|
%I #16 Apr 02 2022 09:23:22
%S 2,5,28,210,1848,18018,188760,2085798,24018280,285817532,3493769552,
%T 43672119400,556315613280,7202300350500,94561966954800,
%U 1256886144107550,16888191533829000,229116465142280100,3135277944052254000,43238333101156993800,600502564063273528800
%N a(n) = C(n)*C(n+2), where C(n) is the n-th Catalan number A000108(n).
%F a(n) = a(n-1)*4*(2*n-1)*(2*n+3)/((n+1)*(n+3)). - _Chai Wah Wu_, Mar 26 2021
%F Sum_{n>=0} a(n)/4^(2*n+2) = 1/2 - 16/(15*Pi). - _Amiram Eldar_, Apr 02 2022
%t a[n_] := CatalanNumber[n] * CatalanNumber[n + 2]; Array[a, 21, 0] (* _Amiram Eldar_, Apr 02 2022 *)
%o (Python)
%o A342288_list = [2]
%o 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
%Y Inspired by A000108, A005568, A014228.
%K nonn
%O 0,1
%A _N. J. A. Sloane_, Mar 26 2021
|
