Let’s assume we have B(n) bitruncatable primes of length (n)
For B(n+2), we have at most 45 times that number (digit added at the front can’t be a zero; digit added at the back must be odd)
The fraction of them that’s prime is about 2/log(10^(n+1)) (factor 2 added because we already dropped all even numbers; n+1 in the denominator as being halfway between n and n+2 digits; neither factors affect the conclusion)
That gives us the recurrence relation
B(n+2) ~= 90B(n)/log(10^(n+1))
= (90/log(10)) B(n)/(n+1)
So, I think it’s very likely there is a largest one.
This line of thought applies to all integer bases > 2 (there isn’t a largest one in base 2 because there isn’t any in base 2)
So finding a new largest bi-truncatable prime is just a matter of trying to add each single digit at both sides of this prime; if any of them is again a prime, we have a new largest and if none of them is (which I guess is the case because the author must have tried this out already), then there exists no larger bi-truncatable prime.
There's no guarantee that the biggest N+2 digit prime comes from the biggest N digit prime. It could come from the smallest N digit prime if that's the only N digit prime that becomes an N+2 digit prime by prepending a 9.
If you mean the sole 97-digit solution given in the comments, the implication that all the bitruncatable primes have been found and enumerated by your method, and the assertion that there are no such 99-digit primes - then yes, that looks pretty legit, though I'd want to see an independent verification before accepting it as a proof, since computer programs have bugs.
I wrote a solver for this problem:
I haven't run it yet, because I don't have a powerful-enough machine at hand. It needs ~6 GB of RAM and ~20 CPU core-hours.
7228828176786792552781668926755667258635743361825711373791931117197999133917737137399993737111177 (97 digits)
is indeed the biggest bi-truncatable prime.
This was done under assumption that zero digits are not allowed. If they are allowed, bigger bi-truncatable primes exists, such as:
90072457733413689120801410250233316614403998951220231333193991731791997911317971131797197339199333933 (101 digits)
Edit: parallelized the code and proved through 19 digits. That's where I'm stopping due to the uint64 space limitations.
Irrationality would be enough to guarantee that you can find a power where your name is right at the start. You do need log(pi)/log(10) (or whatever base you're using) to be irrational for that to work, but I think you can prove that using the fact that pi is transcendental.