Let $\mathbb{N}$ denote the set of positive integers. The *Collatz function* $f:\mathbb{N}\to\mathbb{N}$ is given by $f(n) = n/2$ for $n$ even and $f(n) = 3n+1$ for $n$ odd. Given $k\in\mathbb{N}$ we associate to $k$ its *Collatz sequence* $(c^{(k)}\_n)\_{n\in\mathbb{N}}$ given inductively by $$c^{(k)}(1) = k\text{ and } c^{(k)}\_{n+1} = f(c^{(k)}\_n)\text{ for all } n\geq 1.$$
One version of the [Collatz conjecture](https://en.wikipedia.org/wiki/Collatz_conjecture) states that $$1\in \text{im}(c^{(k)}) \text{ for all }k\in\mathbb{N}.$$
Note that for all $k\in\mathbb{N}$ the sequence $c^{(k)}$ is either injective or eventually periodic. So any $c^{(k)}$ with $1\notin \text{im}(c^{(k)})$ would be either 


(1) injective or 


(2) eventually contain a period not containing $1$. 


**Question.** Can the current state of research exclude one of the two cases above?



Two reasons why the Collatz conjecture could fail
The current state of research excludes none of the two cases. The occurrence of some small cycles has been actually ruled out. You can easily find the relevant references in the [Wikipedia page](https://en.wikipedia.org/wiki/Collatz_conjecture) devoted to Collatz conjecture (the same that you linked), at the section "Cycles".


