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
Examples
sec(x) = 1 + (1/2)*x^2 + (5/24)*x^4 + (61/720)*x^6 + (277/8064)*x^8 + (50521/3628800)*x^10 + ...
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)
- Xuming 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).
- Jan W. H. Swanepoel, A Short Simple Probabilistic Proof of a Well Known Identity and the Derivation of Related New Identities Involving the Bernoulli Numbers and the Euler Numbers, Integers (2025) Vol. 25, Art. No. A50. See p. 4.
- Eric Weisstein's World of Mathematics, Secant
- Eric Weisstein's World of Mathematics, Dirichlet Beta Function
- Eric Weisstein's World of Mathematics, Hyperbolic Secant
Programs
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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
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Mathematica
Numerator[Partition[CoefficientList[Series[Sec[x], {x, 0, 30}], x], 2][[All,1]]]
Formula
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
Comments