login
a(n) = n * 2^n * (n!)^2.
0

%I #24 Jun 01 2024 18:02:56

%S 2,32,864,36864,2304000,199065600,22759833600,3329438515200,

%T 606790169395200,134842259865600000,35895009576222720000,

%U 11277559372311429120000,4129466323494701629440000,1743270091026070964797440000,840505222458998500884480000000

%N a(n) = n * 2^n * (n!)^2.

%C Sum_{n>=1} a(n) / (2*n)! = Pi + 3.

%H Simon Plouffe, <a href="https://arxiv.org/abs/0912.0303">On the computation of the n'th decimal digit of various transcendental numbers</a>, arXiv:0912.0303 [math.NT], 2009.

%e a(2) = 2 * 2^2 * ( 2! )^2 = 2 * 4 * 4 = 32.

%e a(3) = 3 * 2^3 * ( 3! )^2 = 3 * 8 * 36 = 864.

%e Sum_{n=1..10} a(n) / ( 2n )! = 3 + 3.01310...

%e Sum_{n=1..12} a(n) / ( 2n )! = 3 + 3.10046...

%e Sum_{n=1..18} a(n) / ( 2n )! = 3 + 3.14046...

%e Sum_{n=1..20} a(n) / ( 2n )! = 3 + 3.14126...

%e Sum_{n=1..23} a(n) / ( 2n )! = 3 + 3.14154...

%t Table[n*2^n*(n!)^2,{n,20}] (* _Harvey P. Dale_, Jun 01 2024 *)

%o (Rexx)

%o S = 2

%o do N = 2 while length( S ) < 255

%o S = S || ', ' || N * ( 2 ** N ) * ( !( N ) ** 2 )

%o end N

%o say S ; return S

%Y Cf. A001044 ( (n!)^2 ), A010050 ( (2n)! ), A000796 (digits of Pi).

%K nonn

%O 1,1

%A _Frank Ellermann_, Apr 05 2020