strengthening.tex
Realization Calculus
strengthening.tex
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1 % Integrated strengthening, based on the frozen proof revision of 2026-10-06.
2 % No preamble. All valuations have v_p(0)=+infinity.
3 \section{Threshold-free reciprocal transfer at prime-power section orders}
4 \label{sec:prime-power-transfer-final}
5
6 The completion of Nishibuchi's conjecture fixes the requested eventual
7 congruence. The stronger question is whether the sufficient starting row
8 can be removed, and how much precision the difference order and the step
9 actually provide. This section answers the first question at every
10 prime-power section order. Subsequent sections give a factorial improvement
11 and exact results at section order four.
12
13 Throughout these sections, the normalized source and its reciprocal are
14 \begin{equation}
15 G_{N,j}(t)=j!\sum_{n\geq0}\frac{t^{Nn}}{(Nn+j)!},\qquad
16 F_{N,j}(t)=G_{N,j}(t)^{-1}
17 =\sum_{n\geq0}C_{N,j}(n)\frac{t^{Nn}}{(Nn)!},
18 \label{eq:final-normalization}
19 \end{equation}
20 where $N\geq1$ and $0\leq j<N$. In particular, $C_{N,j}(0)=1$.
21 The coefficients in the original convention satisfy
22 \begin{equation}
23 E_{Nn}^{(N,j)}=j!C_{N,j}(n),\qquad
24 E_d^{(N,j)}=0\quad(N\nmid d).
25 \label{eq:final-original-normalization}
26 \end{equation}
27 Thus every valuation of a nonzero difference gains $v_p(j!)$ on returning
28 to the original coefficients. This factor must be retained when stating
29 optimal periods or the final congruence.
30
31 Write $S f(n)=f(n+1)$ and $\Delta_H=S^H-1$. A mixed difference
32 $\Delta_{H_1}\cdots\Delta_{H_k}$ acts on the row variable $n$, rather than
33 on the full Euler subscript. We use $v_p(0)=+\infty$. Orders and starting
34 rows are nonnegative integers. Steps are positive integers unless stated
35 otherwise. An empty mixed difference is the original value, and a zero
36 ordinary step makes every nonempty product containing it zero.
37
38 \subsection{The Hurwitz lattice and translated differential flatness}
39
40 \begin{definition}[Hurwitz integral series]
41 For a prime $p$, let
42 \[
43 \mathcal H_p=\left\{\sum_{d\geq0}f_d\frac{t^d}{d!}:
44 f_d\in\mathbb Z_p\right\}
45 \subset\mathbb Q_p[[t]].
46 \]
47 For $N=p^a$ and $\lambda\in\mathbb Z_p$, put
48 \[
49 D=\frac{d}{dt},\qquad L_\lambda=\frac{D^N-\lambda}{p}.
50 \]
51 A series $A$ is $\lambda$-flat if $L_\lambda^k A\in\mathcal H_p$ for every
52 $k\geq0$.
53 \end{definition}
54
55 Products preserve $\mathcal H_p$ because the coefficient at factorial
56 degree $d$ is
57 \[
58 (AB)_d=\sum_{i=0}^d\binom di a_i b_{d-i}.
59 \]
60 The derivative $D$ acts by the coefficient shift $(DA)_d=a_{d+1}$.
61 Consequently every nonnegative power of $D$ preserves the lattice.
62 If $A\in\mathcal H_p$ has unit constant term, its reciprocal belongs to
63 the same lattice: with $B=A^{-1}$,
64 \begin{equation}
65 b_0=a_0^{-1},\qquad
66 b_d=-a_0^{-1}\sum_{i=1}^d\binom di a_i b_{d-i}.
67 \label{eq:hurwitz-reciprocal-recurrence}
68 \end{equation}
69 If $A$ is supported at degrees divisible by $N$, this recurrence also
70 proves that $B$ has the same support.
71
72 \begin{lemma}[Translated product formula]
73 \label{lem:translated-hurwitz-product}
74 For $N=p^a$, $a\geq1$, and $\lambda,\mu\in\mathbb Z_p$,
75 \begin{equation}
76 L_{\lambda+\mu}(AB)
77 =(L_\lambda A)B+A(L_\mu B)
78 +\sum_{i=1}^{N-1}c_i(D^iA)(D^{N-i}B),
79 \qquad c_i=\frac1p\binom Ni\in\mathbb Z.
80 \label{eq:translated-product}
81 \end{equation}
82 Flatness types add under multiplication.
83 \end{lemma}
84
85 \begin{proof}
86 The identity $(1+X)^{p^a}=1+X^{p^a}$ over $\mathbb F_p$ shows that every
87 interior binomial coefficient $\binom Ni$ is divisible by $p$.
88 Subtract $(\lambda+\mu)AB$ from the ordinary $N$th derivative product
89 formula and divide by $p$ to obtain \eqref{eq:translated-product}.
90
91 For completeness, the iterated formula is an integral multinomial
92 formula, rather than a product rule with unspecified remainders. On the
93 tensor of the two factors use the mutually commuting operators
94 \[
95 L_\lambda\otimes1,\quad 1\otimes L_\mu,\quad
96 c_iD^i\otimes D^{N-i}\quad(1\leq i<N).
97 \]
98 Applying multiplication after the $k$th power of their sum gives exactly
99 $L_{\lambda+\mu}^k(AB)$. Expanding it yields terms indexed by
100 $a,b,e_1,\ldots,e_{N-1}\geq0$, with $a+b+\sum_i e_i=k$:
101 \begin{equation}
102 \frac{k!}{a!b!\prod_i e_i!}\prod_i c_i^{e_i}
103 (D^\alpha L_\lambda^a A)(D^\beta L_\mu^b B),
104 \quad
105 \alpha=\sum_i i e_i,\quad
106 \beta=\sum_i(N-i)e_i.
107 \label{eq:translated-multinomial}
108 \end{equation}
109 Every scalar is integral. If $A$ and $B$ have the stated flatness types,
110 both factors in every term belong to $\mathcal H_p$, as does their product.
111 \end{proof}
112
113 \begin{lemma}[Reciprocal transfer]
114 \label{lem:translated-hurwitz-inverse}
115 If $A$ is $\lambda$-flat and has unit constant term, then $A^{-1}$ is
116 $(-\lambda)$-flat.
117 \end{lemma}
118
119 \begin{proof}
120 Set $B=A^{-1}$. Its order-zero integrality was established in
121 \eqref{eq:hurwitz-reciprocal-recurrence}. Assume by induction that
122 $L_{-\lambda}^bB\in\mathcal H_p$ for $b<k$. Since $AB=1$,
123 \[
124 L_0^k(AB)=\left(\frac{D^N}{p}\right)^k1=0\qquad(k>0).
125 \]
126 Use \eqref{eq:translated-multinomial} with $\mu=-\lambda$.
127 Exactly one term has inverse order $b=k$: it has $a=0$ and all $e_i=0$,
128 and is $A L_{-\lambda}^k B$. In every other term $b<k$, so its second
129 factor is integral by induction and derivative-shift integrality. Its
130 first factor is integral by the flatness of $A$, and its scalar is
131 integral. Hence $A L_{-\lambda}^k B\in\mathcal H_p$. Multiplying by the
132 already integral $B$ proves $L_{-\lambda}^k B\in\mathcal H_p$.
133 This induction takes place in the entire Hurwitz lattice. It places no
134 restriction on the coefficient degree or the starting row.
135 \end{proof}
136
137 \subsection{The actual Euler source}
138
139 The packed factorial coefficient of $G_{N,j}$ is
140 \begin{equation}
141 g_{N,j}(n)=\frac{j!(Nn)!}{(Nn+j)!}
142 =\prod_{i=1}^j\left(1+\frac Ni n\right)^{-1}.
143 \label{eq:actual-prime-power-kernel}
144 \end{equation}
145 The empty product is one. If $N=p^a$ and $1\leq i<N$, then
146 $v_p(i)<a$, so $N/i\in p\mathbb Z_p$. Every affine denominator in
147 \eqref{eq:actual-prime-power-kernel} is therefore a unit on $\mathbb Z_p$.
148
149 For $s\in p\mathbb Z_p$, an exact finite-difference identity is
150 \begin{equation}
151 \Delta_1^k\frac1{1+sn}
152 =\frac{(-s)^k k!}{\prod_{h=0}^k(1+s(n+h))}.
153 \label{eq:affine-reciprocal-exact}
154 \end{equation}
155 It follows by induction, subtracting adjacent rational functions. Each
156 successive subtraction adds the factor $-(k+1)s$ in the numerator.
157 Its denominator is a unit, so it belongs to $p^k\mathbb Z_p$.
158 The discrete Leibniz formula
159 \begin{equation}
160 \Delta_1^k(fg)(n)
161 =\sum_{r=0}^k\binom kr
162 (\Delta_1^r f)(n)(\Delta_1^{k-r}g)(n+r)
163 \label{eq:discrete-leibniz-final}
164 \end{equation}
165 then proves $\Delta_1^k g_{N,j}(n)\in p^k\mathbb Z_p$ for every $k,n$.
166 The coefficient at degree $Nn$ of $L_1^kG_{N,j}$ is precisely
167 $p^{-k}\Delta_1^k g_{N,j}(n)$, and the other residue classes vanish.
168 Thus the actual source is $1$-flat.
169
170 \begin{theorem}[All-row prime-power transfer]
171 \label{thm:all-prime-power-transfer}
172 Let $p$ be prime, $a\geq1$, $N=p^a$, and $0\leq j<N$.
173 The coefficients $C_{N,j}(n)$ belong to $\mathbb Z_p$ for all $n\geq0$.
174 Set $A_{N,j}(n)=(-1)^n C_{N,j}(n)$ and
175 $T_H=S^H-(-1)^H$. For $k\geq1$ and positive $H_1,\ldots,H_k$,
176 \begin{align}
177 v_p(\Delta_{H_1}\cdots\Delta_{H_k}A_{N,j}(n))
178 &\geq k+\sum_{\nu=1}^k v_p(H_\nu),
179 \label{eq:prime-power-signed-bound}\\
180 v_p(T_{H_1}\cdots T_{H_k}E_{Nn}^{(N,j)})
181 &\geq v_p(j!)+k+\sum_{\nu=1}^k v_p(H_\nu).
182 \label{eq:prime-power-original-twist}
183 \end{align}
184 At $p=2$, the ordinary differences of $C_{N,j}$ and of the original
185 coefficients satisfy the same bounds. These dyadic statements also hold
186 for $N=1$.
187 \end{theorem}
188
189 \begin{proof}
190 Reciprocal transfer gives $L_{-1}^kF_{N,j}\in\mathcal H_p$.
191 Extraction of the packed coefficient gives
192 $(S+1)^k C_{N,j}(n)\in p^k\mathbb Z_p$. Moreover,
193 \[
194 \Delta_1^k A_{N,j}(n)=(-1)^{n+k}(S+1)^k C_{N,j}(n),
195 \]
196 so $A_{N,j}$ has the ordinary $p^k$ difference property.
197 At $p=2$, $L_1=L_{-1}-1$. Its integral binomial expansion on
198 $F_{N,j}$ gives $\Delta_1^k C_{N,j}(n)\in2^k\mathbb Z_2$ directly.
199 For $N=1$ the interior sum in the product formula is empty, so the same
200 argument applies without change. Equivalently, $F_{1,0}(t)=e^{-t}$.
201
202 To retain the valuation of a step, let $B$ be any sequence satisfying
203 $\Delta_1^m B\in p^m\mathbb Z_p$ at every order and row. Use the finite
204 operator identity
205 \[
206 \Delta_H=\sum_{i=1}^H\binom Hi\Delta_1^i.
207 \]
208 The relation $i\binom Hi=H\binom{H-1}{i-1}$ implies
209 \[
210 v_p\left(p^i\binom Hi\right)
211 \geq i+v_p(H)-v_p(i)\geq1+v_p(H).
212 \]
213 Multiplying these identities for the $k$ steps, every term of total
214 order $M$ has its scalar multiplied by $p^M$, and is divisible by
215 $p^{k+\sum_\nu v_p(H_\nu)}$. This proves the mixed-step bound.
216 Finally,
217 \[
218 \Delta_H A_{N,j}(n)=(-1)^{n+H}T_H C_{N,j}(n).
219 \]
220 Iteration contributes only a global sign. Multiplying by $j!$ proves
221 \eqref{eq:prime-power-original-twist}.
222 \end{proof}
223
224 \begin{corollary}[Threshold-free dyadic congruence]
225 \label{cor:all-dyadic-final}
226 For $N=2^a$, $a\geq0$, all permitted $j$, and $n\geq0$,
227 \begin{equation}
228 v_2(\Delta_{H_1}\cdots\Delta_{H_k}E_{Nn}^{(N,j)})
229 \geq v_2(j!)+k+\sum_{\nu=1}^k v_2(H_\nu).
230 \label{eq:all-dyadic-final-bound}
231 \end{equation}
232 In particular, $\Delta_H^k$ gains $k(1+v_2(H))$ bits before normalization.
233 For $r>v_2(j!)$, the row displacement
234 $H=2^{\max(0,r-v_2(j!)-1)}$ is a sufficient period modulo $2^r$ from
235 row zero. If $r\leq v_2(j!)$, all original coefficients vanish modulo
236 $2^r$.
237 \end{corollary}
238
239 \begin{remark}[Necessary boundaries]
240 At an odd prime the ordinary step-one difference cannot replace the
241 sign-twisted operator in the general theorem. For $j=0$,
242 $C_{N,0}(0)=1$ and $C_{N,0}(1)=-1$, so their ordinary difference is
243 $-2$, a unit at every odd prime. The proof also does not extend
244 unchanged to non-prime-power $N$, whose interior binomial coefficients
245 need not be divisible by the chosen prime. Neither observation is a
246 counterexample to Nishibuchi's original eventual congruence.
247 \end{remark}
248
249 \section{Factorial precision at section order four}
250 \label{sec:factorial-four-final}
251
252 Write $C_j=C_{4,j}$ and $F_j=F_{4,j}$. The all-row theorem removes the
253 threshold, but at $N=4$ it leaves a further factorial gain unused.
254 The exact algebra below proves that gain for all four addresses and
255 identifies several infinite families where it is sharp.
256
257 \begin{theorem}[Four-address factorial bound]
258 \label{thm:four-factorial-bound}
259 For $j\in\{0,1,2,3\}$, $n\geq0$, $k\geq1$, and positive mixed steps,
260 \begin{equation}
261 v_2(\Delta_{H_1}\cdots\Delta_{H_k}C_j(n))
262 \geq k+v_2(k!)+\sum_{\nu=1}^k v_2(H_\nu)+\beta_j,
263 \quad (\beta_0,\beta_1,\beta_2,\beta_3)=(0,0,3,1).
264 \label{eq:factorial-four-lower}
265 \end{equation}
266 Address two also satisfies
267 \begin{equation}
268 v_2(\Delta_{H_1}\cdots\Delta_{H_k}C_2(n))
269 \geq 2k+v_2(k!)+\left\lfloor\frac{k}{2}\right\rfloor
270 +\sum_{\nu=1}^k v_2(H_\nu).
271 \label{eq:factorial-four-j2-filtration}
272 \end{equation}
273 For the original coefficients, the offsets in
274 \eqref{eq:factorial-four-lower} are $(0,0,4,2)$.
275 No positive-order offset is asserted for the empty difference.
276 \end{theorem}
277
278 The proof of the first two addresses uses one classical arithmetic
279 input: Frobenius's exact congruence for the ordinary Euler numbers in
280 the $\operatorname{sech}$ convention, as stated and derived by
281 Gessel~\cite{gessel}, Section~9. The address-two and address-three
282 bounds below are finite Bernoulli calculations and do not require that
283 input. A divided-power argument is necessary: ordinary integral moments
284 alone do not justify division by $k!$.
285
286 \subsection{Euler moments and an integral divided-power algebra}
287
288 Let $\mathcal L$ be the ordinary Euler evaluation functional on a
289 variable $\mathcal E$, defined by
290 \[
291 \mathcal L(\mathcal E^d)=\operatorname{Euler}_d,
292 \qquad \sum_{d\geq0}\operatorname{Euler}_d\frac{t^d}{d!}
293 =\operatorname{sech}t.
294 \]
295 The functional is linear. It is not a multiplicative evaluation.
296 Independent variables mean that moments of polynomials in separate
297 variables factor. Odd moments vanish. Frobenius's congruence is
298 \begin{equation}
299 v_2\left(\sum_{r=0}^m(-1)^{m-r}\binom mr
300 \operatorname{Euler}_{2n+2r}\right)
301 =m+v_2(m!)\qquad(n,m\geq0).
302 \label{eq:classical-frobenius-exact}
303 \end{equation}
304 For $m=0$, the assertion is that every even Euler number is odd.
305 With $P=(\mathcal E^2-1)/2$ and
306 $\gamma_m(P)=P^m/m!$, the case $n=0$ says
307 \begin{equation}
308 \mathcal L(\gamma_m(P))\in\mathbb Z_2^{\times}
309 \qquad(m\geq0).
310 \label{eq:euler-pd-odd-moments}
311 \end{equation}
312 Introduce an independent Euler variable $\mathcal F$ and
313 $Q=(\mathcal F^2-1)/2$.
314
315 Put $K=\mathbb Q_2(i)$, $R=\mathbb Z_2[i]$,
316 $a=(1+i)/2$, and $b=(1-i)/2$. The exact identity
317 \begin{equation}
318 F_0(t)=\frac{2}{\cosh t+\cos t}
319 =\operatorname{sech}(at)\operatorname{sech}(bt)
320 \label{eq:four-euler-factorization}
321 \end{equation}
322 gives, for $U=a\mathcal E+b\mathcal F$,
323 $\mathcal L(U^{4n})=C_0(n)$. Although $a$ and $b$ have negative
324 valuation, their denominators disappear from the following exact
325 polynomial identities:
326 \begin{align}
327 U^2&=\mathcal E\mathcal F+i(P-Q),\nonumber\\
328 H_0:=\frac{U^4-1}{2}
329 &=P+Q-\frac{P^2}{2}-\frac{Q^2}{2}
330 +3PQ+i\mathcal E\mathcal F(P-Q).
331 \label{eq:four-euler-H}
332 \end{align}
333
334 Let $\Gamma_R(P,Q)$ be the $R$-subalgebra generated by all
335 $\gamma_m(P)$ and $\gamma_r(Q)$. Adjoin $\mathcal E,\mathcal F$ with
336 \[
337 \mathcal E^2=1+2P,\qquad \mathcal F^2=1+2Q,
338 \]
339 and call the resulting algebra $\mathcal A_0$. Its elements are finite
340 expressions. No completion is being presumed. It is generated as a
341 $\Gamma_R(P,Q)$-module by $1,\mathcal E,\mathcal F,\mathcal E\mathcal F$.
342 The functional maps this module to $R$. For an even basis term this is
343 \eqref{eq:euler-pd-odd-moments} and independence. For a basis term with
344 an odd Euler factor the moment is zero, since $P$ and $Q$ are even
345 polynomials in their respective variables.
346
347 Every displayed component of $H_0$ has all its divided powers in
348 $\mathcal A_0$. Besides the generators, the needed identities are
349 \begin{align}
350 \gamma_m(P^2/2)
351 &=\frac{(2m)!}{2^m m!}\gamma_{2m}(P),\nonumber\\
352 \gamma_m(3PQ)&=3^m m!\gamma_m(P)\gamma_m(Q),\nonumber\\
353 \gamma_m(P-Q)&=\sum_{r=0}^m(-1)^r
354 \gamma_{m-r}(P)\gamma_r(Q).
355 \label{eq:components-divided-powers}
356 \end{align}
357 The first scalar counts partitions into pairs and is an integer.
358 For the last term of $H_0$, multiply the third identity by
359 $(i\mathcal E\mathcal F)^m$ and reduce the Euler powers using the
360 integral quadratic relations. Finally,
361 \[
362 \gamma_k(X+Y)=\sum_{r=0}^k\gamma_r(X)\gamma_{k-r}(Y)
363 \]
364 proves $\gamma_k(H_0)\in\mathcal A_0$. Applying the functional and
365 putting $W=U^4-1=2H_0$ gives
366 \begin{equation}
367 \mathcal L(W^m)\in2^m m!R\qquad(m\geq0).
368 \label{eq:euler-W-factorial}
369 \end{equation}
370 These moments are rational. The identity
371 $R\cap\mathbb Q_2=\mathbb Z_2$, obtained from the linear independence
372 of $1,i$ over $\mathbb Q_2$, descends every such divisibility statement
373 to $\mathbb Z_2$. No field property of $R/2R$ is needed.
374
375 For every starting row,
376 \begin{equation}
377 \Delta_1^k C_0(n)
378 =\mathcal L(U^{4n}W^k)
379 =\sum_{s=0}^n\binom ns\mathcal L(W^{k+s}).
380 \label{eq:base-row-finite-expansion}
381 \end{equation}
382 Each summand lies in $2^k k!\mathbb Z_2$, proving the factorial
383 bound for address zero from row zero onward.
384
385 \subsection{A finite step-expansion lemma}
386
387 The conversion from unit steps to arbitrary mixed steps is used several
388 times, so it is useful to make it explicit. Write $\Delta=\Delta_1$.
389 For $H>0$,
390 \begin{equation}
391 \Delta_H=\sum_{m=1}^H\binom Hm\Delta^m.
392 \label{eq:finite-step-expansion}
393 \end{equation}
394 A mixed term with local orders $m_1,\ldots,m_k\geq1$ and
395 $M=\sum_\nu m_\nu$ therefore has valuation at least
396 \[
397 M+v_2(M!)+\beta
398 +\sum_\nu\bigl(v_2(H_\nu)-v_2(m_\nu)\bigr)
399 \]
400 if the unit-order estimate has offset $\beta$ for every positive order.
401 The inequalities
402 \[
403 m-1\geq v_2(m),\qquad M\geq k,\qquad v_2(M!)\geq v_2(k!)
404 \]
405 make this at least $k+v_2(k!)+\beta+\sum_\nu v_2(H_\nu)$.
406 Every expansion is finite. This proves the desired mixed-step transfer
407 without an exchange of limits.
408
409 \subsection{An elementary Bernoulli denominator bound}
410
411 Let $B_m$ denote the Bernoulli numbers with $B_1=-1/2$.
412 The identity obtained by substituting $u=e^t-1$ into
413 $\log(1+u)/u$ is
414 \begin{equation}
415 B_m=\sum_{r=0}^m(-1)^r\frac{r!}{r+1}
416 \left\{\begin{matrix}m\\r\end{matrix}\right\},
417 \label{eq:bernoulli-stirling-denominator}
418 \end{equation}
419 where the braces are Stirling numbers of the second kind. Each
420 $2r!/(r+1)$ is dyadically integral. Indeed, if $s=v_2(r+1)>0$, then
421 $r\geq2^s-1$ and
422 \[
423 v_2(r!)\geq v_2((2^s-1)!)=2^s-1-s\geq s-1.
424 \]
425 The last inequality is $2^s\geq2s$. The case $s=0$ is immediate.
426 Consequently
427 \begin{equation}
428 2B_m\in\mathbb Z_2\qquad(m\geq0).
429 \label{eq:elementary-bernoulli-bound}
430 \end{equation}
431 Thus the transfer factors below require no appeal to the
432 von Staudt--Clausen theorem.
433
434 \subsection{Address one and the scaled-binomial transfer}
435
436 The address-one source is
437 $G_{4,1}(t)=(\sinh t+\sin t)/(2t)$. The identities $a+b=1$,
438 $a-b=i$, and $ab=1/2$ give
439 \begin{equation}
440 D_1(t):=G_{4,1}(t)F_0(t)
441 =\frac12\left(\frac{\tanh(at)}{at}
442 +\frac{\tanh(bt)}{bt}\right).
443 \label{eq:address-one-tangent-factor}
444 \end{equation}
445 For example, its numerator follows from
446 \[
447 \sinh t+\sin t
448 =2b\sinh(at)\cosh(bt)+2a\cosh(at)\sinh(bt),
449 \]
450 while $\cosh t+\cos t=2\cosh(at)\cosh(bt)$.
451 Only degrees divisible by four survive. Its actual Hurwitz packets are
452 \begin{equation}
453 D_1(t)=\sum_{i\geq0}d_i\frac{t^{4i}}{(4i)!},\qquad
454 d_i=\left(-\frac14\right)^i
455 \frac{2^{4i+2}(2^{4i+2}-1)B_{4i+2}}{(4i+2)(4i+1)}.
456 \label{eq:address-one-packets}
457 \end{equation}
458 The constant is one. Since $v_2(4i+2)=1$, the elementary Bernoulli bound
459 shows $v_2(d_i)\geq2i$. Hurwitz inversion therefore gives
460 \[
461 D_1(t)^{-1}=\sum_{i\geq0}h_i\frac{t^{4i}}{(4i)!},
462 \qquad h_0=1,\quad v_2(h_i)\geq2i.
463 \]
464 This follows by induction from the inverse recurrence: for packet index
465 $i$, every product $d_rh_{i-r}$ has valuation at least $2i$, and its
466 binomial multiplier is integral.
467 Because $F_1=F_0D_1^{-1}$,
468 \begin{equation}
469 C_1(n)-C_0(n)
470 =\sum_{i\geq1}h_i\binom{4n}{4i}C_0(n-i).
471 \label{eq:address-one-exact-transfer}
472 \end{equation}
473 At a natural $n$ this is a finite sum, with $i\leq n$.
474
475 Here is the polynomial estimate that supplies one additional bit.
476 For $i\geq1$, put $P_i(x)=4^i\binom{4x}{4i}$. Its constant is zero,
477 and its degree-$d$ coefficient is
478 \begin{equation}
479 [x^d]P_i(x)=\frac{4^i4^d s(4i,d)}{(4i)!},
480 \label{eq:scaled-binomial-coefficients}
481 \end{equation}
482 where $s(m,d)$ is a signed Stirling number of the first kind.
483 In each product contributing to $s(4i,d)$, at most $d-1$ of the
484 $2i-1$ even integers in $1,\ldots,4i-1$ can be omitted. Thus
485 \[
486 v_2(s(4i,d))\geq\max(0,2i-d).
487 \]
488 Writing $s_2(i)$ for the number of ones in the binary expansion of $i$,
489 and using $v_2((4i)!)=4i-s_2(i)$, gives
490 \begin{equation}
491 v_2([x^d]P_i)
492 \geq2d-2i+s_2(i)+\max(0,2i-d)
493 \geq d+s_2(i)\geq d+1\quad(d\geq1).
494 \label{eq:scaled-binomial-degree-bound}
495 \end{equation}
496 For $k\geq1$, $\Delta_1^k x^d$ vanishes when $d<k$ and is an
497 integer polynomial with values divisible by $k!$. To see the latter,
498 expand $x^d=\sum_r\left\{\begin{smallmatrix}d\\r\end{smallmatrix}\right\}
499 r!\binom xr$ and apply
500 $\Delta_1^k\binom xr=\binom{x}{r-k}$.
501 It follows that
502 \begin{equation}
503 \Delta_1^kP_i(x)\in2^{k+1}k!\mathbb Z_2
504 \qquad(k\geq0,\ x\in\mathbb Z_2).
505 \label{eq:scaled-binomial-difference-bound}
506 \end{equation}
507 For $k=0$ the bound follows directly from the coefficient estimate.
508
509 To apply discrete products in \eqref{eq:address-one-exact-transfer}
510 without a moving truncation, first extend $C_0$ to $\mathbb Z_2$ by its
511 uniformly convergent Newton series. The all-row bound
512 $v_2(\Delta_1^m C_0(0))\geq m$ proves this convergence. Its factorial
513 bounds extend to every dyadic starting point by continuity and
514 approximation with natural rows. The sum in
515 \eqref{eq:address-one-exact-transfer} is then uniformly convergent on
516 $\mathbb Z_2$: both the binomial polynomial and $C_0(x-i)$ are
517 integral-valued and $v_2(h_i)\geq2i$.
518
519 For the pointwise product $P_i(x)C_0(x-i)$, the discrete Leibniz formula
520 has scalar divisibility
521 \[
522 \binom kr(2^{r+1}r!)(2^{k-r}(k-r)!)=2^{k+1}k!.
523 \]
524 Multiplication by $h_i/4^i\in\mathbb Z_2$ and uniform summation now give
525 \begin{equation}
526 v_2(\Delta_1^k(C_1-C_0)(n))
527 \geq k+v_2(k!)+1\qquad(k,n\geq0).
528 \label{eq:address-one-extra-bit}
529 \end{equation}
530 The same extra bit holds for positive mixed steps by
531 \eqref{eq:finite-step-expansion}. Combining this with the address-zero
532 bound proves Theorem~\ref{thm:four-factorial-bound} for address one.
533
534 \subsection{Central factorials and integral Bernoulli divided powers}
535
536 \begin{lemma}[Bernoulli divided-power moments]
537 \label{lem:bernoulli-central-pd}
538 Let $Z$ be the ordinary Bernoulli variable,
539 $\mathcal L(Z^r)=B_r$, with exponential transform $t/(e^t-1)$.
540 For $P=Z(Z+1)$ and $\mu_m=\mathcal L(P^m/m!)$,
541 \begin{equation}
542 \mu_m\in\mathbb Z_2,\qquad
543 v_2(\mu_m)\geq s_2(m)-1\quad(m\geq1).
544 \label{eq:bernoulli-central-moment-bound}
545 \end{equation}
546 Moreover, $\mu_m$ is odd exactly when $m=0$ or $m$ is a positive power
547 of two.
548 \end{lemma}
549
550 \begin{proof}
551 Define the central-factorial basis polynomials
552 \[
553 P_r(X)=\prod_{a=0}^{r-1}(X-a(a+1)).
554 \]
555 At $X=Z(Z+1)$, they have the exact factorization
556 \begin{equation}
557 P_r(Z(Z+1))
558 =\prod_{a=0}^{r-1}(Z-a)(Z+a+1)
559 =(2r)!\binom{Z+r}{2r}.
560 \label{eq:central-factorial-substitution}
561 \end{equation}
562 The Bernoulli transform gives
563 $\mathcal L((1+T)^Z)=\log(1+T)/T$, whence
564 $\mathcal L(\binom Zj)=(-1)^j/(j+1)$. Finite Vandermonde expansion
565 therefore gives
566 \begin{align}
567 \mathcal L\left(\binom{Z+r}{2r}\right)
568 &=(-1)^r\sum_{q=0}^r(-1)^q\binom rq\frac1{r+q+1}\nonumber\\
569 &=(-1)^r\frac{(r!)^2}{(2r+1)!}.
570 \label{eq:central-factorial-evaluation}
571 \end{align}
572 The last equality is the rational identity obtained by expanding
573 $\int_0^1t^r(1-t)^r\,dt$. This use of an ordinary finite integral makes
574 no assertion about a positive or bounded $p$-adic measure. In
575 particular,
576 \[
577 \mathcal L(P_r(P))=(-1)^r\frac{(r!)^2}{2r+1}.
578 \]
579
580 Expand $X^m=\sum_{r=0}^mT(m,r)P_r(X)$. Since
581 $XP_r(X)=P_{r+1}(X)+r(r+1)P_r(X)$, these integer coefficients satisfy
582 \begin{equation}
583 T(m+1,r)=T(m,r-1)+r(r+1)T(m,r),
584 \quad T(0,0)=1.
585 \label{eq:central-factorial-recurrence}
586 \end{equation}
587 Every $r(r+1)$ is even. Induction, including the boundary indices,
588 proves $2^{m-r}\mid T(m,r)$ for $0\leq r\leq m$.
589 For $m>0$ the $r=0$ term vanishes. Each remaining summand of
590 $\mathcal L(P^m/m!)$ has valuation at least
591 \begin{equation}
592 m-r+2v_2(r!)-v_2(m!)
593 =r+s_2(m)-2s_2(r)\geq s_2(m)-1\geq0.
594 \label{eq:central-factorial-term-bound}
595 \end{equation}
596 Here $r\geq2s_2(r)-1$ follows from
597 $r\geq2^{s_2(r)}-1$ and $2^u\geq2u$ for $u\geq1$.
598 This proves integrality and the stronger bound.
599
600 If $m$ is not a power of two, $s_2(m)\geq2$ makes $\mu_m$ even.
601 If $m$ is a positive power of two, the only central indices which can
602 contribute an odd residue are $r=1$ and $r=3$, since
603 $r-2s_2(r)=-1$ only at these two positive integers. The $r=1$ term is
604 odd: $T(m,1)=2^{m-1}$ and $v_2(m!)=m-1$. For $m\geq4$ a power of
605 two, the $r=3$ term is even. More explicitly,
606 \[
607 \frac{T(m,3)}{2^{m-3}}=h_{m-3}(1,3,6),
608 \qquad
609 \sum_{d\geq0}h_d(1,3,6)t^d
610 =\frac1{(1-t)(1-3t)(1-6t)}.
611 \]
612 Reduction modulo two gives $(1-t)^{-2}$, whose degree $m-3$
613 coefficient has parity $m-2=0$. At $m=1,2$ the $r=3$ term is absent,
614 and the $r=1$ term is again odd. Finally $\mu_0=1$.
615 \end{proof}
616
617 \subsection{Address two and its stronger filtration}
618
619 The variable $\mathcal B=2Z+1$ has transform
620 \[
621 e^t\frac{2t}{e^{2t}-1}=\frac{t}{\sinh t}.
622 \]
623 Its odd moments vanish, and
624 $P=(\mathcal B^2-1)/4=Z(Z+1)$ has the divided-power moments of
625 Lemma~\ref{lem:bernoulli-central-pd}. Introduce an independent copy
626 $\mathcal D$ and $Q=(\mathcal D^2-1)/4$. The exact factorization
627 \begin{equation}
628 F_2(t)=\frac{t^2}{\cosh t-\cos t}
629 =\frac{at}{\sinh(at)}\frac{bt}{\sinh(bt)}
630 \label{eq:address-two-factorization}
631 \end{equation}
632 shows that $U=a\mathcal B+b\mathcal D$ satisfies
633 $\mathcal L(U^{4n})=C_2(n)$. Direct expansion gives
634 \begin{equation}
635 H_2:=\frac{U^4-1}{4}
636 =P+Q-P^2-Q^2+6PQ+i\mathcal B\mathcal D(P-Q).
637 \label{eq:address-two-H}
638 \end{equation}
639
640 Let $\mathcal A_2$ be the divided-power algebra over $R$ generated by
641 $P,Q$, adjoining $\mathcal B,\mathcal D$ with
642 $\mathcal B^2=1+4P$, $\mathcal D^2=1+4Q$.
643 The functional is integral on its module basis
644 $1,\mathcal B,\mathcal D,\mathcal B\mathcal D$ by the Bernoulli lemma,
645 independence, and vanishing of odd moments. Every component in
646 \eqref{eq:address-two-H} has integral divided powers, by the same
647 finite identities as before. $P^2$ is even easier than $P^2/2$.
648 Consequently
649 \begin{equation}
650 \mathcal L((U^4-1)^m)\in4^m m!\mathbb Z_2,
651 \qquad
652 v_2(\Delta_1^kC_2(n))\geq2k+v_2(k!).
653 \label{eq:address-two-basic-factorial}
654 \end{equation}
655 The starting-row conclusion follows from the finite base expansion
656 \eqref{eq:base-row-finite-expansion}, now with $W=4H_2$.
657
658 The low Newton coefficients of the actual reciprocal are
659 \begin{equation}
660 C_2(0)=1,\quad C_2(1)=-\frac1{15},\quad C_2(2)=\frac{13}{45},
661 \qquad a_1=-\frac{16}{15},\quad a_2=\frac{64}{45},
662 \label{eq:address-two-initial-newton}
663 \end{equation}
664 where $a_m=\Delta_1^m C_2(0)$. Thus $v_2(a_1)=4$ and $v_2(a_2)=6$.
665 For $m\geq3$, \eqref{eq:address-two-basic-factorial} bounds $a_m$ by
666 $2m+v_2(m!)$. The finite shifted Newton formula
667 \begin{equation}
668 \Delta_1^kC_2(n)=\sum_{m=k}^{n+k}a_m\binom n{m-k}
669 \label{eq:shifted-newton-finite}
670 \end{equation}
671 therefore gives
672 $v_2(\Delta_1^kC_2(n))\geq k+v_2(k!)+3$ for every $k\geq1$.
673 For $k\geq3$ this is $2k\geq k+3$. For $k=1,2$ it uses the exact
674 initial coefficients and the higher tail bound. The finite step
675 expansion proves the offset three in
676 \eqref{eq:factorial-four-lower}.
677
678 There is a still stronger filtration. Put
679 \[
680 X=P+Q,\qquad \tau=i\mathcal B\mathcal D,\qquad \epsilon=1+\tau.
681 \]
682 Then the exact identities are
683 \begin{equation}
684 H_2=\epsilon X+2G,\quad
685 G=-\gamma_2(P)-\gamma_2(Q)+3PQ-\tau Q,\quad
686 \epsilon^2=2V,\quad V=\tau-2X-8PQ.
687 \label{eq:address-two-filtration-algebra}
688 \end{equation}
689 Every divided power of $G$ is integral component by component.
690 The finite addition identity gives
691 \begin{equation}
692 \gamma_k(H_2)=\sum_{r=0}^k2^r\gamma_r(G)
693 \epsilon^{k-r}\gamma_{k-r}(X)
694 \in2^{\lfloor k/2\rfloor}\mathcal A_2,
695 \label{eq:address-two-filtration-pd}
696 \end{equation}
697 because the $r$th term belongs to
698 $2^{r+\lfloor(k-r)/2\rfloor}\mathcal A_2$.
699 Applying the functional, rational descent, and the base-index expansion
700 proves the unit-step version of
701 \eqref{eq:factorial-four-j2-filtration}.
702 For a mixed term of total order $M$, its lower bound exceeds the target
703 at order $k$ by at least
704 \[
705 2(M-k)+\left\lfloor\frac M2\right\rfloor
706 -\left\lfloor\frac k2\right\rfloor
707 +v_2(M!)-v_2(k!)-\sum_\nu v_2(m_\nu)\geq0.
708 \]
709 Indeed $\sum_\nu v_2(m_\nu)\leq M-k$. This proves the stronger mixed
710 bound. The earlier offset-three bound is retained, since it is stronger
711 at the lowest orders.
712
713 \subsection{Address three and cancellation of the apparent pole}
714
715 The address-three source is
716 $G_{4,3}(t)=3(\sinh t-\sin t)/t^3$. Addition identities give
717 \[
718 \sinh t-\sin t
719 =2a\sinh(at)\cosh(bt)+2b\cosh(at)\sinh(bt),
720 \qquad \cosh t-\cos t=2\sinh(at)\sinh(bt).
721 \]
722 Consequently
723 \begin{equation}
724 D_3(t):=G_{4,3}(t)F_2(t)
725 =\frac32\left(\frac{\coth(at)}{at}
726 +\frac{\coth(bt)}{bt}\right).
727 \label{eq:address-three-cotangent-factor}
728 \end{equation}
729 This expression is a power series: its two polar terms cancel because
730 $a^{-2}+b^{-2}=0$. Its constant is one, only degrees divisible by four
731 survive, and the Bernoulli expansion gives
732 \begin{equation}
733 D_3(t)=\sum_{i\geq0}d_i\frac{t^{4i}}{(4i)!},\qquad
734 d_i=3\left(-\frac14\right)^i
735 \frac{2^{4i+2}B_{4i+2}}{(4i+2)(4i+1)}.
736 \label{eq:address-three-packets}
737 \end{equation}
738 Again $v_2(d_i)\geq2i$, and the inverse coefficients $h_i$ have the same
739 bound. Since $F_3=F_2D_3^{-1}$,
740 \begin{equation}
741 C_3(n)-C_2(n)=\sum_{i\geq1}h_i\binom{4n}{4i}C_2(n-i).
742 \label{eq:address-three-exact-transfer}
743 \end{equation}
744 The scaled-binomial argument
745 \eqref{eq:scaled-binomial-difference-bound} applies unchanged.
746 The basic bound for $C_2$ includes order zero. Its continuous extension
747 is available from the threshold-free theorem. Thus the shifted sums
748 converge uniformly on $\mathbb Z_2$, and the pointwise product rule
749 proves
750 \begin{equation}
751 v_2(\Delta_1^k(C_3-C_2)(n))\geq k+v_2(k!)+1
752 \qquad(k,n\geq0).
753 \label{eq:address-three-extra-bit}
754 \end{equation}
755 At every positive order $C_2$ has three extra bits, so this gives the
756 one-bit offset for $C_3$. Finite step expansion finishes the proof of
757 Theorem~\ref{thm:four-factorial-bound}.
758
759 \section{Sharp higher-difference families}
760 \label{sec:sharp-higher-final}
761
762 The lower bounds do not mean that every mixed stencil has exactly the
763 stated valuation. Addresses zero and one have a complete classification
764 of when the leading factorial precision is attained. Address two has
765 exact mixed precision at every positive power-of-two order.
766
767 \begin{theorem}[Exact factorial precision at addresses zero and one]
768 \label{thm:zero-one-exact-mixed}
769 Let $j\in\{0,1\}$, $k\geq1$, $n\geq0$, and $H_1,\ldots,H_k>0$.
770 Put
771 \[
772 L=k+v_2(k!)+\sum_{\nu=1}^k v_2(H_\nu),\qquad
773 e=\#\{\nu:2\mid H_\nu\}.
774 \]
775 The mixed difference has valuation exactly $L$ if $k$ is odd or $e$ is
776 even. If $k$ is even and $e$ is odd, it has valuation at least $L+1$.
777 In particular, for every homogeneous positive step,
778 \begin{equation}
779 v_2(\Delta_H^kC_j(n))=k+v_2(k!)+k v_2(H).
780 \label{eq:zero-one-homogeneous-exact}
781 \end{equation}
782 \end{theorem}
783
784 \begin{proof}
785 Return to the Euler variables of address zero and put
786 $X=P+Q$, $\tau=i\mathcal E\mathcal F$, $\epsilon=1+\tau$.
787 Direct rearrangement of \eqref{eq:four-euler-H} gives
788 \begin{equation}
789 H_0=\gamma_2(X)+\epsilon X+2G,\qquad
790 G=-\gamma_2(P)-\gamma_2(Q)+PQ-\tau Q,
791 \qquad \epsilon^2=2(\tau-P-Q-2PQ).
792 \label{eq:zero-parity-decomposition}
793 \end{equation}
794 The divided powers of $G$ are integral by the same component formulas
795 used above. Thus modulo $2\mathcal A_0$, $\gamma_k(H_0)$ equals the
796 divided power of $\gamma_2(X)+\epsilon X$. The identity
797 \[
798 \gamma_r(\gamma_2(X))
799 =\frac{(2r)!}{2^r r!}\gamma_{2r}(X)
800 \]
801 has an odd integer scalar. Every term involving $\epsilon^j$ for
802 $j\geq2$ vanishes modulo two. The surviving zero and first powers give
803 \begin{equation}
804 \gamma_k(H_0)\equiv
805 \gamma_{2k}(X)+\epsilon\gamma_{2k-1}(X)
806 \pmod{2\mathcal A_0}.
807 \label{eq:zero-parity-leading-pd}
808 \end{equation}
809 The multiplication coefficient in the second term is $2k-1$, also odd.
810 By the odd moments \eqref{eq:euler-pd-odd-moments} and independence,
811 \[
812 \mathcal L(\gamma_m(P+Q))\equiv m+1\pmod2.
813 \]
814 The $\tau$-weighted moment vanishes because it contains an odd power of
815 each Euler variable. Therefore the first term of
816 \eqref{eq:zero-parity-leading-pd} evaluates to one modulo two and the
817 second to zero. Hence $\mathcal L(\gamma_k(H_0))$ is odd at every order,
818 including order zero. This proves
819 \[
820 v_2(\Delta_1^k C_0(0))=k+v_2(k!).
821 \]
822 In the base-row expansion \eqref{eq:base-row-finite-expansion}, every
823 $s>0$ term has at least one extra bit after division by $2^k k!$.
824 Thus the same exact valuation holds at every starting row.
825 The extra-bit transfer \eqref{eq:address-one-extra-bit} proves it for
826 address one as well.
827
828 It remains to track possible ties in the mixed-step expansion.
829 For local orders $m_\nu\geq1$ and total $M$, the excess over $L$ is
830 at least
831 \begin{equation}
832 \left(M-k-\sum_\nu v_2(m_\nu)\right)
833 +\left(v_2(M!)-v_2(k!)\right).
834 \label{eq:zero-one-mixed-excess}
835 \end{equation}
836 The first bracket is nonnegative, and equality requires every
837 $m_\nu$ to be one or two. Two or more occurrences of two make the
838 second bracket positive, since the intervening factorial interval then
839 contains an even number. Exactly one occurrence of two can tie only
840 when $k$ is even. Its binomial coefficient can have the minimally
841 allowed valuation only if its associated step is even.
842
843 The all-one multi-index contributes one after normalization by $2^L$.
844 When $k$ is even, each even step supplies exactly one further tied term,
845 again with normalized residue one. For example,
846 $\binom H2=H(H-1)/2$ attains $v_2(H)-1$ exactly when $H$ is even.
847 All remaining factors and the exact normalized order-$k+1$ difference
848 are odd. Thus the total normalized parity is one for odd $k$, and
849 $1+e$ for even $k$. This proves the classification.
850 In the homogeneous case, $e$ is zero or $k$, so it is even whenever
851 $k$ is even, proving \eqref{eq:zero-one-homogeneous-exact}.
852 \end{proof}
853
854 \begin{remark}[A mixed cancellation]
855 The exceptional class can gain more than one bit. The exact values
856 \[
857 \Delta_1\Delta_2C_0(0)=-33728,\qquad
858 \Delta_1\Delta_2C_1(0)=-\frac{150208}{585}
859 \]
860 both have valuation six, whereas their baseline $L$ is four.
861 The theorem classifies attainment of the baseline. It does not give a
862 complete deeper valuation in the cancellation class.
863 \end{remark}
864
865 \begin{theorem}[Exact address-two power orders]
866 \label{thm:two-exact-power-orders}
867 For every starting row $n\geq0$ and all positive mixed steps,
868 \begin{equation}
869 v_2(\Delta_{H_1}\cdots\Delta_{H_k}C_2(n))
870 =V_2(k)+\sum_{\nu=1}^k v_2(H_\nu)
871 \label{eq:two-power-orders-exact}
872 \end{equation}
873 whenever $k$ is a positive power of two, where
874 \[
875 V_2(1)=4,\qquad V_2(2)=6,\qquad
876 V_2(k)=2k+v_2(k!)+\frac{k}{2}\quad(k\geq4).
877 \]
878 For the original coefficients, the right side gains one bit.
879 \end{theorem}
880
881 \begin{proof}
882 For $k=2^r\geq4$, put $q=k/2$, which is even. Normalize the exact
883 expansion \eqref{eq:address-two-filtration-pd} by $2^q$ before reducing
884 modulo two. Its $u$th auxiliary term has excess
885 $u+\lfloor(k-u)/2\rfloor-q=\lfloor u/2\rfloor$.
886 Only auxiliary indices zero and one can therefore survive:
887 \begin{equation}
888 \frac{\gamma_k(H_2)}{2^q}
889 \equiv V^q\gamma_k(X)
890 +\epsilon V^{q-1}G\gamma_{k-1}(X)
891 \pmod{2\mathcal A_2}.
892 \label{eq:two-power-leading-normalized}
893 \end{equation}
894 Here $V\equiv\tau\pmod2$ and $\tau^2\equiv1\pmod2$.
895 Since $q$ is even and $\epsilon\tau\equiv\epsilon\pmod2$, this reduces
896 to $\gamma_k(X)+\epsilon G\gamma_{k-1}(X)$.
897 Moreover, $G\equiv\gamma_2(X)+\tau Q\pmod2$.
898 After evaluation, every term with a single $\tau$ vanishes, so the
899 second contribution reduces to
900 \[
901 \mathcal L\bigl((\gamma_2(X)+Q)\gamma_{k-1}(X)\bigr).
902 \]
903 The $\gamma_2(X)$ product has scalar $\binom{k+1}{2}$, which is even
904 because $4\mid k$. The other contribution expands as
905 \begin{equation}
906 \mathcal L(Q\gamma_{k-1}(P+Q))
907 =\sum_{a=0}^{k-1}(k-a)\mu_a\mu_{k-a}.
908 \label{eq:two-power-Q-evaluation}
909 \end{equation}
910 An odd summand would require $k-a$ to be odd and $\mu_{k-a}$ to be
911 odd. By Lemma~\ref{lem:bernoulli-central-pd}, this requires $k-a=1$.
912 Then $a=k-1$ is not a power of two and $\mu_a$ is even. Hence the
913 second contribution is zero modulo two.
914
915 For the first contribution, independence gives
916 $\mathcal L(\gamma_k(P+Q))=\sum_{a=0}^k\mu_a\mu_{k-a}$.
917 The parity generating series of these convolution moments is
918 \[
919 (1+t+t^2+t^4+t^8+\cdots)^2
920 =1+t^2+t^4+t^8+\cdots\quad\text{in }\mathbb F_2[[t]].
921 \]
922 Its coefficient is one at every $k=2^r\geq4$. Thus the normalized
923 moment in \eqref{eq:two-power-leading-normalized} is odd and
924 \[
925 v_2(\Delta_1^kC_2(0))=2k+v_2(k!)+k/2.
926 \]
927 Inserting $U^{4n}=(1+4H_2)^n$ introduces only corrections with strictly
928 higher valuation, by the filtered bound at orders $k+s$ for $s>0$.
929 The equality consequently holds at every starting row.
930
931 For mixed steps the all-one local multi-index gives the claimed leading
932 term. A higher term of total order $M>k$ has excess at least
933 \[
934 2(M-k)+\left\lfloor\frac M2\right\rfloor-\frac k2
935 +v_2(M!)-v_2(k!)-\sum_\nu v_2(m_\nu)
936 \geq M-k\geq1.
937 \]
938 Hence the leading term cannot cancel.
939
940 At order one, $a_1$ in \eqref{eq:address-two-initial-newton} has
941 valuation four and the Newton tail starts at valuation six. Its shifted
942 value remains exact. In the step expansion a local order two can lose
943 at most one bit through its binomial coefficient, so every correction
944 is still above $4+v_2(H_1)$. At order two, $a_2$ has valuation six and
945 the next coefficient has lower bound eight from the stronger
946 filtration. A total-order-three correction loses at most one bit,
947 so it stays above the leading valuation six, while later orders gain
948 still more. The same argument proves exactness for both mixed steps.
949 \end{proof}
950
951 \begin{remark}[What the power orders do not decide]
952 The exact orders in Theorem~\ref{thm:two-exact-power-orders} form an
953 infinite family. They do not supply a complete valuation formula at
954 non-power-of-two orders in address two, nor an exact higher-difference
955 formula at address three. Numerical patterns at other orders remain
956 bounded observations unless proved separately.
957 \end{remark}
958
959 \section{Exact first differences, least periods, and dyadic images}
960 \label{sec:sharp-periods-final}
961
962 The factorial estimates are stronger than needed for the optimal
963 first-difference law. A general Newton criterion gives the exact law
964 from the threshold-free bound and finitely many rational coefficients.
965 This route also yields all addresses at ranks two and eight, and an
966 infinite family at every larger dyadic rank.
967
968 \begin{lemma}[A first-difference criterion]
969 \label{lem:newton-exact-isometry-criterion}
970 Let $C:\mathbb Z_{\geq0}\to\mathbb Z_2$ have Newton coefficients
971 $a_k=\Delta_1^kC(0)$. Suppose for an integer $b\geq0$ that
972 \begin{equation}
973 v_2(a_1)=b,\qquad
974 v_2(a_k)\geq b+1+\lfloor\log_2k\rfloor\quad(k\geq2).
975 \label{eq:newton-isometry-criterion}
976 \end{equation}
977 Then for every $n\geq0$ and $H>0$,
978 \begin{equation}
979 v_2(C(n+H)-C(n))=b+v_2(H).
980 \label{eq:newton-isometry-conclusion}
981 \end{equation}
982 The Newton interpolation extends continuously to $\mathbb Z_2$ and
983 satisfies the same identity for any two distinct dyadic arguments.
984 \end{lemma}
985
986 \begin{proof}
987 At a natural argument the exact Newton expansion is finite.
988 Vandermonde gives
989 \[
990 \binom{n+H}{k}-\binom nk
991 =\sum_{i=1}^k\binom Hi\binom n{k-i}.
992 \]
993 For $1\leq i\leq H$,
994 $v_2(\binom Hi)\geq v_2(H)-v_2(i)
995 \geq v_2(H)-\lfloor\log_2k\rfloor$. Terms with $i>H$ vanish.
996 Thus each Newton summand with $k\geq2$ has valuation at least
997 $b+v_2(H)+1$. The $k=1$ summand is $Ha_1$, of exact valuation
998 $b+v_2(H)$, proving \eqref{eq:newton-isometry-conclusion}.
999
1000 The coefficient hypothesis implies $a_k\to0$ dyadically, so the
1001 Newton series converges uniformly on $\mathbb Z_2$, where
1002 $\binom xk$ is integral-valued. For distinct $x,y\in\mathbb Z_2$,
1003 approximate the pair by natural integer pairs with their difference
1004 valuation eventually equal to $v_2(x-y)$. The corresponding function
1005 differences have the fixed exact valuation $b+v_2(x-y)$.
1006 A limit of elements with that fixed valuation has the same valuation:
1007 the normalized residues are units, and $\mathbb Z_2^\times$ is closed.
1008 \end{proof}
1009
1010 \begin{theorem}[Exact four-address first differences]
1011 \label{thm:rank-four-exact-periods}
1012 For every $n\geq0$, $H>0$, and $j\in\{0,1,2,3\}$,
1013 \begin{align}
1014 v_2(C_j(n+H)-C_j(n))&=v_2(H)+b_j,
1015 &(b_0,b_1,b_2,b_3)&=(1,1,4,2),\label{eq:rank-four-normalized-isometry}\\
1016 v_2(E_{4(n+H)}^{(4,j)}-E_{4n}^{(4,j)})&=v_2(H)+B_j,
1017 &(B_0,B_1,B_2,B_3)&=(1,1,5,3).
1018 \label{eq:rank-four-original-isometry}
1019 \end{align}
1020 Consequently, the least positive row period modulo $2^r$ is exactly
1021 \begin{equation}
1022 P_j(r)=2^{\max(0,r-B_j)}\qquad(r\geq1).
1023 \label{eq:rank-four-least-period-final}
1024 \end{equation}
1025 The corresponding displacement in the original Euler subscript is
1026 $4P_j(r)$.
1027 \end{theorem}
1028
1029 \begin{proof}
1030 The actual reciprocal recurrence is
1031 \begin{equation}
1032 C_{N,j}(n)=-\sum_{q=1}^n\binom{Nn}{Nq}
1033 g_{N,j}(q)C_{N,j}(n-q),\qquad C_{N,j}(0)=1.
1034 \label{eq:exact-reciprocal-row-recurrence}
1035 \end{equation}
1036 For $N=4$, the four kernels are
1037 \[
1038 1,\qquad\frac1{4q+1},\qquad
1039 \frac1{(4q+1)(2q+1)},\qquad
1040 \frac3{(4q+1)(2q+1)(4q+3)}.
1041 \]
1042 The threshold-free theorem gives $v_2(a_k)\geq k$ in every address.
1043 The following exact values cover all orders where that bound does not
1044 already imply \eqref{eq:newton-isometry-criterion}:
1045 \begin{center}
1046 \begin{tabular}{c r l r}
1047 $j$&$k$&$a_k$&$v_2(a_k)$\\\hline
1048 0&1&$-2$&1\\
1049 0&2&$72$&3\\
1050 1&1&$-6/5$&1\\
1051 1&2&$184/45$&3\\
1052 2&1&$-16/15$&4\\
1053 2&2&$64/45$&6\\
1054 2&3&$-14848/1365$&9\\
1055 2&4&$71426048/69615$&13\\
1056 2&5&$-25509691392/85085$&16\\
1057 2&6&$149113198870528/765765$&22\\
1058 3&1&$-36/35$&2\\
1059 3&2&$256/231$&8\\
1060 3&3&$-65792/35035$&8\\
1061 3&4&$161644544/4849845$&15
1062 \end{tabular}
1063 \end{center}
1064 They are finite rational evaluations of
1065 \eqref{eq:exact-reciprocal-row-recurrence}, followed by
1066 $a_k=\sum_{h=0}^k(-1)^{k-h}\binom kh C_j(h)$.
1067 For addresses zero and one, $k\geq3$ is covered by the generic bound.
1068 For address two, $k\geq7$ is covered. For address three, $k\geq5$ is
1069 covered. In each range the elementary inequality is
1070 $k-\lfloor\log_2k\rfloor\geq b_j+1$.
1071 Lemma~\ref{lem:newton-exact-isometry-criterion} proves the normalized
1072 identity. Multiplication by $j!$ proves the original identity.
1073
1074 A row displacement $H$ preserves the original coefficient modulo
1075 $2^r$ at even one row exactly when $v_2(H)+B_j\geq r$.
1076 The least positive such integer is
1077 $2^{\max(0,r-B_j)}$, and the same displacement works at every row.
1078 This proves both optimality and the absence of a starting threshold.
1079 \end{proof}
1080
1081 \begin{corollary}[Exact image and residue enumeration]
1082 \label{cor:four-exact-images}
1083 Let $\widetilde C_j$ denote the dyadic Newton interpolation.
1084 The map
1085 \[
1086 T_j(x)=\frac{\widetilde C_j(x)-1}{2^{b_j}}
1087 \]
1088 is a bijective isometry $\mathbb Z_2\to\mathbb Z_2$, and
1089 \begin{equation}
1090 \widetilde C_j(\mathbb Z_2)=1+2^{b_j}\mathbb Z_2,
1091 \qquad j!\widetilde C_j(\mathbb Z_2)=j!+2^{B_j}\mathbb Z_2.
1092 \label{eq:four-interpolation-images-final}
1093 \end{equation}
1094 For $r>B_j$, one least period modulo $2^r$ visits every residue in
1095 $j!+2^{B_j}\mathbb Z/2^r\mathbb Z$ exactly once. For $r\leq B_j$,
1096 there is one residue and the least row period is one.
1097 \end{corollary}
1098
1099 \begin{proof}
1100 The exact difference identity gives
1101 $v_2(T_j(x)-T_j(y))=v_2(x-y)$ and $T_j(0)=0$.
1102 For every $s\geq1$, the map induces an injection of
1103 $\mathbb Z_2/2^s\mathbb Z_2$ into itself. It is a bijection because
1104 both sets have $2^s$ elements. These quotient bijections are compatible
1105 with reduction. Inverting them gives a compatible preimage system for
1106 every dyadic integer, proving surjectivity. Multiplication by $j!$
1107 changes the image ideal by exactly $v_2(j!)$.
1108 The finite quotient bijection is also precisely the residue enumeration
1109 in a least row period.
1110 \end{proof}
1111
1112 \subsection{An infinite exact family at all dyadic ranks}
1113
1114 \begin{theorem}[Lower-half exact laws]
1115 \label{thm:lower-half-all-dyadic-isometry}
1116 Let $N=2^a\geq4$ and $0\leq j<N/2$. Then for all $n\geq0$ and $H>0$,
1117 \begin{equation}
1118 v_2(C_{N,j}(n+H)-C_{N,j}(n))=1+v_2(H).
1119 \label{eq:lower-half-all-dyadic-exact}
1120 \end{equation}
1121 Its normalized interpolation is a bijection onto $1+2\mathbb Z_2$.
1122 For the original coefficients, the least row period modulo $2^r$ is
1123 $2^{\max(0,r-1-v_2(j!))}$.
1124 \end{theorem}
1125
1126 \begin{proof}
1127 Every slope $N/i$ in the source has valuation at least two, since
1128 $i\leq j<N/2$. Hence $g(1)\equiv1\pmod4$ and
1129 $g(2)\equiv1\pmod8$. The first two Newton coefficients from the
1130 reciprocal recurrence are
1131 \begin{equation}
1132 a_1=-g(1)-1,\qquad
1133 a_2=\binom{2N}{N}g(1)^2-g(2)+2g(1)+1.
1134 \label{eq:lower-half-initial-newton}
1135 \end{equation}
1136 The first has valuation exactly one. Also
1137 $\binom{2N}{N}\equiv6\pmod8$ for every $N=2^a\geq2$.
1138 To see this, write
1139 \[
1140 B_a=\binom{2^{a+1}}{2^a}
1141 =2\prod_{h=1}^{2^a-1}\left(1+\frac{2^a}{h}\right).
1142 \]
1143 The even-$h$ factors give $B_{a-1}$. The odd-$h$ factors are one
1144 modulo four for $a\geq2$. Since $B_{a-1}$ has valuation one,
1145 $B_a\equiv B_{a-1}\pmod8$, and $B_1=6$.
1146 Therefore $a_2\equiv6-1+2+1=0\pmod8$.
1147 The global Mahler bound covers all $k\geq3$, completing the criterion
1148 with $b=1$. The period and image deductions are the same as above.
1149 The proof also covers $N=2,j=0$, whose kernel is identically one.
1150 \end{proof}
1151
1152 \subsection{Complete exact laws at ranks two and eight}
1153
1154 \begin{proposition}[Small-rank complete isometry table]
1155 \label{prop:two-four-eight-isometry-table}
1156 Every address at $N\in\{2,4,8\}$ satisfies
1157 \[
1158 v_2(C_{N,j}(n+H)-C_{N,j}(n))=b_{N,j}+v_2(H).
1159 \]
1160 The following entries are ordered by $j=0,\ldots,N-1$:
1161 \begin{center}
1162 \begin{tabular}{c l l}
1163 $N$&normalized shifts $b_{N,j}$&original shifts $B_{N,j}=b_{N,j}+v_2(j!)$\\\hline
1164 2&$1,2$&$1,2$\\
1165 4&$1,1,4,2$&$1,1,5,3$\\
1166 8&$1,1,1,1,4,3,2,2$&$1,1,2,2,7,6,6,6$
1167 \end{tabular}
1168 \end{center}
1169 For each entry, the least original row period is
1170 $2^{\max(0,r-B_{N,j})}$, its Euler subscript displacement is $N$ times
1171 that period, and the exact-image and residue-enumeration conclusions
1172 hold with the corresponding shifts.
1173 \end{proposition}
1174
1175 \begin{proof}
1176 Rank four was proved above. At ranks two and eight the finite
1177 criterion is verified by the actual rational recurrence
1178 \eqref{eq:exact-reciprocal-row-recurrence}. The following valuation
1179 lists record every initial Newton coefficient needed in addition to
1180 the generic tail $v_2(a_k)\geq k$:
1181 \begin{center}
1182 \begin{tabular}{c c c l}
1183 $N$&$j$&$b_{N,j}$&$v_2(a_1),v_2(a_2),\ldots$ through the finite cutoff\\\hline
1184 2&0&1&$1,3$\\
1185 2&1&2&$2,5,9,11$\\
1186 8&0&1&$1,3$\\
1187 8&1&1&$1,3$\\
1188 8&2&1&$1,3$\\
1189 8&3&1&$1,3$\\
1190 8&4&4&$4,6,9,13,16,20$\\
1191 8&5&3&$3,6,10,13,17$\\
1192 8&6&2&$2,6,8,13$\\
1193 8&7&2&$2,11,8,15$
1194 \end{tabular}
1195 \end{center}
1196 The first entries can also be read directly from
1197 $a_1=-1-\binom{N+j}{j}^{-1}$. At rank eight they are respectively
1198 \[
1199 -2,\ -\frac{10}{9},\ -\frac{46}{45},\ -\frac{166}{165},\
1200 -\frac{496}{495},\ -\frac{1288}{1287},\
1201 -\frac{3004}{3003},\ -\frac{6436}{6435}.
1202 \]
1203 All later entries in the table are finite rational arithmetic, with
1204 full numerator and denominator records retained in the accompanying
1205 exact certificate \texttt{all\_rank\_extensions\_verified.json}.
1206 Every entry with $k\geq2$ meets
1207 $b_{N,j}+1+\lfloor\log_2k\rfloor$. Beyond the displayed cutoff, the
1208 inequality $k\geq b_{N,j}+1+\lfloor\log_2k\rfloor$ holds at every
1209 order. Thus the finite table and the proved infinite tail establish
1210 the full hypothesis of Lemma~\ref{lem:newton-exact-isometry-criterion}.
1211 This is a finite coefficient proof followed by a universal tail
1212 argument, rather than extrapolation from sampled rows or steps.
1213 \end{proof}
1214
1215 \begin{remark}[Failure of a universal upper-half exact law]
1216 The all-rank lower bound remains valid in every address. The exact
1217 first-difference law does not automatically extend to all upper-half
1218 addresses. Exact rational recurrence calculations give
1219 \begin{center}
1220 \begin{tabular}{c c c c}
1221 $(N,j)$&$v_2(C(1)-C(0))$&$v_2(C(2)-C(0))$&naive $b+v_2(2)$\\\hline
1222 $(16,9)$&5&7&6\\
1223 $(32,18)$&6&6&7\\
1224 $(32,19)$&5&7&6
1225 \end{tabular}
1226 \end{center}
1227 For example, at $(N,j)=(16,9)$ the first difference is
1228 $-2042976/2042975$, whereas the two-row difference is
1229 $-2215603929728/2215923048625$. These exact counterexamples refute
1230 the additional universal isometry proposal. They preserve the
1231 threshold-free lower bound and the original conjecture.
1232 \end{remark}
1233
1234 \section{Analytic interpolation and exact Taylor radii}
1235 \label{sec:analytic-radii-final}
1236
1237 The threshold-free Mahler theorem gives continuous interpolation.
1238 Analytic interpolation needs a stronger estimate, and the factorial
1239 bounds provide it directly. The ordinary Taylor radius is determined
1240 by the coefficient growth, including infinite exact subsequences.
1241
1242 \begin{theorem}[Taylor coefficient bounds and radii]
1243 \label{thm:four-taylor-radii}
1244 For $j=0,1,2,3$, the dyadic interpolation has an ordinary power series
1245 \[
1246 \widetilde C_j(x)=1+\sum_{\ell\geq1}d_{j,\ell}x^\ell,
1247 \qquad v_2(d_{j,\ell})\geq\ell+\beta_j.
1248 \]
1249 Its Taylor series converges for $|x|_2<2$.
1250 At addresses zero and one,
1251 \begin{equation}
1252 v_2(d_{j,\ell})=\ell\quad(\ell\geq1),
1253 \qquad R_0=R_1=2.
1254 \label{eq:zero-one-exact-taylor-radius}
1255 \end{equation}
1256 At address two,
1257 \begin{equation}
1258 v_2(d_{2,\ell})\geq
1259 \max\left(2\ell+\left\lfloor\frac\ell2\right\rfloor,
1260 \ell+3\right),
1261 \qquad R_2=2^{5/2}.
1262 \label{eq:two-exact-taylor-radius}
1263 \end{equation}
1264 The radius at address three is at least two. Its exact radius is not
1265 asserted here. The stated exact radii at addresses zero, one, and two
1266 hold at every center in $\mathbb Z_2$.
1267 \end{theorem}
1268
1269 \begin{proof}
1270 Put $a_m=\Delta_1^mC_j(0)$ and expand the falling factorial in the
1271 Newton series. The ordinary coefficients are
1272 \begin{equation}
1273 d_{j,\ell}=\sum_{m\geq\ell}\frac{a_m}{m!}s(m,\ell).
1274 \label{eq:newton-taylor-stirling}
1275 \end{equation}
1276 Since $s(m,\ell)$ is integral and
1277 $v_2(a_m/m!)\geq m+\beta_j$, the coefficient sum converges and has
1278 valuation at least $\ell+\beta_j$. The sum exchange is justified on
1279 $\mathbb Z_2$ by these uniform geometric coefficient bounds.
1280 The ordinary power series agrees there with the Newton interpolation
1281 and converges whenever $|x|_2<2$.
1282
1283 For $j=0,1$, the term $m=\ell$ in
1284 \eqref{eq:newton-taylor-stirling} has exact valuation $\ell$, since
1285 $s(\ell,\ell)=1$ and the Newton valuation is exact. Every later term
1286 has valuation at least $\ell+1$. Thus
1287 $v_2(d_{j,\ell})=\ell$, and the coefficient radius is exactly two.
1288 At a center $x_0\in\mathbb Z_2$, the coefficient of $y^\ell$ in the
1289 translated series is
1290 \begin{equation}
1291 \sum_{m\geq\ell}d_{j,m}\binom m\ell x_0^{m-\ell}.
1292 \label{eq:taylor-translation-coefficient}
1293 \end{equation}
1294 Its initial term again has exact valuation $\ell$, and all later
1295 terms have at least one more bit. The exact radius is therefore two
1296 at every integral center.
1297
1298 For address two, the stronger factorial filtration gives the lower
1299 bound in \eqref{eq:two-exact-taylor-radius}. At each
1300 $\ell=2^r\geq4$, Theorem~\ref{thm:two-exact-power-orders} makes the
1301 $m=\ell$ term in \eqref{eq:newton-taylor-stirling} have valuation
1302 $2\ell+\ell/2=5\ell/2$. Every $m>\ell$ term has strictly greater
1303 lower bound $2m+\lfloor m/2\rfloor$. Hence
1304 \[
1305 v_2(d_{2,\ell})=5\ell/2\qquad(\ell=2^r\geq4).
1306 \]
1307 The universal lower bound and this unbounded exact subsequence prove
1308 that the coefficient radius is exactly $2^{5/2}$. Translation by an
1309 integral center preserves the same unique leading coefficients at
1310 these degrees in \eqref{eq:taylor-translation-coefficient}, proving
1311 the exact radius at every such center.
1312 \end{proof}
1313
1314 \begin{remark}[Boundary interpretation]
1315 For addresses zero and one, at an argument of absolute value two every
1316 positive-degree Taylor term has valuation zero, so that ordinary Taylor
1317 series diverges. For address two, over a complete extension admitting
1318 arguments of absolute value $2^{5/2}$, the power-of-two-degree terms
1319 have valuation zero and fail to tend to zero. These statements concern
1320 the Taylor expansions just constructed. They do not prohibit a
1321 different analytic continuation. On $\mathbb Q_2$-valued arguments,
1322 open-disc notation must be read with its discrete value group. The
1323 coefficient radius is the precise invariant in the theorem.
1324 \end{remark}
1325
1326 \section{The final congruence in the original Euler convention}
1327 \label{sec:final-original-congruence}
1328
1329 The preceding bounds can be stated as one uniform congruence without
1330 suppressing the address normalization. This is the strengthened formula
1331 for the four-address family, valid from the first row.
1332
1333 \begin{corollary}[Final common-step congruence]
1334 \label{cor:final-euler-congruence}
1335 Let $j\in\{0,1,2,3\}$, $n\geq0$, and $k,H\geq1$. Define
1336 \[
1337 A_0(k)=A_1(k)=k,\qquad
1338 A_2(k)=\max\left(k+4,\ 2k+\left\lfloor\frac k2\right\rfloor+1\right),
1339 \qquad A_3(k)=k+2.
1340 \]
1341 Then
1342 \begin{equation}
1343 \boxed{\displaystyle
1344 \sum_{i=0}^k(-1)^{k-i}\binom ki E_{4(n+iH)}^{(4,j)}
1345 \equiv0\pmod{2^{\,v_2(k!)+k v_2(H)+A_j(k)}}.}
1346 \label{eq:boxed-final-euler-congruence}
1347 \end{equation}
1348 For $j=0,1$, this displayed exponent is the exact valuation for every
1349 $n,k,H$ in the stated range. For $j=2$, the exact positive
1350 power-of-two-order valuations are given by
1351 Theorem~\ref{thm:two-exact-power-orders}, with one added bit.
1352 \end{corollary}
1353
1354 \begin{proof}
1355 The left side is $j!\Delta_H^kC_j(n)$.
1356 At addresses zero and one, $v_2(j!)=0$ and the exact homogeneous
1357 formula gives the exponent. At address two, $v_2(2!)=1$. The maximum
1358 of \eqref{eq:factorial-four-lower} and
1359 \eqref{eq:factorial-four-j2-filtration} yields
1360 $\max(k+4,2k+\lfloor k/2\rfloor+1)$ after removing the common
1361 $v_2(k!)+k v_2(H)$. At address three, the one-bit normalized offset
1362 and $v_2(3!)=1$ give $k+2$.
1363 \end{proof}
1364
1365 For mixed steps the same corollary holds with $k v_2(H)$ replaced by
1366 $\sum_\nu v_2(H_\nu)$ and the binomial stencil replaced by
1367 $j!\Delta_{H_1}\cdots\Delta_{H_k}C_j(n)$.
1368 At addresses zero and one, exactness is then governed by the parity
1369 classification in Theorem~\ref{thm:zero-one-exact-mixed}.
1370 At first order these bounds are exact in every address, with original
1371 valuation offsets $(1,1,5,3)$. The least periods are those in
1372 \eqref{eq:rank-four-least-period-final}.
1373
1374 \subsection{Proof coverage and remaining questions}
1375
1376 The results in these sections are mathematical theorems with detailed
1377 proofs and preserved audit evidence. Their formal status must be read
1378 statement by statement. The recovered Paper~IX release already contains
1379 Lean results for the rank-four threshold-free homogeneous bound, the
1380 sharp first-difference laws, the least periods, and the dyadic isometry
1381 and image conclusions. Those inherited results should not be described
1382 as missing merely because the earlier short completion supplement did
1383 not include their later modules.
1384
1385 The all-rank dyadic and general prime-power transfer proof presented
1386 here, the factorial improvements, the mixed leading-parity
1387 classification, the Bernoulli power-order sharpness, the all-rank
1388 lower-half exact laws, the complete rank-eight extension, and the exact
1389 Taylor-radius arguments have not received new Lean formalization in
1390 this strengthening. A mathematical audit and an exact computation are
1391 neither a Lean compilation nor external peer review. The separate Lean
1392 guide records the recovered formal declarations and any fresh release
1393 checks without extending their coverage to these new claims.
1394
1395 The completed external statement is Nishibuchi's Conjecture~3.9 in
1396 its prescribed parameter family and coefficient normalization.
1397 These stronger congruences enlarge its dyadic branch and remove its
1398 starting threshold there. They do not settle every Euler problem.
1399 In particular, the following questions remain separate:
1400 \begin{itemize}
1401 \item a complete exact higher-difference law at all non-power-of-two
1402 orders in address two;
1403 \item complete higher-difference valuations and the exact ordinary
1404 Taylor radius in address three;
1405 \item a classification of the upper-half dyadic addresses having an
1406 exact first-difference metric law at arbitrary larger ranks;
1407 \item the distinct universal simplicity or nonvanishing question
1408 for the later Euler-tail residuals, and the original native
1409 Paper~IX conjectures beyond the completed external statement.
1410 \end{itemize}
1411 The exact counterexamples above rule out the naive universal upper-half
1412 isometry, and mixed cancellations rule out unconditional baseline
1413 sharpness for arbitrary stencils. No global historical-priority or
1414 novelty assertion is made for these strengthened statements.
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