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:

  [ 0 swap witheach + ]        is sum ( [ --> n )

  [ factors -1 pluck 
    dip sum 
    2dup = iff
      [ 2drop 1 ] done
    < iff 0 else 2 ]           is dpa ( n --> n )

  0 0 0
  20000 times 
    [ i 1+ dpa 
      [ table
        [ 1+ ]
        [ dip 1+ ]
        [ rot 1+ unrot ] ] do ]
  say "Deficient = " echo cr
  say "  Perfect = " echo cr
  say " Abundant = " echo cr