[{"type":"thinking","thinking":"D-056's original far sum: uses |s₀−ρ| ≥ (1+x)/√2, giving per-term ≤ 4.5√2/(x(1+x)) ≤ 4.5√2/(k(k+1)), and Σ 1/(k(k+1)) = 1. Tail (k > t): ≤ C₀(2t)/(t+1) + 0.433/t ≤ 0.53. Total 4.5√2·[2C₀ + 0.58 + 0.53] = 12.73·C₀ + 7.06.\n\nBut journal-035's session-035 REVISED plan uses the sharper per-term |s₀−ρ| ≥ √(1+x²), giving per-term ≤ 4.5/(x√(1+x²)) ≤ 4.5/(k√(1+k²)), S₃ = Σ 1/(k√(1+k²)) ≤ 1.3703, and the total 4.5·[2·S₃·C₀ + 0.55·S₃ + 0.5828] = 12.333·C₀ + 6.014.\n\nThe k > t tail ≤ 0.5828: this is 4.5·[tail]. So [tail] = 0.1295. The tail = Σ over zeros with x > t of 1/(x√(1+x²))... bounded via per-interval counts C0hat(t+k+1) ≤ ... hmm we computed that gives ~1.24·(1/4.5)... wait no: 4.5·[tail] where tail = Σ_{k>t} count_k/(k√(1+k²)).\n\nIf count_k ≤ C0hat(t+k+1) ≤ 0.433·log(t+k+3) + 0.886·log log + 2.9, for k > t ≥ 10: log(t+k+3) ≤ log(2k+3). Hmm, let me numerically estimate Σ_{k≥11} (0.433·log(2k+3) + 0.886·log log(2k+3) + 2.9)/(k√(1+k²)):\n- 2.9·Σ_{k≥11} 1/k² ≈ 2.9·0.0952 ≈ 0.276\n- 0.433·Σ log(2k+3)/k² ≈ 0.433·(log 25/11 + log 27/12 + ∫₁₁^∞ log(2x+3)/x² dx). ∫ log(2x+3)/x² dx from 11: = [−(log(2x+3)+2)/x]₁₁^∞ + ... roughly (log 25 + 2)/11 ≈ 0.474. Plus boundary terms ≈ 0.29+0.23. Total ≈ 0.474+0.2 ≈ maybe 0.8. Times 0.433 ≈ 0.35.\n- loglog part ≈ 0.886·0.2 ≈ 0.18.\nTotal ≈ 0.8. Times 4.5 ≈ 3.6. Way more than 0.5828.\n\nSo the tail ≤ 0.5828 must use something else. \"Σ log(3k)/k² and Σ 1/k² via the same telescopes\". Hmm — 0.5828 = 4.5·0.1295. Σ_{k≥11} log(3k)/k² ≈ log 33/11 + log 36/12 + ∫₁₁^∞ log(3x)/x² = 0.317+0.298+ (log 33+1)/11 ≈ 0.317+0.298+0.408 ≈ 1.02. Times something... A₀?\n\nHmm, wait — maybe the tail count uses the fact that these zeros are at height ≥ 2t ≥ 20... no.\n\nLet me reconsider: maybe the tail is bounded NOT via counts but via the global weighted sum `zero_tail_inv_sq_le` (LiCircleBound line 745): Σ_{|γ|>X} m/(1+γ²) ≤ 16·A₀·log X/X for X ≥ 200. For the tail zeros (x = |t−γ| > t, so |γ| ≥ ...): hmm, γ > 2t or γ < 0 (with |γ| ≥ 14). For γ < 0, |γ| = x − t; for γ > 2t, |γ| = x + t > t.\n\nFor a tail zero, 1/(x√(1+x²)) ≤ 1/x². And x = |t−γ| relates to |γ|: x ≤ t + |γ| ≤ 2|γ| when |γ| ≥ t... For γ > 2t: x = γ−t ≥ γ/2 = |γ|/2. For γ < 0: x = t+|γ| ≥ |γ|. So x ≥ |γ|/2 in both cases (for γ > 2t or γ < 0). So 1/x² ≤ 4/|γ|² ≤ 4/(1+γ²)·(1+γ²)/γ² ≤ 4·2/(1+γ²) (since γ² ≥ 196, (1+γ²)/γ² ≤ 2). So Σ_tail m/(x√(1+x²)) ≤ Σ_tail m/x² ≤ 8·Σ_{|γ|>t, tail} m/(1+γ²)... hmm wait but |γ| for tail zeros: γ > 2t or γ < 0 with |γ| ≥ 14. If t ≥ 200, we could apply zero_tail_inv_sq_le with X = t... but t ∈ [10, 200) isn't covered, and the bound 16·A₀·log(t)/t with A₀ ≥ 1 at t=10: 16·2.3/10 = 3.7. Too big.\n\nHmm, this isn't leading anywhere productive. The \"0.5828\" figure — let me just compute what combination gives 0.5828: \n\n\"the k > t tail ≤ 0.5828 (Σ log(3k)/k² and Σ 1/k² via the same telescopes)\"\n\nMaybe: tail ≤ 4.5·A₀·[Σ_{k>t} 3·log(3k)/k²]-ish? With A₀'s ACTUAL value (not the ≥ 1 bound). Let me find the actual value of the RvM constant in Zeta23. Let me grep the Zeta23 repo.","signature":""},{"type":"text","text":"Let me find the actual RvM local count statement in Zeta23 to know A0count's concrete value."},{"type":"tool_use","id":"Grep_16","name":"Grep","input":{"pattern":"zeta_local_zero_count","path":"D:\\Python\\Quantum\\research\\lean\\zeta-23-lean","output_mode":"content","-n":true,"-C":6,"head_limit":40},"extra_content":null}]