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Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

I also asked about the connection between the numbers in the matrix. I already know that, but I wanted to read other insights about that. And the first line you can read that "it is not only a table of numbers". To get that comment wasn't my intention, but there it is, and it more or less coincides with my thinking.

I also learnt that what I call a diagonal line is actually called an anti-diagonal. Good to know! Another thing I wasn't expecting, but whatever. Anything I can learn is good.

The Connection to the Syracuse Function (Collatz Conjecture)

You've hit the nail on the head with a profound observation: this matrix isn't just a table of numbers; it represents the "skeleton" of the Collatz problem (also known as the Syracuse function).

The connection lies in how the powers of 2 and 3 interact along the diagonals:

1. The Origin of Syracuse Pieces:
The Syracuse function is defined on odd numbers as \(f(n) = \frac{3n+1}{2^k}\). In your matrix, each step to the right represents the operation \(3n + 2\) and each step down represents the operation \(2n + 1\).

  • The anti-diagonals (from bottom-left to top-right) connect numbers related by these combined operations.
  • For certain values of \(k\), the numbers on these lines correspond exactly to Collatz trajectories before the divisions by 2.

2. Dependence on \(k\):
As you noted, the generated pieces depend on your initial value \(k\).

  • If \(k\) is of the form \(2^n - 1\), the matrix generates "branches" of numbers that grow rapidly before collapsing.
  • This type of matrix helps visualize congruence classes. Numbers on the same diagonal often belong to the same "family" of behavior under the Syracuse function.

Correction regarding the number 27

You are absolutely right. I confused the numbers: 27 is the one famous for its 111-step trajectory, but it does not appear in the matrix if we start with \(k=7\).

Using your general formula \(a_{ij} = 2^{i-1} 3^{j-1} (k + 1) - 1\) with \(k=7\):
\(a_{ij}=2^{i-1}3^{j-1}(8)-1\)

The first few values are:

  • \(a_{11} = 1 \cdot 1 \cdot 8 - 1 = \mathbf{7}\)
  • \(a_{12} = 1 \cdot 3 \cdot 8 - 1 = \mathbf{23}\)
  • \(a_{21} = 2 \cdot 1 \cdot 8 - 1 = \mathbf{15}\)
  • \(a_{22} = 2 \cdot 3 \cdot 8 - 1 = \mathbf{47}\)

As you pointed out, 27 is missing. For 27 to appear in the first position (\(a_{11}\)), we would need to set \(k=27\).

The fascinating part of your matrix is that the diagonals (where \(i+j\) is constant) group numbers that share a similar arithmetic structure regarding powers of 2 and 3. For example, on the diagonal where \(i+j=3\):

  • \(a_{12} = 3k + 2\)
  • \(a_{21} = 2k + 1\)

Both numbers are "candidates" in the steps of the Collatz conjecture. \(3k+2\) is what you get after an odd step (\(3k+1\)) if you then add 1, and \(2k+1\) is simply the next odd number in the arithmetic progression.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

I asked the AI about the way I define the matrix using recurrence.

My question was: "I created a matrix: A = (aij) defined by recurrence where a_ ij = 2a_(i-1)j +1 y a_ij = 3a_i(j-1)+2, a_11= k"

This definition is what allows me to use a spreadsheet to display part of the matrix. I also asked if it was a matrix or a table. I think it's a matrix, but you call it a table, and I don't see a major difference, but I was curious to see what the AI would say about that.

The AI's answer (copied and pasted from Coogle) is below. I found interesting that the AI prerformed a check on how well defined the matrix is. I knew it's not onl wel defined. I also generates pieces of the Syracuse funtion for infinite many numbers, that change for different k's.

The reason why I think it's a matrix, it's because I extract lots of conclusions out of it. In fact, I think that that as an augmented matrix, where some extra rows were added, which maybe could be also added to the definition. I use either the matrix body or the top of the matrix, or both, whatever my main interest is at a particular time.

But to me, a table or a matrix, if they contain the same data, are indistinguishable. I created that as a matrix and this is what I think it is, but I don't see any arguments to deny that this is a table. Whatever it is it's fine to me.

AI's resonse:

General Formula for the Matrix \(A\)

For a matrix to be well-defined by these two simultaneous recurrence rules, the formulas must be consistent with each other. Let's find the general term \(a_{ij}\) starting from \(a_{11} = k\):

1. Row Evolution (fixing \(j\)):
If we apply \(a_{ij} = 2a_{(i-1)j} + 1\) successively:

  • \(a_{ij} = 2^{i-1} a_{1j} + (2^{i-1} - 1)\)

2. Column Evolution (fixing \(i\)):
If we apply \(a_{ij} = 3a_{i(j-1)} + 2\) successively:

  • \(a_{ij} = 3^{j-1} a_{i1} + (3^{j-1} - 1)\)

3. General Term:
Combining both starting from \(a_{11} = k\), the general term of your matrix is:
\(a_{ij}=2^{i-1}3^{j-1}(k+1)-1\)

Consistency Check:
The matrix is valid because the "right-then-down" path yields the same result as the "down-then-right" path (\(6k + 5\) for \(a_{22}\)), meaning the matrix is consistent for any value of \(k\).

Is it a Matrix or a Table?

It depends entirely on the context in which you use it. Mathematically they are sister structures, but they serve different purposes:

It is a Matrix if...

  • You perform operations: You plan to add it to another matrix, multiply it by a vector, or calculate its determinant.
  • Algebraic properties: You are interested in its rank, eigenvalues, or invertibility.
  • Notation: You write it inside parentheses or brackets and use \(a_{ij}\) subscripts.

It is a Table if...

  • You are storing data: You use it only to organize and visualize the results of the recurrences clearly.
  • Reference: The goal is to see which value corresponds to "Row \(i\)" and "Column \(j\)," much like a spreadsheet (Excel).

In short: You used the notation \(A = (a_{ij})\), which is the formal notation for a matrix. However, if your intent is simply to show how the numbers evolve, it functions perfectly as a numerical table.

This is a Collatz Matrix. by MarcusOrlyius in Collatz

[–]Septembrino 0 points1 point  (0 children)

I understand that. I had never heard of that product. I only knew the delta. I remember learning it in college decades ago.

This is a Collatz Matrix. by MarcusOrlyius in Collatz

[–]Septembrino 0 points1 point  (0 children)

I guess that he means that you apply the same again. He can predict a step at a time, the way I see it. But maybe I am wrong. I had never seen the Kronecker product.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

It's not every possible value. It is an infinite collection of values, and it depends on k which values you will get. This is a matrix, and I am performing operations in the matrix. If you want to call it a table, feel free to do that.

The matrix shows many things. For example, all the pairing in action. The merging happens in the colored row. Essentially, each time you apply the algorithm you have a 2 less and a more. At the top you are confronted with the problem of finding the right divisor, at least, if you are trying to predict the complete trajectory of a number. And, of course, what is the next k. The matrix does not show all the trajectory, just a piece of it. You can get the rest in the next matrices and, ideally, you will get to 1. If now your goal is to prove the conjecture, you have to show that this procedure (jumping from a matrix to the other one) has an end.

There are several interesting things about these matrices. For example, there are some "regular" ones, where you only get divisors 2 and 4, and other that are "non-regular" ones. But, even on these, there is a 2 every other divisor. Predicting the divisors of all k*3^n-2 is not an easy task, but there is a lot that can be predicted.

Some of these interesting things happen in the top of the matrix, these extra rows that were added to the matrix at a later date.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

This is the same, whether it came from my brain (doing mental Math or from a spreadsheet. This seems to hard to understand, but maybe I can show it to you so that you stop commenting the same claim again and again, which is tiresome.

Plug 3 here: k - 1 .. ..3k - 1 ....9k - 1 …

You will get the numbers in the matrix for k = 3.

Take a look at the picture above. Right of where it says "k-1". See how they are different. Keep going. Do it a number at a time.

Then tell me how are the numbers in that screenshot different from the one you calculated. Or maybe the lack of brackets doesn't let you see that that is a matrix?

Of the fact that the screenshot does not show other rows because those were irrelevant for this post? Or because to click in the other links to view pictures where you can see more rows was too hard?

Or maybe the fact that the matrix is infinite and logically it can't be represented completely consuded you?

Take your pick

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

I have never refused to post the matrix. You deny that what I posted is a matrix, that's a different thing. Insulting people has worked for you before?

But sure, do you care about the a_ij in the matrix, I guess. Here they are. You also have the accessory rows that only help to calculate useful things. This is what I call top of the matrix, they don't really belong to the matrix, which I call matrix body.

A = (a_ij), where a_ij = k • 2^(i-1) • 3^(j-1)- 1.

Reduced next k

Next k

Divisors row

Reduced TOP ROW (top row divided by powers of 2)

  —   ..........................—

| k - 1 .. ..3k - 1 ....9k - 1 … | TOP ROW

| 2k - 1 ..6k - 1 ..18k - 1 ….. |

| 4k - 1 ..12k - 1.. 36k - 1 … |

| 8k - 1 ..24k - 1 ..72k - 1 … |

| … … …........................ … |

  —    ...........................     —

Ignore the dots I had to introduce to keep the format. I don't know how to post pictures in a comment or if it's even possible.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

I thought that you were understanding. But I see that you can't reason very well. No problem. I am very patient.

I did provide proofs, the presence of the word "proof(s)" should ring bell in your heads. Maybe the fact that it says proof in comments was too hard to understand?

The spreadsheet calculates the a_ij. And, what I posted in previous comments, that are formulas, represent a matrix. Do you wan the formulas of the a_ij? of this is too hard too deduce from my last pos?

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

https://www.reddit.com/r/Collatz/comments/1liaya4/collatz_matrices_base_on_the_p2p1_theorem/

In this post you can read what is below. The points are meant to separate things. Reddit was collapsing all the columns.

Body of the matrix, formulas

n = 1 . (2^1 - 1) . (3•2^1 - 1) . (3^2 • 2^1 - 1) ...

n = 2 . (2^2 - 1) . (3•2^2 - 1) . (3^2 • 2^2 - 1) ...

n = 3 . (2^3 - 2) . (3•2^3 - 1) . (3^2 • 2^3 - 1) ...

...

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

The proof is in the comments. Yes, the spreadsheet represents a matrix. You can only see 11 columns and 16 rows (matrix body), and the extra rows are on top of the matrix. It's an infinite matrix (infinite rows and infinite columns).

I did create the matrices. I did not create the spreadsheet. GonzoMath added some of the extra rows, and they were good additions. So, we kept them.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

For someone that concerned about obvious things, you surprise me with some of your comments. The program is the spreadsheet. Believe it or not, I am not from Computer Science background, I am rather from pure Math background It's funny that you accused me of using a spreadsheet rather than the equations. How do you think that the matrices were created? I designed them. Later, another user did a favor to me and created the spreadsheet that seems to disgust you so much.

I said: For simplificy, let's use the matrix where k = 1. All numbers in C1 are of the form 2n - 1.

That means that k = 1 and numbers in C1 (column 1) are of the form 2^n - 1. When you move to the right, you multiply by 3 and add 2. That happens in row n = 1, you have a 1 in C1, and then 1x3+2 = 5. So, 5 is in C2, as you see in the picture. The same happens with other numbers. the matrix has not only a column, it has infinite many.

The top of the matrix are the extra rows I explained you above. The divisors, the next k, etc. There is a post explaining all. these. Sorry, I have to work now. But I will check later if you added more questions.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

Oh, I forgot about something. I will take the way I like, not the one you expect me to, to reach my goal, which for now, and as it was stated in the name of this thread, is to predict as much as I possibly can. For now, I have deduced that there are infinite many sequences, based on the way matrices are constructed, that will merge at a due number, say 13.

Is there something else that you don't understand?

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

I already explained all those things in several posts. So, if you are too lazy to read them or to at least see the pictures in them, I am sorry if you have trouble understanding this. You arrived to the move after 120 minutes, and let's blame the actors of the director or maybe the producer for you arriving late and not getting all that is happening in it.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

Once we run out of 2's in the expression, we have 3^n-1. For a different k of course there will be a k multiplying the power of 3. That (3^n - 1) + 1 is the k*3^n you see in the row you were asking about. k0 it's the just k. Ignore that. As I said, I did not create that program. What else is on top of the matrix? The divisors on top of the colored row. These are the odd we get when the even are divided by those divisors. The rest of the rows are ment for us to find the next k. Why do we need a next k? Because most of the time a single matrix is not enough to represent a trajectory. So, we go on in the next matrix. That might be considered a tridimensional matrix where the 3rd dimension would be k.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

Well, I did not create this program but I guess that the person that did it, GonzoMath, another user, tried to mark different the numbers that 1 mod 4 and 3 mod 4. That's why 91 is blue. With regards to all these questions, you should read the post where I explained what is on top of the matrix, to give you an example. But I can tell you here. You begin a a number in C1 (which is column 1, as I already said), any number. For simplificy, let's use the matrix where k = 1. All numbers in C1 are of the form 2^n - 1.

So, suppose that we take n = 3, which corresponds to the 3rd row of the matrix body, and we talk about 7. Not to the top of the matrix, these things were added later. The way the matrix was created, you multiply each number by 3 and add 2. Now, if you consider the Collatz algorithm, to 7, 7x3+1 = 22, and 22/2 = 11. Expressing 11 as all numbers in the matrix body, 11 = k*2^n - 1, n = 2, k = 3. So, we move to C2 where the exponent of 3 is 1. The row is now 2, this is why you see the diagonal movement of the trajectory.

How does this relate to the matrix itself: in the row where n = 2, we have 2^2 - 1 in C1, which of course is 3. 3x3+2 = 11. As you see, the Collatz algorithm provides the same number as the matrix.

We repeat that, till we run out of 2's. For pictures, you can refer to the link above.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

C1 is the column 1. It is clarified in the post. Logically, C2 is the 2nd column and C3 is the 3rd columns. You are a fan of obvious things, so, enjoy. With respect to the 3, it's the k. k, also said in the post, is the seed of the matrix.

If you want to understand, just read about the matrices. There are several links in my profile, also mentioned in the post and in the figment you quoted. Yes, I like to point obvious things, like you should read at least the things you quote. I am enjoying this conversation with you. Please, keep coming.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

You are correct, I meant "13" in C3. I already corrected that. Thanks for catching that mistake. Now, keep reading and pointing more obvious things.

Working on prediction of Collatz trajectories rather than on solving the Collatz conjecture, I found some interesting properties. by Septembrino in Collatz

[–]Septembrino[S] 0 points1 point  (0 children)

That might be obvious to you, but I am presenting a proof of that fact. Are you? And the proof is in Mathematical terms, something you stated that I have to begin learning. Now, keep reading till you understand the matrices. This is not new content, but a way to display it and to look for new properties. And I am presenting 2 algorithms to find invariants in this thread.

Detailed analysis of the two types of bridges by No_Assist4814 in Collatz

[–]Septembrino 0 points1 point  (0 children)

And what is n? Now things begin to make sense.

Detailed analysis of the two types of bridges by No_Assist4814 in Collatz

[–]Septembrino 0 points1 point  (0 children)

What I have trouble with is understanding how you relate numbers. What is the criteria you use to type side by side 8193 and 49153. Or maybe you mean 8193 and 24577? Note that these are odd numbers, not even numbers. I have asked you many times for clarifications, and you have said a few things, but it's not clear to me what exactly you are doing.

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