Prove/Disprove that there exist positive integers $a$ and $N$ such that the sequence $\overline{N}, \overline{aN}, \overline{aaN}, \overline{aaaN}, \ldots$ contains infinitely many primes.
($\overline{xy}$ refers to concatenating the digits of $x$ and $y$)
This problem was thrown around a maths chat. We ran some computer programs use Miller-Rabin primality test on Python and we got some sequences which seemed to keep having primes (such as $a=2$, $N=3$), and others which we could easily mod-bash to see only have finitely many primes (such as $a=5$, $N=37$).
We couldn't determine whether it would always be the case that there are finitely many primes, or whether $a=5$, $N=37$ was an exception.
EDIT: After a bit more thought, we started considering what sort of conditions would be sufficient for such a sequence to only contain finitely many primes.
There are some obvious ones like $N$ even or divisible by $5$, as well as $p \mid a, N$ for odd primes $p$. However, these don't cover the example of $a=5$, $N=37$.
Are there any more sufficient conditions?
Here's another formulation: Does $a\frac{10^{nd}-1}{10^d-1}+b,n\in \mathbb N$ contain infinitely many primes?
This looks a lot like Mersenne primes except base $10$ instead of $2$. Since no one knows if there are infinitely many Mersenne primes I'd expect this to be very hard to prove either way.