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I have asked the following question on Math.SE some time ago and offered a bounty, yet received no answers nor comments, so I'm posting it here.

The Prouhet-Thue-Morse constant, defined as

$$ \tau =\sum _{{i=0}}^{{\infty }}{\frac {t_{i}}{2^{{i+1}}}}=0.412454033640\ldots $$

where the $t_i$ are elements of the Thue-Morse sequence, is transcendental. But is

$$ \tau_b =\sum _{{i=0}}^{{\infty }}{\frac {t_{i}}{b^{{i+1}}}} $$

also transcendental, for $b>2$?

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    $\begingroup$ Did you try to adapt the proof for the transcendence of $\tau$ to $\tau_b$? $\endgroup$
    – Francesco Polizzi
    Commented Mar 29, 2019 at 10:13
  • $\begingroup$ No, as I was hoping there were already results of that nature available $\endgroup$
    – Klangen
    Commented Mar 29, 2019 at 10:19
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    $\begingroup$ I believe there are papers of Allouche that address this. I will try to check later. $\endgroup$
    – Anthony Quas
    Commented Mar 29, 2019 at 13:53

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Michel Waldschmidt, Words and transcendence, writes in Section 3.1, on page 461, "Mahler also proved in 1929 that the so-called Prouhet–Thue–Morse–Mahler number in base $g\ge2$, given by $$\xi_g=\sum_{n\ge0}{a_n\over g^n}$$ where $(a_n)_{n\ge􏰀0}$ is the Prouhet–Thue–Morse sequence, is transcendental; see [52] and [15, Section 13.4]." [15] is Allouche and Shallit, Automatic sequences: Theory, Applications, Generalizations (Cambridge University Press, 2003). [52] is K. Nishioka, Mahler Functions and Transcendence (Lecture Notes in Mathematics 1631, Springer- Verlag, 1996).

Waldschmidt gives "the idea of proof," and refers to [15, Section 13.4] and [52, Example 1.3.1] for the full proof.

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  • $\begingroup$ Fantastic, thanks a lot! $\endgroup$
    – Klangen
    Commented Mar 31, 2019 at 9:39

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