Guest wrote: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 $x_0=1$.

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

guest

Guest wrote:

Guest wrote:

Guest wrote:...
TL;DR: probabilities are tricky, which is why the "proof" was incorrect.

Your arguments are non-mathematical, illogical, and fictional. What we need is a sound mathematical argument (proof) or a conterexample which explains why the Collatz conjecture is true or false. You have not prove not disprove the conjecture... Your efforts are fruitless. Please stop.

He isn't trying to prove or disprove the conjecture, he is trying to show that mariga's proof is incorrect

Guest wrote:

Guest wrote:...
TL;DR: probabilities are tricky, which is why the "proof" was incorrect.

Your arguments are non-mathematical, illogical, and fictional. What we need is a sound mathematical argument (proof) or a conterexample which explains why the Collatz conjecture is true or false. You have not prove not disprove the conjecture... Your efforts are fruitless. Please stop.

Guest wrote:The "proof" isn't correct. People often say that the Collatz conjecture intuitively ought to be true, because half the time you multiple by 1.5 and half the time you divide by 2, so it seems like on average it should be going down. But that isn't a proof. It's just an intuitive argument. And that kind of argument sometimes gives the wrong result.

Here's how to use a similar "proof" to prove something false. Define f(n) to be n+5 if n is a multiple of 10, and n-1 otherwise. You could say that a random n has only a 10% chance of increasing by 5, and a 90% chance of decreasing by 1, and so on average it will decrease. And so this function shouldn't have any cycles at high numbers. But the truth is that EVERY starting number reaches a small cycle!

Why is the "proof" wrong? Because we are not handling the probabilities correctly.

First, you can't say that a random number is odd 50% of the time, unless you first describe how you plan to generate the random number. You have to say what the probability distribution is. And you can't just say "every number is equally likely". That would be a uniform probability distribution over an infinite set, which is mathematically impossible.

Second, even if you could define the probability distribution of n where you're starting, you'd also having to define the distribution after you've iterated the f() function many times. In my example, after iterating f() many times, you're guaranteed to be at a multiple of 10 or within 5 greater than it. Definitely not a number like 107. So the true probability distribution after many steps isn't as simple as it was on the first or second step.

Third, even if you could somehow prove that a random starting number n has a 100% probability of satisfying Collatz, that wouldn't necessarily prove that all numbers satisfy Collatz. This is something weird about probabilities over infinite sets. Just because something is true 100% of the time, that doesn't mean it's true all the time. Yes, that's strange. In math terms, there's a difference between "true with probability one" and "true". A mathematician would say that if you flip a fair coin repeatedly, forever, then you will eventually see heads, with probability 1 (which is a 100% probability), but that you are not guaranteed to ever flip heads. Because it is theoretically possible to flip only tails forever. Flipping tails forever has a probability of zero, but it is not impossible.

TL;DR: probabilities are tricky, which is why the "proof" was incorrect.