Problem Statement

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.

6   has proper divisors of   1,   2,   and   3. 1 + 2 + 3 = 6,   so   6   is classed as a perfect number.

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

Let's start with the solution:

F sum_proper_divisors(n)
   R I n < 2 {0} E sum((1 .. n I/ 2).filter(it -> (@n % it) == 0))

V deficient = 0
V perfect = 0
V abundant = 0

L(n) 1..20000
   V sp = sum_proper_divisors(n)
   I sp < n
      deficient++
   E I sp == n
      perfect++
   E I sp > n
      abundant++

print(‘Deficient = ’deficient)
print(‘Perfect   = ’perfect)
print(‘Abundant  = ’abundant)