# Generalized Collatz sequences

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 $$k=4a+3$$.)

Questions.

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.

• 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 – zeraoulia rafik Sep 27 '20 at 14:09
• 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. – Goldstern Sep 27 '20 at 14:14
• 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 $4$. – Wojowu Sep 27 '20 at 14:21
• See also my question at Math.SE. I don't think there are any known (provably) divergent trajectories in any of your sequences. – Wojowu Sep 27 '20 at 14:29
• That's right @wojowu, will correct! And thanks everyone for the remarks for $n=5$ – Dominic van der Zypen Sep 27 '20 at 19:35

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).