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%I #15 Jun 20 2022 04:17:57
%S 0,4,32,472,6400,92504,1351616,20060016,300533760,4537591960,
%T 68923172032,1052049834576,16123800489472,247959271674352,
%U 3824345280321920,59132290859989472,916312070170755072
%N a(n) = (1/2)*(binomial(4*n, 2*n) - (-1)^n*binomial(2*n,n)).
%D The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972; Formula (3.74), page 31.
%H G. C. Greubel, <a href="/A037964/b037964.txt">Table of n, a(n) for n = 0..825</a>
%F From _R. J. Mathar_, Feb 20 2015: (Start)
%F n*(2*n-1)*(n-1)*a(n) -12*(n-1)*(4*n^2-11*n+10)*a(n-1) +4*(38*n^3-333*n^2+715*n-435)*a(n-2) +48*(34*n^3-228*n^2+499*n-355)*a(n-3) +16*(4*n-15)*(2*n-7)*(4*n-13)*a(n-4) = 0.
%F n*(n-1)*(2*n-1)*(5*n^2-15*n+11)*a(n) -4*(n-1)*(30*n^4-120*n^3 +161*n^2-82*n+12)*a(n-1) -4*(4*n-7)*(2*n-3)*(4*n-5)*(5*n^2-5*n+1)*a(n-2) = 0. (End)
%F From _G. C. Greubel_, Jun 20 2022: (Start)
%F a(n) = Sum_{k=0..n-1} binomial(2*n, 2*k+1)^2.
%F a(n) = (1/2)*(A001448(n) - (-1)^n*A000984(n)).
%F a(n) = (1/2)*((2*n+1)*A000108(2*n) - (-1)^n*A000108(n)).
%F G.f.: (1/4)*(1/sqrt(1+4*sqrt(x)) + 1/sqrt(1-4*sqrt(x)) - 2/sqrt(1+4*x)). (End)
%p A037964 := proc(n)
%p binomial(4*n,2*n)/2-(-1)^n*binomial(2*n,n)/2 ;
%p end proc: # _R. J. Mathar_, Feb 20 2015
%t With[{C= CatalanNumber}, Table[(1/2)*((2*n+1)*C[2*n] -(-1)^n*(n+1)*C[n]), {n, 0, 30}]] (* _G. C. Greubel_, Jun 20 2022 *)
%o (Magma) [(1/2)*((2*n+1)*Catalan(2*n) -(-1)^n*(n+1)*Catalan(n)): n in [0..30]]; // _G. C. Greubel_, Jun 20 2022
%o (SageMath) [sum(binomial(2*n, 2*k+1)^2 for k in (0..n-1)) for n in (0..30)] # _G. C. Greubel_, Jun 20 2022
%Y Cf. A000108, A000984, A001448.
%K nonn
%O 0,2
%A _N. J. A. Sloane_
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