Let $f(x)=4x(1-x)$.

For which rational numbers $r\in [0,1]$ is the sequence $f^n(r)$, $n\in \mathbb N$, dense in $[0,1]$ ? $(f^n(r)=f\circ f\circ ...\circ f(r)$ n times)

I would be happy to find a single rational number with dense orbit in $[0,1]$, but my guess is that all rational numbers different from k/2^n should work. The numbers 1/3, 1/5, 1/10 are candidates (at least numerically).

It is known that a.e. point x in [0,1] has a dense orbit (w.r.t the Lebesgue measure). This is shown by conjugating f to x->2x mod 1 with the map $x\rightarrow sin^2(\pi x/2)$. So the question can be rephrased as: for which rational r does Arcsin(r) have a dense orbit in [0,1] under the action of x->2x mod 1 ? (this does not seem simpler though)

Let $f(x)=4x(1-x)$.

For which rational numbers $r\in [0,1]$ is the sequence $f^n(r)$, $n\in \mathbb N$, dense in $[0,1]$ ? $(f^n(r)=f\circ f\circ ...\circ f(r)$ n times)

I would be happy to find a single rational number with dense orbit in $[0,1]$, but my guess is that all rational numbers different from k/2^n should work. The numbers 1/3, 1/5, 1/10 are candidates (at least numerically).

It is known that a.e. point x in [0,1] has a dense orbit (w.r.t the Lebesgue measure). This is shown by conjugating f to x->2x mod 1 with the map $x\rightarrow sin^2(\pi x/2)$. So the question can be rephrased as: for which rational r does ${1\over \pi} Arcsin(r)$ have a dense orbit in [0,1] under the action of x->2x mod 1 ? (this does not seem simpler though)

EDIT: from W. Zudilin answer, it seems that the question is open in full generality. But maybe there is a chance of finding just one rational $r$ such that $f^n(r)$, or $Frac({2^n\over \pi} Arcsin\ r ) $, is dense in [0,1] ?