# Collatz Conjecture: Does this argument show if a non-trivial cycle exists, the sum of powers of $2$ must be minimal power of $2 > 3^n$?

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 $$x_i > 1$$:

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

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}

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