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


10 LET n=0
20 FOR i=1 TO 1000
30 LET j=i
40 LET k=0
50 LET k=INT (k+FN m(j,10)^2)
60 LET j=INT (j/10)
70 IF j<>0 THEN GO TO 50
80 LET j=k
90 IF j=89 OR j=1 THEN GO TO 100
95 GO TO 40
100 IF j>1 THEN LET n=n+1
110 NEXT i
120 PRINT "Version 1: ";n
200 DEF FN m(a,b)=a-INT (a/b)*b: REM modulo