# Two reasons why the Collatz conjecture could fail

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 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 cases (1), (2)?

• unfortunately, in the wikipedia-entry they do not show how (and how far) small cycles are excluded aside from that of the specific $k$-cycle structure. With the knowledge of some lower bound $a_0 \approx 2^{60}$ (hope I recall that number correctly) cycles of respectable lengthes but of general structure (not only of $k$-cycles form) can unconditionally be excluded. – Gottfried Helms Oct 27 '18 at 7:20