Equation involving floor function and fractional part function How to solve $\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} = \{x\} + \frac{1}{3}$ , where $\lfloor  \rfloor$ denotes floor function and {} denotes fractional part.  I did couple of questions like this by solving for {x} and bounding it from 0 to 1. So here we will have, $$0\le\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} - \frac{1}{3}\lt1$$  adding throughout by $\frac{1}{3}$ $$\frac{1}{3}\le\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} \lt\frac{4}{3}$$  Now I am stuck.

Let $x=n+f$, where $f$ is the fractional part.
If $f<\dfrac12$,
$$0\le f=\frac1n+\frac1{2n}-\frac13<1.$$
This is achieved for $n=2,3,4.$
If $f\ge\dfrac12$,
$$0\le f=\frac1n+\frac1{2n+1}-\frac13<1,$$ which is not possible.
