Proof for finite number of truncatable primes

How do we prove that there exists a finite number of truncatable primes? It's intuitive that as it gets bigger it has more factors, so less chance of it being primes when truncated, but what is the mathematical proof?

• As far as I know, it's just a case-by-case analysis. Find all the $2$-digit truncatable primes. Then find all the three digits. Note that if you have an $n+1$-digit truncatable prime, then dropping the first or last must also return a truncatable prime. Eventually, you can just list them all. Oct 4, 2016 at 18:38
• So proof be exhaustion is the only way?
– pbsh
Oct 4, 2016 at 18:39
• Well, "only way" is a bit strong, but I suspect it is the best way. Oct 4, 2016 at 18:40
• (It would certainly be interesting if one could prove this true in any base, so finding a non-exhaustive solution might help us generalize.) Oct 4, 2016 at 18:45
• Mathematically, if there are $M$ truncatable primes with $n$ digits in base $b$, then, by chance, we'd expect there to be approximate $2M(b-1)\frac{\pi(b^n)}{b^n} \approx 2M(b-1)\frac{1}{n\log b}$ of $n+1$ digit truncatable primes. That gives a heuristic to expect it is true for all bases, but, obviously, not a proof. Oct 4, 2016 at 18:54

• (For eg. there's no 5-digit truncatable prime, but there's a 6-digit one because there are 5-digit left-truncatable 39397 and right-truncatable 73939 primes that make up the 6-digit truncatable prime 739397 together.) Jun 1, 2018 at 19:59