A004041 Scaled sums of odd reciprocals: a(n) = (2*n + 1)!!*(Sum_{k=0..n} 1/(2*k + 1)).

1, 4, 23, 176, 1689, 19524, 264207, 4098240, 71697105, 1396704420, 29985521895, 703416314160, 17901641997225, 491250187505700, 14459713484342175, 454441401368236800, 15188465029114325025, 537928935889764226500

Offset: 0

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

• nonn

Comments

n-th elementary symmetric function of the first n+1 odd positive integers.

Also the determinant of the n X n matrix given by m(i,j) = 2*i + 2 = if i = j and otherwise 1. For example, Det[{{4, 1, 1, 1, 1, 1}, {1, 6, 1, 1, 1, 1}, {1, 1, 8, 1, 1, 1}, {1, 1, 1, 10, 1, 1}, {1, 1, 1, 1, 12, 1}, {1, 1, 1, 1, 1, 14}}] = 264207 = a(6). - John M. Campbell, May 20 2011

Examples

			(arctanh(x))^2 = x^2 + 2/3*x^4 + 23/45*x^6 + 44/105*x^8 + ...
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..403
• J. Courtiel, K. Yeats, Terminal chords in connected chord diagrams, arXiv:1603.08596 [math.CO], 2016; e.g.f. in Remark 1 B_1(z).

Crossrefs

Cf. A000254, A024199, A049034.

Cf. A002428.

From Johannes W. Meijer, Jun 08 2009: (Start)

Equals second left hand column of A028338 triangle.

Equals second right hand column of A109692 triangle.

Equals second left hand column of A161198 triangle divided by 2.

(End)

Programs

Mathematica

Table[(-1)^(n + 1)* Sum[(-2)^(n - k) k (-1)^(n - k) StirlingS1[n + 1, k + 1], {k, 0, n}], {n, 1, 18}] (* Zerinvary Lajos, Jul 08 2009 *)
FunctionExpand@Table[(2 n + 1)!! (Log[4] + HarmonicNumber[n + 1/2])/2, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 13 2016 *)

Formula

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

a(n) is coefficient of x^(2*n+2) in (arctanh x)^2, multiplied by (n + 1)*(2*n + 1)!!.

a(n) = Sum_{i=k+1..n} (-1)^(k+1-i)*2^(n-1)*binomial(i-1, k)*s1(n, i) with k = 1, where s1(n, i) are unsigned Stirling numbers of the first kind. - Victor Adamchik (adamchik(AT)ux10.sp.cs.cmu.edu), Jan 23 2001

a(n) ~ 2^(1/2)*log(n)*n*(2n/e)^n. - Joe Keane (jgk(AT)jgk.org), Jun 06 2002

E.g.f.: 1/2*(1 - 2*x)^(-3/2)*(2 - log(1 - 2*x)). - Vladeta Jovovic, Feb 19 2003

Sum_{n>=1} a(n-1)/(n!*n*2^n) = (Pi/2)^2. - Philippe Deléham, Aug 12 2003

For n >= 1, a(n-1) = 2^(n-1)*n!*(Sum_{k=0..n-1} (-1)^k*binomial(1/2, k)/(n - k)). - Milan Janjic, Dec 14 2008

Recurrence: a(n) = 4*n*a(n-1) - (2*n - 1)^2*a(n-2). - Vladimir Reshetnikov, Oct 13 2016