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


 
total = 0
d = new dict
var sum
 
for n = 1 to 100 million - 1
{
   sum = n
   do
   {
      if sum < 1000 and d@sum != undef
      {
         sum = d@sum
         break
      }
 
      c = sum
 
      sum = 0
      for digit = integerDigits[c]
         sum = sum + digit^2
   } while (sum != 89) and (sum != 1) 
 
   if (n < 1000)
      d@n = sum
 
   if (sum == 89)
      total = total + 1
}
 
println[total]
 