Let $f(z)=\frac{z}{\mathrm{erf}(z)}$. Find $f^{(k)}(0)$ for all $k$ Consider the following function complex-valued function:
\begin{align}
f(z)=\frac{z}{ \mathrm{erf}(z)}
\end{align}
Here $\mathrm{erf}(z)$ is the error function.
We are interested in finding all order derivatives of $f$ at $z=0$, that is $f^{(k)}(0)$.
It is not difficult to see that  $f^{(k)}(0)=0$ for odd $k$. Therefore, the question is how to find a general formula for even $k$.
Here are some results that I got from hand calculations:
\begin{align}
f^{(2)}(0)&=\frac{\sqrt{\pi}}{3}\\
f^{(4)}(0)&=\frac{2}{15} \sqrt{\pi}\\
f^{(6)}(0)&=-\frac{44}{21} \sqrt{\pi}\\
f^{(8)}(0)&=\frac{232}{45} \sqrt{\pi}\\
\end{align}
 A: Since $f(z)=z/\text{erf}(z)$ is an even function, we know that $f^{(k)}(0)=0$ for odd $k$. So let's define $f^{(2n)}(0)=:\theta_n\frac{\sqrt{\pi}}{2}$. It is easy to check that $\text{erf}(z)=\frac{2}{\sqrt{\pi}}\sum\limits_{m=0}^\infty \frac{(-1)^{m}z^{2m+1}}{(2m+1)m!}$. Then,
$$z = f(z)\text{erf}(z)=\left(\sum_{n=0}^\infty\frac{\theta_n\sqrt{\pi}}{2 (2n)!}z^{2n}\right)\left(\frac{2}{\sqrt{\pi}}\sum\limits_{m=0}^\infty \frac{(-1)^{m}z^{2m+1}}{(2n+1)n!}\right)=\sum_{k=0}^\infty\left(\sum_{j=0}^k\frac{(-1)^{k-j}\theta_j}{(2j)!(2k-2j+1)(k-j)!}\right)z^{2k+1}$$
which leads to the identities $\theta_0=1$ for $k=0$ and, for $k\geq1$,
$$0=\sum_{j=0}^k\frac{(-1)^{k-j}\theta_j}{(2j)!(2k-2j+1)(k-j)!}=\frac{\theta_k}{(2k)!}+\sum_{j=0}^{k-1}\frac{(-1)^{k-j}\theta_j}{(2j)!(2k-2j+1)(k-j)!}$$
$$\therefore \theta_k=\sum_{j=0}^{k-1}\frac{(-1)^{k-j+1}(2k)!\theta_j}{(2j)!(2k-2j+1)(k-j)!}$$
that let's us compute the $k$-th term via recursion. Sadly, it is pretty likely that there's no way of expressing this sequence in a closed manner. Also, if we consider $\tilde \theta_k:=\theta_k/(2k)!$ instead, we get the simpler recursion formula
$$\tilde\theta_k=\sum_{j=0}^{k-1}\frac{(-1)^{k-j+1}\tilde\theta_j}{(2k-2j+1)(k-j)!}=\sum_{j=1}^{k}\frac{(-1)^{j+1}}{2(j+1)!-j!}\tilde\theta_{k-j}$$
A: Let $e(z):=\dfrac{\text{erf}(z)}z$, the Taylor development of which is easy to obtain. This function is even.
As
$$f(z)\,e(z)=1$$ is an identity, by the Leibnitz theorem, for $m>0$,
$$\sum_{k=0}^{m}f^{(2m-2k)}(z)\,e^{(2k)}(z)=0$$
from which we draw the recurrence
$$f^{(2m)}(0)=-\frac1{e(0)}\sum_{k=1}^{m}f^{(2m-2k)}(0)\,e^{(2k)}(0).$$
