TITLE: The sum $\sum_{j=0}^n \binom{n}{j} \left\{ j \atop k \right\} x^j$
QUESTION [10 upvotes]: A recent answer of mine to a question on Math Overflow includes the sum $$S(n,k,x) = \sum_{j=0}^n \binom{n}{j} \left\{ j \atop k \right\} x^j,$$ where $\left\{ j \atop k \right\}$ is a Stirling number of the second kind and $x$ is real.

Questions, in order of interest:

Does this sum simplify?
Can someone give a nice interpretation of this sum?  (Perhaps a combinatorial one?)
Does anyone know any references involving this sum?


Here's are some things I've dug up.

When $x = 1$, we get $$\sum_{j=0}^n \binom{n}{j} \left\{ j \atop k \right\} = \left\{ n+1 \atop k+1 \right\}.$$
This is formula 6.15, p. 265, of Concrete Mathematics.
When $x = 2$, we get sequence A154537 in the OEIS.  There's another formula there.  However, the numbers with $x = 3$ are not in the OEIS.
If my calculations are correct, $S(n,k,x)$ has the bivariate generating function $$\sum_{n,k \geq 0} S(n,k,x) \, y^k \frac{z^n}{n!} = e^{z + y (e^{xz}-1)}.$$
This agrees with the generating function in the OEIS entry for the $x = 2$ case.
The sum satisfies the recurrence $$S(n,k,x) = (xk+1)S(n-1,k,x) + x S(n-1,k-1,x) + [n=k=0].$$

These are all interesting facts, but I would like to put some kind of interpretation on the sum or place it in some larger context.

REPLY [3 votes]: It may interest  the reader that the generating  function $P(z, y)$ of
this  sum  can  be  evaluated  using  the  technique  of  annihilated
coefficient extractors (ACE).

Start by  recalling the bivariate generating function  of the Stirling
numbers of the second kind, which is
$$G(w, u) = \exp(u(\exp(w) -1)).$$

Now  the sum  is $$\sum_{j=0}^n  {n\choose j}  {j\brace  k} x^{n-j}.$$
Therefore the ordinary generating  function of the possible values for
$n$ fixed and $k$ a parameter is
$$\sum_{j=0}^n  {n\choose j} \sum_{k=0}^j y^k {j\brace  k} x^{n-j}.$$

This gives for $P(z, y)$ the form
$$P(z, y) =
\sum_{n\ge 0} \frac{z^n}{n!}
\sum_{j=0}^n  {n\choose j} \sum_{k=0}^j y^k {j\brace  k} x^{n-j}.$$
Using $G(w, u)$ this becomes
$$\sum_{n\ge 0} \frac{z^n}{n!}
\sum_{j=0}^n  {n\choose j} 
\sum_{k=0}^j y^k x^{n-j} \times j! [w^j][u^k] \exp(u(\exp(w) - 1)).$$
This simplifies to
$$\sum_{n\ge 0} z^n
\sum_{j=0}^n  \frac{1}{(n-j)!}
\sum_{k=0}^j y^k x^{n-j} \times [w^j] \frac{(\exp(w) - 1)^k}{k!}$$
or
$$\sum_{n\ge 0} z^n
\sum_{j=0}^n  \frac{x^{n-j}}{(n-j)!}
\sum_{k=0}^j y^k \times [w^j] \frac{(\exp(w) - 1)^k}{k!}.$$
Now reverse the order of the three sums to get
$$\sum_{k\ge 0} y^k 
\sum_{j\ge k} [w^j] \frac{(\exp(w) - 1)^k}{k!}
\sum_{n\ge j} z^n  \frac{x^{n-j}}{(n-j)!}$$
which is
$$\sum_{k\ge 0} y^k 
\sum_{j\ge k} z^j [w^j] \frac{(\exp(w) - 1)^k}{k!}
\sum_{n\ge j} z^{n-j}  \frac{x^{n-j}}{(n-j)!}$$
or
$$\exp(xz) \sum_{k\ge 0} y^k 
\sum_{j\ge k} z^j [w^j] \frac{(\exp(w) - 1)^k}{k!}.$$
The inner  sum is the promised annihilated  coefficient extractor and
transforms into
$$\exp(xz) \sum_{k\ge 0} y^k \frac{(\exp(z) - 1)^k}{k!}$$
which finally yields
$$P(z, y) = \exp(xz) \exp(y (\exp(z) - 1)).$$
Observe that with $x=1$ we obtain
$$\exp(z) \exp(y (\exp(z) - 1))$$
which we immediately recognize as the combinatorial species
$$\mathfrak{P}(\mathcal{Z}) \times 
\mathfrak{P}(\mathcal{Y}\times \mathfrak{P}_{\ge 1}(\mathcal{Z})).$$
This amounts to taking $n$  labelled elements and choosing some subset
of  them, possibly empty,  to mark,  and partition  the rest  into $k$
non-empty subsets which indeed has the count
$$\sum_{j=0}^n  {n\choose j}  {j\brace  k}.$$
There is another annihilated coefficient extractor at this
MSE link.