yo, I'm about to spread some new knowledge about the collatz conjecture. Not sure if this has been shown before or not, but here: https://en.wikipedia.org/wiki/Collatz_conjecture#Cycles it states that all cycles up to 68 have been checked.
Using my formula I've been able to generate every possible cycle of length m that I want, and more specifically I've been able to generate the highest possible number that could be contained in a cycle of length m. (I should state that all these "cycles" use only non-integer positive numbers). Also something important that I found, the lower the cycle number, the lower the largest number in the cycle can be, meaning that if we were to find an integer cycle solution, it would have to contain an extremely large quantity of numbers above 2^60 (since this is as far as we've checked for the collatz conjecture).
Just to put that into perspective real quick, the largest possible number in a loop of 8 would be 82. For 20 it would be 11777. Keep in mind also, these are the largest numbers, the lowest number in that same sequence for 20 is 5, This also shows that for a loop to have a very large number, it must also have a very low number(this is no good).
I can instead find a 20-cycle that consists of numbers between 11777 and 5 and are all very similar in quantity, we can use the smallest number in this cycle to show that a 20-cycle is impossible(We can obtain a cycle which who's smallest number is the largest of all other cycles of that length very easily). In the case of a 20-cycle, that number is 31 (with 129 being the largest).
Now onto the good stuff, since we know that 31 is not a solution, this means that all 20-cycles are not solutions to the collatz conjecture. So this means that all cycle lengths who's greatest, smallest number is less than 2^60 is impossible to be a solution. Here are some examples:
68-cycle: 943
100-cycle: 3,055
200-cycle: 62,869,365
300-cycle: 394,494,065
400-cycle: 4,109,527,163,265
500-cycle: 50,819,440,441,041
600-cycle: 358,161,777,549,373,501
Since we have checked as far as 19 digits, this means that cycles of length 600 or less are impossible. Once more numbers have been checked, this cycle limit can be improved. It is also possible that somebody may be able to prove that no cycles exist using my formula. I do not know how to check larger numbers using my computer, above 2^64. But at least now we know that for a cycle to exist above 2^60 it must have more than 611 odd numbers in its sequence.
So what is the formula? We make up the larger formula by putting 2 simpler formulas together, Doing this will give us a generator formula for all 2-cycles.
$$n_2 = \frac{3n_1+1}{2^a}, n_1 = \frac{2^an_2-1}{3}$$ $$n_1 = \frac{3n_2+1}{2^b} = \frac{2^an_2-1}{3}$$ $$ n_1 = \frac{-2^a-3}{9-2^{a+b}}$$
If we want to extend this to solve for 3-cycles, we must put 3 of the formula's equal to each other, and introduce an $n_3$ and a $c$.
$$n_1 = \frac{3n_3+1}{2^c}, n_3 = \frac{2^cn_1-1}{3}$$ $$n_3 = \frac{3n_2+1}{2^b} = \frac{2^cn_1-1}{3}$$ $$ n_2 = \frac{2^{b+c}n_1-2^b-3}{9} = \frac{3n_1+1}{2^a}$$ $$ n_1 = \frac{-2^{a+b}-3(2^a)-9}{27-2^{a+b+c}}$$
you can now input random values for $a,b,c,d...$ to generate 2-cycles and 3-cycles. Your values for $a,b,c,d...$ must add to $\log_2(3^m)$ where $m$ is your cycle length, this will ensure that your n's are always greater than 1. After solving this for more cycles, you'll eventually stumble across a pattern for generating $n_1$ in your sequence. This example is of a generator formula for the first number in a 5-cycle sequence(The pattern for the generator formula for different cycles should be obvious now):
$$n_1 = \frac{-(3^0)(2^{a+b+c+d}) -(3^1)(2^{a+b+c}) -(3^2)(2^{a+b}) -(3^3)(2^{a}) -(3^4)(2^0)}{3^5-2^{a+b+c+d+e}}$$
This formula, when it has values inserted for [a,b,c,d,e] as described in the paragraph above, it will yield you the value of n1 to find the rest of these values, simply do the following:
$$n_2 = \frac{3n_1+1}{2^a}, n_3 = \frac{3n_2+1}{2^b}, n_4 = \frac{3n_3+1}{2^c}...$$
Something to note is, all values [a,b,c,d...] must have a value of at least 1. If you would like to produce the largest possible number like a showed earlier in this post, you simply set [a] = 1+log2(3^m)-m, with m being the length of your cycle. If you want the greatest, smallest number a cycle length can have (which is what I used to show that cycles of 611 or less are impossible), simply set [a,b,c] = 2, and every second position after it equal to 2 starting from [e], with the rest being equal to 1.
I hope this has been informative to you, and maybe this was easy enough for you all to understand. I believe that this method has potential to go further in this conjecture however I've unable to figure that out if that is the case. Please use this for your own attempts at discovering something and let me know if it was useful. Also ask me any questions you have if something was not clear, and let me know if you have found any issues with what I've stated in this post. I would also like to know if anyone can confirm for me that this does indeed prove that 611-cycles and below are impossible, and if this is a new discovery, since the best I have seen before this is up to 68-cycles.
Thanks for reading
I think that beyond just rendering your results on here, you should consider publishing a paper. Perhaps the rest of the math community will disagree with me, and I have no had the opportunity to look too into detail on your work; but the effort into more neatly organizing your work is desirable. I suggest using overleaf to render your work in LaTex so that it is set in a more presentable manner. It may be useful to others in the future. I cannot be of much assistance for where you should publish something like this, and indeed the stackexchange may be your best bet, but I suggest at least looking into it. If you find somewhere to post this work, I can be of assistance in preparing the document in LaTex if need be, just DM me.
In summary, the information on the stack exchange is, in general, to be sourced from outside websites. Look into more reputable websites where you might present your findings as a source yourself.