I had a few close votes HERE on my post at mathoverflow and before I could ask this commenter to make it an answer so I could accept it, it was closed. I had promised I would show the re-arrangement in my next post, so here it is.
@FredKline As you can see from the wikipedia page, this question has been seriously investigated, but not yet resolved. It's a pretty good guess that any purported proof has errors, but a correct proof that there are no other loops would be a valuable contribution to the field. ? Anton Geraschenko.
So, I have one question and a challenge: Q) What would this re-arrangement of the number line be called? A category, group, partition, translation, etc.
The challenge -- from Anton Geraschenko's comment -- is to find the mistake in the reduction of my logical equation. Like he said, it's most likely I have missed something.
Here is my re-arrangement of the number line that preserves the Collatz interstitial proportionality; $$\text{Let }side=9,\ \{\ 2^m (4 n-2)\ |\ \{1\leq n\leq side\}, \{0\leq m\leq side-1\}\ \},$$ $$ \left( \begin{array}{ccccccccc} 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 \\ 6 & 12 & 24 & 48 & 96 & 192 & 384 & 768 & 1536 \\ 10 & 20 & 40 & 80 & 160 & 320 & 640 & 1280 & 2560 \\ 14 & 28 & 56 & 112 & 224 & 448 & 896 & 1792 & 3584 \\ 18 & 36 & 72 & 144 & 288 & 576 & 1152 & 2304 & 4608 \\ 22 & 44 & 88 & 176 & 352 & 704 & 1408 & 2816 & 5632 \\ 26 & 52 & 104 & 208 & 416 & 832 & 1664 & 3328 & 6656 \\ 30 & 60 & 120 & 240 & 480 & 960 & 1920 & 3840 & 7680 \\ 34 & 68 & 136 & 272 & 544 & 1088 & 2176 & 4352 & 8704 \\ \end{array} \right),$$ where we note that, taken to infinity, we have identified all unique even numbers. By subtracting 1 from each number we identify all unique odd numbers. The left hand column contains all terminating descenders -- those ascenders that when evaluated cause a parity change in the sequence -- important because they cause the path's direction to change.
Now, as promised, we show our proof of no looping except by the known loop 1-4-2-1. First we take the re-arrangement formula (without $2^m$ to restrict it to the left hand column), $(4 n-2)$, and wrap it in our even equivalent to $(3x+1)/2$ thus: $ \frac{3 (4 n-2)-2 }{2(4 n-2)}$ with the original formula also inserted into the denominator. We find the limit of this formula is $\frac{3}{2}$, which we use as the upper bound. Then we create a logical expression:$$n\geq 1\land 1< \frac{3(4 n-2)-2 }{2(4 n-2)}<\frac{3}{2},$$ which reduces to $n>1$, therefore the only loop is within the first category.$\ \ \square$
Below we show the modulus 3 values for the above re-arrangement. We plan to use this information to show the path constraints. $$ \left( \begin{array}{ccccccccc} 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 \\ 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 \\ 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 1 & 2 & 1 & 2 & 1 & 2 & 1 \\ \end{array} \right).$$
Fred, you're making me look bad just for giving you the time of day. Please don't challenge people to find mistakes in your proofs. It's like standing in your driveway challenging passers by to clean your garage ("What's the matter? Don't think you could do it?"). When you ask a question, the onus is on you to try seriously to solve it yourself first. If your question can be answered by a link to the first wikipedia page one would think to go to, it shows you haven't seriously attempted it. If your question is "is anything wrong with this simple proof of a famous open problem?", it shows you haven't bothered to even break the proof into its components to understand how they work. With that out of the way, since I've stopped by your driveway to wag my finger and lecture you on manners, I will pick up this one scrap in your garage.
Your question: call them "integers with the same odd part". The odd part of an integer is its largest odd divisor. The rows of your table are all even integers with the same odd part.
Your challenge: your error is in the final statement "therefore the only loop is within the first category." The preceding argument shows nothing of the sort. The gist of your argument is "the operation $n\mapsto n/2$ perserves the odd part and the operation $n\mapsto \frac{3n+1}{2}$ makes the odd part larger, so since the odd part can never get smaller, there cannot be a loop." If this were correct, there would be no loops at all, clearly contradicted by the trivial loop. Of course the problem is that $n\mapsto \frac{3n+1}{2}$ does not make the odd part of the integer larger; it only makes the integer itself larger.
An alternative formulation of your argument is that when you run the Collatz procedure and follow the number on your table, you can only stay in the same row or move to a row further down. To see that this is wrong, pick any number not in the first row (e.g. 10) and see what happens when you run the Collatz procedure. You'll find that you end up on the first row.