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Post Undeleted by Greg Kuperberg
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Ian Morris has it essentially correct in his comment. If you solve the equation `$x = (\sin \pi y/2)^2$` for $y$, then the orbit of $y \in \mathbb{R}/\mathbb{Z}$ under `$y \mapsto 2y$` is dense if and only if $y$ is 2-normal. In other words, if every finite binary string appears in the binary expansion of $y$. If you look at the Wikipedia page for Now, this is weaker than being normal numbersin base 2, it's clear because that there won't be requires that every binary string appears equally often, not just that it appears. Let's call such a simpler description than number "topologically 2-normal" (or 2-dense could be another name), because the 2-normality condition is equivalent to saying that the orbit of any proven example$y$ is not just dense, but ergodic. You My impression is that not much more is known about topologically normal numbers than about normal numbers. For instance, you can certainly conjecture that any reasonably simple choice of $x$ that makes $y$ irrational also makes $y$ 2-normalalgebraic number is topologically normal in base 2 (or in any other base), but such numbers have generally not been proven to be normalit doesn't look like it is known. In any case, topological normality is the heart of the question.

Post Deleted by Greg Kuperberg
1

Ian Morris has it essentially correct in his comment. If you solve the equation `$x = (\sin \pi y/2)^2$` for $y$, then the orbit of $y \in \mathbb{R}/\mathbb{Z}$ under `$y \mapsto 2y$` is dense if and only if $y$ is 2-normal. In other words, if every finite binary string appears in the binary expansion of $y$. If you look at the Wikipedia page for normal numbers, it's clear that there won't be a simpler description than that of any proven example. You can certainly conjecture that any reasonably simple choice of $x$ that makes $y$ irrational also makes $y$ 2-normal, but such numbers have generally not been proven to be normal.
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