I have a little question. In fact, is too short.

Is infinite sequence of irrational numbers digits mathematically observable?

I would like to explain it by example because the question seems unclear in this way.

A simple example:

$\sqrt 2=1,41421356237309504880168872420969\\807856967187537694807317667973799073247\\846210703885038753432764157273501384623\\091229702492483605585073721264412149709\\993583141322266592750559275579995050115\\278206057147010955997160597027453459686\\201472851741864088 \cdots$

Is it possible to prove that there is no combination of $\left\{0,0,0\right\}$, $\left\{1,1,1\right\}$ or $\left\{2,2,2\right\}$ in this writing?

By symbolic mathematical definition,

Let, $\phi_{\sqrt 2}(n)$ is n'th digit function of $\sqrt 2.$

Question: Is there an exist such a $n\in\mathbb{Z^{+}}$, then $\phi_{\sqrt 2}(n)=0, \phi_{\sqrt 2}(n+1)=0, \phi_{\sqrt 2}(n+2)=0$ ?

Or other combinations can be equal,

$$\phi_{\sqrt 2}(n)=0, \phi_{\sqrt 2}(n+1)=1,\phi_{\sqrt 2}(n+2)=2, \phi_{\sqrt 2}(n+3)=3, \phi_{\sqrt 2}(n+4)=4, \phi_{\sqrt 2}(n+5)=5$$

Here, $\sqrt 2$ is an only simple example. The question is not just $\sqrt 2$.

Generalization of the question is :

For function $\phi _\alpha (n)$, is it possible to find any integer sequence ? where $\alpha$ is an any irrational number or constant ($e,\pi\cdots$ and etc).

I "think" , the answer is undecidability. Because, we can not observe infinity. Of course, I dont know the correct answer.

Sorry about the grammar and translation errors in my English.

Thank you very much.