A086280 Decimal expansion of 3rd Stieltjes constant gamma_3.
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, 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, 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
Offset: 0
Examples
0.0020538...
References
- Steven R. Finch, Mathematical Constants, Cambridge, 2003, p. 166.
Links
- G. C. Greubel, Table of n, a(n) for n = 0..10000
- Krzysztof Maślanka and Andrzej Koleżyński, The High Precision Numerical Calculation of Stieltjes Constants. Simple and Fast Algorithm, arXiv preprint, arXiv:2210.04609 [math.NT], 2022.
- Eric Weisstein's World of Mathematics, Stieltjes Constants
- Wikipedia, Stieltjes constants
Crossrefs
Programs
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Maple
evalf(gamma(3)) ; # R. J. Mathar, Feb 02 2011
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Mathematica
Join[{0, 0}, RealDigits[ N[ -StieltjesGamma[3], 103]][[1]]] (* Jean-François Alcover, Nov 07 2012 *)
Formula
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