OFFSET
1,1
COMMENTS
In general, (n*m)! is divisible by m!^n*n!, cf. A060540 for the quotients. It was asked when it is also divisible by m!^n*(kn)! for some k>1. The present sequence answers this for the special case m=n. For the values m=n=52,69,75,77,78,92,95,... one can take k=3; m=n=120 is the least case where one can take k=4.
PROG
(PARI) max_k(n)=for(k=1, m=n, Mod((n*m)!, m!^n*(k*n)!) && return(k-1)) \\ returns the maximal k for m=n.
for(n=1, 99, a(n)>1&&print1(n, ", ")) \\ prints this sequence
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Aug 23 2014
STATUS
approved