Catalan Constant: source and evidence

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1 \chapter{The determinant that controls the approximants}
2 \label{cb_ch_wronskian}
3 
4 The recurrence produces two rational sequences. Its most useful internal
5 certificate is the determinant of two adjacent coefficient pairs. This
6 determinant can be evaluated without knowing Catalan's constant, without a
7 decimal approximation to it, and without any assumption about its arithmetic
8 nature. It proves that the two sequences remain independent and gives an
9 exact description of the movement of their quotient.
10 
11 \section{A determinant with a closed formula}
12 
13 Write the recurrence in the form
14 \begin{equation}\label{cb_eq_rec_short}
15  a_n w_{n+1}=q(n)w_n+c_nw_{n-1},\qquad n\geq1,
16 \end{equation}
17 where, in this chapter only,
18 \[
19  a_n=(2n+1)^2(2n+2)^2p(n),\qquad
20  c_n=(2n-1)^2(2n)^2p(n+1).
21 \]
22 These are recurrence coefficients, not the integer coefficients of a cleared
23 linear form used later. In particular, both are positive. The initial data are
24 \[
25  u_0=1,\quad u_1=\frac74,\qquad
26  v_0=0,\quad v_1=\frac{13}{8}.
27 \]
28 The polynomial $p(n)=20n^2-8n+1$ is positive for integer $n\geq0$.
29 Define the discrete Wronskian
30 \begin{equation}\label{cb_eq_W_def}
31  W_n=u_{n+1}v_n-u_nv_{n+1},\qquad n\geq0.
32 \end{equation}
33 This convention fixes its sign. Reversing the two terms reverses every sign
34 in the subsequent formulas.
35 
36 \begin{theorem}[Exact Wronskian]\label{cb_thm_W}
37 For every integer $n\geq0$,
38 \begin{equation}\label{cb_eq_W_closed}
39  W_n=\frac{(-1)^{n+1}p(n+1)}{2(2n+1)^2(2n+2)^2}.
40 \end{equation}
41 In particular, $W_n\ne0$.
42 \end{theorem}
43 
44 \begin{proof}
45 At $n=0$ the determinant is $-13/8$, which agrees with the displayed
46 formula because $p(1)=13$. For $n\geq1$, insert
47 \eqref{cb_eq_rec_short} into both terms of \eqref{cb_eq_W_def}. The terms
48 containing $q(n)u_nv_n$ cancel, leaving
49 \[
50  W_n=\frac{c_n}{a_n}(u_{n-1}v_n-u_nv_{n-1})
51      =-\frac{c_n}{a_n}W_{n-1}.
52 \]
53 Iterating this equality gives
54 \[
55  W_n=(-1)^nW_0
56  \prod_{j=1}^n
57  \frac{(2j-1)^2(2j)^2p(j+1)}{(2j+1)^2(2j+2)^2p(j)}.
58 \]
59 There are two independent telescopes. The polynomial factors leave
60 $p(n+1)/p(1)$. The integer factors leave
61 $4/((2n+1)^2(2n+2)^2)$. Substituting $W_0=-13/8$ and $p(1)=13$
62 proves the identity. The empty product gives the same calculation at $n=0$.
63 \end{proof}
64 
65 The proof is universal. A rational arithmetic check of its first 201
66 instances is an implementation check, not its logical foundation. The finite
67 check is still useful because it can detect a shifted index or a reversed
68 determinant convention. The first values make those conventions visible.
69 
70 \begin{center}
71 \begin{tabular}{@{}rrrrr@{}}
72 \toprule
73 $n$ & $0$ & $1$ & $2$ & $3$\\
74 \midrule
75 $W_n$ & $-13/8$ & $65/288$ & $-157/1800$ & $289/6272$\\
76 \bottomrule
77 \end{tabular}
78 \end{center}
79 
80 \begin{figure}[tbp]
81 \centering\includegraphics[width=0.92\linewidth]{../visual/figures/14_wronskian_identity.pdf}
82 \caption{The determinant recurrence loses the middle coefficient. Its
83 remaining product telescopes exactly. The figure illustrates the written
84 identity, which holds at every index.}\label{cb_fig_W_identity}
85 \end{figure}
86 
87 \section{Positive coefficients and decreasing steps}
88 
89 The next argument needs the positivity of $u_n$. It also needs a modest lower
90 bound on it, rather than a full asymptotic theorem. Both follow from the
91 recurrence. To see the relevant sign without relying on a plot, put $t=n-1$.
92 Direct polynomial expansion gives
93 \begin{align*}
94  q(n)-a_n={}&3200t^6+24000t^5+73392t^4+117072t^3\\
95             &+102768t^2+47112t+8827.
96 \end{align*}
97 Thus $q(n)>a_n>0$ for every $n\geq1$. Starting from $u_1>u_0>0$, the
98 recurrence implies
99 \[
100  u_{n+1}>\frac{q(n)}{a_n}u_n>u_n.
101 \]
102 In particular $u_n\geq1$. This elementary bound is enough for the convergence
103 argument below.
104 
105 Set
106 \begin{equation}\label{cb_eq_ratio_steps}
107  r_n=\frac{v_n}{u_n},\qquad
108  d_n=\frac{p(n+1)}{2(2n+1)^2(2n+2)^2u_nu_{n+1}}.
109 \end{equation}
110 The symbol $d_n$ denotes a positive rational step size. It is unrelated to
111 the least common multiples $L_N$ used in denominator clearing. Dividing the
112 Wronskian identity by $u_nu_{n+1}$ yields
113 \begin{equation}\label{cb_eq_alternating_steps}
114  r_{n+1}-r_n=(-1)^n d_n.
115 \end{equation}
116 The movement of the approximants therefore alternates exactly.
117 
118 \begin{lemma}[Strict decrease of the step sizes]\label{cb_lem_d_decrease}
119 The rational numbers $d_n$ are strictly positive, decrease strictly, and tend
120 to zero.
121 \end{lemma}
122 
123 \begin{proof}
124 Positivity is immediate. For $n\geq1$, the determinant recurrence gives
125 \[
126  \frac{d_n}{d_{n-1}}
127  =\frac{c_n}{a_n}\frac{u_{n-1}}{u_{n+1}}<1.
128 \]
129 The strict inequality follows from
130 $a_nu_{n+1}=q(n)u_n+c_nu_{n-1}>c_nu_{n-1}$.
131 Finally $u_nu_{n+1}\geq1$, so
132 \[
133  0<d_n\leq
134  \frac{p(n+1)}{2(2n+1)^2(2n+2)^2}\longrightarrow0.
135 \]
136 No limiting characteristic root is needed for this conclusion.
137 \end{proof}
138 
139 This is a useful example of the distinction between two arguments with
140 different jobs. The exponential rates from Zudilin's theorem describe the
141 quality of large approximants. The simple inequalities above establish
142 convergence and strict brackets directly from rational recurrence data.
143 
144 \section{The limit and its exact rational brackets}
145 
146 \begin{theorem}[Alternating brackets]\label{cb_thm_brackets}
147 There is a real number $L$ such that $r_n\to L$. For every $k\geq0$,
148 \begin{equation}\label{cb_eq_bracket}
149  r_{2k}<L<r_{2k+1}.
150 \end{equation}
151 The even subsequence increases strictly and the odd subsequence decreases
152 strictly. For every $n\geq0$,
153 \begin{equation}\label{cb_eq_error_bounds}
154  d_n-d_{n+1}<|L-r_n|<d_n,
155  \qquad \operatorname{sign}(L-r_n)=(-1)^n.
156 \end{equation}
157 \end{theorem}
158 
159 \begin{proof}
160 Equation \eqref{cb_eq_alternating_steps} and the preceding lemma identify
161 $r_n$ with partial sums of a convergent alternating series:
162 \[
163  r_n=\sum_{j=0}^{n-1}(-1)^j d_j,
164  \qquad L=\sum_{j=0}^{\infty}(-1)^j d_j.
165 \]
166 For the even subsequence,
167 $r_{2k+2}-r_{2k}=d_{2k}-d_{2k+1}>0$.
168 For the odd subsequence,
169 $r_{2k+3}-r_{2k+1}=-d_{2k+1}+d_{2k+2}<0$.
170 Their adjacent separation $d_{2k}$ tends to zero, so they have the same limit.
171 Strictness of both subsequences places that limit strictly between every
172 displayed pair. The alternating tail beginning with $d_n$ gives the sign
173 and upper bound in \eqref{cb_eq_error_bounds}. Pairing its first two terms
174 and then successive pairs gives the strict lower bound. All omitted pairs
175 are positive after multiplication by $(-1)^n$.
176 \end{proof}
177 
178 The endpoints in \eqref{cb_eq_bracket} are exact rationals. For example,
179 the first few approximants are
180 \begin{center}
181 \begin{tabular}{@{}rl@{}}
182 \toprule
183 $n$ & $r_n$\\
184 \midrule
185 $0$ & $0$\\
186 $1$ & $13/14$\\
187 $2$ & $10699/11682$\\
188 $3$ & $8025653/8761950$\\
189 \bottomrule
190 \end{tabular}
191 \end{center}
192 There is no rounding decision in these brackets. Increasing the index
193 nests a narrower interval inside the earlier one. A program can use the
194 same determinant to check the interval width exactly.
195 
196 The recurrence and its initial values establish a limit $L$. The statement
197 that this limit is $G$ is a separate mathematical identification. We use
198 Zudilin's Theorem 1 for that identification, with the normalization recorded
199 in source contract LC01 \cite{Zudilin2003}. After this external input,
200 \eqref{cb_eq_bracket} and \eqref{cb_eq_error_bounds} become exact bounds for
201 Catalan's constant and its rational approximants. In particular,
202 \begin{equation}\label{cb_eq_linear_error_sign}
203  (-1)^n(u_nG-v_n)>0\qquad(n\geq0).
204 \end{equation}
205 
206 This order of reasoning prevents a circular shortcut. Decimal agreement
207 between $r_n$ and a familiar constant does not identify a limit. Nor does
208 the nonzero Wronskian alone exclude equality between one particular
209 approximant and an arbitrary real number. The strict brackets supply
210 nonvanishing for the recurrence limit, and the source theorem then identifies
211 that limit with $G$.
212 
213 \section{What this certificate does and does not settle}
214 
215 The determinant settles several questions completely. Adjacent coefficient
216 pairs never become proportional. The ratio has a unique limit. Its even and
217 odd approximants bracket that limit, and their signs of error are fixed.
218 Each of these conclusions has a written argument valid at every index.
219 
220 The determinant does not classify the limit as rational or irrational. A
221 sequence of distinct rational numbers with alternating errors can converge
222 to a rational number. Irrationality requires an additional comparison
223 between error size and the integer lattice after clearing denominators.
224 The Wronskian contributes exact nonvanishing and useful quantitative bounds,
225 but it does not pay the arithmetic cost of that clearing. The next chapters
226 keep these two requirements visible in the same coefficient construction.
227 
228 \chapter{Integer coefficients and the corrected integral}
229 \label{cb_ch_integrality}
230 
231 The rational recurrence is an efficient way to construct coefficients. It
232 does not by itself explain their denominators. A second description, in
233 terms of half integer binomial coefficients, gives a short universal
234 integrality proof. A third description, as a positive integral, supplies the
235 analytic error. Matching these descriptions requires care with constants.
236 In the source used here, one printed integral factor fails even at the
237 initial index. We record that failure and derive the factor used throughout
238 this edition.
239 
240 \section{A coefficient identity, rather than an irrationality assumption}
241 
242 Krattenthaler and Rivoal use coefficients $a_n,b_n$ for a linear form in
243 Catalan's constant \cite{KR2008}. These letters belong to their source and
244 do not denote the recurrence coefficients of the previous chapter. In the
245 normalization of this book their coefficients satisfy
246 \begin{equation}\label{cb_eq_KR_dictionary}
247  a_n=-4u_n,\qquad b_n=-4v_n.
248 \end{equation}
249 The notation bridge is recorded in source contract LC05. After shifting the
250 source summation variable by one, their hypergeometric series is $-1/2$
251 times Zudilin's series. The source partial fraction expansion identifies
252 the two rational coefficients separately, giving
253 \eqref{cb_eq_KR_dictionary}.
254 
255 It would be invalid to obtain this dictionary merely by equating two real
256 expressions in $G$ and then treating $1$ and $G$ as linearly independent over
257 $\Q$. That independence is exactly the irrationality under investigation.
258 Instead, take a formal variable $X$. The partial fraction coefficient
259 calculation identifies the polynomial $a_nX-b_n$ with
260 $-4(u_nX-v_n)$ in $\Q[X]$. Equality of coefficients in a polynomial ring
261 does not depend on the arithmetic nature of any later substitution for
262 $X$. This distinction also applies to the packet coefficients below.
263 
264 The same source gives the explicit finite sum
265 \begin{equation}\label{cb_eq_halfbinomial}
266  u_n=\sum_{j=0}^n
267  \binom nj\binom{n-\tfrac12}{j}\binom{n+j-\tfrac12}{j},\qquad n\geq0.
268 \end{equation}
269 Here
270 \[
271  \binom{x}{0}=1,\qquad
272  \binom{x}{j}=\frac{x(x-1)\cdots(x-j+1)}{j!}\quad(j\geq1)
273 \]
274 defines the generalized binomial coefficient. Formula
275 \eqref{cb_eq_halfbinomial} is Lemma 3, equation (3.1), after the coefficient
276 normalization above. Its identification with the recurrence is an external
277 source input. The denominator argument that follows is elementary and is
278 written in full.
279 
280 \section{The exact integer sequence}
281 
282 Define
283 \begin{equation}\label{cb_eq_T_def}
284  T_n=16^nu_n.
285 \end{equation}
286 The importance of the power $16^n$ is that it clears each summand of
287 \eqref{cb_eq_halfbinomial}, rather than relying on cancellation between
288 different summands.
289 
290 \begin{lemma}[Half binomial denominators]\label{cb_lem_half_denom}
291 For integers $h,j\geq0$, the number
292 $4^j\binom{h-\tfrac12}{j}$ is an integer.
293 \end{lemma}
294 
295 \begin{proof}
296 Work with formal power series, so no convergence issue arises. The central
297 binomial expansion gives
298 \[
299  (1+4z)^{-1/2}=\sum_{k\geq0}(-1)^k\binom{2k}{k}z^k.
300 \]
301 Every coefficient is an integer. Multiplying by the integer polynomial
302 $(1+4z)^h$ preserves this property. The coefficient of $z^j$ in the product
303 is $4^j\binom{h-\tfrac12}{j}$. More explicitly it is
304 \[
305  \sum_{\ell=0}^{\min(h,j)}
306  \binom h\ell 4^\ell(-1)^{j-\ell}
307  \binom{2(j-\ell)}{j-\ell},
308 \]
309 an integer sum.
310 \end{proof}
311 
312 \begin{theorem}[Universal integrality of $T_n$]\label{cb_thm_T_integer}
313 For every integer $n\geq0$, $T_n$ is a positive integer. In fact, after
314 multiplication by $16^n$, each summand in
315 \eqref{cb_eq_halfbinomial} is an integer.
316 \end{theorem}
317 
318 \begin{proof}
319 For $0\leq j\leq n$, apply the preceding lemma with $h=n$ and $h=n+j$.
320 The scaled summand is
321 \[
322  16^{n-j}\binom nj
323  \left(4^j\binom{n-\tfrac12}{j}\right)
324  \left(4^j\binom{n+j-\tfrac12}{j}\right).
325 \]
326 All four factors are integers. Summing proves integrality. Every original
327 summand is positive: the generalized binomial factors have only positive
328 factors in their defining products when $j\leq n$. The $j=0$ term is one,
329 including at $n=0$. Thus the sum is positive.
330 \end{proof}
331 
332 For example, the $n=2$ sum is
333 \[
334  u_2=1+\frac{15}{2}+\frac{105}{64}=\frac{649}{64}.
335 \]
336 Multiplication by $16^2$ gives the three integers $256$, $1920$, and $420$,
337 whose sum is $T_2=2596$. This illustrates the proof at the level of individual
338 terms. The first sequence values are
339 \[
340  T_0=1,\quad T_1=28,\quad T_2=2596,\quad
341  T_3=311536,\quad T_4=41759524.
342 \]
343 
344 The source theorem $16^na_n\in\Z$, combined with
345 \eqref{cb_eq_KR_dictionary}, would establish only $4T_n\in\Z$ directly.
346 Division by four is not an integrality argument. The half binomial proof
347 above establishes the stronger statement actually needed by the integer
348 packet construction. It removes a possible hidden dependence on a finite
349 exact division experiment.
350 
351 The recurrence for $T_n$ is obtained by multiplying the original recurrence
352 by $16^{n+1}$:
353 \begin{equation}\label{cb_eq_T_recurrence}
354  a_nT_{n+1}=16q(n)T_n+256c_nT_{n-1}\qquad(n\geq1).
355 \end{equation}
356 A program can therefore generate $T_n$ using integer numerators and exact
357 division by $a_n$. A nonzero remainder is a failed implementation check and
358 must terminate the computation. The written theorem establishes why exact
359 division succeeds universally for the correctly normalized recurrence.
360 
361 \begin{figure}[tbp]
362 \centering\includegraphics[width=0.93\linewidth]{../visual/figures/15_scaled_coefficients.pdf}
363 \caption{The recurrence coefficients become an integer sequence after
364 scaling by $16^n$. The finite values shown here accompany a universal
365 half binomial denominator proof.}\label{cb_fig_T}
366 \end{figure}
367 
368 \section{The rational coefficient has a different clearing cost}
369 
370 Put
371 \[
372  L_N=\operatorname{lcm}(1,2,\ldots,N),\qquad L_0=1.
373 \]
374 The source's least common multiple notation is translated to $L_N$ here.
375 The letter $D$ will later denote an actual reduced denominator, which can
376 be much smaller than this universal clearing bound.
377 Krattenthaler and Rivoal's Theorem 2 gives
378 $16^nL_{2n}^2b_n\in\Z$ for positive integer $n$
379 \cite{KR2008}. In the present normalization it implies
380 \begin{equation}\label{cb_eq_v_sharp_clear}
381  4\cdot16^nL_{2n}^2v_n\in\Z\qquad(n\geq0).
382 \end{equation}
383 The case $n=0$ is immediate from $v_0=0$. In particular the historical
384 script's weaker clearing factor, with $8$ in place of $4$, is valid at every
385 index. This is a source theorem about the rational coefficient. The
386 half binomial argument for $u_n$ does not automatically prove it.
387 
388 Earlier asymptotic denominator bounds in Zudilin's remarks establish the
389 appropriate exponential scale with a logarithmic correction in the exponent
390 \cite{Zudilin2002}. They serve a different purpose from exact clearing at
391 every index. An asymptotic expression involving $o(n)$ should not be read
392 as a precise power of two that can be inserted into a finite computation.
393 For the packet arithmetic we use \eqref{cb_eq_v_sharp_clear} and the exact
394 integer theorem for $T_n$.
395 
396 The finite direct product check in this release compares
397 \eqref{cb_eq_halfbinomial} with a separately indexed rational recurrence for
398 every $n=0,\ldots,50$. It checks 1326 summands, including the two half
399 binomial denominator statements and scaled integrality of each term. The
400 independent recurrence check extends through $n=500$ and verifies the weaker
401 factor $8$ clearing there. These are useful finite corroborations. The
402 external identity and the two universal arithmetic arguments have their
403 own evidence and remain distinct from those checks.
404 
405 \section{A printed factor that fails at the initial index}
406 
407 Consider the positive double integral
408 \begin{equation}\label{cb_eq_J_def}
409  J_n=\int_0^1\!\int_0^1
410  \frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}}
411  {(1-xy)^{n+1}}\,\dd x\,\dd y,
412  \qquad n\geq0.
413 \end{equation}
414 Zudilin's Theorem 3 prints $u_nG-v_n=(-1)^nJ_n/4$ in both the authenticated
415 arXiv version and the published journal PDF \cite{Zudilin2003}. With
416 $u_0=1$ and $v_0=0$, this would force $J_0=4G$.
417 The integral itself gives $J_0=8G$. Thus that printed factor cannot be used
418 with these initial conditions. This edition uses
419 \begin{equation}\label{cb_eq_J_correct}
420  J_n=8(-1)^n(u_nG-v_n).
421 \end{equation}
422 No erratum was recovered in the bounded source search. The correction is
423 supported by the explicit derivation below, rather than an assertion that
424 a published correction exists.
425 
426 The initial integral supplies a short diagnostic. Substitute $x=t^2$ and
427 $y=1-s^2$. Then
428 \begin{align*}
429  J_0
430  &=4\int_0^1\!\int_0^1\frac{\dd s\,\dd t}{1-t^2+t^2s^2}\\
431  &=4\int_0^1\frac{\arcsin t}{t\sqrt{1-t^2}}\,\dd t
432   =4\int_0^{\pi/2}\theta\csc\theta\,\dd\theta.
433 \end{align*}
434 Since an antiderivative of $\csc\theta$ is $\log\tan(\theta/2)$, integration
435 by parts yields
436 \[
437  J_0=-4\int_0^{\pi/2}\log\tan(\theta/2)\,\dd\theta
438      =-8\int_0^{\pi/4}\log\tan z\,\dd z=8G.
439 \]
440 Both boundary terms vanish. At zero this uses
441 $\theta\log\tan(\theta/2)\to0$, and at $\pi/2$ the logarithm is zero.
442 
443 Checking one index detects the wrong normalization. It does not by itself
444 prove \eqref{cb_eq_J_correct} for every $n$. The next calculation explains
445 the all index extension and identifies its external transformation input.
446 
447 \section{The all index normalization}
448 
449 Expanding $(1-xy)^{-n-1}$ as its nonnegative binomial series and applying
450 Tonelli's theorem gives
451 \begin{equation}\label{cb_eq_beta_H}
452  J_n=C_nH_n,
453 \end{equation}
454 where
455 \[
456  C_n=\frac{\Gamma(n+1)^2\Gamma(n+\tfrac12)^2}
457  {\Gamma(2n+\tfrac32)^2},
458 \]
459 and
460 \[
461  H_n={}_3F_2\!\left(
462  \begin{matrix}n+1,n+\tfrac12,n+1\\
463  2n+\tfrac32,2n+\tfrac32\end{matrix};1\right).
464 \]
465 Indeed, the two beta integrals in the term with index $k$ are
466 $B(n+\tfrac12+k,n+1)$ and $B(n+1+k,n+\tfrac12)$.
467 Their ratios to the corresponding $k=0$ integrals give exactly the displayed
468 hypergeometric parameters. The excess of the lower parameters over the
469 upper parameters is $n+1/2>0$, so the series converges for every $n\geq0$.
470 
471 Let $E_n$ denote the well poised ${}_6F_5$ series at $-1$ in Zudilin's
472 displayed Whipple transformation, with
473 $a=3n+1$, $b=c=d=n+1/2$, and $e=n+1$. The transformation states
474 \[
475  H_n=\frac{\Gamma(3n+2)\Gamma(n+\tfrac12)}
476  {\Gamma(2n+\tfrac32)\Gamma(2n+1)}E_n.
477 \]
478 Its condition $\operatorname{Re}(1+a-d-e)>0$ becomes $n+1/2>0$ here.
479 The displayed prefactor of the source hypergeometric form $F_n^{\rm src}$
480 is
481 \[
482  F_n^{\rm src}=(-1)^n
483  \frac{\Gamma(n+1)^2\Gamma(3n+2)\Gamma(n+\tfrac12)^3}
484  {\Gamma(2n+\tfrac32)^3\Gamma(2n+1)}E_n.
485 \]
486 Substitution in \eqref{cb_eq_beta_H} therefore gives
487 $F_n^{\rm src}=(-1)^nJ_n$. There is no additional factor two in the gamma
488 cancellation. The formal partial fraction coefficient identification in
489 the same source gives $F_n^{\rm src}=8(u_nG-v_n)$, with the recurrence
490 initial data used here. Combining these two identities proves
491 \eqref{cb_eq_J_correct} at every index.
492 
493 The beta integral expansion and prefactor cancellation are written
494 calculations. The Whipple transformation and the partial fraction
495 identification are external mathematical inputs from the authenticated
496 paper. Their precise roles are recorded in LC03 and the accompanying
497 normalization audit. The independent initial integral agrees with the
498 result, and Eskandari's coefficient normalization uses the same factor $8$
499 \cite{Eskandari2026}.
500 
501 \section{Three descriptions, three different guarantees}
502 
503 The recurrence permits efficient exact computation. The half binomial
504 formula proves that $16^nu_n$ is always an integer. The integral describes
505 the error and makes its sign transparent after the corrected normalization.
506 These descriptions agree, but each answers a different question. Keeping
507 their roles distinct prevents a finite arithmetic check from being mistaken
508 for an analytic theorem, or a positive integral from being mistaken for
509 denominator control. The packet construction will use all three.
510 
511 \chapter{The binomial packet and the integral filter}
512 \label{cb_ch_packets}
513 
514 The packet combines a finite block of recurrence coefficients. Its integer
515 part is a positive binomial sum, while its error part is a signed integral.
516 These two facts coexist because the signs in the original error already
517 alternate. The packet removes those coefficient signs and replaces them
518 with a polynomial filter inside the integral. That filter is useful for
519 approximation, but its changing sign imposes a real restriction on any
520 nonvanishing argument.
521 
522 \section{A finite transform with an exact additive evolution}
523 
524 For integers $n,m\geq0$, define
525 \begin{align}
526  S_{n,m}&=\sum_{i=0}^m\binom miT_{n+i},\label{cb_eq_S_def}\\
527  H_{n,m}&=\sum_{i=0}^m\binom mi16^{n+i}v_{n+i}.
528  \label{cb_eq_H_packet_def}
529 \end{align}
530 The two index symbol $H_{n,m}$ denotes a rational packet coefficient. It is
531 distinct from the hypergeometric function $H_n$ used in the integral
532 normalization derivation. The theorem on $T_n$ proves
533 $S_{n,m}\in\Z_{>0}$ at every pair. The coefficient $H_{n,m}$ is rational
534 and requires separate denominator clearing.
535 
536 \begin{proposition}[Pascal evolution]\label{cb_prop_Pascal}
537 For $n,m\geq0$,
538 \begin{equation}\label{cb_eq_Pascal}
539  S_{n,m+1}=S_{n,m}+S_{n+1,m}.
540 \end{equation}
541 The same identity holds with $H$ in place of $S$.
542 \end{proposition}
543 
544 \begin{proof}
545 Use $\binom{m+1}{i}=\binom mi+\binom m{i-1}$, with out of range binomial
546 coefficients interpreted as zero. The first contribution is $S_{n,m}$.
547 Changing $i$ to $i+1$ in the second contribution gives $S_{n+1,m}$. The
548 argument uses only the finite weights, so it applies to either seed
549 sequence.
550 \end{proof}
551 
552 At $m=0$, the packet is just $S_{n,0}=T_n$. The first two nontrivial rows
553 are
554 \[
555  S_{n,1}=T_n+T_{n+1},\qquad
556  S_{n,2}=T_n+2T_{n+1}+T_{n+2}.
557 \]
558 For example $S_{1,1}=28+2596=2624$. The Pascal identity produces exactly
559 the same integer as direct evaluation of its defining sum. This agreement
560 between an additive row evolution and a direct weighted sum gives two
561 different implementations of the finite transform.
562 
563 If $E$ denotes the shift operator $(Ef)_n=f_{n+1}$, the packet is
564 $(1+E)^mT$ at index $n$. This is an identity about a finite polynomial in
565 one shift. It does not introduce a further assumption about recurrence
566 solutions or convergence. Reduction modulo a power of two commutes with
567 this polynomial, which is the basis of the valuation computation in the
568 next chapter.
569 
570 \begin{figure}[tbp]
571 \centering\includegraphics[width=0.91\linewidth]{../visual/figures/16_pascal_transform.pdf}
572 \caption{Each new packet row is the sum of adjacent entries in the
573 preceding row. This exact identity is also valid after modular reduction.}
574 \label{cb_fig_Pascal}
575 \end{figure}
576 
577 \section{The source filter and the raw form}
578 
579 Eskandari's preprint applies the filter
580 \begin{equation}\label{cb_eq_Jtilde}
581  \widetilde J_{n,m}=\sum_{i=0}^m(-16)^i\binom miJ_{n+i}
582  \qquad(n,m\geq0)
583 \end{equation}
584 to the integral family \cite{Eskandari2026}. The filter and its approximation
585 analysis are external work. This book examines its coefficient arithmetic
586 and the finite evidence recovered from the project's script.
587 
588 Substitute the corrected normalization
589 $J_k=8(-1)^k(u_kG-v_k)$ into \eqref{cb_eq_Jtilde}. The two factors
590 $(-1)^i$ cancel. Thus
591 \begin{equation}\label{cb_eq_Jtilde_coeffs}
592  \widetilde J_{n,m}=8(-1)^n16^{-n}(S_{n,m}G-H_{n,m}).
593 \end{equation}
594 Define the raw packet form
595 \begin{equation}\label{cb_eq_raw_F}
596  F_{n,m}=16^n\widetilde J_{n,m}
597         =8(-1)^n(S_{n,m}G-H_{n,m}).
598 \end{equation}
599 The adjective raw means that the common denominator of $H_{n,m}$ has not
600 yet been cleared. It does not mean that $F_{n,m}$ is an integer linear form.
601 The coefficient of $G$ is an integer, but the other coefficient may be
602 rational.
603 
604 This normalization is important for comparison with historical data.
605 Multiplying or dividing a linear form by a growing power changes its
606 displayed decay rate. The integral filter is $\widetilde J_{n,m}$, while
607 the recorded arithmetic form $F_{n,m}$ is $16^n$ times it. Their signs and
608 zeros agree because $16^n>0$. Their logarithmic size per unit $n$ differs by
609 $\log16$.
610 
611 \section{The measure and its sign changing polynomial}
612 
613 On the open unit square, set
614 \begin{equation}\label{cb_eq_psi_mu}
615  \psi(x,y)=\frac{xy(1-x)(1-y)}{1-xy},\qquad
616  \dd\mu=\frac{\dd x\,\dd y}{\sqrt{x(1-y)}(1-xy)}.
617 \end{equation}
618 The factor $1-xy$ is outside the square root. Moving that factor inside, as in
619 \[
620  \frac{\dd x\,\dd y}{\sqrt{x(1-y)(1-xy)}},
621 \]
622 would represent a different integral. With the measure in \eqref{cb_eq_psi_mu},
623 \[
624  J_n=\int_{(0,1)^2}\psi^n\,\dd\mu,
625  \qquad
626  \widetilde J_{n,m}=\int_{(0,1)^2}\psi^n(1-16\psi)^m\,\dd\mu.
627 \]
628 The second identity follows by the ordinary binomial theorem inside a
629 finite sum. The total measure is $J_0=8G<\infty$.
630 
631 The kernel has a simple maximum that also explains why the filter changes
632 sign. Put $r=\sqrt{xy}$. Since $x+y\geq2r$,
633 \[
634  \psi(x,y)\leq\frac{r^2(1-r)^2}{1-r^2}
635               =\frac{r^2(1-r)}{1+r},\qquad0<r<1.
636 \]
637 Equality is achieved when $x=y=r$. The derivative of the last expression
638 is
639 \[
640  \frac{2r(1-r-r^2)}{(1+r)^2}.
641 \]
642 It has its interior maximum at $r=\phi^{-1}$, where
643 $\phi=(1+\sqrt5)/2$. Since $1-r=r^2$ there, the maximum is
644 \begin{equation}\label{cb_eq_psi_max}
645  \max\psi=\phi^{-5}\approx0.09017.
646 \end{equation}
647 This is greater than $1/16$. One can already see a negative filter value
648 at $x=y=1/2$, where $\psi=1/12$ and $1-16\psi=-1/3$.
649 Near the sides of the square $\psi$ is small and the filter is positive.
650 
651 \begin{theorem}[Nonvanishing for even filter order]\label{cb_thm_even_positive}
652 For every $n\geq0$ and every even integer $m\geq0$,
653 $\widetilde J_{n,m}>0$ and $F_{n,m}>0$.
654 \end{theorem}
655 
656 \begin{proof}
657 For even $m$, the integrand $\psi^n(1-16\psi)^m$ is nonnegative. It is
658 strictly positive on an open subset of the square with $0<\psi<1/16$.
659 The measure has positive density there. Hence its integral is strictly
660 positive. It is finite because $\psi$ and $1-16\psi$ are bounded and
661 $\mu$ has finite total mass. Multiplication by $16^n$ preserves positivity.
662 \end{proof}
663 
664 This theorem includes $m=0$ and $n=0$. It does not require that $G$ be
665 irrational. Indeed, positivity is one of the ingredients that an eventual
666 irrationality criterion would combine with sufficiently small cleared
667 integer forms. The even order restriction is easy to impose along an
668 asymptotic slope by choosing $m(n)$ to be the nearest suitable even integer.
669 The change in $m/n$ then tends to zero.
670 
671 \section{The odd sign hypothesis failed}
672 
673 For odd $m$, the polynomial factor changes sign, so the preceding positivity
674 proof does not apply. A proposed shortcut was the sign pattern
675 $\operatorname{sign}F_{n,m}=(-1)^m$. The finite audit falsified it. The first
676 counterexample is the pair $(n,m)=(1,1)$.
677 
678 Using $u_1=7/4$, $v_1=13/8$, $u_2=649/64$ and $v_2=10699/1152$, one obtains
679 \[
680  S_{1,1}=2624,\qquad H_{1,1}=\frac{21632}{9}.
681 \]
682 Consequently
683 \begin{equation}\label{cb_eq_odd_counterexample}
684  F_{1,1}=\frac{173056}{9}-20992G>0.
685 \end{equation}
686 The inequality is certified by the exact rational enclosure of $G$ recorded
687 in the independent computation. The threshold for it is $G<338/369$.
688 The enclosure lies strictly below that rational number. Thus this form is
689 positive although $m$ is odd, contradicting the proposed pattern.
690 The sign is decided by rational interval endpoints, rather than floating
691 point subtraction between nearly equal numbers.
692 
693 The audit checks 860 pairs with $1\leq n\leq20$ and $0\leq m\leq4n$.
694 Every tested form is certified nonzero, but a number of odd pairs violate
695 the proposed pattern. Both positive and negative odd order forms occur in
696 that finite domain. For example the recorded signs at $(1,1)$ and $(2,1)$
697 are opposite. These observations do not establish nonvanishing for every
698 odd pair. This edition retains the failed hypothesis and its diagnostic
699 record, and uses the universal even order theorem when nonvanishing is
700 needed in a global implication.
701 
702 \begin{figure}[tbp]
703 \centering\includegraphics[width=0.91\linewidth]{../visual/figures/27_filter_signs.pdf}
704 \caption{Certified signs in a bounded packet domain. Even filter orders
705 have a universal positivity proof. Odd filter orders display both signs,
706 and the proposed alternating rule fails.}\label{cb_fig_filter_signs}
707 \end{figure}
708 
709 \section{The exact historical integer pair}
710 
711 Let $N=n+m$ and put
712 \begin{equation}\label{cb_eq_C_clear}
713  C_{n,m}=L_{2N}^2.
714 \end{equation}
715 The pair produced by the historical script is
716 \begin{align}
717  A_{n,m}&=8(-1)^nC_{n,m}S_{n,m},\label{cb_eq_historical_A}\\
718  B_{n,m}&=8(-1)^nC_{n,m}H_{n,m}.
719  \label{cb_eq_historical_B}
720 \end{align}
721 Both coefficients are integers. Integrality of $A$ follows from that of
722 $S$. For $B$, each index $k=n+i$ satisfies $k\leq N$, so $L_{2k}$ divides
723 $L_{2N}$. The sharp source clearing
724 $4\cdot16^kL_{2k}^2v_k\in\Z$ therefore proves integrality term by term
725 after multiplication by $8L_{2N}^2$. The binomial weights introduce no
726 further denominators.
727 
728 The cleared linear form is exactly
729 \begin{equation}\label{cb_eq_cleared_integral}
730  A_{n,m}G-B_{n,m}
731  =C_{n,m}F_{n,m}
732  =16^nL_{2N}^2\widetilde J_{n,m}.
733 \end{equation}
734 Every factor in this identity has a fixed meaning. In particular, the
735 historical pair includes the constant factor $8$ and uses the possibly
736 redundant common clearing $L_{2N}^2$. It is not automatically a primitive
737 integer pair. Dividing out its common content is a separate arithmetic
738 operation examined later.
739 
740 At $m=0$ the construction reduces to the cleared original recurrence form.
741 At $n=0$ all identities remain valid with $T_0=1$, $v_0=0$, and $L_0=1$.
742 The large valuation scan starts at $n=1$ for its stated triangular domain,
743 but that computational choice does not restrict the algebraic identities
744 proved in this chapter.
745 
746 \chapter{What the modular valuation scan proves}
747 \label{cb_ch_modular}
748 
749 The packet coefficients grow rapidly, but their divisibility by two can be
750 decided from a sufficiently precise residue. The large scan uses that fact
751 to evaluate a finite domain efficiently. Its exactness comes from a simple
752 valuation lemma and from the absence of zero residues. Its limitation comes
753 from the finite domain. These are separate issues: modular arithmetic can
754 give an exact answer at millions of pairs without producing a theorem at
755 every pair.
756 
757 \section{The residue lemma}
758 
759 For a nonzero integer $z$, let $v_2(z)$ be the largest nonnegative integer
760 $t$ such that $2^t$ divides $z$. Thus $v_2(12)=2$ and $v_2(1)=0$.
761 The value $v_2(0)$ is infinite rather than a finite integer. Our packet
762 $S_{n,m}$ is strictly positive, so its valuation is always finite, even when
763 a finite precision residue is zero.
764 
765 \begin{lemma}[Exactness below the precision ceiling]\label{cb_lem_mod_v2}
766 Let $K\geq1$, let $z\ne0$ be an integer, and let $\rho$ be its least
767 nonnegative residue modulo $2^K$. If $\rho\ne0$, then
768 \[
769  v_2(z)=v_2(\rho)<K.
770 \]
771 If $\rho=0$, the residue gives only $v_2(z)\geq K$.
772 \end{lemma}
773 
774 \begin{proof}
775 Write $z=\rho+2^Kh$. If $\rho=2^ta$ with $a$ odd and $t<K$, then
776 $z=2^t(a+2^{K-t}h)$ and the parenthesis is odd. Hence the valuation is
777 exactly $t$. If $\rho=0$, the congruence says that $2^K$ divides $z$, but
778 contains no information about further powers of two.
779 \end{proof}
780 
781 For example, the residue of $S_{1,1}=2624$ modulo $2^{128}$ is nonzero and
782 has valuation six. Its residue modulo $16$ is zero, which says only that
783 its valuation is at least four. The integer itself did not change. The
784 precision changed the information available about it.
785 
786 A program that returns a number $K$ for a zero residue must attach a lower
787 bound interpretation to that number. It must not use it as an exact
788 valuation or as an exact maximum of the domain. The historical script
789 counts unresolved residues and includes an explicit exactness flag. The
790 release tests that flag deliberately.
791 
792 \begin{figure}[tbp]
793 \centering\includegraphics[width=0.91\linewidth]{../visual/figures/18_modular_exactness.pdf}
794 \caption{A nonzero residue modulo $2^K$ determines the exact valuation.
795 A zero residue reaches the precision ceiling and supplies only a lower
796 bound.}\label{cb_fig_mod_exactness}
797 \end{figure}
798 
799 \section{The domain and the Pascal computation}
800 
801 For a positive integer $N$, the scanned domain is
802 \begin{equation}\label{cb_eq_scan_domain}
803  \mathcal D_N=\{(n,m):1\leq n\leq N,\ 0\leq m\leq4n\}.
804 \end{equation}
805 The number of scanned pairs is
806 \begin{equation}\label{cb_eq_scan_count}
807  \sum_{n=1}^N(4n+1)=2N^2+3N.
808 \end{equation}
809 The upper slope $4$ is part of the experiment's definition. The
810 scan supplies no information about pairs above this boundary or with
811 $n>N$.
812 
813 The computation first constructs $T_0,\ldots,T_{5N}$ through the exact
814 integer recurrence. It then reduces these integers modulo $2^K$. The
815 resulting array is the packet row with $m=0$. At each subsequent row,
816 Pascal evolution replaces adjacent entries by their sum modulo $2^K$.
817 Induction from \eqref{cb_eq_Pascal} proves that every entry read by the
818 scan is the exact residue of its packet sum.
819 
820 For a fixed row $m$, the allowed indices are
821 \[
822  \max(1,\lceil m/4\rceil)\leq n\leq N.
823 \]
824 Rows are needed through $m=4N$. In row $m$, the seed array is sufficient
825 for every $n\leq5N-m$. At the extreme row $m=4N$ this bound is exactly
826 $n\leq N$. Thus no scanned entry needs an unconstructed seed term.
827 The algorithm evaluates additional row entries to support later rows.
828 Those additions do not enlarge the declared set of scanned pairs.
829 
830 This procedure uses full exact integers to construct the seed sequence,
831 including checks of all division remainders. Only the subsequent packet
832 row evolution is performed at fixed modular precision. Describing the
833 entire calculation as a floating point estimate would therefore be
834 incorrect. Describing its finite result as a uniform asymptotic theorem
835 would also be incorrect.
836 
837 \section{The reproduced 2000 row result}
838 
839 The historical run takes $N=2000$ and $K=128$. The release reran the
840 unchanged historical algorithm and compared its result with the recovered
841 output. The finite arithmetic results agree. In the domain
842 $\mathcal D_{2000}$ there are exactly $8{,}006{,}000$ pairs. Every observed
843 residue is nonzero, so the residue lemma applies at every pair. The exact
844 maximum valuation in this finite domain is
845 \begin{equation}\label{cb_eq_finite_max}
846  \max_{(n,m)\in\mathcal D_{2000}}v_2(S_{n,m})=31.
847 \end{equation}
848 The recorded maximizing pair is $(1466,2113)$. The record reports the
849 first maximum encountered in its scan order and makes no assertion that
850 this maximizing pair is unique.
851 
852 There is a generous precision margin: the largest observed valuation is
853 $31<128$, and there are zero unresolved residues. Consequently the finite
854 maximum is exact, rather than a maximum of truncated lower bounds. No
855 statistical confidence interval is involved. The essential qualification
856 is the finite domain in \eqref{cb_eq_finite_max}.
857 
858 \begin{figure}[tbp]
859 \centering\includegraphics[width=0.93\linewidth]{../visual/figures/19_v2_heatmap.pdf}
860 \caption{An exact valuation grid in the smaller independently checked
861 domain. The pattern can suggest questions about cancellation, but its
862 appearance does not imply an asymptotic law.}\label{cb_fig_v2_grid}
863 \end{figure}
864 
865 \section{A comparison at every pair and a full integer witness}
866 
867 The historical smaller crosscheck reports the same maximum from direct
868 sums through $N=100$. Matching only two summary maxima would be a weak
869 comparison: different entry values can share the same maximum. The release
870 therefore compares every pair in $\mathcal D_{100}$.
871 
872 For each of its $20{,}300$ pairs, a separate implementation evaluates
873 \[
874  \sum_{i=0}^m\binom miT_{n+i}
875 \]
876 as a full positive integer and compares both its residue modulo $2^{128}$
877 and its exact valuation with the Pascal row entry. All pairs agree. There
878 are no zero residues. The finite maximum is $20$, recorded first at
879 $(52,6)$ in the direct comparison's order.
880 
881 The direct summation and Pascal evolution use different operations on the
882 same seed data. This catches errors in the packet transform that agreement
883 between two copies of the same row update could miss. A separately indexed
884 rational recurrence also checks the seed construction through $n=500$.
885 These comparisons still share the intended recurrence formulas and require
886 the literature normalization contract. Their independence is operational,
887 not a claim that all common source risk has disappeared.
888 
889 The pair $(1466,2113)$ was additionally evaluated as a full integer, along
890 with four neighboring pairs. The maximizing witness has bit length
891 $26{,}743$ and exact valuation $31$. Its complete hexadecimal integer is
892 included in the evidence bundle with a hash. The neighboring valuations
893 are much smaller:
894 \begin{center}
895 \begin{tabular}{@{}rrr@{}}
896 \toprule
897 $n$ & $m$ & $v_2(S_{n,m})$\\
898 \midrule
899 $1465$ & $2113$ & $12$\\
900 $1466$ & $2112$ & $13$\\
901 $1466$ & $2113$ & $31$\\
902 $1466$ & $2114$ & $11$\\
903 $1467$ & $2113$ & $11$\\
904 \bottomrule
905 \end{tabular}
906 \end{center}
907 These five full integer computations corroborate a particular spike.
908 They do not establish the maximum over millions of pairs. The exhaustive
909 modular scan with no unresolved residues establishes that finite maximum.
910 
911 \section{A deliberate precision failure}
912 
913 The release includes a diagnostic with $N=3$ and just $K=4$ bits of
914 precision. There are 27 pairs. Eight residues are zero, so those eight
915 valuations are unresolved at this precision. The historical scan records
916 a ceiling value of four and sets its exactness flag to false. Full integer
917 evaluation shows that the actual finite maximum is six.
918 
919 This is a successful diagnostic of a deliberately insufficient precision,
920 not an exact four bit maximum. It demonstrates why the unresolved count
921 is load bearing. The program's numerical field alone would be misleading
922 if read without its flag. In the actual 128 bit release run, the same
923 classification logic reports zero unresolved entries and therefore
924 permits an exact interpretation.
925 
926 One should also distinguish this diagnostic from the failed odd filter
927 sign hypothesis. The low precision test intentionally triggers a known
928 information limitation and verifies that it is reported correctly. The
929 odd sign test disproves a proposed mathematical pattern. The evidence
930 bundle records both, with their different meanings.
931 
932 \section{Why a finite maximum does not settle growth}
933 
934 The largest valuation seen so far is small compared with the largest
935 index. That observation motivates a question such as
936 \[
937  v_2(S_{n,m})=o(n+m)
938 \]
939 along a specified family of packet indices. It does not prove the answer.
940 A universal statement needs quantified control beyond the scanned domain,
941 with the family of $m$ values and the meaning of uniformity stated
942 explicitly. A maximum found on a bounded grid cannot rule out later spikes
943 of a different scale.
944 
945 Even a proved estimate for $v_2(S_{n,m})$ would address only one prime and
946 one coefficient. The common content of the full integer pair involves both
947 coefficients, their clearing factor, and odd prime divisibility. The scan
948 therefore supplies a precise finite arithmetic fact and a well defined
949 research question. It is not, by itself, the missing denominator theorem
950 or an irrationality proof.
951 
952 The next stage should preserve this clarity. A proposed global valuation
953 bound must have a written argument valid at every permitted index. A
954 claimed content improvement must refer to the exact integer pair or to
955 its primitive reduction. A decay claim must include the clearing factors
956 that turn a real error into an integer linear form. The finite experiment
957 can guide those arguments and test their implementations, while the
958 global conclusion remains subject to its own proof.
959 

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