Right-Truncatable Prime Counter

thechao · 2025-05-27T20:06:15 1748376375

Just in case any else is wondering: there are only 83 right-truncatable primes (RTP) and that is it. There's two constraints that let you see this "immediately":

1. An RTP must start with {2,3,5,7,9}; and,

2. An RTP must end with {1,3,7,9}.

So, let's take the largest RTP (73939133) and try to "extend" it: there are only four possible extensions: 73939133[1], 73939133[3], 73939133[7], 73939133[9]. None of these are prime. This holds for the other 8-digit RTPs. Therefore, there is no extension to a 9-or-longer RTP. Thus, the list is exhaustive.

jinwoo68 · 2025-05-27T19:28:40 1748374120

There's a Project Euler problem for finding truncatable prime numbers, from both left and right: https://projecteuler.net/problem=37

throwawaymaths · 2025-05-27T19:16:11 1748373371

Curious about base 2. Obviously if you hit a 0 it's immediately not prime, but maybe adjust the rules so:

- you drill through as many 0's on the right.

- you finish on 1.

3, 5, 7, 11, 13, 15, 17 are all right truncatable, 19 is the first non-truncatable prime in this scheme.

nh23423fefe · 2025-05-27T20:09:05 1748376545

i dont think smaller radixes make the problem more interesting. the problem is interesting because base 10 has a large branching factor

throwawaymaths · 2025-05-27T21:14:05 1748380445

I think in the base2 reformulation I propose we do not know for certain if the list of numbers terminates, as all Fermat primes are in the set and we don't know if there are infinitely many Fermat primes.

For base-10 and the original rules the set is provably closed.

"Drilling through zeros" makes the branching unbounded.

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