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^2\,\partial^2+g^3\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 perhaps there is a simple pattern that I'm failing to see.

The partitionings of the $\partial$ and $g$ are reminiscent of Bell polynomials but the coefficients are more complicated. Perhaps it is useful to make explicit that the general expansion is of the form: $$ (g\,\partial)^n=g^n\sum_{p=0}^{n-1}a_{n,p}(g)\,\partial^{\,n-p} $$ with, $$ a_{n,p}(g)=\sum_{m_1+2m_2+\dots+pm_{p}=p} C_{n,p}(m_1,\dots,m_{p})\Big(\frac{\partial g}{g}\Big)^{m_1}\Big(\frac{\partial^2 g}{g}\Big)^{m_2}\dots \Big(\frac{\partial^{p} g}{g}\Big)^{m_{p}}\qquad (*) $$ and the latter sum is over all non-negative integers, $\{m_1,\dots,m_{p}\}$, subject to: $$ m_1+2m_2+\dots+pm_{p}=p $$

From this viewpoint the objective is to determine the coefficients $C_{n,p}(m_1,\dots,m_{p})$, which in turn depend on *all* integers, $n$, $p$ and $\{m_1,\dots,m_p\}$.

Any ideas?

Many thanks in advance.

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