We have the inequality: $2\left\lfloor x \right\rfloor \leq \left\lfloor 2x \right\rfloor \leq 2 \left\lfloor x \right\rfloor + 1$. Therefore, $2f_n \leq f_{n+1} \leq 2f_n + 2,\forall n\in\mathbb N$. Suppose there exists $n_0\in\mathbb N$ such that $f_n$ is even, for all $n\geq n_0$. Then, from the above inequality, we have $f_{n+1} = 2f_n$ or $f_{n+1} = 2f_n + 2$, $\forall n\geq n_0$. We know that $f_{n+1} = 2f_n\Leftrightarrow \begin{cases} \left\lfloor 2^{n+1}\sqrt{69}\right\rfloor = 2\left\lfloor 2^n\sqrt{69}\right\rfloor \\ \left\lfloor 2^{n+1}\sqrt{96}\right\rfloor = 2\left\lfloor 2^n\sqrt{96}\right\rfloor\end{cases}$. And $f_{n+1} = 2f_n + 2\Leftrightarrow \begin{cases} \left\lfloor 2^{n+1} \sqrt{69}\right\rfloor = 1 + 2\left\lfloor 2^{n} \sqrt{69}\right\rfloor \\ \left\lfloor 2^{n+1} \sqrt{96}\right\rfloor = 1+2\left\lfloor 2^{n} \sqrt{96}\right\rfloor\end{cases}$. Hence, $\left\lfloor 2^{n+1} \sqrt{96} \right\rfloor - \left \lfloor 2^{n+1}\sqrt{69} \right\rfloor = 2\left \lfloor 2^n\sqrt{96}\right\rfloor - 2\left \lfloor 2^n\sqrt{69}\right\rfloor,\forall n\geq n_0$ $\Leftrightarrow \left\{2^{n+1}\sqrt{96}\right\}-\left\{2^{n+1}\sqrt{69}\right\} = 2\left\{2^{n}\sqrt{96}\right\} - 2\left\{2^{n}\sqrt{69}\right\},\forall n\geq n_0$ $\Rightarrow \left\{2^{n+1}\sqrt{96}\right\} - \left\{2^{n+1}\sqrt{69}\right\} = 2^{n+1 - n_0}\left(\left\{2^{n_0}\sqrt{96}\right\} - \left\{2^{n_0}\sqrt{69}\right\}\right)$. Since $\left\{2^{n+1}\sqrt{96}\right\} - \left\{2^{n+1}\sqrt{69}\right\}$ is bounded in $(-1;1)$, $\left\{2^{n_0}\sqrt{96}\right\} - \left\{2^{n_0}\sqrt{69}\right\}$ must be $0$ $\Rightarrow 2^{n_0}\sqrt{96} - 2^{n_0}\sqrt{69} = \left\lfloor 2^{n_0}\sqrt{96}\right\rfloor - \left\lfloor 2^{n_0}\sqrt{69}\right\rfloor$, which gives us a contradiction since $LHS$ is not an integer but $RHS$ is. Thus, there exists infinitely $n\in\mathbb N^*$ such that $f_n$ is odd. I think the even case is quite similar.