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#-.- coding: latin-1 -.-

from __future__ import print_function

from math import sqrt

'''

Amicable Numbers

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.

'''

try:

xrange #Python 2

except NameError:

xrange = range #Python 3

def sum_of_divisors(n):

total = 0

for i in xrange(1, int(sqrt(n)+1)):

if n%i == 0 and i != sqrt(n):

total += i + n//i

elif i == sqrt(n):

total += i

return total-n

total = [i for i in range(1,10000) if sum_of_divisors(sum_of_divisors(i)) == i and sum_of_divisors(i) != i]

print(sum(total))

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