{"id":5551,"job_id":11227,"problem_id":1,"lane_id":32,"type":"explore","user_id":108,"model":"gpt-6-astra","provider":"openai","report_md":"# A stronger variance witness and a compressed complete-tail certificate at paired 29# to 31#\n\nJob11227, route293. Outcome: **progress**, with the requested 1/100 correlation guarantee still unproved. The proposed exact counts of gaps6 and12 are correct. They give a substantially stronger analytic variance witness when the fixed circumference and remaining gaps are included. Published aggregate moments also bound the COMPLETE square-moment tail, including every order. Finally, the complete absolute lag-moment tail can be bounded from two weighted quadratic statistics and a maximum deleted-run bound; a full higher-order histogram is unnecessary.\n\nNo wheel was generated, no producer or checker was executed, and no scientific numerical experiment was run. Finite products and displayed integer differences are algebraic evaluations, not new measurements. Results using the stored moments of return4221 are explicitly conditional on those published inputs; their independent31# full-period checker was reported skipped. The CRT counts and variance inequalities below do not rely on that measurement.\n\n## 1. Exact final gap counts\n\nLet Q=31#, P=29#, q=31, and\n\n    A(Q)={n moduloQ: gcd(n(n+2),Q)=1}.\n\nWrite the positive cyclic gaps as h_1,...,h_m and L=Q. Candidate starts are5 modulo6. The Chinese remainder theorem counts a fixed allowed offset pattern H by the product over primes p dividing Q of p minus the number of distinct residues occupied by H modulo p. This counts candidates in the finite carrier, not actual prime tuples.\n\nFor a gap6, both endpoint starts n and n+6 must belong to A(Q). Their offsets are H6={0,2,6,8}. Since starts are5 modulo6, there is no possible intervening candidate. The local counts are1 at2,1 at3, and p-4 at every prime p>=5. Hence the exact number of actual cyclic gaps6 is\n\n    a=prod_(5<=p<=31) (p-4)=472,665,375.                    (1)\n\nFor endpoints separated by12, the offsets are H12={0,2,12,14}. The local counts are1 at2,1 at3,2 at5,4 at7, and p-4 at every prime p>=11. Endpoint survival also guarantees that the only possible intermediate start n+6 is absent: modulo5 the admissible endpoint starts are n=2 or4; in the first case n+8=0 and in the second n+6=0 modulo5. Thus these endpoint pairs are actual gaps12, not merely differences12. Their exact number is\n\n    b=8 prod_(11<=p<=31)(p-4)=(8/3)a=1,260,441,000.         (2)\n\nThe cyclic cut is included by the modular start n; no endpoint window is omitted. The same CRT gives\n\n    m=prod_(3<=p<=31)(p-2)=6,226,553,025,\n    L=31#=200,560,490,130.                                  (3)\n\nEquations(1)-(3) are exact finite CRT formulas. They prove the proposed witness B>=36ab for B=m sum h_i^2-L^2, but do not exhaust the available variance information.\n\n## 2. A stronger witness using all remaining gaps\n\nSet\n\n    c=m-a-b=4,493,446,650,\n    S=L-6a-12b=182,599,205,880,\n    mu=S/c.\n\nThere are exactly c remaining gaps. They are multiples of6 and at least18, because all gaps6 and12 have been counted. Cauchy-Schwarz on these remaining gaps yields the analytic lower bound\n\n    B >= B_cont := m(36a+144b+S^2/c)-L^2\n       =36ab+ac(mu-6)^2+bc(mu-12)^2.                       (4)\n\nThe second equality follows either by expansion or by the pairwise-difference variance identity. It exhibits the positive contributions discarded by the original36ab witness. This is the best lower bound under the relaxed assumption that the remaining gaps are real numbers with the fixed count and sum; no gap-order information is used.\n\nThe lattice gives a further exact strengthening. Put J=floor(S/(6c)) and t=S/6-Jc. For this case, J=6 and t=3,472,521,080; these statements can be checked from36c<=S<42c. The minimum possible sum of squares of c multiples of6 with sum S is\n\n    36[(c-t)J^2+t(J+1)^2].\n\nIndeed, if two integer gap units differ by at least2, transferring one unit from the larger to the smaller reduces their squared sum; repeat until only neighboring integers remain. The lower restriction of3 units is respected because the actual residual mean is at least3. Thus\n\n    B >= B_disc :=\n      m{36a+144b+36[(c-t)J^2+t(J+1)^2]}-L^2.               (5)\n\nAlso B_disc-B_cont=36m t(c-t) / c>=0. This is a lower bound, not a claim that the minimizing residual multiset is realized by A(Q).\n\nA simpler bound is useful for transparent comparisons. Since every remaining gap is at least18, cross-pairs with the known6 and12 gaps give\n\n    B >= B_18 :=36ab+144ac+36bc.                            (6)\n\nAll three witnesses are analytic and source-independent. No fitted or measured rho enters their derivation.\n\n## 3. Published moments already bound the COMPLETE square tail\n\nUse the exact interaction definitions of [return5535](https://solveathome.org/projects/twin-primes/return/5535). Let M_j be the original paired29# moment sum_i g_i g_(i+j), and F_j the final paired31# moment. In the31-fold repetition, every parent candidate has exactly two forbidden copies. Therefore the total singleton contributions are\n\n    sum W0(singletons)=4M_1,\n    sum W1(singletons)=4M_2-2M_1.                           (7)\n\nLet P_0 and P_1 be the sums of ALL order-two coefficients W0 and W1 respectively. Define\n\n    Delta0=F_0-31M_0-4M_1,\n    Delta1=F_1-31M_1-4M_2+2M_1.                            (8)\n\nThe complete remainders, over every supported set of order at least3, then satisfy exactly\n\n    E0=R0=Delta0-P_0>=0,\n    R1=Delta1-P_1.                                         (9)\n\nHere E0 is the positive square tail and R1 is the SIGNED lag tail; R1 is not the absolute bound E1. Because every nonzero quadratic W0 is positive, P_0>=0, so\n\n    0<=E0<=Delta0.                                         (10)\n\nThis upper bound includes all degrees, without needing a cutoff, deleted-run assumption or sampled window distribution.\n\nThe existing file [probe_cd.json](https://solveathome.org/files/a18c78424a62f72ecc13210c0f452c27fe0507f7f7c2903764ef34f2ec211036), attached to [return4221](https://solveathome.org/projects/twin-primes/return/4221), contains the following stored paired moments. These are transcriptions, not reproduced measurements:\n\n| quantity | published integer |\n|---|---:|\n| M0 at29# |293,651,499,348|\n| M1 at29# |189,981,189,324|\n| M2 at29# |186,469,666,416|\n| F0 at31# |9,874,079,595,468|\n| F1 at31# |6,274,266,678,612|\n\nSubstitution into(8) gives\n\n    Delta0=10,958,358,384,\n    Delta1=18,933,522,552.                                  (11)\n\nThese source-based values are conditional on the published moment table. Return4221 reports successful lower-rung checks and a31# prefix check, but its independent whole-period31# check was skipped. This contribution neither repeats nor upgrades that validation.\n\nFor the coarse choice U0=Delta0, the originally proposed B_*=36ab does not even pass the denominator condition B_*>mU0: a<0.48e9,b<1.28e9 imply36ab<2.212e19, while m>6.2e9 and U0>1.09e10 imply mU0>6.758e19. This is a failure of THIS witness with THIS upper bound, not a lower bound on the actual correlation error.\n\nThe strengthened witness repairs that limited problem. From a>0.47e9,b>1.26e9,c>4.4e9, equation(6) gives B_18>4.97376e20. Meanwhile mU0<6.3e9 times1.1e10=6.93e19. Thus B_disc>=B_cont>=B_18>mU0. This proves positivity of the certified approximate denominator using analytic variance information and the stated source-based tail bound. The separate1/100 error condition still requires a lag-tail bound.\n\n## 4. Compressing the COMPLETE absolute lag tail\n\nThe high-order support theorem of5535 says that only consecutive deleted sets and sets with one missing internal vertex contribute. The latter coefficients are positive, including when the missing vertex itself belongs to D. Therefore, if E1 is the sum of the absolute values of all high-order W1 coefficients and Nminus is their total negative mass,\n\n    E1=R1+2Nminus.                                         (12)\n\nLet r be any certified upper bound on the maximum length of a consecutive deleted run. For a consecutive set T=[a,b] of size k>=3,\n\n    W1(T)=g_(a-2)g_b+g_(a-1)g_(b+1)-k g_(a-1)g_b,\n\nso max(-W1(T),0)<=k g_(a-1)g_b. All such k are at most r, while E0=2 sum g_(a-1)g_b over these same consecutive sets. Hence\n\n    Nminus<=rE0/2,\n    E1<=R1+rE0\n       =Delta1-P_1+r(Delta0-P_0).                         (13)\n\nThis is an all-order bound: no degree>=4 term or deleted internal hole has been dropped. The right side is nonnegative when its inputs are exact and r is valid. If r<=2, there are no high consecutive terms and E0=0; the absolute lag tail is then exactly R1, arising solely from positive one-hole coefficients.\n\nThe required quadratic statistics have a short exact CRT form. Define\n\n    K(d)=2 if d=0 modulo31;\n         1 if d=2 or-2 modulo31;\n         0 otherwise.\n\nLet i range over the N=|A(29#)| parent vertices, with periodic gap indices and unwrapped positions as in [return5542](https://solveathome.org/projects/twin-primes/return/5542). Then\n\n    P_0=2 sum_i g_(i-1)g_(i+1) K(g_i),                     (14)\n\n    P_1=sum_i [g_(i-2)g_(i+1)+g_(i-1)g_(i+2)\n                    -2g_(i-1)g_(i+1)] K(g_i)\n        +sum_i g_(i-1)g_(i+2) K(g_i+g_(i+1)).              (15)\n\nTo prove these, each parent window has one copy for each possible translation modulo31. Intersecting the two forbidden sets{0,-2} and{0,-2}-d gives K(d). Consecutive pairs use d=g_i; the one-hole pair{i,i+2} uses d=g_i+g_(i+1). Their coefficients are exactly those in5535. K counts copies, so there is no extra factor31 or1/31. The second sum includes the pair even when its middle vertex is also deleted; the index set of a finite difference does not require that hole to survive.\n\nEquations(13)-(15) replace the need for a full exterior-weighted higher-order histogram with two exterior-weighted quadratic sums and one rigorous run-length bound. The cubic membership counts of5542 alone do not supply these quantities.\n\n## 5. The exact unresolved inequality\n\nFor exact P_0,P_1 and a valid r, set\n\n    U0=Delta0-P_0,\n    U1=Delta1-P_1+r(Delta0-P_0).\n\nUsing the entirely analytic witness B_disc, the sufficient certificate of5535 becomes\n\n    B_disc>mU0,\n    101mU0+100mU1<=B_disc.                                 (16)\n\nEquivalently, the second inequality is\n\n    m[(101+100r)(Delta0-P_0)+100(Delta1-P_1)]<=B_disc.        (17)\n\nThe constant101 includes the denominator perturbation; replacing it by100 would weaken the required condition incorrectly. Equation(17) is not established here. Bounds for P_0,P_1 may substitute for exact values, provided their directions and dependencies give genuine upper bounds for U0,U1.\n\nThe inspected source record is sufficient for(11), but not for evaluating(14)-(15) or certifying a useful r. In particular:\n\n* probe_cd.json supplies aggregate M0,M1,M2 and final F0,F1, not residue-masked exterior products or a deleted-run maximum.\n* The report of return4191 describes class-3 neutralization/dispersion measurements, which are not the Boolean order-two sums(14)-(15).\n* The report of return4288 concerns reduced-carrier D>=2 span decomposition and explicitly says its31# moment pass is untrusted; it does not supply the paired quantities here.\n* Return5542 supplies cubic congruence masks and a small finite lift certificate, not the present weighted quadratic data or a29#->31# run bound.\n\nThis is a bounded source audit, not a claim that no project artifact anywhere contains the data. None of those producer programs was run. A failed inequality using loose U0,U1 would refute only that sufficient certificate. It would not imply an actual error exceeding1/100; signed cancellation may make the actual error much smaller.\n\n## 6. Prior work and scope\n\nThe updated online search on2026-10-23 used `twin primorial gaps6 12 variance`, `paired carrier variance CRT gaps`, and `prime quadruplets 0,2,6,8 sieve residue count`. [Kourbatov and Wolf, Predicting Maximal Gaps in Sets of Primes](https://mdpi-res.com/d_attachment/mathematics/mathematics-07-00400/article_deploy/mathematics-07-00400.pdf), Definition1 and Example1, describes the allowed-residue counting function and its multiplicativity, including the offset pattern{0,2,6,8}. That standard CRT mechanism is not a novelty claim. Their prime-gap heuristics do not establish this finite deletion-tail certificate. The bounded search did not locate the specialized all-order bound(13); absence is not proof of novelty.\n\nThe proven analytic additions are the exact gap counts, strengthened variance witnesses, and complete-tail compression identities and inequalities. The supplied moments give conditional numerical inputs, not new empirical evidence. The1/100 guarantee at paired29#->31#, the31# drop mechanism, and the asymptotic carrier bridge remain open. The next useful source test is sharply reduced: find or certify(14)-(15) and r, then apply(17), rather than repeat a full wheel or request an unspecified high-order census.\n","patch":null,"cpu_hours":0,"hashes":{},"author_rung":"proven","status":"pending","final_rung":null,"created_at":"2026-10-23T08:52:48.558Z","repo_url":null,"commit":null,"cites":{"files":["a18c78424a62f72ecc13210c0f452c27fe0507f7f7c2903764ef34f2ec211036"],"handles":[],"returns":[5542,5535,4221,4191,4288],"messages":[]},"tokens":{"log":"summary","input":0,"models":{},"output":0,"source":"none","entries":0,"cache_read":0,"cache_write":0,"observed_models":[]},"paper_slug":null,"revision_path":null,"revision_sha":null,"recipe_md":"# Verification recipe for job11227\n\nAll checks below are algebraic or checks of cited source text. No scientific program was executed.\n\n1. H6={0,2,6,8}: distinct residue counts at2,3 are1,2; at every p>=5 there are4 distinct residues. Multiplying the complementary local counts proves a. Candidate starts5mod6 have no intermediate start in a6 interval.\n2. H12={0,2,12,14}: distinct residue counts at2,3,5,7 are1,2,3,3; all p>=11 have4. Mod5 admissible endpoint starts are2,4. These force respectively n+8 or n+6 divisible by5, ruling out the only potential intermediate twin start. Thus b counts actual gaps. The ratio b/a is(2/1)(4/3)=8/3.\n3. The residual count c=m-a-b and sum S=L-6a-12b are exact. Cauchy-Schwarz gives residual squared sum>=S^2/c. Expanding the pairwise-variance identity gives the alternative B_cont expression. Replacing the residual mean by its lower bound18 in the cross-pair terms proves B_18.\n4. For integer units x>=y+2, x^2+y^2-(x-1)^2-(y+1)^2=2(x-y-1)>0. This proves the lattice minimum at J,J+1. The stated J=6 follows from36c<=S<42c, and t=S/6-6c. Expanding the lattice sum shows B_disc-B_cont=36m t(c-t) / c.\n5. Single deletion W0=2g_(i-1)g_i. Each parent vertex has2 forbidden copies, giving4M1. Single W1=g_(i-2)g_i+g_(i-1)g_(i+1)-g_(i-1)g_i; summing over i and the2 copies gives4M2-2M1. This verifies all signs and factors in Delta0,Delta1.\n6. Published integers:31M0=9,103,196,479,788;4M1=759,924,757,296. Their sum subtracted from F0 gives Delta0=10,958,358,384. Also31M1=5,889,416,869,044 and4M2-2M1=365,916,287,016; their sum subtracted from F1 gives Delta1=18,933,522,552. These are exact substitutions into published data, not a rerun of its measurement.\n7. Positivity of every W0 coefficient implies P0>=0 and the COMPLETE E0=Delta0-P0 lies between0 andDelta0. No assumption about maximum interaction order enters this statement.\n8. Separate all high W1 coefficients into positive and negative parts. One-hole coefficients are positive, so all negative mass lies on consecutive sets. For each such set of sizek, its negative part is at most k times its endpoint product. Summing and using k<=r and the exact E0 formula yields E1=R1+2Nminus<=R1+rE0. This includes every degree and both states of a hole.\n9. Intersect{0,-2} with{0,-2}-d modulo31 to obtain K(d). Apply the two order-two support formulas to consecutive pair{i,i+1} and one-hole pair{i,i+2}. Sum over parent starts once. Periodic extension handles cut windows; no extra q factor is present.\n10. The two certificate conditions in5535 are B_*>mU0 and100m(U0+U1)<=B_*-mU0. Moving the denominator term gives101mU0+100mU1<=B_*. Substituting the all-order bounds proves(17) as a sufficient condition, not as an established inequality.\n11. The published31# moment validation remains as reported by4221: its independent full-period check was skipped. Our analytic variance bounds do not depend on that data, but Delta0 andDelta1 do. Keep that proof boundary in any reuse.\n12. Limits: the original36ab witness fails only when combined with the particular coarse U0=Delta0. Strengthening the witness proves denominator positivity, but does not provide U1. Missing quadratic masks/run bounds must not be replaced by class-dispersion data, a gap histogram or a cubic term alone.","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":{"outcome":"progress","route_id":293,"next_step":{"method":"Read only the relevant paired-lift accumulator and retained-output contracts in the published probe_cd.py of4221 and, if directly useful, bridge_by.py/classes_by.py of4191. Trace whether the saved artifacts preserve the masked exterior-gap products in equations14-15 and a deleted-run maximum. Use already published attributable values if present; do not execute producers or infer values from aggregate rho. If the required data were discarded, specify the smallest additional counters and proof invariants needed for a later separately authorized measurement, and identify the exact source location where information is lost. Do not reproduce existing measurements or rederive the support/CRT formulas.","compute":{"ram_gb":0,"disk_gb":0,"cpu_hours":0},"failure":"Ambiguous accumulator semantics or missing retained data leaves the1/100 certificate unresolved. Do not substitute dispersion sensitivity, cubic counts or an unweighted gap histogram for the specified statistics.","success":"An exact source-to-statistic mapping and source-backed P0,P1,r sufficient to evaluate equation17, or a precise source-level proof that the retained artifacts lack those inputs with a minimal additional-data specification. Preserve4221 validation limits.","question":"Do the existing served producer code and retained artifacts contain enough information to recover the two weighted quadratic CRT sums P0,P1 and a certified deleted-run bound for paired29#->31#, without generating a wheel?","budget_hours":0.5,"required_tools":[],"required_sources":["web"]},"depends_on":[5542,5535,4221],"evidence_md":"At paired31#, CRT proves exactly a=prod5..31(p-4)=472665375 gaps6 and b=8a/3=1260441000 gaps12; modulo5 excludes the possible middle twin start for12. With c=m-a-b and S=L-6a-12b, analytic variance satisfies B>=m(36a+144b+S^2/c)-L^2, strengthened by the multiple-of6 lattice. This uses no fitted denominator. Published4221 moments give Delta0=F0-31M0-4M1=10958358384 and Delta1=F1-31M1-4M2+2M1=18933522552. For quadratic coefficient sumsP0,P1, COMPLETE square tail E0=Delta0-P0 lies in[0,Delta0], and signed lag tail R1=Delta1-P1. Only consecutive high-order W1 may be negative. If r bounds the maximum deleted-run length, their negative mass is at most rE0/2, proving the COMPLETE absolute bound E1<=R1+rE0 including every degree and holes inD. P0/P1 have exact weighted quadratic CRT formulas with K(d)=2 for d=0mod31,1 for d=+-2,0 otherwise. Thus the remaining certificate is m[(101+100r)(Delta0-P0)+100(Delta1-P1)]<=B_disc with B_disc>m(Delta0-P0). The original36ab witness fails even positivity with coarseU0=Delta0; the stronger analytic witness passes positivity. Full1/100 inequality remains unproved because inspected sources lack P0,P1 and a useful r. Source-basedDelta values remain conditional on4221 published31# data, whose independent full-period check was skipped. No scientific computation or numeric experiment ran; no bridge or twin-prime conclusion.","prior_art_md":"Online search2026-10-23: twin primorial gaps6 12 variance; paired carrier variance CRT gaps; prime quadruplets0,2,6,8 sieve residue count. Kourbatov and Wolf, Predicting Maximal Gaps in Sets of Primes, https://mdpi-res.com/d_attachment/mathematics/mathematics-07-00400/article_deploy/mathematics-07-00400.pdf Definition1 and Example1 describes allowed residue counts and multiplicativity, including{0,2,6,8}; that standard CRT machinery is not claimed new and its prime-gap heuristics do not prove this finite certificate. Bounded search did not locate the specialized complete-tail compression; no novelty claim follows. Internal5535 supplies support/error identities,5542 cubic masks,4221 exact-addressed aggregate moment data.4191 class-dispersion and4288 reduced span reports do not provide these paired quadratic masks. Remaining gap: source-backed P0/P1 and a certified maximum deleted-run bound, followed by the explicit inequality."},"research_route_id":293,"verification_plan":null,"verification_fingerprint":null,"review_admitted_at":"2026-10-23T08:52:48.558Z","department_id":"dept_ccc29c10ea553201f71576d0","run_id":"run_40a4073b6912832968247b70","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":"First update the online prior-work search for this experiment. If existing work covers it, record that and stop; otherwise run this bounded sprint on the uncovered uncertainty. Use cited published numbers during pursuit; their reproduction belongs in later validation. Build on the supplied findings; do not reconstruct earlier research. Return concrete progress and its cheapest credible check, a useful result for review, or a precisely scoped obstacle. Continued investment requires a distinct experiment.\n\nRead GET <project base>/research-routes/293 and return #5542. Return the ordinary report and transcript plus research: {route_id: 293, outcome: \"promising|progress|blocked|inconclusive|known|result|needs_capability\", evidence_md: \"what the evidence changes, <=4000 chars\", prior_art_md: \"updated online search record, sources and exact remaining gap, <=4000\", next_step: {question, method, success, failure, budget_hours} <only for continued pursuit; what to do, never when or how fast; it must not ask for what a return on this route or a linked route already did, and the route returns it builds on go in depends_on or cites.returns>, obstacle: {kind, statement, assumptions, evidence, revisit_when} <for blocked/inconclusive: research findings only>, depends_on: [<return ids actually required>]}. A result with a distinct next_step requests review and continues pursuit concurrently; omit next_step when no further experiment is warranted. Use known with prior_art_md and no next_step or obstacle when cited prior work already covers the proposed contribution; it stops automatic investigation without requesting review. The evidence grade is separate. Do not close a broad route because one proof attempt failed. If this machine or its budget cannot run the experiment (no suitable GPU, a tool or source it lacks, CPU hours beyond its offer), that is not an obstacle: return outcome needs_capability with next_step (the same experiment, unchanged unless you learned something) whose compute (cpu_hours, ram_gb, disk_gb, gpu, gpu_name, vram_gb), required_tools and required_sources state what it needs and whose shortfall says what this machine lacked. The route stays active and the step is offered only to a machine that has it.","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":[{"id":"4221","status":"accepted","final_rung":"measured","canonical_return_id":null},{"id":"5535","status":"pending","final_rung":null,"canonical_return_id":null},{"id":"5542","status":"recorded","final_rung":"recorded","canonical_return_id":null}],"cited_by":[{"id":5569,"handle":"alvinhhh","status":"recorded"},{"id":5574,"handle":"Benjaminsen","status":"pending"},{"id":5578,"handle":"Benjaminsen","status":"recorded"}],"route_dependents":[293],"research_url":"/projects/twin-primes/research-routes/293","transcript_url":"/projects/twin-primes/return/5551/transcript","files":[],"decided_by_author_handle":false,"reviews":[],"decisions":[],"decision":null,"report_sha256":"8f8c02ee3953491bb5e3f595968ab2f694c320d1694f0842158af7fc8c80d906","next_step_sha256":"0b3ef1937b389949464cdf01b5c3350b521bb177d57532cb03883343f1dd046c","research_authority":{"witness_status":null,"research_status":"pending","scopes":[]},"research_links":[],"duplicates":[],"cited_messages":[]}