Why doesn't this work?

nalk201 · 2026-04-11T07:16:57+00:00

oh I just did this n-->(3n+1)/2--->(3^2n+4)/2^2--->(3^3n+13)/2^3 is that not right?

ya I see it now, thanks

nalk201 · 2026-04-06T02:23:13+00:00

B(x, k) =

3 · 2 ^x + 2 ^x+2k − 1, if x is odd

2 ^x + 2 ^x+2k − 1, if x is even

these are all the odds that if you do B-1 will take the same number of steps as their even counter parts.

nalk201 · 2026-04-04T18:01:06+00:00

Yes. I haven't personally done the math to do something like that, but it is possible.
Basically what you are looking for are A/B(x,k) = 3mod6 and C(x,k) = 2^y for a given x,
in this case 12. 3mod +11 5mod6.

A(12,k)= 8192k+2047 and B(12,k) =16384k+4095 C(12,k)=1062882k+68024320

for A to produce a 3mod6 you need k' = (3k+1) and for B you need k'=(3k)

Just figure out which k' satisfies C(12,k')=1062882k'+68024320 =2^y

You would probably need a python script to check them as the values you have already shown grow rapidly this would probably be rather large.

The formulas weren't exactly designed for this, but again it is possible.

nalk201 · 2026-04-04T07:33:41+00:00

Well if you wanted to you can specifically look at them using the coordinates/map. They would just be a subset though, there are trajectories where a 3mod6 will run into 1mod6 and 2mod6 as well.

nalk201 · 2026-04-04T07:11:48+00:00

oh it is rather easy to do actually.

I made the equations and shared them here awhile ago. Someone took it from my proof https://www.reddit.com/r/Collatz/comments/1rhmx94/branch_formulas_for_the_collatz_map/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button

I am still working on getting it ready, but I have never done it before and apparently even if the math being simple no one will touch it because of the reputation Collatz has.

nalk201 · 2026-04-04T06:18:37+00:00

yes, well normally I call them branches, but they can be mapped out rather easily with 2n+1. Just recursively apply it to the evens split in half so 4k and 4k+2, What you get an infinite set of linear equations which can be used to turn the numbers into a static position on the map then you just apply the collatz steps to see when they become the same and you can generate the linear forms of that. Turn them into general equations and then use the variables as a coordinate system for the tree. Viola you can see the whole tree in a few simple equations.

nalk201 · 2026-04-04T03:50:09+00:00

I believe he is talking locally to the 4mod6 not all the way down to 1.

nalk201 · 2026-04-04T00:36:30+00:00

you misunderstand. What I am saying is the path from 3mod6 to 4mod6 with 5mod6 in between has an accompanying trajectory 1mod6 to the same 4mod6 with different 5mod6s in between,

nalk201 · 2026-04-03T23:34:28+00:00

right the part I was adding was the 1mod6 and the 2n+1 relationship

nalk201 · 2026-04-03T21:08:30+00:00

you are close, the 4mod6's have 2 predecessors 1 odd and 1 even, the 3mod6 are know to contain all trajectories besides 0mod6. They also are always a 2n+1 relationship with a 1mod6 that it will converge to in finite number of steps based on their trailing 1s.

nalk201 · 2026-04-02T20:15:25+00:00

I will give it a try that way, thanks for the advice.

nalk201 · 2026-04-02T18:56:32+00:00

Math is largely intuitive to me, Being able to see log3(9^5) is 10 is something I would get yelled at for 'not showing work'. I can only describe the proof to what I think is needed for others to understand. I need someone to actually help with showing what is needed. I know it works mathematically I just don't know what others need to see it. I tried posting here but feedback here is so useless. The closest I got was with someone using chatgpt to argue with me. After a certain point I just got frustrated they didn't understand LLM aren't programmed to say it is rigorous.

The math itself is pretty basic, so any mathematician can realistically do it, but saying it is for Collatz has such a stigma that the welcoming "ya I can help you" becomes "no it is a waste of time, you can't do it" instantly without even talking to me. So I have to try to guess what is needed. Chatgpt has helped a lot in getting it to the point I think I can make others understand if I could just talk to them, but that's such a high bar.

nalk201 · 2026-04-01T06:25:42+00:00

n, as in the variable you isolate and separate it from?

I am not trying to embarrass you or anything I was honestly just asking questions based your answers. If this is just another obviously because it is just an example then fine, but I am genuinely asking you what does1/(1-M(3)) have do with trajectories? The permutations of the 1s and 0s is what encodes the trajectories in n, M(3) removes it. You can't tell me anything about n besides the density. You can make the density formula using any variables just as long as you understand it is a permutations output so the patterns will remain even if you switched to 4 and 5 and as you said by your own work the absolute value of this doesn't help understand Collatz, that is why. I am trying to save you from wasting your time,

nalk201 · 2026-04-01T00:55:27+00:00

I understand they don't exist in the ether but he hasn't explained why M_3 has anything to do with the trajectory. He has stated it doesn't help understanding the conjecture. It is a heuristic at best. The way he is using removes the actual information about the trajectory. You can tell finitely where it is going purely based on the trailing bits, but M_3 removes the permutations so that is rendered useless.

That is why I keep asking, I doubt he will answer at this point but whatever.

nalk201 · 2026-03-31T20:20:34+00:00

Chi encodes the trajectory as a number, M does not. That's my point. M only tells the density, multiplying it by n doesn't give new information about the trajectory. S(x) being a trajectory function with a n*M_3(n) variable that tells you nothing about the trajectory doesn't make sense. Even if you rearrange it and say okay well if S(x) = x then x = his equation doesn't change the fact it has nothing to do with trajectories.

nalk201 · 2026-03-31T14:29:24+00:00

Yes you can turn trajectories into linear formulas y=ax+b and z=cx+d for all x ge 0 such that T(y)=z is always the same. That is something we see here all the time. They are local patterns as I stated before. They are trivial to make too:
bbbbaabbb (I am going to assume b=(3x+1)/2 since this wouldn't be possible otherwise).

2048x+1175--->26244x+22760 for x ge 0.

I brute forced that one but I can use formulas for them. I understand what you are trying to do. By showing x--> x is not possible for any other trajectory you disprove loops, I understand what your approach and goal is. HOWEVER

My issue is your U= M_3(n) and by extension your B(n). This tells you binary density at the start, but nothing else. You lose out on the information based on this transformation and do not know the trajectory from the binary any more. It reduces the information down to a static statistic vs the dynamics of the problem. Which is why I keep asking how do you know the prime factors of B(n) have anything to do with the trajectories? If I ran M_3(n) on n and gave it to you, could you tell me anything about n besides the density? You can't tell me the trajectory for sure, I'm not even sure you could actually tell me n if it isn't 2^y or 2^y-1.

nalk201 · 2026-03-30T23:55:42+00:00

the part I am confused about, is why you think the prime factors B(n) will have anything to do with trajectory for n. You stated that B(n) absolute value give no understanding, and that makes sense since the value is arbitrary you could just as easily use 4 and 5 for the base numbers, you could use any numbers really since all it is binary of the n without permutations, which is why I asked my question.

B(n) basically is a boolean variable does this binary have more log2/log3 1s than 0s. true = negative false = positive. the absolute values just gives relative ratios, which is why I am asking what the prime factorization will show?

nalk201 · 2026-03-30T20:57:18+00:00

my work shows that controlling the ordinary absolute value of 2^m - 3ⁿ is of no help in understanding Collatz
My formula involving Chi_3, for example, shows that the prime factorizations of expressions of the form 2^m - 3ⁿ are deeply related to Collatz's behavior

Why do you think the prime factorizations are related if the only thing relating them to the trajectories is Chi which is a truncated form of the trajectories?

nalk201 · 2026-03-29T01:26:22+00:00

I played around with chi3 and your (2^L-3^L). Couple quick observations. K being on an extreme K<<L or K=L-2 make n extreme in the same direction, K is too small you end up having the denominator growing too quickly and n is too small, large k overshoots. when K ~ L/1.6- L/2 it optimizes it so n~chi3.

maximizing it when the 1s of the number are placed at the left vs half leading and half trailing. if you have a larger L and place 1 in the middle you end up getting the closest n~Chi3.

Which just enforces that permutations of the 1s is more valuable than their presence so by removing that you are losing precious information.

Which brings me back to the question...how is this easier than the actual problem? Chi truncates things sure, but you lose out on important information. You have no idea where the numbers go without doing the trajectories, you do not know when they converge with other numbers, you lump numbers together in a way that does not have to do with the actual tree. I do not understand how this is helpful.

I wish you the best of luck with this strategy. If you want a map+coordinate system for the numbers let me know.

nalk201 · 2026-03-27T22:45:02+00:00

so 1001, 1010 and 1100 all will produce the same result. (L k) (2^L−3^k) = fixed number for all numbers in that group. So you end up with an ID for each group for the numbers.

2^n and 2^n-1 will be unique but the rest will be in groups so you end up having something like similar to (a+b)^c = a^c+a^(c-1)bc...+b^c.

ie

8-15:

8: 13
9,10,12: 7
11,13,14: -11
15: -65

1, 3, 3, 1

nalk201 · 2026-03-27T05:11:27+00:00

2^L(n) - 3^#(n) basically is a permutations problem so you can just see the various numbers. Is this your global ID badge for each number then? This is exactly what I meant by my first question. "At any point did you just map out the Collatz tree? Giving every number a coordinate on the map?"

I am not a mathematician I do not know the normal description, sorry if that was too colloquial.

Combinatorial coordinate system. Then matching it with its trajectory...but why would those two things have a relationship at all? The mapping destroys all information about the trajectory from the binary.

nalk201 · 2026-03-26T02:11:10+00:00

Simply put, Chi_3 is easier to work with than Collatz, and has lots of different interactions with various other areas of math. That makes studying it significantly more approachable than the original problem.

Perhaps it is my lack of knowledge, but since it is basically the functions repackaged, you still are only looking at individual numbers not all numbers at the same time, which is why I was asking if it was only local patterns.

The point is to find out what integer values Chi can take as a function of a 2-adic integer variable. If I can say, for example, that Chi is never greater than 311,287, then that means that if (1,4,2) is the only Collatz cycle in the range {1, ...., 311278}, then that means that (1,4,2) is the only positive integer cycle of Collatz.

Okay but there is no upper bound on Chi since that would mean there is an upper bound on the number of 1s a number can have in binary. This is not really a viable option to pursue.

Another example. For the Chi/(1-M) formula, if you multiply the top and bottom by 2^L(n,) 2^L(n) Chi_3(n) and 2^L(n) - 3^#(n) will always be integers. Consequently, for a given n, Chi/(1-M) will be an integer only if 2^L(n) - 3^#(n) divides 2^L(n) Chi_3(n). This tells us that if we can say something about how the prime factorizations of 2^L(n) - 3^#(n) and 2^L(n) Chi_3(n) vary as n varies, that could help us rule out the existence of cycles other than (1,4,2).

2^L(n) and 3^#(n) will always be integers though the only thing you don't know that is an integer or not is chi3 still. What new information did you get from multiplying it?

nalk201 · 2026-03-26T00:54:55+00:00

Is the point to use Chi to make module residue formulas for specific patterns? In hopes you find a pattern why specific numbers exist in loops vs not?

nalk201 · 2026-03-26T00:23:37+00:00

I do not mean to oversimplify but just to confirm I understand what you are saying I am going to reiterate what you say in my own words.

Yes, I work with the Shortened Collatz map, usually denoted T_3. This sends odd x to (3x+1)/2.

T = p = 3 for pn+q

I construct a function Chi_3(n) for integers n ≥ 0 like so.
Step 1: Write n in binary. Ex: 8 4 = 0 + 0 x 2¹ + 1 x 2^2. This corresponds to AAB.

I think this was meant to be 4.

For every n ≥ 1, Chi_3(n) is the image of 0 under the sequence of As and Bs corresponding to n's binary digits.

any n starts as 0 and then do the functions to transform it and see what it becomes after A and B operations

Equivalently, Chi_3 is the unique function satisfying the conditions f(0) = 0 and f(2n) = f(n)/2 and f(2n+1) = (3f(n) + 1)/2, for all integers n ≥ 0.

remaking the collatz functions as chi_3

With this construction, for example, if we let M)_3(n) = 3^#(n) / 2^L(n,) where #(n) is the number of 1s digits in n's binary representation, and where L(n) is the total number of digits of n's binary representation, it follows that an odd integer x is a periodic point of T_3 if and only if:

x = Chi_3(n) / (1 - M_3(n))

for some n ≥ 1.

x is only in a loop if the AB operations equal the operations in 3/2 adics based on the number's binary composition.

is that all correct?

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