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In this post, we are interested in the Rimenann zeta function $\zeta(s)$ in $s > 1$ only where it is strictly decreasing rather than $s$ in the entire complex plane. We have the Stieltjes series expansion of the Riemann Zeta function. I inverted the first few terms of this series using series reversion and showed that if $s > 1$ and $\zeta(s) = a$, then,

$$ s = \zeta^{-1}(a) = 1 + \frac{1}{a - \gamma_0} - \frac{\gamma_1}{1!(a - \gamma_0)^2} + \frac{\gamma_2}{2!(a - \gamma_0)^3} - \frac{\gamma_3 - 12\gamma_1}{3!(a - \gamma_0)^4} + \mathcal O(a^{-5}) $$

It seems that $\zeta^{-1}(a)$ can be expressed in the form

$$ 1 + \sum_{n=0}^{\infty} (-1)^n\frac{f(\gamma_1, \gamma_2, \ldots, \gamma_n)}{n!(a - \gamma_0)^{n+1}} $$

where $f(\gamma_1, \gamma_2, \ldots, \gamma_n)$ is some polynomial function of the Stieltjes constants $\gamma_n$.

Question: I am looking for a closed or a recurrence formula for $f$? Also any reference in this series in literature?

Note: The question was posted in MSE. It got upvotes but no answers hence posting in MO.

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    $\begingroup$ This problem is not substantially different from general series reversion, so you should not have hopes for closed formulas more simple then for the reverse series itself. $\endgroup$
    – Alex Gavrilov
    Aug 31, 2020 at 11:25
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    $\begingroup$ I believe this problem is still more important than at first may seem. Since $\zeta(s)$ is monotone and smooth for real $s>1$ it has a smooth inverse and so is analytic, at least as a real "one-dimensional" function. The question is whether or not this inverse has an analytic continuation and in particular what its properties and limitations are. I would much like to have answers and insights to this more general question. I worry about the Stieltjes constants -- they behave terribly. $\endgroup$
    – AndreyF
    Dec 11, 2020 at 18:24

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