Just for fun I was trying to find a formula that calculates the value of the sum of the Riemann zeta non trivial roots raised to a power n $Z(n)$.

$$Z(n) = \sum_{\rho} ' \frac{1}{\rho ^n}$$

I managed to find one monster of an equation after a while and it seems to work fine for n=1 but when I try n=3 the sum I have is almost exactly the same as the one given on Wolfram but with a negative sign on the $3\gamma \gamma_1 $ term.

According to the relation I derived there should be alternating signs in the gamma terms but Wolfram disagrees.
If it is needed I can type down my working but basically I worked out that for n>1

$$Z(n) = 1 - \frac{2^n - 1}{2^n} \zeta (n) + \sum_{k=1}^{n} \frac{(-1)^{n-1-k} (k-1)!}{(n-1)!} B_{n,k} ((-1)^{n-k+1}(n-k+1)\gamma _{n-k}$$

For n=1 the equation it's slightly different but I have confirmed that case to be true.

Plugging in n=3 for this equation gives

$$Z(3)=1+\frac{3}{2} \gamma _2 -3\gamma \gamma_1 +\gamma^3 - \frac{7}{8} \zeta (3)$$

Wolfram gives the value

$$Z(3)=1+\frac{3}{2} \gamma _2 +3\gamma \gamma_1 +\gamma^3 - \frac{7}{8} \zeta (3)$$

Instead. Can anybody prove the Wolfram version so at least I can try to find where I went wrong?

The $B_{n,k}$ are the Bell polynomials that I used in Faa di Bruno's formula to calculate $Z(n)$ and I shorthanded the notation slightly because it was too long.

Link to the page:

https://mathworld.wolfram.com/RiemannZetaFunctionZeros.html