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Consider an algebraic irrational number in $(0,1)$ with binary expansion $x = \sum_{n\ge 1} \frac{a_n}{2^n}$. Is it possible that the number $\sum_{n\ge 1}\frac{a_{2n}}{2^n}$ is again algebraic irrational ?

My opinion is no. Heuristics: it should be difficult to determine binary expansions of other algebraic irrationals, even if we have a supply of other expansions, except for trivial operations. My previous question (see link) is of a similar flavor.

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  • $\begingroup$ Do you really mean what you've just written down, or $\sum_{n\geq 1}\frac{a_{2n}}{2^{2n}}$ ? $\endgroup$
    – Loïc Teyssier
    Commented Oct 16, 2017 at 18:34
  • $\begingroup$ Most likely the answer is not known... $\endgroup$
    – Wojowu
    Commented Oct 16, 2017 at 18:35
  • $\begingroup$ @Loïc Teyssier: Correct, what is written down. $\endgroup$
    – orangeskid
    Commented Oct 16, 2017 at 18:43
  • $\begingroup$ Ok, only to check (though I guess it wouldn't make the question any easier). $\endgroup$
    – Loïc Teyssier
    Commented Oct 16, 2017 at 18:45
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    $\begingroup$ It might be fun to do some experiments here. Take some such algebraic irrational, maybe $\sqrt2-1$, apply your operation to it, and then feed the result into the Inverse Symbolic Calculator or into an integer-relation finder like PSLQ, and see whether anything interesting comes out. $\endgroup$
    – Gerry Myerson
    Commented Oct 16, 2017 at 22:46

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