Collatz Conjecture: If a non-trivial cycle exists, would the sum of powers of $2$ be less than $2n$?

Question

For the Collatz Conjecture, it seems to me that if $m$ is the sum of the powers of $2$ for a non-trivial $n$ cycle (where each $x_1, \dots, x_n $ is odd and $x_i > 1$), it follows that $m < 2n$

Is my reasoning correct?

Let:

• $\nu_2(x)$ be the 2-adic valuation of $x$
• $x_1, x_2, \dots, x_n$ be $n$ distinct integers such that:
  • $x_{i+1} = \dfrac{3x_i + 1}{2^{\nu_2(3x_i+1)}}$
  • $x_i > 1$

Observations:

• $\left(3 + \dfrac{1}{x_{i}}\right) = \left(\dfrac{x_{i+1}}{x_{i}}\right)2^{\nu_2(3x_{i} + 1)}$ since:

  • $x_{i+1} = \dfrac{3x_{i}+1}{2^{\nu_2(3x_{i}+1)}}$
  • $2^{\nu_2(3x_{i}+1)}x_{i+1} = 3x_{i} + 1$

• $\prod\limits_{i=1}^{n}\left(3 + \frac{1}{x_i}\right) = \left(\dfrac{x_{n+1}}{x_1}\right)\prod\limits_{i=1}^n2^{\nu_2(3x_i + 1)}$

This follows directly from the previous observation.

• $\left(3 + \dfrac{1}{x_{\text{max}}}\right)^{n} \le \left(\dfrac{x_{n+1}}{x_1}\right)\prod\limits_{i=1}^n2^{{\nu}_2(3x_i + 1)} \le \left(3 + \dfrac{1}{x_{\text{min}}}\right)^{n}$

This follows directly from the previous observation.

Claim:

If $n \ge 1$, $x_1, x_2, \dots, x_n$ forms a cycle, then $\sum\limits_{i=1}^n \nu_2(3x_i+1) < 2n$

Argument:

(1) Assume $x_1, x_2, \dots, x_n$ form a cycle such that $x_{i+n} = x_i$

(2) Let $m = \sum\limits_{i=1}^n \nu_2(3x_i+1)$ so that:

$$\left(3 + \dfrac{1}{x_{\text{max}}}\right)^{n} \le 2^{m} \le \left(3 + \dfrac{1}{x_{\text{min}}}\right)^{n}$$

(3) Clearly, $2^m > 3^n$ so that: $2^{\frac{m}{n}} > 3$

(4) It follows:

• $$3 + \frac{1}{x_{\text{max}}} \le 2^{\frac{m}{n}} \le 3 + \frac{1}{x_{\text{min}}}$$
• $$\frac{1}{x_{\text{max}}} \le 2^{\frac{m}{n}} - 3 \le \frac{1}{x_{\text{min}}}$$
• $$x_{\text{max}} \ge \frac{1}{2^{\frac{m}{n}} - 3} \ge x_{\text{min}}$$

(5) $2^{\frac{m}{n}} - 3 < 1$ since $x_{\text{min}} > 1$ and $2^{\frac{m}{n}} > 3$

(6) It follows:

• $$2^{\frac{m}{n}} < 2^2$$
• $$m < 2n$$

Answer 1

MikeB almost pinned it.

In a cycle, it is believed (but not proved) that $m$ is the first exponent of $2$ that make $2^m$ larger than $3^n$, in other word, $m=\lceil n\cdot log_2(3)\rceil$: $$n\cdot log_2(3)<m<n\cdot log_2(3)+1$$

Answer 2

[Correction: this argument contains a mistake]

Actually, a stronger result can be shown using the following argument:

Looking at both even and odd numbers in the cycle, it contains $n$ odd numbers and $m$ even numbers.

Since

$$ \frac{3^{m}}{2^{m+n}} < 1$$

it follows that

$$ 3^{m} < 2^{m+n} $$ $$ (\frac{3}{2})^m < 2^n $$ $$ m \cdot \ln (3/2) < n \cdot \ln 2 $$ $$ m < \frac{\ln 2}{\ln(3/2)}\cdot n < 1.71n $$