I have the proof to Collatz conjecture.

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There you go nerds I found a number that surpasses the four two one trap I'm not sure how far it goes because I found it using the calculator on my phone well over an hour and it seems to be continuing to rise well past the quadrillion mark and has a creeping gain of numbers in small sections. It looks like it could continue on this path to infinity so if you have a program to run the collatz conjecture try this input. Sincerely Anthony Ryan xvaver

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reached 1 after 2744 steps, this is quite fast for such a big number.

btw, if there is a counterexample, then it needs at least "17 087 915" steps (see https://en.wikipedia.org/wiki/Collatz_conjecture)

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In Collatz Conjecture
If x = ⅓ please check
It prove's it as 0

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Full proof -> https://www.researchgate.net/publication/351347153_COLLATZ_CONJECTURE_-THE_PROOF

[Nobody Knows]

Full proof -> https://www.researchgate.net/publication/351347153_COLLATZ_CONJECTURE_-THE_PROOF

[Guest]

Hi,
According to your definition of xn, it is always more than x0. Why are any further steps needed?

Let $x_n$ be an odd positive integer. From the sequence’s formula, $x_{n+1}=\frac{3x_n+1}{2^k}$, $x_{n+2}=\frac{3x_{n+1}+1}{2^m}$ and so forth, $m,k\in \mathbb{Z^+}$
For there to exist a cycle in the sequence, there must exist an odd integer $x_0$ such that $x_n=x_0$.
Let $x_1$, the next odd integer after $x_0$,be given by
$$x_1=\frac{3x_0+1}{2^{a_1}}$$
$x_2$ will be given by
$$x_2=\frac{3x_1+1}{2^{a_2}}$$
$x_2$ in terms of $x_0$ will be
$x_2=\frac{3(\frac{3x_0+1}{2^{a_1}})+1}{2^{a_2}}=\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}}$
$x_3$ will be given by
$x_3=\frac{3x_2+1}{2^{a_3}}=\frac{3(\frac{3x_1+1}{2^{a_2}})+1}{2^{a_3}}=\frac{3(\frac{9x_0+3+2^{a_1}}{2^{a_1+a_2}})+1}{2^{a_3}}=\frac{27x_0+9+3.2^{a_1}+2^{a_1+a_2}}{2^{a_1+a_2+a_3}}$
From the three examples, we can generate a formula for $x_n$ in terms of $x_0$,which will be
$$x_n=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$
Let $x_n=x_0$, hence
$$x_0=\frac{3^nx_0+3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}}{2^{\sum\limits_{n=1}^na_i}}$$
Let $\sum\limits_{n=1}^na_i$ be $k$
$$2^kx_0-3^nx_0=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$
$$x_0(2^k-3^n)=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}$$
$2^k$ can be expressed as $(2^{\frac{k}{n}})^n$
$$x_0[(2^{\frac{k}{n}})^n-3^n]=3^{n-1}+3^{n-2}2^{a_1}+3^{n-3}2^{a_1+a_2}+\cdots +3.2^{\sum\limits_{n=1}^{n-2}a_i}+2^{\sum\limits_{n=1}^{n-1}a_i}\tag{1}$$
When factoring,
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +a.b^{n-2}+b^{n-1})$$
This is same as
$$a^n-b^n=(a-b)(b^{n-1}+b^{n-2}a+b^{n-3}a^2+\cdots +b.a^{n-2}+a^{n-1})$$
We therefore notice the polynomial in equation (1) is regular.
Hence, $x_0[(2^{\frac{k}{n}})^n-3^n]$ can be expressed as,
$$x_0[(2^{\frac{k}{n}})^n-3^n]=x_0(2^{\frac{k}{n}}-3)(3^{n-1}+3^{n-2}2^{\frac{k}{n}}+3^{n-3}(2^{\frac{k}{n}})^2+\cdots +3(2^{\frac{k}{n}})^{n-2}+(2^{\frac{k}{n}})^{n-1})$$
$$=3^{n-1}(2^{\frac{k}{n}}-3)x_0+3^{n-2}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)x_0+3^{n-3}(2^{\frac{k}{n}})^2(2^{\frac{k}{n}}-3)x_0+\cdots +3(2^{\frac{k}{n}})^{n-2}(2^{\frac{k}{n}}-3)x_0+(2^{\frac{k}{n}})^{n-1})(2^{\frac{k}{n}}-3)x_0$$
Each term in this polynomial corresponds to the terms in the polynomial in equation (1).
$\begin{align*}
3^{n-1}\quad &:\quad 3^{n-1}(2^{\frac{k}{n}}-3)x_0\\
3^{n-2}2^{a_1}\quad &:\quad 3^{n-2}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)x_0\\
3^{n-3}2^{a_1+a_2}\quad &:\quad 3^{n-3}(2^{\frac{k}{n}})^2(2^{\frac{k}{n}}-3)x_0\\
\vdots\qquad &:\qquad \vdots \\
3.2^{\sum\limits_{n=1}^{n-2}a_i}\quad &:\quad 3(2^{\frac{k}{n}})^{n-2}(2^{\frac{k}{n}}-3)x_0\\
2^{\sum\limits_{n=1}^{n-1}a_i}\quad &:\quad (2^{\frac{k}{n}})^{n-1})(2^{\frac{k}{n}}-3)x_0
\end{align*}$
Hence, each of these corresponding terms will be equal. Hence,
$3^{n-1}=3^{n-1}(2^{\frac{k}{n}}-3)x_0\\ 1=(2^{\frac{k}{n}}-3)x_0\\ x_0=\frac{1}{(2^{\frac{k}{n}}-3)} $
Substituting this in the next pair of terms,
$3^{n-2}2^{a_1}=3^{n-2}2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)\frac{1}{(2^{\frac{k}{n}}-3)}\\2^{a_1}(2^{\frac{k}{n}}-3)=2^{\frac{k}{n}}(2^{\frac{k}{n}}-3)\\2^{a_1}=2^{\frac{k}{n}}$
Substituting this in equation (2),
$$x_0=\frac{1}{(2^{a_1}-3)}$$
The only value that satisfies this equation such that $x_0$ and $a_1$ are both positive integers is $a_1=2$ to give $x_0=1$.