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%I #14 Feb 04 2021 10:41:39
%S 5,45,390,3315,27846,232050,1922700,15862275,130423150,1069469830,
%T 8750207700,71460029550,582674087100,4744631852100,38589672397080,
%U 313541088226275,2545215892660350,20644528907133950,167329339563085700
%N a(n) = (2^(3*n-1))/(integral_{x=0..1} (1-x^4)^n dx).
%F The integral is equal to n!*Pi*sqrt(2)/(4*GAMMA(3/4)*GAMMA(n+5/4)). - _N. J. A. Sloane_
%F GAMMA(3/4)*GAMMA(n+5/4) is Pi*sqrt(2)*A007696(n+1)/4^(n+1), so the integral is n!*4^n/A007695(n+1) and a(n) = 2^(n-1)*A007696(n+1)/n!. - _R. J. Mathar_, Feb 04 2021
%F D-finite with recurrence n*a(n) +2*(-4*n-1)*a(n-1)=0. - _R. J. Mathar_, Feb 04 2021
%e a(3)=390 (a(0) would be 1/2, so the sequence begins at n=1).
%t f[n_] := 2^(3n - 1)/Integrate[(1 - x^4)^n, {x, 0, 1}]; Table[ f[n], {n, 1, 19}] (* _Robert G. Wilson v_, Feb 26 2004 *)
%o (PARI) a(n)=round(2^(3*n-1)/(n!*Pi*sqrt(2)/(4*gamma(3/4)*gamma(n+5/4))))
%K nonn
%O 1,1
%A Al Hakanson (hawkuu(AT)excite.com), Nov 13 2003
%E More terms from _Benoit Cloitre_, Nov 14 2003
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