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I have this problem involving a differential operator that I encountered while studying an enduring problem in chemical kinetics. However, I am unable to find previous results or develop a solution myself. Here is my problem:

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be sufficiently differentiable and define operator $T_n(x) := (x\partial_x)^n$ for $x \in \mathbb{R}$ that acts on the function $f$ an $n \in \mathbb{N}$ number of times. So explicitly,

$$ T_3(x)f(x) = x\partial_x(x\partial_x(x\partial_xf(x))) $$

I have been attempting to find some sort of recurrence relation that tells me what the operator should be in terms of $x$ and $n$ with no derivatives. Can this be done? As a preliminary result to help, if $f(x) = e^x$, then $T_n(x)$ will be polynomials, but I can't seem to develop a formula for those polynomials in terms of $n$. Can anyone provide some insight to this?

I have tried using the sequence

$$ T_n(x) = x\partial_xT_{n-1}(x) = x(n-1)T_{n-2}(x)[x\partial^2_x + \partial_x] = \cdot\cdot\cdot $$

but that seems to get me nowhere as I just get another build up of $(x\partial^2_x + \partial_x)^{some\,power}$ terms.

An explicit expression helps as this sequence describes a probability distribution. A distribution is difficult to describe and model in terms of derivatives.

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    Solution found. Polynomials follow the Sterling numbers of the second kind. See my other question [here](http://math.stackexchange.com/questions/2164578/sequence-of-numbers-representing-probability-distribution)2017-02-28

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Turns out this is the Grünert's Polynomials done infinity years ago, just like everything in mathematics. See this document.