1.Take a random positive integer, divisible by three.
2.Consider its base 10 digits.
3.Take the sum of their cubes.
4.Go back to step 2.
Rem: You may limit your research for n<1000.
See The Solution  Submitted by Ady TZIDON 
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re: nonanalytic solution 

(In reply to nonanalytic solution by Paul)
don't know enough about ruby to determine where you went wrong with your code. However, consider 25
2^3+5^3=133
1^3+3^3+3^3=55
5^3+5^3=250
2^3+5^3=133
and thus it will cycle indefinitely and never reach 153 as you proposed.
Posted by Daniel on 20150918 00:52:10 