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%I #9 Aug 05 2019 04:07:06
%S 0,0,1,1,1,1,2,2,2,1,1,2,3,1,5,2,1,1,3,5,2,3,2,3,5,4,4,7,1,5,5,7,7,6,
%T 8,6,6,5,6,3,5,4,8,6,4,5,6,6,12,8,15,17,7,10,8,11,10,8,9,10,7,18,6,15,
%U 4,9,5,10,10,8
%N a(n) = |{0 < k < n: prime(k) is a primitive root modulo prime(n) and also a primitive root modulo prime(2*n)}|.
%C Conjecture: (i) a(n) > 0 for all n > 2.
%C (ii) For any integer n > 4, there is a primitive root 0 < g < prime(n) modulo prime(n) which is also a primitive root modulo prime(n+1).
%H Zhi-Wei Sun, <a href="/A243847/b243847.txt">Table of n, a(n) for n = 1..6000</a>
%H 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 a(3) = 1 since prime(1) = 2 is a primitive root modulo prime(3) = 5 and also a primitive root modulo prime(2*3) = 13. Note that prime(2) = 3 is not a primitive root modulo prime(2*3) = 13 since 3^3 == 1 (mod 13).
%t dv[n_]:=Divisors[n]
%t Do[m=0;Do[Do[If[Mod[(Prime[k])^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}]; Do[If[Mod[(Prime[k])^(Part[dv[Prime[2n]-1],j]),Prime[2n]]==1,Goto[aa]],{j,1,Length[dv[Prime[2n]-1]]-1}];m=m+1;Label[aa];Continue,{k,1,n-1}];Print[n," ",m];Continue,{n,1,70}]
%Y Cf. A000040, A242345, A243755, A243837, A243839.
%K nonn
%O 1,7
%A _Zhi-Wei Sun_, Jun 12 2014
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