def abundant_numbers(limit=None):
from maths.misc import find_divisors
if limit is None:
# Infinite sequence
import itertools
for i in itertools.count(1,1):
divisors = find_divisors(i, True)
if sum(divisors) > i:
yield i
else:
# Finite sequence
for i in xrange(1, limit, 1):
divisors = find_divisors(i, True)
if sum(divisors) > i:
yield i
def reduce_fraction(num, denom):
from maths.misc import find_divisors
# Find divisors of num & denom
proper_divisors = True
div_n = find_divisors(num, proper_divisors)
div_d = find_divisors(denom, proper_divisors)
# Find intersection between divisors of numerator and denominator
intersection = list(div_n.intersection(div_d))
# Sort descending
intersection.sort()
intersection.reverse()
for i in intersection:
# Check that num and denom are still evenly divisible by i
# Need to check this as we're not only reducing by prime numbers
# We don't want to "over-reduce"
if num % i == 0 and denom % i == 0:
num = num / i
denom = denom / i
return num, denom
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 d.n.e. 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.
"""
if __name__ == "__main__":
from maths.misc import find_divisors
amicable_numbers = {}
for a in xrange(1, 10001, 1):
if a not in amicable_numbers:
proper_divisors_a = list(find_divisors(a, True))
divisor_sum_a = sum(proper_divisors_a)
# Ensure that d(a) != a
if divisor_sum_a == a:
continue
proper_divisors_b = list(find_divisors(divisor_sum_a, True))
divisor_sum_b = sum(proper_divisors_b)
if divisor_sum_b == a:
amicable_numbers[a] = divisor_sum_a
amicable_numbers[divisor_sum_a] = a
numbers = [nums[1] for nums in amicable_numbers.items()]
