# Numbers whose sum of divisors is smaller than the sum of the divisors of any larger number

A positive integer $$n$$ is called highly abundant if for all positive integers $$m, $$\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 $$n, then $$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.

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 $$\sqrt{x}$$.
• 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. Commented Apr 25, 2022 at 14:05
• 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. Commented Apr 25, 2022 at 19:12