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%I #11 Jan 21 2016 14:38:32
%S 0,3,7,7,0,7,6,7,2,9,8,4,8,4,7,5,4,4,0,1,1,3,0,4,7,8,2,2,9,3,6,5,9,9,
%T 1,4,8,2,2,6,0,1,3,1,9,4,1,5,2,7,7,5,2,4,0,1,2,6,4,5,0,7,7,8,0,3,9,1,
%U 0,9,3,8,7,5,5,5,0,7,2,1,9,8,9,1,3,8,3,6,0,2,9,8,1,9,0,7,7,0,8,6
%N Decimal expansion of the multiple zeta value (Euler sum) zetamult(5,3).
%H Eric Weisstein's MathWorld, <a href="http://mathworld.wolfram.com/MultivariateZetaFunction.html">Multivariate Zeta Function</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Multiple_zeta_function">Multiple zeta function</a>
%F zetamult(5,3) = Sum_{m>=2} (sum_{n=1..m-1} 1/(m^5*n^3)).
%F Equals Sum_{m>=2} (H(m-1, 3)+polygamma(2,1)/2+zeta(3))/m^5, where H(n,3) is the n-th harmonic number of order 3.
%F Also equals Sum_{m>=2} (polygamma(2,m)+zeta(3))/(2m^5).
%F Also equals 5*zeta(3)*zeta(5) - (147/24)*zeta(8) - (5/2)*zetamult(6, 2), where zetamult(6,2) is A258947.
%e 0.03770767298484754401130478229365991482260131941527752401264507780391...
%t digits = 99; zetamult[6, 2] = NSum[HarmonicNumber[m-1, 2]/m^6, {m, 2, Infinity}, WorkingPrecision -> digits+20, NSumTerms -> 200, Method -> {"NIntegrate", "MaxRecursion" -> 18}]; zetamult[5, 3] = 5*Zeta[3]*Zeta[5] - (147/24)*Zeta[8] - (5/2)*zetamult[6, 2]; Join[{0}, RealDigits[zetamult[5, 3], 10, digits] // First]
%o (PARI) zetamult([5,3]) \\ _Charles R Greathouse IV_, Jan 21 2016
%Y Cf. A072691, A197110, A258947.
%K nonn,cons
%O 0,2
%A _Jean-François Alcover_, Jun 16 2015
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