We have $\lfloor x_n\sqrt{5}\rfloor =x_{n+1}-3x_n$, so \begin{align*} &x_{n+1}-3x_n\leq x_n\sqrt{5}<x_{n+1}-3x_n+1\\ \Leftrightarrow \ & 3x_n-x_{n+1}\geq -x_n\sqrt{5}>-x_{n+1}+3x_n-1\\ \Leftrightarrow \ & (3+\sqrt{5})x_n\geq x_{n+1}>(3+\sqrt{5})x_n-1\\ \Leftrightarrow \ & 4x_n\geq x_{n+1}(3-\sqrt{5})>4x_n-1\\ \Leftrightarrow \ & 3x_{n+1}-4x_n \leq \sqrt{5}x_{n+1}<3x_{n+1}-4x_n+1 \end{align*}Therefore $\lfloor x_n\sqrt{5}x_{n+1}\rfloor=3x_{n+1}-4x_n$. So from the sequence with the original general formula we can convert it back to: $$\begin{cases} x_0=1\\ x_{n+2}=6x_{n+1}-4x_n \end{cases}$$So: \[x_n=2^{n-2}\dfrac{1}{\sqrt{5}}\left(\dfrac{(1+\sqrt{5})^{2n+1}}{2^{2n+1}}-\dfrac{(1-\sqrt{5})^{2n+1}}{2^{2n+1}}\right)\]Like $x_n=2^{n-2}F_{2n+1}$. So from this formula we will prove this problem easy