Does the sum-of-divisors contain each divisor only once?

Question

To calculate perfect numbers, is each divisor added to the sum-of-divisors only once or as often as it appears in the factorisation of the number?

Answer 1

The definition of perfect numbers counts each divisor once. For example, $28$ is a perfect number because $$ 28=1+2+4+7+14 \, . $$ Note that, e.g., $2$ appears only once in this sum, despite appearing twice in the prime factorization of $28$.

In fact there are a few different possible ways you could define "number of times $a$ divides $b$" when $a$ is not prime. The two that occur to me are:

• "$a$ divides $b$ $n$ times" means $a^n$ divides $b$, and $a^{n+1}$ does not. Under this definition, $6$ divides $72$ twice (because $6^2=36$ divides $72$ and $6^3=216$ doesn't). This is probably the most natural definition, but it doesn't lead to a good sum-of-divisors function because of problems when $a=1$; $1^n$ divides $b$ for all $n$, so $1$ divides $b$ infinitely many times.
• "$a$ divides $b$ $n$ times" means that, if you consider each prime factor of $b$ to be distinguishable, there are $n$ possible ways to form $a$ out of $b$'s prime factors. Under this definition, $6$ divides $72$ $6$ times (because $72=2^33^2$, so we can get $6$ by independently choosing any one of $3$ 2's and either of $2$ 3's). This might be a little less natural, but it does give a sum-of-divisors function that works sort of like the standard one (e.g., it's multiplicative). Unfortunately, the analogue of perfect numbers under this definition is entirely trivial: it turns out that $6$ is the only one!