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The L- series 1 + 1/3^(2*n+1) - 1/5^(2*n+1) - 1/7^(2*n+1) + + - - ... = sqrt(2)*(Pi/4)^(2*n+1)*a(n)/(2*n)! (see Shanks), which gives a(n) ~ (1/sqrt(2))*(2*n)!*(4/Pi)^(2*n+1). (End)
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Matthieu Josuat-Vergès and Jang Soo Kim, <a href="http://arxiv.org/abs/1101.5608">Touchard-Riordan formulas, T-fractions, and Jacobi's triple product identity</a>, arXiv:1101.5608 [math.CO] (2011).
D. Shanks, <a href="http://dx.doi.org/10.1090/S0025-5718-1968-0227093-9">Corrigenda to: "Generalized Euler and class numbers"</a>, Math. Comp. 22 (1968), 699.
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The L- series 1 + 1/3^(2*n+1) - 1/5^(2*n+1) - 1/7^(2*n+1) + + - - ... = sqrt(2)*(Pi/4)^(2*n+1)*a(n)/(2*n)!, (see Shanks), which gives a(n) ~ 1/sqrt(2)*(2*n)!*(4/Pi)^(2*n+1). (End)
a(n-1) is approximately 2^(4*n-3)*(2*n-1)!*sqrt(2)/((Pi^(2*n-1))*(2*n-1)). The approximation is quite good a(250) is of the order of 10^1181 and this formula is accurate to 238 digits. - Simon Plouffe, Jan 31 2007
a(n) = (-1)^n*4^(2*n)*E(2*n,1/4), where E(n,x) denotes the n-th Euler polynomial.
From Peter Bala, Nov 08 2019: (Start)
a(n) = sqrt(2)*4^n*Integral_{x = 0..inf} x^(2*n)*cosh(Pi*x/2)/cosh(Pi*x) dx. Cf. A002437.
The L- series 1 + 1/3^(2*n+1) - 1/5^(2*n+1) - 1/7^(2*n+1) + + - - ... = sqrt(2)*(Pi/4)^(2*n+1)*a(n)/(2*n)!, which gives a(n) ~ 1/sqrt(2)*(2*n)!*(4/Pi)^(2*n+1). (End)
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G. C. Greubel, <a href="/A000281/b000281_1.txt">Table of n, a(n) for n = 0..215</a> (terms 0..50 from T. D. Noe)