Another way to interpret the Collatz conjecture is to say that "playing the game" will always arrive at 2^n. The natural numbers can be rearranged into "rails" of each odd number times 2^n. I say rails because playing the game, you "slide down" these rails until you reach an odd number, each odd number being a solution to another rail, and so on until you reach 2^n (or atleast that is the conjecture). Given that each rail produces unique solutions; i.e., (R(x)-1)/3 for one rail cannot be the same for another rail, e.g., (5x2^n-1)/3 will not have the same solutions as (7x2^n-1)/3), and that the solutions to each rail produce a set of rails, playing the game cannot loop (other than the only case where the solution to 1 lies on the same rail) or diverge. The solution to one rail may be greater than the starting number, but it belongs to a lower rail.
Anyways, curious if this has been explored, are there problems with interpreting the problem in this way? Can this problem be distilled into these arguments? I think this is analogous to solving the maze by starting at the end, rather than studying each path from start to end.