{"id":109,"job_id":247,"problem_id":1,"lane_id":5,"type":"explore","user_id":1,"model":"claude-opus-5","provider":"anthropic","report_md":"# Job #247 (explore, `Q-signed-moment`)\n\n**Caveat first.** Nothing here adds a region, lowers the block budget at the corner (19/25, 19/20) or at the target box (8/25, 9/20), or bears on twin-prime infinitude. What is new:\n- a bound for the **whole R=0 class**, derived here and not reviewed (*rung: heuristic until reviewed*);\n- a structural reading of why it stops at the corner;\n- a prose slip in §4.\n\nEverything else in the note re-checks.\n\n## 1. Rechecked, no error found (rung: the note's own)\n\n- **Lemma A.**\n - Pairs per residue r = m̄₁ − m̄₂ mod u are at most M(M/u + 1).\n - Σ_r min(A, ‖θr/u‖⁻¹) ≪ A + uv + u log A.\n - The bracket in (4), with top-band terms x, Mx/N, x²/N, x²/M, hence x^(2−min(a,b)) in (5).\n- **Lemma B.**\n - P = m̄₁ℓ₂ − m̄₂ℓ₁ runs over a coset of size c/ℓ₁ = u₂.\n - The bracket N²M² + NM²A + MN³ + MN²A.\n - Top-band terms √(MNx), M√N, N√x and N√M have exponents (1+a+b)/2, a + b/2, b + 1/2 and b + a/2. The §3 sentence lists the same *set* of exponents in a different order from the terms (M√N goes with a + b/2, N√x with b + 1/2); the set, and so (8), is right.\n- **The §4 table.** All six rows of the Lemma B and \"Lemma A on the square\" columns reproduce in exact rationals: 3/2, 103/100, 209/200, 41/40, 21/20, 3/2 and 1, 3/2, 31/20, 151/100, 19/10, 38/25.\n- **§5(i).** The bracket (M² + MN)(N + A) gives (1 + max(a,b))/2: 39/50 at the target, below one on exactly the 36100 interior lattice boxes.\n- **§5 ceiling algebra.** BCf√(MN/A)(√N + √A), which is x^((1+b)/2) on the top band.\n- **§6 joint Cauchy.** Budgets 2 at the corner and 103/100 at the target; floor M√N = 81/100.\n- **§1 counterexample** to the general Hölder floor (M = 4, p = 1): left side 1, right side 2.\n\n## 2. Verified\n\n`node research/signed-moment-validation.js` exits 0 on Node v25.2.0 (the embedded run used v22.21.0). Its stdout sha256 is e85405abc7c2aaad9fe65ad58de77625f34622ab70b3d3de4c26e1d8f36c62fd, equal to the embedded out-sha256.\n\n## 3. Prose slip in §4 (rung: proven)\n\n§4 says \"(8) is below one exactly when a+b<1 and b<1/2, i.e. delta+nu<71/100 and nu<2/5\". Since b = ν + 1/20, b < 1/2 is equivalent to **ν < 9/20**.\n\nOn the validator's 191×191 lattice (δ ∈ [6/25, 19/25], ν ∈ [1/20, 19/20]):\n- the budgets themselves give 6906 usable boxes;\n- δ + ν < 71/100 with ν < 9/20 also gives 6906;\n- with ν < 2/5 it would give 6717.\n\nThe validator computes usability from the four budgets directly, so its 6906 and its \"0 outside the controlled region\" stand. The conclusion is unaffected either way, since δ + ν < 71/100 already lies inside δ + ν < 19/25. Only the prose condition needs correcting.\n\n## 4. The whole R=0 class (derived here, unreviewed)\n\n**Definition.** Review 20 F4 defines the R=0 class as the pairs (u₁,h₁), (u₂,h₂) with h₁/u₁ = h₂/u₂. Group the (u,h) by reduced ratio p/q, (p,q) = 1, writing u = tq and h = tp.\n\n**Exact structure.**\n- e_u(σθh m̄) = e(σθ·tp·m̄/(tq)) = e_q(σθp m̄), because m̄ mod u reduces to the inverse mod q.\n- Φ_{u,h}(m) = e(hz₀′/(gmu)) − e(hz′/(gmu)) depends on h/u only, so Φ_{u,h} = Φ_{q,p}.\n- 1_{(m,u)=1} = 1_{(m,q)=1} Σ_{s|(m,t)} μ(s).\n\nHence the R=0 part of (2), which contains (a)+(b), is exactly\n\n R0 = Σ_{(p,q)=1} |Σ_t b_{tq} c_{tp} G(tq,tp)|² = Σ_{(p,q)=1} |Σ_s μ(s) β_s(q,p) G^{(s)}_{q,p}|²,\n\nwhere β_s(q,p) = Σ_{t: s|t, tq∼N, tp∈H} b_{tq} c_{tp}, and G^{(s)}_{q,p} = Σ_{m∈I_m, (m,q)=1, s|m} A_left(gm) e_q(σθp m̄) Φ_{q,p}(m).\n\nIt is a sum of squares, so the R=0 class is nonnegative.\n\n**Bound.**\n1. **Coefficients.** |β_s| ≤ (2N/(qs))·BC/A, and β_s = 0 for s > 2N/q.\n2. **Weighted Cauchy in s.** |Σ_s μ(s)β_s G^{(s)}|² ≤ (Σ_s s|β_s|²)(Σ_s |G^{(s)}|²/s), with Σ_s s|β_s|² ≤ (2NBC/(qA))² log(4N).\n3. **The range of p.** A class exists only if p ∈ P_q := [qA/(2N), 2qA/N]. So q ≥ N/(2A) and |P_q| ≤ 5qA/N. After Cauchy the summand is nonnegative, so extend to all p ∈ P_q.\n4. **Geometric series in p.** Both e_q(σθp m̄) and Φ_{q,p}(m) are pure exponentials in p. The endpoint perturbation is at most 2x/(gMq), and times |P_q| that is at most 4v/g.\n - Group (m₁,m₂) by r = m̄₁ − m̄₂ mod q. There are at most M/s + 1 values of m₁, and for each r at most M/(sq) + 1 values of m₂, since s | m₂ and (s,q) = 1 fix one class mod sq.\n - Σ_r min(|P_q|, ‖·‖⁻¹) ≪ |P_q| + q(1+v) + q log q.\n - So Σ_{p∈P_q} |G^{(s)}_{q,p}|² ≪ x^ε f²(1+v)(M²/s² + Mq/s + q)(A/N + 1).\n5. **Sum over s, then over q ≥ N/(2A).** Σ_{q≥N/2A} q⁻² ≤ 4A/N, and Σ_q q⁻¹ ≪ log N. For A ≤ N:\n\n R0 ≪_ε B²C² x^ε f²(1+v) [M²N/A + MN²/A²]. (★)\n\n**On the top band** (A ≍ MN/x, f = 1, v ≍ 1), R0 ≪ x^(1+a+ε) + x^(2−a+ε).\n\n| box | a | Lemma A (diagonal only), square | (★), whole R=0 class, square |\n|---|---|---|---|\n| corner (19/25, 19/20) | 1 | 1 | **2** |\n| target (8/25, 9/20) | 14/25 | 3/2 | 39/25 |\n| benchmark (2/5, 2/5) | 16/25 | 31/20 | 41/25 |\n| (8/25, 11/25) | 14/25 | 151/100 | 39/25 |\n| d-edge (19/25, 1/20) | 1 | 19/10 | **2** |\n| e-edge (6/25, 19/20) | 12/25 | 38/25 | 38/25 |\n\n**Readings.**\n\n1. **Away from the edge.** For a < 1 (δ < 19/25), the whole R=0 class, including the unequal proportional pairs of F4, is a power below x². That holds at 36290 of the 36481 lattice boxes. It closes F4's scope gap at those boxes. It is an upper bound on one piece of |T|², not a bound on |T| or on piece (c) with R ≠ 0.\n\n2. **On the edge.** On the a = 1 edge (δ = 19/25: 191 boxes, including the corner and the d-edge), (★) gives x^(2+ε), which is no saving.\n\n3. **Why it stops there** (heuristic). The term M²N/A comes from classes with q ≍ N/A = x^(1−a). At a = 1 these have bounded denominators.\n - For q = p = 1 (h = u, which exists when A ≍ N, i.e. M ≍ x) the phase is trivial: e_u(σθu m̄) = 1, and Φ_{u,u} depends on m only.\n - That class contributes Σ_u b_u c_u G(u,u) to T, where G(u,u) = Σ_{m∈I_m, (m,u)=1} A_left(gm) Φ_{u,u}(m) is an unoscillating interval sum of the Möbius-type coefficient of `grouped-divisor-moment.md` (13): μ(ℓ)ℓ^(−s), or its log/Λ variant.\n - With only |b_u| ≤ B and |c_h| ≤ C/A, the signs can align and give (BC/A)Σ_u |G(u,u)|. That is x^(1−η) only if |G(u,u)| ≪ M·x^(−η) uniformly in u.\n - Such a power saving for a Möbius sum over an interval is not known unconditionally; the zero-free region gives only exp(−c√log x). Bounded q > 1 gives the same sums twisted by e_q(σθp m̄).\n\n So at the corner, any cancellation from the R=0 class has to come from the actual coefficients: the correlation along multiples β_s(q,p) = Σ_{s|t} b_{tq} c_{tp}, or signs inside G^{(s)}_{q,p}. Positivity plus a geometric series in the class variable reaches exactly x² there. This agrees with the §5 heuristic ceiling (block x¹ at the corner) and does not contradict any proven bound.\n\n**Falsifiers.**\n- An error in the per-residue pair count, the p-range (q ≥ N/(2A)), or the perturbation step.\n- A choice of |b| ≤ B, |c| ≤ C/A and an admissible H for which R0 exceeds (★) by a power at a box with a < 1.\n\n## 5. What remains open\n\n- The corner, and the whole a = 1 edge, for the R=0 class.\n- Piece (c) with R ≠ 0.\n- A signed estimate of β_s, or of the bounded-q class sums.\n- The target-box deficit of 1/40 (41/40 from left-divisor-signs).\n\nNo region is added.\n\n## 6. Suggested record changes (for the owner)\n\n- **§4 prose:** \"nu<2/5\" should read \"nu<9/20\".\n- **Ledger or OUTCOMES addition:** \"The full R=0 class h1/u1=h2/u2 is ≪ x^ε B²C²f²(1+v)[M²N/A+MN²/A²] for A≤N (unreviewed, job #247): below x² on the top band for δ<19/25, x^(2+ε) on the δ=19/25 edge, where bounded-denominator classes have trivial phase.\"\n\n## Channel\n\nClaim: msg 339. Found: msg 349.\n\n## Sources\n\nServed documents, snapshot `main`, no local-only sources:\n- `research/signed-moment.md` (served sha256 f83eb30929ecdfffa484af3e0d0e6cce07f08418e063155a8089abce1bfe2222).\n- `research/signed-moment-validation.js`: embedded out-sha256 e85405ab….\n- `research/history/reviews-0906/20-independent-handoff-review.md` §F4.\n- `research/grouped-divisor-moment.md` (13).\n- `research/OUTCOMES.md`: entry \"Signed-moment — Alternatives to the first Cauchy inequality\".\n- `research/QUESTIONS.md`: row `Q-signed-moment`.\n\nThe statement about unconditional Möbius sums is standard: the zero-free region of ζ and Dirichlet L-functions gives only exp(−c√log x) savings. No specific source was re-read for it.\n\n## Transcript\n\nRemoved or redacted: the bearer token, platform and harness session ids, account and organisation identifiers, UUIDs, absolute local paths, the person's email, the contents of local memory, the local notebook and user-context reminders, and every line from earlier assignments (jobs #220 and #233).\n","patch":null,"cpu_hours":0,"hashes":{"signed-moment-validation.stdout":"e85405abc7c2aaad9fe65ad58de77625f34622ab70b3d3de4c26e1d8f36c62fd"},"author_rung":"heuristic","status":"recorded","final_rung":"recorded","created_at":"2026-09-11T15:47:35.907Z","repo_url":null,"commit":null,"cites":{"files":[],"handles":[],"returns":[],"messages":[]},"tokens":{"log":"claude-code","input":256,"models":{"claude-opus-5":47690},"output":47690,"source":"claude-jsonl","entries":8,"cache_read":2627501,"cache_write":67416},"paper_slug":null,"revision_path":null,"revision_sha":null,"recipe_md":"1. `node research/signed-moment-validation.js | shasum -a 256` should give e85405abc7c2aaad9fe65ad58de77625f34622ab70b3d3de4c26e1d8f36c62fd, the embedded out-sha256. Takes 0.4 s.\n2. Lattice check of the §4 prose slip and of the R=0 exponents: exact fractions, under 1 s.\n```python\nfrom fractions import Fraction as F\nn=191; c={'budgets':0,'nu<9/20':0,'nu<2/5':0}; r0=0\nfor i in range(n):\n d=F(6,25)+F(13,25)*i/(n-1)\n for j in range(n):\n v=F(1,20)+F(9,10)*j/(n-1); a=d+F(6,25); b=v+F(1,20)\n if max((1+a+b)/2,b+F(1,2),a+b/2,b+a/2)<1: c['budgets']+=1\n if d+v