I asked my last question on the topic of representing numbers by decimals which was intimately linked to another matter: namely, the relationship between the size of a decimal expansion of a fraction and it's numerator/denominator. This in itself led to some other inquiries and then the following question popped up:

Is it true that there exist an infinite number of primes whose decimal string has no consecutive repeated digits?

This seems to be an even more difficult question and I am curious if one can hope of giving this an answer.

  • $\begingroup$ This is not a site about numerology, but about real mathematical problems of al sorts. $\endgroup$ Nov 10, 2017 at 18:47
  • $\begingroup$ Looking just at the sequence $s = 127, 12127, 1212127\dots$, where $s_i = 120(100^i - 1)/99 + 7$, if the conjecture is true, then for $s$ and all other similar sequences, then all $s_i$ are composite after a certain $i$. $\endgroup$ Nov 10, 2017 at 21:58
  • $\begingroup$ Do I understand it right that you want to know whether there are infinite many primes such that none of the digit-pairs $00,11,22,\cdots,99$ appears in the decimal expansion ? This is almost certain the case, but also almost certain is a proof out of reach. $\endgroup$
    – Peter
    Nov 11, 2017 at 0:11
  • $\begingroup$ I rolled back my edit to your Question, which was made with the understanding that you were asking about repeated digits in the decimal fraction corresponding to prime reciprocals. Since you present no example, it is at least as likely that you mean the digits of the prime itself not repreating (which Ross addressses in his Answer). You don't seem to have responded to any of the Comments or the Answer, so it's not clear what you intended to ask about. $\endgroup$
    – hardmath
    Nov 30, 2017 at 20:38

1 Answer 1


We can give a heuristic answer. Consider all the $n$ digit numbers. There are $9\cdot 10^{n-1}$ of them, of which $9^n$ have no repeated digits. The density of primes around $N$ is $\frac 1{\log N}$. We will take the maximum $N$ among our numbers, giving the minimum chance for a prime, which is $10^{n}$. We therefore expect more than $\frac {9^n}{\log \left(10^{n}\right)}$ primes in the decade with no repeated digits. This is huge compared with $1$, so summing over $n$ will give an infinite number of expected primes with no repeated digits.

  • $\begingroup$ It would take a bit of work, but you could probably use Dirichlet's theorem to make that sort of argument rigorous. $\endgroup$
    – anomaly
    Nov 30, 2017 at 20:37
  • $\begingroup$ @anomaly: I don't see how to find an arithmetic progression within the numbers with no repeated digits, but I haven't thought much about it. $\endgroup$ Nov 30, 2017 at 20:42

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