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A prime number $p=\overline{a_na_{n-1}\ldots a_1a_0}$ is called a two sided prime number if its reverse representation $q=\overline{a_0a_1\ldots a_{n-1}a_n}$ is a prime number too.

Are there infinitely many two sided prime numbers?

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    $\begingroup$ Those are called emirps and your question is a (presumably very difficult) open problem. en.wikipedia.org/wiki/Emirp $\endgroup$
    – Wojowu
    Sep 23, 2019 at 21:19
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    $\begingroup$ @Wojowu the OEIS reserves the term emirp for primes whose reversal is a different prime, so not including $2,3,5,7,11,101$ and others. oeis.org/A006567 The numbers Ali asks about, OEIS calls reversible primes. oeis.org/A007500 $\endgroup$ Sep 23, 2019 at 23:07
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    $\begingroup$ @GerryMyerson Good catch, my bad. $\endgroup$
    – Wojowu
    Sep 24, 2019 at 8:03
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    $\begingroup$ More generally, are there infinitely many primes such that at least one non-trivial permutation of their digits preserves the primality? $\endgroup$ Sep 24, 2019 at 12:06
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    $\begingroup$ Pretty much any statement of the form "there are infinitely many primes $p$ such that $f(p)$ is also prime" for a given function $f$ is beyond current technology to establish if $f$ is not something totally degenerate (e.g. a constant function or the identity function). The case $f(p)=p+2$ being the most famous, of course. $\endgroup$
    – Terry Tao
    Sep 25, 2019 at 4:54

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For many variants of this question the answer seems to be not known but at least this question in the comments

More generally, are there infinitely many primes such that at least one non-trivial permutation of their digits preserves the primality? – Sylvain JULIEN

in binary has the answer Yes. Thanks to @AlexeiKulikov for the improved argument.

Use

  • (i) Prime number theorem
  • (i ii) Pigeonhole principle

By (i) there are at least $c (2^n/n)$ primes in $[2^{n-1},2^n)$ whose number of 1s is in the interval $[1,n]$.

By (ii) there are two primes in $[2^{n-1},2^n)$ with the same number of 1s.

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    $\begingroup$ Sorry, what you did in the second step? Most numbers are not primes as well so it may well be that for primes their digit sums are always very small. In any case you don't need anything like this because there are at most $n$ digit sums and the muber of primes is much, much more (asymptotically $2^n/n$) and this actually works for any base -- there are at most $n^{b-1}$ different sets of digits. $\endgroup$ Sep 24, 2019 at 18:12
  • $\begingroup$ @AlekseiKulikov you are right... fixed now $\endgroup$ Sep 24, 2019 at 18:30
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    $\begingroup$ Can Maynard's work about primes with missing digits be of any use to show a similar result for base ten? $\endgroup$ Sep 24, 2019 at 18:41
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    $\begingroup$ @SylvainJULIEN I guess I wrote it in my comment -- you don't need anything, there are at most $n^9$ sets of $n$ digits and the number of primes is $\sim 10^n/n$ which is much larger. $\endgroup$ Sep 24, 2019 at 18:47

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