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. – Winther Feb 16 '15 at 8:01
• Unclear question! How about the pattern after 123456789? – Harry Peter Feb 16 '15 at 17:15
• It would be followed by 12345678910. – dalastboss Feb 16 '15 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. – kennytm Feb 28 '15 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. – Thomas Andrews Feb 28 '15 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.

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 Bland Oct 15 '15 at 2:53

protected by Community♦Oct 14 '15 at 18:58

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