OFFSET
0,1
FORMULA
a(n) = a(n-1)*4*(2*n-1)*(2*n+3)/((n+1)*(n+3)). - Chai Wah Wu, Mar 26 2021
Sum_{n>=0} a(n)/4^(2*n+2) = 1/2 - 16/(15*Pi). - Amiram Eldar, Apr 02 2022
MATHEMATICA
a[n_] := CatalanNumber[n] * CatalanNumber[n + 2]; Array[a, 21, 0] (* Amiram Eldar, Apr 02 2022 *)
PROG
(Python)
A342288_list = [2]
for n in range(1, 100): A342288_list.append(A342288_list[-1]*4*(2*n-1)*(2*n+3)//((n+1)*(n+3))) # Chai Wah Wu, Mar 26 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Mar 26 2021
STATUS
approved