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Independently check #4439/#4442: halo exponents 12, 26, 28, 28, 6 under eps = 1/(16p^2) (no new κ) - #4471
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Mister-Meeseeks merged 1 commit intoOct 22, 2026
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…6, 28, 28, 6 under eps = 1/(16p^2) (no new κ) Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
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Summary
This adds
research/eps-halo-exponent-check/. It is an independent check of jwr404's #4439 (within one cell) and #4442 (across cells, downward reading), the halo exponents under the relative budgetε = 1/(16p²). It is the ε-route counterpart of my #4460.Claim type: an independent check by separate code, with exact rationals. No new κ.
Results
All statements check out.
off ≤ (16p²−2)/(16p²−1), so the gap is exactly1/(16p²−1).pin 2..64, with all-parguments.p = 2; 25 atp = 2(certified); 27 atp = 2(loose count) and forp ≥ 3(sharp count).p = 2tightness63·63 > 62·64and the bare-kernel example (square gap 32/16 against 20/16) both hold.Evidence
Head:
0cf8e2b0722379e9195f855c6c28aeaffa39aebd. Base: canonical main2a13eb18.Credit: James Reed (jwr404, #4439 and #4442). Prepared by Ryanson Jonathan with substantial Claude Code (Claude Opus 5.5) assistance.
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