Three observations: 1) The Collatz mapping can be thought of as just going from odd integers to odd integers, via f(x) = 2K+1 for any odd integer x, where 3x+1 = 2^n * (2K+1) for integers n >= 0 and K >= 0. This seems much more natural than dividing by 2 only once per "time step". Then the Collatz conjecture can be phrased as saying that for odd x > 0, iterating f to get f(x), f(f(x)), ... must end up in the (1 ) cycle. 2) No reason we can't look at this for odd x < 0. In that case the cycles I know of are (-1 ), (-5 -7 ), and (-17 -25 -37 -55 -41 -61 -91 ). 3) It seems even more natural to look, more generally, at Collatz on rational numbers of the form x = odd/odd. Let x be such a rational. Then much as in 1) set f(x) = p/q in lowest terms where 3x+1 = 2^n * p/q for n, p, q integers with p and q odd. There seem to be lots of cycles, but they don't seem easy to find. And it's not clear if studying this more general situation will shed any light on the original conjecture. —Dan