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%I #16 Jun 06 2021 09:01:44
%S 1,11,249,8747,369241,17110731,840221217,42944901219,2260581606657,
%T 121714776747971,6671749658197129,371062413164972955,
%U 20887218937200347281,1187720356043817041843,68124474120573747125529
%N a(n) = 3^n * Sum_{k=0..n} binomial(2*k,k)^3 / 3^k.
%C The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
%C The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).
%H Vincenzo Librandi, <a href="/A167868/b167868.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) = 3^n * Sum_{k=0..n} binomial(2*k,k)^3 / 3^k.
%F Recurrence: n^3*a(n) = (67*n^3 - 96*n^2 + 48*n - 8)*a(n-1) - 24*(2*n-1)^3*a(n-2). - _Vaclav Kotesovec_, Aug 13 2013
%F a(n) ~ 2^(6*n+6)/(61*(Pi*n)^(3/2)). - _Vaclav Kotesovec_, Aug 13 2013
%t Table[3^n Sum[Binomial[2k,k]^3/3^k,{k,0,n}],{n,0,20}] (* _Vincenzo Librandi_, Mar 26 2012 *)
%Y Cf. A079727, A167867, A167868, A167869, A167870, A167872.
%Y Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.
%K nonn
%O 0,2
%A _Alexander Adamchuk_, Nov 14 2009
%E More terms from _Sean A. Irvine_, Apr 27 2010
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