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 $4$, which implies that $\text{Coll}_{k,k}$ is unbounded.

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.

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.

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 $n\mapsto c_k(n-1)$ 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 $4$, which implies that $\text{Coll}_{k,k}$ is unbounded.

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?

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 $4$, which implies that $\text{Coll}_{k,k}$ is unbounded.

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.