closed formula for: $(g\partial)^n$ The objective is to obtain a closed formula for:
$$
\boxed{A(n)=\big(g(z)\,\partial_z\big)^n,\qquad n=1,2,\dots}
$$
where $g(z)$ is smooth in $z$ and $\partial_z$ is a derivative with respect to $z$. 
I think the first few terms are,
\begin{equation}
\begin{aligned}
A(1) &= g\,\partial\\
A(2)&= g\,(\partial g)\,\partial+g^2\,\partial^2\\
A(3)&= \big[(\partial^2g)g^2+(\partial g)^2g\big]\partial+3(\partial g)g\,\partial^2+g^2\partial^3\\
A(4) &= \big[(\partial^3g)g^3+4(\partial^2g)(\partial g)g^2+(\partial g)^3g\big]\partial\\
&\quad +\big[4(\partial^2g)g^3+7(\partial g)^2g^2\big]\partial^2+6(\partial g)g^3\partial^3+g^4\partial^4\\
&\,\,\vdots
\end{aligned}
\end{equation}
and presumably there is a simple pattern that I'm failing to see. The coefficients do not seem (to me) to be associated to a special function (such as a Bell polynomial) in a simple way.
Any ideas? Perhaps there is a standard formula? 
Thanks!
 A: It seems a closed formula with general $g$ is not within reach.

The coefficients of 
  \begin{align*}
A(1) &= \color{blue}{1}g\,\partial\\
A(2)&= \color{blue}{1}g\,(\partial g)\,\partial+\color{blue}{1}g^2\,\partial^2\\
A(3)&= \big[\color{blue}{1}(\partial^2g)g^2+\color{blue}{1}(\partial g)^2g\big]\partial+\color{blue}{3}(\partial g)g\,\partial^2+\color{blue}{1}g^2\partial^3\\
A(4) &= \big[\color{blue}{1}(\partial^3g)g^3+\color{blue}{4}(\partial^2g)(\partial g)g^2+\color{blue}{1}(\partial g)^3g\big]\partial\\
&\quad +\big[\color{blue}{4}(\partial^2g)g^3+\color{blue}{7}(\partial g)^2g^2\big]\partial^2+\color{blue}{6}(\partial g)g^3\partial^3+\color{blue}{1}g^4\partial^4\\
&\,\,\vdots
\end{align*}
are
  \begin{align*}
&1;\\
&1;1;\\
&1,1;3;1;\\
&1,4,1;4,7;6;1;\\
&\ldots
\end{align*}
  They are archived as A139605. Since there is no closed formula stated, it indicates that no one is available.

But we know some special cases. For instance $g(z)=z$ can be written as
\begin{align*}
\big(z\,\partial_z\big)^n=\sum_{k=0}^n {n\brace  k}z^k\partial_z^k\qquad\qquad n\geq 0
\end{align*}
with ${n\brace  k}$ the Stirling numbers of the second kind. See for instance formula (1.3) in Stirling Operators by L. Carlitz and M.S. Klamkin.
A: Two ways to write his are:
$$(g\partial)^N = \sum_{p+n+m+...=N} a^N_{p,nm..}(\partial^n g)(\partial^mg) ...\partial^p$$ 
or
 $$(g\partial)^N =\sum_{n+m+o+..=N} b^N_{p,nmo...} g^n (\partial g)^m (\partial^2 g)^o ...\partial^p$$
Shouldn't be too hard to find a recursive formula in either case. Probably the best one can do.
