So, is there a truly largest one?

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))
            = 90B(n)/((n+1)log(10))
            = (90/log(10)) B(n)/(n+1)

That number gets smaller and smaller once n > 90/log(10) (about 39)

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)