Let $\mathbb{N}$ denote the set of positive integers. For $k\in\mathbb{N}$ let $c_k:\mathbb{N}\to\mathbb{N}$ be defined by $x\mapsto x/2$ for $x$ even and $x\mapsto kx+1$ otherwise. The Collatz sequence of $x\in \mathbb{N}$ with respect to $k$, denoted by $\text{Coll}_{x,k}:\mathbb{N}\to\mathbb{N}$ is defined by $1\mapsto x$ and $n\mapsto c_k(\text{Coll}_{x,k}(n-1))$ for $n\in\mathbb{N}\setminus\{1\}$.

The famous Collatz conjecture states that $$1\in\text{im(Coll}_{x,3})$$ for every $\in\mathbb{N}$.

For $k$ even, the behavior of $\text{Coll}_{x,k}$ is uninteresting, and it is easy to see that for every $x\in\mathbb{N}$, the sequence $\text{Coll}_{x,1}$ eventually periodic. Moreover, if $k>1$ and $k=4a+1$ for some $a\in\mathbb{N}$, we get that no member of $\text{im}(\text{Coll}_{k,k})$ is divisible by $4$ ... (Edit: apologies, this last statement is false as pointed out by user @wojowu! So I erroneously thought only $k=4a+1$ is uninteresting, so the questions below focus on $k=4a+3$.)


  1. Is there $a\in\mathbb{N}$ such that there is a positive integer $x$ such that $\text{Coll}_{x,4a+3}$ is unbounded? (The smallest known value of $a$ satisfying this would be of interest.)

  2. Is there $a\in\mathbb{N}$ such that there is a positive integer $x$ such that $\text{Coll}_{x,4a+3}$ is bounded, but $1\notin \text{im}(\text{Coll}_{x,4a+3})$, or in other words, $\text{Coll}_{x,4a+3}$ is eventually periodic, but $1$ is not involved in the period?

Edit. I corrected the inductive definition of $\text{Coll}_{x,k}$. Thanks to user @wojowu for spotting my error.

  • 1
    $\begingroup$ I think the answer is negative because it was shown by Terras that for almost all ${N}$ (in the sense of natural density), one has ${\mathrm{Col}_{\min}(N) < N}$ and this was then improved by Allouche and if there is a small a the investigation for it would be beyond current tecknology $\endgroup$ Sep 27, 2020 at 14:09
  • 4
    $\begingroup$ Note that $k=4a+1$ is not completely uninteresting. For example, $k=5$ gets you the cycle 13,66,33,166,83,416,208,104,52,26,13, which does not contain 1. $\endgroup$
    – Goldstern
    Sep 27, 2020 at 14:14
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    $\begingroup$ Are you sure you mean $n\mapsto c_k(n-1)$ and not $n\mapsto c_k(Coll_{x,k}(n-1))$? Your claim about $k=4a+1$ is incorrect, since for $k=5$ the sequence starts with $5,26,13,66,33,166,83,416$ which is divisible by $4$. $\endgroup$
    – Wojowu
    Sep 27, 2020 at 14:21
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    $\begingroup$ See also my question at Math.SE. I don't think there are any known (provably) divergent trajectories in any of your sequences. $\endgroup$
    – Wojowu
    Sep 27, 2020 at 14:29
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    $\begingroup$ That's right @wojowu, will correct! And thanks everyone for the remarks for $n=5$ $\endgroup$ Sep 27, 2020 at 19:35

1 Answer 1


This long comment might be helpful:

I think the answer is negative as I pointed out in the comments because almost all Collatz orbits attain almost bounded values, the result which is shown by Terras and was proven by Allouch which states that ${\mathrm{Col}_{\min}(N) < N}$ For almost all $N$ (in the sense of natural density), and there is an improvement here by Terry Tao in the sense of logarithmic density (see Theorem 2).


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