We start with a prime number. Now, we append one of the digits $1,3,7,9$ to get another prime. We continue until it is not possible anymore to append a digit without getting a composite number.

Can we generate arbitary long prime-number sequences this way by choosing a suitable start-prime and suitable digits ?


The truncatable primes have a property stronger than the property I demand. For example, $$\color\red {409} \color\green {9339193933}$$ is not a truncatable prime because $40$ (and also $4$) is not prime, but my condition is satisfied because the start prime is arbitary and beginning with $409$ if we append the green digits we have a prime in each step. Appending another digit makes the number composite, so we cannot continue.

I think a similar argument that there are only finite many truncatable prime will show that we can append only finite many digits , no matter what the starting prime is and which digits we choose.

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    $\begingroup$ Can we append the next digit on the left or right ? Or are you just asking if there are arbitarily large primes whose decimal representation only contain these digits ? $\endgroup$ – Donald Splutterwit Sep 28 '17 at 23:16
  • $\begingroup$ @DonaldSplutterwit Only at the right. For the second question : No $\endgroup$ – Peter Sep 28 '17 at 23:20
  • $\begingroup$ So we can only start with $3$ or $7$ and at each stage after that we have $4$ possible digits we could add ... & so we build a tree of primes? $\endgroup$ – Donald Splutterwit Sep 28 '17 at 23:24
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    $\begingroup$ @DonaldSplutterwit The starting prime is arbitary, it can contain any digits, a start with $23$ is valid. $\endgroup$ – Peter Sep 28 '17 at 23:26
  • $\begingroup$ what remainders do you get when you start dividing by a certain value ? that may help $\endgroup$ – user451844 Sep 28 '17 at 23:33

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