A258982 Decimal expansion of the multiple zeta value (Euler sum) zetamult(5,3).

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

Links

Table of n, a(n) for n=0..99.

Eric Weisstein's MathWorld, Multivariate Zeta Function

Wikipedia, Multiple zeta function

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

Cf. A072691, A197110, A258947.

Sequence in context: A363146 A013564 A009467 * A227336 A373862 A353049

Adjacent sequences: A258979 A258980 A258981 * A258983 A258984 A258985

Keyword

nonn,cons

Author

Jean-François Alcover, Jun 16 2015

Status

approved