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90 lines
2.2 KiB
90 lines
2.2 KiB
8 years ago
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# A perfect number is a number for which the sum of its proper
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# divisors is exactly equal to the number. For example, the sum
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# of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28,
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# which means that 28 is a perfect number.
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#
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# A number n is called deficient if the sum of its proper divisors
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# is less than n and it is called abundant if this sum exceeds n.
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#
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# As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16,
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# the smallest number that can be written as the sum of two abundant
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# numbers is 24. By mathematical analysis, it can be shown that all
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# integers greater than 28123 can be written as the sum of two
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# abundant numbers. However, this upper limit cannot be reduced any
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# further by analysis even though it is known that the greatest
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# number that cannot be expressed as the sum of two abundant numbers
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# is less than this limit.
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#
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# Find the sum of all the positive integers which cannot be written
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# as the sum of two abundant numbers.
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# This program is essentially n^3 in the search and n^2 for the divisor count,
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# so it runs reaalllyy slowly
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import math
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import pickle
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def get_divisor_sum(input):
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divisor_sum = 0
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cutoff = math.ceil(input / 2)
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for idx, val in enumerate(range(cutoff), 1):
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if input % idx == 0:
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divisor_sum += idx
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return divisor_sum
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def get_abundants():
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list = []
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for i in range(28123):
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sum = get_divisor_sum(i)
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if sum > i:
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list.append(i)
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with open('assets/outfile', 'wb') as fp:
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pickle.dump(list, fp)
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return list
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# uncomment on any subsiquent runs to retrieve the cached data
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# with open('assets/outfile', 'rb') as fp:
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# abundant_list = pickle.load(fp)
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# comment this line out after the first run
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abundant_list = get_abundants()
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abundant_list.sort()
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seen = {}
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abundant_list = [seen.setdefault(x, x) for x in abundant_list if x not in seen]
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summation = 0
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for i in range(28123):
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found = False
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for first_number in abundant_list:
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if first_number > i:
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break
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search = i - first_number
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if search in abundant_list:
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found = True
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break
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if not found:
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summation += i
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print(summation)
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print(abundant_list)
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