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%I #22 Aug 05 2019 02:48:46
%S 0,0,1,1,1,1,3,2,1,3,2,1,2,2,5,6,3,4,3,5,4,5,7,9,3,5,2,10,7,7,7,7,9,5,
%T 10,4,5,7,12,11,14,6,7,5,10,9,8,5,12,15,14,8,12,11,16,12,16,9,12,10,
%U 10,14,15,10,12,14,9,10,21,9,22,21,11,9,18,24,20,17,17,16
%N Number of primes p < prime(n)/2 such that 2^p - 1 is a primitive root modulo prime(n).
%C Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there is a prime q < p/2 with the Mersenne number 2^q - 1 a primitive root modulo p.
%C We have verified this for all n = 3, ..., 530000.
%C See also the comment in A234972.
%H Zhi-Wei Sun, <a href="/A236966/b236966.txt">Table of n, a(n) for n = 1..1200</a>
%H Z.-W. Sun, <a href="http://arxiv.org/abs/1405.0290">New observations on primitive roots modulo primes</a>, arXiv preprint arXiv:1405.0290 [math.NT], 2014.
%e a(12) = 1 since 17 is a prime smaller than prime(12)/2 = 37/2 with 2^(17) - 1 = 131071 a primitive root modulo prime(12) = 37.
%t f[k_]:=2^(Prime[k])-1
%t dv[n_]:=Divisors[n]
%t Do[m=0;Do[If[Mod[f[k],Prime[n]]==0,Goto[aa],Do[If[Mod[f[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}]];m=m+1;Label[aa];Continue,{k,1,PrimePi[(Prime[n]-1)/2]}];Print[n," ",m];Continue,{n,1,80}]
%Y Cf. A000040, A001348, A001918, A234972, A235709, A235712, A236306, A236308.
%K nonn
%O 1,7
%A _Zhi-Wei Sun_, Apr 22 2014
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