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


import std.stdio, std.algorithm, std.range, std.functional;
 
uint step(uint x) pure nothrow @safe @nogc {
    uint total = 0;
    while (x) {
        total += (x % 10) ^^ 2;
        x /= 10;
    }
    return total;
}
 
uint iterate(in uint x) nothrow @safe {
    return (x == 89 || x == 1) ? x : x.step.memoize!iterate;
}
 
void main() {
    iota(1, 100_000_000).filter!(x => x.iterate == 89).count.writeln;
}