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)
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
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
a(n)/A053005(n) = ((-1)^n * E_{2n}) / (2^(2n+2) * (2n)!), where E_k are the Euler numbers (A122045). - Jwalin Bhatt, May 12 2026
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]]]
PROG
(Python)
from sympy import euler, factorial
def a(n): return (abs(euler(2*n)) / ((2**(2*n+2))*factorial(2*n))).numerator # Jwalin Bhatt, May 12 2026
CROSSREFS
KEYWORD
nonn,frac,nice,easy
AUTHOR
STATUS
approved