Collatz Conjecture: For a cycle where the maximum odd integer is $x_{max}$, does it follow that $x_{max} < 3^n$ I am working on understanding the upper limit in the case where a non-trivial cycle exists for the Collatz Conjecture.
Is the following reasoning valid for establishing that the maximum odd integer in a cycle consisting $n$ odd integers is less than $3^n$?
Let:

*

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

*$n > 5$ be an integer

*$x_1, x_2, \dots, x_n$ be the distinct odd integers that make up a cycle of length $n$ with:

*

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

*$x_{i+n} = x_i$

*$x_{max}$ is the highest odd integer in the cycle



*$p_0, p_1, \dots p_n$ be integers such that:

*

*$p_0 = 0$

*$p_{i+1} = p_i + \nu_2(3x_{i+1}+1)$

*$p_n > p_{n-1} > \dots > p_1 > p_0 = 0$
(1)  Since $x_1, \dots x_n$ is a cycle, we can assume that $x_1 = x_{max}$
(2)  It follows from reasoning in step 2 here:
$$2^{p_n}x_{n+1} - 3^{n}x_1 = x_{max}\left(2^{p_n} - 3^n\right) = \sum\limits_{i=0}^{n-1}3^{n-1-i}2^{p_i}$$
(3)  From reasoning here for a cycle:
$$p_n < 2n$$
(4)  $\sum\limits_{i=0}^{n-1}3^{n-1-i}2^{p_i} < 6^n$ since:
$$\sum\limits_{i=0}^{n-1}3^{n-1-i}2^{p_i} < 3^{n-1} + 4^{n-1} + (2^{2n-n+1})(\sum\limits_{i=1}^{n-2}3^{n-i-i}2^{i}) < 3^{n-1} + 4^{n-1} + 2^{n+1}(3^{n-1}-2^{n-1}) = 3^{n-1} + 4^{n-1} + 4(6^{n-1} - 4^{n-1}) < 6^{n-1} + 6^{n-1} + 4\times6^{n-1} = 6^n$$
(5) $2^{p_n} \ge 2^{\lceil{n}\log_2{3}\rceil} > 3^n$

*

*From reasoning here, since $n > 5$, it follows that $2^{p_n} - 3^n > 2^n$


*It follows that $x_{max} < \dfrac{6^n}{2^n} = 3^n$
Did I make a mistake?  Is there any step that is unclear?  Is there a simpler argument to reach the same conclusion?

Edit:
There is no advantage to the 2 cases in step 5.  I have updated to simplify based on suggestions in the comment.

The start of your reasoning is quite good, although there are some issues with your later steps. Otherwise, your arguments are basically correct.
In your (4), there are several relatively minor mistakes, but your overall upper bound still applies. Using your result of $p_n \lt 2n$, then each $p_{i+1} \gt p_{i}$ for $1 \le i \le n - 1$ means each lower index $p$ value must be at least one less than the one above, with this giving that for all $0 \le j \le n - 1$
$$p_{n-j} \lt 2n - j \tag{1}\label{eq1A}$$
Thus, for example, $p_{n - 1} \lt 2n - 1$ and $p_i \lt 2n - (n - i) = n + i$ for all $1 \le i \le n$ (thus, for example, $p_1 \lt n + 1$). This then, along with $n \gt 5$, gives
$$\begin{equation}\begin{aligned}
\sum_{i=0}^{n-1}3^{n-1-i}2^{p_i} & \lt 3^{n-1} + 2^{2n-1} + (2^n)\left(\sum_{i=1}^{n-2}3^{n-1-i}2^{i}\right) \\
& = 3^{n-1} + 2(2^{2(n-1)}) + (2^n)((2)(3)(3^{n-2} - 2^{n-2})) \\
& = 3^{n-1} + 2(4^{n-1}) + (2^2)((2)(3))(6^{n-2} - 4^{n-2}) \\
& \lt 6^{n-1} + 6^{n-1} + (4)(6)(6^{n-2}) \\
& = 6^{n-1} + 6^{n-1} + 4(6^{n-1}) \\
& = 6^n
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
You made a mistake with using $4^{n-1}$ instead of $2(4^{n-1})$. Also, since the starting point for $i = 1$ has a factor of $2^i = 2$ instead the summation, the common factor outside is $2^n$, not $2^{n+1}$. Finally, assuming you tried to use the actual summation instead of performing several steps at once, then note the summation has a factor of $3$ outside, with the exponents inside being $n - 2$, not $n - 1$. To see this, using the sum of a geometric series formula gives
$$\begin{equation}\begin{aligned}
\sum_{i=1}^{n-2}3^{n-1-i}2^{i} & = 2(3^{n-2})\left(\frac{1 - \left(\frac{2}{3}\right)^{n-2}}{1 - \frac{2}{3}}\right) \\
& = 2(3^{n-2})\left(\frac{\frac{3^{n-2} - 2^{n-2}}{3^{n-2}}}{\frac{1}{3}}\right) \\
& = 2(3)(3^{n-2} - 2^{n-2})
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Regarding your step (5), here are a few more details. With $m = \left\lceil n\log_2 3 \right\rceil$ being the positive exponent $m$ which allows $2^m - 3^n \gt 0$, then since $2^{p_n} \gt 3^n$, we have $p_n \ge m$. This then gives from your linked answer for $n \gt 5$ that
$$2^{p_n} - 3^n \ge 2^{\left\lceil n\log_2 3 \right\rceil} - 3^n \gt 2^n \tag{4}\label{eq4A}$$
