1

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!

Larry Freeman
  • 10,433
  • What is the point of the values $p_i$ in the second bullet point? Do they have something to do with the values $p_i$ that get introduced later on? – Asinomás Apr 28 '21 at 02:09
  • 1
    Yes. $p_t > p_{t-1} > \dots > p_1 > 1$ for the sequence that consists of $x_{max} = x_1, x_2, \dots, x_t = x_{min}$ – Larry Freeman Apr 28 '21 at 02:12
  • $t$ is the distance in the sequence between $x_{max}$ and $x_{min}$. Because it is a cycle and repeats, we can set $x_1 = x_{max}$. $0 < t < k$ is the distance between. It is less than $k$ since if $x_1 = x_{max}$, then $x_{1+k} = x_{max}$. If $x_1=x_{max}$, then $x_{t+1}=x_{min}$ if I calculated it correctly. – Larry Freeman Apr 28 '21 at 02:20
  • @LarryFreeman FYI, one minor formatting issue is that since "min" and "max" are not sets of $3$ variables but, instead, are specific names, you may wish to use something like \text{min} and \text{max} with those subscripts, so you get $x_{\text{min}}$ and $x_{\text{max}}$ instead of $x_{min}$ and $x_{max}$. – John Omielan Apr 28 '21 at 03:09

1 Answers1

2

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 $(3)$.

John Omielan
  • 47,976