TITLE: "Rectangularity" of integers
QUESTION [35 upvotes]: We can sort of think of a number $n$ as "almost square" if $n = kl$ where $$\frac{k}{l} \approx 1.$$
More generally, we can talk about the "rectangularity" of an integer $n$ as
$$
\max_{k \leq l | kl = n} \frac{k}{l}.
$$
(We might also look at the inverse of that number and call it the​ "eccentricity." We also might somehow look at the eccentricity for more than 2 factors.)
Then the "most rectangular" numbers, the squares, have rectangularity 1, and the "least rectangular" numbers, the primes, have rectangularity 1/p.
Some questions are naturally suggested. Does the expected rectangularity of a number go to 1 as $n \rightarrow \infty$? How fast does it grow? What does the probability density function of different rectangularities for $n$ look like, as a function of the size of $n$? What about over the p-adics?
What is known about this property? Has it been investigated? Does it have a real name?

REPLY [4 votes]: We will show the following fact:
Fact: Let $\epsilon > 0$. The set of natural numbers $n$ that have a divisor $d$ such that $n^{1/2}\ge d \ge \epsilon n^{1/2}$ has natural density zero. 
A number with no divisor between $\epsilon n^{1/2}$ and $n^{1/2}$ has rectangularity at most $\epsilon^2$, so by making $\epsilon$ arbitrarily small, together with the fact that rectangularity is bounded above by one, this implies the expected value of rectangularity is zero.
To prove the fact we use Erdös-Kac theorem, which states, say, that almost every integer up to $x$ has $\sim \log \log x$ prime factors.
The rough idea is that a typical integer up to $x^{1/2}$ still has around the same number $\log \log x^{1/2} = \log \log x +O(1)$ of prime factors. But if $n$ were to break down as $n=d_1 d_2$ with both $d_1,d_2$ close to $x^{1/2}$, if all $n$, $d_1,d_2$ were typical then  $\log \log x \sim 2 \log \log x$ which can't happen. So typically $n$ won't break down this way. A bit more precisely:
Proof of Fact: Call an integer less than $x$ typical if its number of prime factors is between $0.9 \log \log x$ and $1.1 \log \log x$ (so almost every number is typical).
Let's now count how many $n$ up to $x$ have a divisor between $\epsilon n^{1/2}$ and $n^{1/2}$. There are two cases:
$n$ is not typical. We already know there are $o(x)$ of these.
$n$ is typical. In which case, $n = d_1d_2$ where $n^{1/2} \ge d_1 \ge \epsilon n^{1/2}$, therefore both $d_1,d_2 \le \frac{1}{\epsilon} x^{1/2}$. Notice that at least one of $d_1, d_2$ are not typical (since $n$ is typical so one of these has less than $0.55 \log \log x$ factors), but by Aplying Erdös-Kac again there are at most $o(x)$ such pairs $(d_1,d_2)$, hence again at most $o(x)$ possibilities for $n$.
This completes the proof of our fact.
With a bit more care I believe this argument should give an upper bound $O\left(\frac{1}{(\log \log x)^{2/3}}\right)$ to the average rectangularity of integers up to $x$.