3
$\begingroup$

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 two cases above?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Is there any interesting modular information known about possible cycles, like: there cannot be a cycle of length $n$ if $n\equiv 3 (\mathrm{mod} 8)$, for example? $\endgroup$
    – Somatic Custard
    Commented Nov 7, 2019 at 22:28
  • $\begingroup$ Great question @SomaticCustard! $\endgroup$
    – Dominic van der Zypen
    Commented Nov 7, 2019 at 23:55

1 Answer 1

Reset to default
19
$\begingroup$

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 devoted to Collatz conjecture (the same that you linked), at the section "Cycles".

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – Gottfried Helms
    Commented Oct 27, 2018 at 7:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .