Help finding a closed form I have the following function:
$$\frac{2e^x}{e^{2x}+1+2x}=\sum_{n=0}^\infty \varepsilon_n\frac{x^n}{n!}$$
I would like to find a closed form for the $\varepsilon_k$.  One thing that I do know is that the $\varepsilon_n$ satisfy the following recurrence relation:
$$\varepsilon_n=1-2n\varepsilon_{n-1}-\sum_{k=0}^{n-2}\binom{n}{k}2^{n-k-1}\varepsilon_k$$
The first 11 numbers are below:
$$1,-1,3,-15,93,-725,6815, -74627, 933849, -13148361, 205690779$$
I've tried slowly plugging in small values of $n$, expanding the recurrence relation and then condensing it back to something smaller.  I try to find patterns then amongst the summands and products but this method is very long and tedious, and the size of the summands expands quickly.  What are some of the methods the have been used for closed forms similar to this and how can I better approach the problem?
 A: $$\varepsilon_n=n![x^n]\left(\frac{2e^{x}}{e^{2x}+1+2x}\right)$$
$$=n![x^n]\left(\frac{2e^{x}}{e^{2x}-1+2+2x}\right)$$
$$=n![x^n]\left(\frac{\frac{2e^{x}}{2+2x}}{1-\frac{1-e^{2x}}{2+2x}}\right)$$
$$=n![x^n]\left(\left(\frac{2e^{x}}{2(1+x)}\right)\sum_{r=0}^{\infty}\left(\frac{1-e^{2x}}{2(1+x)}\right)^{r}\right)$$
$$=n![x^n]\left(2e^{x}\sum_{r=0}^{\infty}\frac{(1-e^{2x})^{r}}{\left(2(1+x)\right)^{r+1}}\right)$$
$$=n![x^n]\left(2e^{x}\sum_{r=0}^{\color{red}{n}}\frac{(1-e^{2x})^{r}}{\left(2(1+x)\right)^{r+1}}\right)$$
$$=n![x^n]\left(e^x\sum_{r=0}^{n}\sum_{k=0}^{r}\binom{r}{k}(-e^{2x})^{k}\left(\frac{1}{2^r}\right)\sum_{m=0}^{\infty}\binom{r+m}{m}(-x)^m\right)$$
$$=n![x^n]\left(\sum_{r=0}^{n}\sum_{k=0}^{r}\binom{r}{k}(-1)^{k}\left(e^{(2k+1)x}\right)\left(\frac{1}{2^r}\right)\sum_{m=0}^{\infty}\binom{r+m}{m}(-x)^m\right)$$
$$=n![x^n]\left(\sum_{r=0}^{n}\sum_{k=0}^{r}\binom{r}{k}(-1)^{k}\sum_{h=0}^{\infty}\frac{(2k+1)^{h}x^{h}}{h!}\left(\frac{1}{2^r}\right)\sum_{m=0}^{\infty}\binom{r+m}{m}(-x)^m\right)$$
$$=n![x^n]\left(\sum_{r=0}^{n}\sum_{k=0}^{r}\sum_{h=0}^{\infty}\sum_{m=0}^{\infty}\binom{r}{k}(-1)^{k+m}\left(\frac{1}{2^r}\right)\frac{(2k+1)^{h}}{h!}\binom{r+m}{m}x^{h+m}\right)$$
$$=n!\sum_{r=0}^{n}\sum_{k=0}^{r}\sum_{m=0}^{n}\binom{r}{k}\binom{r+m}{m}(-1)^{k+m}\left(\frac{1}{2^r}\right)\frac{(2k+1)^{n-m}}{(n-m)!}$$
$$=n!\sum_{r=0}^{n}\sum_{k=0}^{r}\sum_{m=0}^{n}\frac{(r+m)!{\cdot}(2k+1)^{n-m}{\cdot}(-1)^{m+k}}{m!{\cdot}(r-k)!{\cdot}k!{\cdot}(n-m)!{\cdot}2^{r}}$$.
