That's a school exercise, but I can't resist adding a haskell version - perfects = [n | n <- [1..], n == sum (divisors n)] where divisors n = [k |

That's a school exercise, but I can't resist adding a haskell version - perfects = [n | n <- [1..], n == sum (divisors n)] where divisors n = [k | k <- [1..(n `div` 2)], n `mod` k == 0]