GPT 6 Astra Ultra attempt on Math-792

Problem 386: a prime-gap condition in the thin boundary layer

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. I read the complete 386.tex, whose target is to classify binomial coefficients $\binom nk$, $2\le k\le n/2$, that are products of distinct consecutive primes. Its entropy argument confines any infinite family to $k=o(n)$. I investigate a smaller part of that remaining layer and prove a further necessary condition. The supplied numerical searches are not inherited as independently verified computations.

2) WORK

For all sufficiently large $n$, if $2\le k\le n^{1/3}$ and $\binom nk$ is a product of consecutive primes, then the interval \[ (n-k,n] \] contains no prime.

Indeed, by the prime number theorem there is a prime $q$ in $(n/2,2n/3]$ for every sufficiently large $n$. Our range has $k\le n/3$, so $q\le n-k$, $q>k$, and $q^2>n$. Legendre's formula gives \[ v_q\binom nk=1-0-1=0. \] If a prime $p$ lies in $(n-k,n]$, the same formula gives $v_p\binom nk=1$. Consecutivity of the prime support, together with the missing $q<p$, then forbids all prime factors at most $q$. Every prime between $q$ and $n-k$ also has valuation zero. Therefore all prime factors lie in $(n-k,n]$.

There are at most $\lceil k/2\rceil$ primes in those $k$ consecutive integers, since they are odd for large $n$. Squarefreeness would give \[ \binom nk\le n^{\lceil k/2\rceil}\le n^{2k/3}, \] where the last inequality holds for every $k\ge2$. But \[ \binom nk=\prod_{i=0}^{k-1}\frac{n-i}{k-i} >\left(\frac nk\right)^k\ge n^{2k/3}. \] The strict inequality follows because $n>k$ and at least the factor $i=1$ is strictly greater than $n/k$. This contradiction proves the claim.

3) ADVERSARIAL VERIFICATION

The result is necessary, not sufficient. Prime-free intervals of fixed length occur often enough that excluding intervals containing a prime does not resolve even the fixed-$k$ problem. The proof uses an ordinary fixed-proportion consequence of the prime number theorem, not an unproved statement about primes in intervals of length $k$. The odd-prime count is valid because the interval is far above two. Repeated prime factors would invalidate the product bound, so the original distinct-prime hypothesis is essential.

4) FINAL

UNRESOLVED. In the regime $2\le k\le n^{1/3}$, every sufficiently large candidate must end inside a prime-free interval of length $k$. No exclusion of all such intervals or full classification is obtained.

5) SOURCES CHECKED

The complete supplied version, read 2026-09-23. Its prime number theorem input is used only to obtain a prime between $n/2$ and $2n/3$; the additional parity/product argument is proved above. No claim of literature priority is made.

6) COMPLETION ESTIMATE

Subjective progress toward classifying all consecutive-prime binomial coefficients. COMPLETION: 15%