GPT 6 Astra Ultra attempt on Math-723

Erdős Problem #234: independent continuation

Date: 23 September 2026. Source versions read in full: 234.tex in gpt_pro_5.4.

1) OBJECTIVE AND SOURCE AUDIT

For $a_n=(p_{n+1}-p_n)/\log n$, the problem asks that the empirical distribution converge to a continuous distribution. The source supplies a sieve-based small-gap upper bound and correctly leaves convergence open. Its assertion that there are no prime pairs of odd difference is false: $2$ and $2+h$ can both be prime. There is at most one such pair for each odd $h$. Consecutive gaps for $n\ge2$ are even, so restricting that argument to even $h$ removes the error without changing the intended estimate.

2) WORK: WHAT COMPACTNESS DOES AND DOES NOT GIVE

Assume the mean estimate already used in the source, \[ \frac1{N-1}\sum_{n=2}^N a_n\longrightarrow1. \] Define probability measures $\mu_N=(N-1)^{-1}\sum_{n=2}^N\delta_{a_n}$. For every $R>0$, Markov's inequality gives $\mu_N([R,\infty))\le(1+o(1))/R$. These measures are tight. One can prove subsequential compactness directly: choose successively subsequences on which $\mu_N([0,q])$ converges for each nonnegative rational $q$, use a diagonal subsequence, then take right limits to obtain a nondecreasing right-continuous distribution function. Tightness gives total mass one; squeezing between rational endpoints gives convergence at every continuity point of the limiting function.

Every resulting probability measure $\mu$ satisfies \[ \int_0^\infty t\,d\mu(t)\le1. \] For proof, apply weak convergence to $\min(t,R)$, whose average is at most $1+o(1)$, and let $R$ increase. Equality of the limiting mean is not justified without uniform integrability: a small amount of mass can escape to large values while preserving total first moment.

If one additionally retains the source's small-gap estimate $\limsup_N\mu_N([0,c))\le Cc$, then the open-set inequality for weak convergence gives $\mu([0,c))\le Cc$. Letting $c$ decrease to zero shows $\mu(\{0\})=0$. This last conclusion uses the stated sieve input; the compactness and moment arguments above do not require it.

3) ADVERSARIAL VERIFICATION

Neither tightness nor the absence of an atom at zero implies uniqueness of the subsequential measure. They also do not exclude an atom at a positive point. For example, $\delta_1$ and $(\delta_{1/2}+\delta_{3/2})/2$ both have mean one and no atom at zero. This is an illustration of the logical gap, not a model of the actual prime gaps.

7) FINAL AND REFERENCE

UNRESOLVED. The precise remaining tasks are uniqueness of all subsequential limits and absence of positive atoms. The supplied elementary bounds do not supply either. No new prime-gap theorem is claimed. Exact target checked at https://www.erdosproblems.com/234 through its indexed primary record.

8) COMPLETION ESTIMATE

COMPLETION: 5% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.