Yes. We prove that every natural number can be reached from $1$ which is equivalent to the required result We also note the series of transformations $n\rightarrow 2n\rightarrow 4n\rightarrow 12n+1\rightarrow 36n+4\rightarrow 18n+2\rightarrow 9n+1\rightarrow 3n$ We use strong induction. We start off from $1$ so that the base case is true We assume that every natural less than $k$ can be reached from $1$ If $k\equiv 0(mod3), k\rightarrow \frac{k}{3}$ If $k\equiv 1(mod3), k\rightarrow \frac{k-1}{3}$ If $k\equiv 2(mod3), k\rightarrow 2k\rightarrow \frac{2k-1}{3}$

Make it a very good bit harder by only having a portal between $3n+1$ and the other two of $n$ is odd (but still always a portal between $n$ and $2n$)