Timeline for answer to Numbers with known irrationality measures? by Gerry Myerson

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Feb 27, 2012 at 15:43	comment	added	Kevin O'Bryant		A result of Florian Luca et al is that if the scf of $\alpha$ has exactly 2 values of partial quotients, and the indices where one of the values happens is automatic (in the sense of the Allouche & Shallit) book, then $\alpha$ is transcendental. So, for example, if you take $a_i=3$ if the binary expansion of $i$ has an even number of 1's, and $a_i=7$ otherwise, you get a provably transcendental number.
Feb 27, 2012 at 15:40	comment	added	Kevin O'Bryant		Fix an irrational $\alpha>1$, $A=\\{ \lfloor n \alpha \rfloor: n\geq 1\\}$, and let $a_i=2$ if $i\in A$ and $a_i=1$ if $i\not\in A$. The number $[a_0;a_1,a_2,\dots]$ is known to be transcendental. The idea of the proof is that if the scf of $\alpha$ is almost periodic, then there are quadratic irrationals too close to $\alpha$: by a theorem of Schmidt (analogous to Liousville's Theorem) algebraic numbers cannot be super-well-approximated by other algebraics.
Feb 27, 2012 at 5:59	comment	added	Gerry Myerson		To the best of my knowledge, no.
Feb 27, 2012 at 5:28	comment	added	Noam D. Elkies		Is any explicit number with this property known to be transcendental, or to be definable by a closed form other than a continued fraction with a given sequence of bounded partial quotients (other than a quadratic irrationality, which is known to happen iff the sequence is eventually periodic)?
Feb 27, 2012 at 5:22	history	answered	Gerry Myerson	CC BY-SA 3.0