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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
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.
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
KEYWORD
nonn
AUTHOR
M. F. Hasler, Aug 23 2014
STATUS
approved