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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?

MPW
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Jeffrey Sun
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  • How do you define expected rectangularity of $n$? Would that be something like the mean of the rectangularities of $1, 2, ...n$? – Solomon Slow May 31 '17 at 20:51
  • Good question. I hadn't thought that out. Maybe it can be defined this way. We are looking for a function $f(n)$ such that $$ \int_{0}^{n}f(x)dx \sim \sum_{k=1}^n{r(k)} $$ where $r(k)$ is the rectangularity of $k$. – Jeffrey Sun May 31 '17 at 21:07
  • That is, $f$ asymptotically looks like the derivative of (some smoothed version of) the function giving the sum of all rectangularities of integers from $1$ to $n$. – Jeffrey Sun May 31 '17 at 21:13
  • In fact, instead of using an integral, we might just require that $$ \sum_{k=1}^n{f(k)} \sim \sum_{k=1}^n{r(k)} $$ – Jeffrey Sun May 31 '17 at 21:17
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    In the long run , for $f(k)$ to be smooth, we can bound $\frac{0.193147}{\ln k} \leq f(k) \leq 0.193147$ where $0.193147$ is approximation for $-\frac{1}{2}+\ln 2$ , but even this bound is not sufficient to tell if $f(k)$ will thin out in large number or we have "fixed dense". –  May 31 '17 at 22:33
  • This upper bound looks surprisingly low. How do you get it? – Jeffrey Sun May 31 '17 at 22:51
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    every number can be written as $k=a^2-b^2 =(a+b)(a-b)$ so its "rectangularity" $\frac{a-b}{a+b}$ (not necessarily the maximum one), and it could be unique representation which correspond to $-\frac{1}{2}+\ln 2$ or at most $\ln x$ representation for all numbers less than $k$ which correspond to $\frac{-\frac{1}{2}+\ln 2}{\ln k}$ if you take the sum of all such representation less or equal to $k$. –  May 31 '17 at 23:23
  • I did some heuristic calculations and I'm not convinced that the expected rectangularity of a number is nonzero. $\frac{1}{n} \sum_{k=1}^n r(k)$ seems to decrease very slowly. I can't rule out it converges to something nonzero, but if it does the convergence is very slow. – Dark Malthorp Jul 12 '18 at 23:38

1 Answers1

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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 $x$.

Rodrigo
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