inequality – Collatz Conjecture: Are the following valid properties of a cycle?

Let:

  • $nu_2(x)$ be the 2-adic valuation of $x$
  • $x_1, x_2, dots, x_k$ be distinct odd integers that make up a cycle of length $k$ with:
    • $x_{i+1} = dfrac{3x_i+1}{2^{nu_2(3x_i+1)}}$
    • $x_{i+k} = x_i$
  • $p_1, p_2, dots, p_k$ be positive integers associated with the above cycle in the following way:
    • Each $p_i = sumlimits_{t=1}^i nu_2(3x_t+1)$
    • $p_k > p_{k-1} > dots > p_2 > p_1 > 0$
    • $2^{p_k} > 3^k$
    • $2^{p_k}x_{k+1} – 3^{k}x_1 = 3^{k-1} + sumlimits_{m=1}^{k-1}3^{k-1-m}2^{p_m}$

Note: Details for this last equation can be found here.

  • $s_1, s_2, dots, s_{k+1}$ be positive integers that are not a cycle but are also characterized by $p_1, p_2, dots, p_k$ so that:

    • $s_1 < x_1$
    • $s_{i+1} = dfrac{3s_i+1}{2^{nu_2(3s_i+1)}}$
    • $s_{k+1} ne s_1$
    • $2^{p_k}s_{k+1} – 3^{k}s_1 = 3^{k-1} + sumlimits_{m=1}^{k-1}3^{k-1-m}2^{p_m}$
  • $t_1, t_2, dots, t_{k+1}$ be positive integers that are not a cycle but are also characterized by $p_1, p_2, dots, p_k$ so that:

    • $t_1 > x_1$
    • $t_{i+1} = dfrac{3t_i+1}{2^{nu_2(3t_i+1)}}$
    • $t_{k+1} ne t_1$
    • $2^{p_k}t_{k+1} – 3^{k}t_1 = 2^{p_k}s_{k+1} – 3^{k}s_1 = 2^{p_k}x_{k+1} – 3^{k}x_1$

Question:

In the case of the trivial cycles, $s_1, dots, s_{k+1}$ may not exist. Since there are an infinite number of non-cycles for any combination of $p_1, dots, p_k$, there are always an infinite number of instances of $t_1, dots, t_{k+1}$

It seems to me that if a cycle exists, then for each $s_1, dots, s_{k+1}$ that exists:
$$s_{k+1} > s_1$$

For each $t_1, dots, t_{k+1}$ that exists
$$t_1 > t_{k+1}$$

Am I right? Did I make a mistake? Is any point in the question unclear?

Here’s the argument:

(1) $2^{p_k}(x_{k+1} – s_{k+1}) = 3^k(x_1 – s_1)$

(2) Since $2^{p_k} > 3_{k}$, it follows that:

$$x_{k+1} – s_{k+1} < x_1 – s_1$$

$$s_1 – s_{k+1} < x_1 – x_{k+1} = 0$$

(3) $2^{p_k}(t_{k+1} – x_{k+1}) = 3^k(t_1 – x_1)$

(4) So:

$$t_{k+1} – x_{k+1} < t_1 – x_1$$

$$t_{k+1} – t_1 < x_{k+1} – x_1 = 0$$