Hyper Birthday Paradox? There are $N$ buckets.
Each second we add one new ball to a random bucket - so at $t=k$, there are a total of $k$ balls collectively in the buckets.
At $t=1$, we expect that at least one bucket contains one ball.
At $t=\sqrt{2N\ln{2}}$, due to birthday paradox, we expect that at least one bucket contains two balls.
.
.
At $t=f(m)$, we expect at least one bucket to contain $m$ balls.
What is the function $f(m)$?
 A: Firstly, for $m = 2$, the expected time $f(2)$ at which at least one bucket contains two balls is not $t = \sqrt{2N\ln 2}$. That is the time $t$ at which the probability that there is at least one bucket with two balls crosses $\frac12$. The actual expected time $t$ at which at least one bucket contains two balls is instead given by $$\begin{align}
1 + Q(N) 
&= 1 + 1 + \frac{N-1}{N} + \frac{(N-1)(N-2)}{N^2} + \cdots + \frac{(N-1)(N-2) \cdots 1}{N^{N-1}} \\
&\sim \sqrt{\frac{\pi N}{2}} + \frac{1}{3}+\frac{1}{12}\sqrt{\frac{\pi}{2N}}-\frac{4}{135N}+\cdots
\end{align} $$
where the $\sim$ denotes that these are the precise asymptotics.

(Mildly related: this question and the methods of asymptotic analysis at this one.)
For $m = 3$, see this question, where it is shown that the answer is 
$$
f(3) = \int_0^\infty   \left(1+{x\over N}+{x^2\over2N^2}\right)^N \,e^{-x}\,dx
$$
which has asymptotics
$$
f(3) \sim 6^{1/3}\,\Gamma(4/3)\, N^{2/3} \approx 1.6226\,N^{2/3}.
$$

For general $m$, the above question also says that the above answer generalizes to
$$f(m) \sim \sqrt[m]{m!}\ \Gamma(1 + 1/m)\ N^{1-1/m}$$
(for fixed $m$ and asymptotically as $N \to \infty$).
A: Just as a sidenote.
Interestingly, there is a neat formula for expectation of number of buckets containing exactly $m$ balls after $n$ balls were thrown in.
Say, we have $N$ buckets.
We define 
$$
\alpha = \frac{N - 1}{N}.
$$
The expectation $E_n^m$ of number of buckets containing exactly $m$ balls after $n$ balls were thrown in can be written as
$$
E_n^m = \frac{\alpha^n}{1 - \alpha}
        \begin{pmatrix} n\\ m \end{pmatrix}
        \left(
            \frac{1 - \alpha}{\alpha}
        \right)^m.
$$
The proof a bit tedious (but quite simple), so I just leave a link here to the original work it appears:  "On the average number of birthdays in birthday paradox setup"
