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A046976
Numerators of Taylor series for sec(x) = 1/cos(x).
6
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
KEYWORD
nonn,frac,nice,easy
STATUS
approved