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In this question

First $k$ digits of $\pi^n$ and compositeness

it is asked for some $\ n\ $ giving late or possible never a prime number. A good condidate is $\ n=18\ $. According to my calculations with PARI/GP, we do not get a prime after more than $\ 11\ 000$ digits. Note that the digits before the comma are used as well and we also do not arrive at a prime before reaching the comma.

Is $\ \lfloor \pi^{18}\cdot 10^k \rfloor\ $ ever prime ?

Heuristically, we can expect that a prime will eventually occur , if we assume that the digits of $\ \pi^{18}\ $ behave like a pseudo-random-generator. Motivation for the $\ 18\ $ is that it is the first tough case (see the table in the answer).

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  • $\begingroup$ What did I say ? The downvote is there :) $\endgroup$
    – Peter
    Commented Mar 14, 2020 at 13:45
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    $\begingroup$ Is it clear that a (uniformly) randomly generated string yields a prime at some point with probability $1$? Is there an expected density of primes produced in this way? $\endgroup$
    – lulu
    Commented Mar 14, 2020 at 13:55
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    $\begingroup$ @lulu Even if clear if it´s random, primes are not, only in random models about them. $\endgroup$
    – user757601
    Commented Mar 14, 2020 at 13:57
  • $\begingroup$ @lulu This has a point. Even, if the probability is not exactly as the 1/ln(n)-model predicts it is probably large enough to ensure that we must arrive at a prime eventually, if the digits are actually random. Since we have no reason to doubt that the digits of $\pi^{18}$ behave like a random-generator, I highly exect not only one prime, but infinite many. But the smallest could well be too large to be ever detected ! $\endgroup$
    – Peter
    Commented Mar 14, 2020 at 14:05
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    $\begingroup$ In the mean time, I passed $\ 14\ 000\ $ digits $\endgroup$
    – Peter
    Commented Mar 14, 2020 at 14:07

1 Answer 1

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$ \large \left \lfloor \pi ^{18} \cdot 10^{16718} \right \rfloor $ is a probable prime.

Pari/GP code:

\p 16800
ispseudoprime(floor(Pi^18*10^16718))
time = 1min, 22,156 ms.
%# = 1
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  • $\begingroup$ I see, Peter already found it about an hour before I posted. $\endgroup$ Commented Mar 14, 2020 at 16:23

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