A060851 a(n) = (2n-1) * 3^(2n-1).

3, 81, 1215, 15309, 177147, 1948617, 20726199, 215233605, 2195382771, 22082967873, 219667417263, 2165293113021, 21182215236075, 205891132094649, 1990280943581607, 19147875284802357, 183448998696332259, 1751104078464989745, 16660504517966902431

Offset: 1

Author

Frank Ellermann, May 03 2001

Keywords

• nonn
• easy

Comments

Denominators of odd terms in expansion of arctanh(s/3); numerators are all 1. - Gerry Martens, Jul 26 2015

References

• Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 28-40.

Links

• Harry J. Smith, Table of n, a(n) for n = 1..200
• Xavier Gourdon and Pascal Sebah, Riemann's zeta function.
• Pablo A. Panzone, Formulas for the Euler-Mascheroni constant, Rev. Un. Mat. Argentina, Vol 50, No. 1 (2009), pp. 161-164.
• Simon Plouffe, Other interesting computations at numberworld.org.
• Index entries for linear recurrences with constant coefficients, signature (18,-81).

Crossrefs

Cf. A002162 (log(2)), A001620 (Euler's constant).

Programs

Magma

[ (2*n-1) * (3^(2*n-1)): n in [1..100]]; // Vincenzo Librandi, Apr 20 2011

Maple

A060851:=n->(2*n-1)*3^(2*n-1); seq(A060851(n), n=1..20); # Wesley Ivan Hurt, Dec 02 2013

Mathematica

Table[(2*n - 1)*3^(2*n - 1), {n, 20}] (* Wesley Ivan Hurt, Dec 02 2013 *)
a[n_] := 1/SeriesCoefficient[ArcTanh[s/3],{s,0,n}]
Table[a[n], {n, 1, 40, 2}] (* Gerry Martens, Jul 26 2015 *)

PARI

a(n) = (2*n - 1)*(3^(2*n - 1)) \\ Harry J. Smith, Jul 13 2009

Formula

Sum_{n>=1} 2/a(n) = log(2).

Sum_{n>=1} (2/a(n) - zeta(2n+1)/(2^(2n)*(2n+1))) = gamma (Euler's constant).

Sum_{n>=1} ((4n+2)/a(n) - zeta(2n+1)/2^(2n))/(2n+1) = gamma (Euler's constant).

Sum_{n>=1} ((4n+2)/a(n) - zeta(2n+1)/2^(2n)) = 7/4.

Sum_{n>=1} ((2n+1)/a(n) - zeta(2n+1)/2^(2n+1)) = 7/8.

From R. J. Mathar, May 07 2013: (Start)

G.f.: 3*x*(1+9*x) / (9*x-1)^2.

a(n+1) = 3*A155988(n). (End)

Sum_{n>=1} (-1)^(n+1)/a(n) = arctan(1/3). - Amiram Eldar, Feb 26 2022

E.g.f.: (1 + exp(9*x)*(18*x - 1))/3. - Stefano Spezia, Dec 26 2024

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), May 07 2001