A167588 The second column of the ED4 array A167584.

1, 6, 41, 372, 4077, 53106, 795645, 13536360, 257055705, 5400196830, 124170067665, 3104906420700, 83818724048325, 2431059231544650, 75354930324303525, 2486926158748693200, 87036225272850632625, 3220532233879435917750, 125594424461427237941625

Offset

1,2

Links

G. C. Greubel, Table of n, a(n) for n = 1..150

Formula

a(n) = (1/2)*(-1)^(n)*(2*n-3)!!*(n+(4*n^2-1)*Sum_{k=0..n-1} ((-1)^(k+n)/(2*k+1))).

From Peter Bala, Nov 01 2016: (Start)

a(n) = (2*n + 1)!! * Sum_{k = 0..n-1} (-1)^(k-1)/((2*k - 1)*(2*k + 1)*(2*k + 3)).

a(n) ~ Pi * 2^(n-3/2) * ((n+1)/e)^(n+1).

E.g.f.: (4*x*sqrt(1 - 4*x^2) + 2*arcsin(2*x))/(8*(1 - 2*x)^(3/2)).

a(n) = 6*a(n-1) + (2*n - 5)*(2*n - 1)*a(n-2) with a(0) = 0, a(1) = 1.

The sequence b(n) := (2*n + 1)!! = (2*n + 2)!/((n + 1)!*2^(n+1)) satisfies the same recurrence with b(0) = 1 and b(1) = 3. This leads to the continued fraction representation a(n) = b(n)*[ 1/(3 - 3/(6 + 5/(6 + 21/(6 + ... + (2*n - 5)*(2*n - 1)/(6))))) ] for n >= 2.

As n -> infinity, a(n)/(A001147(n+1)) -> 1/2!*Pi/4 = 1/(3 - 3/(6 + 5/(6 + 21/(6 + ... + (2*n - 5)*(2*n - 1)/(6 + ...))))). Compare with the generalized continued fraction representation Pi = 3 + 1^2/(6 + 3^2/(6 + 5^2/(6 + ...))). See A142970. (End)

Mathematica

Table[(1/2)*(-1)^(n)*(2*n - 3)!!*((n) + (4*n^2 - 1)*Sum[(-1)^(k + n)/(2*k + 1), {k, 0, n - 1}]), {n, 1, 50}] (* G. C. Greubel, Jun 17 2016 *)

Crossrefs

Equals the second column of the ED4 array A167584.

Other columns are A024199 and A167589.

Cf. A007509 and A025547 (the sum((-1)^(k+n)/(2*k+1), k=0..n-1) factor), A001147, A142970.

Sequence in context: A094869 A178824 A006198 * A323573 A230134 A007130

Adjacent sequences: A167585 A167586 A167587 * A167589 A167590 A167591

Keyword

easy,nonn

Author

Johannes W. Meijer, Nov 10 2009

Status

approved