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Collatz Conjecture Ideas #2

Anonymous Fri 22 Apr 07:28:21 2022 No.14417749 View Reply Original Report

Quoted By: >>14417751 >>14418398 >>14418428

Its the second version
Previous thread
>>14412442

Anonymous

Anonymous Fri 22 Apr 2022 07:29:16 No.14417751 Report

Quoted By: >>14417785

>>14417749
I proved it but lost the paper. :DD

Anonymous

Anonymous Fri 22 Apr 2022 07:56:43 No.14417785 Report

Quoted By:

>>14417751
xD...

Anonymous

Anonymous Fri 22 Apr 2022 15:35:46 No.14418384 Report

Quoted By: >>14418387

Proof of Collatz conjecture: let $f(n)[\math] be collatz function. Then <script type="math/tex">f(2n)[\math] is divisible by <script type="math/tex">(3^2-1)=<script type="math/tex">13. So n=3k+1=1,2,3,6 or 9 for some integer <script type="math/tex">k.$

We should be careful that the Collatz function $f(n)[\math] is well defined on the set of numbers satisfying the conditions for a function, so there is a certain degree of arbitrariness as the "natural definition" of the function <script type="math/tex">f(n)[\math].<br /><br />Since its definition is so complex, I will take as my definition of the Collatz function<br /><br /><script type="math/tex">f(n)=\begin{cases}2n/3^2 & n\equiv0\bmod 3\\ \frac{2n}{3^2}+\frac23 & n\equiv1\bmod 3\\ \frac{2n+1}{3^2}+\frac23 & n\equiv2\bmod 3\end{cases}$

I take as an axiom that Collatz is well defined on the integers.

As for the Collatz conjecture, I will take this as an axiom of Collatz. I know that it was proposed many times, so to me it's not a conjecture anymore but a fact, therefore its negation should also be a fact. So the Collatz conjecture should be read as "any number n that has exactly one prime factor of the form 3k+1 is not of the form 4n, 3n+1, 6n or 7n"

To prove that a number is not of the form 4n, I must take its two prime factors and show that either it is not divisible by 4 or not divisible by either prime.

It can also be shown that

So the case of 4n is proven.

For the other cases, you will have to make a case by case analysis, which will lead to a rather complex proof. It can be avoided by making some strong assumptions that Collatz should be well defined.

Here are two such assumptions.

Collatz is well defined for the naturals. For that to be true, all rationals have to be defined as well. Therefore, we need a well defined Collatz function on rationals.

Anonymous

Anonymous Fri 22 Apr 2022 15:36:53 No.14418387 Report

Quoted By:

>>14418384
Collatz is well defined on the integers. For that to be true, the converse of the Collatz function has to be well defined as well.

Let's go through the proofs for these two cases.

Assume first that Collatz is well defined for the naturals. As I mentioned above, Collatz is not well defined for the rationals, and we also have to define the converse function of Collatz. The converse of Collatz for a positive integer is

$\frac{\mathit{3}^{2\mathit{k+2}}+\mathit{3}^{2\mathit{k+1}}}{2\mathit{3}^{2\mathit{k+2}}+2\mathit{3}^{2\mathit{k+1}}+1}=\frac{\mathit{3}^{2\mathit{k+2}}+\mathit{3}^{2\mathit{k+1}}}{\mathit{3}^{2\mathit{k+2}}-1}$

Similarly, the converse of Collatz for a negative integer is
$\frac{\mathit{3}^{2\mathit{k+2}}+\mathit{3}^{2\mathit{k+1}}}{2\mathit{3}^{2\mathit{k+2}}+2\mathit{3}^{2\mathit{k+1}}-1}=\frac{\mathit{3}^{2\mathit{k+2}}+\mathit{3}^{2\mathit{k+1}}}{\mathit{3}^{2\mathit{k+2}}-1}$

Note that we can't write the converse of Collatz as

$\frac{\mathit{3}^{2\mathit{k+2}}+\mathit{3}^{2\mathit{k+1}}}{2\mathit{3}^{2\mathit{k+2}}+2\mathit{3}^{2\mathit{k+1}}+\mathit{3}^{2\mathit{k+1}}+\mathit{3}^{2\mathit{k+1}}}$

because that would not define a function for negative integers.

I think it's now a good time to make an axiom that Collatz is well defined on the naturals. The converse of Collatz is defined as $\frac{\mathit{3}^{2\mathit{k+2}}+\mathit{3}^{2\mathit{k+1}}}{2\mathit{3}^{2\mathit{k+2}}+2\mathit{3}^{2\mathit{k+1}}+1}$ for positive integers and as $\frac{\mathit{3}^{2\mathit{k+2}}+\mathit{3}^{2\mathit{k+1}}}{\mathit{3}^{2\mathit{k+2}}-1}$ for negative integers.

It is now possible to prove that 4n is not of the form 4n

Anonymous

Anonymous Fri 22 Apr 2022 15:45:45 No.14418398 Report

Quoted By: >>14418830

>>14417749
you aren't gonna solve it with elementary school math

Anonymous

Anonymous Fri 22 Apr 2022 16:00:20 No.14418428 Report

Quoted By:

>>14417749
I once tried to explore the problem like this: If a number c disproves the collatz conjecture, then there exists a minimum number C that disproves the conjecture. Therefore, you only need to prove that all numbers eventually become smaller than themselves, as this would prove there is no minimum, as it reaches numbers which themselves reach 1.

You can begin to eliminate certain forms by starting with the odd numbers 2n + 1 and realizing that 4n + 1 numbers easily shrink. So then we can only consider numbers of the form 4n + 3. But then you get 16n + 7, 16n + 11, 16n + 15, and it becomes increasingly complex. I once wrote a program to do the work for me, and it just seemed like a waste of time. However, given a powerful computer this could be an efficient way of finding a minimum number for which all numbers below it seem to prove the conjecture.

I wondered if there could be a way to prove that you could generate these forms forever. But in the case that n is 0, perhaps we come across some form (2^k)n + m, where m is actually C.

Anonymous

Anonymous Fri 22 Apr 2022 18:58:42 No.14418830 Report

Quoted By:

>>14418398
ik so

Anonymous

Anonymous Sat 23 Apr 2022 04:12:39 No.14420087 Report

Quoted By:

https://www.researchgate.net/publication/330786705_A_topological_approach_to_the_Ulam-Kakutani-Collatz_conjecture
> In fact we prove that if $(\mathbb{N}, \mathcal{T}_f)$ is not a $w–R_0$ space is equivalent to saying that the conjecture is true, $\mathcal{T}_f$ being the topology on $\mathbb{N}$ given by the open sets as those subsets $/theta$ of $\mathbb{N}$ such that $f^{-1}(\theta) \subset \theta$ , where $f$ is the Collatz function.

Anonymous

Anonymous Sat 23 Apr 2022 08:02:17 No.14420367 Report

Quoted By:

Why do mathtards care so much about exploring the properties of what's essentially a very shitty PRNG? Is it autism? Do they really have nothing better to work on?

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