www.mikescher.com/www/statics/euler/Euler_Problem-092_explanation.md

My approach to this problem is pretty crude. Perhaps I will later come back and try to find a better algorithm. Currently we iterate (brute-force) through all possible numbers and count the chains that end with one. The next() function is implemented like this:

0\>:55+%:*\ :#v_$$
  ^/+55g05+p05<   

We also remember in an 8x71 cache all previously found numbers so we can abort some sequences before we reach 1 or 89. This is the main optimization from pure brute-force in this program.

We can prove that an 568-element cache is enough because no number in the sequence (except the first) can be greater than 9^2 * 7 ( = 567)