ProjectEuler Problem 23 answer. From http://codereview.stackexchange.com/questions/39946/optimizing-solution-for-project-euler-problem-23-non-abundant-sums

pe023.py

def is_abundant(n):
    max_divisor = int(n / 2) + 1
    sum = 0
    for x in range(1, max_divisor):
        if n % x == 0:
            sum += x  
    return sum > n

abundants = list(x for x in range(1, 28123) if is_abundant(x))

sums = 0
for i in range(12, 28123):
    for abundant in abundants:
        if abundant >= i and is_abundant(i + abundant):
            sums += i
print(sums)