<s>[INST] 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 $$\lfloor \frac{2n+2p+1}{3}\rfloor +\lfloor \frac{\lfloor4n+4p\rfloor+2}{3}\rfloor=1\\0\leq 2p<2 \\0\leq 4p<4 $$
 [/INST] If $x<0$ then $a=\lfloor \frac{2x+1}3\rfloor$ and $b=\lfloor \frac{\lfloor 4x\rfloor+2}3\rfloor$ are both inferior or equal to $0$, so the sum cannot be $1$.
Now since $a$ is smaller than $b\quad$  [ $(\lfloor 4x\rfloor+2)-(2x+1)\ge 4x-1+2+2x-1\ge 2x\ge 0$ ]

The only way to get $a+b=1$ with $0\le a\le b\quad$ integers is $a=0$ and $b=1$.
$\begin{cases} a=\lfloor \frac{2x+1}3\rfloor=0\\ b=\lfloor \frac{\lfloor 4x\rfloor+2}3\rfloor=1\end{cases}\iff
\begin{cases} 0\le2x+1<3\\ 3\le\lfloor 4x\rfloor+2<6 \end{cases}\iff
\begin{cases} -\frac 12\le x<1\\ 1\le\lfloor 4x\rfloor<4 \end{cases}\iff
\begin{cases} -\frac 12\le x<1\\ \frac 14\le x<1 \end{cases}$
Finally we get $\bbox[5px,border:2px solid]{x\in[\frac 14,1[}$
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