Collatz conjecture pattern (3n + 1 problem). I have a pattern I found in Collatz Conjecture I want to share. Afterwards, I would like to know, if I could try harder at this pattern (I am stuck), if it could lead to a proof. Or it would just be other thing to waste my time here.
I made this spreadsheet, which you can download and play and see it for yourself, how the pattern shows itself.
At "end of column" I calculate a number that, when plugged at the odd function $3n + 1$, will produce a number that is only divisible by 2. That formula is given by $S=\frac{4^k-1}{3}$. S would be the number I will choose to stop at each column. And K would be a natural number in sequence. If I start with k=0, then I have only 0 at the end of the column, which is good, zero has a column all by himself, as equal as 1. So, I could plug 2, and get 5. So, I have a column of just 0, one for 1, and column 2, would have 3 and 5. Then column 3 would have 7, 9, 11, 13, 15, 17, 19 and 21. As 21 is S, when k = 3.
The funny thing is that S is also a formula for partial sum of $4^k$. That also means that each column is 4 times greater than the one before.
I then, because I am only interested by what happens with odd natural numbers, start by putting all natural numbers ordered in sequence. Then, I stop when I find a number calculated by the previous formula ("End of column" sheet).
Then, I continue the sequence of odd natural numbers in the next column of the "collatz sheet".
I can make this sheet as big as I want (natural numbers are infinite). The only limits to that would be (as I used my own Excel sheet, running on my 2011 computer) hard disk space, memory and cpu power. If you make it big, it will run in these issues. Which you can make it less agravating, if you choose to transform the formulas in values, after you calculated the numbers you wanted. I don't know how this would turn out in google spread sheet I am sharing with you.
The pattern is: if you take these odd numbers, compute the next odd numbers, you will see that the numbers will repeat as a clockwork in the next columns (in position-wise).
Examples are colored in the spreadsheet I shared. Example: 21 and 85 are the last one of their colums, and they turn directly to 1, as the formula would give, of course. Then, 19, and 77 are the last part of a 7/8 position [explaining, there are numbers (7, 9, 11, 13, 15, 17, 19, 21 and 19 is the 7th); (23, 25, 27, ... , 75, 77, 79, 81, 83, 85 and 77 is at the end of 7/8 position block). So, this pattern shows that each 7/8 position will cook the number 19 after just the first iteration.
The numbers are coloured for you to see them, how they happen. First, I coloured the background colour, to show that the numbers follow their position. Then, I coloured the font, to show that new numbers of the next column fit in the next column afterwards and so on and so on. Explaining (look at 23, 25, 27, they produce 35, 19, 41, that are also produced by 93, 101, 109). That is, the same position percentile-wise one number occupies in the position of the natural numbers distributed as I created and showed in this spreadsheet, produces a pattern that not only repeats itself, as it grows, to fill the new spaces. I mean. The first pattern has 8 numbers, the next one has the same 8 numbers (each in their position) + 24 new ones (which will keep their positions on the next column.
Note, if you want to create a bigger sheet, just plug in whatever the size size you want. And then, just copy-paste the formula, it will just self-referentiate nicely from each one cell you choose.
Now, the question is: can I keek this work to develop a proof of the Collatz Conjecture, knowing this will have to have lots of work of arithimetic and geometric progression and mathematic induction, all mixed and much more that I don't know yet or is this already fruitless endeavour?
Edit:
@NickGuerrero
The numbers in this example sheet, from cell A1 to E128, are just the natural numbers, odd ones, which continue above when they encounter, 5, 21, 85, 341, 1365 and so on (given by $\frac{4^k-1}{3}$).
The next set of numbers are given by a spreadsheet formula, that uses "IF" functions inside other "IF" funcions (nested "IF"s). There is a limit about how much you do it. I just, in my example, custructed the "IF" condition, to ask if the result from $3n+1$ would be divided by $2^{24}$, then, divide it by $2^{24}$, if not, ask if it was divided by $2^{23}$, then divide it by it, if not... till I divide it and it rests a number that is not divided by 2, that is, it is next odd number from iteration from odd function and even function of Collatz conjecture algorithm.
edit:
@GottfriedHelms
Let me try to explain some other way.
So, I have some numbers, let me say end of column of a table.
This colum is made by the S number above, with given K.
So I have for K = 2, I get S = 5, the first column of the all natural numbers that are odd, are 3 and 5. Then I get K = 3, I get S = 21. So, my next column is from 7, 9, 11, 13, 15, 17, 19, and last 21. The next one would be 85 the last number, so it would be 23, 25, 27, ..., 81, 83, and 85, and so on, so on.
The next columns of the excel sheet would be the next iteration of that table of the above paragraph. In terms of the next odd number (it means that it would not matter how many n/2 even functions would be).
If you make this thing in excel sheet, just remember that to make a big spreadsheet it will take much space and much memory and also processing of the CPU.
But it will allow you to see, the very next step of the next odd number get always in the same positons they occupy from the starting configuration (of course, adapted to the fact, that every column grows by a factor of 4). So once a first iteration of the previous column is made, it is somehow carried to the next colum. And there will be created 3 new spaces for new results, that will have their space, the next time the column grows, it growns in a way as to always behave in an ordered way, to give the same results, growing in to the infinity.
But how is the order carried to the next configuration, and how to prove that it will be always be this way is that is the hard thing.
 A: I will answer your question: I bet you won't like my answer but it is answer so I hope you keep that in mind. 
Your question was: "I would like to know, if I could try harder at this pattern (I am stuck), if it could lead to a proof. Or it would just be other thing to waste my time here."
When I was a younger math nerd I attempted to take the P vs NP  problem very seriously and after burning out a few times a mathematician and mentor of mine told me that "This is not to be done." He wanted me to focus on acquiring some mathematical tools and some experience that would help me become a more competent math nerd and wanted to give me a warning about diving into exceptionally difficult mysteries without sufficient experience. 
Questions like the Collatz conjecture are good for getting mathematicians to wake up but not a good place to spend your time. 
I mean that this is the type of thing that mathematicians as a culture should be on the same page about: when you see a young and-up-and-coming mathematician (Or really any non pro) making serious efforts towards a problem that has been open for a century we should dissuade them from getting bogged down in these mental traps. 
This is a great puzzle. And it's very accessible which makes it a nice tool to get students excited about open mathematics. But also very very impressive mathematicians have said things along the lines of "we're not quite ready to solve such mathematical mysteries. We don't have all the tools yet."  
The reasons you should not spend more time on this puzzle.
1) With probablity approaching 1, better mathematicians have already attempted a route similar to the one above. Which (and without any insult intended here) doesn't look particularly novel. 
2) The problem isn't of particular interest. We have more important work to be doing: work that we can actually make real progress on. 
3) It's probably a dead end. I mean... thus far all the numbers we have tested head to one... and all of the attempts we have made(for a century) have led to either undecidable generalizations or just a flat failure to say anything interesting. 
You should enter the mathematical community with burning questions like this one... and then when you seen the human effort put toward this question and have read Conway's book on the 3x+1 problem and checked out Lagarias and whoever else you need to convince yourself that very impressive minds have spent considerable energy on this... 
You should put down these puzzles and help make genuine mathematical progress. It will be better for you and the mathematical community at large. 
A: 
Now, the question is: can I keek this work to develop a proof of the Collatz Conjecture, knowing this will have to have lots of work of arithimetic and geometric progression and mathematic induction, all mixed and much more that I don't know yet or is this already fruitless endeavour?            

It might not be fruitless - at least it helps to learn much about such structures.     
I've come across many attempts to look at such trees and their patterns in the hope to prove/disprove the Collatz conjecture. But that alone does not say enough: there are other problem-configurations having the same tree-structure but having cycles other than the "trivial" one. So there must be added one more ingredient to the soup: why do some problems ($5x+1$,$3x-1$, $181x+1$) have more than one tree of this structure - not connected!- and have nontrivial cycles but the $3x+1$ possibly not?           
Why does the $3x+1$ might have only one tree rooted on $1$ and yet it covers the whole positive numbers? Why in the negative numbers $3$ trees instead (and $3$ cycles) and likely they cover the whole negative numbers?    
The answer on this "why's" is possibly the crucial one.
A: It is actually trivial to show that numbers of the form  $\frac{4^k+1}{3}$ lead to 1, going from there to a proof of the collatz conjecture is very hard, if not impossible.
