If you add the square of the digits of a Natural number (an integer bigger than zero), you always end with either 1 or 89:

15 -> 26 -> 40 -> 16 -> 37 -> 58 -> 89
7 -> 49 -> 97 -> 130 -> 10 -> 1
An example in Python:

>>> step = lambda x: sum(int(d) ** 2 for d in str(x))
>>> iterate = lambda x: x if x in [1, 89] else iterate(step(x))
>>> [iterate(x) for x in xrange(1, 20)]
[1, 89, 89, 89, 89, 89, 1, 89, 89, 1, 89, 89, 1, 89, 89, 89, 89, 89, 1]


Task

 Count how many number chains for integers 1 <= n < 100_000_000 end with a value 89.
Or, for much less credit - (showing that your algorithm and/or language is slow):

 Count how many number chains for integers 1 <= n < 1_000_000 end with a value 89.
This problem derives from the Project Euler problem 92.

For a quick algorithm for this task see the talk page



Related tasks

 Combinations with repetitions
 Digital root
 Digital root/Multiplicative digital root


use warnings;
use strict;
 
my @sq = map { $_ ** 2 } 0 .. 9;
my %cache;
my $cnt = 0;
 
sub Euler92 {
    my $n = 0 + join( '', sort split( '', shift ) );
    $cache{$n} //= ($n == 1 || $n == 89) ? $n : 
    Euler92( sum( @sq[ split '', $n ] ) )
}
 
sub sum {
   my $sum;
   $sum += shift while @_;
   $sum;
}
 
for (1 .. 100_000_000) {
   ++$cnt if Euler92( $_ ) == 89;
}
 
print $cnt;