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a(n) = (2/3) * 4^n * binomial(3*n, n).
3

%I #31 Apr 25 2024 16:22:35

%S 8,160,3584,84480,2050048,50692096,1270087680,32133218304,

%T 819082035200,21002987765760,541167892561920,13999778090188800,

%U 363391162981023744,9459706464902840320,246865719056498950144,6456334894356662059008,169176689745174567321600,4440485304168581976555520

%N a(n) = (2/3) * 4^n * binomial(3*n, n).

%D Identity (3.116) in H. W. Gould, Combinatorial Identities, Morgantown, 1972, page 35.

%H G. C. Greubel, <a href="/A036909/b036909.txt">Table of n, a(n) for n = 1..695</a>

%F Sum_{k=0..n} binomial(4*n, 2*(n-k))*binomial(k+n, n) = (2/3)*4^n*binomial(3*n, n) = (2/3)*4^n*A005809(n) = 2*4^n*A025174(n).

%F G.f.: (2/3) * 2F1([1/3, 2/3], [1/2], 27*x) = 2*(cos((1/6)*arccos(1-54*x))/sqrt(1-27*x) - 1) /(3*x). - _Harvey P. Dale_, Mar 26 2012

%F D-finite with recurrence n*(2*n-1)*a(n) = 6*(3*n-1)*(3*n-2)*a(n-1). - _R. J. Mathar_, Feb 08 2021

%F G.f.: (2/3)*(cos((1/3)*Arcsin(3*sqrt(3*x)))/sqrt(1-27*x) - 1). - _G. C. Greubel_, Jun 22 2022

%F a(n) ~ 3^(3*n)/sqrt(3*n*Pi). - _Stefano Spezia_, Apr 25 2024

%t Table[2/3 4^n Binomial[3n,n],{n,20}](* _Harvey P. Dale_, Mar 26 2012 *)

%o (Magma) [(2/3)*4^n*Binomial(3*n,n): n in [1..30]]; // _G. C. Greubel_, Jun 22 2022

%o (SageMath) [(2/3)*4^n*binomial(3*n, n) for n in (0..30)] # _G. C. Greubel_, Jun 22 2022

%Y Cf. A005809, A025174.

%K nonn

%O 1,1

%A _N. J. A. Sloane_