A positive integer $n$ is called highly abundant if for all positive integers $m<n$, $\sigma(m)<\sigma(n)$, where $\sigma$ is the sum of divisors function.

Now, consider the property that $\sigma(m)>\sigma(n)$ for all $m>n$. Then, does this property hold for $n>1$ if and only if $n$ is a prime number? Henceforth, this question will be referred to as the main question.

The "if" direction is easily verified to be true (if $n>p$, where $p$ is prime, then it is clear that $\sigma(n)>p+1=\sigma(p)$).

For the "only if" direction, if $n$ is a composite number and there is a prime $p$ for which $n<p<\sigma(n)$, then $p>n$, but $\sigma(p) \le \sigma(n)$. If $n$ is perfect or abundant, then by Bertrand's postulate, there is a prime $p$ for which $n < p < 2n \le \sigma(n)$. So, any counterexample to the main question (if one exists) must be a deficient number.

There would not be any counterexamples if there was always a prime strictly between $n$ and $\sigma(n)$ for any composite number $n$. Does such a prime always exist for any composite number $n$? If so, then that would also answer "true" for the main question.


1 Answer 1


Obviously for a prime $p$, the sum of divisors is $p+1$, which is less than the sum of divisors of any larger number.

Let $x = pq$, and $y$ the smallest prime $> x$. The sum of divisors of $x$ is at least $1 + x + p + q \leq x + 1 + 2 \sqrt{x}$. We’d need this to be less than $y + 1$. This can only be true if the gap between consecutive primes can be larger than twice the square root of the smaller prime. I would bet this is not possible but I’m not sure if there is a proof.

Correction: If $x$ is the square of a prime, $x= p^2$, then the sum of divisors is only $x + p + 1$. So the gap need to be only slightly larger than $\sqrt{x}$.

  • $\begingroup$ I believe the current state of the art is a gap of $x^{0.525}$ (Baker et al, 2001), but $x^{0.5}$ has remained elusive. If we assume the Riemann Hypothesis, the gap has been shrunk to $(1+\epsilon)\sqrt{x}\log(x)$ for sufficiently large $x$ (Dudek, 2014). But even that is not good enough to make your proof work. $\endgroup$
    – dshin
    Apr 25, 2022 at 14:05
  • $\begingroup$ Incredible difference to numerical / heuristic arguments. Average gap about log x, occasional gaps of log^2 x, and maximum gap maybe slightly larger, maybe not. $\endgroup$
    – gnasher729
    Apr 25, 2022 at 19:07
  • $\begingroup$ To be clear, when I say the state of the art is a gap of $x^{0.525}$, I mean that it has been proven that there is always a prime in the range $[x, x+x^{0.525}]$. This represents an improvement on Bertrand’s postulate, which asserts the existence of a prime in the looser range [x, x+x^1]$. It is not a statement about average gaps. $\endgroup$
    – dshin
    Apr 25, 2022 at 19:12
  • $\begingroup$ Sorry if I wasn’t clear: I understand that proving an upper bound is very hard. Proving that there is a prime between p and p^2 is already hard. It’s just interesting how big the difference between the largest known gaps and the proven upper bound is. $\endgroup$
    – gnasher729
    Apr 26, 2022 at 0:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .