Proof of the sum of the reciprocal non trivial zeros cubed

Question

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 work, 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

Answer 1

This comes directly from the Hadamard product given in the Wolfram page you refer to by taking logarithmic derivatives and identifying powers of $s$. However, following Harold Stark, the classical formula given in Wolfram should be replaced by the much simpler formula $s(s-1)\Lambda(s)=\prod_{\rho}(1-s/\rho)$, where the product is to be understood as the limit as $T\to\infty$ of $\prod_{|\Im(\rho)|<T}$, and as usual $\Lambda(s)=\pi^{s/2}\Gamma(s/2)\zeta(s)$. Now take logarithmic derivatives and the formula follows.

One can also easily obtain a recursion (equivalent to your use of Bell polynomials): define by induction $$\delta_{k+1}=(k+1)\dfrac{\gamma_k}{k!}+\sum_{j=0}^{k-1}\dfrac{\gamma_j\delta_{k-j}}{j!}\quad\text{($\delta_1=\gamma_0=\gamma)$}\;.$$ Then $Z(k)=1-(1-1/2^k)\zeta(k)+\delta_k\;.$