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:

USING: fry math.primes.factors math.ranges ;
: psum     ( n -- m )   divisors but-last sum ;
: pcompare ( n -- <=> ) dup psum swap <=> ;
: classify ( -- seq )   20,000 [1,b] [ pcompare ] map ;
: pcount   ( <=> -- n ) '[ _ = ] count ;
classify [ +lt+ pcount "Deficient: " write . ]
         [ +eq+ pcount "Perfect: "   write . ]
         [ +gt+ pcount "Abundant: "  write . ] tri