<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 $x \geqslant 1,$ 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 $\frac{1}{4} \leqslant x < 1.$
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