$2017 /1= 2017$, a prime;
$2018/2 = 1009$, a prime; and
$2019/3 = 673$, a prime.
Call such a sequence a trio.
If $2020/4 = 505$ were prime, then we would have a quartet and similarly for a quintet, etc. But $505$ is not prime so this is not a quartet. I haven't found any quartets (yet) although trios are not hard to find.
Question: How many quartets are there?
Let's identify a sequence by its initial prime.
Observe that a prime can be a trio prime only if it is $13$ or it is $=1$ or $37$ mod $60$.
And it can be a quartet prime only if it is $=1$ mod $120$.
My calculations, with assistance from Alexa, indicate that the trio primes less than $2017$ are:
$13, 37, 157, 421, 541, 877, 1201, 1381, \text { and } 1621.$
EDIT: $421$ is not a trio prime. I can't blame this on Alexa.
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$\begingroup$The trio primes are A036570 on OEIS (it's always a good idea to look up a sequence on the OEIS if you've managed to compute its first few terms) and that entry has a link to the the quartet primes, which are A278585 plus one; the smallest one is apparently $12721$, where
- $12721$ is prime,
- $12722 = 2 \cdot 6361$ and $6361$ is prime,
- $12723 = 3 \cdot 4241$ and $4241$ is prime, and
- $12724 = 4 \cdot 3181$ and $3181$ is prime.
Probably there are infinitely many of these although this is quite beyond current technology; it would follow from a suitable generalization of the prime k-tuple conjecture, and would imply that there are infinitely many primes $p$ such that $2p-1$ is prime, which is just as open as the infinitude of the Sophie Germain primes.
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