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I have already found a summation formula here: https://math.stackexchange.com/a/22723, and also a very interesting recursive formula here: https://math.stackexchange.com/a/22744. Any ideas on how to reduce either of these to a closed form expression would be greatly appreciated.

Edit: Essentially, I want a finite expression using elementary functions that gives the sum of the proper divisors of an integer n in terms of n (and not the prime decomposition).

Community
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avatan
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1 Answers1

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There is no computationally efficient one at least. Let $\sigma(n)$ be the sum of all divisors, which behaves better analytically, the sum of proper divisors is just $\sigma(n)-n$. Consider the simplest case $n=pq$, then $\sigma(n)=n+p+q+1$. So if you know $n$ and can find $\sigma(n)$ efficiently then you know $pq$ and $p+q$, and can factor $n$ just by solving a quadratic equation. So the formula you want can't do much better than prime factoring, which is believed to be a 'hard' problem.

It would also mean that you can have an explicit formula that gives you prime factors of $n$ as elementary functions of $n$ itself, at least when $n$ only has two factors. No such formula is known, and it is unlikely to exist.

Conifold
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  • Great answer (+1), but not everyone believes factoring is 'hard'. – RghtHndSd Jun 21 '14 at 01:06
  • The first thing I thought when I saw the question is "that would give a really neat way to test for primality ($\sigma(n)=n+1$)". But this argument is even more convincing. To be precise the problem it would solve is: factoring a number known to be a product of two distinct primes, which is indeed thought to be hard (RSA cryptography is based on that assumption). – Marc van Leeuwen Jun 21 '14 at 05:18
  • this is a simple explanation and a convincing answer. There is a confusing answer (2020) here https://math.stackexchange.com/a/2334535/542895 – Black Mild Dec 30 '23 at 08:06