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 $$\lfloor \frac{2n+2p+1}{3}\rfloor +\lfloor \frac{\lfloor4n+4p\rfloor+2}{3}\rfloor=1\\0\leq 2p<2 \\0\leq 4p<4 $$
 A: 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...
A: If $x \geqslant 1,$ then $\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor \geqslant 1+2>1.$ If $x<.25,$ then $\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor < 0+1,$ so solutions will not be found in these intervals. Conversely, if $.25 \leqslant x < 1,$ then $\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor = 1.$
For a more direct solution, consider the possible (integer) values of $\lfloor{4 x \rfloor}.$
Case $\lfloor{4 x \rfloor} \leqslant -1:$ This means $x <0.$ In this case, 
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor \leqslant 0+0=0$
Case $\lfloor{4 x \rfloor}=0:$ This means $0 \leqslant x <\frac{1}{4}.$ In this case, 
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor = 0+0=0$
Case $\lfloor{4 x \rfloor}=1:$ This means $\frac{1}{4} \leqslant x <\frac{1}{2}.$ In this case, 
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor = 0+1=1$
Case $\lfloor{4 x \rfloor}=2:$ This means $\frac{1}{2} \leqslant x <\frac{3}{4}.$ In this case, 
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor =0+1=1 $
Case $\lfloor{4 x \rfloor}=3:$ This means $\frac{3}{4} \leqslant x <1.$ In this case, 
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor =0+1=1$
Case $\lfloor{4 x \rfloor} \geqslant 4:$ This means $1 \leqslant x.$ In this case, 
$\lfloor{\frac{2 x+1}{3}} \rfloor + \lfloor{\frac{\lfloor{4 x \rfloor}+2}{3}} \rfloor \geqslant 1+2=3.$
Thus the solution is $\frac{1}{4} \leqslant x < 1.$
A: 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 $\bbox[5px,border:2px solid]{x\in[\frac 14,1[}$
A: Clearly it is not question of negative numbers. Looking at $x\ge0$,
$$\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor=\begin{cases}0\text{ for } 0\le x\lt\frac14\\1\text{ for }\frac14\le x\lt1\end{cases}$$
$$\lfloor \frac{2x+1}{3}\rfloor=\begin{cases}0\text{ for } 0\le x\lt1\\1\text{ for }\frac14\le x\lt\frac52\end{cases}$$
Since you want $$\lfloor \frac{2x+1}{3}\rfloor +\lfloor \frac{\lfloor4x\rfloor+2}{3}\rfloor=1$$ you just have to see at the only compatibility $$\color{red}{\frac14\le x\lt1}$$
A: The function
$$f(x)=\left\lfloor2x+1\over3\right\rfloor+\left\lfloor\lfloor4x\rfloor+2\over3\right\rfloor$$
is the sum of two non-decreasing and right semi-continuous step functions, so we need to find the smallest values for $a$ and $b$ such that $f(a)=1$ and $f(b)\gt1$, which will give an interval $[a,b)$ within which $f(x)=1$. That is, we need to consider where $f$ takes its steps.
The first part of $f$, $\lfloor(2x+1)/3\rfloor$, steps from $0$ to $1$ at $x=1$, while the second part steps from $0$ to $1$ at $x={1\over4}$ and then from $1$ to $2$ at $x=1$. Thus $a={1\over4}$ and $b=1$ (with $f(b)=3$), so the solution set is $[{1\over4},1)$.
