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My question is in the title:

Is $a=0.248163264128…$ a transcendental number? The number $a$ is defined by concatenating the powers of $2$ (in base $10$).


It is possible to express $a$ as a series :

$$a = \sum\limits_{n=1}^{\infty} 2^n \cdot 10^{ -\sum\limits_{k=1}^{n} (\lfloor{ k \cdot \log_{\,10}\,(2) }\rfloor + 1) } \tag{*}$$

I know that $a$ is irrational.

I know that if I consider the powers of $10$ instead the powers of $2$, i.e. if I consider $b=0.10100100010000...$, this number is transcendental.

Looking at the series (*), it seems very difficult to establish the transcendence of $a$. However, it is known (thanks to Kurt Mahler) that numbers as:

$$c = 0.149162536… = \sum\limits_{n=1}^{\infty} n^2 \cdot 10^{ -\sum\limits_{k=1}^{n} (\lfloor{ 2 \cdot \log_{\,10}\,(k) }\rfloor + 1) } \tag{**} $$

are transcendental ($c$ is the concatenation of the square numbers in base $10$ ; the same holds for third powers and so on).

I am aware that this could be a difficult problem. Similar numbers, as Copeland-Erdős constant, are not known to be transcendental. I would really appreciate if anyone had a reference about this number $a$, because I didn't find anything that could help me to determine whether $a$ is transcendental, or whether it is still unknown.

Thank you very much!

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    $\begingroup$ It is almost surely trancendental. That's all I can tell. $\endgroup$ – Arthur Feb 17 '16 at 10:05
  • $\begingroup$ It's most likely not Liouville though, so that rules out one potential solution $\endgroup$ – vrugtehagel Feb 17 '16 at 10:15
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    $\begingroup$ @Arthur: Do you mean that the probability of it being algebraic is $\frac{\aleph_0}{2^{\aleph_0}}=0$? $\endgroup$ – barak manos Feb 17 '16 at 10:44
  • $\begingroup$ The simple continued fraction starts with $[0;4,33,1,3,2,565,3,5,...]$. I would guess it's transcendental $\endgroup$ – Yuriy S Feb 17 '16 at 11:53
  • $\begingroup$ @barakmanos Yes, that is what I mean. $\endgroup$ – Arthur Feb 17 '16 at 11:55
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Yann Bugeaud "Distribution Modulo One and Diophantine Approximation", page 221:

For integers $b \geq 2$ and $c \geq 2$, let $(c)_b$ denote the sequence of digits of $c$ in its representation in base $b$. Mahler [471] proved that the real number $0 (c)_{10}(c^2)_{10} \dots$ is irrational. This was subsequently reproved and extended to every base $b \geq 2$ by Bundschuh [170] and Niederreiter [539]; see also [69, 172, 647, 652].

Problem 10.48. With the above notation, prove that, for arbitrary integers $b \geq 2$ and $c \geq 2$, Mahler's number $0 (c)_{b}(c^2)_{b} \dots$ is transcendental and normal to base $b$.

The question of normality of $0.248163264 \dots$ to base $10$ was already posed by Pillai [561].

Some of the links I was able to recover:

[172] P.Bundschuh, P. J.-S. Shiue and X.Y. Yu, Transcendence and algebraic independence connected with Mahler type numbers, Publ. Math. Debrecen 56 (2000), 121-130.

[647] Z.Shan, A note on irrationality of some numbers, J. Number Theory 25 (1987), 211-212.

[652] Irrationality criteria for numbers of Mahler's type. In: Analytic Number Theory (Kyoto, 1996), 343-351, London Math. Soc. Lecture Note Ser., 247, Cambridge University Press, Cambridge, 1997.


Some data on this number from me. More digits:

$$0.2481632641282565121024204840968192163843276865536\dots$$

Simple continued fraction:

$$[0; 4, 33, 1, 3, 2, 565, 3, 5, 1, 10, 1, 43, 1, 1, 1, 1, 3, 1, 4, 1, 1, 3, 2, 3, 3, 2, 1, 1, 3, 5, 1, 16, 1, 15, 1, 2, 1, 3, 1, 3, 3, 327, \dots]$$

Euler type continued fraction:

$$\cfrac{1}{5-\cfrac{5}{6-\cfrac{5}{6-\cfrac{5}{51-\cfrac{50}{51-\cfrac{50}{51-\cfrac{50}{501-...}}}}}}}$$

The probability of a bigger partial quotent to occur after a smaller one in this fraction is equal to:

$$\frac{\ln 2}{\ln 10}=0.30103 \dots$$

Note that this fraction always approaches the number from below, fot example this truncation is exactly equal to $0.248163264128$

Unfortunately, general continued fractions do not afford any insight in the trancendentality of a number, as far as I know.

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  • $\begingroup$ Well, thank you very much! Maybe it can be useful to notice that your first quote belongs to the chapter "Conjectures and open questions"… ! $\endgroup$ – Watson Feb 18 '16 at 11:00
  • $\begingroup$ @Watson, yes, of course. But google books did not allow me to view every page, so I only knew it was chater 10 $\endgroup$ – Yuriy S Feb 18 '16 at 11:02
  • $\begingroup$ No problem! Do you have a reference for "The probability of a bigger partial quotient to occur after a smaller one is equal to $\log(2)$"? Is this about the simple continued fraction or the Euler type continued fraction? Have you found this in Y. Bugeaud's book or one of the references you gave? $\endgroup$ – Watson Feb 18 '16 at 11:05
  • $\begingroup$ @Watson, no, that one is my own conclusion. It is rather obvious, since the number of $2^k$ terms having exactly $n$ digits in their base $10$ representation is either $3$ or $4$ for any $n$, and the number of digits for any $k$ is equal to the whole part ot $\log_{10}2^k$ $\endgroup$ – Yuriy S Feb 18 '16 at 11:10
  • $\begingroup$ @Watson, again, in the context of your question Euler CF is useless, unless someone can connect the properties of general continued fractions (since they are not unique) to the properties of a number $\endgroup$ – Yuriy S Feb 18 '16 at 11:27

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