(A sequence of explicitly constructed independent sets in strong powers of C_11, with successive root improvements certified by exact integer comparisons, using cross-powers when dimensions differ. The current bound is Θ(C_11) ≥ 5.295526013632343, from an independent set in C_11^ 213 (R10, 9 September 2026), improving the R3 result of 5.295492477500681 in dimension 207 that this entry was first listed for, and the Buys-Polak-Zuiddam baseline of 5.295492315784620. The intermediate steps R5 and R6 and the superseded R9 are recorded on the frontier. The exact capacity remains open, and no upper bound is claimed. The R5, R6, R9 and R10 builds carry disclosed native-evaluation dependencies rather than being kernel-only.)
Zero-error information theory; graph capacity; combinatorics
Determine the Shannon capacity of the eleven-cycle C11, or improve its best explicit lower bound. The preceding BPZ construction, updated on 10 August 2026, gives an independent set of cardinality N0 in dimension 207 and the lower bound Θ(C11)≥N01/207=5.29549231578462014255…. The exact capacity remains open.
Posed by Claude Shannon (1956), underlying capacity problem; Buys, Polak and Zuiddam (2026), preceding C11 record, 1956·Open 70y·Model Astra 6 Pro; Codex GPT-6 Astra Extra-High (OpenAI)·Solved 2026-09-08
Lean-checked, statement unauditedSignificance 14Submitted by Matthew Protti on 8 Sep 2026
(Yes. Theorem 2.1: there are a smooth odd initial density _in∈ C^∞( T^2) of zero spatial mean, an odd force F∈ C^∞([0,1]× T^2), and a solution smooth on [0,T] for every T<1 with (t)→_* in C^ for every 0≤<1, yet ‖(t)‖_∞ and ‖D_xu_ T((t))‖_∞ both diverging as t1. The advance over Córdoba and Martínez-Zoroa is precisely the force class, from L^∞_t C^∞_x to uniformly space-time smooth, on the torus rather than the plane. Not the Clay Millennium problem: that problem is Navier-Stokes with viscosity, and Fefferman's official description states that the Euler equation "is not on the Clay Institute's list of prize problems".)
Fluid dynamics; incompressible porous media equation
Córdoba and Martínez-Zoroa proved finite-time singularity formation for the two-dimensional incompressible porous media equation from smooth initial data with a force smooth in space but merely bounded in time, that is in Lt∞Cx∞. Their Remark 1 anticipates joint smoothness in space and time but does not prove it. Can the force be taken uniformly smooth in space and time?
Posed by Diego Córdoba and Luis Martínez-Zoroa, Remark 1 of arXiv:2410.22920, where joint space-time smoothness is anticipated but not part of the theorem, 2024·Open 2y·Model Claude, Codex with GPT-5.6 Sol (Anthropic / OpenAI)·Solved 2026-09-08
(For a stationary finite-alphabet law , let n=_ R H_p be the intrinsic real Hankel dimension of its cylinder-probability function. If has any finite Markov order, then ()≤ n2. The proof passes to a reduced n-dimensional linear representation and uses Holland's criterion that k-step Markovity is equivalent to every length-k transition product having rank at most one. Applying ^2 turns this into vanishing of products on a space of dimension n2; a uniform nilpotence argument then forces vanishing after n2 steps. Sharpness is attained for every n2. The construction gives a stationary sofic process with a nonnegative rational presentation of minimal dimension n and exact order n2. One realization uses n2-1+n^2 output symbols.)
Sofic measures
Let n be the real Hankel dimension of the cylinder probabilities of a stationary finite-alphabet process. If its Markov order is finite, thenord(μ)≤(2n).For every n≥2, there exists a stationary sofic process with a nonnegative rational presentation of minimal real dimension n and exact Markov order(2n).Thus the bound is sharp when the alphabet is allowed to grow.
Posed by Béal, Jugé, Mairesse and Perrin, 2026·Open 0y·Model GPT 6 Astra (OpenAI)·Solved 2026-09-08
Lean-checked, statement unauditedSignificance 12Submitted by VibeGene on 8 Sep 20262comments
(Yes. Theorem 1.1: for every r_0>0 and z_0 there are a time T_*>0, a divergence-free axisymmetric u_0∈ C_c^∞ supported in a fixed solid torus with nonzero swirl and zero meridional velocity, and an axisymmetric force f∈ C^∞( R^3×[0,T_*]) supported in that torus, with a solution smooth on [0,T_*) for which circulation and meridional velocity stay bounded while ‖(t)‖_∞ and ‖(t)‖_∞ tend to infinity and _0^T_*‖(t)‖_∞ dt=∞, so the blowup is genuine by Beale-Kato-Majda. It is unique among divergence-free locally space-time Lipschitz solutions with the same data, and competitors need not be axisymmetric. Not the Clay Millennium problem: that problem is Navier-Stokes with viscosity, and Fefferman's official description states that the Euler equation "is not on the Clay Institute's list of prize problems".)
Fluid dynamics; singularity formation for incompressible flow
Can a solution of the three-dimensional incompressible Euler equations on R3, started from smooth data and driven by a force that is smooth in space and time up to and including the blowup time, lose regularity in finite time? Finite-time singularity formation from genuinely smooth data is the central open question for the equations. Elgindi obtained blowup for C1,α solutions in 2021, and Córdoba and Martínez-Zoroa built a multiscale program producing forced blowup for related equations with forces of limited regularity, but no construction reached three-dimensional Euler with a space-time smooth force. Note on the forced formulation, since it is easy to misread: alternatives (C) and (D) of Fefferman's Clay problem description do permit a smooth force obeying rapid space-time decay, so forcing is not a dodge and the forced route is a genuine path to the prize. It is a path for Navier-Stokes with viscosity, however, and not for Euler, which Fefferman's description excludes from the prize list.
Posed by Leon Lichtenstein (1925) and Nikolai Gunther (1927), whose local existence left global regularity open; Elgindi states the smooth-force form as his Question 1.1, 1925·Open 101y·Model Claude, Codex with GPT-5.6 Sol (Anthropic / OpenAI)·Solved 2026-09-08
(For any family of real n× n matrices generating a finite entire product monoid of minimum rank s, some word attains rank s within B(n,s)=n2^n-s-(n+1)2+(s-1)2 letters. The same holds over Q, without requiring a finite generating alphabet. For mortality, s=0, giving a bound of order n2^n. Over Q, this improves Kiefer–Ryzhikov's (2026) 3^n^2 bounds for mortality and minimum-rank diameter, and Almeida–Steinberg's (2009) mortality bound (2n-1)^n^2-1 for n>1. The proof uses rank-dependent sandwich descent. Further results give the sharp planar threshold four, additive invariant-flag bounds with sharp small-block examples, and cubic-size compressed witness existence with correct evaluation. Boundedness and individually finite-power generators alone admit no uniform planar mortality bound. General optimality and polynomial-time synthesis are not claimed.)
Linear algebra
Let n>0, and let a family of real n×n matrices generate a finite entire product monoid S. If s=minX∈SrankX, then some word attains rank s with length at mostB(n,s)=n2n−s−2n(n+1)+2s(s−1).In particular, if S contains zero, a zero word has length at mostB(n,0)=n2n−2n(n+1)=Θ(n2n).Over the rationals, these bounds improve Kiefer–Ryzhikov's (2026) 3n2 bounds for mortality and minimum-rank diameter. The mortality bound also improves Almeida–Steinberg's (2009) universal rational bound (2n−1)n2−1 for n>1. No finiteness assumption on the generating alphabet is needed. The empty word is allowed and suffices when s=n.
Posed by Jorge Almeida and Benjamin Steinberg, 2009·Open 17y·Model GPT-6 Astra (OpenAI)·Solved 2026-09-08
Lean-checked, statement unauditedSignificance 12Submitted by VibeGene on 8 Sep 20262comments
(The paper proves that the reciprocal Fermat constant is disjunctive but nonnormal in binary, explicitly locates every finite word, and establishes positive lower word frequencies. It extends these properties to a broad family of arithmetic series. Behavior in unrelated bases remains open.)
Theory of Normal Numbers and Digit Expansions
Posed by —·Open —·Model ChatGPT-6 Astra (OpenAI)·Solved 2026-09-08
Lean-checked, statement unauditedSignificance 18Submitted by nufrogcaca on 8 Sep 2026
(Yes. Blowup for the inviscid Boussinesq system on R^2 with forces in C^∞( R^2×[0,T_*]) in both equations, supported in one fixed spatial ball, from smooth compactly supported initial temperature and zero initial velocity. The temperature stays bounded while ‖(t)‖_∞→∞ and the vorticity norm has infinite limsup as t T_*. The solution is smooth on every closed interval before blowup and unique in a finite-energy Lipschitz class. This lifts the force from barely C^1 to fully smooth, and is the construction the Euler paper then builds on. Not the Clay Millennium problem: that problem is Navier-Stokes with viscosity, and Fefferman's official description states that the Euler equation "is not on the Clay Institute's list of prize problems".)
Fluid dynamics; singularity formation for incompressible flow
Does the inviscid Boussinesq system on R2 admit finite-time blowup from smooth data with forces that are smooth in both space and time? Córdoba, Laín-Sanclemente and Martínez-Zoroa obtained finite-time singularity for the two-dimensional Boussinesq equation with a force only of class C1,4/3−1−ϵ∩L2, leaving the smooth-force case open.
Posed by Diego Córdoba, Antonio Laín-Sanclemente and Luis Martínez-Zoroa, whose multiscale construction reached a force of limited Hölder regularity, 2025·Open 1y·Model Claude, Codex with GPT-5.6 Sol (Anthropic / OpenAI)·Solved 2026-09-08
(Proved f(732)=f(731) unconditionally as a structural finite theorem. The isolated component G=122,183,244,366,732 has exactly the edges 122,183,366, 183,244,732, 244,366,732. Every admissible selection in G has size at most 3; replacement by 122,183,244 preserves admissibility and cardinality or increases it while removing 732. Combining this with the published f(731)=606 gives the candidate new table term f(732)=606. The historical asymptotic problem remains open. A general large-prime recurrence is proved separately in prose, not in Lean; it may be folklore.)
Let f(n) be the maximum size of a subset of {1,...,n} containing no three distinct a,b,c satisfying 1/a = 1/b + 1/c. A published finite frontier for Erdős problem 302 asks whether f(732) is 606 or 607, using the existing OEIS value f(731)=606.
Posed by Finite frontier: Erdős Frontier Atlas P302, recorded July 2026. Parent problem: Erdős and Graham (1980); the parent asymptotic problem is not solved., 2026·Open 0y·Model OpenAI ChatGPT/Codex (exact model identifier unavailable) (OpenAI)·Solved 2026-09-07
Lean-checked, statement unauditedSignificance 4Submitted by ZestyDingo473 on 7 Sep 2026
(Koizumi constructs a representable simple rank-6 matroid M for which (M;q) has a pole of order 4 at q=-1 but a pole of order 5 at q=i. After writing A_M(t)=(M;-t), the higher-order pole at t=i forces the Taylor coefficients of A_M(t) to fail eventual nonnegativity. Consequently, (-1)^[q^](M;q)<0 for infinitely many . The matroid is representable by twelve vectors in R^6, so it yields an actual real central hyperplane arrangement and therefore directly disproves Koizumi–Liu's conjecture for real arrangements.)
Matroid theory
Koizumi and Liu conjectured that for every real hyperplane arrangement A, the coefficients ofMag(A;−t)are eventually nonnegative, equivalently that the coefficients of Mag(A;q) eventually alternate in sign.
The conjecture is false. There exists a rank-6 real hyperplane arrangement A such that(−1)ℓ[qℓ]Mag(A;q)<0for infinitely many ℓ.
Posed by Junnosuke Koizumi and Ye Liu, 2026·Open 0y·Model GPT-5.6 Sol; GPT-6 Astra (OpenAI)·Solved 2026-09-07
PreprintSignificance 12Submitted by VibeGene on 9 Sep 2026
(This work proves that the Erdős–Borwein constant's binary expansion contains every finite binary sequence as a consecutive block, each occurring infinitely often. This implies that the Erdős–Borwein constant is 2-dense.)
Analytic number theory; digit distribution of constants
Is the Erdos-Borwein Constant 2-Dense?
Posed by Richard E. Crandall, 2002·Open 24y·Model ChatGPT-6 Astra (OpenAI)·Solved 2026-09-07
Lean-checked, statement unauditedSignificance 25Submitted by nufrogcaca on 7 Sep 2026
(The manuscript proves the upper bound |A|1280 for every admissible subset of the fixed ambient code D. Together with Ho’s existing construction, this determines the exact maximum. The proof partitions D into 256 cosets of a 16-word subspace whose induced forbidden-distance graph is the Clebsch graph; each coset contributes at most five words. This does not determine the unrestricted kissing number k(19) or improve the known 11948-point configuration. Independent expert review is pending.)
Let D⊆F219 be the fixed 4096-word ambient binary linear code used in Ho’s 19-dimensional improvement of the Cohn–Li kissing construction. Is every subset A⊆D with minimum Hamming distance at least 5 of size at most 1280? No linearity assumption is imposed on A.
Posed by Gonzalez, Conjecture 16 (preprint, version 3, 2026), after Ho's 1280-word construction, 2026·Open 0y·Model ChatGPT (OpenAI, model version unstated) (OpenAI)·Solved 2026-09-06
Site-confirmedSignificance 8Submitted by cheptil on 6 Sep 2026
(The manuscript gives an explicit centrally symmetric simplicial 7-polytope with 88 integer vertices. Its antipodal boundary quotient is a 44-vertex triangulation of real projective 6-space, with f-vector (44, 938, 7024, 22555, 34936, 25914, 7404). This improves the 45-vertex construction of Guyer, Steinerberger and Yang and gives an affirmative answer to their Question 3.2. Integer coordinates, facet lists and exact-arithmetic verification code accompany the manuscript. No claim that 44 is vertex-minimal is made.)
Combinatorial topology; triangulations of manifolds; convex polytopes
Does real projective 6-space admit a simplicial triangulation with fewer than 45 vertices? This is Question 3.2 of Guyer, Steinerberger and Yang, “An Efficient Triangulation of RP^5”, arXiv:2603.07808 (2026).
The question asks for an improvement on their 45-vertex construction, not for the exact minimum number of vertices.
Posed by Dan Guyer, Stefan Steinerberger, Yirong Yang — Question 3.2, arXiv:2603.07808, 2026·Open 0y·Model ChatGPT (OpenAI, model version unstated) (OpenAI)·Solved 2026-09-06
Site-confirmedSignificance 10Submitted by cheptil on 6 Sep 2026
(Answered in the negative for every n≥ 2. The preprint constructs a bijection F: R^n→ R^n that maps every connected set to a connected set, is continuous exactly off the closed ray [0,∞)×\0\^n-1, and pulls the straight segment \(1,0,…,0)\×[0,1] back to the middle-thirds Cantor set on that ray, so F^-1 is not connectedness-preserving. The construction extends a thin solid tube by finger moves so its cross-sections recur near every point of the complementary compactum, collapses the ray onto the tube's ideal end, and certifies arbitrary connected sets by a separation argument; F and F^-1 can be taken Borel. The same author's companion note on Darboux injections from closed manifolds (Banakh-Banakh Problems 1.7 and 1.8) is a separate result and belongs in its own entry.)
General Topology
Willie Wong asked on MathOverflow in April 2016: if f:Rn→Rn is a bijection that maps every connected set to a connected set, must f−1 do the same? By Tanaka's theorem and invariance of domain this is equivalent to asking whether every connectedness-preserving bijection of Rn is continuous. For n=1 the answer is yes. For n≥2 the question stayed open for a decade: the top-voted answer constructs such a bijection only from R to R2, and Banakh and Banakh (2020) proved continuity in several compact settings while calling Wong's problem still open.
Posed by Willie Wong, MathOverflow question 235893, 2016·Open 10y·Model GPT-6 (Codex, Ultra effort), Claude Fable 5.1 (OpenAI, Anthropic)·Solved 2026-09-05
PreprintSignificance 22Submitted by WittyHeron892 on 5 Sep 2026
(OpenAI claims to construct a smooth finite-energy solution of the three-dimensional incompressible Navier–Stokes equations, starting from a fluid at rest and driven by smooth forcing, that develops a finite-time singularity. The construction is claimed both on R^3 and in the periodic setting, thereby establishing statements C and D of Fefferman's official Clay formulation. If the correspondence between the released proof/formalization and the Clay statements survives independent review, this resolves the Navier–Stokes Millennium Prize problem. It does not establish finite-time blow-up for the unforced Navier–Stokes equations; the Clay resolution comes through the smooth-forcing alternatives C/D.)
PDEs; fluid dynamics; singularity formation
For the three-dimensional incompressible Navier–Stokes equations with positive viscosity, do there exist smooth divergence-free initial data and smooth external forcing for which a global smooth physically reasonable solution does not exist? The Clay Millennium Prize formulation allows this to be established either on R^3 (statement C) or on the periodic three-torus (statement D).
Posed by Jean Leray (1934), whose weak solutions left smoothness open; stated as Millennium alternatives (C) and (D) by Charles Fefferman for the Clay Mathematics Institute in 2000, 1934·Open 92y·Model Unnamed internal OpenAI model (OpenAI)·Solved 2026-09-05
(Astra constructs a strictly stationary real process X_t=_t+K(_t-1,_t-2,…), where the innovations _t are i.i.d. Gaussian variables convolved with a symmetric rare-spike law, and K is bounded, continuous, and odd. The process is -mixing, centered, square-integrable, and satisfies (S_n)→∞. Nevertheless there are times n_j→∞ such that _n_j√()(S_n_j)0 in probability. Therefore the normalized sums cannot converge in distribution to N(0,1). The same example also rules out Iosifescu's stronger weak invariance-principle conjecture, since Brownian convergence would imply the CLT at time 1.)
Stationary stochastic processes
Ibragimov conjectured that if (Xn) is a strictly stationary, φ-mixing sequence withEX0=0,EX02<∞,andσn2=Var(Sn)→∞,Sn=j=0∑n−1Xj,thenσnSn⇒N(0,1).GPT-6 Astra constructs a counterexample satisfying all these hypotheses for which, along a subsequence nj→∞,σnjSnj→0in probability. Hence the conjectured central limit theorem fails.
Posed by I. A. Ibragimov, 1971·Open 55y·Model GPT-6 Astra (pre-release) (OpenAI)·Solved 2026-09-05
Lean-checked, statement unauditedSignificance 40Submitted by VibeGene on 6 Sep 2026
(Let H be the Berlekamp–van Lint–Seidel graph on 243 vertices, the Cayley graph of Z_3^5 with strongly regular parameters (243,22,1,2), and let G= H. The proof establishes γ(G)=3 because every pair has a common neighbor in H, while an H-triangle gives a dominating triple in G. It then proves γ^∞(G)=3 by showing that the family of all dominating triples is closed under a legal response to every attack: after moving one guard to the attacked vertex, another dominating triple can always be obtained. Finally, a double-counting argument shows that H is not 3-colorable, hence (G)=(H)>3. Therefore γ(G)=γ^∞(G)=3<(G).)
Graph theory
The γ–θ conjecture asserts that for every finite graph G,γ(G)=γ∞(G)⟹γ(G)=θ(G),where γ is the domination number, γ∞ is the eternal domination number in the one-guard-moves model, and θ is the vertex clique-cover number.
The conjecture is false. The complement G of the 243-vertex ternary Golay graph satisfiesγ(G)=γ∞(G)=3<θ(G).
Posed by William F. Klostermeyer and C. M. Mynhardt, 2014·Open 12y·Model GPT-6 Astra (pre-release) (OpenAI)·Solved 2026-09-05
Lean-checked, statement unauditedSignificance 12Submitted by VibeGene on 6 Sep 2026
(We prove HG_P(K_8-e)=14 using an explicit PGL(2,13)-equivariant strategy. A compact formula fixes 990 normalized twin decisions, and a 48,510-edge residual-orbit matching induces consistent rules on 53,460 labelled clique-view orbits. Independent implementations regenerate the certificate and check all 138,378,240 proper fourteen-colourings with zero failures. Unlike the n=5,6,7 constructions, residual right degrees reach eight, so the explicit global matching is load-bearing. A separately labelled companion exhibits one fixed clique completion compatible with 2^380 equivariant twin-rule pairs and proves 380 optimal within the stated independent whole-tail reversal model; that companion's independent review is pending. The release does not solve the general K_n-e family or K_9-e.)
Graph theory; hat-guessing games; finite geometry; matching theory
Determine the exact proper hat-guessing number of the complete graph on eight vertices with one edge removed. The general bounds leave HGP(K8−e)∈{13,14}.
Posed by Adriaensen et al., 2026·Open 0y·Model GPT-5.6 Pro (OpenAI)·Solved 2026-09-04
AnnouncedSignificance 7Submitted by Matthew Protti on 4 Sep 2026
(The preprint claims an exact classification. Write C_5(G) for the residues modulo five represented by cycle lengths in G. Let E_5=K_6,K_5,5_5,n;t:2≤ t5<n, where H_5,n;t is obtained from K_5,n by deleting 5-t edges incident with one vertex in the part of size n. For every finite simple graph G with minimum degree at least five, exactly one alternative holds: C_5(G)= Z_5; or every end-block belongs to E_5 and every non-end-block contains no cycle of length congruent to two modulo five. Every member of E_5 has cycle-residue spectrum 0,1,3,4. The proof combines structural arguments with finite computational checks. It uses the separately established Dean–5 theorem and its weak-graph strengthening as inputs. The contribution is the stronger stability classification; independent expert review remains pending.)
Graph Theory
Classify the finite simple graphs of minimum degree at least five whose cycle lengths fail to represent every residue class modulo five. Is residue two the only possible missing residue, and can all such graphs be characterized through an explicit family of exceptional end-blocks together with a condition on the remaining blocks?
Posed by Luo, Ma and Zhao, whose stability theorem covers every k >= 6 and leaves k = 5, 2026·Open 0y·Model GPT-5.6 Sol, GPT-6 Astra (OpenAI)·Solved 2026-09-04
PreprintSignificance 20Submitted by eli on 4 Sep 2026
(The formal proof constructs a complex polynomial p such that p(0)=0, p'(0)=1, and for every critical point c of p, |p(c)/c|>1. Thus at z=0 there is no critical point satisfying |p(0)-p(c)|/|c|≤ |p'(0)|=1, which disproves Smale's conjectured universal constant K=1. The counterexample has very large unspecified degree and violates the bound only by a small margin, so it is consistent with Smale's original K=4 theorem, the known low-degree positive cases, and previous asymptotic improvements toward 1.)
Complex polynomials
Posed by Stephen Smale, 1981·Open 45y·Model GPT-6 Astra (pre-release) (OpenAI)·Solved 2026-09-03
Lean-verifiedSignificance 45Submitted by VibeGene on 4 Sep 2026
(Assuming three explicit input statements, the project proves DHL[40,2]: every admissible 40-tuple contains at least two primes infinitely often after translation. An explicit admissible 40-tuple of diameter 186 then gives _n→∞(p_n+1-p_n)186. The Lean proof of the implication from the three inputs to the final theorem is kernel-checked. Two inputs are Kloosterman-type estimates cited to Katz/Deligne and Fouvry--Kowalski--Michel; the third consists of finitely many numerical integral and cap inequalities backed by a Python/FLINT certificate.)
Analytic number theory
For the sequence of primes pn, the project derivesn→∞liminf(pn+1−pn)≤186.More precisely, assuming three explicit analytic/numerical inputs, it proves DHL[40,2]: every admissible set of forty integer shifts has infinitely many translates containing at least two primes. Applying this to an explicit admissible 40-tuple of diameter 186 yields infinitely many consecutive prime gaps of size at most 186. The Lean development verifies the deduction from the stated inputs; the two Kloosterman-type estimates and the finite physical-integral/cap bounds remain external assumptions.
Posed by Alphonse de Polignac (the twin prime conjecture); the bounded form since Goldston, Pintz and Yıldırım, 1849·Open 177y·Model GPT 6 Astra (OpenAI)·Solved 2026-09-03
Lean-checked, statement unauditedSignificance 62Submitted by VibeGene on 3 Sep 2026
(We prove HG_P(K_7-e)=12. One lower-bound proof uses two explicit block-disjoint S(5,6,12) Witt designs. A second uses orbit maps from an explicitly regenerated sharply five-transitive twelve-point permutation group, combining one set-symmetric and one order-sensitive rule. Both satisfy a general coordinate-line twin-completion criterion, and Hall's theorem completes the clique strategy. The release also proves a disjoint completion-design theorem, an even-n obstruction scoped to set-symmetric line-permutation twins in this sufficient framework, and a prime-admissibility theorem for the design parameters. It does not solve the general K_n-e problem or K_8-e.)
Graph theory; hat-guessing games; Steiner systems; permutation groups
Posed by Adriaensen et al., 2026·Open 0y·Model GPT-5.6 Pro (OpenAI)·Solved 2026-09-03
AnnouncedSignificance 7Submitted by Matthew Protti on 4 Sep 2026
(H_1 ≤ 212, improving Stadlmann's 240 of three days earlier and the 246 of Polymath8b that had stood since 2014. The twin prime conjecture, H_1 = 2, is untouched. Held the record for hours at most: OpenAI's paper claiming 186 is dated 30 August, four days before this one, though its Lean development appeared on 2 September.)
Analytic number theory
Write H1=liminfn→∞(pn+1−pn). Stadlmann had recently proved H1≤240, improving the bound 246 of Polymath8b. Building on her work, this paper proves H1≤212: infinitely many pairs of consecutive primes are at most 212 apart.
Posed by Alphonse de Polignac (the twin prime conjecture); the bounded form since Goldston, Pintz and Yıldırım, 1849·Open 177y·Model AxiomProver (Axiom Math)·Solved 2026-09-03
(We prove HG_P(K_6-e)=10. The lower bound uses two order-sensitive twin-player rules obtained by deleting and repairing one point of an explicit sharply four-transitive eleven-point permutation group. On every coordinate line the repaired rules are derangement permutations, are pointwise unequal, and have fixed-point-free composition. Hall's theorem completes the strategy on the four clique vertices. The release also classifies all repairable orbit labels and proves an even-n obstruction for set-symmetric line-permutation twin rules. It does not solve the general K_n-e family.)
Graph theory; hat-guessing games; permutation groups
Posed by Adriaensen et al., 2026·Open 0y·Model GPT-5.6 Pro (OpenAI)·Solved 2026-09-03
AnnouncedSignificance 7Submitted by Matthew Protti on 4 Sep 2026
(The paper proves G(X)≫ log X (log_2 X)^2 log_4 X/(log_3 X)^2 for all sufficiently large X. Its main new ingredient is a short-translates theorem: for any sufficiently small set S⊆[1,H] with |S|≤δ x, one can find a short translate making every corresponding linear form composite. Combining this with an Erdős--Rankin covering argument produces prime-free intervals of the claimed length. This directly and asymptotically improves the August 2026 GPT-5.6 Sol bound G(X)≫log Xlog_2 X/log_4 X by the unbounded factor log_2 X(log_4 X)^2/(log_3 X)^2.)
Analytic number theory
Let G(X) denote the largest gap between consecutive primes not exceeding X, and let logj denote the j-fold iterated logarithm. The paper proves that, for all sufficiently large X,G(X)≫(log3X)2logX(log2X)2log4X.Equivalently, there is an absolute constant c>0 such that G(X) is at least c times the quantity above for all sufficiently large X. This improves Rankin's classical lower bound by a factor of log2X.
Posed by Paul Erdős, 1955·Open 71y·Model GPT 6 Astra (OpenAI)·Solved 2026-09-03
Site-confirmedSignificance 60Submitted by VibeGene on 4 Sep 2026