This is a subsequence of primes $p$ for which $p^2-1$ has at most 6 prime divisors counted with multiplicity.


This sequence described in the question is the sequence [A079153 in OEIS](https://oeis.org/A079153). 


I could not find references on the first sequence, but I found a mention of the second sequence in another MO question:
[Number of prime factors of the order of a finite non-abelian simple group](https://mathoverflow.net/questions/67671/number-of-prime-factors-of-the-order-of-a-finite-non-abelian-simple-group) with a reference to a survey article by [Solomon from 2001](http://www.ams.org/journals/bull/2001-38-03/S0273-0979-01-00909-0/S0273-0979-01-00909-0.pdf): are there infinitely many primes $p$ such that $|PSL(2,p)|=(p-1)\*p\*(p+1)/2$ is a product of six prime factors? In the survey article it is said this problem is akin to the twin prime conjecture, but no references are given.



Are there infinitely many primes p such that both p-1 and p+1 have at most 3 prime factors, counted with multiplicity?
At the present time, the best (general) result is due to James Maynard, who has proven that any admissible 3-tuple takes at most 7 prime factors infinitely often (see [this paper](http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=9030782&fileId=S0305004113000339)). In particular, (using a certain admissible 3-tuple related to what Robert Israel wrote) we can get that $n(n^2-1)$ has at most 11 prime factors infinitely often.


I don't know if this number can be improved for the specific case you are interested in. However, the general consensus is that Dickson's conjecture is true, so both $p-1$ and $p+1$ *should* consist of exactly 3 prime factors, for infinitely many $p$. But this conjecture is currently out of reach of any known techniques, and appears to be harder than the twin prime conjecture, since we must control admissible 3-tuples, but the twin prime problem deals with only admissible 2-tuples.


