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%I #30 Mar 24 2024 02:28:31
%S 1,1,5,61,277,50521,540553,199360981,3878302429,2404879675441,
%T 14814847529501,69348874393137901,238685140977801337,
%U 4087072509293123892361,13181680435827682794403,441543893249023104553682821,2088463430347521052196056349
%N Numerators of Taylor series for sec(x) = 1/cos(x).
%C Also numerator of beta(2n+1)/Pi^(2n+1), where beta(m) = Sum_{k>=0} (-1)^k/(2k+1)^m.
%D J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 384, Problem 15.
%D G. W. Caunt, Infinitesimal Calculus, Oxford Univ. Press, 1914, p. 477.
%H Seiichi Manyama, <a href="/A046976/b046976.txt">Table of n, a(n) for n = 0..243</a> (terms 0..100 from T. D. Noe)
%H X. Chen, <a href="http://dx.doi.org/10.2307/2687382">Recursive formulas for zeta(2*k) and L(2*k-1)</a>, Coll. Math. J. 26 (5) (1995) 372-376. See numerators of D_(2k-1).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Secant.html">Secant</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DirichletBetaFunction.html">Dirichlet Beta Function</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HyperbolicSecant.html">Hyperbolic Secant</a>
%F a(n)/A046977(n) = A000364(n)/(2n)!.
%F 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
%e sec(x) = 1 + (1/2)*x^2 + (5/24)*x^4 + (61/720)*x^6 + (277/8064)*x^8 + (50521/3628800)*x^10 + ...
%p ZBS := z -> (Zeta(0,z,1/4) - Zeta(0,z,3/4))/(2^z-2):
%p R := n -> (-1)^floor(n/2)*(2^n-4^n)*ZBS(1-n)/(n-1)!:
%p seq(numer(R(2*n+1)), n=0..16); # _Peter Luschny_, Aug 25 2015
%t Numerator[Partition[CoefficientList[Series[Sec[x], {x, 0, 30}], x], 2][[All,1]]]
%Y Cf. A000364, A046977, A053005, A099612.
%K nonn,frac,nice,easy
%O 0,3
%A _N. J. A. Sloane_
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