# Collatz conjecture: $2^{m-1}(6n-3)$ is not part of any cycle

My original method was different from the method shown here. Instead of working my way backward through the iterations as below, I worked my way forward. I choose against doing that here despite of it giving more insight into the problem in a more general case. The post simply got to long.

Letting $$k,m,n,p\in\mathbb{N}$$, we start with a refresher of what Collatz conjecture states:

Given the function $$f:\mathbb{N}\rightarrow\mathbb{N}$$,

\begin{align} &f(n)= \begin{cases} 3n+1 & \quad \text{if } n \text{ is odd}\\ n/2 & \quad \text{else}\\ \end{cases}\\ \end{align}

there exist for any number $$n$$ a number $$p$$ such that $$p$$ iterations of $$f$$ evaluated at $$n$$ is equal to $$1$$.

As I find it easy to get blind with my own work, my questions are simple; is this known, and is what's done here valid?

Given a function $$g:\mathbb{N}\rightarrow\mathbb{N}$$, a number $$n$$ is said to be part of a cycle if there exist a number $$p$$ such that $$p$$ iterations of $$g$$ evaluated at $$n$$ is equal to $$n$$.

The result in the title is quite easy to verify using the function $$f$$ in reverse, which is why I'm surprised I haven't been able to find it written down anywhere. If we evaluate $$6k-3$$ we get:

\begin{align} &o_1,o_2\in\mathbb{O}\\ \\ &\frac{3o_1+1}{2^m}=o_2 \implies o_1=\frac{2^mo_2-1}{3}\\ \\ &\frac{2^m(6k-3)-1}{3}=2^{m+1}k-2^m-\frac{1}{3}\\ \end{align}

We can see that this will never be an integer. It's therefore impossible to iterate the function $$f$$ evaluated at any number $$n$$ and get $$6k-3$$ with the exception of $$2^m(6k-3)$$, showing that $$2^{m-1}(6k-3)$$ cannot be part of any cycle.

• No multiple of 3 can be part of a cycle because the prime factorisation of 3n+1 does not contain any "3" factor (and dividing by $2^k$ only remove "2" factors). That's why multiple of 3 are leaves in the collatz tree. – Collag3n May 24 at 17:42

If $$2^{m-1}(6k-3)$$ is part of a cycle, then there must exist a largest $$r$$ such that with the same $$k$$, $$2^r(6k-3)$$ is part of that cycle. What can its predecessor be?
It can't be $$2^{r+1}(6k-3) > 2^r(6k-3)$$ because $$r$$ was the largest exponent with that property. So it would have to be some $$n$$ such that $$2^r(6k-3) = 3n+1$$ But the LHS of that equation is divisible by $$3$$ and the RHS is not.
Now if you showed that $$2^m(8k-3)$$ cannot be part of a cycle, that would be non-trivial enough to be worth mentioning in the literature.
More generally no multiple of $$3$$ sits within any cycle, because a) no multiple of $$3$$ is in the image of $$3x+1$$, and b) $$x/2$$ is nullipotent over the property of being a multiple of $$3$$.