%I #15 Aug 24 2014 16:56:28

%S 6,12,14,15,21,22,24,26,28,30,35,38,39,42,44,45,46,48,50,51,52,54,55,

%T 56,57,58,60,62,63,69,70,74,75,76,77,78,82,84,85,86,87,88,90,91,92,93,

%U 94,95,96,98,99

%N Numbers for which (n^2)! is divisible by n!^n*(2n)!.

%C 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.

%C Farideh Firoozbakht observes that all terms are composite numbers. The comment in A057599 and conjecture in A096126 seem to confirm that there are no primes nor powers of primes in this sequence.

%o (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.

%o for(n=1,99,a(n)>1&&print1(n,",")) \\ prints this sequence

%K nonn

%O 1,1

%A _M. F. Hasler_, Aug 23 2014