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?
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.