problem21.py

'''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.'''

import math

cache = {}

def divisors(n):
    if n not in cache:
        divs = [1]
        for i in range(2, int(math.ceil(math.sqrt(n)))):
            if n % i == 0:
                divs.append(i)
                divs.append(n//i)
        divs.sort()
        cache[n] = divs
    return cache[n]

if __name__ == '__main__':
    cache = {}
    result = []
    for a in range(2,10001):
        b = sum(divisors(a))
        if a != b and a == sum(divisors(b)):
            result.append((a,b))
    print(sum([a + b for a, b in result if a < b]))