A number-theory problem
Let n be an odd number. For what values of k is 2kn-1 divisible by 3?
Edited to add: Of course, this post is inspired by the recent xkcd #710 on the Collatz Problem :-)
Suppose that we find a counterexample: a number C1 such that the Collatz iterates {Ck}k>0 starting from C1 does not include 1. Note that the iterates must have a minimum value greater than 1. If there are 1+ distinct counterexamples, define m to be the minimum over all the minimums-of-iterates --- otherwise define m to be infinity.
Assuming that m is finite, what can we say about it? It is not even: the Collatz iterates starting from an even number immediately decrease, so the minimum is not even. So m (if it is finite) must be odd.
We can consider which iterates preceded m --- in other words, run the Collatz formulas backwards. We know that m is the minimum of a set of iterates, so the first k>0 Collatz steps immediately preceding m must be of the type number-is-even-so-divide-by-2. Also k is finite so the (k+1)th step must be of the type number-is-odd-so-multiply-by-3-and-increment.
Thus for each odd number n, we are interested in an equivalence class that contains all numbers of the form 2kn = 3*o+1 for an odd number o. ... I think this is the same as all values of k>0 such that 2k n ≅ 1 (mod 3) but I haven't convinced myself.