Let $(a_k)^{\infty}_{k=0}$ be a sequence of real numbers such that if $k$ is a non-negative integer, then
$$a_{k+1} = 3a_k - \lfloor 2a_k \rfloor - \lfloor a_k \rfloor.$$Definitely all positive integers $n$ such that if $a_0 = 1/n$, then this sequence is constant after a certain term.
Fortunately all terms of the sequence are rational.
Suppose the ultimate value of the sequence is $a$ then $a=3a-[2a]-[a]\iff 2a=[2a]+[a]\iff \{2a\}=[a]\iff \{2a\}=[a]=0\iff a=0 \vee a=1/2$.
Now suppose that the prime $p\not=3$ divides the numerator of $a_0$. Then this is true for all terms $a_k$ and $a_k\not\in Z$ for all $k$. Contradiction. In the same way 4 cannot divide the numerator.
Hence $n=3^s$ or $n=2\cdot 3^s$ for some $s\ge 0$. All these values work.
@below: thank you for the correction.
@above, you missed $a=\frac{1}{2}$ as solution of $\{2a\}=[a]$
Using same logic, can be proved that $n=3^s$ or $n=2*3^s$ because if $p \neq 2,3$ is prime or $p=4$ and $p|n$ then $p$ divides denominator of $a_k$ for every $k$