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:
<pre>15 -> 26 -> 40 -> 16 -> 37 -> 58 -> 89
7 -> 49 -> 97 -> 130 -> 10 -> 1</pre>

An example in Python:

<lang 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]</lang>


;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 [https://projecteuler.net/problem=92 Project Euler problem 92].

For a quick algorithm for this task see [[Talk:Iterated_digits_squaring|the talk page]]


;Related tasks:
* [[Combinations with repetitions]]
* [[Digital root]]
* [[Digital root/Multiplicative digital root]]
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