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


 
#!/bin/gawk -f
function sumprop(num,   i,sum,root) {
if (num == 1) return 0
sum=1
root=sqrt(num)
for ( i=2; i < root; i++) {
    if (num % i == 0 )
    { 
    sum = sum + i + num/i
    }
    }
if (num % root == 0) 
   {
    sum = sum + root
   }    
return sum
}
 
BEGIN{
limit = 20000
abundant = 0
defiecient =0 
perfect = 0
 
for (j=1; j < limit+1; j++)
    {
    sump = sumprop(j)
    if (sump < j) deficient = deficient + 1
    if (sump == j) perfect = perfect + 1
    if (sump > j) abundant = abundant + 1
    }
print "For 1 through " limit
print "Perfect: " perfect
print "Abundant: " abundant
print "Deficient: " deficient    
}
 