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I've been fascinated by perfect numbers ever since I learned about them, but I don't think 1 should be counted in the sum. Every integer is divisible by itself and 1, so why is the number itself excluded and not 1? Are there any numbers that are the sum of all their proper divisors excluding 1 (that is, they are 1 less than the sum of all the proper divisors)?

Cain Goldhardt
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    I think they would need to be one $\textit{less}$ than the sum of their proper divisors based on what you wrote. In that case the powers of two mentioned below would not work – Dave Oct 07 '23 at 00:13
  • I don't think any exist. Select[Range[10000], DivisorSigma[1, #] - # - 1 == # &] I would guess there is an easy proof for it. – qwr Oct 08 '23 at 00:03

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Jonathan Vos Post comments on OEIS033880, the abundance of numbers:

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).

Cain Goldhardt
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