Is $ 0.112123123412345123456\dots $ algebraic or transcendental?

Question

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 $1234567891011$; ... )

Answer 1

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

Answer 2

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.

Answer 3

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!