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a(n) = A000108(n)*A001764(n).
1

%I #11 Dec 31 2022 15:06:38

%S 1,1,6,60,770,11466,188496,3325608,61866090,1199333850,24030289140,

%T 494663027040,10414559269296,223487031938800,4874879691748800,

%U 107852781825352080,2415945569351185530,54714061423541554650,1251237165698155135500,28864572348777684057000

%N a(n) = A000108(n)*A001764(n).

%C G.f. of A000108, C(x), satisfies: C(x) = 1 + x*C(x)^2;

%C G.f. of A001764, F(x), satisfies: F(x) = 1 + x*F(x)^3.

%F Using the Stirling approximation for n! we get the asymptotic expansion a(n) ~ 3^(3*n+1/2)/(2*Pi*n*(n+1)*(2*n+1) = A086201*3^(3*n+1/2)/(n*(n+1)*(2*n+1)). - _A.H.M. Smeets_, Dec 31 2022

%e G.f.: A(x) = 1 + x + 6*x^2 + 60*x^3 + 770*x^4 + 11466*x^5 + 188496*x^6 +...

%o (PARI) {a(n)=binomial(2*n,n)/(n+1)*binomial(3*n,n)/(2*n+1)}

%o for(n=0,25,print1(a(n),", "))

%o (Maxima) A218441[n]:=binomial(2*n, n)/(n+1)*binomial(3*n, n)/(2*n+1)$

%o makelist(A218441[n],n,0,30); /* _Martin Ettl_, Oct 29 2012 */

%Y Cf. A000108, A001764.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 28 2012