A099617 Denominators of the coefficients in the Taylor expansion of sec(x) + tan(x) around x=0.

1, 1, 2, 3, 24, 15, 720, 315, 8064, 2835, 3628800, 155925, 95800320, 6081075, 87178291200, 638512875, 4184557977600, 10854718875, 6402373705728000, 1856156927625, 97316080327065600, 194896477400625, 1124000727777607680000, 2900518163668125, 9545360026665222144000

Offset: 0

Author

N. J. A. Sloane, Nov 19 2004

Keywords

• nonn
• frac

Examples

			1 + x + (1/2)*x^2 + (1/3)*x^3 + (5/24)*x^4 + (2/15)*x^5 + (61/720)*x^6 + (17/315)*x^7 + ...
1, 1, 1/2, 1/3, 5/24, 2/15, 61/720, 17/315, 277/8064, 62/2835, 50521/3628800, 1382/155925, 540553/95800320, ... = A099612/A099617
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..451
• L. Euler, On the sums of series of reciprocals, par. 13, arXiv:math/0506415 [math.HO], 2005-2008.
• L. Euler, De summis serierum reciprocarum, E41, par. 13.
• Peter Luschny, The Euler-Bernoulli diamond and the lost Bernoulli numbers.

Crossrefs

Cf. A099612, A279107.

Programs

Maple

# From Peter Luschny, Aug 03 2017: (Start)
S := proc(n, k) option remember; if k = 0 then `if`(n = 0, 1, 0) else
S(n, k - 1) + S(n - 1, n - k) fi end: A099617 := n -> denom(S(n, n)/n!):
seq(A099617(n), n=0..24); # version 1
P := proc(n,x) local k,j; add(add((-1)^j*2^(-k)*binomial(k,j)
*(k-2*j)^n* x^(n-k), j=0..k), k=0..n) end: R := n -> `if`(n = 0, 1, P(n-1, -I)/ n!): seq(denom(R(n)), n=0..24); # version 2
ep := n -> `if`(n=0,1,2^n*abs(euler(n,1/2) - euler(n,1))):
a := n -> denom(ep(n)/n!): seq(a(n), n=0..24); # version 3 (End)

Mathematica

nn = 24; Denominator[CoefficientList[Series[Sec[x] + Tan[x], {x, 0, nn}], x]] (* T. D. Noe, Jul 24 2013 *)
Table[If[n==0,1,2 I ^(n+1) PolyLog[-n, -I] / n!], {n,0,24}] // Denominator (* Peter Luschny, Aug 03 2017 *)
Table[2 (1 + Mod[n, 2]) LerchPhi[(-1)^(n+1), n+1, 1/2] / Pi^(n+1), {n, 0, 24}] // Denominator (* Peter Luschny, Aug 03 2017 *)

Formula

A099612(n)/a(n) = A000111(n)/n!. - Seiichi Manyama, Jan 27 2017

From Peter Luschny, Aug 03 2017: (Start)

a(n) = denominator(2*i^(n+1)*PolyLog(-n, -i)/n!) for n > 0.

a(n) = denominator(2^n*|Euler(n,1/2) - Euler(n,1)|/n!) for n > 0 where Euler(n,x) are the Euler polynomials. (End)

Conjecture: For n >= 0, (-1)^n * a(n+1)/(n+1) is the denominator of the n-th term of the Taylor expansion of 1/(1 + sin(x)) around x = 0. [This is based on the fact that (sec(x) + tan(x))' = 1/(1 + sin(-x)). Cf. also the comments in A099612 and A279107.] - Petros Hadjicostas, Oct 06 2019