%I
%S 2,3,4,28
%N Numbers k such that (prime(k)1)! + prime(k)^3 is prime.
%C k = {2, 3, 4, 28} yields primes p(k) = {3, 5, 7, 107}. There are no more such k up to k=100. Verified by _Ray Chandler_.
%C a(5) > 600.  _Jinyuan Wang_, Apr 10 2020
%F Numbers k such that (prime(k)1)! + prime(k)^3 is prime, where prime(k) is the kth prime.
%e a(3) = 4 because (prime(4)1)! + prime(4)^3 = (71)! + 7^3 = 720 + 343 = 1063 is the 3rd prime of this form.
%t lst={};Do[p=Prime[n];If[PrimeQ[(p1)!+p^3], AppendTo[lst, n]], {n, 10^2}];lst (* _Vladimir Joseph Stephan Orlovsky_, Sep 08 2008 *)
%o (PARI) is(k) = ispseudoprime((prime(k)1)! + prime(k)^3); \\ _Jinyuan Wang_, Apr 10 2020
%Y Cf. A100605, A100858.
%K nonn,hard,more
%O 1,1
%A _Jonathan Vos Post_, Nov 30 2004
