A034405 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).

1, 2, 14, 216, 5976, 262800, 16945200, 1511395200, 178458940800, 26959810348800, 5071861902240000, 1162523770531200000, 318880083535896960000, 103120648805872938240000, 38820554918130896951040000, 16829499728777665273344000000, 8323409867177396185818624000000, 4657912954052653582049258496000000

Offset: 0

Author

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

Keywords

• nonn

Comments

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

Links

• Robert Israel, Table of n, a(n) for n = 0..238

Crossrefs

Cf. A001147.

Programs

Maple

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):
map(f, [$0..30]); # Robert Israel, Mar 14 2018

Mathematica

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 *)
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 *)

Formula

a(n+2) = (3*n+5)*(n+2)*a(n+1) + (2*n+3)*(n+1)^2*a(n). - Robert Israel, Mar 14 2018

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

a(n) ~ sqrt(Pi) * 2^(n+2) * n^(2*n + 1/2) / exp(2*n). - Vaclav Kotesovec, Jan 02 2020

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

Extensions

Edited by, and more terms from Robert Israel, Mar 14 2018