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May 25, 2016 at 16:35 comment added Terry Tao Actually the discussion on page 2 of the Maynard paper in my previous comment suggests that one can take k=11 for the original problem.
May 25, 2016 at 16:28 comment added Terry Tao As $p^2-1$ is going to be divisible by 24 for large $p$, one should reduce $k$ by $4$ and work with $(p+1)(p-1)/24$. In view of arxiv.org/abs/1205.5021 it might be possible to get $k$ down to 6+4=10 with current methods, though in that paper one cannot specify one of the three linear forms $L_1(n), L_2(n), L_3(n)$ to be prime, so probably one has to make do with something slightly worse than 10. In any event Maynard's result certainly shows that $\Omega(n(n^2-1)) \leq 11$ infinitely often, with $n$ coprime to any fixed finite number of primes.
May 25, 2016 at 15:27 comment added Lucia Any sieve method will show this for some value of $k$ -- e.g. $k=20$ would probably be pretty easy to show (and probably something like $k=10$ would be known with some effort).
May 25, 2016 at 11:51 answer added joro timeline score: 3
May 25, 2016 at 8:46 history asked Buturlakin Alexander CC BY-SA 3.0