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$n$, $Z(n)$.
$$Z(n) = \sum_{\rho} ' \frac{1}{\rho ^n}$$
I managed to find one monster of an equation after a while and it seems to work fine for n=1$n=1$ but when I try n=3$n=3$ 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 workingwork, 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$n>1$ $$ Z(n) = 1 - \frac{2^n - 1}{2^n} \zeta (n) + \sum_{k=1}^{n} \frac{(-1)^{n-1-k} (k-1)!}{(n-1)!} B_{n,k} ((-1)^{n-k+1}(n-k+1)\gamma _{n-k}. $$ For $n=1$ the equation it's slightly different but I have confirmed that case to be true.
Plugging in n=3$n=3$ for this equation gives
$$Z(3)=1+\frac{3}{2} \gamma _2 -3\gamma \gamma_1 +\gamma^3 - \frac{7}{8} \zeta (3)$$
Wolfram $$ Z(3)=1+\frac{3}{2} \gamma _2 -3\gamma \gamma_1 +\gamma^3 - \frac{7}{8} \zeta (3) $$ Wolfram gives the value
$$Z(3)=1+\frac{3}{2} \gamma _2 +3\gamma \gamma_1 +\gamma^3 - \frac{7}{8} \zeta (3)$$
Instead $$ Z(3)=1+\frac{3}{2} \gamma _2 +3\gamma \gamma_1 +\gamma^3 - \frac{7}{8} \zeta (3) $$ 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:
