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A291657 Primes p such that p is a primitive root modulo prime(p).

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%I A291657 #17 Aug 30 2017 04:28:42
%S A291657 2,3,7,11,13,41,71,79,83,107,109,131,139,157,163,173,179,191,211,223,
%T A291657 229,263,271,277,293,311,313,317,337,353,359,367,373,389,419,431,439,
%U A291657 449,457,463,479,521,547,569,577,593,607,641,661,709,719,727,743,757,761,769,787,811,823,827
%N A291657 Primes p such that p is a primitive root modulo prime(p).
%C A291657 The conjecture in A291615 implies that the current sequence has infinitely many terms. In fact, if there are only finitely many primes p with p a primitive root modulo prime(p) and we let P denote the product of all such primes, then by Dirichlet's theorem there is a prime q == 1 (mod 4*P) and hence any prime p with p a primitive root modulo prime(p) is a quadratic residue modulo q and hence not a primitive root modulo q.
%C A291657 Conjecture: a(n)/(n*log(n)) has a positive limit as n tends to the infinity. Equivalently, all the terms in this sequence form a subset of the set of all primes with positive asymptotic density.
%H A291657 Zhi-Wei Sun, <a href="/A291657/b291657.txt">Table of n, a(n) for n = 1..10000</a>
%H A291657 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1405.0290">New observations on primitive roots modulo primes</a>, arXiv:1405.0290 [math.NT], 2014.
%e A291657 a(1) = 2 since the first prime 2 is a primitive root modulo prime(2) = 3.
%e A291657 a(2) = 3 since the prime 3 is a primitive root modulo prime(3) = 5.
%t A291657 p[n_]:=p[n]=Prime[n];
%t A291657 n=0;Do[Do[If[Mod[p[k]^(Part[Divisors[p[p[k]]-1],i])-1,p[p[k]]]==0,Goto[aa]],{i,1,Length[Divisors[p[p[k]]-1]]-1}];
%t A291657 n=n+1;Print[n," ",p[k]];Label[aa],{k,1,145}]
%Y A291657 Cf. A000040, A242345, A243164, A243403, A291615, A291690.
%K A291657 nonn
%O A291657 1,1
%A A291657 _Zhi-Wei Sun_, Aug 28 2017