Maximum occupancy balls in bins with limited independence Throw $n$ balls into $n$ bins and let $X_n$ be the maximum occupancy. That is the maximum number of balls found in any bin.
If you throw the balls uniformly and independently it is known that $\mathbb{E}(X_n) = \Theta(\log{n}/\log{\log{n}})$.  If the process is merely pairwise independent (but still uniform) then it is known that  $\mathbb{E}(X_n)$ can grow as quickly as $\Theta(\sqrt{n})$.  
If the random process is $k$-wise independent, then for each $k$ there will be some tight asymptotic upper bound for $\mathbb{E}(X_n)$.  

How do the asymptotics of a tight upper bound for $\mathbb{E}(X_n)$ depend on $k$?

This question may be hard to answer precisely but approximate answers and even conjectures would be very helpful too.  Sometimes $k=4$ is a transition point as you can then use more powerful moment methods but I am not sure how to do that usefully here.

What are the asymptotics for a tight upper bound for $\mathbb{E}(X_n)$ when $k=4$?

 A: I think https://arxiv.org/abs/1502.05729 might be at least a partial answer to my question. It shows "a $k$-independent family of functions that imply [heaviest loaded bin] size is $\Omega(n^{1/k})$".
A: Consider the probability that $k$ particular balls go into the same bin. By $k$-wise independence, this is $n^{-k}$ for each bin, or $n^{-k+1}$ when we sum over all bins. On the other hand, if we choose a random $k$-tuple of indices, the probability that these are all sent to the same bin is at least the probability that all $k$ are sent to the bin with the highest load $X_n$, so if we condition on $X_n$, the probability is at least ${X_n \choose k} / {n \choose k}$. 
Unfortunately, $x \choose k$ is not always convex as a function of $x$, so we have to modify it slightly to use convexity. For $k \in \mathbb{N}, x\in \mathbb{R}$, define ${x \choose k}^+ = {\begin{cases} {x \choose k}, x \gt k-1 \newline  0, x \le k-1\end{cases}} = \frac{1}{k!}\prod_{i=0}^{k-1} \max (0,x-i).$ Since each factor $\max (0,x-i)$ is convex and nonnegative, the scaled product ${x \choose k}^+$ is convex.
By Jensen's inequality, the probability $k$ balls go to the same bin is at least ${E[X_n] \choose k}^+ / {n \choose k}$. So, $k$-wise independence implies
$$\begin{eqnarray}{E[X_n] \choose k}^+  &\le & {n \choose k}n^{-k+1} \newline &\le & \frac{n}{k!} \newline \prod_{i=0}^{k-1} \max(0,X_n-i)  &\le & n \newline \max(0,E[X_n]-k+1)^k &\le & n \newline E[X_n] &\le & \sqrt[k]{n} + k-1. \end{eqnarray}$$
