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%I #25 Dec 14 2022 08:58:12
%S 0,0,9,6,9,0,3,6,3,1,9,2,8,7,2,3,1,8,4,8,4,5,3,0,3,8,6,0,3,5,2,1,2,5,
%T 2,9,3,5,9,0,6,5,8,0,6,1,0,1,3,4,0,7,4,9,8,8,0,7,0,1,3,6,5,4,5,1,8,5,
%U 0,7,5,5,3,8,2,2,8,0,4,1,4,1,7,1,9,7,8,1,9,7,3,8,1,3,7,4,5,3,7,3,1,9
%N Decimal expansion of 2nd Stieltjes constant gamma_2 (negated).
%D S. R. Finch, Mathematical Constants, Cambridge, 2003, p. 166.
%H G. C. Greubel, <a href="/A086279/b086279.txt">Table of n, a(n) for n = 0..10000</a>
%H Krzysztof Maślanka and Andrzej Koleżyński, <a href="https://arxiv.org/abs/2210.04609">The High Precision Numerical Calculation of Stieltjes Constants. Simple and Fast Algorithm</a>, arXiv preprint (2022). arXiv:2210.04609 [math.NT]
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/StieltjesConstants.html">Stieltjes Constants</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Stieltjes_constants">Stieltjes constants</a>
%F Using the abbreviations a = log(z^2 + 1/4)/2, b = arctan(2*z) and c = cosh(Pi*z) then gamma_2 = -(Pi/3)*Integral_{0..infinity}(a^3-3*a*b^2)/c^2. The general case is for n >= 0 (which includes Euler's gamma as gamma_0) gamma_n = (-Pi/(n+1))* Integral_{0..infinity} sigma(n+1)/c^2, where sigma(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n,2*k)*b^(2*k)*a^(n-2*k). - _Peter Luschny_, Apr 19 2018
%e -0.0096903...
%p evalf(gamma(2)); # _R. J. Mathar_, Feb 02 2011
%t Join[{0, 0}, RealDigits[N[-StieltjesGamma[2], 101]][[1]]] (* _Jean-François Alcover_, Oct 23 2012 *)
%t N[4*EulerGamma^3 + Residue[Zeta[s]^4 / 2 - 2*EulerGamma*Zeta[s]^3, {s, 1}], 100] (* _Vaclav Kotesovec_, Jan 07 2017 *)
%Y Cf. A001620, A082633, A086280, A086281, A086282, A183141, A183167, A183206, A184853, A184854.
%K nonn,cons
%O 0,3
%A _Eric W. Weisstein_, Jul 14 2003
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