Erdős Problem #393
No proof or disproof is known, and it isn't known whether a finite computation could settle it.
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`f n` is \(f(n)\), the minimal \(m \geq 1\) such that \(n! = a_1 \cdots a_t\) with \(a_1 < \cdots < a_t = a_1 + m\). The factors are recorded as a finite set \(s \subseteq [a, a + m]\) of natural numbers that contains \(a = a_1\) and \(a + m = a_t\). Since \(m \geq 1\) there are at least two factors, and they are positive because \(n! \neq 0\). Such a factorisation exists for \(n \geq 2\). For \(n \leq 1\) there is none, since \(n! = 1\), and `f n` is the junk value `0`.
Attempts
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Exact exhaustive computation extends A388302 from n=79 through n=100 (in particular f(80)=76), with a complete geometric-mean reduction and standalone verifier; the uniform asymptotic problem remains
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