A246048 Numbers for which (n^2)! is divisible by n!^n*(2n)!.

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

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

Table of n, a(n) for n=1..51.

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

Sequence in context: A325700 A360551 A306999 * A348251 A290648 A336549

Adjacent sequences: A246045 A246046 A246047 * A246049 A246050 A246051

Keyword

nonn

Author

M. F. Hasler, Aug 23 2014

Status

approved