Question

A positive integer n is “perfect” if the sum of its positive factors, excluding itself, equals n.  Write a perfect function in Haskell that takes a single integer argument and returns the list of all perfect numbers up to that argument. Report all of the perfect numbers up to 1000 (i.e. call 1000)

Answer 1

isPerfect :: Int -> Bool
isPerfect n =
  let lows = filter ((0 ==) . rem n) [1 .. floor (sqrt (fromIntegral n))]
  in 1 < n &&
     n ==
     quot
       (sum
          (lows ++
           [ y
           | x <- lows 
           , let y = quot n x 
           , x /= y ]))
       2
 
main :: IO ()
main = print $ filter isPerfect [1 ..

isPerfect :: Int -> Bool
isPerfect n =
  let lows = filter ((0 ==) . rem n) [1 .. floor (sqrt (fromIntegral n))]
  in 1 < n &&
     n ==
     quot
       (sum
          (lows ++
           [ y
           | x <- lows 
           , let y = quot n x 
           , x /= y ]))
       2
 
main :: IO ()
main = print $ filter isPerfect [1 .. 10000]

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