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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.

Jack
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    Yes, people have noticed that Collatz is easy to solve for those numbers which reach a power of $2$ under iterations of the Collatz mapping. – lulu Feb 18 '23 at 21:23
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    If you like, you can modify the standard Collatz map by considering $n\mapsto \frac n{2^{v_2(n)}}$ for even $n$ where, as usual, $v_2(n)$ denotes the order to which $2$ divides $n$. That spares you the steps of "sliding down the rails", as you put it. At each stage, you simply extract the odd part (if the number is even). – lulu Feb 18 '23 at 21:40
  • This is utter equivalent to the usual Collatz conjecture. It only reduces the number of steps by omitting all even numbers – Peter Feb 19 '23 at 12:30
  • I don't know whether I understand you correctly. Are you considering the *even* numbers ($b_k \in \mathbb 2 \times N^+$) and the transformation $ b_{k+1} = {b_k}_2 \cdot 3+1 $ (where the expression with braces and index $ {n}_p$ mean removing the primefactor $p$ completely from number $n$) ? – Gottfried Helms Feb 19 '23 at 15:37

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