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A082633
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Decimal expansion of the 1st Stieltjes constant gamma_1 (negated).
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102
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0, 7, 2, 8, 1, 5, 8, 4, 5, 4, 8, 3, 6, 7, 6, 7, 2, 4, 8, 6, 0, 5, 8, 6, 3, 7, 5, 8, 7, 4, 9, 0, 1, 3, 1, 9, 1, 3, 7, 7, 3, 6, 3, 3, 8, 3, 3, 4, 3, 3, 7, 9, 5, 2, 5, 9, 9, 0, 0, 6, 5, 5, 9, 7, 4, 1, 4, 0, 1, 4, 3, 3, 5, 7, 1, 5, 1, 1, 4, 8, 4, 8, 7, 8, 0, 8, 6, 9, 2, 8, 2, 4, 4, 8, 4, 4, 0, 1, 4, 6, 0, 4
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OFFSET
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0,2
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COMMENTS
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The Stieltjes constants are named after the Dutch mathematician Thomas Joannes Stieltjes (1856-1894). - Amiram Eldar, Jun 16 2021
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REFERENCES
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Steven R. Finch, Mathematical Constants, Cambridge, 2003, p. 166.
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LINKS
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FORMULA
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Equals lim_{y->infinity} y*(Im(zeta(1+i/y))+y).
Equals lim_{n->infinity} (((log(n))^2)/2 - Sum_{k=2..n} (log(k))/k). - Warut Roonguthai, Aug 04 2005
Equals Integral_{0..infinity} (coth(Pi*x)-1)*(x*log(1+x^2)-2*arctan(x))/(2*(1+x^2)) dx. - Jean-François Alcover, Jan 28 2015
Using the abbreviations a = log(z^2 + 1/4)/2, b = arctan(2*z) and c = cosh(Pi*z) then gamma_1 = -(Pi/2)*Integral_{0..infinity} (a^2 - 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
Equals log(2)^2/6 - log(2)*gamma/2 + (1/(2*log(2)) * Sum_{k>=1} (-1)^k * log(k)^2/k, where gamma is Euler's constant (A001620) (Hardy, 1912). - Amiram Eldar, Jun 09 2023
Equals Sum_{j>=1} Zeta'(2*j + 1) / (2*j + 1). - Peter Luschny, Jun 16 2023
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EXAMPLE
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-0.0728158454836767248605863758749...
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MAPLE
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MATHEMATICA
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Prepend[RealDigits[c=N[StieltjesGamma[1], 120], 10][[1]], 0]
N[EulerGamma^2 - Residue[Zeta[s]^3, {s, 1}]/3, 100] (* Vaclav Kotesovec, Jan 07 2017 *)
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PROG
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(PARI) intnum(x=0, oo, (1/tanh(Pi*x)-1)*(x*log(1+x^2)-2*atan(x))/(2*(1+x^2))) \\ Charles R Greathouse IV, Mar 10 2016
(PARI) Stieltjes(n)=my(a=log(2)); a^n/(n+1)*sumalt(k=1, (-1)^k/k*subst(bernpol(n+1), 'x, log(k)/a))
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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