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I was looking at the Collatz Conjecture and I thought of something:

If we denote two operators $a_n = 2n$ and $b_n = \frac{n-1}{3}$, then every number that converges using the Collatz Conjecture can be represented in the form of a continued subscript, like 5: $$b_{a_{a_{a_{a_{1}}}}}$$

And to make it easier to read, I'll remove the subscripts, making it:

$$baaaa1$$

If we remove the $1$, giving $baaaa$, and then replace $b$ as $0$ and $a$ as $1$, we get a psuedo-binary system. Thus $5$ is $01111$ and $7$ is $0101011011101111$. The longer it takes for a number to converge, the longer the number is written down.

I realize that this has an information density much much less than binary, but I was wondering: Given this number system, and given two numbers, $x$ and $y$, how can we determine what their sum, $x+y$ will be, or if it will even exist in this number system?

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    I'm not sure of the answer to your question, but I think your continued subscript notation reflects the fact that the Collatz function relates any endpoint and any startpoint by a continued fraction.2017-03-27

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After looking at it for a little bit, it does not make too much sense to "add" or "subtract" these numbers together in the way one normally adds and subtracts. As for checking what values are real or not, I have a few ideas.

I tested adding by using the numbers 5 (01111) and 4 (11). By adding normally, I received the answer (10010), which is a nonsense answer because you can not multiply an even number by 3 and add 1. I then tried simply moving the (11) in front of (01111). It made (1101111), which is 20. Doing the reverse (0111111) created 21. I then tried 'adding' 5 (01111) and 13 (011101111). Putting 5 in front of 13 created 69, however the reverse created a non-existent number because 80 does not produce a whole number after ${(n-1)}/{3}$.

Overall, concatenating numbers seems to work sometimes, although I am unsure if this counts as "addition" by any means.

To check if the numbers are real: if there are no easy consecutive zeros to rule out, for instance (11001111) is a fake number, then the best way I found to check it is through brute force. Follow the number from right to left and follow the process you mentioned earlier to find if it works or not.

There may be more interesting properties about these numbers or better ways to work with them. I would think it could be interesting to look further into this!

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    The natural operation on those "numbers" are $a,b$ and $a^{-1},b^{-1}$ (their inverse operators). We can also try simplification rules as $(aba)^{-1}bab = ?$2017-08-09