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A274181
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Decimal expansion of Phi(1/2, 2, 2), where Phi is the Lerch transcendent.
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6
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3, 2, 8, 9, 6, 2, 1, 0, 5, 8, 6, 0, 0, 5, 0, 0, 2, 3, 6, 1, 0, 6, 2, 5, 2, 8, 0, 6, 3, 8, 7, 2, 0, 4, 3, 4, 9, 7, 6, 7, 9, 3, 8, 9, 9, 2, 2, 4, 5, 0, 5, 7, 0, 1, 7, 3, 7, 3, 8, 8, 1, 9, 1, 4, 9, 2, 6, 8, 4, 1, 7, 6, 2, 8, 6, 7, 3, 2, 8, 0, 3, 2, 6, 7, 3, 6, 1, 2, 7, 4, 3, 5, 1, 6, 6, 3, 4, 2, 8, 7, 4
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OFFSET
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0,1
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COMMENTS
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The exponential integral distribution is defined by p(x, m, n, mu) = ((n+mu-1)^m * x^(mu-1) / (mu-1)!) * E(x, m, n), see A163931 and the Meijer link. The moment generating function of this probability distribution function is M(a, m, n, mu) = Sum_{k>=0}(((mu+k-1)!/((mu-1)!*k!)) * ((n+mu-1) / (n+mu+k-1))^m * a^k).
In the special case that mu=1 we get p(x, m, n, mu=1) = n^m * E(x, m, n) and M(a, m, n, mu=1) = n^m * Phi(a, m, n), with Phi the Lerch transcendent. If n=1 and mu=1 we get M(a, m, n=1, mu=1) = polylog(m, a)/a = Li_m(a)/a.
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REFERENCES
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William Feller, An introduction to probability theory and its applications, Vol. 1. p. 285, 1968.
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LINKS
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FORMULA
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Equals Phi(1/2, 2, 2) with Phi the Lerch transcendent.
Equals Sum_{k>=0}(1/((2+k)^2*2^k)).
Equals 4 * polylog(2, 1/2) - 2.
Equals Pi^2/3 - 2*log(2)^2 - 2.
Equals Integral_{x=0..oo} x*exp(-x)/(exp(x)-1/2) dx. - Amiram Eldar, Aug 24 2020
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EXAMPLE
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0.32896210586005002361062528063872043497679389922...
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MAPLE
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Digits := 101; c := evalf(LerchPhi(1/2, 2, 2));
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MATHEMATICA
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N[HurwitzLerchPhi[1/2, 2, 2], 25] (* G. C. Greubel, Jun 19 2016 *)
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PROG
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(PARI) Pi^2/3 - 2*log(2)^2 - 2 \\ Altug Alkan, Jul 08 2016
(Python)
from mpmath import mp, lerchphi
mp.dps=102
print([int(d) for d in list(str(lerchphi(1/2, 2, 2))[2:-1])]) # Indranil Ghosh, Jul 04 2017
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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