Let $x \geq 0$. If $x \in (0,2)$ then $f(x)=1+5 f(0)-6 f(0)=1$.
If $x \in [2,4)$ then $f(x)=1+5 f(1)-6 f(0)=6=a_{1}$.
If $x \in [4,8)$ then $\left\lfloor\frac{x}{2}\right\rfloor \in [2,4)$ and $\left\lfloor\frac{x}{4}\right\rfloor \in [1,2)$, and therefore $f(x)=1+5 \cdot 6-6 \cdot 1=25=a_{2}$.

Assume for $n \geq 1$, that function $f$ is constant on $[2^{n}, 2^{n+1})$ taking a value $a_{n}$, and constant on $[2^{n+1}, 2^{n+2})$ taking a value $a_{n+1}$, and let $x \in [2^{n+2}, 2^{n+3})$. Because $\left\lfloor\frac{x}{2}\right\rfloor \in [2^{n+1}, 2^{n+2})$ and $\left\lfloor\frac{x}{4}\right\rfloor \in [2^{n}, 2^{n+1})$, we deduce that $f(x)= 1+5 a_{n+1}-6 a_{n}$. Therefore, function $f$ is also constant on $[2^{n+2}, 2^{n+3})$ taking the value $a_{n+2}=1+5 a_{n+1}-6 a_{n}$, which can be rewritten
$$a_{n+2}-\frac{1}{2}=5\left(a_{n+1}-\frac{1}{2}\right)-6\left(a_{n}-\frac{1}{2}\right) .$$
Because the roots of the characteristic polynomial $X^{2}-5 X+6$ are $2$ and $3$, there exist two real numbers $u, v$ such that
$$a_{n}=\frac{1}{2}+2^{n} u+3^{n} v, \quad \text{ for all } n \geq 1$$
Because $a_{1}=6$ and $a_{2}=25$, we have $u=-4$ and $v=\frac{9}{2}$. We deduce that
$$f(x)= \begin{cases}0 & \text{ if } x=0 \\ 1 & \text{ if } x \in (0,2) \\ -2^{n+2}+\frac{3^{n+2}+1}{2} & \text{ if } x \in [2^{n}, 2^{n+1}) \text{ and } n \geq 1\end{cases}$$