I am trying to understand the dynamics of the aliquot sum.

I am wondering if a recurrence relation exists?

For example, would this work:

  • Let $s(x)$ be the aliquot sum for $x$
  • Let $p$ be a prime
  • If $x$ is $1$, then $s(x)=0$
  • If $x$ is prime, then $s(x)=1$
  • If $p \nmid x$, then $s(px) = s(x) + ps(x) + x$
  • If $p | x$, then $s(px) = s(x) + x$


Edit: Made updates based on comments received by Mason.

  • $\begingroup$ Can you give an example? What's $s(25)$? And are we to assume $p$ is prime? Typically the aliquot sum is defined such that $s(1)=0$. $\endgroup$ – Mason Dec 3 '19 at 17:57
  • $\begingroup$ Yes, $p$ is prime. I will add. Thanks for calling that out. $s(5) = 1$. $s(25) = 1 + 5 = 6 = s(5) + 5s(5)$ $\endgroup$ – Larry Freeman Dec 3 '19 at 17:59
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    $\begingroup$ Also you can probably get away with modifying Euler's recurrence formula for the divisor function. See this $\endgroup$ – Mason Dec 3 '19 at 18:00
  • $\begingroup$ Awesome! Thanks very much. That's exactly what I was looking for. :-) $\endgroup$ – Larry Freeman Dec 3 '19 at 18:01
  • $\begingroup$ @LarryFreeman: I suppose that you are interested in perfect numbers (odd ones, in particular). To broaden your horizons, note that an odd perfect number $N$ given in the so-called Eulerian form $N = q^k n^2$ satisfies $$D(q^k)D(n^2)=2s(q^k)s(n^2)$$ where $D(x)=2x-\sigma(x)$ is the deficiency of $x$ and $s(x)=\sigma(x)-x$ is the aliquot sum of $x$. $\endgroup$ – Arnie Bebita-Dris Dec 17 '19 at 9:19

Define $f(x)+x=\sigma(x)=\sum_{d|x}d$. Then $f(x)$ is our aliquot function.

For coprime numbers $a,b$ Then $\sigma(a b)= \sigma(a)\sigma(b) $ so


This implies that

$f(ab)= f(a)f(b)+bf(a)+af(b)$ for coprime $a,b$.

While we're talking about recurrence we should mention this amazing recurrence formula from Euler that can be found here.

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  • $\begingroup$ and $p+q=2m$ so $2(p+q)=4m$ so they are alway different by a multiple of 4. $\endgroup$ – user645636 Dec 3 '19 at 20:46
  • $\begingroup$ @RoddyMacPhee. I think your comment must have got cut off. $\endgroup$ – Mason Dec 4 '19 at 1:54
  • $\begingroup$ nope just relating in another unsolved problem. $$\varphi(n)+2(p+q) = \sigma_1(n)$$ with $p$ and $q$ prime is saying the difference ( admittedly in this case) is a multiple of 4 (see Goldbach). $\endgroup$ – user645636 Dec 4 '19 at 3:21
  • $\begingroup$ Actually I think this write up is a better resource. A little clearer. $\endgroup$ – Mason Dec 5 '19 at 6:49

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