Does the average primeness of natural numbers tend to zero?

Question

This question was posted in MSE. It got many upvotes but no answer hence posting it in MO.

---

A number is either prime or composite, hence primality is a binary concept. Instead I wanted to put a value of primality to every number using some function $f$ such that $f(n) = 1$ iff $n$ is a prime otherwise, $0 < f(n) < 1$ and as the number divisors of $n$ increases, $f(n)$ decreases on average. Thus $f(n)$ is a measure of the degree of primeness of $n$ where 1 is a perfect prime and 0 is a hypothetical perfect composite. Hence $\frac{1}{N}\sum_{r \le N} f(r)$ can be interpreted as a the average primeness of the first $N$ integers.

After trying several definitions and going through the ones in literature, I came up with:

Define $f(n) = \dfrac{2s_n}{n-1}$ for $n \ge 2$, where $s_n$ is the standard deviation of the divisors of $n$.

One reason for using standard deviation was that I was already studying the distribution of the divisors of a number.

Question 1: Does the average primeness tend to zero? i.e. does the following hold?

$$ \lim_{N \to \infty} \frac{1}{N}\sum_{r = 2}^N f(r) = 0 $$

Question 2: Is $f(n)$ injective over composites? i.e., do there exist composites $3 < m < n$ such that $f(m) = f(n)$?

---

My progress

• $f(4.35\times 10^8) \approx 0.5919$ and decreasing so the limit if it exists must be between 0 and 0.5919.
• For $2 \le i \le n$, computed data shows that the minimum value of $f(i)$ occurs at the largest highly composite number $\le n$.

Note: Here standard deviation of $x_1, x_2, \ldots , x_n$ is defined as $\sqrt \frac{\sum_{i=1}^{n} (x-x_i)^2}{n}$. Also notice that even if we define standard deviation as $\sqrt \frac{\sum_{i=1}^{n} (x-x_i)^2}{n-1}$ our questions remain unaffected because in this case in the definition of $f$, we will be multiplying with $\sqrt 2$ instead of $2$ to normalize $f$ in the interval $(0,1)$.

Answer 1

The answer to Question 1 is "yes". To see this, notice that $s_n$ is at most square root of the average square of divisor, i.e. $$ s_n\leq \sqrt{\frac{\sum_{d\mid n}d^2}{\sum_{d\mid n} 1}}=\sqrt{\frac{\sigma_2(n)}{\sigma_0(n)}}, $$

where $\sigma_k(n)$ is the sum of $k$-th powers of divisors of $n$. Now,

$$ \sigma_2(n)=n^2\sigma_{-2}(n), $$

so

$$ \sigma_2(n)<\frac{\pi^2}{6}n^2 $$

for all $n$. Therefore we have

$$ f(n)\leq \frac{2}{n-1} \sqrt{\frac{\pi^2}{6}n^2/\sigma_0(n)}\leq \frac{5.14}{\sqrt{\sigma_0(n)}} $$

for all $n$. Now, almost all $n\leq N$ have at least $0.5\ln\ln N$ distinct prime factors. In particular, for almost all $n\leq N$ we have $\sigma_0(n)\geq 0.5\ln\ln N$. Therefore, our bound for $f(n)$ together with the trivial observation that $0\leq f(n)\leq 1$ gives

$$ \sum_{n\leq N} f(n)\leq \sum_{n\leq N, \sigma_0(n)\geq 0.5\ln\ln N} \frac{5.14}{\sqrt{\sigma_0(n)}}+\sum_{n\leq N, \sigma_0(n)<0.5\ln\ln N} 1= o(N), $$

as needed.

Using contour integration method one can even prove something like

$$ \sum_{n\leq N} f(n)=O(N(\ln N)^{1/\sqrt{2}-1}) $$