A034405 - OEIS

%I #42 May 23 2022 04:03:21

%S 1,2,14,216,5976,262800,16945200,1511395200,178458940800,

%T 26959810348800,5071861902240000,1162523770531200000,

%U 318880083535896960000,103120648805872938240000,38820554918130896951040000,16829499728777665273344000000,8323409867177396185818624000000,4657912954052653582049258496000000

%N Let f(x) = (Pi - 2*arctan(1/(sqrt(x)*sqrt(x+2))))/(2*sqrt(x)*sqrt(x+2)), take (-1)^n*(n-th derivative from right at x=0) and multiply by A001147(n+1).

%C This is related to the solution of Problem 12150 of American Math. Monthly, vol. 126 (2019), page 946. - _Stephen J. Herschkorn_, Dec 14 2019

%H Robert Israel, <a href="/A034405/b034405.txt">Table of n, a(n) for n = 0..238</a>

%F a(n+2) = (3*n+5)*(n+2)*a(n+1) + (2*n+3)*(n+1)^2*a(n). - _Robert Israel_, Mar 14 2018

%F Empirical observation: a(n-1) = (2*n - 1)! / 4^(n-1) * Integral_{t=0..Pi/4} sec(t)^(2*n). - _Stephen J. Herschkorn_, Dec 14 2019

%F a(n) ~ sqrt(Pi) * 2^(n+2) * n^(2*n + 1/2) / exp(2*n). - _Vaclav Kotesovec_, Jan 02 2020

%F a(n) = (1/2^n)*Sum_{i=0..n} binomial(n,i) * (2*(n-i))! * (2*i)!. - _Håvar Andre Melheim Salbu_, May 22 2022

%p f:= gfun:-rectoproc({-(3*n+5)*(n+2)*a(n+1)+a(n+2)+(2*n+3)*(n+1)^2*(n+2)*a(n),a(0)=1,a(1)=2},a(n),remember):

%p map(f, [$0..30]); # _Robert Israel_, Mar 14 2018

%t Table[FullSimplify[-(2*n + 1)! * Hypergeometric2F1[1, n + 3/2, n + 2, 2]/ ((n + 1)*2^n) - I*n!^2], {n, 0, 20}] (* _Vaclav Kotesovec_, Jan 02 2020 *)

%t Table[FullSimplify[-I*Gamma[1 + n]^2 + I*2^(-1 - 2 n) * Beta[2, 1 + n, 1/2] * Gamma[2 + 2 n]], {n, 0, 20}] (* _Vaclav Kotesovec_, Jan 02 2020 *)

%Y Cf. A001147.

%K nonn

%O 0,2

%A James R. FitzSimons (cherry(AT)neta.com)

%E Edited by, and more terms from _Robert Israel_, Mar 14 2018