According to the rules of the Collatz Conjecture if $n$ is odd then execute $3n + 1$ and then when $n$ is even execute $n/2$ and then we are supposed to repeat, however for the purpose of this experiment I have just initiated the process of staring with an odd number until it becomes odd again, examples:
$1 → 4 → 2 → 1$ with the end result being $1 → 1$ (starting with $1$, ending with $1$)
$13 → 40 → 20 → 10 → 5$ with the end result being $13 → 5$ (starting with $13$, ending with $5$)
Let us denote it as $X → Y$
I have then decided to sort the data by $X$ ending in a specific digit, and doing so has revealed many patterns.
In order to better present this question , I will focus on a consecutive list where $X$ ends in $1$, such as $1 , 11, 21 , 31 …$ (all the way to $1001$).
I will present the list as an Image to include different colors, to better visualize the patterns (with their repetitions being evenly distributed). My question in regards to these patterns is broken into 3 parts:
A) As you will notice each color represents increments of $+30$ in $Y$, and its repetition becomes evenly distributed through out the list, but some how there is never an overlap of colors (more than 1 color in a row). Could there ever be one (a row that has more than 1 color)?
B) Do the listings that aren't colored yet, represent an upcoming pattern (additional colors)?
C) Will there be a finite or infinite of patters possibilities (colors) ?
Note: The best way that came to my mind in order to present this question was by presenting the image and using the colors to ask for logic. If anyone thinks there could be a better way, feel free to edit.
Note: To better understand the patterns, please notice that Yellow repeats every $2$ rows, Red every $4$ rows. …
Best Answer
Answer to Question A: All your patterns will continue indefinitely and there won't be any overlapping. As @GottfriedHelms pointed out, you are simply grouping the results by how many divisions by $2$ is required to yield an odd number.
Answer to Question B: The listings that aren't colored yet do indeed represent upcoming patterns.
Answer to Question C: There will be an infinite list of colors since each division by $2^k$ where $k\in\mathbb{N}$ will need a different color, that is, one color for $k=1$ (yellow in your case), another color for $k=2$ (red in your case), and so on.