Let:
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$k > 0$ be an integer
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$\nu_2(x)$ be the 2-adic valuation of $x$
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$x_1, x_2, \dots, x_n$ 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
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$x_1, x_2, \dots, x_k$ be called a cycle of length $k$ if:
$$x_1 = x_{1+k}$$ -
$p_m, p_{m-1}, \dots, p_2, p_1$ be $m$ integers such that:
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$p_m = \sum\limits_{i=1}^m \nu_2(3x_i + 1)$
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$p_m > p_{m-1} > \dots > p_2 > p_1 > 0$
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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:
I am interested in understanding if there is a relationship between the ratio $\frac{p_i}{i}$ and $x_{i+1}$
If $x_1$ is the beginning of a cycle of length $k$, does it follow that if $\frac{p_i}{i} > \frac{p_j}{j}$, then $x_{i+1} < x_{j+1}$?
It seems to me that the answer would be yes so that the minimum odd integer $x_{l} = x_{min}$ would necessarily be related to the maximum ratio which would then be $\frac{p_{l-1}}{l-1}$
Establishing this has been a bit tricky for me. I thought that an argument similar to John Omielan's here would do the trick but I am finding it less than straight forward to apply.
Am I right in my assumption? If so, is there a simple argument to show that it is true?
The same argument should hold in a non-cycle for $x_i$ and $x_j$ that have the same starting point where $i \ne j$. It may be possible to find a counter example in this case. I have not yet found a counter example.
Best Answer
Regarding your question about, even in a non-cycle with $i \neq j$,
A counter-example is $x_1 = 45$, with $i = 1$ and $j = 2$. Since $3x_1 + 1 = 136 = 2^3(17)$, then $p_1 = 3$ and $x_2 = 17$. Next, $3x_2 + 1 = 52 = 2^2(13)$ means $p_2 = 3 + 2 = 5$ and $x_3 = 13$. Finally, this gives
$$\frac{p_1}{1} = 3 \gt \frac{p_2}{2} = \frac{5}{2} \tag{1}\label{eq1A}$$
However, we also have
$$x_2 = 17 \gt x_3 = 13 \tag{2}\label{eq2A}$$