A floor function equation I want to solve below equation analytically $$\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor=1$$ but I have no idea to start . 
Implicit: I solve it by graphing and the solution was $x\in[\frac {1}{4},1)$
  I am thankful for any idea in advance .
I try to use $x=n+p \\0\leq p<1 $ but can't go further $$\lfloor \frac{2n+2p+1}{3}\rfloor +\lfloor \frac{\lfloor4n+4p\rfloor+2}{3}\rfloor=1\\0\leq 2p<2 \\0\leq 4p<4 $$

Use the fact that for each $x$ we have:
 $$x-1<[x]\leq x$$
So $$ \frac{2x+1}{3}+ \frac{\lfloor4x\rfloor+2}{3}-2<\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor \leq \frac{2x+1}{3}+ \frac{\lfloor4x\rfloor+2}{3}$$
So $$ \frac{2x+1}{3}+ \frac{4x+1}{3}-2<1 \leq \frac{2x+1}{3}+ \frac{4x+2}{3}$$
So So $$ \frac{6x+2}{3}-2<1 \leq \frac{6x+3}{3}$$ 
Thus: $6x-4<3\leq 6x+3$ and so $0\leq x<{7\over 6}$ and this is not the end...
