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a(n) = (2*n)! / CatalanNumber(n - 1) for n >= 1 and a(0) = 1.
2

%I #16 May 12 2021 20:38:35

%S 1,2,24,360,8064,259200,11404800,660441600,48771072000,4477184409600,

%T 500391198720000,66920738734080000,10554356508917760000,

%U 1938789402181632000000,410402940653807861760000,99180710658003566592000000,27141314475238493257728000000

%N a(n) = (2*n)! / CatalanNumber(n - 1) for n >= 1 and a(0) = 1.

%F a(n) = A010050(n)/A000108(n-1) for n>=1. - _Michel Marcus_, May 12 2021

%F a(n) = 2*n*(2*n-1)*(n-1)!*n! = 2*n^2*(2*n-1)*((n-1)!)^2 for n > 0. a(n) = a(n-1)*n^2*(2*n - 1)/(2*n - 3) for n > 1. - _Chai Wah Wu_, May 12 2021

%t a[n_] := (2 n)! / CatalanNumber[n - 1]; a[0] := 1; Table[a[n], {n, 0, 16}]

%o (Python)

%o from math import factorial

%o def A344057(n):

%o return 1 if n == 0 else 2*n**2*(2*n-1)*factorial(n-1)**2 # _Chai Wah Wu_, May 12 2021

%Y Cf. A000108, A010050.

%K nonn

%O 0,2

%A _Peter Luschny_, May 12 2021