# Is $0.112123123412345123456\dots$ algebraic or transcendental?

Let $$x=0.112123123412345123456\dots$$ Since the decimal expansion of $$x$$ is non-terminating and non-repeating, clearly $$x$$ is an irrational number.

Can it be shown whether $$x$$ is algebraic or transcendental over $$\mathbb{Q}$$ ? I think $$x$$ is transcendental over $$\mathbb{Q}$$. But I don't know how to formally prove it. Could anyone give me some help ? Any hints/ideas are much appreciated. Thanks in advance for any replies.

My Number:

$$x=0.\underbrace{1}_{1^{st}\text{ block}}\overbrace{12}^{2^{nd}\text{ block}}\underbrace{123}_{3^{rd}\text{ block}}\overbrace{1234}^{4^{th}\text{ block}}\dots \underbrace{12\dots n}_{n^{th}\text{ block}}\dots$$ where $$n^{th}$$ block is the first $$n$$ positive integers for each $$n\in \mathbb{Z}^+$$.

(That is the 10th block of $$x$$ is $$12345678910$$; The 11th block is $$1234567891011$$; ... )

• This question is related and has some references which might be useful. Feb 16, 2015 at 8:01
• Unclear question! How about the pattern after 123456789? Feb 16, 2015 at 17:15
• It would be followed by 12345678910. Feb 16, 2015 at 17:23
• @tomi: The "better form" may change OP's answer, for comparison Champernowne constant 0.12345678910111213... is transcendental while $\sum_{i=1}^\infty i/10^i$ = 0.123456790123... = 10/81. Feb 28, 2015 at 14:33
• @barakmanos Actually, probability is measure theory, not cardinality, and probability zero doesn't mean much when dealing with a specific real number. Feb 28, 2015 at 14:55

Obviously, it can be outputed by Turing Machine in real time. Thus under the Hartmanis-Stearns conjecture, it is a transcendental number.

• ams.org/journals/tran/1965-117-00/S0002-9947-1965-0170805-7/… Jan 25, 2018 at 2:37
• Which theorem in the paper gives this result? More importantly, for me, what does "in real time" mean? Jan 25, 2018 at 2:44
• @CarlMummert rjlipton.wordpress.com/2012/06/15/… Jan 25, 2018 at 3:11
• You are misstating the conjecture, the number could be rational (think about $0.0000000...$). Jan 26, 2018 at 3:22
• @MoisheCohen Hmmmm, but is this number periodical? I have not misstated anything Jan 26, 2018 at 4:05

I believe your number can be written as

$$\sum _{j=1}^\infty 10^{-\sum _{m=1}^{j+\frac{1}{2}} \sum _{n=1}^m \left\lceil \log _{10}(n+1)\right\rceil } \left\lfloor c 10^{\sum _{n=0}^j \left\lceil \log _{10}(n+1)\right\rceil }\right\rfloor$$

where $c$ is the Champernowne constant.

Don't know if that helps.

Your number can be written with the following formula: $$\sum_{n=1}^{\infty} \frac{ \sum_{r=1}^n r(10)^{n-r}}{10^{\frac{n(n+1)}{2}}}$$ I don't know how to prove it is transcendental, but I hope this helps!

• I just realized this isn't correct, I didn't account for blocks with two or three decimal digits like 1234567891011...
– Rob
Oct 15, 2015 at 2:53
• If I may ask : What is the procedure that you made to arrive to that expression ? Aug 26, 2019 at 13:24