What's the property of this series? Is it special? Coefficients of $\left(x\frac{d}{dx}\right)^n f(x) $ I am think about this expression : $e^{\lambda x \frac{d}{dx}}f(x)$. Let us look at each term in the expansion of the exponential operator $e^{\lambda x \frac{d}{dx}}$,
 $$\left(x\frac{d}{dx}\right)^n f(x) $$
For example, (denote $\partial = \frac{d}{dx}$)
\begin{align*} 
  n&=1 : \quad x \partial  \\
   n&=2 : \quad x^2 \partial^2 + x\partial \\
  n&=3 : \quad x^3 \partial^3 + 3x^2\partial^2 + x\partial \\
  n&=4 : \quad x^4 \partial^4+6x^3 \partial^3 + 7x^2\partial^2 + x\partial \\
 n&=5 : \quad x^5 \partial^5+ 10x^4 \partial^4+25\ x^3 \partial^3 + 15x^2\partial^2 + x\partial \\
n&=6 : \quad x^6 \partial^6+15x^5 \partial^5+ 65x^4 \partial^4+90\ x^3 \partial^3 + 31x^2\partial^2 + x\partial \\
\end{align*}
I wonder if there is a general expression for the coefficient. I have expanded it out and it seems that they are related to binomial coefficients though its kind of tedious. If there is a general expression of these coefficients, what's its properties?  Thank you very much.
 A: You have :
$$(x\frac{d}{dx})^nf(x)=\sum_{k=1}^nS(n,k)x^k\partial^k f(x) $$
Where $S(n,k)$ is the Stirling number of the second kind. See : http://oeis.org/A008277/table.
Whereas $S(n,k)$ is the number of partitions into $n$ sets of a set of cardinal $k$, I don't think there is a combinatorial way to prove the identity. What you can do is to show that the Stirling numbers of the second kind and the coefficients of $(x\frac{d}{dx})^nf(x)$ follow the same induction formula.
One can use this to make a link between Newton sums ($1^k+...+n^k$), the Stirling numbers and some binomials (using generating functions machinery).
A: There’s a purely combinatorial explanation of Clément Guérin's answer. Suppose $f(x)$ is the generating function for some combinatorial class $\mathcal{F}$, where the exponent of $x$ records the size of (i.e. the number of atoms in) an object in $\mathcal{F}$, and the coefficient of $x^m$ is the number of elements of size $m$ in $\mathcal{F}$.
Then $x\partial_x f(x)$ is the generating function for the class consisting of elements of $\mathcal{F}$ with a single distinguished atom. Flajolet and Sedgewick (p.102) call this pointing, i.e. “pointing at a distinguished atom”.
Moreover, the operator $x^k\partial_x^k$ corresponds to pointing at $k$ distinct atoms, whereas $(x\partial_x)^n$ corresponds to pointing at $n$ atoms that are not necessarily distinct.
Thus $(x\partial_x)^n f(x) = \sum_{k=1}^n s(n,k) x^k\partial_x^k f(x)$, where the sum distinguishes between the number of distinct distinguished atoms and $s(n,k)$ is the number of ways of assigning $n$ pointers to $k$ distinct atoms.
