{
  "file_path": "D:\\Python\\Quantum\\research\\derivations\\D-072-finding-F-D-right-tail-stepping.md",
  "content": "# D-072 \u2014 Finding F-D: the (6.12) stepping in Lagarias's Thm 6.1 is broken (right-tail of the horizontals)\nDate: 2026-08-24 | Session: 038 (ii\u2032-6-c discovery, before any Lean work) | Status: DOCUMENTED, direction OPEN (user opted to pause and think)\nNumerical evidence: E-031 (ALL CHECKS PASS).\n\n## 1. The claim that fails\n\nLagarias, \"Li coefficients for automorphic L-functions\" (arXiv:math/0404394,\nAnn. Inst. Fourier 57 (2007) 1689\u20131740), Theorem 6.1's proof, page 29 (the\nright part of the horizontal segments):\n\n> \"Now we step across the interval (II) towards the right, in segments of\n> length 1, starting from \u03c3 = 3. At the initial point we have |L\u2032/L| = O(1)\n> ... the estimate (6.10) gains a factor of 2 going one unit to the right.\n> In comparison |k_n(s+1)+1|/|k_n(s)+1| \u2264 e [via (6.12)]. We obtain an upper\n> bound for |k_n(s)L\u2032/L(s,\u03c0)| that **decreases geometrically at each step**,\n> and after O(log n) steps it becomes O(1) so |I_II| = O(\u221an + (log n)\u00b3).\"\n\nThe contour: integrand k_n(s)\u00b7(\u2212L\u2032/L(s+1)), k_n(s) = (1+1/s)\u207f \u2212 1, on the\nrectangle \u03c3\u2080 \u2208 (\u22123, \u22122), \u03c3\u2081 = 2\u221an, im = \u00b1T, T = \u221an + \u03b5.\n\n## 2. Why it fails (the mechanism)\n\nPer unit step right from \u03c3 = 3:\n- the zeta bound halves: |L\u2032/L(\u03c3+1+iT)| \u2264 2^{\u2212(\u03c3\u22122)}\u00b7|L\u2032/L(2)| ((6.10));\n- the kernel ratio is \u2264 e: |k_n(s+1)+1|/|k_n(s)+1| \u2264 (1+1/T\u00b2)\u207f \u2264 e^{n/T\u00b2}\n  \u2264 e ((6.12), since T\u00b2 \u2265 n).\n\nThe product ratio is therefore \u2264 e/2 \u2248 1.359 > 1 \u2014 **the product GROWS**,\nit does not decrease. The kernel's own per-step log-increment is \u2248 1\u20132 for\n\u03c3 \u226a \u221an (from log|1+1/s| \u2248 \u03c3/(\u03c3\u00b2+T\u00b2), increment n\u00b7d/d\u03c3[\u03c3/(\u03c3\u00b2+T\u00b2)] \u2248\nn(T\u00b2\u2212\u03c3\u00b2)/(\u03c3\u00b2+T\u00b2)\u00b2 \u2248 1 at \u03c3 small, T\u00b2 = n), which exceeds log 2 \u2248 0.693\nuntil \u03c3 \u2248 0.36\u221an. So the product rises from ~O(1) at \u03c3 = 3 to a peak\n~e^{0.066\u221an} at \u03c3 \u2248 0.36\u221an, then falls to e^{\u221an/2}\u00b72^{\u2212(2\u221an\u22122)} \u2248 0 at\n\u03c3\u2081 = 2\u221an. The stepping's own geometric series \u03a3(e/2)^j with ratio > 1\ndiverges; the published conclusion O(\u221an) requires the ratio < 1.\n\nThe same bound works with a taller horizontal: T = c\u221an gives kernel ratio\n\u2264 e^{1/c\u00b2}, which is < 2 (hence product ratio e^{1/c\u00b2}/2 < 1) whenever\nc > 1/\u221a(log 2) \u2248 1.2012 \u2014 verified numerically (E-031 V3: the product\ndecreases monotonically from \u03c3 = 3 at c = 1.5 and c = 2).\n\n## 3. Numerical evidence (E-031, direct computation, dps 25)\n\nThe product |k_n(w\u22121)|\u00b7|\u03a3_{k<300} \u039b(k)/k^w| at w = \u03c3 + i\u221an:\n\n| n | peak product | peak at \u03c3 | \u03c3/\u221an | (log peak)/\u221an |\n|---|---|---|---|---|\n| 10\u00b3 | e^{1.44} \u2248 4.2 | 12 | 0.38 | 0.046 |\n| 10\u2074 | e^{6.05} \u2248 430 | 30 | 0.30 | 0.061 |\n| 10\u2075 | e^{20.98} \u2248 1.3\u00d710\u2079 | 120 | 0.38 | 0.066 |\n\nEnvelope integral of the right tail \u222b_{3}^{2\u221an} (step 2):\n- n = 10\u2074: \u2248 1.05\u00d710\u2074 (peak 494 at \u03c3 = 35);\n- n = 10\u2075: \u2248 **5.87\u00d710\u00b9\u2070** (peak 1.56\u00d710\u2079 at \u03c3 = 111).\n\nDirect factor evaluation at the peak (n = 10\u2075, \u03c3 = 100, T = \u221an = 316.2):\n|k_n(100+317i)| = |(1+1/(100+317i))^{100000}| = 4.39\u00d710\u00b3\u2079 (mpmath, dps 30);\n|\u03b6\u2032/\u03b6(101+317i)| \u2248 1.8\u00d710\u207b\u00b3\u00b9 (mpmath finite difference + Dirichlet series\ncross-check). Product \u2248 8\u00d710\u2078.\n\nRepair verification (n = 10\u2075, product log P at \u03c3 = 3, 10, 30, 60, 100, 200):\n- T = \u221an: \u22120.03, +2.17, +8.08, +15.51, +20.86, +3.71 (GROWS to \u03c3 \u2248 120);\n- T = 1.5\u221an: \u22121.91, \u22123.10, \u22128.10, \u221215.92, \u221227.32, \u221263.66 (decreases);\n- T = 2\u221an: \u22121.59, \u22124.91, \u221213.80, \u221227.21, \u221245.41, \u221293.63 (decreases).\n\n## 4. Inheritance into R-004\n\nD-036 \u00a73 and D-037 \u00a74 inherited the stepping verbatim: the right part of\nthe horizontals is charged \"3.3\" (D-037: \"right part [3, 2\u221an] decays\ngeometrically by the stepping argument (6.12): |I_II|+|I_IV| \u2264\n2(e\u2074+1)[7H(T)+3.3]\"). The true right-tail is ~5.9\u00d710\u00b9\u2070 at n = 10\u2075 \u2014 the\ncharge is off by ~10\u2077\u00d7, and the error grows like e^{0.066\u221an}.\n\nWith the honest right-tail, B(n) must charge ~e^{0.066\u221an}\u00b7poly, which\novertakes S_\u221e(n) = \u00bdn log n + O(n) at n \u2248 3\u00d710\u2074 \u2014 **the margin argument\nfor [N\u2080, N\u2081] collapses** (margin(n) < 0 for n \u2273 3\u20134\u00d710\u2074).\n\n## 5. What survives\n\n- **The theorem statement** (\u03bb_n > 0 for n \u2264 N\u2081): directly verified to\n  n = 5\u00d710\u2075 (Misguich\u2013Voros + E-011/E-015 to 3000); the measured |S_f| is\n  ~13\u201320 (E-021, Ma\u015blanka). The closed contour identity is exact \u2014 the\n  sides cancel dramatically ((II)/(IV) are conjugates: real parts cancel;\n  plus phase-winding of arg(k_n\u00b7L\u2032/L) ~ \u2212300 rad over the peak region).\n  Lagarias's Theorem 6.1 CONCLUSION (Sf = \u03bb_n(\u221an) + O(\u221an log n)) may well\n  hold via that cancellation \u2014 but the paper's PROOF of the horizontal\n  bound is broken, and the side-wise triangle bounds (what D-036/037/R-004\n  use) are unusable as charged.\n- **All Lean work to date**: the honest-Bn correction (4F, (2/3)Z2hat),\n  margin convexity, base points, the sides identity, the bridge\n  (4\u00b7FAlt), the perimeter Props, the wide pigeonhole \u2014 the bug lives in\n  the horizontal piece (ii\u2032-6-c), which had NOT been built yet.\n- **The residue-identity framework** (rectInt'_Jlag) \u2014 exact, unaffected.\n\n## 6. Repair path (R) \u2014 verified numerically (E-031 V3)\n\nRaise the horizontal height to T = c\u221an with c > 1/\u221a(log 2) \u2248 1.2012\n(kernel ratio \u2264 e^{1/c\u00b2} < 2 \u27f9 the product decreases per step \u27f9 the\nstepping works; the right-tail becomes genuinely small \u2014 at T = 2\u221an the\nkernel at \u03c3 = 100 is e^{24.6} and the product is ~e^{\u221243}).\n\nCosts (c = 2 shown; c = 1.5 halves the range-cost):\n- **N\u2081 \u2192 (T\u2080/c)\u00b2**: c = 2 gives (T\u2080/2)\u00b2 \u2248 2.25\u00d710\u00b2\u2074; c = 1.5 gives\n  (T\u2080/1.5)\u00b2 \u2248 4\u00d710\u00b2\u2074 (from 9\u00d710\u00b2\u2074 \u2014 still criterion-3).\n- **Bridge** 4\u00b7FAlt(T) grows by \u2248 c\u00d7 (4\u00b7FAlt(2\u221an) \u2248 0.64\u221an log n).\n- **I_III** length doubles: 2T = 2c\u221an (bound 4c\u221an\u00b7[(2/3)Z2 + 10.5 log +\n  48] \u2248 42\u221an log n at c = 2).\n- **Horizontal kernel charge drops**: |k_n| \u2264 (1+4/(16+4c\u00b2n/c\u00b2))\u207f+1 \u2264\n  e^{4/c\u00b2}+1 \u2248 e+1 at c = 2 (from e\u2074+1).\n- **I_I** (right vertical) shrinks (kernel \u2264 e^{\u221an/(2c)}+1).\n- The zero-avoiding T pigeonhole and the wide pigeonhole carry over\n  verbatim (window [c\u221an, c\u221an+1]).\n- The margin survives comfortably: B(n) ~ 4c\u221an\u00b7(0.4+10.5)\u00b7log\u00b7... \u2248\n  21c\u00b7\u221an log n \u226a \u00bdn log n for n \u2273 10\u2074\u201310\u2075 (crossover of the same order\n  as before, re-verified by the margin machinery).\n\n## 7. Literature status (session 038, \u00a74A.3 checks)\n\n- arXiv math/0404394 version history (v1\u2013v4, 2004\u20132005): the only noted\n  correction is \"to remainder term in Theorem 1.1\" \u2014 Thm 6.1 untouched.\n- Web searches (DuckDuckGo, Bing): no erratum or discussion of a gap in\n  Thm 6.1/(6.12) found. MathOverflow search not accessible from this\n  environment. **The issue may be undocumented** \u2014 or read charitably\n  with an unwritten fix (the T = c\u221an repair is essentially free and\n  restores the O(\u221an) horizontal bound with the same proof).\n\n## 8. Open questions (for the return)\n\n1. Adopt repair (R) (which c: 2 for margin comfort, 1.5 for range) and\n   rework Bn/constants accordingly (the D-071 protocol re-applies:\n   convexity re-proof + base-point re-certification)?\n2. Or attempt a salvage of T = \u221an via phase-winding/oscillation (the\n   sides DO cancel \u2014 research-level, no existing machinery, open-ended)?\n3. Or a deeper literature/community check first (MathSciNet review,\n   MathOverflow direct, treatments of (6.12) in citing papers \u2014\n   e.g., Ma\u015blanka's, Voros's, the explicit finite-height follow-ups)?\n4. The theorem statement R-004 itself: with repair (R), Paper A's\n   headline becomes \u03bb_n > 0 for n \u2264 (T\u2080/c)\u00b2 (\u2248 2.25\u20134\u00d710\u00b2\u2074) \u2014 the\n   finite-verification range (5\u00d710\u2075) and the criterion-3 status are\n   unchanged.\n\n## 9. Honesty notes\n\n- The envelope integral (E-031 V2) is an UPPER bound on the true side\n  integral |I_II's right part| \u2264 \u222b|f| \u2014 the true complex integral could\n  be smaller via oscillation; but the envelope is what every side-wise\n  assembly (D-036/037/R-004) charges, so it is the right measure for the\n  fold's provability.\n- The zeta factor was bounded by the Dirichlet series \u03a3_{k<300} \u039b(k)/k^w\n  (absolute convergence, re w \u2265 4) and cross-checked against mpmath's\n  direct \u03b6\u2032/\u03b6 at \u03c3 = 100 (1.8\u00d710\u207b\u00b3\u00b9 vs series 2.7\u00d710\u207b\u00b3\u00b9 \u2014 same order).\n  The tail k \u2265 300 contributes < 4\u00d710\u207b\u2075 at \u03c3 = 3 and less elsewhere.\n- No Lean code was harmed: ii\u2032-6-c had not begun; the finding was caught\n  at the design-check stage (the D-071 fold analysis).\n"
}

{
  "type": "create",
  "filePath": "D:\\Python\\Quantum\\research\\derivations\\D-072-finding-F-D-right-tail-stepping.md",
  "content": "# D-072 \u2014 Finding F-D: the (6.12) stepping in Lagarias's Thm 6.1 is broken (right-tail of the horizontals)\nDate: 2026-08-24 | Session: 038 (ii\u2032-6-c discovery, before any Lean work) | Status: DOCUMENTED, direction OPEN (user opted to pause and think)\nNumerical evidence: E-031 (ALL CHECKS PASS).\n\n## 1. The claim that fails\n\nLagarias, \"Li coefficients for automorphic L-functions\" (arXiv:math/0404394,\nAnn. Inst. Fourier 57 (2007) 1689\u20131740), Theorem 6.1's proof, page 29 (the\nright part of the horizontal segments):\n\n> \"Now we step across the interval (II) towards the right, in segments of\n> length 1, starting from \u03c3 = 3. At the initial point we have |L\u2032/L| = O(1)\n> ... the estimate (6.10) gains a factor of 2 going one unit to the right.\n> In comparison |k_n(s+1)+1|/|k_n(s)+1| \u2264 e [via (6.12)]. We obtain an upper\n> bound for |k_n(s)L\u2032/L(s,\u03c0)| that **decreases geometrically at each step**,\n> and after O(log n) steps it becomes O(1) so |I_II| = O(\u221an + (log n)\u00b3).\"\n\nThe contour: integrand k_n(s)\u00b7(\u2212L\u2032/L(s+1)), k_n(s) = (1+1/s)\u207f \u2212 1, on the\nrectangle \u03c3\u2080 \u2208 (\u22123, \u22122), \u03c3\u2081 = 2\u221an, im = \u00b1T, T = \u221an + \u03b5.\n\n## 2. Why it fails (the mechanism)\n\nPer unit step right from \u03c3 = 3:\n- the zeta bound halves: |L\u2032/L(\u03c3+1+iT)| \u2264 2^{\u2212(\u03c3\u22122)}\u00b7|L\u2032/L(2)| ((6.10));\n- the kernel ratio is \u2264 e: |k_n(s+1)+1|/|k_n(s)+1| \u2264 (1+1/T\u00b2)\u207f \u2264 e^{n/T\u00b2}\n  \u2264 e ((6.12), since T\u00b2 \u2265 n).\n\nThe product ratio is therefore \u2264 e/2 \u2248 1.359 > 1 \u2014 **the product GROWS**,\nit does not decrease. The kernel's own per-step log-increment is \u2248 1\u20132 for\n\u03c3 \u226a \u221an (from log|1+1/s| \u2248 \u03c3/(\u03c3\u00b2+T\u00b2), increment n\u00b7d/d\u03c3[\u03c3/(\u03c3\u00b2+T\u00b2)] \u2248\nn(T\u00b2\u2212\u03c3\u00b2)/(\u03c3\u00b2+T\u00b2)\u00b2 \u2248 1 at \u03c3 small, T\u00b2 = n), which exceeds log 2 \u2248 0.693\nuntil \u03c3 \u2248 0.36\u221an. So the product rises from ~O(1) at \u03c3 = 3 to a peak\n~e^{0.066\u221an} at \u03c3 \u2248 0.36\u221an, then falls to e^{\u221an/2}\u00b72^{\u2212(2\u221an\u22122)} \u2248 0 at\n\u03c3\u2081 = 2\u221an. The stepping's own geometric series \u03a3(e/2)^j with ratio > 1\ndiverges; the published conclusion O(\u221an) requires the ratio < 1.\n\nThe same bound works with a taller horizontal: T = c\u221an gives kernel ratio\n\u2264 e^{1/c\u00b2}, which is < 2 (hence product ratio e^{1/c\u00b2}/2 < 1) whenever\nc > 1/\u221a(log 2) \u2248 1.2012 \u2014 verified numerically (E-031 V3: the product\ndecreases monotonically from \u03c3 = 3 at c = 1.5 and c = 2).\n\n## 3. Numerical evidence (E-031, direct computation, dps 25)\n\nThe product |k_n(w\u22121)|\u00b7|\u03a3_{k<300} \u039b(k)/k^w| at w = \u03c3 + i\u221an:\n\n| n | peak product | peak at \u03c3 | \u03c3/\u221an | (log peak)/\u221an |\n|---|---|---|---|---|\n| 10\u00b3 | e^{1.44} \u2248 4.2 | 12 | 0.38 | 0.046 |\n| 10\u2074 | e^{6.05} \u2248 430 | 30 | 0.30 | 0.061 |\n| 10\u2075 | e^{20.98} \u2248 1.3\u00d710\u2079 | 120 | 0.38 | 0.066 |\n\nEnvelope integral of the right tail \u222b_{3}^{2\u221an} (step 2):\n- n = 10\u2074: \u2248 1.05\u00d710\u2074 (peak 494 at \u03c3 = 35);\n- n = 10\u2075: \u2248 **5.87\u00d710\u00b9\u2070** (peak 1.56\u00d710\u2079 at \u03c3 = 111).\n\nDirect factor evaluation at the peak (n = 10\u2075, \u03c3 = 100, T = \u221an = 316.2):\n|k_n(100+317i)| = |(1+1/(100+317i))^{100000}| = 4.39\u00d710\u00b3\u2079 (mpmath, dps 30);\n|\u03b6\u2032/\u03b6(101+317i)| \u2248 1.8\u00d710\u207b\u00b3\u00b9 (mpmath finite difference + Dirichlet series\ncross-check). Product \u2248 8\u00d710\u2078.\n\nRepair verification (n = 10\u2075, product log P at \u03c3 = 3, 10, 30, 60, 100, 200):\n- T = \u221an: \u22120.03, +2.17, +8.08, +15.51, +20.86, +3.71 (GROWS to \u03c3 \u2248 120);\n- T = 1.5\u221an: \u22121.91, \u22123.10, \u22128.10, \u221215.92, \u221227.32, \u221263.66 (decreases);\n- T = 2\u221an: \u22121.59, \u22124.91, \u221213.80, \u221227.21, \u221245.41, \u221293.63 (decreases).\n\n## 4. Inheritance into R-004\n\nD-036 \u00a73 and D-037 \u00a74 inherited the stepping verbatim: the right part of\nthe horizontals is charged \"3.3\" (D-037: \"right part [3, 2\u221an] decays\ngeometrically by the stepping argument (6.12): |I_II|+|I_IV| \u2264\n2(e\u2074+1)[7H(T)+3.3]\"). The true right-tail is ~5.9\u00d710\u00b9\u2070 at n = 10\u2075 \u2014 the\ncharge is off by ~10\u2077\u00d7, and the error grows like e^{0.066\u221an}.\n\nWith the honest right-tail, B(n) must charge ~e^{0.066\u221an}\u00b7poly, which\novertakes S_\u221e(n) = \u00bdn log n + O(n) at n \u2248 3\u00d710\u2074 \u2014 **the margin argument\nfor [N\u2080, N\u2081] collapses** (margin(n) < 0 for n \u2273 3\u20134\u00d710\u2074).\n\n## 5. What survives\n\n- **The theorem statement** (\u03bb_n > 0 for n \u2264 N\u2081): directly verified to\n  n = 5\u00d710\u2075 (Misguich\u2013Voros + E-011/E-015 to 3000); the measured |S_f| is\n  ~13\u201320 (E-021, Ma\u015blanka). The closed contour identity is exact \u2014 the\n  sides cancel dramatically ((II)/(IV) are conjugates: real parts cancel;\n  plus phase-winding of arg(k_n\u00b7L\u2032/L) ~ \u2212300 rad over the peak region).\n  Lagarias's Theorem 6.1 CONCLUSION (Sf = \u03bb_n(\u221an) + O(\u221an log n)) may well\n  hold via that cancellation \u2014 but the paper's PROOF of the horizontal\n  bound is broken, and the side-wise triangle bounds (what D-036/037/R-004\n  use) are unusable as charged.\n- **All Lean work to date**: the honest-Bn correction (4F, (2/3)Z2hat),\n  margin convexity, base points, the sides identity, the bridge\n  (4\u00b7FAlt), the perimeter Props, the wide pigeonhole \u2014 the bug lives in\n  the horizontal piece (ii\u2032-6-c), which had NOT been built yet.\n- **The residue-identity framework** (rectInt'_Jlag) \u2014 exact, unaffected.\n\n## 6. Repair path (R) \u2014 verified numerically (E-031 V3)\n\nRaise the horizontal height to T = c\u221an with c > 1/\u221a(log 2) \u2248 1.2012\n(kernel ratio \u2264 e^{1/c\u00b2} < 2 \u27f9 the product decreases per step \u27f9 the\nstepping works; the right-tail becomes genuinely small \u2014 at T = 2\u221an the\nkernel at \u03c3 = 100 is e^{24.6} and the product is ~e^{\u221243}).\n\nCosts (c = 2 shown; c = 1.5 halves the range-cost):\n- **N\u2081 \u2192 (T\u2080/c)\u00b2**: c = 2 gives (T\u2080/2)\u00b2 \u2248 2.25\u00d710\u00b2\u2074; c = 1.5 gives\n  (T\u2080/1.5)\u00b2 \u2248 4\u00d710\u00b2\u2074 (from 9\u00d710\u00b2\u2074 \u2014 still criterion-3).\n- **Bridge** 4\u00b7FAlt(T) grows by \u2248 c\u00d7 (4\u00b7FAlt(2\u221an) \u2248 0.64\u221an log n).\n- **I_III** length doubles: 2T = 2c\u221an (bound 4c\u221an\u00b7[(2/3)Z2 + 10.5 log +\n  48] \u2248 42\u221an log n at c = 2).\n- **Horizontal kernel charge drops**: |k_n| \u2264 (1+4/(16+4c\u00b2n/c\u00b2))\u207f+1 \u2264\n  e^{4/c\u00b2}+1 \u2248 e+1 at c = 2 (from e\u2074+1).\n- **I_I** (right vertical) shrinks (kernel \u2264 e^{\u221an/(2c)}+1).\n- The zero-avoiding T pigeonhole and the wide pigeonhole carry over\n  verbatim (window [c\u221an, c\u221an+1]).\n- The margin survives comfortably: B(n) ~ 4c\u221an\u00b7(0.4+10.5)\u00b7log\u00b7... \u2248\n  21c\u00b7\u221an log n \u226a \u00bdn log n for n \u2273 10\u2074\u201310\u2075 (crossover of the same order\n  as before, re-verified by the margin machinery).\n\n## 7. Literature status (session 038, \u00a74A.3 checks)\n\n- arXiv math/0404394 version history (v1\u2013v4, 2004\u20132005): the only noted\n  correction is \"to remainder term in Theorem 1.1\" \u2014 Thm 6.1 untouched.\n- Web searches (DuckDuckGo, Bing): no erratum or discussion of a gap in\n  Thm 6.1/(6.12) found. MathOverflow search not accessible from this\n  environment. **The issue may be undocumented** \u2014 or read charitably\n  with an unwritten fix (the T = c\u221an repair is essentially free and\n  restores the O(\u221an) horizontal bound with the same proof).\n\n## 8. Open questions (for the return)\n\n1. Adopt repair (R) (which c: 2 for margin comfort, 1.5 for range) and\n   rework Bn/constants accordingly (the D-071 protocol re-applies:\n   convexity re-proof + base-point re-certification)?\n2. Or attempt a salvage of T = \u221an via phase-winding/oscillation (the\n   sides DO cancel \u2014 research-level, no existing machinery, open-ended)?\n3. Or a deeper literature/community check first (MathSciNet review,\n   MathOverflow direct, treatments of (6.12) in citing papers \u2014\n   e.g., Ma\u015blanka's, Voros's, the explicit finite-height follow-ups)?\n4. The theorem statement R-004 itself: with repair (R), Paper A's\n   headline becomes \u03bb_n > 0 for n \u2264 (T\u2080/c)\u00b2 (\u2248 2.25\u20134\u00d710\u00b2\u2074) \u2014 the\n   finite-verification range (5\u00d710\u2075) and the criterion-3 status are\n   unchanged.\n\n## 9. Honesty notes\n\n- The envelope integral (E-031 V2) is an UPPER bound on the true side\n  integral |I_II's right part| \u2264 \u222b|f| \u2014 the true complex integral could\n  be smaller via oscillation; but the envelope is what every side-wise\n  assembly (D-036/037/R-004) charges, so it is the right measure for the\n  fold's provability.\n- The zeta factor was bounded by the Dirichlet series \u03a3_{k<300} \u039b(k)/k^w\n  (absolute convergence, re w \u2265 4) and cross-checked against mpmath's\n  direct \u03b6\u2032/\u03b6 at \u03c3 = 100 (1.8\u00d710\u207b\u00b3\u00b9 vs series 2.7\u00d710\u207b\u00b3\u00b9 \u2014 same order).\n  The tail k \u2265 300 contributes < 4\u00d710\u207b\u2075 at \u03c3 = 3 and less elsewhere.\n- No Lean code was harmed: ii\u2032-6-c had not begun; the finding was caught\n  at the design-check stage (the D-071 fold analysis).\n",
  "structuredPatch": [],
  "originalFile": null
}