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Claim. For f(n)=min⁡1<k≤n/2gcd⁡ ⁣(n,(nk))f(n)=\min_{1<k\le n/2}\gcd\!\left(n,\binom nk\right) as in Problem 700, there are infinitely many composite nn with f(n)>n1/2f(n)>n^{1/2}. The write-up, the Overleaf document linked above, proves more: there are fixed even integers 0<a<c0<a<c and infinitely many primes pp for which p+ap+a and p+a+cp+a+c are also prime, and for n=p(p+a)(p+a+c)n=p(p+a)(p+a+c) with pp sufficiently large one has

f(n)=p(p+a)=n2/3(1+o(1)),f(n)=p(p+a)=n^{2/3}(1+o(1)),

so every sufficiently large member of this family is squarefree with three prime factors and satisfies f(n)>nf(n)>\sqrt n. The identity is asserted only for large pp, and the write-up needs the order a<ca<c of the two shifts. The write-up's Lemma 4 gets the prime triples from Maynard's theorem on primes in admissible tuples, applied to the admissible set {2i:1≤i≤K}\{2^i:1\le i\le K\} with KK so large that infinitely many translates contain three primes; some triple of positions i<j<ui<j<u then recurs infinitely often, and its gaps satisfy c=2u−2j≥2j>2j−2i=ac=2^u-2^j\ge2^j>2^j-2^i=a. Its Proposition 3 gives f(pqr)=pqf(pqr)=pq for primes p<q<rp<q<r with c>ac>a and p>2(a+c)2p>2(a+c)^2, by an elementary argument from Lucas's theorem. For the large members with f(n)=p(p+a)f(n)=p(p+a), since p+a+cp+a+c is the largest prime factor of nn, the equality f(n)=n/P(n)f(n)=n/P(n) holds, the equality the first question asks to characterize.

Submission note. Posted to erdosproblems.com as a proof claim by Liam Price (account Leeham) on 27 July 2026, giving "GPT 5.6 Sol Pro" as the AI used, which the site marks as accepted as correct:

GPT 5.6 Sol Pro proves the stronger result that there exist fixed positive even integers a<ca<c and infinitely many primes pp such that p+ap+a and p+a+cp+a+c are also prime. Writing

n=p(p+a)(p+a+c),n=p(p+a)(p+a+c),

one has

>f(n)=p(p+a)=n2/3(1+o(1)).> f(n)=p(p+a)=n^{2/3}(1+o(1)).

Thus nn is squarefree with exactly three

distinct prime factors, and f(n)>nf(n)>\sqrt{n} for all sufficiently large members of this family. This answers the second question in the affirmative.

Covers. The second question, answered yes: infinitely many composite nn have f(n)>n1/2f(n)>n^{1/2}, with f(n)∼n2/3f(n)\sim n^{2/3} along the family. The first question (which composite nn have f(n)=n/P(n)f(n)=n/P(n)) and the third (whether f(n)≪An/(log⁡n)Af(n)\ll_A n/(\log n)^A for every AA) remain open, and the problem's label is unchanged.

Depends on. No page of this wiki.

Claimant and system. Liam Price submitted the claim on 2026-07-27 and attributes the proof to GPT 5.6 Sol Pro; the site credits the answer to GPT 5.6 Sol Pro, prompted by Price. A commenter on the claim reported having checked the write-up in detail the next day.

Standing. Pending. The site's proof-claims tab marks the claim as accepted by the site, and Thomas Bloom, the site's curator, records on the problem page (edited 28 August 2026) that GPT 5.6 Sol Pro, prompted by Price, gave a positive answer to the second question. The site labels the problem OPEN, the first and third questions are unanswered, and the problem has no parts, so this credit is not acceptance. No refereed publication exists, the claim lists no formalization, and this corpus has not checked the write-up.