Consider the recurrence $(a,b,c)\mapsto \left(\frac{3a+4b}{5}, \frac{4a-3b}{5}, c\right)$ and suppose that we start with $(2,3,2)$, and allow swapping positions before applying the recurrence again.
If we start with $(2,3,2)$, will we be able to obtain $(4,2,0)$?