Maximum area of rectangle in a young tableau. Given a positive integer $n$, is there a formula or a bound on the maximum area of rectangles contained in the Young diagrams of the partitions of $n$.
For example when $n=5$, the Young diagrams for the partitions of $5$ are
$\Box\Box\Box\Box\Box$
$\Box\Box\Box\Box$$\Box$
$\Box\Box\Box$$\Box\Box$
$\Box\Box\Box$$\Box$$\Box$
$\Box\Box$$\Box\Box$$\Box$
$\Box\Box$$\Box$$\Box$$\Box$
$\Box$$\Box$$\Box$$\Box$$\Box$
which contain a rectangle of maximum area $5,4,4,3,4,4,5$ respectively. So the quantity I want for when $n=5$ is $3$.
 A: Consider the rectangular hyperbola $xy=A$ in the quadrant $x, y \gt 0$.
Any rectangle below the hyperbola has area $\le A$, and clearly the tightest hyperbola to any configuration passes through at least one corner, so that for the limiting case there is a rectangle of area $A$.
The question then is how to find the minimum value of $A$ for a given value of $n$. The number of squares which fit under a such hyperbola is given by
$$n=\sum_{i=1}^{\lfloor A \rfloor}\lfloor \frac Ai\rfloor\sim A(\ln A +\gamma)$$
Others will do the estimates better than me to get a bound, but it looks as though $A\sim \cfrac n {\ln n}$.
A: Not a full solution, but an upper bound and a guess at a lower one:  It seems like you want the minimum over partitions of the maximum area of a rectangle of the tableaux of total size $n$.  It is clearly no larger than $\lceil \frac {n+1}2 \rceil$, given by the partition $\lceil \frac {n+1}2 \rceil$ and the rest $1$'s.  For a triangle number $n=\frac 12m(m+1)$it seems likely it will be asymptotic to $(\frac m2)^2 \approx \frac n2$ given by the triangle partition.  These are very close.
A: I worked out the first few cases which turned out to be much easier than listing all configurations for each $n$. Then I found the sequence on OEIS as http://oeis.org/A061017.
