A359945 Largest k < n such that n! / k! = m! = A000142(m) for some m.

0, 1, 1, 1, 1, 5, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 23, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset

1,6

Comments

For n = 0 there is no k < 0 for which k! would be defined, therefore the sequence starts at offset n = 1.

Surányi conjectured (cf. Erdős and Habsieger references and A003135) that a(10) = 7 corresponding to 10! = 7! * 6! is the only nontrivial solution, i.e., other than a(n) = n-1 for n = m! and a(n) = 1 otherwise.

Links

Table of n, a(n) for n=1..93.

Paul Erdős, Problems and results on number theoretic properties of consecutive integers and related questions, Proc. 5th Manitoba Conf. Numerical Math., Congress. Num. 16 (1975), 25-44.

Laurent Habsieger, Explicit bounds for the diophantine equation A!B! = C!, arXiv:1903.08370, March 2019.

Formula

a(n) > n/2 unless a(n) = 1 or n = 1.

a(n) = n-1 iff n is in A000142 = factorial numbers.

Example

For n = 1, the largest k < n is k = 0 and indeed, 1! / 0! = 1! is a factorial number, so a(1) = 0.
Similarly, for all n in A000142, i.e., n = m!, the largest k < n is k = n-1 and n! / (n-1)! = n = m!, so a(n = m!) = n-1.
For n = 10, 10! / 9! = 10 and 10! / 8! = 90 aren't factorial numbers, but 10! / 7! = 10*9*8 = 2*3*4*5*6, so a(10) = 7.

Prog

(PARI) a(n)={my(m=1, f=n!); while(n-->m, while(m!*n!<f, m++); n!*m!==f && return(n)); f>1; }

Crossrefs

Cf. A000142 (factorial numbers).

Cf. A003135 (n! is a nontrivial product of factorials).

Sequence in context: A035316 A293718 A068316 * A284252 A284254 A309206

Adjacent sequences: A359942 A359943 A359944 * A359946 A359947 A359948

Keyword

nonn

Author

M. F. Hasler, Jan 19 2023

Status

approved