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 $1234567891011$; … )

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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!

$\lim_{n\to \infty}$ of the “nth block” will be equal to the exact same digits as the Champernowne constant. The Champernowne constant is transcendental, so no matter what comes before its digits your number will be transcendental.