Collatz Conjecture: Properties of odd integers that make up a cycle

Question

I came across an argument to show that there are no non-trivial cycles in collatz conjecture which is too simple to be correct. My interest is to find the flaw.

I believe that the first step is correct. I will post the next step as a separate question. I believe that there is a flaw in the next step but I have not yet found it.

Here is the first step:

Let:

• $\nu_2(x)$ be the 2-adic valuation of $x$
• $L_k = \sum\limits_{i=1}^k \nu_2(3x_i+1)$

(1) Let $x_1, x_2, \dots, x_k$ be a sequence of $k$ odd integers with:

• $x_{i+1} = \dfrac{3x_i+1}{2^{\nu_2(3x_i+1)}}$

(2) $2^{L_k}x_{i+k} = 3^kx_i + 3^{k-1} + \sum_{s=1}^{k-1}3^{k-1-s}2^{\sum_{t=0}^s\nu_2(3x_{i+t}+1)}$

• Base Case: $k=2$

$$2^{\nu_2(3x_{i+1}+1)}x_{i+2} = 3x_{i+1}+1 = 3\left(\frac{3x_i + 1}{2^{\nu_2(3x_i+1)}}\right)+1 = \frac{3^2x_i + 3}{2^{\nu_2(3x_i+1)}}+1$$

• Assume that up to $k$

$$2^{L_k}x_{i+k} = 3^kx_i + 3^{k-1} + \sum_{s=1}^{k-1}3^{k-1-s}2^{\sum_{t=0}^s\nu_2(3x_{i+t}+1)}$$

• Inductive Step:

$$x_{i+k+1} = \frac{3x_{i+k}+1}{2^{\nu_2(3x_{i+k}+1)}}=\frac{3(2^{L_k}x_{i+k})+2^{L_k}}{2^{\nu_2(3x_{i+k}+1)+L_k}} = \frac{3^{k+1}x_i + 3^{k} + \sum\limits_{s=1}^{k}3^{k-s}2^{\sum_{t=0}^s\nu_2(3x_{i+t}+1)}}{2^{\nu_2(3x_{i+k}+1)+L_k}}$$

(3) Assume that a cycle exists such that $x_{i+k} = x_i$

(4) It follows from (2) that for $k\ge2$:

$$(2^{L_k} - 3^k)x_i = 3^{k-1} + \sum_{s=1}^{k-1}3^{k-1-s}2^{\sum_{t=0}^s\nu_2(3x_{i+t}+1)}$$

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Edit 1: I found the mistake in the next step of the argument.

Based on the answer given, I realize that the order of the arguments confuses the argument. I have changed the order since (2) does not assume that a cycle exists. It should be true for all sequences of odd integers where each integer $x_{i+1} = \dfrac{3x_i+1}{2^{\nu_2(3x_i+1)}}$

Edit 2: I made a typo in my statement of the summation that have been identified by Hagen von Eitzen and Collag3n.

Step (2) should be:

(2) $2^{L_k}x_{i+k} = 3^kx_i + 3^{k-1} + \sum_{s=1}^{k-1}3^{k-1-s}2^{\sum_{t=0}^{s-1}\nu_2(3x_{i+t}+1)}$

Then, using Hagen von Eitzen's example:

$$9 \times 27 + 3 + 2 = 248$$ $$2^{2+1}x_3 = 8\times 31 = 248$$

Answer 1

You make an induction on the length of a cycle. And then in the induction step you even assume that the same numbers $x_i$ of a $k$-cycle are also in a $(k+1)$-cycle. Thus you essentially assume that the cycle has length $1$.

Edit after revision of the question: The claim $$ 2^{L_k}x_{i+k} = 3^kx_i + 3^{k-1} + \sum_{s=1}^{k-1}3^{k-1-s}2^{\sum_{t=0}^s\nu_2(3x_{i+t}+1)}$$ is verifiably wrong, e.g., for $k=2$, $x_1=27$, $x_2=\frac{3\cdot 27+1}2=41$, the right hand side evaluates to $$9\cdot 27+3+2^{\nu_2(3\cdot 27+1)+\nu_2(3\cdot 41+1)} =254,$$ which is not even a multiple of $x_3=\frac{3\cdot 41+1}{4}=31$.