Collatz Conjecture: Reasoning about the upper limit for the maximum odd integer in a collatz cycle Let:

*

*$k > 0$ be an integer

*$p_k, p_{k-1}, \dots, p_2, p_1$ be $k$ integers such that:
$$p_k > p_{k-1} > \dots > p_2 > p_1 > 0$$

*$\nu_2(x)$ be the 2-adic valuation of $x$

*$x_1, x_2, \dots, x_n$ be the next $n$ odd integers in a sequence such that:

*

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

*$x_1 > 1$ can be any odd integer

*$x_{max}$ be the largest of the $n$ integers

*$x_{min}$ be the smallest of the $n$ integers



*$x_1, x_2, \dots, x_k$ be called a cycle of length $k$ if:
$$x_1 = x_{1+k}$$
Observation:

*

*For any $k$ odd integers in the sequence above there exists $p_1, p_2, \dots, p_k$ such that:
$$2^{p_k}x_{k+1} = 3^kx_1 + 3^{k-1} + \sum\limits_{s=1}^{k-1}3^{k-1-s}2^{p_s}$$

Note:  Details for this can be found in step 2 here

Question:
Does it now follow that there exists an integer $k > t> 0$ such that:
$$x_{max} < \frac{2^{p_t}}{3^t}x_{min}$$
Here's my thinking:
(1)  Assume that $x_1, x_2, \dots x_k$ form a cycle of length $k$ so that $x_1 = x_{1+k}$
(2)  From the observation, there exists an integer $0 < t < k$ with $3^{t-1} > 0$ and $S \ge 0$ such that:
$$2^{p_t}x_{min} = 3^tx_{max} + 3^{t-1} + S$$
where $S=\begin{cases}
0, && \text{if }t=1\\
\sum\limits_{s=1}^{t-1}3^{t-1-s}2^{p_s}, && \text{if }t>1\\
\end{cases}$
(3) $2^{p_t}x_{min} > 3^{t}x_{max}$ so that:
$$x_{max} < \frac{2^{p_t}}{3^{t}}x_{min}$$

Edit 1:  Updating some mistakes in the question.  Thanks to John Omielan for his answer!
 A: Your reasoning is basically correct, except there are a few relatively minor issues with it. First, with your observation, as stated in your linked question and you also used later in your $(2)$ thinking step, there's a missing middle term of $3^{k-1}$. Also, the observation is more general than just for the $k$ terms in your cycle and, in fact, it's used to define your $p_i$ for all $1 \le i \le k$. Thus, you should use a new variable here, e.g., say it holds for all integers $1 \le m \le k$.
You're correct you can set $x_1 = x_{\text{max}}$ and $x_t = x_{\text{min}}$ since it's assumed to be cyclic, but you should state this explicitly in your question text instead of later in a comment. As the question reads now, it's not clear which value is assumed to be $x_{\text{max}}$ and $x_{\text{min}}$ among the $x_{i}$ values.
Next, in your thinking part, with your $(2)$, the $S$ value can be combined into $1$ part since the summation where the upper limit is $1$ less than the start is, by default, $0$. Also, the summation limit you use is $k$, but I believe it was meant to be $t$ instead. Thus, you could just set it to be
$$S = \sum_{s=1}^{t-1}3^{t-1-s}2^{p_s} \tag{1}\label{eq1A}$$
Finally, with your relationship,
$$2^{p_t}x_{\text{min}} = 3^tx_{\text{max}} + 3^{t-1} + S \tag{2}\label{eq2A}$$
although it should be fairly obvious, you may wish to explicitly state that $3^{t-1} \gt 0$ and $S \ge 0$. This then gives your $(3)$.
