{
 "what": "The iteration GNNW v2 prints as \"preliminary, unverified\" (G_AI, proposed by ChatGPT 5.6 Sol), decided: Theorem 14 of the same paper checked on all of (0, 1] for F = h + G_AI, with a continuous M chosen here and Y = Y_f(X) from Lemma 15 for the proved bound f = F_0.03 of Theorem 1. The conclusion is the paper's: R(k, l) <= e^{F(l/k)k + o(k)} for k >= l, so R(k, k) <= c^(k+o(k)), c = e^F(1). Re-run: python3 verify/verify_gnnw_gai.py certs/gnnw-certificate.json",
 "generated": "2026-09-29",
 "source": {
  "paper": "arXiv:2407.19026v2",
  "sha256": "2d5be61d04424f096ab1e4f6d94aefaacda71189276f281807ee5e8650c2eae9",
  "remark": "We asked ChatGPT 5.6 Sol to perform an additional iteration of optimization in Theorem 14. A preliminary, unverified iteration suggests that Theorem 1 holds with G_AI(λ) = e^{−λ}(−0.3864λ + 0.8347λ^2 − 2.0156λ^3 + 2.7171λ^4 − 1.7541λ^5 + 0.4522λ^6). If verified it would improve the upper bound on the diagonal Ramsey numbers to R(k, k) ≤ (3.78233 . . .)^{k+o(k)}. Further improvements by performing additional iterations are possible, but we expect that lowering the base of the exponent below 3.7 and, likely, even below 3.75 would require new ideas.",
  "uses": [
   "Theorem 14 (v2; Theorem 13 in v1): the three conditions and the conclusion R(k, l) <= e^{F(l/k)k + o(k)} for k >= l; M continuous",
   "Theorem 1: R(k, l) <= e^{G(l/k)k + o(k)} binom(k+l, l), G = (-0.25 l + 0.03 l^2 + 0.08 l^3) e^-l, i.e. the bound F_0.03 = h + G",
   "Lemma 15 and Remark 17: (x, Y_f(x)) in R for every x in (0, 1), f = F_0.03 (f strictly concave by (34), f' > 0 by (35); A < B because (1+t)f'(t) - f(t) decreases and is positive at t = 1)"
  ]
 },
 "q": [
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  "0.8347",
  "-2.0156",
  "2.7171",
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  "0.4522"
 ],
 "m": {
  "N": 100,
  "m0": "1.505717",
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 },
 "mWhat": "m(l) = M(l)/l at l = j/100, j = 0..100; M is l times the piecewise-linear interpolant (continuous, as GNNW Theorem 14 asks). Chosen by this program: at each node the float value of M maximising the slack, rounded to six decimals; at 0 the root 1.505717 of 1/mu + 1/(E + mu) = 1, E = e^0.3864, which maximises the slack's limit slack/l as l -> 0. A proposer's choice: any continuous M in (0, 1) that passes is a witness.",
 "verifier": {
  "file": "tools/verify_gnnw_gai.py",
  "sha256": "996c82c5500cde1b7ca72e55ee38b0ac27dd54be92387b4230117637e596a2f7"
 },
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 },
 "decided": {
  "verdict": "CERTIFIED",
  "claim": "GNNW Theorem 14 holds for F = h + q e^-l with the certificate's q and M and Y = Y_f(X), f = F_0.03, on all of (0, 1]; so R(k,k) <= e^{F(1)(k+o(k))}",
  "c": [
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  ],
  "printedBase": "3.78233",
  "printedDigitsHold": true,
  "printedIs": "its rounding",
  "stats": {
   "tailIntervals": 79,
   "mainIntervals": 336,
   "minTailS": 4.716462266515588e-06,
   "minMainLower": 9.55470225858566e-07,
   "minFp": 0.7560740516384318,
   "maxM": 0.3928025,
   "minX": null,
   "maxX": null,
   "worst": [
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    0.69
   ]
  },
  "seconds": 7.6
 }
}