OPENdatabase updated 2025-08-31
No proof or disproof is known, and it isn't known whether a finite computation could settle it.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
`f n = min_{1 < k ≤ n/2} gcd(n, C(n,k))`. (The infimum is `0` when the range is empty, i.e. `n < 4`.)
`P n` is the largest prime factor of `n` (and `0` if `n ≤ 1`).
\(Q(n)\) is the largest prime-power divisor of \(n\): the maximum of \(p^{v_p(n)}\) over prime divisors \(p\) of \(n\). It is `0` when `n ≤ 1`.
Let \(f(n) = \min_{1 < k \le n/2} \gcd(n, \binom{n}{k})\) and let \(P(n)\) be the largest prime dividing \(n\). **(a)** Characterise those composite \(n\) such that \(f(n) = n/P(n)\). Erdős–Szekeres [ErSz78] note that \(f(n) = n/P(n)\) when \(n\) is a product of two primes (`erdos_700.variants.prime_mul`), with \(n = 30\) a further example. The characterisation itself is open; we state it as the (unknown) set of all composite `n` with `f n = n / P n`. This follows the largest-prime reading on erdosproblems.com. In [ErSz78, p. 98], the first prose line below (7) reads: "where \(P(n)\) is the greatest prime power which divides \(n\)." The paper's reading is stated with \(Q(n)\) in `erdos_700.variants.prime_power_divisor`.
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research open · 5 further variants / known partial results in the Lean file. The authoritative wording is on erdosproblems.com/700.
FULL WRITE-UP BY THIS SITE · JUL 24, 2026
A real theorem for close prime triples, plus Maynard's sieve theorem, appears to unconditionally answer 'yes, infinitely many' to one of the page's three sub-questions. We verified everything we could — 650 sampled cases, zero disagreements — but this needs outside eyes before we call it settled.
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