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I read a couple days ago about the Collatz conjecture. Considering that this problem has been around since the '30s, I'm guessing that my idea written below has a few problems. Nonetheless, I want to know if this sounds like a good approach: trying to analyze how the Collatz sequence grows.

Suppose we have a starting number (obviously, the starting number must itself be large) with a Collatz sequence that does not eventually decay to $1$. It follows that the sequence is infinite. Assume further that there are no infinite cycles away from the cycle which leads to $1$: $4,2,1,4,2,1,...$. Then, for such a starting value, it should be true that its Collatz sequence diverges to $\infty$.

The geometric mean of a terminating Collatz sequence of a number $m$ can be written as $$\mu = m\sqrt[n](\prod_{i=1}^{n-1} a_i)$$, where there are $n$ terms in the Collatz sequence, and the $a_i$s are the terms of the sequence with $a_{n-1} = 1$. Now, suppose $m$ is the starting value of an infinite, diverging Collatz sequence. We can define the $n^{th}$ geometric mean of the diverging sequence to be $$\mu_n = m\sqrt[n](\prod_{i=1}^{n-1} a_i)$$ with $a_{n-1} \neq 1$. If the sequence diverges, it should be true that its $n^{th}$ geometric mean also diverges. In particular, there should exist an $n$ such that $$m\sqrt[n](\prod_{i=1}^{n-1} a_i) > m'$$, where $m'<m$ is a lower bound of the sequence, but also large. Observe that if term $a_k$ is odd, then $a_{k+1} = 3a_k+1 \implies \frac{a_{k+1}}{a_k} = \frac{3a_k+1}{a_k}$. In the limit that $a_k \rightarrow \infty$, $\frac{3a_k+1}{a_k} \rightarrow 3$. For $a_k$ even, $\frac{a_{k+1}}{a_k} = \frac{1}{2}$.

Now, for the $n^{th}$ Collatz sequence, we will have a ratio $p$ of the $n$ terms which are odd, and the remainder even. The number of odd terms, excluding the starting value $m$ is approximately $(n-1)p$ because $n$ is large, and the number of even terms is approximately $(n-1)(1-p)$. So, in the limit that $n$ is large, the $n^{th}$ geometric mean can be approximated by $$\mu_n \approx m\sqrt[n]((3)^{(n-1)p}(\frac{1}{2})^{(n-1)(1-p)}) > m'$$ because we will multiply by approximately $3$ $(n-1)p$ times, and we will multiply by $\frac{1}{2}$ $(n-1)(1-p)$ times.

Now, it is obvious that for any Collatz sequence, the proportion of terms which are odd is less than $\frac{1}{2}$. This is because, given odd term $a_k = 2n+1$, $a_{k+1} = 6n+4$, which is even, meaning we cannot have two consecutive odd terms. If we do some algebra to simplify the above expression for $\mu_n$, we find that $$\frac{1}{2} > p > \frac{(\frac{m'}{m})^n}{(n-1)\log(6)} + \frac{\log(2)}{\log(6)}$$.

Because we can take $n>>m$ and $n>>m'$, then this expression can be approximated by $$\frac{1}{2} > p > \frac{\log(2)}{\log(6)}$$

So, for divergence $p > 0.387$ approximately. So, for a diverging Collatz sequence, a minimum of about $38\%$ of its terms must be odd.

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  • $\begingroup$ when you are assuming the contrary, you can only deduce that the $\lim\sup a_n = \infty$, which is much weaker than diverging to infinity i.e, $\forall M > 0, \exists N\in\mathbb{N}: n>N\implies a_n>M.$ $\endgroup$ – dezdichado Apr 30 '18 at 1:43
  • $\begingroup$ No, it would diverge. If the sequence is nonrepeating, it can only visit numbers below M finitely many times. $\endgroup$ – Nathaniel Mayer Apr 30 '18 at 1:57
  • $\begingroup$ But there must be some M' lower bound, right? Because if there is some a_k < M', the Collatz would converge to 1. $\endgroup$ – D.B. Apr 30 '18 at 2:00
  • $\begingroup$ Nice try. I like to see a bit of ambition and original thought. $\endgroup$ – samerivertwice May 1 '18 at 15:22
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The geometric mean is $$\sqrt[n]{m \prod_{i=1}^{n-1} a_i},$$ but your approximation seems to be using $$\sqrt[n]{m \prod_{i=1}^{n-1} \frac{a_i}{a_{i-1}}} = \sqrt[n]{a_{n-1}},$$ which does not need to diverge.

It doesn't seem like taking the geometric mean is helping you. As you note, the geometric mean will diverge if and only if the sequence itself diverges. You have observed that, on the assumption the sequence diverges, $$a_n \approx m (3)^{np} \left(\frac{1}{2}\right)^{n(1-p)} = m \left(\frac{1}{2} \cdot 6^p\right)^n.$$ This does in fact diverge so long as $$p > \frac{\log 2}{\log 6} \approx 0.387.$$

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  • $\begingroup$ Oh. Sorry, I screwed up the geometric mean. That makes sense. $\endgroup$ – D.B. Apr 30 '18 at 2:15
  • $\begingroup$ Maybe it could still be useful. I calculated values of p for the Collatz sequences of number 1 to 10,000 and the max of p was only 0.37. $\endgroup$ – D.B. Apr 30 '18 at 2:45
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    $\begingroup$ Who knows, could be! I'm no expert on the Collatz conjecture. I figure it's hard because it's been open and famous for so long, but if you're excited about it then run with your ideas! Also it's probably worth reading up on the literature to see what's been done before. $\endgroup$ – Nathaniel Mayer Apr 30 '18 at 23:42

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