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


REM Version 1: Brute force
 
T = TIMER
n1 = 0
FOR i = 1 TO 10000000
    j = i
    DO
        k = 0
        DO
            k = INT(k + (j MOD 10) ^ 2)
            j = INT(j / 10)
        LOOP WHILE j <> 0
        j = k
    LOOP UNTIL j = 89 OR j = 1
    IF j > 1 THEN n1 = n1 + 1
NEXT i
PRINT USING "Version 1: ####### in ##.### seconds."; n1; (TIMER - T)
END