Collatz-Matrix Equations(Concept by me)

[Unity+]

Well, this is probably thought of already, but might as well give people the concept.

This work is a small fraction of Collatz Theory

So, basically, the idea is that it is an "equation" that lies on multiple dimensions(two dimensional). It is an evolving equation matrix, meaning on an infinite scale it involves infinitely.

Here is the notation of a Collatz Matrix equation:

[math]C(x)_{k\times d}\begin{Bmatrix}a_{f}&b_{f}\\u_{f}&v_{f}\end{Bmatrix},s(k_p,d_p) = A_{k\times d}[/math]

In this case, the four variables that end with the subscript f are the formulas used in the equation to calculate the matrix. In each case, if the number were to be moved up the matrix, the original number stays at its coordinate while, when moving up the value, the number has to follow the rules of the [math]a_f[/math] variable formula. If the number is to be moved left, the value must follow the formula within the [math]b_f[/math] slot, if the number were to be moved right it must follow the rule or formula in the [math]u_f[/math] slot, and then when moving down the number must follow the rule or formula in the [math]v_f[/math] slot. The starting coordinate is the end coordinate in the notation. The x variable refers to the number being used in the whole entire Collatz Matrix equation.

Now, the reason why I developed this is it originally was a game I made, where it follow the same rules. It used some of the rules of the Collatz Conjecture.

http://en.wikipedia.org/wiki/Collatz_conjecture

What I found interesting about the idea was the fact that within the Collatz Matrix equation the numbers next to a number in a slot(not in the diagonal directions) were approximately equal. I thought maybe a Collatz Perfect Matrix, which is a matrix following the Collatz Matrix Equation that has values working together equally being equal to the one that is either up, down,left, or right of it.

I still have not answered the question, but the question is:

Does a Collatz Perfect Matrix exist?

This question probably isn't important, so I just want thoughts on this concept.

If your a bit confused, here is a link to the game: http://www.thescienceforum.com/general-discussion/35844-collatzs-matrix-game.html

The question or problem only applies for Collatz's rules, no other rules.

[math]C(x)_{k\times d}\begin{Bmatrix}\frac{x}{2}&\frac{x-1}{3}\\3x+1&2x\end{Bmatrix},s(k_p,d_p)\ = A_{k\times d}[/math]

X in the formula rules refers to the evolving x value, while the x next to the C refers to the initial x value.

And the question also refers to a matrix that [math]k=d[/math] and [math]k > 2, d>2[/math]

There are also Collatz Numbers, which are numbers that in all directions work for the Collatz Matrix Equation, though this does not mean that they are numbers that can form Collatz Perfect Matrices.

Here are a few Collatz Numbers:

[math]4,10,16,22...[/math]

Here is a formula to find Collatz numbers if given a Collatz number:

[math]C_{n} = c_{n} + 6[/math]

Also....

[math]c_{n} = 6n-2[/math]

Here are some conjectures that I put up about Collatz numbers.

Collatz Number Conjecture: All Collatz Numbers are even

Collatz Differential Number Conjecture: All Collatz Numbers can be found by adding 6 to another Collatz number.

Collatz Sequence Number Conjecture: The Collatz Numbers will always be found when applying the Collatz Rules(Collatz Conjecture) to a Collatz Matrix Equation.

Collatz Initial State Conjecture: For any number being processed through the Collatz Rules, the number as a result as the first step being an odd value will always be a Collatz Number.

Collatz End Number Conjecture: For the first step value, if [math]C_{n} = x - 6,C_{n}\geq 4[/math] is done repeatedly until it equals 4, than the Collatz process will reach 1.

Collatz Rule Perfect Matrix Hypothesis: If a Collatz Perfect Matrix can be formed, one can only be formed with the Collatz rules.

Collatz 3x3 Perfect Matrix Conjecture(Questionable): A Collatz Perfect Matrix can only be formed in a 3x3 matrix

Here is some Mathematica code(thanks to another user).

Clear["Global'*"]
out = ""
up[x_] := If[Divisible[x, 2], x/2, 0];
left[x_] := If[Divisible[x - 1, 3], (x - 1)/3, 0];
right[x_] := 3 x + 1;
down[x_] := 2 x;
dimension = 4;
x = 2;
c = Table[0, {a, 1, dimension}, {b, 1, dimension}]
MatrixForm[c]
row = 2; col = 2
c[[row]][[col]] = x;
MatrixForm[c]
iterate[c_, row_, col_] := 
  Module[{x, deadEnd, upM, downM, leftM, rightM},
   x = c[[row]][[col]];
   (*Print["here it is:",MatrixForm[c]];*)
   upM = c; downM = c; leftM = c; rightM = c;
   deadEnd = 1;
   If[row != 1 && c[[row - 1]][[col]] == 0 && up[x] != 0,
    (*If we're not at the top, 
    and we haven't filled in the cell above, 
    and the 'up' calculation gives an integer, 
    construct the new matrix, 
    and pass it recursively to the algorithm.*)
     upM[[row - 1]][[col]] = up[x];
    deadEnd = 0;
    iterate[upM, row - 1, col]];
   If[row != dimension && c[[row + 1]][[col]] == 0 && down[x] != 0, 
    downM[[row + 1]][[col]] = down[x];
    deadEnd = 0;
    iterate[downM, row + 1, col]];
   If[col != 1 && c[[row]][[col - 1]] == 0 &&  left[t] != 0, 
    leftM[[row]][[col - 1]] = left[x];
    deadEnd = 0;
    iterate[leftM, row, col - 1]];
   If[col != dimension && c[[row]][[col + 1]] == 0 && right[x] != 0, 
    rightM[[row]][[col + 1]] = right[x];
    deadEnd = 0;
    iterate[rightM, row, col + 1]];
   (*If we were at a dead-end up, down, left, and right, 
   then reveal the matrix*)
   If[deadEnd == 1, Print[MatrixForm[c]];
    out = out <> ""
    ];
   ];
iterate[c, row, col]
out

Thanks to the code, I made some fascinating findings.

I made a Collatz-Matrix algorithm that can take a prime product and break it up into its two different factors. Right now, it is inefficient because there is currently no equation formed yet to retrieve the values of specific coordinates related to the value inputed. Also, the increase in the size of the prime product affects how large the matrix solutions must be in order to retrieve the prime factors involved in the prime products. Another inefficiency is the size of the matrix solutions at the moment.

However, if the equations can be found to describe the values within the matrix solutions this maybe a new find of breaking down prime products.

For example, for the sake of simplicity, let us take the two prime numbers 13 and 3 and multiply them together(the product is 39). Now, the algorithmic equation would look like the following:

[math]C(39)_{5 \times 5}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1&2x \end{Bmatrix},s(3,3)[/math]

In many of the solutions, one of the prime factors comes up once or twice. This means this will take less time to find the prime factor within the "jumble" of numbers that have resulted from the solutions. With my Collatz determinant equations, I may be able to form an equation that will predict the locations of the factors within the matrix solutions. If so, this may be revolutionary.

Another thing to do to clean up the mess would be to set up a set that consists only one of each value from all the solutions(and then only include prime numbers into the set). And then you could eliminate all values that are not prime.

Here would be the result:

[math]S_{1} \left \{ 29,3,19,117,353,89,59,1423,157,13,229,119,79,241,11,103,7,101,67,17,709 \right \}[/math]

Primes larger than the initial value also can be eliminated:

[math]S_{2} \left \{ 29,3,19,13,11,7,17\right \}[/math]

And one could run through these possibilities to test for the factors.

There is also something I need to clarify.

The zeros in the solutions do not necessarily mean they are 0(I know it is kind of paradoxical, but for the sake of argument). Though 0 can refer to the fact that element was a result of the parameters within the equations, the 0 can also mean a null-zero element(meaning it is non-existent as a part of the matrix solution). More definition will be provided later.

Here is a file with mathematical properties of the concept: https://www.dropbox.com/s/n8o6dauuxq8i8s0/Collatz%20Theory.odf(requires Open Office Math).

Here is another problem to be addressed by the Collatz-Matrix Prime Factorization algorithm.

Since, by prediction, the size of the matrix solutions will increase as the size of the composite number gets larger the amount of matrix solutions that are produced by the Collatz-Matrix equation will increase because an increase in the size of the matrix solutions will bring a larger amount of solutions available(though there are exceptions). However, the problem is still pending because I haven't measured to a point where if the ratio of the size of the number to the size of the matrix solutions to the amount of matrix solutions will be linear, exponential, or a wave function.

[math]C(p_{1} \times p_{2})_{k\times d}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1&2x \end{Bmatrix},s(k-3,d-3)[/math]

[math]x=P=p_{1}\times p_{2}[/math]

[math]k=d[/math]

[math]A = k\times d=k^{2}=d^{2}[/math]

[math]A:P[/math]

[math]k^{2}: P[/math]

[math]d^{2}: P[/math]

[math]k^{2}:x[/math]

[math]d^{2}:x[/math]

[math]k^{2}: p_{1}p_{2}[/math]

[math]d^{2}: p_{1}p_{2}[/math]

[math]k:\sqrt{P}[/math]

[math]d:\sqrt{P}[/math]

[math]k:\sqrt{x}[/math]

[math]d:\sqrt{x}[/math]

[math]r=\frac{A}{P}=\frac{A}{p_{1}p_{2}}=\frac{k^{2}}{p_{1}p_{2}}=\frac{d^{2}}{p_{1}p_{2}}=\frac{k^{2}}{P}=\frac{d^{2}}{P}[/math]

Where the function for these proportions is defined by

[math]\Gamma (x) [/math]

I am still working on the function, but this function is defined as such. One to keep in mind is this function does not refer to the proportion between the size of the matrix solutions and how many that will occur, but the composite and the minimal size of matrix solution needed, where k = d, to find the factors of the composite.

Though, here are a few questions that must be asked:

1. Is the relation between the size of the matrix solution and the number of matrix solutions exponential, linear, wave-like, all, or none?

2. For the Collatz-Matrix Prime Factorization algorithm, is the relation between the size(or magnitude) of the composite number and the size of the matrix solution linear, exponential, wave-like, or neither?(If neither, then the function of [math]\Gamma (x) [/math] does not exist). Simply put, the question asks whether there is a function [math]\Gamma (x) [/math] to describe the relationship between [math]P[/math] and [math]A[/math].

3. In relation with the first question, is growth dependent on the initial position of x?

I forgot to mention that x refers to the size of the composite made up of two primes and two primes only, where a composite is referred to as [math]s_{n}[/math], where n is the index of the composite number made up of two primes.

[math]C(2\times 2)_{2\times 2}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1 & 2x \end{Bmatrix},s(2,2)[/math]

[math]\begin{bmatrix} 0 &2 \\ 0 &4 \end{bmatrix}[/math]

[math]\begin{bmatrix} 0 &0 \\ 1 &4 \end{bmatrix}[/math]

Now, using the proportions given above, here is the result:

[math]P = 4[/math]

[math]A = 4[/math]

[math]r=\frac{4}{4}[/math]

[math]r=1[/math]

Here is another, which is 6:

[math]C(2\times 3)_{2\times 2}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1 & 2x \end{Bmatrix},s(2,2)[/math]

[math]\begin{bmatrix} 0 &3 \\ 0 &6 \end{bmatrix}[/math]

Now, using the proportions given above, here is the result:

[math]P = 6[/math]

[math]A = 4[/math]

[math]r=\frac{4}{6}[/math]

[math]r=\frac{2}{3}[/math]

Here is a problem, however, that arises from proving the second question, whether it is yes or no.

Since with function [math]\Gamma (x) [/math] there requires an introduction of [math]s_{n}[/math] which represents a composite of two primes, where n represents the index of such a composite, and since there are infinite primes that exist, which is proven by Euler: Euler?s analytic proof of the infinitude of the primes | Beyond the Numbers(This is just a source out of many that give the article of his proofs), there would be no known way to make an analysis of the growth of the matrix solution sizes with the composite numbers of two primes.

Since there are an infinitesimal set of prime numbers, [math]\Gamma (x) [/math] would be hard to be proven to exist because there is no known way(as I know of) to prove that an exponential, linear, or wave-like growth in the relationship between [math]P[/math] and [math]A[/math] exists. The only way to prove such a relationship would be to use Calculus(I think) and the concept of infinity in some fashion.

Instead of looking at each semi-prime in order, it seems more consistent to observe each prime factor with another prime factor that is larger than it. For example, the first prime to analyze would be 2. Now, the tricky thing with this function is since there are infinite primes in analyze with the prime 2 this causes a problem with trying to analyze the primes precisely.

Here is a graph of the ratios(not the function that I am looking for though).

[math]P_n = \frac{1}{p_n},\frac{1}{p_{n+1}}, \frac{1}{p_{n+2}}, ...[/math] where [math](p)_\mathbb{N}[/math] is the normal sequence of prime numbers.

Then [math]\Gamma: \mathbb{N} \to \mathbb{Q}^\mathbb{N}: x \mapsto \Gamma(x) = \frac{A}{p_x} P_{x}[/math].

[math]\Gamma(1) = \frac{A}{p_{1}p_{1}}, \frac{A}{p_{1}p_{2}}, \frac{A}{p_{1}p_{3}}, ...[/math]

[math]\Gamma(2) = \frac{A}{p_{2}p_{2}}, \frac{A}{p_{2}p_{3}}, \frac{A}{p_{2}p_{4}}, ...[/math]

If the [math]\Gamma (x)[/math] function can be found, an equation can be formed to predict the amount of solutions for the Collatz-Matrix Prime Factorization equation because the amount of matrix solutions is dependent on the size of the matrix solutions. This can also be used to predict the efficiency of the equation.

The following, if this is true, would be the equation for finding the area of a matrix solution from a Collatz-Matrix equation(this only applies to Collatz-Matrix Prime Factorization equations).

[math]A = \Gamma p_{x}p_{n}, x\geq n[/math]

Which is also...

[math]A = \Gamma P[/math]

[math]P = \frac{A}{\Gamma}[/math]

Here are a few more equations that can be formed, though they are irrelevant(at the moment).

[math]p_{x} = \frac{A}{\Gamma p_{n}}, x\geq n[/math]

[math]p_{n} = \frac{A}{\Gamma p_{x}}, x\geq n[/math]

One thing I noticed while graphing this function is how as the semi-primes get larger for the Collatz-Matrix Prime Factorization equation the more primes that occur through out the matrix solutions that exist before the prime numbers that make up the semi-prime within the equation.

Edited July 23, 2013 by Unity+

[Unity+]

Is anyone going to comment? More work coming.

[studiot]

 

Is anyone going to comment? More work coming.

 

 

Well I'd like to thank you for bringing something I'd never heard of to my attention.

 

There really are some fascinating relationships in number theory.

[Unity+]

 

Well I'd like to thank you for bringing something I'd never heard of to my attention.

 

There really are some fascinating relationships in number theory.

Thank you for commenting, I thought no one was paying attention to it.

 

Collatz Theory(or this part of it) was originally meant to try to solve the Collatz Conjecture, but once I found more there seemed to be more relationships between primes and other types of numbers that it expanded.

[Unity+]

Here is a more precise graph to replace the graph above:

 

Though it seems as if y is approaching 0 and that there is a function to describe the slope as seen in the graph, there is still no proof of whether this will continue onto infinity. This proof would include a function that would describe this graph. The one problem with developing this function of [math]\Gamma (x)[/math] is being able to describe the exception of the prime number 2, which is described by [math]p_{x}[/math], where [math]x=1[/math].

 

Now, one could argue that the reason for the exception of 2 is due to the parameters given by the Collatz-Matrix equation, which is that there is a parameter for [math]\frac{x}{2}[/math]. Though this is a simple explanation, I still find it quite interesting as it being the exception.

 

Here is something Daedalus brought into mind, which sparked an idea with Hailstone sequences. Here are examples:

 

{1, 4, 2, 1} -> 1 HE: 1

{2, 1} -> 0 HE: 1

{3, 10, 5, 16, 8, 4, 2, 1} -> 3 HE: 1

{5, 16, 8, 4, 2, 1} -> 2 HE: 1

{6, 3, 5, 16, 8, 4, 2, 1} -> 2 HE: 2

{7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} -> 7 HE: 2

{9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} -> 8 HE: 3

{11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} -> 6 HE: 2

{13, 40, 20, 10, 5, 16, 8, 4, 2, 1} -> 4 HE: 1

{15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1} -> 8 HE: 1

{17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} -> 5 HE: 2

{19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} -> 9 HE: 2

{21, 64, 32, 16, 8, 4, 2, 1} -> 3 HE: 1

{23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1} -> 9 HE: 1

{25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} -> 10 HE: 3

{27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1, } -> 49 HE: 12

{29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} -> 8 HE: 2

 

(HE: Hailstone Exceptions, Hailstone remainders)

 

Now, notice here that each Collatz number is placed on an even number element within each Hailstone sequence, though there are some that are placed on odd elements. Each Collatz element has a pattern(where a Collatz number will be found next to another number that is next to a Collatz number).

 

However, there are what are known as Hailstone exceptions, which are sequences of two numbers that break the sequence pattern. For example, {26, 13} is a Hailstone exception. Hailstone exceptions will always be located in sequences that have an odd number length. That is one property of Hailstone sequences(of course this set of Hailstone sequences are prime numbers). These Hailstone exceptions are important because they do something interesting with the placement of Collatz numbers within these sequences.

 

For example, {7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1} has an odd number of elements, therefore the Hailstone exception {26, 13} is found(there is currently no equation to predict the elements of the Hailstone exceptions). Collatz numbers, before a Hailstone exception, will always be located on even numbered indexes. However, after an exception there is a switch. Collatz numbers will begin to be located on odd-numbered placements while the other numbers are located on even numbered placements. If another Hailstone exception occurs within the sequence then the properties are flipped again.

 

Here is dealing with Hailstone exceptions:

 

{25, 76, 38} [math]\rightarrow[/math] {19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26} [math]\rightarrow[/math] {13, 40, 20, 10, 5, 16, 8, 4, 2} = {1} HE: 3

 

{9, 28, 14} [math]\rightarrow[/math] {7, 22, 11, 34, 17, 52, 26} [math]\rightarrow[/math] {13, 40, 20, 10, 5, 16, 8, 4, 2} = {1} HE: 3

 

{27, 82, 41, 124, 62}[math]\rightarrow[/math]{31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242} [math]\rightarrow[/math]{121, 364, 182, 91, 274, 137, 412, 206}[math]\rightarrow[/math]{103, 310, 155, 466, 233, 700, 350}[math]\rightarrow[/math]{175, 526, 263, 790, 395, 1186, 593,1780, 890} [math]\rightarrow[/math]{445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850} [math]\rightarrow[/math]{425, 1276, 638}[math]\rightarrow[/math]{319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154}[math]\rightarrow[/math]{577, 1732, 866}[math]\rightarrow[/math]{433, 1300, 650}[math]\rightarrow[/math]{325, 976, 488, 244, 122}[math]\rightarrow[/math]{61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2} = {1} HE: 12

 

The Hailstone remainder determines how many sub sequences exist within a Hailstone sequence.

Edited July 24, 2013 by Unity+

[Unity+]

Though this is work-in-progress, there is a new development called Set-Sequence Mathematics(Collatz-Raymond(since it was inspired by Daedalus) Arithmetic). Here are some findings from the use of Hailstone sequences.

 

Here are some examples:

 

{2, 1} -> 0 HE: 1

{4, 2, 1} -> 1 HE: 1

{6, 3, 10, 5, 16, 8, 4, 2, 1} -> 3 HE: 2

{8, 4, 2, 1} -> 1 HE: 1

{10, 5, 16, 8, 4, 2, 1} -> 3 HE: 1

{12, 6, 3, 10, 5, 16, 8, 4, 2, 1} -> 3 HE: 2

{14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1}-> 7 HE: 2

{16, 8, 4, 2, 1}-> 2 HE: 1

{18, 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 } -> 8 HE: 3

{20, 10, 5, 16, 8, 4, 2, 1} -> 3 HE: 1

 

Let us say we have two sequences:

 

A{} and B{}

 

Now, formatting the sequences for Hailstone equations:

 

 

Where H represents the amount of hailstone remainders or hailstone exceptions within a given sequence and n represents the amount of Collatz numbers within a given sequence.

 

Now, let us compare this to normal multiplication:

 

[math]A \times B = C[/math]

 

Now make this for sequences:

 

 

Here is what it would look like:

 

 

For example:

 

 

Now carry out the operations:

 

 

 

And this statement is true:

 

{12, 6, 3, 10, 5, 16, 8, 4, 2, 1} -> 3 HE: 2

 

Here's another example:

 

{2} = {1} -> 0 HE: 1

{9, 28, 14} [math]\rightarrow[/math] {7, 22, 11, 34, 17, 52, 26}[math]\rightarrow[/math] {13, 40, 20, 10, 5, 16, 8, 4, 2} = {1} -> 8 HE: 3

 

{18} [math]\rightarrow[/math]{9, 28, 14} [math]\rightarrow[/math]{7, 22, 11, 34, 17, 52, 26}[math]\rightarrow[/math]{13, 40, 20, 10, 5, 16, 8, 4, 2} = {1} -> 8 HE: 3

 

This mathematical construct, however, is incomplete because there are still some loops that need to be filled with the arithmetic form.

 

These Hailstone equations are a predictive concept to the Hailstone formulas as shown in the previous post. This ability to predict Hailstone formulas with Hailstone equations shows that there are specific sequences known as identity sequences that are located in all Hailstone sequences. This ability to predict them would give a better understanding of the Collatz conjecture.

Edited July 24, 2013 by Unity+

[Unity+]

Based on my analysis of numbers within Set-Sequence arithmetic, there are some special properties that need to be addressed.

 

1. Identity numbers: These are numbers like 1 and 2. Within Set-Sequence, they would be addressed as [math]1\left \{ \right \}^{1}_{1}[/math] and [math]2\left \{ \right \}^{1}_{0}[/math]. Identity numbers are what give other numbers their identity as Hailstone sequences. In fact, if A or B is equal to 1, then multiplication applies differently, where the following would occur.
2. Non-Communicative: All numbers within this arithmetic are not flexible as they are in regular multiplication. There are unique factors for unique values.
3. Property of powers: If one is to increase a value A by the power of n, here would be the following equation to represent the change in B, being the resulting value: . However, this specific equation of the property only applies to the identity of 2.

More to come.

Edited July 25, 2013 by Unity+

[Unity+]

Here is a conjecture to present for Set-Sequence arithmetic.

 

Collatz Number Hailstone Conjecture: For all Collatz numbers, defined by 6n - 2, H of a Hailstone sequences will always be equal to 1. Disproven at 22.

 

Rewritten Collatz Number Hailstone Conjecture: For any Collatz number that has an index within the sequence of Collatz numbers of a Collatz number, the Hailstone sequence will have an HE bigger than 1.

 

Collatz-Raymond Hailstone Conjecture: For any Hailstone equation, iff C is not a Collatz number.

Edited July 26, 2013 by Unity+

[Unity+]

Here is an extension of the Collatz-Raymond Hailstone Conjecture: Iff C is a Collatz number and A is the identity 2, here would be the equation:

 

[math]2\left \{ \right \}^{1}_{0} \times B\left \{ \right \}^{H_{2}}_{n_{2}} = C_{n}\left \{ \right \}^{H_{2} + n_{1}}_{H_{1}+ n_{2}}[/math]

 

Here is an equation for division:

 

[math]\frac{A\left \{ \right \}^{H_{1}}_{n_{1}}}{B\left \{ \right \}^{H_{2}}_{n_{2}}} = C\left \{ \right \}^{H_{1} \times n_{2}}_{H_{2}+ n_{1}}[/math]

 

And this is the equation for powers so far:

 

 

And for square roots:

 

[math]\sqrt{A\left \{ \right \}^{H}_{[\frac{n}{2}]}} = B\left \{ \right \}^{H}_{[\frac{n}{2}]+[\frac{n}{2}]}[/math]

Edited July 28, 2013 by Unity+

[Unity+]

Combining Collatz-Matrix equations and Raymond Arithmetic:

 

[math]C(x)_{k\times d}\begin{Bmatrix} a_{f} &b_{f} \\ u_{f}& v_{f} \end{Bmatrix},s(k_{p},d_{p}),M_{k\times d}\left \{ \right \}^{H}_{n}[/math]

 

Here is an example:

 

[math]C(1)_{2\times 2}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(1,1),A_{2\times 2}\left \{ \right \}^{H}_{n} = \left \{1,2,7 \right \}^{1}_{0}[/math]

 

Here is an example of addition:

 

[math]N\left \{ \right \}^{2}_{1}=C(1)_{2\times 2}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(1,1),A_{2\times 2}\left \{ \right \}^{1}_{0}+C(1)_{2\times 2}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(1,1),B_{2\times 2}\left \{ \right \}^{0}_{1} = \left \{ 1,2,4,7,8 \right \}^{2}_{1}[/math]

 

One thing to add is one can shorten the equation length by doing the following:

 

[math]a_{f} = \frac{x}{2}[/math]

[math]b_{f} = \frac{x-1}{3}[/math]

[math]u_{f} = 3x+1[/math]

[math]v_{f} = 2x[/math]

 

[math]N\left \{ \right \}^{2}_{1}=C(1)_{2\times 2},s(1,1),A_{2\times 2}\left \{ \right \}^{1}_{0}+C(1)_{2\times 2},s(1,1),B_{2\times 2}\left \{ \right \}^{0}_{1}=\left \{ 1,2,4,7,8 \right \}^{2}_{1}[/math]

 

Here is another example:

 

[math]N\left \{ \right \}^{2}_{4}=C(4)_{3\times 3}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(1,1),A_{3\times 3}\left \{ \right \}^{2}_{3}+[/math] [math]C(4)_{3\times 3}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(1,1),B_{3\times 3}\left \{ \right \}^{1}_{2}=\left \{ 8, 4, 151, 16, 25, 50, 49, 148, 74, 37, 12, 24 \right \}^{2}_{4}[/math]

 

And a more simplified version:

 

[math]a_{f} = \frac{x}{2}[/math]

[math]b_{f} = \frac{x-1}{3}[/math]

[math]u_{f} = 3x+1[/math]

[math]v_{f} = 2x[/math]

 

[math]N\left \{ \right \}^{2}_{4}=C(4)_{3\times 3},s(1,1),A_{3\times 3}\left \{ \right \}^{2}_{3}+[/math] [math]C(4)_{3\times 3},s(1,1),B_{3\times 3}\left \{ \right \}^{1}_{2}=\left \{ 8, 4, 151, 16, 25, 50, 49, 148, 74, 37, 12, 24 \right \}^{2}_{4}[/math]

 

The following is the equation for solving:

 

[math]N\left \{ \right \}^{H_{1}}_{H_{1}+n_{2}}=C(x)_{k\times d}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(k_{p},d_{p}),A_{k\times d}\left \{ \right \}^{H_{1}}_{n_{1}}+C(x)_{k\times d}\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(k_{p},d_{p}),B_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}}[/math]

 

Simplified:

 

[math]a_{f} = \frac{x}{2}[/math]

[math]b_{f} = \frac{x-1}{3}[/math]

[math]u_{f} = 3x+1[/math]

[math]v_{f} = 2x[/math]

 

[math]N\left \{ \right \}^{H_{1}}_{H_{1}+n_{2}}=C(x)_{k\times d},s(k_{p},d_{p}),A_{k\times d}\left \{ \right \}^{H_{1}}_{n_{1}}+C(x)_{k\times d},s(k_{p},d_{p}),B_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}}[/math]

Edited July 28, 2013 by Unity+

[Unity+]

Here is a continuation of the arithmetic:

[math]a_{f} = \frac{x}{2}[/math]

[math]b_{f} = \frac{x-1}{3}[/math]

[math]u_{f} = 3x+1[/math]

[math]v_{f} = 2x[/math]

[math]N\left \{ \right \}^{H_{1}}_{H_{1}-n_{2}}=C(x)_{k\times d},s(k_{p},d_{p}),A_{k\times d}\left \{ \right \}^{H_{1}}_{n_{1}}-C(x)_{k\times d},s(k_{p},d_{p}),B_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}}[/math]

Here is an example of multiplication:

[math]a_{f} = \frac{x}{2}[/math]

[math]b_{f} = \frac{x-1}{3}[/math]

[math]u_{f} = 3x+1[/math]

[math]v_{f} = 2x[/math]

[math]N\left \{ \right \}^{H}_{n}=C(x)_{k\times d},s(k_{p},d_{p}),A_{k\times d}\left \{ \right \}^{H_{1}}_{n_{1}}\times C(x)_{k\times d},s(k_{p},d_{p}),B_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}}= A_{k\times d}\left \{ \right \}^{H_{1}}_{n_{1}}\rightarrow B_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}}[/math]

And here his division:

[math]N\left \{ \right \}^{H}_{n}=\frac{C(x)_{k\times d},s(k_{p},d_{p}),A_{k\times d}\left \{ \right \}^{H_{1}}_{n_{1}}}{C(x)_{k\times d},s(k_{p},d_{p}),B_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}}}[/math]

The main difference between Raymond subtraction and division is with division all the elements of [math]B_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}}[/math] must be located within [math]A_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}}[/math] where [math]A_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}} \setminus \ B_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}} \neq \left \{ \right \} \wedge B_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}} \setminus A_{k\times d}\left \{ \right \}^{H^{2}}_{n_{2}} \neq \left \{ \right \}[/math]

This rule solves Russel's paradox because it points out that the division develops an undefined set. If it develops an undefined set, this undefined set never existed in the first place, therefore the set never becomes a set of itself.

Edited July 29, 2013 by Unity+

[Unity+]

Raymond arithmetic, in fact, gives a definition to 0 and null for the Collatz-Matrix equations.

 

They are represented as [math]0\left \{ \right \}^{0}_{0}[/math], which is zero, and [math]\left \{ \right \}^{0}_{0}[/math], which is null.

 

Null can also be denoted as [math]\O \left \{ \right \}^{0}_{0}[/math].

 

In a Collatzian view of Collatz Theory, all sets or matrix solutions of a given Collatz-Matrix equation have a finite amount of null elements. However, in a non-Collatzian view of Collatz Theory, all matrix solutions and sets have an infinite amount of null elements. Because of these properties, the null element is an undefined element, which is a result of undefined sets from the development of Raymond division.

 

Null elements that are within sets as a part of division do not affect the division result, unless there are only null elements within each set, because each set would have an equal infinite amount of null elements.

Edited July 29, 2013 by Unity+

[Unity+]

Here is a special property I found while investigating Mersenne primes.

 

Take any Mersenne prime and apply it to the Raymond arithmetic and Collatz-Matrix equations.

 

[math]M\left \{ \right \}^{H}_{n}=\overset{C}{\leftarrow}(2^{P} - 1)_{\infty \times \infty }\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(1,1),M_{\infty \times \infty }\left \{ \right \}^{H}_{n}[/math]

(The new notation is that the bottom arrows will always refer to horizontal change and top arrow will refer to be the vertical change).

 

The hailstone formula will always consist of two sections, the Mersenne section and the 2-power section.

 

[math]\left \{M, b, c \right \}\rightarrow \left \{ 2^{n}, \frac{2^{n}}{2},\frac{2^{n}}{4},...1 \right \}[/math]

 

[math]\left \{2^{P}-1, b, c \right \}\rightarrow \left \{ 2^{n}, \frac{2^{n}}{2},\frac{2^{n}}{4},...1 \right \}[/math]

 

Here are some interesting things to be aware of:

 

[math]M-b-c=2^{n}[/math]

 

[math]M-2^{n} = 2^{n} - 1[/math]

 

[math]M-(M-b-c) = (M-b-c) - 1[/math]

 

[math]b+c = M-b-c - 1[/math]

 

[math]2b+2c = M - 1[/math]

 

Also, the length of the Mersenne section will always be 3 while the length of the 2-power section will always be [math]P-1[/math]. HE for Mersenna primes for these Collatz-Matrix equations will always be 2. The amount of Collatz numbers within the 2-power section will always be [math]\frac{P-1}{2}[/math].

 

However, none of these rules apply to the Mersenne prime 3.

 

These rules do not just apply to prime Mersenne numbers, but Mersenne numbers in general.

 

[math]M\left \{ \right \}^{H}_{n}=\overset{C}{\leftarrow}(2^{12P-7} - 1)_{\infty \times \infty }\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(1,1),M_{\infty \times \infty }\left \{ \right \}^{H}_{n}[/math]

 

[math]b\in C_{n}[/math]

 

And

 

[math]M\left \{ \right \}^{H}_{2P}=\overset{C}{\leftarrow}(2^{4P+1} - 1)_{\infty \times \infty }\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(1,1),M_{\infty \times \infty }\left \{ \right \}^{H}_{2P}[/math]

 

Predicting [math]2^{n}[/math]

 

[math]M\left \{ \right \}^{H}_{n}=\overset{C}{\leftarrow}(2^{2P+1} - 1)_{\infty \times \infty }\begin{Bmatrix} \frac{x}{2} &\frac{x-1}{3} \\ 3x+1& 2x \end{Bmatrix},s(1,1),M_{\infty \times \infty }\left \{ \right \}^{H}_{n}[/math]

 

[math]2^{n}=4^{P}[/math]

 

Here is how to find primes using Collatz numbers:

 

[math]P_{C}=2c_{n}-1[/math]

I am still developing a way to predict(easily) how to determine if the number is prime or not.

Edited July 30, 2013 by Unity+

[Unity+]

Correction, the 2-power section will be the length of [math]P-1[/math].

 

And the amount of Collatz numbers within the 2-power section will be [math]\frac{P-1}{2}[/math]

Edited July 31, 2013 by Unity+

[Unity+]

Here is some more work with the hailstone formulas and Mersenne primes:

 

Mersenne Number: [math]2^{P}-1[/math]

 

Collatz Number: [math]C_{n} = 6n - 2[/math]

 

iff [math]P[/math] is odd and [math]P > 5[/math]

 

[math]\left \{ N + 7, b_{1}, c_{1} \right \}[/math]

[math]\left \{ N + 1, b_{2}, c_{2} \right \} \rightarrow c_{2} - 5 = c_{1}C_{1} - C_{1}[/math]

 

[math]\left \{ N + 1, b_{1}, c_{1} \right \}[/math]

[math]\left \{ N + 7, b_{2}, c_{2} \right \} \rightarrow c_{2} - 1 = c_{1}C_{1}[/math]

 

iff [math]P[/math] is even and [math]P > 2[/math]

 

[math]\left \{ N + 5, b_{1}, c_{1} \right \}[/math]

[math]\left \{ N + 3, b_{2}, c_{2} \right \} \rightarrow c_{2} - 5 = c_{1}C_{1} + C_{0}[/math]

 

[math]\left \{ N + 3, b_{1}, c_{1} \right \}[/math]

[math]\left \{ N + 5, b_{2}, c_{2} \right \} \rightarrow c_{2} - 3 = c_{1}C_{1}[/math]

 

 

 

Here is an example of a Hailstone sequence tree for if [math]P[/math] is even:

 

 

This equation to describe [math]b[/math] is the following:

 

[math]C(x)_{k} \begin{Bmatrix} 4x+6\\ \frac{x-6}{4} \end{Bmatrix},s(k_{p})[/math]

This equation to describe [math]M[/math] is the following:

 

[math]C(x)_{k} \begin{Bmatrix} 4x+3\\ \frac{x-3}{4} \end{Bmatrix},s(k_{p})[/math]

This equation to describe [math]c[/math] is the following:

 

[math]C(x)_{k} \begin{Bmatrix} 4x+3\\ \frac{x-3}{4} \end{Bmatrix},s(k_{p})[/math]

Here is an example of a Hailstone sequence tree for if [math]P[/math] is odd:

This equation to describe [math]b[/math] is the following:

 

[math]C(x)_{k} \begin{Bmatrix} 4x+2\\ \frac{x-2}{4} \end{Bmatrix},s(k_{p})[/math]

This equation to describe [math]M[/math] is the following:

 

[math]C(x)_{k} \begin{Bmatrix} 4x+3\\ \frac{x-3}{4} \end{Bmatrix},s(k_{p})[/math]

This equation to describe [math]c[/math] is the following:

 

[math]C(x)_{k} \begin{Bmatrix} 4x+1\\ \frac{x-1}{4} \end{Bmatrix},s(k_{p})[/math]

 

Edited August 1, 2013 by Unity+

[Unity+]

Adding on more to the hailstone proportions:

 

[math]\{ M, b, c\}[/math]

 

[math]M:b:c[/math]

 

Here is the identity hailstone sequence for the even of [math]P[/math]

 

[math]\left \{ 0, 1, \frac{1}{2} \right \}[/math]

 

[math]\begin{bmatrix} \frac{1}{2}&0 \\ 1& 0\end{bmatrix}[/math]

 

Here is the identity hailstone sequence for the odd of [math]P[/math]

 

[math]\left \{ 1, 0,0 \right \}[/math]

 

[math]\begin{bmatrix} 0&0 \\ 1& 0 \end{bmatrix}[/math]

Edited August 2, 2013 by Unity+

[Unity+]

Adding on more to the hailstone proportions:

 

[math]\{ M, b, c\}[/math]

 

[math]M:b:c[/math]

 

Here is the identity hailstone sequence for the even of [math]P[/math]

 

[math]\left \{ 0, 1, \frac{1}{2} \right \}[/math]

 

[math]\begin{bmatrix} \frac{1}{2}&0 \\ 1& 0\end{bmatrix}[/math]

 

Here is the identity hailstone sequence for the odd of [math]P[/math]

 

[math]\left \{ 1, 0,0 \right \}[/math]

 

[math]\begin{bmatrix} 0&0 \\ 1& 0 \end{bmatrix}[/math]

Based on the identity sequences, I speculate that there may be more types of numbers besides evens and odds(maybe even and odd perfect numbers?)

Edited August 3, 2013 by Unity+

[Unity+]

There was something that came to my mind while doing an analysis of the geometric nature of Collatz-Matrix equations.

 

One speculation that Collatz Theory makes is that all single-formed equations are just a point within a space of dimensions(i.e one and two dimensions). Each equation represents such a point and only one point. Collatz Theory states that Collatz-Matrix equations are what represent theoretical geometric shapes using different equations, both the equation and it's inverse. Using this system, a symmetrical, theotrical geometric shape is developed. The matrix solutions of a Collatz-Matrix equation are what construct the geometric shape that is derived from the Collatz-Matrix equations and each element within the matrix solutions represent a point within it. The values correspond to the specific points within the shape that is produced by the Collatz Matrix equation.

 

There are also asymmetrical Collatz-Matrix equations, which consist of an equation and another equation that is not its inverse. They produce almost the same results except decimal values are allowed, which increases the amount of matrix solutions that exist.

 

This brings into mind another speculation about the equation for gravity and the equations for electromagnetism. The reason why the equations for gravity could be so similar to the equations for electromagnetism is one of them is an inverse system to the other. Though, this is mere speculation.

Edited August 3, 2013 by Unity+

[Unity+]

Here are some examples of the applied speculation:

 

 

EDIT: A part of the notation is the arrows about the coordinates. These arrows indicate the limited direction due to the equations that exist within the Collatz-Matrix equation.

This form of theoretical geometry is an indicator of a new way to interpret mathematical information. Since there are infinitely many dimensions that exist(geometrically), there are an infinite spectrum of levels of number systems that exist(though this is only a conjecture). I postulate that number systems geometrically fractalate as the amount of theoretical dimensions increases.

Edited August 4, 2013 by Unity+

[Unity+]

Any comments? Questions? I will be releasing more work on this later on.

[Sato]

You should write a more organized and self-contained paper on this, which one who is interested can look through and discuss, rather than scrolling through multiple not-so-organized forum posts. Looks interesting, but it seems more like a wall of text and images at this point which induces apathy (at least in me).

[Unity+]

You should write a more organized and self-contained paper on this, which one who is interested can look through and discuss, rather than scrolling through multiple not-so-organized forum posts. Looks interesting, but it seems more like a wall of text and images at this point which induces apathy (at least in me).

I apologize about my inability to organize well. More ideas are found and I thought having them in this topic would help me organize it into one single paper. The problem with writing a paper on it is the many conjectures that I have yet to prove.

[Unity+]

Before I complete the paper I am going to release, here is some more properties and conjectures to present.

 

Well, I actually found the Collatz conjecture useful for primes.

 

I began analyzing the numbers of the variable p within the Mersenne formula for finding this prime candidate. Here is what I did(here is the equation).

 

[math]p = \sum_{n=0}^{k}(2^{n+1} - E(k,d))(10^{n})[/math]

 

Where [math]E(k,d)[/math] is an element within a matrix solution of a Collatz-Matrix equation.

 

Here are some patterns I found:

p=257885161
(2^1 - 1) Odd
(2^2 + 2) (2^2 + (2*1)) Even
(2^3 - 7) (2^2 - ((2*2)+3)) Odd
(2^4 - 11) (2^2 - ((2*7)-3)) Odd
(2^5 - 24) (2^2 - ((2*11)+2)) Even
(2^6 - 56) (2^2 - ((2*24)+8)) Even
(2^7 - 121) (2^2 - ((2*56)+9)) Odd
(2^8 - 251) (2^2 - ((2*121)+9)) Odd
(2^9 - 510) (2^2 - ((2*251)+8)) Even


(2^3 - 7) (2^2 - ((2*2)+3))
(2^4 - 11) (2^2 - ((2*7)-3))
(2^7 - 121) (2^2 - ((2*56)+9))
(2^8 - 251) (2^2 - ((2*121)+9))

(2^2 + 2) (2^2 + ((2*1)+0))
(2^5 - 24) (2^2 - ((2*11)+2)) 0-> |6(0) - 2|
(2^6 - 56) (2^2 - ((2*24)+8)) 1 |6(-1) - 2}|
(2^9 - 510) (2^2 - ((2*251)+8))1 |6(-1) - 2}| 

I found these same patterns while working on Raymond Arithmetic. Also, I noticed that the numbers can be produced using Collatz-Matrix equations. If I can some how form an equation, I may be able to form an efficient equation to predict Mersenne primes.

 

I have a conjecture. This will be the Collatz-Prime Predictive Conjecture.

 

Let us say you have the paramters: [math]\frac{x}{a}[/math], [math]bx+c[/math]

 

A prime formula can be written using those two parameters: [math]a^{p}-c[/math], where the predict equation would be:

 

[math]p = \sum_{n=m}^{k}(a^{n+1} - E(k,d))(10^{n-m})[/math]

 

Where k is a specific magnitude.

 

Where m is an even number or 0.

 

The conjecture is that this equation for a specific set of parameters can predict the numbers needed for primality.

 

[math]2^{\sum_{n=m}^{k}(2^{n+1} - E(k,d))(10^{n-m})} - 1[/math]

 

[math]a^{\sum_{n=m}^{k}(a^{n+1} - E(k,d))(10^{n-m})} - 1[/math]

 

Sub-conjecture: Let us say there is this equation:

 

 

And the formula for the prime is the following:

 

 

Then, , which this notation implies that the number will be located in any of the matrix solutions produced by this Collatz-Matrix equation.

 

This is proven slightly by the work above.

 

Here is some more work I did that is the beginning of the other equations provided:

 

[math]\left \{ \frac{x}{a} \right \}\rightarrow \left \{ bx+c \right \}[/math]

 

[math]\left \{ 1 \right \}\rightarrow \left \{ a^{x} \right \}\rightarrow \left \{ a^{2}x+bax+c \right \}\rightarrow \left \{ C \right \}[/math] Where C is the variance in the hailstone sequences.(equations based off of hailstone sequences).

 

[math]\left\{ C \right \}\leftarrow \left \{ q+a^{3}b, w+a^{2}b, e+a^{1}b, p+a^{0}b \right \} \leftarrow \left \{ q,w,e,p \right \}\leftarrow \left \{ 1 \right \}[/math]

 

[math]q = b^{2} +bp + c[/math]

 

[math]a^{\sum_{n=m}^{k}(a^{n+1} - E(k,d))(10^{n-m})} - c[/math]

 

Which can be also written as:

 

[math]a^{\sum_{n=m}^{k}(a^{n+1} - E(k,d))\times 10^{n-m}} - c[/math]

 

[math]\sum_{n=m}^{\infty}(a^{n+1} - E(k,d))\times 10^{n-m} = \infty[/math]

 

Where m is an even number or 0.

 

Which also is referenced in this way:

 

 

 

 

Where the variables [math]d_{i}[/math] and [math]d_{e}[/math] are a part of the Collatzian ratio.

 

[math]d_{r}=\frac{d_{i}}{d_{e}}[/math]

 

[math]d_{c}=\frac{d_{e}x-d_{e}-d_{i}x}{d_{i}} = (a_{f}v_{f})-(b_{f}u_{f})[/math]

 

 

 

Hailstone-Exception Conjecture:

 

Let us say there is the Hailstone-exception:

 

[math]\left \{ a,b \right \}[/math]

 

The variable b will always be prime.

The variable a will always be a semi-prime.

 

Hailstone-Sequence Default Conjecture:

 

If the initial value of a Hailstone sequence(applying the Collatz conjecture) is equal to [math]d_{i}[/math], where the [math]bx+c[/math] is to be [math]\frac{d_{i}x + d_{e}}{d_{e}}[/math], then the Hailstone sequence will look like the following.

 

[math]\left \{ d_{i} \right \}\rightarrow\left \{ q+r^{3}d_{i},w+r^{2}d_{i},e+r^{1}d_{i},p+r^{0}d_{i} \right \}\rightarrow \left \{ q,w,e,p \right \}\rightarrow \left \{ 1 \right \}[/math]

 

Hailstone-Sequence Complete Conjecture:

 

If the initial value of a Hailstone sequence(applying the Collatz conjecture) is equal to [math]d_{i}[/math], where the [math]bx+c[/math] is to be [math]\frac{d_{i}x + d_{e}}{d_{e}}[/math], then the Hailstone sequence will look like the following.

 

[math]\left \{ d_{i} \right \}\rightarrow\left \{ q+r^{3}d_{i},w+r^{2}d_{i},e+r^{1}d_{i},p+r^{0}d_{i} \right \}\rightarrow \left \{ q,w,e,p \right \}\rightarrow \left \{ d_{e} \right \}[/math]

 

These hailstone equations show the relationship between [math]d_{i}[/math] and [math]d_{e}[/math].

Edited August 11, 2013 by Unity+

[Unity+]

Hyper-Dimensional Number System Problem:

 

There exists number systems below above the natural number system(whole numbers), however the question is whether there exists number systems above below the known natural number line.

 

For example,

 

Though one can increase the amount of dimensions to increase the amount of number systems, but can one develop a number system provable below the 1st dimension?

EDIT: This assumes that there is no number system for the 0th dimension.

Here is a continuation of Raymond Arithmetic.

Now, in order to properly carry out the operations, the Collatzian ratio must be found. The actual operations will be carried out on the Collatzian ratio and the variable r.

,[math]\frac{r}{n}[/math]

Which then , would result in the following, like the two dimensional Collatz-Matrix equation.

Edited August 12, 2013 by Unity+

[Unity+]

And here is for multiplication:

[math]n\times C(x)_{r}\begin{Bmatrix} \cdots \end{Bmatrix},s(r_{p})[/math]

[math]\frac{d_{i}n}{d_{e}}[/math],[math]rn[/math]

The next set of operations are the simple of addition and subtraction.

[math]n+ C(x)_{r}\begin{Bmatrix} \cdots \end{Bmatrix},s(r_{p})[/math]

[math]\frac{d_{i}}{d_{e}}+\frac{nd_{e}}{d_{e}}[/math], [math]r+n[/math]

And then there is the subtraction of Raymond Arithmetic:

[math]n- C(x)_{r}\begin{Bmatrix} \cdots \end{Bmatrix},s(r_{p})[/math]

[math]\frac{d_{i}}{d_{e}}-\frac{nd_{e}}{d_{e}}[/math], [math]r-n[/math]

Now onto powers:

[math](C(x)_{r}\begin{Bmatrix} \cdots \end{Bmatrix}, s(r_{p}))^{n}[/math]

[math](\frac{d_{i}}{d_{e}})^{n}[/math], [math]r^{n}[/math]

And here is rooting Collatz-Matrix equations:

[math]\sqrt[n]{C(x)_{r}\begin{Bmatrix} \cdots \end{Bmatrix}, s(r_{p})}[/math]

[math]\sqrt[n]{\frac{d_{i}}{d_{e}}}[/math], [math]\sqrt[n]{r}[/math]

Here are a few properties to notice.

For multiplication, you may never divide or multiply by 0.

For exponentiation, where n is equal to 0...

Where [math]d_{c} = -1[/math].

Here would be an example of a Collatz-Matrix equation to the 0th that is 5x5 and x equals 1.

[math]\begin{bmatrix} 1 & 2 & 3 & 4 & 5\\ 0 & 0& 0& 0 & 5\\ 0 & 0& 0& 0 &5 \\ 0& 0& 0& 0&5 \\ 0 & 0 & 0& 0 &5 \end{bmatrix}[/math]

A Collatz-Matrix equation, like this, could be simplified to a one dimensional Collatz-Matrix equation:

[math]C(x)_{k}\begin{Bmatrix} x-1 & x+1 \end{Bmatrix},s(k_{p})[/math]

This would apply to all dimensions of Collatz-Matrix equations:

In this case, the variable [math]s_{p}[/math] is a special case dimension, which refers to this Collatz-Matrix equation as a trans-linear equation.

It is not proven yet, but I postulate that the if for any determinant of a Collatz-Matrix equation if solving for x when y equals -1, x will always equal 1.

A formula that is used for finding the numbers that will work in all directions in a matrix solution for Collatz-Matrix equations is the following:

[math]d_{i}xr-d_{e}r[/math]

As a side note, if a parameter is simply equal to x, then it is not necessary to include it within the notation. However, if you are determining the determinant of the Collatz-Matrix equation it will be necessary to include it within the parameter brackets to show the existence of these parameters.

Edited August 12, 2013 by Unity+