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.

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?)

Well, a proof could be derived from it. Since the two variables of U and L are prime(assuming), then it could try to prove that after a certain amount of numbers that are derived from the Mersenne formula that it will be prime or not prime. For example, I used the new formula to get more primes than the Mersenne formula, however it took the increase of the exponent n for the 4 values to do so. After a certain while, you have to increase n in order to get more primes as you increase the size of the primes.

Collatz Theory is a newer form of mathematics(recent). I developed it originally to try to solve the Collatz conjecture, but I found more than I bargained for with Collatz Theory. For example, I have developed algorithms for the factorizations of composites consisting of two primes, the Mersenne hailstone sequences, hailstone series equations, and Collatz-Matrix equations. The most helpful for this problem are the Mersenne hailstone sequences, hailstone series equations, and Collatz-Matrix equations. I am getting really close to finding some equation to easily prove a Mersenne number to be a prime. It will just take some more analysis.

The modified equation that you presented is actually more efficient than the Mersenne formula based on this sample: While, in this set, 56% of all using the new method are prime while for the other it is 22%. Well I applied Collatz Theory to Mersenne primes and have made some progress, but I am trying to find properties right now of Mersenne primes that separate them from the Mersenne numbers, which are numbers that just come out of the Mersenne formula whether prime or not.

Well one can try to find common relationships of each product when two Mersenne primes are added to 1 and divided by each other. [math]U = 4^{10^{8}}L + 4^{10^{8}} -1[/math] It would be interesting if this equation were true for all Mersenne primes.

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]

So, is there a big chance that the candidate that is presented is a prime? Well, that brings up success for my research.

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]

Using mathematica I haven't find any proof that it is composite, though I could be wrong. I am currently trying to use other methods(it is taking a while).

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]

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.

There is a misunderstanding of conception with your argument. Children do not have a hard time determining right from wrong, rather they have a hard time determining how to react to such concepts. For example, a child may know that doing something may be wrong, however how they react to it is determined by how society teaches them. Parents teach their children how to react to certain feelings, emotions, and even situations. Everyone is born with such forms of morality(whether instinctive or of some other form), however how you react to such moral concepts is up to the society that the child lived in . But the problem with your argument is you would have argued the same way if he did not sacrifice himself for the others. The problem with the survival argument is how broad it is. Yes, evolution deals with survival of the fittest, but if either decision would have lead to the same conclusion then your argument is inconsistent. If he did not die for the others and survived himself he was for his own survival and tried to carry off his own offspring in one form or another. You could also argue that he helped the others for the survival of his species. Also, I find that this scientific research(at the moment) should remain inconclusive about whether humans have a free choice mechanism or not. Yes they may have been able to predict the thoughts and actions of a person within a 10 second range, but you must realize that we are built in with a unnoticeable ability to predict actions in the first place based on previous actions that had occurred. I think the experiments are inconclusive until further results with less area for error are presented.

I have tested many factors with this candidate and so far it is showing to be prime. I have even tested the value with [math]2^{57885161}- 1[/math].

I am using Mathematica to check this. It is a pretty useful tool, especially with prime numbers. Though it takes a few minutes to do, it is efficient. I will try that. I'll tell what I find. Well until I heard about what John was talking about I was going to use a ratio system. I don't really care if this is a prime number, though of it is then okay. This is merely just to the benefit of learning about prime numbers in general. EDIT: 4126162577 is not a factor of the number. I'll keep searching for one.

Though thank you for the information, it does not prove this number I have here not to be prime(unless someone has proven that it is not prime, which then I will discontinue my tests).

Merely gut instinct more than anything. However, I have been making an analysis of the prime numbers through out history and I noticed a pattern of ratios. The ratio between the variable p of Mersenne primes and the amount of digits they have seems to be close to 3.32. I am trying to determine if there is a constant, and coincidentally this value seems to follow the ratio. UPDATE: I am still going up the prime list to see if it remains prime, and it is still succeeding(slowly, very very slowly).

Well, so far I have tried it with factor such as 3, 11, and 13 and so far it is passing the test. I don't know what I am accomplishing by doing this, though if this is truly a prime then it is the largest known prime now.

So, I tried to determine if this is a prime number: [math]2^{257885161} - 1[/math] If this is a prime number, it will be the largest known prime, but if not then okay. Is there a quick way to tell? I tried dividing it by other numbers, yet after a long, long set of calculations I have found no factors yet(though I did this in Mathematica so it may be wrong).

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.

It is probably a too long of title, but this question loomed in my mind. We all know that 3 dimensional geometric volumes have a finite size(unless told otherwise), and 2 dimensional shapes are made up of lines that are infinitely thin. However, 3 dimensional volumes such as cylinders are made up(theoretically) of 2 dimensional circles that are stacked upon each other until a given height, yet 2 dimensional shapes with lines that are infinitely thin are able to make up a certain height. How is this possible? I mean the formulas for determining surface area and volume are based on these principles, though I could be wrong.

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.

The problem with that is no one would be willing to do that(because children have been mainly the shooters in all the shootings that do occur). I agree that armed security guards should be used instead of armed teachers, though teachers should be allowed if given approval by the state, to have a concealed fire arm for the protection of students. I think we are getting to hyped on the gun killings to be honest. More people are killed by other objects(such as cars, whether incidental or not), than gun murder. It is merely political fighting that drives hysteria within our communities. And also, the higher murder rates could be due to many factors, not just weaponry. As economic downfall occurs, there is more "anger". You can look at statistics and compare the economic progression to the amount of killings that year or span of years. Again, I feel this is all due to hysteria. I am not ignoring the issue, however I am simply pointing out that these aren't the only issues.

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]

Is it because he was shipwrecked and since he starved he ate his ship mates by making them into soup and ever since he ate the turtle soup he remembers something about the taste and feels guilty?