A046977 Denominators of Taylor series for sec(x). Also denominators of Taylor series for sech(x) = 1/cosh(x).
1, 2, 24, 720, 8064, 3628800, 95800320, 87178291200, 4184557977600, 6402373705728000, 97316080327065600, 1124000727777607680000, 9545360026665222144000, 403291461126605635584000000, 3209350995912777478963200000, 265252859812191058636308480000000
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 + ... sech(x) = 1 - 1/2 *x^2 + 5/24 *x^4 - 61/720 *x^6 + 277/8064 *x^8 - ...
References
- G. W. Caunt, Infinitesimal Calculus, Oxford Univ. Press, 1914, p. 477.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..225 (terms 0..100 from T. D. Noe)
- Eric Weisstein's World of Mathematics, Secant
- 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(denom(R(2*n+1)), n=0..16); # Peter Luschny, Aug 25 2015
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Mathematica
Table[ EulerE[n]/n! // Denominator, {n, 0, 30, 2}] (* Jean-François Alcover, Oct 04 2012 *)
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) = denominator(R(2*n+1)) and A046976(n) = numerator(R(2*n+1)). - Peter Luschny, Aug 25 2015