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The reciprocal gamma function has the following Taylor series.

$$\frac{1}{\Gamma(z)}=\sum_{k=1}^{\infty}a_kz^k,$$

where the $a_k$ coefficient are given by the followint recursion.

$a_1=1$, $a_2=\gamma$, the Euler–Mascheroni constant and for $k>2$ the coefficient $a_k$ is

$$a_k = \frac{{- a_2 a_{k-1} + \sum_{j=2}^{k-1} (-1)^j \, \zeta(j) \, a_{k-j}}}{1-k}.$$

I computed some of them with Maple.

$$\begin{align} a_1 & = 1 \\ a_2 & = \gamma \\ a_3 & = \frac{1}{2} \gamma^2 - \frac{1}{12} \pi^2 \\ a_4 & = \frac{1}{6} \gamma^3 - \frac{1}{12} \pi^2 \gamma + \frac{1}{3} \zeta(3) \\ a_5 & = \frac{1}{24} \gamma^4 - \frac{1}{24} \pi^2 \gamma^2 + \frac{1}{3}\zeta(3)\gamma + \frac{1}{1440}\pi^4 \\ a_6 & = \frac{1}{120}\gamma^5 - \frac{1}{72}\pi^2\gamma^3+\frac{1}{6}\zeta(3)\gamma^2+\frac{1}{1440}\pi^4\gamma-\frac{1}{36}\pi^2\zeta(3)+\frac{1}{5}\zeta(5)\\ \dots \end{align}$$ With Maple we could calculate as many as we want. Numerical values of $a_k$ coefficients are in this wiki section.

Question. Can we solve the recursion and give an explicit formula of $a_k$? In other words: is there a non-recursive closed-form of $a_k$?

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