23
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is your second statement about the minimum of $f$ obvious? I'm unable to come up with an argument. $\endgroup$ – user113102 Apr 5 at 21:30
  • 4
    $\begingroup$ Why involve the standard deviation? Why not something simpler, like $2/d(n)$, where $d(n)$ is the number of divisors of $n$? $\endgroup$ – Gerry Myerson Apr 6 at 4:06
  • 3
    $\begingroup$ @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. $\endgroup$ – Nilotpal Kanti Sinha Apr 6 at 7:27
  • 1
    $\begingroup$ 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 $0.404801$. $\endgroup$ – TheSimpliFire Apr 6 at 19:23
  • $\begingroup$ @TheSimpliFire I have added my Sagemath code above $\endgroup$ – Nilotpal Kanti Sinha Apr 7 at 3:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.