# Does the average primeness of natural numbers tend to zero?

Note 1: This questions requires some new definitions, namely "continuous primeness" which I have made. Everyone is welcome to improve the definition without altering the spirit of the question. Click here for a somewhat related question.

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 measure of 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 advantage of using standard deviation is that even if two numbers have the same number of divisor their value of $$f$$ appears to be different hence their measure of primeness will be different.

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 $$2 \le i \le n$$, the minimum value of $$f(i)$$ occurs at the largest highly composite number $$\le n$$.

Note 2: 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 $$(0,1)$$.

Note 3: Posted this question in MO and got answer for question 1. Indeed the imit tends to zero. Question 2 is still open.

• Is your second statement about the minimum of $f$ obvious? I'm unable to come up with an argument. – user113102 Apr 5 at 21:30
• Why involve the standard deviation? Why not something simpler, like $2/d(n)$, where $d(n)$ is the number of divisors of $n$? – Gerry Myerson Apr 6 at 4:06
• @GerryMyerson: Here is a more technical answer why standard deviation. If two numbers have the same number of divisors then the value $2/d(n)$ is same for both but the values of $f(n)$ is different. So under my definition, I will consider the number with smaller value of $f(n)$ to have a greater primness because we are not just measuring how many divisors a number has but also how scattered these divisors are. At the moment, I don't know if $f(n)$ is unique. I will add this to the question. – Nilotpal Kanti Sinha Apr 6 at 7:27
• I have written some code for this in R: stanfun <- function(n){sd(divisors(n))/(n-1)};funstan <- function(m){sum(sapply(2:m, function(i){stanfun(i)}))/m}, so $\frac1{10000}\sum\limits_{r=1}^{10000}f(r)$ is given by funstan(10000) which outputs $0.404801$. – TheSimpliFire Apr 6 at 19:23
• @TheSimpliFire I have added my Sagemath code above – Nilotpal Kanti Sinha Apr 7 at 3:23