Criticism of proof attempt: Collatz Conjecture I'm going to start out by saying I'm an amateur, I'm sure I missed something. I know there are lots of people who claim to have solved this. I just want to find out where I'm wrong and what I missed. There's been far more brilliant minds that have worked on this than me. That said, I've tried to write out as formal a proof as I can. It's 9 pages long. I'll link it here, I've saved it on my github.
To summarize it though. I start out by building a function that can give us some odd number down from a starting odd number. That equation is given by.
$$O_{1+m} = 
\frac{ 3^{m}O_{1}}{2^{p_{1} + p_{2} + \dots + p_{m-2} + p_{m-1} + p_{m}}}
 + \frac{ 3^{m-1}}{2^{p_{1} + p_{2} + \dots + p_{m-2} + p_{m-1} + p_{m}}}
 + \frac{ 3^{m-2}}{2^{p_{2} + \dots + p_{m-2} + p_{m-1} + p_{m}}}
 + ...
 + \frac{3^{1}}{2^{p_{m-1} + p_{m}}} 
 + \frac{3^{0}}{2^{p_{m}}}
$$
Where $O_{1+m}$ is the odd number m odd numbers out from where we started. $O_1$ is the odd number we start at. and $p_{i}$ is the number of times a number will have to be divided by two. For example, consider
$$7 \rightarrow 22 \rightarrow 11 \rightarrow 32 \rightarrow 16 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1$$
Then our starting odd number, $O_1$ is $7$, our second odd number $O_2$ is $11$, the number of times we had to divide by two to get from $7$ to $11$ is once, so $p_1 = 1$. The third odd number is $1$ so $O_3 = 1$, because we had to go through $11$ to get there and the number of times we had to divide from $11$ to get to $1$ is $4$, therefore $p_1 = 1, p_2 = 4$.
I then showed that we have three cases, the number increases indefinitely, it loops, or it decreases.
Case 1: Indefinitely Increasing
I showed that the only way we could have the number increase indefinitely from one odd number to some odd number down the line and then some odd number after that, is if our starting odd number $O_1$ is negative. But $O_1 \geq 1$, therefore there are no indefinitely increasing numbers.
Case 2: Looping Numbers
I showed that we can construct an equation which will give us our starting odd number after m transformations. That equation is given by.
$$O_{1+m} = \frac{3^{m-1}2^{0} + 3^{m-2}2^{p_{1}} + ... + 3^{1}2^{p_{1} + p_{2} + \dots + p_{m-2}} + {3^{0}2^{p_{1} + p_{2} + \dots + p_{m-2} + p_{m-1}}}}{(2^{p_{1} + p_{2} + \dots + p_{m-1} + p_{m}} - 3^{m})}$$
I then used prime decomposition and the fundamental theorem of algebra to show that $O_{1+m} = 1$ and $p_{i} = 2$ for all $i$. Therefore the only looping number is $1$.
Case 3: Decreasing numbers
This follows from case 2. Because $1$ is the only looping number and it's also the smallest usable number, any number which consistently decreases over $m$ transformations, must eventually end up at $1$, where it loops indefinitely.
Therefore the Collatz Conjecture is true. Every number will inevitably transform down to $1$.
Sorry for the difficult to read LaTeX, kind of a new user, but feel free to criticize away about anything! And thanks all for everyone's help.
 A: I believe there is a problem in (14)-(15). Firstly I don't understand how the equation (14) holds. What do you mean by plugging in $_2O$ for $_1O$ ? The only equations you have so far are
$$ _iO = A_iO_1 + B_i$$
and I don't see how this enables you to deduce that equation 14 holds. When reading what you wrote it looks like you think that the following equality is true for all $m$
$$ _{m+1}O =  {_n}O A_n + B_n$$
but you haven't proven this (and I don't believe that this equality actually holds). and even if it did hold it is not immediately clear that $(15)$ is correct. The statement "doing the same up to infinity" is vague. You can however do the procedure $m$ times
$$O_1 = \underbrace{\text{negative terms}}_{:= N_m} + \frac{{_mO}}{A_1...A_m} $$
and take the limit for $m \to\infty$. However it does not look easy to show that the  right hand side would converge to something negative since the term  $\frac{{_mO}}{A_1...A_m}$ is positive with $_mO$ diverging to $+\infty$.

And important thing to note is that you in your proof of "Case 1 : odd numbers that transform up forever" you never use the hypothesis that the sequence $_1O,...,{_mO}$ is transforming up forever. That is suspicious and should lead you to believe that you've made a mistake somewhere.
A: May I first commend you on your approach. You wrote down a rather legible proof and then sought out criticism with an open mind. That's how it should be done.
Unfortunately, and as you probably feared, the proof doesn't seem to work. First, note that, if you are trying to prove there are no looping numbers except $1$, you must use the positivity of the starting number somewhere, because $-5$ also loops. I don't see, at first glance, where you use that, so that is a warning sign.
I immediately skipped to section 4 in your paper because, as you say yourself, that's where the meat of the matter is. Your jump from equation 27 to 28 & 29 seems dubious. Why can $1+3O_{3}$ and $2^{p_{2}}O_{3}-1$ not have some prime factors in common? In this case, you can probably still prove that that's impossible with some simple calculations, but when you make the analogous jump from 37 to 38 & 39, this becomes very much non-trivial.
A: Using the more general iterate $mx + a$, where $m,a$ are odd positive integers ($3x+1$ is the classic Collatz iterate), the next odd integer $x_i$ is given by
$$x_i = \frac{mx_{i-1}+a}{2^{p_i}}$$
where $p_i$ is the unique positive integer such that $x_i$ is odd. Note that I am using $x_i$ instead of $O_i$ and that the $i$ index of the $p_i$ matches the index of the next odd integer and not the current odd integer. This makes things cleaner IMHO. If we define $P_i = \sum\limits_{k=1}^{i}p_k$ for $i\gt0$ and $P_0 = 0$ then we obtain the equation:
$$x_n = {x_0 m^n + a \sum\limits_{i=0}^{n-1} m^{i}2^{P_{n - 1 - i}} \over 2^{P_n}}$$
Note that the sum can be written two ways:
$$\sum\limits_{i=0}^{n-1} m^{i}2^{P_{n - 1 - i}} = \sum\limits_{i=0}^{n-1} m^{n - 1 - i}2^{P_{i}}$$
This is the equation you obtained with $m=3$ and $a=1$.
A loop of length $n$ occurs when $x_n = x_0$ which gives the loop condition:
$$x_0 = {a \sum\limits_{i=0}^{n-1} m^i 2^{P_{n-1-i}} \over {2^{P_n} - m^n}}$$
A loop occurs whenever there exist an $n$ and $p_i$ and hence $P_i$ such that the denominator divides the numerator. See Böhm and Sontacchi: https://mathscinet.ams.org/mathscinet-getitem?mr=551509 For $3x+1$ a sequence starting at $1$ has $p_i = 2$ so $P_i = 2i$ which gives a numerator:
$$\sum\limits_{i=0}^{n-1}3^{n-1-i}2^{2i} = 2^{2n} - 3^n$$
which is equal to the denominator so $x_0 = 1$.
So are there any other $n$ and $p_i$ and hence $P_i$ where the numerator is divisible by the denominator? So far not for $3x+1$ but definitely yes for $mx+a$ in general. For example $3x+5$ has several loops. $a \gt 1$ can supply some factors to help with the divisibility.
I need to read your paper more carefully to comment further. Thought it might help to show you some better (IMHO) notation. You may also want to consult https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/
