#!/usr/bin/python
#
# Project Euler 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.

sopd_cache = {}
def sopd(n):
    if not sopd_cache.has_key(n):
        sum = 1 # 1 is a proper divisor...
        for i in range(2, int(n**.5)+1): # but n is not so start at 2
            if n%i==0:
                sum+=i
                if i!=(n/i):
                    sum+=n/i
        sopd_cache[n] = sum
    return sopd_cache[n]

sum = 0
for n in range(10000):
    d = sopd2(n)
    if sopd2(d) == n and d != n:
        sum += n
print sum