In my post The formula for $g\frac{d}{dg}g\frac{d}{dg}...g\frac{d}{dg}f(g)$ I found a formula given by a finite sum with Sitrling numbers of the second kind. $$ \sum_{j=1}^{n}{n \brace j}g^{j}\frac{d^{j}}{dg^{j}}f(g) = g\frac{d}{dg}g\frac{d}{dg}...g\frac{d}{dg}f(g)$$
Stirling numbers of the second kind are given by $$ {n + 1\brace j} = j{n \brace j} + {n \brace j - 1}. $$
Now, consider $g^{2}\frac{d}{dg}g^{2}\frac{d}{dg}...g^{2}\frac{d}{dg}f(g)$. First of all, it appears to be given by $$ \sum_{j=1}^{n}a_{j}g^{j+n}\frac{d^{j}}{dg^{j}}f(g). $$
Now, redefine the Stirling numbers to be $$ {n + 1\brace j} = (j+n){n \brace j} + {n \brace j - 1}. $$ Then notice that I can conduct only a slightly different proof by induction to the one I linked. Generally, to obtain a formula for $g^{z}\frac{d}{dg}g^{z}\frac{d}{dg}...g^{z}\frac{d}{dg}f(g)$ one needs to introduce numbers of the form $$ {n + 1\brace j} = (j+(z-1)n){n \brace j} + {n \brace j - 1}. $$
Are there Stirling numbers of $z$-kind?