Is there a number $n$ such that it equals the sum of its non-trivial divisors (i.e. all of its divisors except 1 and $n$)? If yes, what are such numbers called and what are some examples of them?

I have not found any answers on the internet.

  • 1
    $\begingroup$ do you mean the perfect numbers? $\endgroup$ Feb 14, 2018 at 13:33
  • 1
    $\begingroup$ I was going to say that, but I note that OP excludes 1 from his definition. $\endgroup$
    – Matt
    Feb 14, 2018 at 13:34
  • $\begingroup$ have you tried a computer search? $\endgroup$
    – lulu
    Feb 14, 2018 at 13:37
  • $\begingroup$ @lulu I'm not a programmer. $\endgroup$
    – vomadaxela
    Feb 14, 2018 at 13:38
  • $\begingroup$ even so. Shouldn't be hard to check up to $100$, at least. Easy to see that $pq$, product of distinct primes can't work, for example. Maybe the product of three primes. $\endgroup$
    – lulu
    Feb 14, 2018 at 13:39

1 Answer 1


Congratulations, you've walked into an open problem. Any number $n$ would satisfy $\sigma(n)-n-1=n$ or $\sigma(n)=2n+1$, i.e. have an abundance of 1. But a note on the relevant OEIS entry, A033880 (abundance of $n$), states:

For no known $n$ is $a(n)=1$. If there is such an $n$ it must be greater than $10^{35}$ and have seven or more distinct prime factors (Hagis and Cohen 1982). - Jonathan Vos Post, May 01 2011

These are called quasiperfect numbers.


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