This question is an offshoot of this earlier post.

It is known that the abundancy index $$I(n) = \frac{\sigma(n)}{n}$$ is a multiplicative function. (Note that the divisor sum $\sigma$ is also multiplicative.)

How about $$J(n) = \frac{\sigma(n)}{2n}?$$


Suppose that $\gcd(x,y)=1$. Then $$J(xy) = \frac{\sigma(xy)}{2xy} = \frac{\sigma(x)\sigma(y)}{2xy} \neq \frac{\sigma(x)}{2x}\cdot\frac{\sigma(y)}{2y}.$$

In general, we have $$2J(x)J(y) = \frac{\sigma(x)\sigma(y)}{2xy} = J(xy).$$

This implies that $$J(xy) = 2J(x)J(y) > J(x)J(y),$$ for all $x,y$ such that $\gcd(x,y)=1$. This means that $$J(n) = \frac{\sigma(n)}{2n}$$ is supermultiplicative.


Is my understanding of the notion of a supermultiplicative function correct?

  • 1
    $\begingroup$ It is correct. And generally, by the same argument, if $\mu(x)$ is multiplicative then $a\mu(x)$ is supermultiplicative when $0<a<1$, and submultiplicative when $a>1$. $\endgroup$
    – Conifold
    Aug 31, 2020 at 5:02
  • $\begingroup$ Thank you for confirming, @Conifold! Please write your last comment out as an actual answer so that I can upvote and accept it. =) $\endgroup$ Aug 31, 2020 at 6:24
  • $\begingroup$ I honestly do not understand why this was downvoted. Any form of feedback, hopefully constructive, would go a long way towards improving future questions/posts. As it is, I am totally clueless. $\endgroup$ Sep 19, 2021 at 12:31

1 Answer 1


The argument is correct and can be generalized. Suppose $\mu(x)$ is any non-negative multiplicative function and $0\leq a\leq1$. Then we have $a\geq a^2$, and therefore $$a\mu(xy)\geq a^2\mu(xy)=a^2\mu(x)\mu(y)=a\mu(x)\,a\mu(y).$$ In other words, $J(x)=a\mu(x)$ is supermultiplicative. By analogous argument, it is submultiplicative when $a\geq1$.


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