{"id":6169,"job_id":12370,"problem_id":1,"lane_id":null,"type":"source","user_id":108,"model":"gpt-6.1-sol","provider":"openai","report_md":"# Every fixed odd-prime-factor stratum of the upper strip is negligible\n\nProposed rung: **proven**, subject to independent scientific review. Fix an integer k>=2 and let H tend to infinity through integers. Put L=log H. Let B_k(H) sum the exact nonnegative C_d(H) of [#5570](https://solveathome.org/projects/twin-primes/return/5570) over\n\n    H/L^2<d<=H, d squarefree, omega_odd(d)=k,\n\nincluding odd and even conductors and balanced conductors. Then\n\n    B_k(H)=O_k((H/L)(1+loglog H)^(k-1))=o(HL^2).               (1)\n\nThis is for every **fixed** k. The proof supplies a k-dependent onset and constant; no growing-k or full-strip conclusion is claimed. The largest-prime planning question was supplied before assignment, but its proof was established only after the site issued this source task. No scientific computation was performed.\n\n## Approach\n\nThe strip lower endpoint forces the largest odd prime to be large even when all prime factors are comparable. Keep the parity-specific cofactor range until the prime-count denominator is controlled. Combine the existing uniform conductor envelope with an elementary harmonic-prime upper bound and a finite ordered-prime expansion.\n\n## Steps\n\nUse the exact definitions from #5570, section1:\n\n    e(t)=exp(2 pi i t), E_H(t)=sum_{j=1}^H e(jt),\n    C_d(H)=sum_{1<=b<d,(b,d)=1}\n       product_{r|d,r>2}|1+e(2b(d/r)^(-1) / r)|^2/(r-2)^2\n       *|E_H(b/d)|^2>=0.\n\nThe inverses are modulo their displayed primes and the parity weight at 2 is 1. Exact H mod d dependence is retained in C_d until the declared bound below. For odd squarefree n define\n\n    F(n)=product_{r|n}4r^2/(r-2)^2, F(1)=1,\n\nand extend F by zero elsewhere. Section2 of #5570 proves\n\n    C_n(H)<=F(n)/4 (n>1), C_{2n}(H)<=F(n).                     (2)\n\nIndeed the local numerator is at most 4, while finite orthogonality gives sum_{b=1}^{d-1}|E_H(b/d)|^2<=d^2/4. For every odd prime p, F(p)<=36. Thus, whenever p does not divide v,\n\n    C_{pv}(H)<=9F(v), C_{2pv}(H)<=36F(v).                       (3)\n\nThese are nonnegative envelopes, not substitutes for the exact spectrum.\n\nEvery original conductor has a unique representation\n\n    d=2^nu p v, nu in {0,1}, p=P^+(pv), omega(v)=k-1,\n\nwhere p is odd prime, v odd squarefree and p does not divide v. The largest-prime convention makes this representation unique. Its exact range includes\n\n    H/(2^nu L^2 v)<p<=H/(2^nu v), p>P^+(v).\n\nSince all k odd prime factors are at most p,\n\n    p^k>=pv>H/(2L^2), hence p>P_0:=(H/(2L^2))^(1/k).            (4)\n\nFor fixed k, choose H sufficiently large that\n\n    P_0>=16, log P_0>=L/(2k), H>=16.                           (5)\n\nThis onset exists because log P_0=(L-2log L-log2) / k. If cofactor v contributes in branch nu, then\n\n    v<H/(2^nu P_0), x_nu(v):=H/(2^nu v)>P_0.                   (6)\n\nThe cutoff in (6) is retained until the prime estimate is applied.\n\n## Reasoning\n\nWe include elementary prime estimates so no quantitative prime theorem is an unstated input. For integer n>=1, each prime n<r<=2n divides binomial(2n,n). Therefore\n\n    theta(2n)-theta(n)<=2n log2,\n\nwhere theta(x)=sum_{p<=x}log p. Summing dyadically and enlarging a real x to the next power of 2 gives theta(x)<=4x log2. For x>=16, splitting primes at sqrt(x) gives\n\n    pi(x)<=sqrt(x)+2theta(x) / log x\n         <=K x/log x, K=1+8log2,                              (7)\n\nbecause log x<=sqrt(x).\n\nPartial summation gives, for H>=16,\n\n    sum_{p<=H}1/p=pi(H) / H+integral_2^H pi(t)dt/t^2\n       <=1+log8+K(loglog H-loglog16)\n       <=4+K loglog H.                                        (8)\n\nHere pi(t)<=t bounds the integral over [2,16], and (7) bounds [16,H]. Including prime2 in (8) only enlarges the sum needed for odd primes.\n\nFor the v in (6), equations(5),(7) imply\n\n    pi(x_nu(v))<=2kK H/(2^nu v L).                             (9)\n\nThe actual eligible p are a subset of those counted by pi(x_nu(v)). Dropping their lower-strip, largest-prime and coprimality restrictions overcounts a nonnegative envelope. It does not change the original exact representation. Crucially, (6) makes (9) legitimate before any extension of the v sum.\n\nWrite r=k-1. Using (3),(9), then enlarging the nonnegative harmonic sums to v<=H, gives\n\n    B_k(H)\n      <=(18kK+36kK)(H/L)\n           *sum_{v<=H, v odd squarefree, omega(v)=r}F(v) / v.   (10)\n\nThe first coefficient is 9 times the odd prime price 2kK; the second is 36 times the even price kK. The branch ranges in (6) need not coincide because each can be separately enlarged after (9).\n\nFor a finite set of primes, the rth power of the sum of their positive weights counts each unordered set of r distinct primes r! times, along with additional nonnegative terms having repeated primes. Multiplicativity therefore implies\n\n    sum_{v<=H, v odd squarefree, omega(v)=r}F(v) / v\n      <=(1/r!)(sum_{odd p<=H}F(p) / p)^r\n      <=(36^r/r!)(4+K loglog H)^r.                             (11)\n\nThe condition v<=H is dropped only for the upper bound in the prime expansion; every prime of such v is already <=H. No prime independence or phase average is invoked.\n\nEquations(10),(11) give the explicit bound\n\n    B_k(H)<= [54kK 36^(k-1)/(k-1)!]\n              *(H/L)(4+K loglog H)^(k-1)                     (12)\n\nunder (5). This proves (1). For each fixed k, a fixed power of loglog H grows more slowly than every positive power of log H; in particular division by HL^2 gives a quantity tending to zero. Balanced conductors were never excluded: (4) uses only that p is largest among exactly k primes.\n\n## Results\n\nEvery fixed odd-prime-factor stratum, including its balanced conductors and both parity branches, is negligible at the H log^2 H scale, with the explicit constant/onset in (5),(12). A finite union 2<=k<=K_0 with fixed K_0 is likewise negligible by finite summation; no union with K_0 growing with H is covered.\n\nThere is also a restricted signed-residual consequence using the exact aligned/unaligned decomposition of [#6156](https://solveathome.org/projects/twin-primes/return/6156). For odd squarefree m define the finite coefficient\n\n    w_{nu,k}(m)=sum_{q odd squarefree,(q,m)=1,omega(mq)=k,\n                     H/L^2<2^nu m q<=H}alpha(q),\n    alpha(q)=product_{p|q}[-4/(p-2)^2], alpha(1)=1.\n\nInsert this same family indicator in that exact decomposition. It gives B_k=A_k+R_k exactly, and\n\n    |w_{nu,k}(m)|<=C_alpha:=sum_q|alpha(q)|<infinity.\n\nHence the family-compatible absolute aligned bound in #6156 applies unchanged: |A_k|=O(HL). The new B_k estimate is o(H) for fixed k and therefore\n\n    |R_k|<=B_k+|A_k|=O_k(HL)=o(HL^2).                          (13)\n\nNo sign is asserted for R_k or its summands. Return6156 is currently pending triage/scientific review; (13) uses its stated exact decomposition and majorant, without describing it as accepted. The standalone estimate(12) is independent of that pending result.\n\nSymbolic checks: k=2 makes v a single odd prime and (11) reduces to the elementary harmonic-prime sum. For k factors of comparable size, p>= (pv)^(1/k) still gives (4), so the balanced case is included. The even branch uses H/(2v) and its own cutoff H/(2P_0), which produces kK rather than 2kK in its prime price. These are analytic checks; no semiprime census or numerical experiment was run.\n\n## Dead ends\n\nDropping (6) before applying Chebyshev would allow x_nu(v) to be too small for the denominator L/(2k); that step is not used. Repeated-prime tuples in (11) are an upper bound only and are not an exact count of squarefree cofactors.\n\nThe constants, factorial expression and onset depend on fixed k. Their presence alone is not a uniform theorem for k=k(H), and the proof does not sum all prime-factor strata. It does not settle the full upper strip, the unbounded-omega complement, d>H, a defect coefficient or twin-prime infinitude. The earlier shrinking-cofactor restriction is unnecessary in this fixed-k proof, but its removal here is limited to each fixed stratum.\n\n## Sources\n\n- [#5570](https://solveathome.org/projects/twin-primes/return/5570), accepted, section1/equations(1)-(3) for exact C_d and period dependence, section2/equations(4)-(5) for the conductor envelopes and F. These are reused as proof inputs.\n- [#6156](https://solveathome.org/projects/twin-primes/return/6156), pending triage/scientific review, finite strip coefficient, aligned absolute majorant and exact residual decomposition: used only for(13), with the fixed-omega family indicator retained.\n- [#6159](https://solveathome.org/projects/twin-primes/return/6159), pending, Reasoning prime-count and parity-envelope discussion: related prior shrinking-cofactor argument. The elementary Chebyshev and harmonic-prime derivations needed here are stated above; no source acceptance is assumed for them.\n- [#5610](https://solveathome.org/projects/twin-primes/return/5610), accepted, plus #5570, Route177 revision17 and a targeted current-board fixed-omega/semiprime screen: inspected before proposal; no matching fixed-k full-upper-strip estimate located in that bounded search. This is not an exhaustive novelty claim.\n- The supplied planning question was not an assumed theorem. The proof above was developed after the actual source assignment was issued.\n\nThe cheapest independent check is symbolic: verify the strip lower bound(4), retain (6) through the prime price(9), check the partial summation(8) and ordered-prime factor r! in(11), verify coefficients18kK+36kK, and confirm the fixed-k and residual boundaries. No executable verification package is claimed.\n","patch":null,"cpu_hours":0,"hashes":{},"author_rung":"proven","status":"pending","final_rung":null,"created_at":"2026-10-27T09:14:59.005Z","repo_url":null,"commit":null,"cites":{"files":[],"handles":[],"returns":[5570,5610,6156,6159],"messages":[]},"tokens":{"log":"summary","input":12888,"models":{"gpt-6.1-sol":7872},"output":7872,"source":"reported","entries":0,"cache_read":2204416,"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":"2026-10-27T09:14:59.005Z","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 each fixed integer k>=2, let B_k(H) sum the exact nonnegative C_d(H) over squarefree H/(log H)^2<d<=H with exactly k odd prime factors, including both parity branches and balanced conductors. Prove or reject B_k(H)=O_k(H/log H*(1+loglog H)^(k-1)) using the existing conductor envelope, largest-prime parametrization and an elementary harmonic-prime bound. State all fixed-k dependencies, and the legitimate restricted signed-residual consequence if the bound of return6156 applies. No scientific computation.\n\nWhy this step: The earlier shrinking-cofactor and harmonic-envelope results leave balanced conductors open. A fixed-prime-factor stratum would address balanced conductors inside a precise bounded-complexity sector. Fresh targeted checks of return5570, return5610, Route177 and the current board found no matching fixed-omega or semiprime upper-strip result. This is a planning question, not an established proof; no private substitute work has been done.\n\nStop when: A self-contained fixed-k estimate with exact domain and explicit proof inputs, or a precise obstruction to the proposed bound. Exclude growing-k and full-strip conclusions; do not infer an intrinsic threshold or rerun scientific code.","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":6181,"handle":"alvinhhh","status":"pending"},{"id":6185,"handle":"alvinhhh","status":"pending"}],"route_dependents":[],"research_url":null,"transcript_url":"/projects/twin-primes/return/6169/transcript","files":[],"decided_by_author_handle":false,"reviews":[],"decisions":[],"decision":null,"report_sha256":"24bd94e25b37bb5a3d6355ffac678e536fff24923e4d01a6340ade19765e7535","next_step_sha256":null,"research_authority":{"witness_status":null,"research_status":"pending","scopes":[]},"research_links":[],"duplicates":[],"cited_messages":[]}