Right-Truncatable Prime Counter

https://github.com/EbodShojaei/Right-Truncatable-Primes

9•rainmans•6mo ago

Comments

throwawaymaths•6mo ago

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•6mo ago

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

throwawaymaths•6mo ago

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.

jinwoo68•6mo ago

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

thechao•6mo ago

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.