Question
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.
Haskell
import Data.List (sort, group, union)
import Data.Array
pairwise :: (a -> a -> a) -> [a] -> [a]
pairwise f (xs:ys:t) = f xs ys : pairwise f t
pairwise _ t = t
primes :: [Int]
primes = 2 : _Y ((3 :) . gaps 5 . _U . map (\p-> [p*p, p*p+2*p..]))
where
_Y g = g (_Y g) -- recursion, Y combinator
_U ((x:xs):t) = x : (union xs . _U . pairwise union) t -- ~= nub.sort.concat
gaps k s@(x:xs)
| k < x = k : gaps (k+2) s -- ~= [k,k+2..]\\s, when
| otherwise = gaps (k+2) xs -- k <= head s && null(s\\[k,k+2..])
factorize :: Int -> [Int]
factorize n = primeFactors n primes where
primeFactors 1 _ = []
primeFactors _ [] = []
primeFactors m (p:ps) | m < p * p = [m]
| r == 0 = p : primeFactors q (p:ps)
| otherwise = primeFactors m ps
where (q, r) = quotRem m p
primePowers :: Int -> [(Int, Int)]
primePowers n = [(head x, length x) | x <- group $ factorize n]
divisors :: Int -> [Int]
divisors n = filter (<n) $ map product $ sequence
[take (k+1) $ iterate (p*) 1 | (p, k) <- primePowers n]
upperBound :: Int
upperBound = 20161
abundant :: Int -> Bool
abundant n = (sum . divisors) n > n
abundantsArray :: Array Int Bool
abundantsArray = listArray (1, upperBound) $ map abundant [1..upperBound]
abundants :: [Int]
abundants = filter (abundantsArray !) [1..upperBound]
remainders :: Int -> [Int]
remainders x = map (x-) $ takeWhile (<= x `quot` 2) abundants
sums :: [Int]
sums = filter (any (abundantsArray !) . remainders) [1..upperBound]
main :: IO ()
main = print $ sum [1..upperBound] - sum sumsPython
#!/usr/bin/env python3
import math
from collections import defaultdict
from itertools import *
from functools import reduce
def factorize(n):
if n < 1:
raise ValueError('fact() argument should be >= 1')
if n == 1:
return [] # special case
res = []
# iterate over all even numbers first.
while n % 2 == 0:
res.append(2)
n //= 2
# try odd numbers up to sqrt(n)
limit = math.sqrt(n+1)
i = 3
while i <= limit:
if n % i == 0:
res.append(i)
n //= i
limit = math.sqrt(n+i)
else:
i += 2
if n != 1:
res.append(n)
factors = sorted(res)
histogram = defaultdict(int)
for factor in factors:
histogram[factor] += 1
return list(histogram.items())
def divisors(n):
factors = factorize(n)
nfactors = len(factors)
f = [0] * nfactors
while True:
yield reduce(lambda x, y: x*y, [factors[x][0]**f[x] for x in range(nfactors)], 1)
i = 0
while True:
f[i] += 1
if f[i] <= factors[i][1]:
break
f[i] = 0
i += 1
if i >= nfactors:
return
def proper_divisors(n):
return list(divisors(n))[:-1]
def classify(n):
total = sum(proper_divisors(n))
if total == n:
# perfect
return 0
elif total > n:
# abundant
return 1
else:
# deficient
return -1
def main():
abundant = set(number for number in range(2, 30000) if classify(number) == 1)
sums = sorted(set(sum(c) for c in combinations_with_replacement(abundant, 2)))
print((sum(number for number in range(1,30000) if number not in sums)))
if __name__ == "__main__":
main()Ruby
#!/usr/bin/env ruby
require 'mathn'
class Integer
def divisors
return [1] if self == 1
primes, powers = self.prime_division.transpose
exponents = powers.map{|i| (0..i).to_a}
divisors = exponents.shift.product(*exponents).map do |powers|
primes.zip(powers).map{|prime, power| prime ** power}.inject(:*)
end
divisors.take divisors.length - 1
end
def abundant?
self.divisors.reduce(:+) > self
end
end
abundants = (1..28213).select { |n| n.abundant? }
i = 0
sums = []
abundants.each do |x|
abundants[i..abundants.length].each do |y|
sum = x + y
sums << sum unless sum > 28213
end
i += 1
end
sums.uniq!
puts (1..28213).reject { |n| sums.include? n }.reduce(:+)Rust
fn sum_divisors(n: usize) -> usize {
let mut result = 0;
let max = 1 + (n as f64).sqrt() as usize;
for i in 2..max {
if n % i == 0 {
let x = n / i;
if x == i {
result += i;
} else {
result += i + x;
}
}
}
1 + result
}
fn main() {
let max = 28123;
let abundant: Vec<usize> = (2..max + 1).filter(|&n| sum_divisors(n) > n).collect();
let mut abundant_sums = vec![false; 2 * max + 1];
for i in 0..abundant.len() {
for j in i..abundant.len() {
abundant_sums[abundant[i] + abundant[j]] = true;
}
}
let sum: usize = (1..max + 1).filter(|&i| !abundant_sums[i]).sum();
println!("{}", sum);
}