closed formula for: $(g\partial)^n$

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.