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!
It seems a closed formula with general $g$ is not within reach.
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.