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.



Generalized Collatz sequences
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](https://mathscinet.ams.org/mathscinet-getitem?mr=568274) and was proven by [Allouch](https://mathscinet.ams.org/mathscinet-getitem?mr=567874) which states that ${\mathrm{Col}\_{\min}(N) < N}$ For almost all $N$ (in the sense of natural density), and there is an improvement [here](https://terrytao.wordpress.com/2019/09/10/almost-all-collatz-orbits-attain-almost-bounded-values/) by Terry Tao in the sense of logarithmic density (**see Theorem 2**).


