These define three classifications of positive integers based on their   proper divisors.

Let   P(n)   be the sum of the proper divisors of   n   where the proper divisors are all positive divisors of   n   other than   n   itself.

   if    P(n) <  n    then  n  is classed as  deficient  (OEIS A005100).
   if    P(n) == n    then  n  is classed as  perfect    (OEIS A000396).
   if    P(n) >  n    then  n  is classed as  abundant   (OEIS A005101).



Example

6   has proper divisors of   1,   2,   and   3.

1 + 2 + 3 = 6,   so   6   is classed as a perfect number.



Task

Calculate how many of the integers   1   to   20,000   (inclusive) are in each of the three classes.

Show the results here.



Related tasks

   Aliquot sequence classifications.   (The whole series from which this task is a subset.)
   Proper divisors
   Amicable pairs


void setup() {
  int deficient = 0, perfect = 0, abundant = 0;
  for (int i = 1; i <= 20000; i++) {
    int sum_divisors = propDivSum(i);
    if (sum_divisors < i) {
      deficient++;
    } else if (sum_divisors == i) {
      perfect++;
    } else {
      abundant++;
    }
  }
  println("Deficient numbers less than 20000: " + deficient);
  println("Perfect numbers less than 20000: " + perfect);
  println("Abundant numbers less than 20000: " + abundant);
}
 
int propDivSum(int n) {
  int sum = 0;
  for (int i = 1; i < n; i++) {
    if (n % i == 0) {
      sum += i;
    }
  }
  return sum;
}