Problem Euler #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.
import time from math import sqrt ts = time.clock() def get_div(num): for n in xrange(2, int(sqrt(num)) + 1): if num % n == 0: yield n yield num/n def get_sum(divisors): return sum([n for n in divisors]) + 1 if __name__ == '__main__': nums = [n for n in range(1, 10000) if n == get_sum(get_div(get_sum(get_div(n)))) \ and n != get_sum(get_div(n))] print sum(nums) print time.clock() - ts