Looking for references essentially corroborating (to authoritatively satisfy some editors) the sketch below of the relationship between *even power* (2,4,...) sums (traces) of the *imaginary part* of the complex zeros above the real axis of the Riemann zeta function $\zeta(s)$ and the derivatives evaluated at $t =0$ of $\Omega(t) = \xi(1/2+it)/\xi(1/2)$ where $\xi(s)$ is the Landau Riemann xi function. Please inform me if you feel there are unleapable gaps or flaws in the analysis below.


The [Landau Riemann function](https://en.m.wikipedia.org/wiki/Riemann_Xi_function) $\xi(s)$ investigated in "Relations and positivity results for the derivatives of the Riemann ξ function" by [Coffey](https://doi.org/10.1016/j.cam.2003.09.003) can be used to define the real-valued, entire, even (recall $\xi(s)=\xi(1-s)$) function 


$$\Omega(t) = \xi(1/2 + it)/\xi(1/2)$$


which can be expanded in the Taylor series


$$\Omega(t) = \sum\_{n \geq 0} (-1)^n \frac{\xi^{(2n)}(1/2)}{\xi(1/2)} \frac{t^{2n}}{(2n)!}.$$


The numerical values for the first few derivatives are given in Coffey (as well as some ways to compute them and others).


The log of the Hadamard product [(Weierstrass factorization](https://en.m.wikipedia.org/wiki/Weierstrass_factorization_theorem)) for $\xi$ allows the derivatives of $\Omega(t)$ to be expressed in terms of the even power sums (traces $Tr\_{2n}$) of the reciprocals of the imaginary parts of the zeros of the Riemann zeta above the real axis, and conversely, the power sums can be calculated in terms of the derivatives, all through the Newton identities.


For example, for a polynomial


$$p(x) = \sum\_m c\_m x^m = \prod\_m (1-x/x\_m),$$


$$=\exp[\log(p(x))] = \exp[\sum\_m\log(1-x/x\_m)]$$


$$=\exp[\sum\_{k \geq1} (-\sum\_m \frac{1}{x\_m^k})\frac{x^k}{k}] =\exp[\log(1-Tr.x)]$$


$$= \sum\_n ST1\_n(Tr\_1,Tr\_2,..,Tr\_n) \frac{x^n}{n!},$$


and this is precisely the exponential generating function for the signed Stirling partition polynomials of the first kind, a.k.a., the cycle index partition polynomials for the symmetric groups, given in [A036039](https://oeis.org/A036039), known more commonly as the [Newton identity](https://en.m.wikipedia.org/wiki/Newton%27s_identities) expressing the elementary symmetric polynomials in terms of the power sums. The Faber polynomials give the Newton identity expressing the power sums in terms of the elementary symmetric polynomials.


Now for some sanity checks:


$Tr\_{2n} = \sum\_m 1/z\_m^{2n}$ for $n=1,2,3,4$ for the first several thousands of the zeros were calculated and presented by Gottfried Helms in an [MSE question](https://math.stackexchange.com/questions/3693429/sums-of-reciprocals-of-powers-of-the-imaginary-part-of-the-nontrivial-zeros-of-t/3693530#3693530). $m$ indexes the zeros above the real axis and $z\_m$ denotes the imaginary part.


Using the Newton identities with all odd-indexed indeterminates nulled, I have the numerical consistency checks between Coffey's and Helms' values:


A) From Coffey's derivative estimates, 


$$Tr\_2 = \xi^{(2)}(1/2) / (2\xi(1/2)) \simeq .022972/(2 \cdot .497) \simeq .02311,$$


B) and Helms' sum is


$$Tr\_2 = .02307.$$


C) From Coffey's estimates, 


$$\xi^{(4)}(1/2)/\xi(1/2) \simeq .0.002963/.497 \simeq .005962,$$


D) and from Helms,


$$3(-2Tr\_2)^2 + 6(-2Tr\_4) = 12(Tr\_2^2-Tr\_4) \simeq 12((.02311)^2-.0000372) \simeq.005962.$$


E) From Coffey's estimates,


$$2 Tr\_4 = -4 \Omega^{(4)}(0)/4! + 2 (\Omega^{(2)}(0)/2)^2$$


$$\simeq -4 \cdot .005962/4! + 2 (-.04622/2)^2 \simeq .00007448,$$


F) and from Helms


$$2Tr\_4 \simeq 2 \cdot .000037173 = .00007435.$$


If there is doubt about the order of $\xi$ and whether $Tr\_2$ converges, the analysis here, since the sums are truncated in the computation by Helms, can be regarded as for a corresponding truncation of the Hadamard product for $\xi$. Also, the power sum of order two over the full complex zeros is convergent and can be checked with estimates of derivatives of $\xi(s)$ at $s=0$ by using precisely the Newton identities as above.


Edit (June 5, 2020):


Affirming the convergence, see the comments in the MSE question, extracted from the Titchmarsh references.



Derivatives of Riemann $\xi$ and traces of zeros
It is not a good idea to compute 
$$\sum\_{n=1}^\infty \gamma\_n^{-k}$$ by computing a partial sum of several thousands of terms. The series converges but too slowly for this. 
(see "Computation of the secondary zeta function" is on arXiv now: <https://arxiv.org/abs/2006.04869>).


For simplicity assume that the Riemann Hypothesis is true, then $\gamma\_n$ are the ordinates of non-trivial zeros $\rho$ of $\zeta(s)$ with $\textrm{Im}(\rho)>0$. 


Here I can not give a complete account of how to compute these sums efficiently. But I implemented this in mpmath some years ago. You can use this in Sage. For example writing


from mpmath import \*


mp.dps=50


for n in range(2,6):


$\quad$ print secondzeta(n)


You get the approximate values with 50 digits (it is true the last four of five will not be good, as can be seen computing to more precision)


$$\begin{array}{c}
0.023104993115418970788933810430339014003381760397422\\
0.00072954827270970421587551856909397050335150570355395\\
0.000037172599285269686164866262471740578453650889730014\\
0.0000022311886995021033286406286918371933760764310879243
\end{array}
$$ 


I will try to revise my old paper about this, and put it on arXiv in a few days.


In mpmath the function
secondzeta(s)$=\sum\_n \gamma\_n^{-s}$. This extends to a meromorphic function on the complex plane. So secondzeta(1) is the value of this extended function. The series do not converge. For comparison we should give the values for second zeta(2n) these are:


$$\begin{array}{c}
0.000037172599285269686164866262471740578453650889730014\\
0.00000014417393140097327969538155609482090703688300853254\\
0.00000000066303168025299086987327208196135724847369284211165\\
0.0000000000032136641506166012161021165998346551415628219091519
\end{array}$$


