zeta(...) の 0 < x < y < ... という条件を取り除き、単に全部異なるとしたときの和を f(...) と呼ぶ。

z(a,a,...,a) = f(a,a,...,a) / k!


z(a, a) = (z(a)^2 - z(2a)) / 2
z(a)^3 = \sum_{x, y, z} 1/x^ay^az^a = 6 z(a,a,a) (x, y, z distinct) + 3 z(2a,a) + 3 z(a,2a) (two of x, y, z equal) + z(3a) (all equal)
z(2a,a) + z(a, 2a) = \sum_{x,y, x!=y} 1/x^(2a)y^a = z(2a)z(a) - z(3a)
Therefore, z(a)^3 = 6z(a,a,a) + 3 z(2a)z(a) - 2 z(3a)
f(a,a,a) = z(a)^3 - 3 f(2a, a) - f(3a)
f(2a, a) = z(2a)z(a) - f(3a)

f(a,a,a,a) = z(a)^4 - 4 f(3a, a) - 6 f(2a,a,a) - 3 f(2a, 2a) - z(4a)
f(3a,a) = z(3a)z(a) - z(4a)
f(2a,a,a) = z(2a)z(a)z(a) - 2 f(3a,a) - f(2a,2a)
f(2a,2a) = z(2a)^2 - z(4a)

f(a,a,a,a) = z(a)^4 + 8 z(3a)z(a) - 6 z(2a)z(a)z(a) + 3 z(2a)^2 - 12 z(4a)