Expected number of distinct integers in a random multiset 
Consider a random multiset $X_n$ of size $n$ sampled uniformly from the integers $\{1,\dots,n\}$.   What is the expected number of distinct integers in $X_n$?

To clarify, each multiset of size $n$ should occur with equal probability.
From computer simulations it seems that the expected value tends to $n/2$ as $n$ tends to infinity. Is that correct?
 A: We  get  from  first  principles the  following  bivariate  generating
function  with $A_q$  marking  the  value $q$  being  sampled and  $u$
marking different types being seen:
$$G(z, u) =
\prod_{q=1}^n \left(1+uzA_q+uz^2A_q^2+uz^3A_q^3+\cdots\right)
\\ = \prod_{q=1}^n \left(1+u\frac{zA_q}{1-zA_q}\right).$$
Extracting the count of multisets we find
$$[z^n]\left. G(z, u)\right|_{u=1, A_1=1, A_2=1, \ldots}
= [z^n] \left(\frac{1}{1-z}\right)^n = {2n-1\choose n-1}.$$
Counting the number of distinct $A_q$ present we obtain
$$[z^n]\left. \frac{\partial}{\partial u}
G(z, u)\right|_{u=1, A_1=1, A_2=1, \ldots}
\\ = [z^n] \left.
\prod_{q=1}^n \left(1+u\frac{zA_q}{1-zA_q}\right)
\sum_{q=1}^n \left(1+u\frac{zA_q}{1-zA_q}\right)^{-1}
\frac{zA_q}{1-zA_q}
\right|_{u=1, A_1=1, A_2=1, \ldots}
\\ = [z^n] \left(\frac{1}{1-z}\right)^n
\sum_{q=1}^n (1-z) \frac{z}{1-z}
= [z^n] nz \left(\frac{1}{1-z}\right)^n
= n [z^{n-1}] \left(\frac{1}{1-z}\right)^n
\\ = n \times {2n-2\choose n-1}.$$
This yields for the expectation
$$n \times {2n-2\choose n-1} \times {2n-1\choose n-1}^{-1}
= n \times
\frac{n}{2n-1} {2n-1\choose n-1} \times {2n-1\choose n-1}^{-1}
\\ = \frac{n^2}{2n-1}.$$
We obtain for the desired answer the closed form
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2}n + \frac{1}{4} + \frac{1}{4}\frac{1}{2n-1}
\sim \frac{1}{2}n}$$
confirming the claim.
