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What is wrong with this argument?

I am sure that I am misunderstanding something or there is a mistake in this argument. This argument is taken from the answer given to one of my questions about the Collatz Conjecture.

Let:

  • $v_2(x)$ be the 2-adic valuation of $x$
  • $C(x) = \dfrac{3x+1}{2^{v_2(3x+1)}}$
  • $x_1>1, x_2>1, \dots, x_n>1$ be the sequence of $n$ distinct odd integers for each application of $C(x_i)$ so that for each $x_i$:
  • for $i > 1$, $x_i = C(x_{i-1})$
  • $x_i > 1$
  • $x_{\text{min}}, x_{\text{max}}$ be the minimum and maximum value of $x_1, x_2, \dots, x_n$
  • $C_1(x) = C(x)$
  • $C_n(x) = C(C_{n-1}(x))$

Observations:

  • $\left(3 + \dfrac{1}{x_{i-1}}\right) = \left(\dfrac{x_i}{x_{i-1}}\right)2^{v_2(3x_{i-1} + 1)}$
  • $x_i = \dfrac{3x_{i-1}+1}{2^{v_2(3x_{i-1}+1)}}$
  • $2^{v_2(3x_{i-1}+1)}x_i = 3x_{i-1} + 1$
  • $\prod\limits_{k=1}^{n}\left(3 + \frac{1}{x_k}\right) = \left(\dfrac{x_{n+1}}{x_1}\right)\prod\limits_{k=1}^n2^{v_2(3x_k + 1)}$

This follows directly from the previous observation.

  • $\left(3 + \dfrac{1}{x_{\text{max}}}\right)^{n} \le \left(\dfrac{x_{n+1}}{x_1}\right)\prod\limits_{k=1}^n2^{v_2(3x_k + 1)} \le \left(3 + \dfrac{1}{x_{\text{min}}}\right)^{n}$

This follows directly from the previous observation.

  • if a non-trivial cycle exists, $n > 1$

$x = \dfrac{3x+1}{2^{v_2(3x+1)}}$ implies $x(2^{v_2(3x+1)} - 3) = 1$ which implies that $x=1$

Claim:

If there is a non-trivial cycle, the sum of the powers of $2$ in the cycle are the minimal integer power of $2$ greater than $3^n$

Argument:

(1) Assume that $x_1>1, x_2>1, \dots, x_n>1$ form an $n$-cycle such that:

  • $x_i = C(x_{i-1})$
  • $x_i = C_n(x_i)$ if $i \ge 1$
  • Each $x_i$ is distinct. If $j < n$, $x_{i+j} \ne x_i$

(2) Let $m = \sum\limits_{k=1}^{n} v_2(3x_k + 1)$

(3) From the third observation and since each $x_i$ in the cycle is distinct and repeats:

$$2^m = \left(\dfrac{x_{\text{i+n}}}{x_{i}}\right)2^m < \left(3 + \dfrac{1}{x_{\text{min}}}\right)^{n}$$

(4) Assume that $2^{m-1} > 3^n$

(5) $2\times3^n < 2^m < \left(3 + \dfrac{1}{x_{\text{min}}}\right)^{n}$

(6) But then we have a contradiction because $x_{\text{min}} < 1$ which is impossible since all $x_i > 1$:

  • $2^{\frac{1}{n}}3 < 3+ \dfrac{1}{x_{\text{min}}}$
  • $x_{\text{min}}\left(3(2^{\frac{1}{n}} - 1)\right) < 1$
  • $x_{\text{min}} < \dfrac{1}{3(2^{\frac{1}{n}} - 1)} < \dfrac{1}{3}$
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You wrote in your last line

$x_{\text{min}} < \dfrac{1}{3(2^{\frac{1}{n}} - 1)} < \dfrac{1}{3}$

However, note for $n \gt 1$ that $2^{1/n} \lt 2 \implies 2^{1/n} - 1 \lt 1$, so $3(2^{1/n} - 1) \lt 3$ and, thus, $\frac{1}{3\left(2^{1/n} - 1\right)} \gt \frac{1}{3}$. For example, $n = 10$ gives

$$\frac{1}{3\left(2^{0.1} - 1\right)} \approx 4.64 \tag{1}\label{eq1A}$$

Using

$$2^{1/n} = e^{\ln(2)(1/n)} \tag{2}\label{eq2A}$$

and the first few terms of the exponential Taylor series expansion, gives

$$\begin{equation}\begin{aligned} \frac{1}{3\left(2^{1/n} - 1\right)} & = \frac{1}{3\left(\left(1 + \frac{\ln(2)}{n} + \frac{\ln(2)^2}{2n^2} + O\left(n^{-3}\right)\right) - 1\right)} \\ & = \frac{1}{3\left(\frac{\ln(2)}{n} + \frac{\ln(2)^2}{2n^2} + O\left(n^{-3}\right)\right)}\\ & = \frac{1}{3\left(\frac{\ln(2)}{n}\right)\left(1 + \frac{\ln(2)}{2n} + O\left(n^{-2}\right)\right)}\\ & = \frac{n}{3\ln(2)}\left(1 - \frac{\ln(2)}{2n} + O\left(n^{-2}\right)\right) \\ & = \frac{n}{3\ln(2)} - \frac{1}{6} + O\left(n^{-1}\right) \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

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  • $\begingroup$ Thanks. I knew that I was making a mistake! That was it. $\endgroup$ Sep 28 '20 at 0:30
  • $\begingroup$ Very nice (+1) I didn't detect that glitch in the last inequality in the OP although I read it somehow carefully... $\endgroup$ Oct 1 '20 at 19:58

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