OFFSET
1,2
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..150
FORMULA
a(n) = (1/8)*(-1)^(n)*(2*n-5)!!*((4*n^3-11*n)+(16*n^4-40*n^2+9)*(Sum_{k=0..(n-1)} ( (-1)^(k+n)/(2*k+1) ) ).
From Peter Bala, Nov 01 2016: (Start)
a(n) = 3*(2*n + 3)!! * Sum_{k = 0..n-1} (-1)^k/((2*k - 3)*(2*k - 1)*(2*k + 1)*(2*k + 3)*(2*k + 5)).
a(n) ~ Pi*2^(n - 5/2)*((n + 2)/e)^(n + 2).
E.g.f.: (6*arcsin(2*x) + 4*x*sqrt(1 - 4*x^2)*(5 - 8*x^2))/(32*(1 - 2*x)^(5/2)).
a(n) = 10*a(n) + (2*n - 7)*(2*n + 1)*a(n-2) with a(0) = 0, a(1) = 1.
The sequence b(n) := (2*n + 3)!! = (2*n + 4)!/((n + 2)!*2^(n+2)) = A001147(n+2) satisfies the same recurrence with b(0) = 3 and b(1) = 15. This leads to the continued fraction representation a(n) = 1/3*b(n)*( 1/(5 - 15/(10 - 7/(10 + 9/(10 + 33/(10 + ... + (2*n - 7)*(2*n + 1)/(10)))))) ) for n >= 2.
As n -> infinity, 3*a(n)/(A001147(n+2)) -> 9/4!*Pi/4 = 1/(5 - 15/(10 - 7/(10 + 9/(10 + 33/(10 + ... + (2*n - 7)*(2*n + 1)/(10 + ...)))))). (End)
MATHEMATICA
Table[(1/8)*(-1)^(n)*(2*n - 5)!!*((4*n^3 - 11*n) + (16*n^4 - 40*n^2 + 9)*(Sum[(-1)^(k + n)/(2*k + 1), {k, 0, n - 1}])), {n, 1, 50}] (* G. C. Greubel, Jun 17 2016 *)
CROSSREFS
Equals the third column of the ED4 array A167584.
KEYWORD
nonn,easy
AUTHOR
Johannes W. Meijer, Nov 10 2009
STATUS
approved