A046976 Numerators of Taylor series for sec(x) = 1/cos(x).

1, 1, 5, 61, 277, 50521, 540553, 199360981, 3878302429, 2404879675441, 14814847529501, 69348874393137901, 238685140977801337, 4087072509293123892361, 13181680435827682794403, 441543893249023104553682821, 2088463430347521052196056349

Offset

0,3

Comments

Also numerator of beta(2n+1)/Pi^(2n+1), where beta(m) = Sum_{k>=0} (-1)^k/(2k+1)^m.

References

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 384, Problem 15.

G. W. Caunt, Infinitesimal Calculus, Oxford Univ. Press, 1914, p. 477.

Links

Seiichi Manyama, Table of n, a(n) for n = 0..243 (terms 0..100 from T. D. Noe)

X. Chen, Recursive formulas for zeta(2*k) and L(2*k-1), Coll. Math. J. 26 (5) (1995) 372-376. See numerators of D_(2k-1).

Eric Weisstein's World of Mathematics, Secant

Eric Weisstein's World of Mathematics, Dirichlet Beta Function

Eric Weisstein's World of Mathematics, Hyperbolic Secant

Formula

a(n)/A046977(n) = A000364(n)/(2n)!.

Let ZBS(z) = (HurwitzZeta(z,1/4) - HurwitzZeta(z,3/4))/(2^z-2) and R(z) = (cos(z*Pi/2)+sin(z*Pi/2))*(2^z-4^z)*ZBS(1-z)/(z-1)!. Then a(n) = numerator(R(2*n+1)) and A046977(n) = denominator(R(2*n+1)). - Peter Luschny, Aug 25 2015

Example

sec(x) = 1 + (1/2)*x^2 + (5/24)*x^4 + (61/720)*x^6 + (277/8064)*x^8 + (50521/3628800)*x^10 + ...

Maple

ZBS := z -> (Zeta(0, z, 1/4) - Zeta(0, z, 3/4))/(2^z-2):
R := n -> (-1)^floor(n/2)*(2^n-4^n)*ZBS(1-n)/(n-1)!:
seq(numer(R(2*n+1)), n=0..16); # Peter Luschny, Aug 25 2015

Mathematica

Numerator[Partition[CoefficientList[Series[Sec[x], {x, 0, 30}], x], 2][[All, 1]]]

Crossrefs

Cf. A000364, A046977, A053005, A099612.

Sequence in context: A201848 A087871 A242194 * A092838 A196296 A196214

Adjacent sequences: A046973 A046974 A046975 * A046977 A046978 A046979

Keyword

nonn,frac,nice,easy

Author

N. J. A. Sloane

Status

approved