This question is prompted by thinking about this question and the answer I gave there.
Consider the family of functions of the form \begin{equation} f_a(n):=\begin{cases} n/2 & \text{if $n$ is even}\\ an+1 & \text{if $n$ is odd}\\ \end{cases} \end{equation}
where $a$ is fixed and odd. The standard Collatz conjecture is about $f_3(n)$. Standard heuristics predict that for any odd $a>3$, $f_a(n)$ will have orbits will go to infinity, but this is open.
We can however, create a Collatz like situation where we allow multiple numbers and include new numbers in a new set. For example, consider the following: Let $S_0(n)=n$, and for $i>0$, define $S_i(n)$ as follows:
$S_i(n)= \{x \in {\mathbb{N}} | y \in S_{i-1}(n) \,\,\, \mathrm{where} \,\,\, \mathrm{and} \,\,\, x = 2y \,\,\, \mathrm{or} \,\,\, x=3y+1 \,\,\, \mathrm{or} \,\,\, 3y-1 \}$.
That is, if $x \in S_{i-1}(n)$ we put $x/2$ into $S_i(n)$, and if $x$ is odd then we put both $\frac{3x+1}{2}$ and $\frac{3x-1}{2}$ into $S_i(n)$. This is essentially then a variant of Collatz where at odd step we are allowed to "choose" either $\frac{3x+1}{2}$ or $\frac{3x-1}{2}$. It is easy to see that this if one is allowed this choice then for any $n$, there is some $i$ where $1 \in S_i(n)$ since for any odd $x$, one of $\frac{3x+1}{2}$ or $\frac{3x-1}{2}$ will always be even, and thus for any odd $x \in S_i$ we will have a number which is at most about $\frac{3}{4}x$ in $S_{i+1}(n)$.
However other similar examples seem less well behaved. Consider in particular the similar recursion $T_i(n)$ where $T_0(n)=n$, and where for $i >0$, $T_i(n)$ is defined as follows:
If $x \in T_{i-1}(n)$ we put $x/2$ into $T_i(n)$, and if $x$ is odd then we put both $\frac{3x+1}{2}$ and $\frac{5x+1}{2}$ into $T_i(n)$. It is not hard to show that $T_i(n)$ contains arbitrarily large elements, since for any odd $x$, one of $\frac{3x+1}{2}$ and $\frac{5x+1}{2}$ will be odd.
However, one suspects that the following is also true:
Conjecture 1: For any $n$, there is an $i$ where $1 \in T_i(n)$.
Conjecture is strictly weaker than the Collatz conjecture since we have more options to reach $1$).
The naive approach to A is to use the same sort of argument as we did for $S_i$ but this doesn't by itself work because although of our two numbers $\frac{3x+1}{2}$ and $\frac{5x+1}{2}$ one will still be even, but $\frac{5x+1}{4}$ is in general greater than $x$ so we don't get a reduction in size there.
We can weaken our conjecture even further:
Fix a positive odd integer $a \geq 3$, and define $T_{i,a}$ to be the same as $T_i$ except we now throw into $T_{i,a}(n)$ all of $3x+1$, $5x+1$, $\cdots ax+1$.
Question: Can we show that there is some odd $a$ such that for any $n$ for sufficiently large $i$, $1 \in T_{i,a}(n)$?
Two notes: First, this is different than the relaxation proposed by Max Alekseyev in this question where the new choice occurs when $x$ is even.
Second, it is not hard to at least show that for any $k$, there is an $a$ such that if $n \not \equiv -1 \pmod {2^k}$, then $T_{i,a}(n)$ will always contain at least one element smaller than $n$.
