The rules are simple:
Take any number $n$. If $n$ is even divide it by two, if $n$ is odd triple it and subtract one. Repeat indefinitely. (Note that this is a variation, in the original Collatz conjecture you add one.)
Like the original Collatz conjecture seems to always get to one, this variation always seems to get to either $1, 7$ or $17$. I checked that it does for initial values of $n$ up to 443 million.
Can you give a number that doesn't get to $1$, $7$ or $17$, or if not, at least show that such number exists?
Your function (let's call it $g: \mathbb Z \rightarrow \mathbb Z$) is just $g(x)=-f(-x)$ with $f$ being the normal Collatz function.
Your cycles are known, they correspond to $\{-1\rightarrow-2\}$, $\{-5\rightarrow-14\rightarrow-7\rightarrow-20\rightarrow-10\}$ and $\{-17\rightarrow-50\rightarrow-25\rightarrow-74\rightarrow-37\rightarrow-110\rightarrow-55\rightarrow-164\rightarrow-82\rightarrow-41\rightarrow-122\rightarrow-61\rightarrow-182\rightarrow-91\rightarrow-272\rightarrow-136\rightarrow-68\rightarrow-34\}$ for $f$.
It is unknown whether the Collatz hypothesis is true on the naturals, even less is known about extensions. See also https://en.wikipedia.org/wiki/Collatz_conjecture#Iterating_on_all_integers