Problem Statement

In number theory, the home prime HP(n) of an integer n greater than 1 is the prime number obtained by repeatedly factoring the increasing concatenation of prime factors including repetitions. The traditional notation has the prefix "HP" and a postfix count of the number of iterations until the home prime is found (if the count is greater than 0), for instance HP4(2) === HP22(1) === 211 is the same as saying the home prime of 4 needs 2 iterations and is the same as the home prime of 22 which needs 1 iteration, and (both) resolve to 211, a prime. Prime numbers are their own home prime; So: If the integer obtained by concatenating increasing prime factors is not prime, iterate until you reach a prime number; the home prime.

Let's start with the solution:

USING: formatting kernel make math math.parser math.primes
math.primes.factors math.ranges present prettyprint sequences
sequences.extras ;

: squish ( seq -- n ) [ present ] map-concat dec> ;

: next ( m -- n ) factors squish ; inline

: (chain) ( n -- ) [ dup prime? ] [ dup , next ] until , ;

: chain ( n -- seq ) [ (chain) ] { } make ;

: prime. ( n -- ) dup "HP%d = %d\n" printf ;

: setup ( seq -- n s r ) unclip-last swap dup length 1 [a,b] ;

: multi. ( n -- ) chain setup [ "HP%d(%d) = " printf ] 2each . ;

: chain. ( n -- ) dup prime? [ prime. ] [ multi. ] if ;

2 20 [a,b] [ chain. ] each