One possibility: the second digit is the number of non-zero digits, the third digit is the number of zeroes. Then by the same argument as above we always end up with a three digit number, except that in this case we alternate between 330 and 321 (cycle of length 2). Seb On Fri, Oct 15, 2010 at 16:49, Victor Miller <victorsmiller@gmail.com> wrote:
So how about the following generalization?
We're working in base B, we have some fixed partition of the digit set {0,...,B-1} into r pieces, numbered from 1 to r.
For any base B number we make a new number the same way:
first write down the base B representation of the number of digits, concatenating the base B counts of the digits for the r pieces given above. It's clear by the same reasoning that eventually we'll reach a cycle. So the challenge is to describe other situations where the cycle is length 1.
Victor