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
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
KEYWORD
nonn
AUTHOR
M. F. Hasler, Jan 19 2023
STATUS
approved