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


shared void run() {
 
	function digitsSquaredSum(variable Integer n) {
		variable value total = 0;
		while(n > 0) {
			total += (n % 10) ^ 2;
			n /= 10;
		}
		return total;
	}
 
	function lastSum(variable Integer n) {
		while(true) {
			n = digitsSquaredSum(n);
			if(n == 89 || n == 1) {
				return n;
			}
		}
	}
 
	variable value eightyNines = 0;
	for(i in 1..100M - 1) {
		if(lastSum(i) == 89) {
			eightyNines++;
		}
	}
	print(eightyNines);
}