Let $I_n$ be the interval $[2^n - 1, 2^{n+1} - 1)$ for $n = 0, 1, 2, \dots$. Then the $I_n$ are disjoint and cover the non-negative reals. Also $x \to 2x + 1$ maps $I_n$ onto $I_{n+1}$. Thus $f$ is determined by the values it takes on $I_0 = [0, 1)$. These can be arbitrary, but the simplest is to take $f(x) = 0$ on $[0, 1)$. Then we get $f(x) = 5$ on $[1, 3)$, $20$ on $[3, 7)$, $65$ on $[7, 15)$ and by a simple induction $\frac{5(3^n-1)}{2}$ on $I_n$.