Jtdcftul

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

• 
  $f(4.35times 10^8) approx 0.5919$ 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 $(0,1)$.

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.
$$
s_nleq sqrt{frac{sum_{dmid n}d^2}{sum_{dmid n} 1}}=sqrt{frac{sigma_2(n)}{sigma_0(n)}},
$$

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

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

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

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

as needed.

Using contour integration method one can even prove something like

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