# Is there a recurrence relation that describes the aliquot sum?

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 $$s(px) = s(x) + x$$

Thanks.

• 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 $s(1)=0$. – Mason Dec 3 '19 at 17:57
• Yes, $p$ is prime. I will add. Thanks for calling that out. $s(5) = 1$. $s(25) = 1 + 5 = 6 = s(5) + 5s(5)$ – Larry Freeman Dec 3 '19 at 17:59
• Also you can probably get away with modifying Euler's recurrence formula for the divisor function. See this – Mason Dec 3 '19 at 18:00
• Awesome! Thanks very much. That's exactly what I was looking for. :-) – Larry Freeman Dec 3 '19 at 18:01
• @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 $x$. – 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

$$f(ab)+ab=(f(a)+a)(f(b)+b)$$

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.

• and $p+q=2m$ so $2(p+q)=4m$ so they are alway different by a multiple of 4. – user645636 Dec 3 '19 at 20:46
• @RoddyMacPhee. I think your comment must have got cut off. – Mason Dec 4 '19 at 1:54
• 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). – user645636 Dec 4 '19 at 3:21
• Actually I think this write up is a better resource. A little clearer. – Mason Dec 5 '19 at 6:49