A156769 a(n) = denominator(2^(2*n-2)/factorial(2*n-1)).

1, 3, 15, 315, 2835, 155925, 6081075, 638512875, 10854718875, 1856156927625, 194896477400625, 49308808782358125, 3698160658676859375, 1298054391195577640625, 263505041412702261046875, 122529844256906551386796875, 4043484860477916195764296875

Offset: 1

Author

Johannes W. Meijer, Feb 15 2009

Keywords

• easy
• nonn

Comments

Resembles A036279, the denominators in the Taylor series for tan(x). The first difference occurs at a(12).

The numerators of the two formulas for this sequence lead to A001316, Gould's sequence.

Stephen Crowley indicated on Aug 25 2008 that a(n) = denominator(Zeta(2*n)/Zeta(1-2*n)) and here numerator((Zeta(2*n)/Zeta(1-2*n))/(2*(-1)^(n)*(Pi)^(2*n))) leads to Gould's sequence.

This sequence appears in the Eta and Zeta triangles A160464 and A160474. Its resemblance to the sequence of the denominators of the Taylor series for tan(x) led to the conjecture A156769(n) = A036279(n)*A089170(n-1). - Johannes W. Meijer, May 24 2009

Links

• G. C. Greubel, Table of n, a(n) for n = 1..250

Crossrefs

Cf. A036279 Denominators in Taylor series for tan(x).

Cf. A001316 Gould's sequence appears in the numerators.

Cf. A000265, A036279, A089170, A117972, A160464, A160469 (which resembles the numerators of the Taylor series for tan(x)), A160474. - Johannes W. Meijer, May 24 2009

Programs

Magma

[Denominator(4^(n-1)/Factorial(2*n-1)): n in [1..25]]; // G. C. Greubel, Jun 19 2021

Maple

a := n ->(2*n-1)!*2^(add(i,i=convert(n-1,base,2))-2*n+2); # Peter Luschny, May 02 2009

Mathematica

a[n_] := Denominator[4^(n-1)/(2n-1)!];
Array[a, 15] (* Jean-François Alcover, Jun 20 2018 *)

Sage

[denominator(4^(n-1)/factorial(2*n-1)) for n in (1..25)] # G. C. Greubel, Jun 19 2021

Formula

a(n) = denominator( Product_{k=1..n-1} 2/(k*(2*k+1)) ).

G.f.: (1/2)*z^(1/2)*sinh(2*z^(1/2)).

From Johannes W. Meijer, May 24 2009: (Start)

a(n) = abs(A117972(n))/A000265(n).

a(n) = A036279(n)*A089170(n-1). (End)

a(n) = A049606(2*n-1). - Zhujun Zhang, May 29 2019