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Claim. Let $u_1<u_2<\cdots$ be the integers with at most two prime factors, counted with multiplicity. The write-up claims that $\limsup_k (u_{k+1}-u_k)/\log k=\infty$, the affirmative answer to Problem 1139. The route, as the claimant's thread comment outlines it: choose a modulus $M_0=\prod_{p\le z}p^2\cdot\prod_{z<p\le Y/z}p$ and, by the Chinese remainder theorem, an integer $N$ such that $N+n$ has at least three prime factors for every offset $1\le n\le Y$; the Green--Tao theorem on linear equations in primes supplies the moment estimates behind the covering, and the $o(Y/\log Y)$ offsets left uncovered are handled with a reserve of primes. No term of the sequence then lies in $(N,N+Y]$, so if $u_k\le N<u_{k+1}$ the gap $u_{k+1}-u_k$ is at least $Y$, while $\log k\le\log N=o(Y)$, so the ratio tends to infinity along these gaps.

Submission note. Posted to erdosproblems.com as a proof claim by Liam Price (account Leeham) on 15 July 2026, giving "GPT Pro" as the AI used:

GPT Pro proves this problem in the affirmative. Notes: There is a short proof summary of the claim in a comment from me on the respective problem page.

Posted to the site's forum by Liam Price on 19 June 2026:

GPT Pro claims a solution here. I have attempted to distil the proof strategy below:

The proof chooses an $N$ so that every offset $n\in[1,Y]$ makes $N+n$ have at least three prime factors. It sets

$$> M_0:=\prod_{p\le z}p^2\prod_{z<p\le Y/z}p, >$$

and imposes $N\equiv 0\pmod{M_0}$. Hence, every prime $p\leq Y/z$ dividing $n$ also divides $N+n$, and if $p\leq z$ satisfies $p^2\mid n$, then $p^2\mid(N+n)$. This handles all offsets for which these forced small-prime divisors already contribute at least two prime factors. The remaining underlying target integers are mainly large primes $q$, represented by two coloured vertices, and semiprimes $sq$, represented by one vertex. For each modulus prime $P$, a residue class $A\pmod P$ induces the finite arithmetic progression $A+dP$, and imposing $N\equiv -A\pmod{P}$ makes $P\mid(N+(A+dP))$ for every selected index $d$. The proof chooses a residue $b\pmod W$, a set of selected indices $d$, and a "type" $\kappa(d)$ for each selected index. The type records whether $A+dP$ is intended to be a prime target $q$, or a semiprime target $sq$ with prescribed small factor $s$. The residue $b$ is chosen so that $A+dP=s_dQ_d(P,C)$, where $s_d=1$ or $s$, and $Q_d(P,C)$ is an integral primitive linear form with no forced small-prime divisor. Whenever $P$ and all the associated forms $Q_d(P,C)$ are prime, the resulting admissible hyperedge covers several target vertices at once.

The Main Theorem (page 13) of Green and Tao on linear equations in primes then gives asymptotic counts for the simultaneous prime values of $P$ and the forms $Q_d(P,C)$. Applying it both to individual hyperedges and to pairs of hyperedges yields the required first- and second-moment estimates: The total fractional weight above a typical modulus prime is close to a fixed value below one, while the fractional load on almost every target vertex is close to a large prescribed value. This allows at most one admissible hyperedge to be chosen independently for each regular modulus prime, leaving only $o(Y/\log Y)$ vertices uncovered. Each remaining uncovered vertex, together with every offset in the sparse exceptional set, is assigned one or two distinct reserve primes from $(Y,2Y]$, according to how many additional forced prime factors it still needs. The Chinese remainder theorem then combines the congruence $N\equiv0\pmod{M_0}$, the selected residue classes modulo the modulus primes, and the reserve-prime congruences into one integer $N$. Every $N+n$ has two forced prime factors whose product is smaller than $N+n$, so a third prime factor remains. Since $(N,N+Y]$ contains no integer with at most two prime factors, if $u_k\le N<u_{k+1}$, then

$$> u_{k+1}-u_k\ge Y,\qquad\frac{u_{k+1}-u_k}{\log k}\geq\frac{Y}{\log N}\to\infty, >$$

because the construction ensures $\log N=o(Y)$.

Standing. Liam Price posted the claim as a comment on the site's problem page on 19 June 2026, naming GPT Pro as the AI system that produced the proof, and filed it on the proof-claims tab on 15 July 2026 with the same write-up. A thread comment of 19 June 2026 reports that a screening check found no issues and notes a typo on p. 12; that is a screening report, not a review by a named expert, and no one on the thread has endorsed the argument. The site's label is unchanged (OPEN; page last edited 23 January 2026), the tab entry carries no comments, and no refereed publication, formalization or outside review is known (thread accessed 2026-10-07). The claim stays claimed.

Depends on. Nothing on the wiki; the comment names the Green--Tao theorem on linear equations in primes as the outside input.