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Show the smallest Neumann coefficient is 9 (no new κ) - #4493
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Mister-Meeseeks merged 1 commit intoOct 22, 2026
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Co-authored-by: Cursor <cursoragent@cursor.com>
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Summary
For every integer
p ≥ 2, the within-cell Neumann ceilingsatisfies
8 p^2 < U < 9 p^2. The smallest integercwithU < c p^2for every suchpis therefore9. Every integerc ≤ 8fails becauseU > 8 p^2. The same9 p^2is strictly sharper than16 p^2 - 1andp^6.This is a property of the formula. It is an inverse bound only when the #4452 gauge holds. The ideal gaps stay hypotheses. No new κ. The selected record remains
23/10000.This answers the open question of #4492. It does not replay #4452 or #4492. The tensor absorption and the Schur margin still use
10.Comparison
With a positive denominator,
U < 9 p^2is equivalent to8(D - 4) < 9(D - 32 p^2). The difference isSo it is enough that
D > 288 p^2 - 32. Forp ≥ 2,2^{2p} ≥ p^2 + 1, hence288 p^2 < 2^{2p+9}. The exponent gap32760 p - (2p + 9) = 32758 p - 9is 65507 atp = 2and increases, so2^{2p+9} < D. The same estimate givesD > 32 p^2.Test plan
python3.12 -B scripts/research_package.py --package research/gaussian-neumann-ninePASS_GAUSSIAN_NEUMANN_NINEandoutcome=completedBase
2a13eb1827784206202db90e5dc8327cf8a250db, priority revision 7. Upstream fetched at 2026-10-22T01:42:53Z was still that commit, and the pre-submit fetch was unchanged.Prepared by James Reed with substantial Cursor Grok 4.7 assistance.