If we choose some prime, say $11$, we can concatenate one digit to the left and one to the right to obtain another prime, for example, $2113$, we can do the same with $2113$ and obtain $121139$, a prime, and so on...

Let us call some prime (written in decimal notation as $a_1...a_k$) a $1$-concatenable prime if there exist digits $b_1$ and $c_1$ such that $b_1a_1...a_kc_1$ is prime and digits $b_2$ and $c_2$ such that $b_2b_1a_1...a_kc_1c_2$ is prime and...and digits $b_l$ and $c_l$ such that $b_l...b_2b_1a_1...a_kc_1c_2...c_l$ is prime, and so on...

That is, a prime number is $1$-concatenable prime if, starting from it, we can build larger and larger primes in such a way that at each new step we concatenate one digit from the left and one from the right, and if we can repeat that process an infinite number of times.

Is there at least one $1$-concatenable prime?

If there is none, are there chains of arbitrary length, that is, is it true that for every $n \in \mathbb N$ there exists some prime $p$ such that $p$ is $n$ times concatenable in the described way, that is, by concatenating at each step one digit from the left and one from the right?

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    $\begingroup$ Posted to mse but deleted by author, math.stackexchange.com/questions/2890598/1-concatenable-primes $\endgroup$ – Gerry Myerson Aug 22 '18 at 13:23
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    $\begingroup$ Now posted to mse again, math.stackexchange.com/questions/2891014/1-concatenable-primes. PLEASE DON'T DO THIS. $\endgroup$ – Gerry Myerson Aug 22 '18 at 13:34
  • $\begingroup$ Here is some more concrete advice on posting questions to both math.stackexchange and mathoverflow: meta.mathoverflow.net/a/2638 $\endgroup$ – j.c. Aug 22 '18 at 13:40
  • $\begingroup$ It is not a priori clear to me that this question is decidable $\endgroup$ – Stanley Yao Xiao Aug 22 '18 at 16:02
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    $\begingroup$ For any two digit pattern $bc$ from $00$ to $99$ there is a prime $abcd $ with $a \gt 0$. The same is true for three digit patterns (in fact $909$ and $962$ can be extended two ways and everything else at least three ways.) On average there are $8$ ways each. That won't persist forever but by my back of the envelope calculations only for $n$ around $40$ digits does the expected number of completions $anz$ of $n$ to a prime fall below 1. $\endgroup$ – Aaron Meyerowitz Aug 22 '18 at 19:36

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