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


USING: kernel math math.ranges math.text.utils memoize
prettyprint sequences tools.time ;
IN: rosetta-code.iterated-digits-squaring
 
: sum-digit-sq ( n -- m ) 1 digit-groups [ sq ] map-sum ;
 
MEMO: 1or89 ( n -- m )
    [ dup [ 1 = ] [ 89 = ] bi or ] [ sum-digit-sq ] until ;
 
[
    0 1
    [
        dup sum-digit-sq 1or89 89 = [ [ 1 + ] dip ] when
        1 + dup 100,000,000 <
    ] loop
    drop .
] time