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 $\xi(s)$ investigated in "Relations and positivity results for the derivatives of the Riemann ξ function" by Coffey 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) 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, known more commonly as the Newton identity 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. $m$ 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.