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?
Also asked on MO a few hours ago, here.
I will treat primality as random, I expect arbitrarily long but not infinite strings.
Take all thousand-digit numbers. One number is every 2000 or so is prime. So of the 90 ways to extend, there is less than 1 chance in 10 that one of the 90 is prime.
Of the roughly $10^{1000}$ starting points, I expect
• fewer than $10^{999}$ to work for one step,
• fewer than $10^{998}$ to work for two steps,
• maybe one to work for $1000$ steps, and
• one chance in $10^{1000}$ of any of them to work for $2000$ steps
Similarly, for million-digit numbers, you lose more than 9999 out of 10000 at every step, leading to fewer than 1000000/4 steps for numbers of this size.