%I #28 May 24 2017 02:36:57
%S 1,4,27,208,1700,14364,123970,1085760,9612108,85795600,770755843,
%T 6960408624,63127818572,574609830760,5246348656500,48027225765120,
%U 440671237120764,4051508174260272,37315784743418332
%N a(n) = T(4*n,n), where T = Catalan triangle (A008315).
%H G. C. Greubel, <a href="/A026005/b026005.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = (2n+2)/(3n+2) * C(4n+1, n). - _Ralf Stephan_, Apr 30 2004
%F a(n) = C(4n,n)-C(4n,n-2)=A039598(2n,n). - _Paul Barry_, Apr 21 2009
%F G.f.: (g-2)*g^2/(3*g-4) where g = 1+x*g^4 is the g.f. of A002293. - _Mark van Hoeij_, Nov 11 2011
%F Conjecture: 9*n*(3*n+2)*(3*n+1)*a(n) +12*(-55*n^3-59*n^2+65*n-11)*a(n-1) -32*(4*n-5)*(4*n-3)*(2*n-3)*a(n-2)=0. - _R. J. Mathar_, May 22 2013
%F a(n) = Sum_{k=0..n}((n+k+1)*binomial(n+k,k)*binomial(3*n-k,n-k))/(2*n+1). - _Vladimir Kruchinin_, Dec 02 2016
%F a(n) ~ 2^(8*n+7/2)*3^(-3*n-5/2)/sqrt(Pi*n). - _Ilya Gutkovskiy_, Dec 02 2016
%t Table[(2 n+2)/(3 n+2) Binomial[4 n+1, n], {n, 0, 20}] (* _Vaclav Kotesovec_, Dec 02 2016 *)
%o (PARI) a(n) = (2*n+2)/(3*n+2)*binomial(4*n+1,n)
%K nonn
%O 0,2
%A _Clark Kimberling_
%E More terms from _Ralf Stephan_, Apr 30 2004