The primes such that removing digits from the right end leaves another prime

Question

The number 73,939,133 is prime. Keep removing a digit from the right end. Each of the remaining numbers is prime.

How to find other numbers with this property?

Answer 1

I don't think there's any particularly elegant way to solve this. So I wrote an algorithm which started with the primes $1$ through $9$ and, at each step, it took the previous list, and made a new list out of the previous list, with elements for each of the digits $0$ through $9$ (though, really only $1$, $3$, $7$ and $9$ need to be checked since even digits and $5$ never end a prime in decimal), and checked this new list for primality. When the list was empty, I stopped iterating, since this meant that there were no primes of a certain length that satisfied the property, which means that no higher length prime could exist, certainly. I got the following list:

2, 3, 5, 7, 23, 29, 31, 37, 53, 59, 71, 73, 79, 233, 239, 293, 311, 313, 317, 373, 379, 593, 599, 719, 733, 739, 797, 2333, 2339, 2393, 2399, 2939, 3119, 3137, 3733, 3739, 3793, 3797, 5939, 7193, 7331, 7333, 7393, 23333, 23339, 23399, 23993, 29399, 31193, 31379, 37337, 37339, 37397, 59393, 59399, 71933, 73331, 73939, 233993, 239933, 293999, 373379, 373393, 593933, 593993, 719333, 739391, 739393, 739397, 739399, 2339933, 2399333, 2939999, 3733799, 5939333, 7393913, 7393931, 7393933, 23399339, 29399999, 37337999, 59393339, 73939133