$\pi$ is irrational, and of course its digits are not a repeating sequence. However I cannot wrap my head around the following...

Is there a digit $x$ in $\pi$ such that the digits before are repeated in reverse order after that digit? $$\pi = 3.14159\ldots zyxxyz \ldots 951413 [\ldots]$$

In other words, is there a positive integer $n$ such that $\lfloor 10^n\pi\rfloor$ is palindromic?

$\pi$ has infinite digits, but it is unknown whether any sequence of digits will appear at some point. Moreover, this isn't just a requirement for $n$ digits appearing at some position, but for a sequence of $n$ digits has to appear at position $n$.

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    "Pi has infinite digits, so any sequence of digits will appear at some point." Why is that? – Randall Aug 05 '20 at 16:54
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    Regarding what @Randall said (I was about to comment myself, then saw his/her comment), note that the digit $2$ does not appear in the following irrational number: $0.101001000100001000001\ldots$ (pattern of $1$'s and $0$'s continues in the suggested pattern) – Dave L. Renfro Aug 05 '20 at 16:57
  • @an4s well my question was based on a false premise, and as the answer to the other question is "not really" it also does answer this one... – 463035818_is_not_a_number Aug 05 '20 at 16:58
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    You can restate your question as: *"Is there a positive integer $n$ such that $\lfloor 10^n\pi\rfloor$ is [palindromic](https://en.wikipedia.org/wiki/Palindromic_number)?"* ... The answer to this is presumably independent of whether "any sequence of digits" appears in $\pi$, so this is a perfectly reasonable stand-alone question. (So, I don't believe this question should have been closed as a duplicate.) In any case, I'd be pretty surprised if the answer were "Yes". – Blue Aug 05 '20 at 17:07
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    For those interested, I collected together a lot of Mathematics Stack Exchange questions (made before 9 April 2018) relating to digits of $\pi$ (among other related topics) in my answer to [Normal Numbers as members of a larger set?](https://math.stackexchange.com/a/2729156/13130) – Dave L. Renfro Aug 05 '20 at 17:12
  • @Blue you are right, I edited accordingly. Anyhow, I'm still be fine with the duplicate, that wrong assumption aside I don't believe anymore that such digit exists – 463035818_is_not_a_number Aug 05 '20 at 17:32
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    A heuristic argument, assuming $\pi$ has "random" digits: since $\sum_{n\ge1}10^{-n/2}=\frac{1}{10-\sqrt{10}}\ll1$, (i) there probably is no such $n$ & (ii) there almost certainly are no more than a few such $n$. – J.G. Aug 05 '20 at 17:56

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