Expansion of Confluent Hypergeometric Function in terms of $\operatorname{erfi}(x)$ I have the following confluent hypergeometric function: $_1F_1\left(2(m+1),\frac{1}{2},-x^2\right)$.
By using Mathematica, I know that for values of $m=0,1,2,...$ this function expands into a power series involving $\operatorname{erfi}(x)$, that can be further simplified by introducing the Dawson function.
For example, for $m=0$ $$_1F_1\biggl(2,\frac{1}{2},-x^2\biggr)=1-\frac{3}{2}xe^{-x^2} \sqrt[]{\pi} \operatorname{erfi}(x)-x^2+x^3e^{-x^2} \sqrt[]{\pi} \operatorname{erfi}(x)$$ $$= 1-3xD_+(x)-x^2+2x^3D_+(x)$$
Here are two more examples, for $m=1$: 
and $m=2$: 
What is the general term of this expansion, and how can I obtain it by hand?
EDIT
As Aaron Hendrickson suggested in the comments, Mathematica expands the function by this formula, so I've found the general term of the sum. But how can I get this expansion starting from the definition of the particular confluent hypergeometric function $_1F_1\left(n,\frac{1}{2},z\right)$?
 A: So as mentioned in the comments this identity provides a "closed form" for your hypergeometric function in terms of Laguerre polynomials. Here is another option that may be considered a more elementary form:
Using DLMF 13.3.17 for $n=2m+1$ and $a=1$ you may write
$$
\begin{aligned}
{_1F}_1\left({2m+2\atop 1/2};z\right)%
&=\frac{1}{(2m+1)!}\partial_z^{2m+1} z^{2m+1} {_1F}_1\left({1\atop 1/2};z\right)\\
&=\frac{1}{(2m+1)!}\partial_z^{2m+1} z^{2m+1} (1+\sqrt{\pi z} e^z\operatorname{erf}(\sqrt z))\\
&=\frac{1}{(2m+1)!}\left((2m+1)!+\partial_z^{2m+1} z^{2m+1}\sqrt{\pi z} e^z\operatorname{erf}(\sqrt z)\right)\\
&=1+\frac{1}{(2m+1)!}\partial_z^{2m+1} z^{2m+1}\sqrt{\pi z} e^z\operatorname{erf}(\sqrt z)\\
\end{aligned}
$$
You may then apply the General Leibniz rule to evaluate the derivative. Once the derivatives are evaluated you may then make the substitution $z\mapsto-x^2$ to obtain the final result.
There are many other different forms you can put your hypergeometric function into using similar methods and variants of this method. What its going to come down to is what form is useful for your application.
