PyPE solution to PE problem 21

    ProjectEuler.net problem 21
    ===========================

    Let d(n) be defined as the sum of proper divisors of n (numbers less than n
    which divide evenly into n). If d(a) = b and d(b) = a, where a != b, then a
    and b are an amicable pair and each of a and b are called amicable numbers.

    For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44,
    55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4,
    71 and 142; so d(284) = 220.

    Evaluate the sum of all the amicable numbers under 10000.

    Solution
    --------

    TODO: See Project Euler PDF solution.
    TODO: Develop a function PyPE.pemath.divisors(n) which returns (or
            generates) a list of divisors of n.
    TODO: Develop a function PyPE.pemath.proper_divisors(n) which returns
            (or generates) a list of proper divisors of n (likely using
            PyPE.pemath.divisors(n)).

    Answer
    ------

    31626