Let $\{0,1,\dots,a\}$ be the interval in which you search for the numbers.
The first number may be chosen arbitrary. For the second number we can say, that there is a probability of $1\over a$ to choose a number already chosen (the number of picks becomes $2$) and a probability of $a-1\over a$ to choose a new number.
In the second case, there are $2$ number already chosen, so the probability of chosing on of them again becomes $2\over a$ and the probability of chosing a new number becomes $a-2\over a.$ Finally, we get an equation like this, where $E$ is your expected value:
$E = \frac1a\times1 + \frac{a-1}a\times a\left(\frac2a\times2+\frac{a-2}a\dots\left(\frac{a-1}a\times(a-1)+\frac1a\times a\right)\dots\right)$
The term of $E$ can be generated using a simple recurrence relation:
$\begin{eqnarray} E\hphantom{_n} &=& E_1\\ E_a &=& a\\ E_n &=& \frac{n^2}a+\frac{a-n}aE_{n+1} \end{eqnarray}$
Using a simple Haskell program I could calculate that $E\approx 197.833.$