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A258982
Decimal expansion of the multiple zeta value (Euler sum) zetamult(5,3).
9
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, 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, 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
OFFSET
0,2
FORMULA
zetamult(5,3) = Sum_{m>=2} (sum_{n=1..m-1} 1/(m^5*n^3)).
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.
Also equals Sum_{m>=2} (polygamma(2,m)+zeta(3))/(2m^5).
Also equals 5*zeta(3)*zeta(5) - (147/24)*zeta(8) - (5/2)*zetamult(6, 2), where zetamult(6,2) is A258947.
EXAMPLE
0.03770767298484754401130478229365991482260131941527752401264507780391...
MATHEMATICA
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]
PROG
(PARI) zetamult([5, 3]) \\ Charles R Greathouse IV, Jan 21 2016
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
STATUS
approved