Бесконечный список совершенных чисел

factor n = [m | m <- [1..n `div` 2], mod n m == 0]

factor n = [m | m <- [1..n `div` 2], mod n m == 0]
 
summ l = foldr (+) 0 l --сумма всех элеменов в массиве
 
isSuper n = n == (summ (factor n)) --проверяем на крутость
 
supers = [m | m <- [1..], isSuper m] --бесконечный список совершенных
 
main = do
    putStrLn $ show $ take 3 supers --показываем три первых совершенных числа

factorSum = sum . factor

isPerfect n = factorSum n == n

perfectList = filter isPerfect [1..]

perfects = filter (\ n -> sum (factors n) == n) [1..]
 
factors n = [i | i <- [1..n `div` 2], n `mod` i == 0]

let f a = a + 1

perfects = filter (\ n -> sum (factors n) == n) [2,4..]
 
factors n = [i | i <- [1..n `div` 2], n `mod` i == 0]

[2,4..]

perfects = filter (\ n -> chk' (factors n) n) [2,4..]
  where chk' fs n = chk fs 0  
            where chk []     s = n == s
                  chk (x:xs) s = 
                        case x+s of
                            s'  | s' > n    -> False
                                | otherwise -> chk xs s' 
 
factors n = [i | i <- [1..n `div` 2], n `mod` i == 0]

module Main where
 
primes :: Integral a => [a] -> [a]
primes [] = []
primes (x:xs) = x : primes (filter (\p -> rem p x /= 0) xs)
 
-- Lucas-Lehmer primality test.
llt :: Integer -> Bool
llt p = p == 2 || loop 3 (4 :: Integer)
  where loop n s = if n > p then s == 0 else loop (succ n) (rem (s*s - 2) $ pred(2^p))
 
task :: [Integer]
task = foldr (\x acc -> if llt x then 2^(pred x) * pred(2^x) : acc else acc) [] $ primes [2..]
 
main :: IO ()
main = print $ take 10 task
 
-- [6,28,496,8128,33550336,8589869056,137438691328,2305843008139952128,2658455991569831744654692615953842176,191561942608236107294793378084303638130997321548169216]