For $x\in\mathbb{R}$ define
\begin{equation} g(x)=1+\tfrac{3}{2}\sum_{k=0}^{\infty}\left(\frac{\left\lfloor2^{2k}x\right\rfloor}{2^{2k}}-\frac{\left\lfloor2^{2k+1}x\right\rfloor}{2^{2k+1}}\right) \end{equation}
Desmos link to first four convergents of $g(x)$
Question: Does $g$ map $\mathbb{R}$ onto the middle-thirds Cantor set of $[0,1]$?
One need only consider the question whether $g$ maps the interval $[0,1)$ onto the Cantor set since for $x\in\mathbb{R}$,
\begin{equation} g(x)=g(x-\lfloor x\rfloor) \end{equation}
\begin{align*} g(x) &= g(\lfloor x\rfloor+(x-\lfloor x\rfloor))\\ &= 1+\tfrac{3}{2}\sum_{k=0}^{\infty}\left(\frac{\left\lfloor2^{2k}(\lfloor x\rfloor+(x-\lfloor x\rfloor))\right\rfloor}{2^{2k}}-\frac{\left\lfloor2^{2k+1}(\lfloor x\rfloor+(x-\lfloor x\rfloor))\right\rfloor}{2^{2k+1}}\right)\\ &= 1+\tfrac{3}{2}\sum_{k=0}^{\infty}\left(\frac{2^{2k}\lfloor x\rfloor+\left\lfloor2^{2k}(x-\lfloor x\rfloor)\right\rfloor}{2^{2k}}-\frac{2^{2k+1}\lfloor x\rfloor+\left\lfloor2^{2k+1}(x-\lfloor x\rfloor)\right\rfloor}{2^{2k+1}}\right)\\ &=1+\tfrac{3}{2}\sum_{k=0}^{\infty}\left(\frac{\left\lfloor2^{2k}(x-\lfloor x\rfloor)\right\rfloor}{2^{2k}}-\frac{\left\lfloor2^{2k+1}(x-\lfloor x\rfloor)\right\rfloor}{2^{2k+1}}\right)\\ \end{align*}
More about the Cantor set at the following link: