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To write it symbolically Let $\sqrt{2}=1.41421356...=1.x_{1}x_{2}x_{3}...x_{n}...$

so my question does there exist at least one digit $x_{m}$ in the decimal representation of $\sqrt{2}$ such that there exist fixed positive integer $k$ such that:

$$x_{m}=x_{m+Lk}, \forall L \in \{0,1,2,...\}$$ that is $$x_{m}=x_{m+k}=x_{m+2k}=x_{m+3k}=x_{m+4k}=....$$

Math Admiral
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    Are you asking "can we" (per the question title) or "do we" per the question body? – Chris Lewis Nov 28 '24 at 08:38
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    I don't think that question is easily answered. There exist irrational numbers where all of the even-positioned digits are 0. – Calvin Lin Nov 28 '24 at 09:11
  • @ChrisLewis I will fix it now. Sorry for the confusion I caused – Math Admiral Nov 28 '24 at 09:43
  • @CalvinLin Do you mean irrational numbers as Liouville's irrational numbers and similar to them? – Math Admiral Nov 28 '24 at 09:47
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    Right, for example, if $x_i = 1$ when $i=2^k+1$ and 0 otherwise, then this is irrational and all of $x_{2k} = 0 $. $\quad$ Conversely, for the number $0.101100111000111100001111100000\ldots$, it is clear that there is no $(m, k)$ for which $x_{m+Lk}$ is a constant. $\quad$ Thus, irrationality alone is insufficient. – Calvin Lin Nov 28 '24 at 09:55
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    I'm fairly certain that for no "naturally occurring" irrational number (e.g. quadratic or cubic irrationals or even any algebraic irrational, trig. values at rational multiples of $\pi,$ rational $e^x$ where $x$ is a nonzero rational, etc.) do we even know of a single specific digit that appears infinitely many times. FYI, for each irrational number we do know that there are at least $2$ digits that appear infinitely often. However, we don't explicitly know for any "naturally occurring" irrational what at least one of these digits is. – Dave L. Renfro Nov 28 '24 at 10:56

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