

A246048


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


2



6, 12, 14, 15, 21, 22, 24, 26, 28, 30, 35, 38, 39, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 69, 70, 74, 75, 76, 77, 78, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99
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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.
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.


LINKS



PROG

(PARI) max_k(n)=for(k=1, m=n, Mod((n*m)!, m!^n*(k*n)!) && return(k1)) \\ returns the maximal k for m=n.
for(n=1, 99, a(n)>1&&print1(n, ", ")) \\ prints this sequence


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



