show by induction that $f$ is constant on each $I_n=\big[2^n, 2^{n+1}\big)$, with $\alpha_n=f(I_n)\Rightarrow~\alpha_{n+2}=1+5\alpha_{n+1}-6\alpha_n, \; \alpha_0=1, \alpha_1=0$, hence $\alpha_n=(3^n-2^{n+1}+1)/2, \boxed{f(x)=(3^{[\ln_2x]}-2^{[\ln_2x]+1}+1)/2, \; x>1}$

Hello_Kitty wrote: show by induction that $f$ is constant on each $I_n=\big[2^n, 2^{n+1}\big)$, with $\alpha_n=f(I_n)\Rightarrow~\alpha_{n+2}=1+5\alpha_{n+1}-6\alpha_n, \; \alpha_0=1, \alpha_1=0$, hence $\alpha_n=(3^n-2^{n+1}+1)/2, \boxed{f(x)=(3^{[\ln_2x]}-2^{[\ln_2x]+1}+1)/2, \; x>1}$ I think $ \alpha_1 = 6 $, so $ \alpha_n = \frac {3^{n+2} -2^{n+3} +1} {2} $

Determine all functions $f:[0,\infty)\rightarrow\mathbb{R}$ such that $f(0)=0$ and $$f(x)=1+5f\left( \left\lfloor \frac{x}{2} \right \rfloor\right)-6f\left(\left \lfloor\frac{x}{4} \right \rfloor \right)$$for all $x>0$.

PCO please come on