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%I #27 Mar 29 2017 08:55:24
%S 1,2,21,455,14950,658008,36288252,2404808340,186087894300,
%T 16466440817750,1639875152957965,181513211783531667,
%U 22105238818111121361,2937279723907795168000,422879027090638497044400,65572027180886570401929144,10894880155433107077641916792
%N a(n) = binomial(n*(3*n+1)/2, n).
%C a(5*n+3) is divisible by 5; a(5*n+4) is divisible by 5.
%H G. C. Greubel, <a href="/A199747/b199747.txt">Table of n, a(n) for n = 0..300</a>
%F a(n) = C(n*(3*n+1)/2, n*(3*n-1)/2) = C(A005449(n), A000326(n)), where A000326 and A005449 are pentagonal numbers.
%F a(n) ~ n^(n-1/2) * exp(n) * 3^n / (sqrt(Pi) * 2^(n+1/2)). - _Vaclav Kotesovec_, Apr 23 2015
%t Table[Binomial[(n(3n+1))/2,n],{n,0,20}] (* _Harvey P. Dale_, Apr 14 2015 *)
%o (PARI) a(n)=binomial(n*(3*n+1)/2, n)
%Y Cf. A199748, A000326, A005449, A001318.
%K nonn,easy
%O 0,2
%A _Paul D. Hanna_, Nov 09 2011
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