def is_amicable(a):
"""
Determines if a number `a` has an amicable pair, defined by
d(a) = b, where b is sum of divisors of a, where a != b such that d(b) = a
"""
global FOUND
if a in FOUND:
return True
b = sum_of_divisors(a)
if b == a:
return False
if sum_of_divisors(b) == a:
FOUND.update([a, b])
return True
return False
A perfect number is a number for which the sum of its proper divisors is exactly equal
to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28,
which means that 28 is a perfect number.
A number n is called deficient if the sum of its proper divisors is less than n and it is
called abundant if this sum exceeds n.
As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can
be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown
that all integers greater than 28123 can be written as the sum of two abundant numbers. However,
this upper limit cannot be reduced any further by analysis even though it is known that the greatest
number that cannot be expressed as the sum of two abundant numbers is less than this limit.
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
"""
from itertools import (combinations_with_replacement as combos,
ifilter,
starmap)
from operator import add
from util import sum_of_divisors
min, max = 1, 28123 # Exclude 0, specify upper bounds per problem
all = set(xrange(min, max)) # Set of all numbers [min, max)
# Find all abundant numbers <= max
abundant_nums = list(ifilter(lambda n: sum_of_divisors(n) > n, xrange(min, max)))
# Find the set of all unique sums of abundant numbers where sum(n) <= max. Exclude from all nums
print sum(all - set(ifilter(lambda n: n <= max, set(starmap(add, combos(abundant_nums, 2))))))
#!/usr/bin/env python3
import sys
sys.path.insert(0, '../../')
import util
if __name__ == '__main__':
limit = 28123
is_sum = [False for _ in range(limit + 1)]
abd_num = []
ans = 0
for i in range(1, limit + 1):
if not is_sum[i]:
ans += i
if util.sum_of_divisors(i) > i:
abd_num.append(i)
for abd in abd_num:
if abd + i <= limit:
is_sum[abd + i] = True
else:
break
print(ans)
