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Claim. David Turturean's thread post of 24 April 2026 links a write-up, a live Overleaf project that Turturean describes as what they have so far, and reports that it proves $h(x)\gg(\log x)^{1/3}$ for the function of Problem 852, by a route they say corroborates Chojecki's note of the same day. The result came from a scaffold on top of ChatGPT-5.5-Pro, run the day before the post. The argument, as the post describes it: among blocks of $H$ consecutive gaps whose total length is at most a constant multiple of $H\log N$, which a positive proportion of starting positions satisfy, a block with two equal gaps $d_i=d_j=a$ at positions $i<j$ gives, with $p=p_i$ and $b=p_j-p_i$, four primes $p$, $p+a$, $p+b$, $p+a+b$ arranged in a rectangle, or the triple $p,p+a,p+2a$ when $j=i+1$. For $a,b\le W$ the Selberg or Brun upper-bound sieve bounds each pair's count by the expected $N(\log N)^{-4}$ up to the singular series, and Gallagher's mean-value estimate averages the singular series over $(a,b)$, so the number of rectangles is $\ll NW^2(\log N)^{-4}$, with a smaller contribution from the triples. Each lies in at most $H$ starting blocks; balancing the resulting count of bad short-span starts against the first-moment count of long-span blocks at $W\asymp(\log N)^{4/3}$ leaves a bad count $\ll xH(\log N)^{-1/3}$, smaller than the number of starts once $H$ is a small multiple of $(\log N)^{1/3}$. This uses four primes and two averaged parameters where Chojecki's argument uses five and three. The write-up also derives the heuristic constant: in a model of independent geometric gaps a saddle-point computation gives $\log\Pr(H\text{ distinct})=-L\,I_0(c)+o(L)$ with an explicit $I_0$, and $c_0=1.32322827686395\ldots$ solves $I_0(c_0)=1$, matching the guess $h(x)\sim1.32323\log x$ reported on the thread.

The post further describes a conditional argument: under a strong uniform Hardy-Littlewood $k$-tuples conjecture on average, in lower-bound form with a power saving, combined with a parity-correct Bonferroni minorant identity, one would have $h(x)\ge(c-o(1))\log x$ for small $c>0$, which would refute $h(x)=o(\log x)$. The passage from the independent model to the primes replaces the rate $I_0(c)$ by $I_0(c)+J(c)$, where $J$ is a singular-series correction from a Markov chain on residue classes modulo odd primes, with $J(c)=C_*c^2+O(c^3)$ and $C_*=\tfrac12\bigl(\prod_{p\ge3}(1+(p-1)^{-3})-1\bigr)$, so the combined rate is below $1$ for small $c$. The post presents this as a hypothesis the author has, not as a theorem, so no claim page records it. A post of 26 August 2026 recomputes $c_0$ and $C_*$ with interval arithmetic, confirms $c_0$, and corrects $C_*$ from its twelfth significant digit to $0.0752403861783092\ldots$, a floating-point slip that the post says leaves the argument unchanged. The write-up is a live document that may change.

Submission note. Posted to the site's forum by David Turturean on 24 April 2026:

I am able to corroborate the findings, using a slightly different route. The writeup of what I have so far is at this Overleaf link.

Via a scaffold on top of ChatGPT-5.5-Pro that I ran yesterday, $h(x) \gg (\log x)^{1/3}$ comes out of a four-prime rectangle count: we again look at blocks of $H$ consecutive prime gaps whose total length is at most a constant multiple of $H \log N$, and a positive proportion of starting positions have this property.

If such a block has two equal gaps $d_i = d_j = a$ at positions $i < j$, set $p = p_i$ and $b = p_j - p_i$; then the four integers $p, p+a, p+b, p+a+b$ are all prime, arranged in a rectangle (or the degenerate triple $p, p+a, p+2a$ when $j = i+1$). This uses only four primes instead of five, so one averages over only two parameters rather than three. For $a, b \leq W$, the standard Selberg/Brun upper-bound sieve gives an upper bound of the expected order $N(\log N)^{-4}$ per pair, up to the singular series; averaging the singular series over $(a, b)$ using Gallagher's mean-value estimate, the number of rectangles is $\ll N W^2 (\log N)^{-4}$, with a smaller contribution from the degenerate triples.

Since each rectangle or triple lies in at most $H$ starting blocks, the number of bad small-span starts is $\ll x H W^2 (\log N)^{-3}$. Balancing this against the first-moment count $\ll xHL/W$ of large-span blocks yields $W \asymp (\log N)^{4/3}$ and bad count $\ll x H (\log N)^{-1/3}$: this is smaller than the number of available starts once $H$ is a sufficiently small multiple of $(\log N)^{1/3}$.

The same writeup derives the $1.32323$ constant explicitly. In the iid geometric-gap model with $q = e^{-2/L}$, a saddle-point calculation on the distinctness generating function gives $\log \Pr(H \text{ distinct}) = -L \cdot I_0(c) + o(L)$ with

$$> I_0(c) = c + c \log\!\left(\frac{e^{2c}-1}{2c}\right) + \tfrac{1}{2} \operatorname{Li}_2(1 - e^{2c}), >$$

expansion $I_0(c) = c^2/2 + c^3/18 - c^5/900 + O(c^7)$, and $c_0$ defined by $I_0(c_0) = 1$ is $1.32322827686395\ldots$.

On the conditional side, I have a hypothesis based on a strong (uniform) Hardy-Littlewood k-tuples conjecture on average, in lower-bound form with power-saving, combined with a parity-correct Bonferroni minorant identity (see write-up for details), that would imply $h(x) \geq (c - o(1)) \log x$. The rough idea is that passing from the iid model above to actual primes incurs a correction: primes avoid residue classes modulo small primes, so the model rate $I_0(c)$ must be replaced by $I_0(c) + J(c)$, where $J(c)$ is an odd-prime singular-series pressure coming from a Markov chain on the residue-class trajectories of prefix sums modulo each odd prime. For the conditional bound to give a block of length $cL$ with distinct consecutive prime gaps, one needs $I_0(c) + J(c) < 1$. The relevant fact is that $J$ has the explicit leading expansion

$$> J(c) = C_* c^2 + O(c^3), \qquad C_* = \tfrac{1}{2}\!\left(\prod_{p \geq 3} \left(1 + \tfrac{1}{(p-1)^3}\right) - 1\right) = 0.0752403861777\ldots, >$$

so the combined rate $I_0(c) + J(c) = (\tfrac{1}{2} + C_*) c^2 + O(c^3) \approx 0.5752 , c^2$ is strictly less than $1$ for all sufficiently small $c > 0$: this is then exactly what drives the conditional lower bound $h(x) \geq (c - o(1)) \log x$, and in particular rules out $h(x) = o(\log x)$.

I think it is interesting GPT-5.5-Pro was able to be elicited to give the same outcome, by quite similar methods, at about the same time, right after its release. Looks like a moderate step jump in the direction of analytic number theory from GPT-5.4-Pro. Unless it is shown that h(x) = o(log x) is highly tied to a notoriously difficult conjecture (granted, such as the first Hardy-Littlewood conjecture...), I am optimistic GPT-5.5-Pro itself can eventually resolve $h(x) = o(log x)$ in the negative.

Covers. The first of the problem's two particular questions, answered yes: $h(x)>(\log x)^c$ for every fixed $c<1/3$. The write-up gives no upper bound and does not estimate $h(x)$; its argument toward refuting $h(x)=o(\log x)$ rests on an unproved hypothesis and settles nothing.

Depends on. Nothing in this wiki: the inputs are the standard upper-bound sieve and Gallagher's mean-value estimate for singular series.

Standing. Claimed. The write-up is unrefereed and lives in a live Overleaf project; the site's label is OPEN, its commentary records only that Brun's sieve gives $h(x)\to\infty$, and its proof-claims tab lists nothing for the problem, so the curator records no acceptance. A reply on the thread the same day reports that a check found two minor issues, without naming them or saying which of the two write-ups it checked. No independent review of the argument is recorded.