"Rectangularity" of integers

Solution 1:

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 $x$.