1
$\begingroup$

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.


Edit: Made updates based on comments received by Mason.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.