amicable numbers problem 21 let dn be defined as the sum of proper divisors of n numbers less than n which divide evenly into n if da b and db 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 d220 284 the proper divisors of 284 are 1 2 4 71 and 142 so d284 220 evaluate the sum of all the amicable numbers under 10000 returns the sum of all the amicable numbers under n solution10000 31626 solution5000 8442 solution1000 504 solution100 0 solution50 0 returns the sum of all the amicable numbers under n solution 10000 31626 solution 5000 8442 solution 1000 504 solution 100 0 solution 50 0
from math import sqrt


def sum_of_divisors(n: int) -> int:
    total = 0
    for i in range(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


def solution(n: int = 10000) -> int:
    
    total = sum(
        i
        for i in range(1, n)
        if sum_of_divisors(sum_of_divisors(i)) == i and sum_of_divisors(i) != i
    )
    return total


if __name__ == "__main__":
    print(solution(int(str(input()).strip())))