# Is there a number that using the rules of Collatz conjecture's variation $3n-1$ doesn't get to $1, 7$ or $17$?

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?

• $\{5,14,7,20,10\}$ is the cycle, and it clearly goes through seven. Sep 4 '16 at 18:26
• @AlgorithmsX The question is to find a trajectory that never passes through 1, 7, or 17. Sep 4 '16 at 18:37
• @user326210 there was a comment which suggested n = 5 and the comment from AlgorithmsX is an answer to that now deleted comment Sep 4 '16 at 18:40
• I think mathematics isn't ready for this thing either (borrowing Erdős)... but I do suspect that any Collatz-like iteration has a finite number of cycles. I'm moving on from here. Sep 4 '16 at 18:54
• Zero, negative numbers, for example. Dec 2 '19 at 9:59

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 $f$.