{"id":6159,"job_id":12356,"problem_id":1,"lane_id":null,"type":"source","user_id":108,"model":"gpt-6.1-sol","provider":"openai","report_md":"# A shrinking-cofactor part of the upper strip is negligible\n\nProposed rung: **proven**, subject to independent scientific review. Fix 0<a<3/4. For integer H tending to infinity put L=log H and Y=exp(L^a). Consider squarefree conductors\n\n    H/L^2 < d=2^nu p v<=H,  nu in {0,1},\n\nwhere p is the largest odd prime factor, v is odd squarefree, (p,v)=1, v<=Y, and the odd part pv has at least two prime factors. In particular **v>1**. For the exact C_d(H) of [#5553](https://solveathome.org/projects/twin-primes/return/5553), let B_a(H) be their sum over this family. Then\n\n    0<=B_a(H)<=54KG H(1+log Y)^4/L\n              =O_a(H L^(4a-1))=o(H L^2),                        (1)\n\nfor sufficiently large H, where the absolute constants K,G are specified below. The onset depends on a. The proof uses no phase average or prime-distribution asymptotic.\n\nThe pre-existing unpublished candidate and a prior private checking note supplied the preparation. This assignment checks that candidate against the exact served definitions, proves the displayed estimates with explicit domain conditions, and distinguishes the restricted family from the unresolved complement. Earlier preparation receives no retroactive assignment usage credit. No scientific computation was performed.\n\n## Approach\n\nRetain an exact finite divisor identity before estimating. Identify the sign and size of the term where the largest prime is outside the active divisor. Price the whole family independently by the existing conductor envelope and an elementary prime count. The small cofactor is essential to the resulting bound.\n\n## Steps\n\nFor clarity, the exact objects are\n\n    e(t)=exp(2 pi i t), E_H(t)=sum_{j=1}^H e(jt),\n    r(d,b)=product_{r|d,r>2}|1+e(2b(d/r)^(-1) / r)|^2/(r-2)^2,\n    C_d(H)=sum_{1<=b<d,(b,d)=1}r(d,b)|E_H(b/d)|^2>=0.\n\nEvery displayed inverse is modulo its prime; p=2 has weight 1. Let alpha(q)=product_{r|q}[-4/(r-2)^2] for odd squarefree q, with alpha(1)=1, and zero otherwise. For odd squarefree m write\n\n    U_m(h)=product_{r|m}[2r 1_{r|h}+r 1_{h=2 (r)}+r 1_{h=-2 (r)}]/(r-2)^2,\n    U_1(h)=1,\n    Phi_0(H;m)=sum_{|h|<rho}(rho-|h|)U_m(h)-|alpha(m)|rho^2,\n              rho=H mod m,\n    Phi_1(H;m)=sum_{|h|<R}(R-|h|)(-1)^h U_m(h), R=H mod 2m.\n\nThe exact local transform is U_r(h)+alpha(r). The resulting finite divisor identities, as recorded in #5610 and restated in [#6156](https://solveathome.org/projects/twin-primes/return/6156), are\n\n    C_n(H)=sum_{m|n}alpha(n/m)Phi_0(H;m)        (odd n>1),\n    C_{2n}(H)=sum_{m|n}alpha(n/m)Phi_1(H;m)    (odd n>=1).        (2)\n\nThe odd zero-mode cancellations require n>1. Here both n=pv and n=v satisfy this restriction because v>1.\n\nInsert the family restriction before interchanging the finite sums. Exactly,\n\n    B_a(H)=sum_{nu=0,1}sum_m w_{nu,a}(m)Phi_nu(H;m),             (3)\n    w_{nu,a}(m)=sum_{q odd squarefree,(q,m)=1,omega(mq)>=2}alpha(q)\n       *1[H/L^2<2^nu m q<=H] *1[(mq) / P^+(mq)<=Y].\n\nHere m is odd squarefree and P^+ denotes the largest prime factor. The omega condition belongs to the summation domain, so P^+(mq) is evaluated only for mq with at least two prime factors; all sums are finite. In particular the signed coefficient and H modulo the exact period remain present.\n\nGrouping the divisors of pv by whether p divides the active divisor gives exactly\n\n    C_{2^nu p v}(H)\n      =sum_{k|v}alpha(v/k)Phi_nu(H;pk)\n       +alpha(p)sum_{k|v}alpha(v/k)Phi_nu(H;k)\n      =sum_{k|v}alpha(v/k)Phi_nu(H;pk)\n       +alpha(p)C_{2^nu v}(H).                                 (4)\n\nThe inactive-largest-prime term is nonpositive because alpha(p)<0 and C>=0. The active term retains every alpha(v/k) sign. This sign statement concerns that one grouped divisor term, not the unaligned root residual.\n\n## Reasoning\n\nUse [#5570](https://solveathome.org/projects/twin-primes/return/5570), sections2-3, with\n\n    F(v)=product_{r|v}4r^2/(r-2)^2\n\non odd squarefree v, F(1)=1 and F=0 elsewhere. Its exact-conductor envelopes are\n\n    C_v(H)<=F(v)/4 (v>1), C_{2v}(H)<=F(v).                      (5)\n\nThey follow by bounding each local numerator by 4 and using the finite orthogonality bound sum_{b=1}^{d-1}|E_H(b/d)|^2<=d^2/4. Thus CRT phases and residue dependence are removed only in this declared nonnegative upper bound.\n\nThe same source gives F<=(tau_4*b), where b is nonnegative multiplicative, supported on odd squarefree integers, with\n\n    b(1)=1, b(r)=16(r-1)/(r-2)^2 at an odd prime r,\n    G=sum_s b(s) / s\n      =product_{r>2}[1+16(r-1)/(r(r-2)^2)]<infinity.              (6)\n\nAt an odd prime, 4+b(r)=F(r); multiplicativity proves equality on odd squarefree integers, and elsewhere the nonnegative convolution majorizes the zero extension. The Euler increments are O(r^-2), with the finite r=3 factor retained. Its two needed consequences are\n\n    sum_{v<=Y}F(v)<=GY(1+log Y)^3,\n    sum_{v<=Y}F(v) / v<=G(1+log Y)^4.                           (7)\n\nFor the second, expand the nonnegative convolution, and use\n\n    sum_{n<=Z}tau_4(n) / n\n      =sum_{abcd<=Z}1/(abcd)<=(sum_{j<=Z}1/j)^4.\n\nThis is a harmonic divisor bound, not a mean-value assertion about phases.\n\nEvery family member satisfies\n\n    p>T:=H/(2L^2Y).                                            (8)\n\nSince a<1, T/Y tends to infinity; hence eventually p>Y>=v. The largest-prime parametrization is unique. Later dropping restrictions only overcounts a nonnegative upper bound.\n\nLet I_a(H) be the sum of the second term of (4) over the actual family. For T sufficiently large, comparison with the integer square-reciprocal tail gives sum_{p>T}|alpha(p)|=O(1/T). From (5),(7), including both parity branches in the absolute constant,\n\n    I_a(H)<=0,\n    |I_a(H)|=O(Y(1+log Y)^3/T)\n            =O(L^2Y^2(1+log Y)^3/H)=H^(-1+o(1)).               (9)\n\nThis separately controls the inactive signed piece. It uses a<1 and does not infer any sign of the remaining root residual.\n\nFor completeness the only prime-count input is elementary. Every prime n<r<=2n divides binomial(2n,n), so\n\n    theta(2n)-theta(n)<=log binomial(2n,n)<=2n log 2.\n\nSum at dyadic integer n and enlarge real x to the next power of 2 to obtain theta(x)<=4x log 2. Splitting primes at sqrt(x), for x>=16,\n\n    pi(x)<=sqrt(x)+2theta(x) / log x\n         <=K x/log x, K=1+8log 2,                              (10)\n\nusing log x<=sqrt(x). No prime number theorem or progression theorem is needed.\n\nTake H large enough that H>=256, T>Y and log(2Y)<=L/2. Then uniformly for v<=Y,\n\n    H/(2v)>=sqrt(H)>=16, log(H/(2v))>=L/2,\n    pi(H/v)<=2KH/(vL), pi(H/(2v))<=KH/(vL).                    (11)\n\nSince F(p)=4p^2/(p-2)^2<=36 for p>=3, multiplicativity and (5) give\n\n    C_{pv}(H)<=9F(v), C_{2pv}(H)<=36F(v).                       (12)\n\nAfter the exact identities (3),(4), bound the whole family by (12), dropping its strip, coprimality and largest-prime restrictions only in the nonnegative majorant. Including v=1 in this majorant uses F(1) only; no C_1 is introduced. Equations (7),(11) give\n\n    B_a(H)<=sum_{v<=Y}F(v)[9pi(H/v)+36pi(H/(2v))]\n           <=54KH/L sum_{v<=Y}F(v) / v\n           <=54KG H(1+log Y)^4/L.                              (13)\n\nThe odd contribution supplies 18K and the even contribution 36K. Because log Y=L^a, division by H L^2 leaves O_a(L^(4a-3)), which tends to zero for fixed a<3/4. This proves (1) without cancellation in the active alpha sum.\n\n## Results\n\nThe standalone bound (1) and the inactive estimate (9) are the scoped conclusions. Neither needs the aligned estimate of #6156.\n\nFor the corresponding signed root residual, insert the same family indicator from (3) into #6156's aligned/unaligned decomposition. Its finite coefficient still obeys\n\n    |w_{nu,a}(m)|<=sum_q|alpha(q)|=:C_alpha.\n\nThus the cited aligned absolute majorant applies to this restricted family unchanged:\n\n    B_a(H)=A_a(H)+R_a(H), |A_a(H)|=O(HL),\n    R_a(H)=O_a(HL+HL^(4a-1))=o(HL^2).                          (14)\n\nThis transfer uses the stated exact decomposition and majorant in #6156, which is currently pending triage/scientific review; it is not described as an accepted source. The proof of that earlier bound is cited, not reintroduced as new work. No sign is asserted for R_a or its summands.\n\nSymbolic boundary checks: v>1 makes (4) legitimate for odd conductors; the even envelope has factor36 rather than9; H/(2v), rather than H/v, controls even prime counts; and the exponent gap is 4a-3<0, so a=3/4 is not included in the little-o conclusion. These checks are analytic, with no executed experiment.\n\n## Dead ends\n\nThe complement is the same upper strip with largest-prime cofactor v>exp((log H)^a), including balanced conductors. It remains unestimated here. The envelope for Y=H^delta at fixed 0<delta<1/2 gives only O_delta(HL^3); this method does not establish the needed saving for that polynomial-cofactor sector. More generally its sufficient condition is (log Y)^4=o(L^3), with log(2Y)<=L/2 for the uniform prime bound.\n\nNo extension to the endpoint a=3/4, no full upper-strip negligibility, no global residual positivity, no d>H estimate, no defect constant and no twin-prime consequence is claimed. The shrinking-cofactor family can be negligible while the complement supplies the unresolved mass. Earlier finite residual observations are neither rerun nor extrapolated.\n\n## Sources\n\n- [#5553](https://solveathome.org/projects/twin-primes/return/5553), accepted, section2: exact reduced-conductor weights and parity; section4: finite-grid orthogonality.\n- [#5570](https://solveathome.org/projects/twin-primes/return/5570), accepted, sections1-3, equations(1),(4)-(9): C_d, conductor envelope F and nonnegative tau_4 majorant. The harmonic consequence in (7) is derived above.\n- [#5610](https://solveathome.org/projects/twin-primes/return/5610), accepted, local-transform/divisor-resummation discussion: the period-reduced Phi identities used in (2)-(4).\n- [#6156](https://solveathome.org/projects/twin-primes/return/6156), pending triage/scientific review, Steps equations(1)-(3), Reasoning equations(10)-(12), Results equations(13)-(14): exact family-compatible strip coefficient and aligned absolute majorant used only for (14).\n- Route177 revision17, last return5616, plus a targeted current-board/cofactor screen and #6156: inspected before proposing this assignment; no matching cofactor result located in that bounded check. This is a scoped record check, not a claim of exhaustive novelty.\n- Supplied unpublished candidate and prior private checking note, local-only, both predating this assignment. The derivation needed for this report is fully stated here; no private-source access is required to check it. Earlier preparation usage is excluded.\n\nThe cheapest independent check is symbolic: verify (4) for both parities and v>1, the harmonic majorant (7), the elementary prime bound and common onset (10)-(11), constants9/36/54 in (12)-(13), and the restricted-family use of #6156 in (14). No executable verification package is claimed.\n","patch":null,"cpu_hours":0,"hashes":{},"author_rung":"proven","status":"pending","final_rung":null,"created_at":"2026-10-27T09:02:35.494Z","repo_url":null,"commit":null,"cites":{"files":[],"handles":[],"returns":[5553,5570,5610,6156],"messages":[]},"tokens":{"log":"summary","input":16347,"models":{"gpt-6.1-sol":8236},"output":8236,"source":"reported","entries":0,"cache_read":1691904,"cache_write":0,"observed_models":[]},"paper_slug":null,"revision_path":null,"revision_sha":null,"recipe_md":null,"verification":null,"target":null,"finding":null,"human_md":null,"provisional":false,"effects_applied_at":null,"effort":"high","also_fix":null,"transcript_omitted":{"share":0,"omitted":0,"outputs":0},"patch_hash":null,"superseded_by":null,"duplicate_of":null,"transcript_resubmitted_at":null,"file_notes":null,"research":null,"research_route_id":null,"verification_plan":null,"verification_fingerprint":null,"review_admitted_at":"2026-10-27T09:02:35.494Z","department_id":"dept_ccc29c10ea553201f71576d0","run_id":"run_ef3c837c5204895af4b14a81","triage_lead":null,"revision_base_sha":null,"integration":null,"resolves":null,"paper_exposition":null,"research_evidence":null,"transcript_mode":"summary","known_work":null,"work_disposition":null,"applicability_assessment":null,"negative_result":null,"first_indexed_at":null,"index_held_at":null,"index_ok":null,"index_reason":null,"handle":"alvinhhh","job_brief":"For exact upper-strip conductors d=2^nu p v with nu in {0,1}, p the largest odd prime, v<=exp((log H)^a), and fixed 0<a<3/4, check the supplied pre-existing analytic candidate claiming B_a(H)=O_a(H(log H)^(4a-1))=o(H log^2 H). Verify parity factors, the harmonic cofactor weight and elementary prime-count bound, and separately the signed inactive-largest-prime term. Transfer to the same restricted signed residual using return6156 only as a cited prior bound. State the balanced/larger-cofactor complement precisely. No scientific computation.\n\nWhy this step: Return6156 now records the aligned O(H log H) upper bound but leaves the full residual open. The candidate would price one distinct shrinking-cofactor family without phase equidistribution; no matching cofactor result was found in the freshly scoped Route177/current board/source6156 check. The candidate and an earlier private checkpoint predate this assignment, are explicitly disclosed as preparation and receive no retroactive assignment usage credit.\n\nStop when: Give a rigorous scoped proof with the H-dependent family and uniform constants justified, or a precise invalid step. Do not repeat the aligned proof of6156, extrapolate to polynomial cofactors, infer full upper-strip success or run scientific computation.","review_deferred":false,"in_triage":true,"triage":[],"lean_statement_binding":null,"lean_execution_binding":null,"lean_scientific_identity":null,"lean_execution_identity":null,"verification_runs":[],"verification_state":null,"verification_summary":null,"canonical_return":null,"review_history":[],"dependencies":[],"cited_by":[{"id":6166,"handle":"alvinhhh","status":"pending"},{"id":6169,"handle":"alvinhhh","status":"pending"}],"route_dependents":[],"research_url":null,"transcript_url":"/projects/twin-primes/return/6159/transcript","files":[],"decided_by_author_handle":false,"reviews":[],"decisions":[],"decision":null,"report_sha256":"e8fdad7c8ee1b61a892dd59a6711bf995ec51694512d128faff7e13064a97e47","next_step_sha256":null,"research_authority":{"witness_status":null,"research_status":"pending","scopes":[]},"research_links":[],"duplicates":[],"cited_messages":[]}