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Let $\text{fd}(k,x)$ be first $k$ digits of some real number $x$.

For $\pi=x$ we have the sequence $,3,31,314,3141,3141,31415,...$ (in base $10$)

For $\pi^2=x$ we have $9,98,986,9869,...$ (in base $10$)

And so on.

I came to an idea of thinking does there exists $m \in \mathbb N$ such that $k \to \text{fd}(k,\pi^m)$ are all composite numbers?

This seems to be highly unlikely, and I do not know how to provide a proof.

This question on MO.

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  • $\begingroup$ @Peter This is about all being composites, not about finite sequences of primes. $\endgroup$
    – user757601
    Mar 14, 2020 at 12:44
  • $\begingroup$ Sorry, but what do you think about Haran's suggestion that we should look at the fractional part ? $\endgroup$
    – Peter
    Mar 14, 2020 at 12:45
  • $\begingroup$ Heuristically, a prime should occur at some point, but I think we can find very tough cases. $\endgroup$
    – Peter
    Mar 14, 2020 at 12:45
  • $\begingroup$ @Peter That´s just a different question, very similar. $\endgroup$
    – user757601
    Mar 14, 2020 at 12:45
  • $\begingroup$ @Peter How much tough? $\endgroup$
    – user757601
    Mar 14, 2020 at 12:46

1 Answer 1

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If we consider all digits (as suggested in the question), the first tough case is for $\ m=18\ $. The search is still in progress. I passed $\ 9\ 000\ $ digits without finding a prime.

The smallest $\ n\ $ such that $$\lfloor x\cdot 10^n \rfloor$$ is prime can be seen in the following table. A positive entry means that this number of digits after the comma is needed. A non-positive number indicates that we get a prime already before the comma is reached :

1  0
2  106
3  -1
4  0
5  -2
6  1
7  -3
8  2
9  -4
10  27
11  -5
12  0
13  -6
14  -1
15  -7
16  6
17  -8
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  • $\begingroup$ A probable prime occured for $\ n=16\ 718\ $ $\endgroup$
    – Peter
    Mar 14, 2020 at 15:04
  • $\begingroup$ The next possibly tough case is $\ \pi^{237}\ $ which passed $\ 6\ 000\ $ digits. $\endgroup$
    – Peter
    Mar 14, 2020 at 17:05

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