{
 "what": "Detached certificates for the two constants of Erdős problem #852 (erdosproblems.com/852): c0, the unique positive root of I0(c) = 1 with I0(c) = c + c·log((e^{2c}-1)/(2c)) + Li2(1-e^{2c})/2, and C* = (1/2)(prod_{p>=3}(1 + 1/(p-1)^3) - 1). Both were published in the problem thread (2026-04-24) as bare decimals with no error bound. The claims here: c0 lies strictly inside bracket40 (so its first 40 decimals are as stated), the published c0 decimal is correct only as a rounding, and the published C* decimal is REFUTED — the true value differs from the 12th significant digit on. Verify with tools/verify_erdos852.py — Python stdlib only, no code from this repo.",
 "generatedBy": "tools/export-erdos852-certificate.js @ git 1d2b452",
 "sourcePins": {
  "erdos852_page.html": "a32562ab8f2444520775020d92c5673fbe1ed5410539129b1088b87431920a89",
  "erdos852_thread.html": "ad5968022abd22ed0e1176b081a5ca87059739e3e4571997b820a4e4b12726e6"
 },
 "c0": {
  "definition": "the unique positive root of I0(c) = 1, I0(c) = c + c·log((e^{2c}-1)/(2c)) + Li2(1-e^{2c})/2; for c in the bracket use Li2(-x) = -pi^2/6 - log(x)^2/2 - Li2(-1/x) with x = e^{2c}-1 > 1 (Lewin 1.12)",
  "certifiedDigits": "1.3232282768639494690289693932974634613586535126004759968489856",
  "bracket40": {
   "lo": "1.3232282768639494690289693932974634613586",
   "hi": "1.3232282768639494690289693932974634613587",
   "claim": "I0(lo) < 1 < I0(hi), both margins ~5e-41 — decidable by any 100+ digit evaluation of I0"
  },
  "uniqueness": "I0'(c) = log((e^{2c}-1)/(2c)) (elementary calculus) > 0 for c > 0 since e^{2c} > 1 + 2c; so the bracket contains THE root",
  "published": {
   "value": "1.32322827686395",
   "verdict": "VERIFIED_ROUNDED",
   "note": "correct as a rounding to 14 places; its trailing ellipsis is wrong — the expansion continues ...9469"
  },
  "instrument": {
   "precisionBits": 320,
   "bisections": 200
  }
 },
 "cstar": {
  "definition": "(1/2)(prod over primes p>=3 of (1 + 1/(p-1)^3) - 1)",
  "enclosure": {
   "lo": "0.0752403861783092455893",
   "hi": "0.0752403861783095651674",
   "method": "directed-rounding partial product over 1857858 odd primes to 30000000 at 192 bits; tail: sum_{p>L} 1/(p-1)^3 <= 1/(2(L-1)^2) = S, prod(1+a_p) <= e^S <= 1+S+S^2"
  },
  "published": {
   "value": "0.0752403861777",
   "verdict": "REFUTED",
   "mechanism": "the published decimal equals the naive IEEE-754 double product: 1 + 1/(p-1)^3 rounds to 1.0 once (p-1)^3 >= 2^53 (p-1 >= 208064), so ~87% of the factors vanish and the float product self-truncates at p ~ 2e5 regardless of the loop bound"
  },
  "refutation": {
   "limit": 400000,
   "statement": "the EXACT partial product N/D = prod_{3<=p<=400000, p prime}((q^3+1)/q^3), q = p-1, is a strict lower bound of the full product and already satisfies (N/D - 1)/2 > 752403861778/10^13, the upper edge of the published claim's window under BOTH truncation and rounding readings. One integer inequality: 5·(N-D)·10^12 > 752403861778·D. No tail bound, no rounding, no trust."
  }
 }
}