XV OBM

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 $I_n$.