Solve $f(x) = f(\lfloor x/2 \rfloor) + f(\lfloor x/3 \rfloor)$ 
Solve the following recurrence relation 
  $$f(1)=1, f(2)=2\\ 
f(x) = f\left(\left \lfloor \frac{x}{2} \right \rfloor\right) + f\left(\left\lfloor \frac{x}{3} \right\rfloor\right), \forall x \in \mathbb{N}, x \geq 3 $$    

I tried the simple ways to solve a recurrence relation but got things messed up. Any Hint will be helpful.  

Is $x$ constrained in some particular set?
Note that:
$$
\begin{aligned}
f(2) &= f\left(\left\lfloor \frac{2}{2} \right\rfloor\right) + f\left(\left\lfloor \frac{2}{3} \right\rfloor\right)\\
&=f(1) + f(0)
\end{aligned}
$$
Therefore, $f(0) = f(2) - f(1) =  1$. At the same time, for $x=1$, 
$$
\begin{aligned}
f(1) &= f\left(\left\lfloor \frac{1}{2} \right\rfloor\right) + f\left(\left\lfloor \frac{1}{3} \right\rfloor\right)\\
&=2f(0),
\end{aligned}
$$
therefore, $f(x) = \tfrac{1}{2}f(1) = \tfrac{1}{2}$, which is a contradition. Therefore, $f$ cannot be defined (at least) at $x=0$.
