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

Question

Let:

• $\nu_2(x)$ be the 2-adic valuation of $x$
• $x_1, x_2, \dots, x_k$ be 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 $p_i = \sum\limits_{t=1}^i \nu_2(3x_t+1)$
  • $p_k > p_{k-1} > \dots > p_2 > p_1 > 0$
  • $2^{p_k} > 3^k$
  • $2^{p_k}x_{k+1} - 3^{k}x_1 = 3^{k-1} + \sum\limits_{m=1}^{k-1}3^{k-1-m}2^{p_m}$

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:

  • $s_1 < x_1$
  • $s_{i+1} = \dfrac{3s_i+1}{2^{\nu_2(3s_i+1)}}$
  • $s_{k+1} \ne s_1$
  • $2^{p_k}s_{k+1} - 3^{k}s_1 = 3^{k-1} + \sum\limits_{m=1}^{k-1}3^{k-1-m}2^{p_m}$

• $t_1, t_2, \dots, t_{k+1}$ be positive integers that are not a cycle but are also characterized by $p_1, p_2, \dots, p_k$ so that:

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

Example:

Here is a cycle of length $3$:

• $k=3$, $x_1=1, x_2=1, x_3=1$ with $p_1=2, p_2=4, p_3=6$

It follows that: $$2^{6}x_4 - 3^{3}x_1 = 64(1) - 27(1) = 3^{2} + 3\times2^{2} + 2^{4} = 37$$

We can find $t_1, t_2, t_3$ in the following way:

• I start with $t_4=7$ where $t_3=\dfrac{4\times7-1}{3} = 9$
• I next try $t_4=19$ where $t_3=\dfrac{4\times19-1}{3} = 25$ and $t_2=\dfrac{4\times25-1}{3}= 33$
• I next try $t_4=37$ where $t_3=\dfrac{4\times37-1}{3} = 49$ and $t_2=\dfrac{4\times49-1}{3}=65$
• I next try $t_4=55$ where $t_3=\dfrac{4\times55-1}{3} = 73$ and $t_2=\dfrac{4\times73-1}{3}=97$ and $t_1=\dfrac{4\times97-1}{3}=129$

So that:

$$2^{6}55 - 3^{3}129 = 2^{6}1 - 3^{3}1 = 37$$

Note: I can add more examples if it is helpful.

Question:

In the case of the known cycles, $s_1, \dots, s_{k+1}$ does 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:

$$t_{k+1} - x_{k+1} < t_1 - x_1$$

$$t_{k+1} - t_1 < x_{k+1} - x_1 = 0$$

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Update:

It may be that the existence of an infinite number of non-cycles also characterized by the $p_1, p_2, \dots, p_k$ may not be clear. Apologies on this.

I have added an example of $t_1, t_2, \dots, t_{k+1}$

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Update 2:

Make corrections based on John Omielan's answer.

Answer 1

Your argument is correct regarding if $s_{k+1} \gt s_1$ and $t_{k+1} \gt t_1$. You're first just basically using that the right sides' expression of $3^{k-1} + \sum_{m=1}^{k-1}3^{k-1-m}2^{p_m}$ is the same for, and independent of, each of the $x_i$, $s_i$ and $t_i$ left side expressions. Thus, when subtracting the expression values, you get a $0$ on the right side, so you moved the left side terms with a factor of $3^k$ to the right side and then factored to get your results of $2^{p_k}(x_{k+1} - s_{k+1}) = 3^k(x_1 - s_1)$ and $2^{p_k}(x_{k+1} - t_{k+1}) = 3^k(x_1 - t_1)$. This is something, though, you perhaps could have provided some explanation.

The other mathematical aspect you're using next is that for real, positive $a$, $b$, $c$ and $d$, with $ab = cd \iff \frac{a}{c} = \frac{d}{b}$, then $a \gt c \iff \frac{a}{c} \gt 1 \iff \frac{d}{b} \gt 1 \iff b \lt d$. This, I believe, is unnecessary to state since it should be fairly obvious.

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With your specific example, to find a valid value of $t_4$ (and, thus, $t_3$, $t_2$ and $t_1$ by the reverse formula you used), you appear to be doing something similar to trial & error. However, there's an easier way, where all possible values of $t_1$ and $t_4$ (and, correspondingly, $s_1$ and $s_4$) are determined. To handle the general case, start with your results from subtraction, and use $u_i$ to represent $s_i$ or $t_i$ to keep it a bit more general, to get

$$2^{p_k}(u_{k+1} - x_{k+1}) = 3^k(u_1 - x_1) \tag{1}\label{eq1A}$$

Dividing both sides by $2^{p_k}(3^k)$ results in

$$\frac{u_{k+1} - x_{k+1}}{3^k} = \frac{u_1 - x_1}{2^{p_k}} \tag{2}\label{eq2A}$$

Using \eqref{eq1A}, $\gcd(2^{p_k}, 3^k) = 1$ means $3^k \mid u_{k+1} - x_{k+1}$ and $2^{p_k} \mid u_1 - x_1$. Thus, both fractions in \eqref{eq2A} are an integer. Since $u_{k+1} - x_{k+1}$ is even, this integer must be even, so call it $2j, \; j \in \mathbb{Z}$. Using this in \eqref{eq2A} then gives

$$\frac{u_{k+1} - x_{k+1}}{3^k} = 2j \implies u_{k+1} = x_{k+1} + 2j(3^k) \tag{3}\label{eq3A}$$

$$\frac{u_{1} - x_{1}}{2^{p_k}} = 2j \implies u_{1} = x_{1} + 2j(2^{p_k}) \tag{4}\label{eq4A}$$

It's easy to verify that, for any integer $j$, \eqref{eq3A} and \eqref{eq4A} satisfy \eqref{eq1A}. Thus, \eqref{eq3A} and \eqref{eq4A} represent all possible solutions. This means that, from \eqref{eq4A}, the positive $j$ give your $u_1 = t_1 \gt x_1$, and negative $j$ give your $u_1 = s_1 \lt x_1$. Note this also explicitly shows, as you stated,

... there are always an infinite number of instances of $t_1, \dots, t_{k+1}$

With $k = 3$ and $p_k = 6$, using $j = 1$ in \eqref{eq3A} gives $u_1 = t_1 = 1 + 2(2^{6}) = 129$, and $j = 1$ in \eqref{eq4A} gives $u_4 = t_4 = 1 + 2(3^{3}) = 55$. In particular, this matches your specific example values.