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>



Proof of the sum of the reciprocal non trivial zeros cubed
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 $Z(k)=1-(1-1/2^k)\zeta(k)+\delta\_k\;.$


