I have the following recursive equation $$l_{n}=b_{n-1}+\sum_{k=0}^n l_k a_k b_{n-k},\ n\geq 1$$ where $b_n=1/n!,\ a_n=S_n(-1)$ where $S_n(x)=\sum_{k=0}^n \frac{x^k}{k!}\ \forall x\in \mathbb{R}$ and $l_0=0$. How to solve this?
I proceeded a little bit to solve it and here is what I did
\begin{equation*}
\begin{split}
f(z)=&\sum_{n=0}^\infty l_n a_n z^n \\
=& \sum_{n=1}^\infty \left(b_{n-1}a_n z^n+\sum_{k=0}^n l_k a_k a_n z^n \right)\\
=& \phi(z,1)+\sum_{m=1}^\infty l_m a_m \phi(z,m)
\end{split}
\end{equation*}
Where $$\phi(z,m)=\sum_{n=m}^\infty b_{n-m}a_n z^n $$ Which I have shown to respect the identity $$ \phi(z,m) = z^m \left(e^{z-1}-\sum_{n=m+1}^\infty J_n(-2\sqrt{z})z^{-n/2}\right)$$ Where $J_n(x)$ is the $n$ th order Bessel function. But after that, I don't know how to proceed. Any idea?
Let us aim to find first $a_nl_n$ and then to get $l_n$ we just divide by $a_n$, which is given explicitly.
Let $f(z):=\sum_{k=0}^{\infty}l_ka_kz^k$.
Multiply the recurrence by $a_nz^n$ and add from $n=0$ to $\infty$
We get
$$f(z)=\sum_{k=0}^{\infty}\frac{a_nz^n}{(n-1)!}+f(z)e^{z}.$$
We can solve for $f$ here to get
$$f(z)=\frac{\sum_{n=0}^{\infty}\frac{a_nz^n}{(n-1)!}}{1-e^z}$$
Now, it would be useful to sum the numerator, if possible.
$$\sum_{n=0}^{\infty}\frac{a_nz^n}{(n-1)!}=\sum_{n=0}^{\infty}\sum_{i=0}^{n}\frac{(-1)^iz^n}{i!(n-1)!}$$
Uhm ... I don't see if it has a closed form. Maybe it doesn't. If it doesn't then the formula for $f$ above is the best I can give you.