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a(n) = (2*n+1)!*(2*n+3)/3.
3

%I #25 Sep 08 2022 08:45:47

%S 1,10,280,15120,1330560,172972800,31135104000,7410154752000,

%T 2252687044608000,851515702861824000,391697223316439040000,

%U 215433472824041472000000,139600890389978873856000000

%N a(n) = (2*n+1)!*(2*n+3)/3.

%H G. C. Greubel, <a href="/A165457/b165457.txt">Table of n, a(n) for n = 0..223</a>

%F a(n) = 2*n*(2*n+3)*a(n-1).

%F Sum_{k>=0} 1/a(k) = 3/e = A135003.

%F G.f.: 3F0(1,1,5/2;;4x). - _R. J. Mathar_, Oct 15 2009

%F Sum_{k>=0} (-1)^k/a(k) = 3*(sin(1)-cos(1)) = (-3)*A143624. - _Amiram Eldar_, Apr 12 2021

%p seq(factorial(2*n+1)*(2*n+3)/3,n=0..12); # _Muniru A Asiru_, Oct 21 2018

%t Table[(2*n + 1)!*(2*n + 3)/3, {n, 0, 30}] (* _G. C. Greubel_, Oct 20 2018 *)

%o (PARI) a(n)=(2*n+1)!*(2*n+3)/3

%o (Magma) [Factorial(2*n+1)*(2*n+3)/3: n in [0..30]]; // _G. C. Greubel_, Oct 20 2018

%o (GAP) List([0..12],n->Factorial(2*n+1)*(2*n+3)/3); # _Muniru A Asiru_, Oct 21 2018

%o (Python)

%o import math

%o for n in range(0, 12): print(int(math.factorial(2*n+1)*(2*n+3)/3), end=', ') # _Stefano Spezia_, Oct 21 2018

%Y Cf. A133942, A143624.

%Y Cf. A135003. [_Jaume Oliver Lafont_, Oct 03 2009]

%K nonn

%O 0,2

%A _Jaume Oliver Lafont_, Sep 20 2009

%E frac keyword removed by _Jaume Oliver Lafont_, Nov 02 2009