At the source's approximately optimal slope $\\alpha=3.31765$, the two branches of $M$ are near $0.006040$. The resulting exponent $\\kappa_0+\\log M$ is about $14.93384$ per unit $n$. These are rounded evaluations of explicit formulas, not certified extrema or an estimate for $\\log|\\Lambda_n|/n$. They show how large a content rate would need to be for this particular upper bound to force decay. The five observed gcd ratios are reported per $n+m$ and cannot be substituted as asymptotic rates in this calculation.","$\\alpha=3.31765$, the two branches of $M$ are near $0.006040$. The resulting exponent $\\kappa_0+\\log M$ is about $14.93384$ per unit $n$. These are rounded evaluations of explicit formulas, not certified extrema or an estimate for $\\log|\\Lambda_n|/n$. They show how large a content rate would need to be for this particular upper bound to force decay. The five observed gcd ratios are reported per $n+m$ and cannot be substituted as asymptotic rates in this calculation.","Eskandari's coefficient growth statement is\n\\begin{equation}\n \\lim_{n\\to\\infty}\\frac{\\log|\\widetilde U_{n,m(n)}|}{n}\n =\\gamma(\\alpha)\n =-\\log16+\\log R+\\alpha\\log(1+R),\n \\label{eq:gamma-alpha}\n\\end{equation}\nwhere $\\widetilde U=8(-1)^n16^{-n}U$. This is an external asymptotic result in the preprint, not a conclusion of the finite recurrence audit. Together with the filter bound it controls the approximation error\n\\begin{equation}\n \\left|G-\\frac VU\\right|\n =\\frac{|\\widetilde J|}{|\\widetilde U|}.\n \\label{eq:approximation-error}\n\\end{equation}\nBefore any additional quantified gcd saving, the reduced denominator $q$ is at most the absolute cleared coefficient $|A|$. Its available exponent is\n\\begin{equation}\n c_A(\\alpha)=4(1+\\alpha)+\\log R+\\alpha\\log(1+R).\n \\label{eq:cA-alpha}\n\\end{equation}\nThe source's resulting guaranteed approximation exponent is\n\\begin{equation}\n \\eta(\\alpha)=\n \\","Eskandari's coefficient growth statement is\n\\begin{equation}\n \\lim_{n\\to\\infty}\\frac{\\log|\\widetilde U_{n,m(n)}|}{n}\n =\\gamma(\\alpha)\n =-\\log16+\\log R+\\alpha\\log(1+R),\n \\label{eq:gamma-alpha}\n\\end{equation}\nwhere $\\widetilde U=8(-1)^n16^{-n}U$. This is an external asymptotic result in the preprint, not a conclusion of the finite recurrence audit. Together with the filter bound it controls the approximation error\n\\begin{equation}\n \\left|G-\\frac VU\\right|\n =\\frac{|\\widetilde J|}{|\\widetilde U|}.\n \\label{eq:approximation-error}\n\\end{equation}\nBefore any additional quantified gcd saving, the reduced denominator $q$ is at most the absolute cleared coefficient $|A|$. Its available exponent is\n\\begin{equation}\n c_A(\\alpha)=4(1+\\alpha)+\\log R+\\alpha\\log(1+R).\n \\label{eq:cA-alpha}\n\\end{equation}\nThe source's resulting guaranteed approximation exponent is\n\\begin{equation}\n \\eta(\\alpha)=\n \\frac{\\log R+\\alpha\\log(1+R)-\\log16-\\log M(\\alpha)}","tatement is\n\\begin{equation}\n \\lim_{n\\to\\infty}\\frac{\\log|\\widetilde U_{n,m(n)}|}{n}\n =\\gamma(\\alpha)\n =-\\log16+\\log R+\\alpha\\log(1+R),\n \\label{eq:gamma-alpha}\n\\end{equation}\nwhere $\\widetilde U=8(-1)^n16^{-n}U$. This is an external asymptotic result in the preprint, not a conclusion of the finite recurrence audit. Together with the filter bound it controls the approximation error\n\\begin{equation}\n \\left|G-\\frac VU\\right|\n =\\frac{|\\widetilde J|}{|\\widetilde U|}.\n \\label{eq:approximation-error}\n\\end{equation}\nBefore any additional quantified gcd saving, the reduced denominator $q$ is at most the absolute cleared coefficient $|A|$. Its available exponent is\n\\begin{equation}\n c_A(\\alpha)=4(1+\\alpha)+\\log R+\\alpha\\log(1+R).\n \\label{eq:cA-alpha}\n\\end{equation}\nThe source's resulting guaranteed approximation exponent is\n\\begin{equation}\n \\eta(\\alpha)=\n \\frac{\\log R+\\alpha\\log(1+R)-\\log16-\\log M(\\alpha)}\n {4(1+\\alpha)+\\log R+\\alpha\\log(1+R)}.\n \\label{eq:eta-alpha}\n\\end{equation}\nThe source reports a maximum near $0.62335$ at $\\alpha\\approx3.31765$ and states its theorem in the rounded form …","idetilde U=8(-1)^n16^{-n}U$. This is an external asymptotic result in the preprint, not a conclusion of the finite recurrence audit. Together with the filter bound it controls the approximation error\n\\begin{equation}\n \\left|G-\\frac VU\\right|\n =\\frac{|\\widetilde J|}{|\\widetilde U|}.\n \\label{eq:approximation-error}\n\\end{equation}\nBefore any additional quantified gcd saving, the reduced denominator $q$ is at most the absolute cleared coefficient $|A|$. Its available exponent is\n\\begin{equation}\n c_A(\\alpha)=4(1+\\alpha)+\\log R+\\alpha\\log(1+R).\n \\label{eq:cA-alpha}\n\\end{equation}\nThe source's resulting guaranteed approximation exponent is\n\\begin{equation}\n \\eta(\\alpha)=\n \\frac{\\log R+\\alpha\\log(1+R)-\\log16-\\log M(\\alpha)}\n {4(1+\\alpha)+\\log R+\\alpha\\log(1+R)}.\n \\label{eq:eta-alpha}\n\\end{equation}\nThe source reports a maximum near $0.62335$ at $\\alpha\\approx3.31765$ and states its theorem in the rounded form $\\eta\\geq0.62$~\\cite{Eskandari2026}. The present edition records that as a preprint result with its stated dependencies. It does not upgrade it to a peer reviewed theorem, a certified …","asymptotic result in the preprint, not a conclusion of the finite recurrence audit. Together with the filter bound it controls the approximation error\n\\begin{equation}\n \\left|G-\\frac VU\\right|\n =\\frac{|\\widetilde J|}{|\\widetilde U|}.\n \\label{eq:approximation-error}\n\\end{equation}\nBefore any additional quantified gcd saving, the reduced denominator $q$ is at most the absolute cleared coefficient $|A|$. Its available exponent is\n\\begin{equation}\n c_A(\\alpha)=4(1+\\alpha)+\\log R+\\alpha\\log(1+R).\n \\label{eq:cA-alpha}\n\\end{equation}\nThe source's resulting guaranteed approximation exponent is\n\\begin{equation}\n \\eta(\\alpha)=\n \\frac{\\log R+\\alpha\\log(1+R)-\\log16-\\log M(\\alpha)}\n {4(1+\\alpha)+\\log R+\\alpha\\log(1+R)}.\n \\label{eq:eta-alpha}\n\\end{equation}\nThe source reports a maximum near $0.62335$ at $\\alpha\\approx3.31765$ and states its theorem in the rounded form $\\eta\\geq0.62$~\\cite{Eskandari2026}. The present edition records that as a preprint result with its stated dependencies. It does not upgrade it to a peer reviewed theorem, a certified global numerical maximum, or an irrationality pro…","sion of the finite recurrence audit. Together with the filter bound it controls the approximation error\n\\begin{equation}\n \\left|G-\\frac VU\\right|\n =\\frac{|\\widetilde J|}{|\\widetilde U|}.\n \\label{eq:approximation-error}\n\\end{equation}\nBefore any additional quantified gcd saving, the reduced denominator $q$ is at most the absolute cleared coefficient $|A|$. Its available exponent is\n\\begin{equation}\n c_A(\\alpha)=4(1+\\alpha)+\\log R+\\alpha\\log(1+R).\n \\label{eq:cA-alpha}\n\\end{equation}\nThe source's resulting guaranteed approximation exponent is\n\\begin{equation}\n \\eta(\\alpha)=\n \\frac{\\log R+\\alpha\\log(1+R)-\\log16-\\log M(\\alpha)}\n {4(1+\\alpha)+\\log R+\\alpha\\log(1+R)}.\n \\label{eq:eta-alpha}\n\\end{equation}\nThe source reports a maximum near $0.62335$ at $\\alpha\\approx3.31765$ and states its theorem in the rounded form $\\eta\\geq0.62$~\\cite{Eskandari2026}. The present edition records that as a preprint result with its stated dependencies. It does not upgrade it to a peer reviewed theorem, a certified global numerical maximum, or an irrationality proof.","lde J|}{|\\widetilde U|}.\n \\label{eq:approximation-error}\n\\end{equation}\nBefore any additional quantified gcd saving, the reduced denominator $q$ is at most the absolute cleared coefficient $|A|$. Its available exponent is\n\\begin{equation}\n c_A(\\alpha)=4(1+\\alpha)+\\log R+\\alpha\\log(1+R).\n \\label{eq:cA-alpha}\n\\end{equation}\nThe source's resulting guaranteed approximation exponent is\n\\begin{equation}\n \\eta(\\alpha)=\n \\frac{\\log R+\\alpha\\log(1+R)-\\log16-\\log M(\\alpha)}\n {4(1+\\alpha)+\\log R+\\alpha\\log(1+R)}.\n \\label{eq:eta-alpha}\n\\end{equation}\nThe source reports a maximum near $0.62335$ at $\\alpha\\approx3.31765$ and states its theorem in the rounded form $\\eta\\geq0.62$~\\cite{Eskandari2026}. The present edition records that as a preprint result with its stated dependencies. It does not upgrade it to a peer reviewed theorem, a certified global numerical maximum, or an irrationality proof.","fficient $|A|$. Its available exponent is\n\\begin{equation}\n c_A(\\alpha)=4(1+\\alpha)+\\log R+\\alpha\\log(1+R).\n \\label{eq:cA-alpha}\n\\end{equation}\nThe source's resulting guaranteed approximation exponent is\n\\begin{equation}\n \\eta(\\alpha)=\n \\frac{\\log R+\\alpha\\log(1+R)-\\log16-\\log M(\\alpha)}\n {4(1+\\alpha)+\\log R+\\alpha\\log(1+R)}.\n \\label{eq:eta-alpha}\n\\end{equation}\nThe source reports a maximum near $0.62335$ at $\\alpha\\approx3.31765$ and states its theorem in the rounded form $\\eta\\geq0.62$~\\cite{Eskandari2026}. The present edition records that as a preprint result with its stated dependencies. It does not upgrade it to a peer reviewed theorem, a certified global numerical maximum, or an irrationality proof.","$. Its available exponent is\n\\begin{equation}\n c_A(\\alpha)=4(1+\\alpha)+\\log R+\\alpha\\log(1+R).\n \\label{eq:cA-alpha}\n\\end{equation}\nThe source's resulting guaranteed approximation exponent is\n\\begin{equation}\n \\eta(\\alpha)=\n \\frac{\\log R+\\alpha\\log(1+R)-\\log16-\\log M(\\alpha)}\n {4(1+\\alpha)+\\log R+\\alpha\\log(1+R)}.\n \\label{eq:eta-alpha}\n\\end{equation}\nThe source reports a maximum near $0.62335$ at $\\alpha\\approx3.31765$ and states its theorem in the rounded form $\\eta\\geq0.62$~\\cite{Eskandari2026}. The present edition records that as a preprint result with its stated dependencies. It does not upgrade it to a peer reviewed theorem, a certified global numerical maximum, or an irrationality proof.","pha)+\\log R+\\alpha\\log(1+R).\n \\label{eq:cA-alpha}\n\\end{equation}\nThe source's resulting guaranteed approximation exponent is\n\\begin{equation}\n \\eta(\\alpha)=\n \\frac{\\log R+\\alpha\\log(1+R)-\\log16-\\log M(\\alpha)}\n {4(1+\\alpha)+\\log R+\\alpha\\log(1+R)}.\n \\label{eq:eta-alpha}\n\\end{equation}\nThe source reports a maximum near $0.62335$ at $\\alpha\\approx3.31765$ and states its theorem in the rounded form $\\eta\\geq0.62$~\\cite{Eskandari2026}. The present edition records that as a preprint result with its stated dependencies. It does not upgrade it to a peer reviewed theorem, a certified global numerical maximum, or an irrationality proof.","The denominator index in \\eqref{eq:eta-alpha} is $n$, while the filter depth grows like $\\alpha n$. This is why the lcm term is $4(1+\\alpha)$ rather than $4$. Replacing $n$ by $n+m$ consistently rescales every rate. Rescaling just one term changes the statement and is invalid.","An approximation exponent below one cannot contradict rationality. If $G=a/b$ were rational and $p/q\\ne a/b$ were reduced, then\n\\begin{equation}\n \\left|\\frac ab-\\frac pq\\right|\n =\\frac{|aq-bp|}{bq}\\geq\\frac1{bq}.\n \\label{eq:rational-approx-lower}\n\\end{equation}\nA bound of order $q^{-0.62}$ is larger than this obstruction for large $q$. Even rational numbers have many distinct approximations of that quality. For example $a/b+1/(bk)$ has error $1/(bk)$ and, after reduction, denominator at most $bk$. Nothing in such a bound isolates irrationality.","t below one cannot contradict rationality. If $G=a/b$ were rational and $p/q\\ne a/b$ were reduced, then\n\\begin{equation}\n \\left|\\frac ab-\\frac pq\\right|\n =\\frac{|aq-bp|}{bq}\\geq\\frac1{bq}.\n \\label{eq:rational-approx-lower}\n\\end{equation}\nA bound of order $q^{-0.62}$ is larger than this obstruction for large $q$. Even rational numbers have many distinct approximations of that quality. For example $a/b+1/(bk)$ has error $1/(bk)$ and, after reduction, denominator at most $bk$. Nothing in such a bound isolates irrationality.","radict rationality. If $G=a/b$ were rational and $p/q\\ne a/b$ were reduced, then\n\\begin{equation}\n \\left|\\frac ab-\\frac pq\\right|\n =\\frac{|aq-bp|}{bq}\\geq\\frac1{bq}.\n \\label{eq:rational-approx-lower}\n\\end{equation}\nA bound of order $q^{-0.62}$ is larger than this obstruction for large $q$. Even rational numbers have many distinct approximations of that quality. For example $a/b+1/(bk)$ has error $1/(bk)$ and, after reduction, denominator at most $bk$. Nothing in such a bound isolates irrationality.","/q\\ne a/b$ were reduced, then\n\\begin{equation}\n \\left|\\frac ab-\\frac pq\\right|\n =\\frac{|aq-bp|}{bq}\\geq\\frac1{bq}.\n \\label{eq:rational-approx-lower}\n\\end{equation}\nA bound of order $q^{-0.62}$ is larger than this obstruction for large $q$. Even rational numbers have many distinct approximations of that quality. For example $a/b+1/(bk)$ has error $1/(bk)$ and, after reduction, denominator at most $bk$. Nothing in such a bound isolates irrationality.","The identity $c_A-\\gamma=\\kappa_0$ gives another way to read the threshold. The unreduced exponent satisfies $\\eta>1$ exactly when $\\kappa_0+\\log M<0$. Proposition~\\ref{prop:positive-cleared-upper} shows why the displayed source estimate does not reach that regime. A genuine content saving changes the denominator and the primitive error together. It cannot be inserted in just one side of the comparison.","\\begin{theorem}[Sufficient content margin]\n\\label{thm:conditional-completion}\nChoose even integers $m(n)\\geq0$ such that $m(n)/n\\to\\alpha\\geq0$. Use the coefficient clearing theorem, the corrected filter identity, and the prime number theorem as specified above. If there exists a real number $\\delta_0$ with\n\\begin{equation}\n \\liminf_{n\\to\\infty}\\frac{\\log g_{n,m(n)}}{n}\\geq\\delta_0\n \\quad\\text{and}\\quad\n \\delta_0>\\kappa_0(\\alpha)+\\log M(\\alpha),\n \\label{eq:conditional-content-hypothesis}\n\\end{equation}\nthen $G$ is irrational.\n\\end{theorem}","h that $m(n)/n\\to\\alpha\\geq0$. Use the coefficient clearing theorem, the corrected filter identity, and the prime number theorem as specified above. If there exists a real number $\\delta_0$ with\n\\begin{equation}\n \\liminf_{n\\to\\infty}\\frac{\\log g_{n,m(n)}}{n}\\geq\\delta_0\n \\quad\\text{and}\\quad\n \\delta_0>\\kappa_0(\\alpha)+\\log M(\\alpha),\n \\label{eq:conditional-content-hypothesis}\n\\end{equation}\nthen $G$ is irrational.\n\\end{theorem}","\\begin{proof}\nThe coefficient clearing theorem and division by the exact gcd make $\\Lambda_n=a_nG-b_n$ with $a_n,b_n\\in\\Z$. Evenness of $m(n)$ makes $\\widetilde J_{n,m(n)}>0$, so \\eqref{eq:Lambda-bottleneck} gives $\\Lambda_n>0$. Choose $\\varepsilon>0$ so that $3\\varepsilon<\\delta_0-\\kappa_0(\\alpha)-\\log M(\\alpha)$. For all sufficiently large $n$, the clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive","\\begin{proof}\nThe coefficient clearing theorem and division by the exact gcd make $\\Lambda_n=a_nG-b_n$ with $a_n,b_n\\in\\Z$. Evenness of $m(n)$ makes $\\widetilde J_{n,m(n)}>0$, so \\eqref{eq:Lambda-bottleneck} gives $\\Lambda_n>0$. Choose $\\varepsilon>0$ so that $3\\varepsilon<\\delta_0-\\kappa_0(\\alpha)-\\log M(\\alpha)$. For all sufficiently large $n$, the clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at le","\\begin{proof}\nThe coefficient clearing theorem and division by the exact gcd make $\\Lambda_n=a_nG-b_n$ with $a_n,b_n\\in\\Z$. Evenness of $m(n)$ makes $\\widetilde J_{n,m(n)}>0$, so \\eqref{eq:Lambda-bottleneck} gives $\\Lambda_n>0$. Choose $\\varepsilon>0$ so that $3\\varepsilon<\\delta_0-\\kappa_0(\\alpha)-\\log M(\\alpha)$. For all sufficiently large $n$, the clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda","\\begin{proof}\nThe coefficient clearing theorem and division by the exact gcd make $\\Lambda_n=a_nG-b_n$ with $a_n,b_n\\in\\Z$. Evenness of $m(n)$ makes $\\widetilde J_{n,m(n)}>0$, so \\eqref{eq:Lambda-bottleneck} gives $\\Lambda_n>0$. Choose $\\varepsilon>0$ so that $3\\varepsilon<\\delta_0-\\kappa_0(\\alpha)-\\log M(\\alpha)$. For all sufficiently large $n$, the clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda_n\\geq1/b$. This contradicts the","\\begin{proof}\nThe coefficient clearing theorem and division by the exact gcd make $\\Lambda_n=a_nG-b_n$ with $a_n,b_n\\in\\Z$. Evenness of $m(n)$ makes $\\widetilde J_{n,m(n)}>0$, so \\eqref{eq:Lambda-bottleneck} gives $\\Lambda_n>0$. Choose $\\varepsilon>0$ so that $3\\varepsilon<\\delta_0-\\kappa_0(\\alpha)-\\log M(\\alpha)$. For all sufficiently large $n$, the clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda_n\\geq1/b$. This contradicts the displayed convergence.\n\\end{proof}","rem and division by the exact gcd make $\\Lambda_n=a_nG-b_n$ with $a_n,b_n\\in\\Z$. Evenness of $m(n)$ makes $\\widetilde J_{n,m(n)}>0$, so \\eqref{eq:Lambda-bottleneck} gives $\\Lambda_n>0$. Choose $\\varepsilon>0$ so that $3\\varepsilon<\\delta_0-\\kappa_0(\\alpha)-\\log M(\\alpha)$. For all sufficiently large $n$, the clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda_n\\geq1/b$. This contradicts the displayed convergence.\n\\end{proof}","of $m(n)$ makes $\\widetilde J_{n,m(n)}>0$, so \\eqref{eq:Lambda-bottleneck} gives $\\Lambda_n>0$. Choose $\\varepsilon>0$ so that $3\\varepsilon<\\delta_0-\\kappa_0(\\alpha)-\\log M(\\alpha)$. For all sufficiently large $n$, the clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda_n\\geq1/b$. This contradicts the displayed convergence.\n\\end{proof}","} gives $\\Lambda_n>0$. Choose $\\varepsilon>0$ so that $3\\varepsilon<\\delta_0-\\kappa_0(\\alpha)-\\log M(\\alpha)$. For all sufficiently large $n$, the clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda_n\\geq1/b$. This contradicts the displayed convergence.\n\\end{proof}","all sufficiently large $n$, the clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda_n\\geq1/b$. This contradicts the displayed convergence.\n\\end{proof}","$n$, the clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda_n\\geq1/b$. This contradicts the displayed convergence.\n\\end{proof}","clearing asymptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda_n\\geq1/b$. This contradicts the displayed convergence.\n\\end{proof}","mptotic gives $K_n\\leq\\exp((\\kappa_0+\\varepsilon)n)$, the filter bound gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda_n\\geq1/b$. This contradicts the displayed convergence.\n\\end{proof}","d gives $|\\widetilde J_{n,m(n)}|\\leq\\exp((\\log M+\\varepsilon)n)$, and the liminf assumption gives $g_{n,m(n)}\\geq\\exp((\\delta_0-\\varepsilon)n)$. Therefore\n\\begin{equation*}\n 0<\\Lambda_n\\leq\n \\exp\\bigl((\\kappa_0+\\log M-\\delta_0+3\\varepsilon)n\\bigr)\n \\longrightarrow0.\n\\end{equation*}\nIf $G=a/b$ with integers $a,b$ and $b>0$, then $b\\Lambda_n=aa_n-bb_n$ is a positive integer. It is at least one, so $\\Lambda_n\\geq1/b$. This contradicts the displayed convergence.\n\\end{proof}","The proof of this implication is complete. Its content hypothesis is not established here. The theorem is therefore conditional, and cannot be listed as a proof of the conjecture. It supplies a concrete target for a future arithmetic result. A stronger analytic bound could replace $M$, provided it were proved for the same infinite family and retained the nonvanishing input.","There is also a formulation independent of the universal clearing choice. With $D_n,h_n$ as in Chapter~\\ref{chap:primitive-content}, the primitive error is $(-1)^n(D_n/h_n)(S_{n,m(n)}G-H_{n,m(n)})$. An estimate directly on that expression would suffice. If denominator reduction yields a provably smaller exponential cost, the excess factor $Q_n/D_n$ should simply be removed from the accounting. The irrationality criterion itself is unchanged.","There is a simple logical reason to resist extrapolation. Given any finite list of positive values $x_1,\\ldots,x_N$, it has a positive extension that tends to zero and another positive extension that tends to infinity. Keep the list fixed and set the tail to $e^{-n}$ in the first case or $e^n$ in the second. Similarly, any finite list of positive integer gcds can be extended by $1$ forever or by integers with an arbitrarily prescribed positive exponential rate.","A positive exponent in an upper bound has the same logical limitation. If $x_n\\leq e^{cn}$ with $c>0$, both $x_n=e^{-n}$ and $x_n=e^{cn/2}$ satisfy the bound eventually. One sequence decays and the other grows. Thus the failure of the bound to force decay is a proved limitation of the estimate, not proof of divergence of the object it bounds.","The same caution applies to the modular scan. Its $8\\,006\\,000$ exact valuations establish a large finite statement. They do not prove $v_2(S_{n,m})=o(n+m)$ outside the scanned domain, a uniform bound on odd common factors, or the lower bound on $\\log g$ in Theorem~\\ref{thm:conditional-completion}. The achievement is reproducible bounded evidence and a sharply stated remaining implication.","The universal arguments written in this edition include the elementary half-binomial denominator lemma, the Wronskian induction, the Pascal identity, the modular valuation criterion, the primitive content decomposition, and the conditional irrationality implication. Some of these arguments start from externally supplied identities. For example, the integrality proof for $T_n$ uses the authenticated half-binomial expression for $u_n$. Its universal arithmetic step is written here, while the identity identifying that expression with the recurrence is an external input.","in this edition include the elementary half-binomial denominator lemma, the Wronskian induction, the Pascal identity, the modular valuation criterion, the primitive content decomposition, and the conditional irrationality implication. Some of these arguments start from externally supplied identities. For example, the integrality proof for $T_n$ uses the authenticated half-binomial expression for $u_n$. Its universal arithmetic step is written here, while the identity identifying that expression with the recurrence is an external input.","The integral normalization receives special attention because the printed source contains a factor discrepancy. In the historical recurrence normalization, the correct relation is\n\\begin{equation}\n u_nG-v_n=\\frac{(-1)^n}{8}J_n.\n \\label{eq:status-normalization}\n\\end{equation}\nThe printed factor $1/4$ is incompatible with $u_0=1$ and the exact evaluation $J_0=8G$. The source audit also checks the all-index normalization through the displayed hypergeometric transformation and partial-fraction coefficient identification. This is an explicit correction with stated dependencies. It is not a silently altered quotation, nor a claim that a published erratum was recovered.","The large modular scan covers $1\\leq n\\leq2000$ and $0\\leq m\\leq4n$, a total of $8\\,006\\,000$ pairs. It finds maximum valuation $31$, with a witness at $(1466,2113)$ in the historical traversal order. No residue is zero modulo $2^{128}$, so every scanned valuation is below the precision ceiling and is exact. The direct witness calculation reconstructs the full integer sum and verifies the same valuation. The evidence does not assert that this maximizing pair is unique.","The smaller direct grid contains $20\\,300$ pairs with $n\\leq100$. It agrees with the modular Pascal computation at every pair and has maximum valuation $20$ at $(52,6)$ in its own traversal order. A deliberate low precision test at modulus $2^4$ produces eight zero residues in the much smaller domain $n\\leq3$. These are correctly marked unresolved. The true direct maximum there is $6$, while the clipped residue report is $4$. This negative control checks that the program does not mistake a precision ceiling for an exact valuation.","ntains $20\\,300$ pairs with $n\\leq100$. It agrees with the modular Pascal computation at every pair and has maximum valuation $20$ at $(52,6)$ in its own traversal order. A deliberate low precision test at modulus $2^4$ produces eight zero residues in the much smaller domain $n\\leq3$. These are correctly marked unresolved. The true direct maximum there is $6$, while the clipped residue report is $4$. This negative control checks that the program does not mistake a precision ceiling for an exact valuation.","Independent rational recurrence calculations agree through $n=500$. The integer recurrence for $T_n$ has zero remainder at every division through $n=10000$. The half-binomial check verifies all $1\\,326$ summands through $n=50$. These finite checks support the implementation and normalization. Their universal counterparts still depend on the written arguments and external coefficient identity.","The five gcd computations in Chapter~\\ref{chap:primitive-content} are exact rational calculations, including complete numerator and denominator strings. The displays of $\\log g/(n+m)$ are ordinary rounded logarithms. The exact factorization and decomposition are the arithmetic evidence. The display ratios are explanatory summaries.","A diagnostic guess proposed during this audit asserted $\\operatorname{sign}F_{n,m}=(-1)^m$ for the raw packet form\n\\begin{equation}\n F_{n,m}=8(-1)^n\\bigl(S_{n,m}G-H_{n,m}\\bigr).\n \\label{eq:status-raw-F}\n\\end{equation}\nIt fails. Already at $(n,m)=(1,1)$,\n\\begin{equation}\n F_{1,1}=\\frac{173056}{9}-20992G>0,\n \\label{eq:status-sign-counterexample}\n\\end{equation}\nwhile the guess predicts a negative sign. A certified rational interval for $G$ determines the sign. Among the $860$ tested pairs with $1\\leq n\\leq20$ and $0\\leq m\\leq4n$, there are $138$ mismatches with the guess. All $860$ actual forms are nevertheless certified nonzero.","s audit asserted $\\operatorname{sign}F_{n,m}=(-1)^m$ for the raw packet form\n\\begin{equation}\n F_{n,m}=8(-1)^n\\bigl(S_{n,m}G-H_{n,m}\\bigr).\n \\label{eq:status-raw-F}\n\\end{equation}\nIt fails. Already at $(n,m)=(1,1)$,\n\\begin{equation}\n F_{1,1}=\\frac{173056}{9}-20992G>0,\n \\label{eq:status-sign-counterexample}\n\\end{equation}\nwhile the guess predicts a negative sign. A certified rational interval for $G$ determines the sign. Among the $860$ tested pairs with $1\\leq n\\leq20$ and $0\\leq m\\leq4n$, there are $138$ mismatches with the guess. All $860$ actual forms are nevertheless certified nonzero.","n}F_{n,m}=(-1)^m$ for the raw packet form\n\\begin{equation}\n F_{n,m}=8(-1)^n\\bigl(S_{n,m}G-H_{n,m}\\bigr).\n \\label{eq:status-raw-F}\n\\end{equation}\nIt fails. Already at $(n,m)=(1,1)$,\n\\begin{equation}\n F_{1,1}=\\frac{173056}{9}-20992G>0,\n \\label{eq:status-sign-counterexample}\n\\end{equation}\nwhile the guess predicts a negative sign. A certified rational interval for $G$ determines the sign. Among the $860$ tested pairs with $1\\leq n\\leq20$ and $0\\leq m\\leq4n$, there are $138$ mismatches with the guess. All $860$ actual forms are nevertheless certified nonzero.","e raw packet form\n\\begin{equation}\n F_{n,m}=8(-1)^n\\bigl(S_{n,m}G-H_{n,m}\\bigr).\n \\label{eq:status-raw-F}\n\\end{equation}\nIt fails. Already at $(n,m)=(1,1)$,\n\\begin{equation}\n F_{1,1}=\\frac{173056}{9}-20992G>0,\n \\label{eq:status-sign-counterexample}\n\\end{equation}\nwhile the guess predicts a negative sign. A certified rational interval for $G$ determines the sign. Among the $860$ tested pairs with $1\\leq n\\leq20$ and $0\\leq m\\leq4n$, there are $138$ mismatches with the guess. All $860$ actual forms are nevertheless certified nonzero.","egin{equation}\n F_{n,m}=8(-1)^n\\bigl(S_{n,m}G-H_{n,m}\\bigr).\n \\label{eq:status-raw-F}\n\\end{equation}\nIt fails. Already at $(n,m)=(1,1)$,\n\\begin{equation}\n F_{1,1}=\\frac{173056}{9}-20992G>0,\n \\label{eq:status-sign-counterexample}\n\\end{equation}\nwhile the guess predicts a negative sign. A certified rational interval for $G$ determines the sign. Among the $860$ tested pairs with $1\\leq n\\leq20$ and $0\\leq m\\leq4n$, there are $138$ mismatches with the guess. All $860$ actual forms are nevertheless certified nonzero.","1)^n\\bigl(S_{n,m}G-H_{n,m}\\bigr).\n \\label{eq:status-raw-F}\n\\end{equation}\nIt fails. Already at $(n,m)=(1,1)$,\n\\begin{equation}\n F_{1,1}=\\frac{173056}{9}-20992G>0,\n \\label{eq:status-sign-counterexample}\n\\end{equation}\nwhile the guess predicts a negative sign. A certified rational interval for $G$ determines the sign. Among the $860$ tested pairs with $1\\leq n\\leq20$ and $0\\leq m\\leq4n$, there are $138$ mismatches with the guess. All $860$ actual forms are nevertheless certified nonzero.","abel{eq:status-raw-F}\n\\end{equation}\nIt fails. Already at $(n,m)=(1,1)$,\n\\begin{equation}\n F_{1,1}=\\frac{173056}{9}-20992G>0,\n \\label{eq:status-sign-counterexample}\n\\end{equation}\nwhile the guess predicts a negative sign. A certified rational interval for $G$ determines the sign. Among the $860$ tested pairs with $1\\leq n\\leq20$ and $0\\leq m\\leq4n$, there are $138$ mismatches with the guess. All $860$ actual forms are nevertheless certified nonzero.","The edition date matters because recent primary preprints discuss the irrationality of $G$. Zhi-Wei Sun's preprint, submitted on 3 September 2026, claims a proof~\\cite{Sun2026}. The checked arXiv history contained only version 1 on 9 October 2026. A subsequent note by Dane Wachs, submitted on 16 September and dated 15 September internally, identifies a gap in a displayed asymptotic derivation~\\cite{Wachs2026}. It describes a quadratic contribution from the prime $2$ that it says the derivation does not account for.","e irrationality of $G$. Zhi-Wei Sun's preprint, submitted on 3 September 2026, claims a proof~\\cite{Sun2026}. The checked arXiv history contained only version 1 on 9 October 2026. A subsequent note by Dane Wachs, submitted on 16 September and dated 15 September internally, identifies a gap in a displayed asymptotic derivation~\\cite{Wachs2026}. It describes a quadratic contribution from the prime $2$ that it says the derivation does not account for.","The work can therefore be advanced in distinct, testable directions. An arithmetic route should distinguish $Q/D$ from $h$ and control both dyadic and odd factors where necessary. An analytic route should prove a bound for the actual integral along an infinite family, rather than infer decay from a plot. A formal route should encode the exact statements and mark external asymptotic inputs explicitly. More computation can expose candidates or counterexamples, but it must continue to state its range.","\\chapter*{The question behind the calculations}\n\\addcontentsline{toc}{chapter}{The question behind the calculations}\nCatalan's constant has a simple definition,\n\\[\nG=\\sum_{k=0}^{\\infty}\\frac{(-1)^k}{(2k+1)^2}.\n\\]\nIt is easy to locate on the real line. It is much harder to classify arithmetically. An accurate decimal approximation does not decide whether the value is rational. A recurrence that produces astonishingly close rational approximations does not decide that question either, unless the denominators and the nonzero errors satisfy the right bounds.","\\section*{How the route would finish}\nIf $G=a/b$ were rational, every nonzero form $AG-B$ with integers $A,B$ would have absolute value at least $1/b$. Thus nonzero integer forms tending to zero would contradict rationality. The recurrence gives small real errors. Clearing its rational coefficients multiplies those errors by an arithmetic cost. Common divisors can reduce that cost, but a divisor created by an unnecessarily large clearing factor is not itself a new saving. The crucial quantity is the error after the coefficients have been reduced."]}