Problem 21

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 ≠ 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.

Solution

def sum_divisors(n, divisor):
  sqrtN = n**0.5    
  i = divisor
  t = 0
  while i <= sqrtN:
      if not n%i:
          divisor = i
          if i == n//i or i == 1:
              t+= i
              break
          t += sum([i,n//i])
          break 
      i += 1
  if i < sqrtN:
      divisor += 1
      t += sum_divisors(n, divisor)
  return t

n = 100000
li = list(range(2,n))
for i in range(2,n):
    di = sum_divisors(i,1)
    if di == i:
        try:
            li.remove(i)
        except:
            pass
        continue
    ddi = sum_divisors(di,1)
    if ddi!=i:
        try:
            li.remove(i)
        except:
            pass

print(sum(li))