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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asked Sep 23, 2019 at 21:09
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10$\begingroup$ Those are called emirps and your question is a (presumably very difficult) open problem. en.wikipedia.org/wiki/Emirp $\endgroup$– WojowuCommented Sep 23, 2019 at 21:19
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7$\begingroup$ @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 $\endgroup$– Gerry MyersonCommented Sep 23, 2019 at 23:07
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2$\begingroup$ @GerryMyerson Good catch, my bad. $\endgroup$– WojowuCommented Sep 24, 2019 at 8:03
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5$\begingroup$ More generally, are there infinitely many primes such that at least one non-trivial permutation of their digits preserves the primality? $\endgroup$– Sylvain JULIENCommented Sep 24, 2019 at 12:06
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3$\begingroup$ 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 TaoCommented Sep 25, 2019 at 4:54
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1 Answer
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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.
answered Sep 24, 2019 at 14:18
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1$\begingroup$ 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. $\endgroup$ Commented Sep 24, 2019 at 18:12
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$\begingroup$ @AlekseiKulikov you are right... fixed now $\endgroup$ Commented Sep 24, 2019 at 18:30
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1$\begingroup$ Can Maynard's work about primes with missing digits be of any use to show a similar result for base ten? $\endgroup$ Commented Sep 24, 2019 at 18:41
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1$\begingroup$ @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. $\endgroup$ Commented Sep 24, 2019 at 18:47