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


fn digit_square_sum(mut num: usize) -> usize {
    let mut sum = 0;
    while num != 0 {
        sum += (num % 10).pow(2);
        num /= 10;
    }
    sum
}
 
fn last_in_chain(num: usize) -> usize {
    match num {
        1 | 89 => num,
        _ => last_in_chain(digit_square_sum(num)),
    }
}
 
fn main() {
    let count = (1..100_000_000).filter(|&n| last_in_chain(n) == 89).count();
    println!("{}", count);
}