Clojure Exercise: Calculate Perfect Numbers

  perfects = [n | n <- [1..], n == sum (divisors n)]
    where
      divisors n = [k | k <- [1..(n `div` 2)], n `mod` k == 0]

    divisors = lambda n: (k for k in range(1, n / 2 + 1) if n % k == 0)
    perfects = (n for n in itertools.count(1) if n == sum(divisors(n)))

  (define (perfect-numbers integers)
    (stream-filter perfect?
        (stream-cdr integers)))

  (define (perfect? n)
    (= n (sum-stream (stream-filter (lambda (x) (zero? (remainder n x)))
       (stream-range 1 (+ (/ n 2) 1)))))))

Nice simple lisp intro though.