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Decimal expansion of 3rd Stieltjes constant gamma_3.
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%I #24 Dec 14 2022 08:58:25

%S 0,0,2,0,5,3,8,3,4,4,2,0,3,0,3,3,4,5,8,6,6,1,6,0,0,4,6,5,4,2,7,5,3,3,

%T 8,4,2,8,5,7,1,5,8,0,4,4,4,5,4,1,0,6,1,8,2,4,5,4,8,1,4,8,3,3,3,6,9,1,

%U 3,8,3,4,4,9,2,1,1,2,9,7,0,0,5,3,5,7,0,5,5,7,1,6,6,2,2,8,5,6,6,7,0,2

%N Decimal expansion of 3rd Stieltjes constant gamma_3.

%D S. R. Finch, Mathematical Constants, Cambridge, 2003, p. 166.

%H G. C. Greubel, <a href="/A086280/b086280.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_3 = -(Pi/4)*Integral_{0..infinity} (a^4 - 6*a^2*b^2+b^4)/c^2. gamma_4 = -(Pi/5)*Integral_{0..infinity} (a^5 - 10*a^3*b^2 + 5*a*b^4) / 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 0.0020538...

%p evalf(gamma(3)) ; # _R. J. Mathar_, Feb 02 2011

%t Join[{0, 0}, RealDigits[ N[ -StieltjesGamma[3], 103]][[1]]] (* _Jean-François Alcover_, Nov 07 2012 *)

%Y Cf. A001620, A082633, A086279, A086281, A086282, A183141, A183167, A183206, A184853, A184854.

%K nonn,cons

%O 0,3

%A _Eric W. Weisstein_, Jul 14 2003