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(PARI) zetamult([4, 2]) \\ Charles R Greathouse IV, Jan 21 2016
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zetamult(4,2) = Sum_{m>=2..infinity} (Sum_{n=1..m-1} 1/(m^4*n^2)) = zeta(3)^2 - (4/3)*zeta(6).
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allocated for Jean-François Alcover
Decimal expansion of the multiple zeta value (Euler sum) zetamult(4,2).
0, 8, 8, 4, 8, 3, 3, 8, 2, 4, 5, 4, 3, 6, 8, 7, 1, 4, 2, 9, 4, 3, 2, 7, 8, 3, 9, 0, 8, 5, 7, 6, 0, 4, 5, 6, 6, 4, 7, 9, 7, 8, 7, 5, 2, 3, 8, 6, 7, 5, 0, 5, 9, 1, 6, 7, 4, 8, 8, 9, 2, 7, 6, 5, 5, 9, 4, 7, 4, 2, 7, 8, 9, 2, 8, 7, 4, 3, 5, 7, 1, 4, 5, 5, 8, 2, 7, 7, 9, 4, 6, 0, 0, 4, 7, 0, 5, 8, 6, 6, 1, 9, 5, 5, 9, 6, 6, 7
0,2
Eric Weisstein's MathWorld, <a href="http://mathworld.wolfram.com/MultivariateZetaFunction.html">Multivariate Zeta Function</a>
Wikipedia, <a href="https://en.wikipedia.org/wiki/Multiple_zeta_function">Multiple zeta function</a>
zetamult(4,2) = Sum_{m=2..infinity} (Sum_{n=1..m-1} 1/(m^4*n^2)) = zeta(3)^2 - (4/3)*zeta(6).
0.088483382454368714294327839085760456647978752386750591674889276559474...
Join[{0}, RealDigits[Zeta[3]^2 - (4/3)*Zeta[6], 10, 107] // First]
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nonn,cons,easy
Jean-François Alcover, Jun 16 2015
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allocated for Jean-François Alcover
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