Mind-boggling property of a prime

Solution 1:

A very heuristic argument says that there should be an infinite number. If we assume the primes are randomly distributed with $p$ having $\frac 1{\ln p}$ chance of being prime, we can ask what is the chance there is a Magical Emirp with $n^2$ decimal digits. You need two numbers of $n^2$ digits and $4n+2$ numbers of $n$ digits to be prime. We will take all the probabilities for $n$ digits to be $\frac 1{\ln 10^{n-1/2}}=\frac 1{(n-1/2)\ln 10}$ For $n$ large, we can ignore the $1/2$ and say the chance is $\frac 1{n \ln 10}$ This says the chance a random number of $n^2$ digits is a Magical Emirp is about $\frac 1{n^{4n+2}(n^2)^210^{2n+2}}$ As there are about $10^{n^2}$ numbers with $n^2$ digits we would expect $\frac {10^{n^2}}{n^{4n+2}(n^2)^210^{2n+2}}$ with $n^2$ digits. This grows without bound, so we would expect more and more of them each square number of digits. They still become rarer and rarer, so finding them is rather difficult. Plugging in $n=10$, there should be $10^{32}$ hundred digit ones. That is a lot, but a small fraction of the hundred digit numbers.