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%I #23 Aug 05 2019 02:49:27
%S 0,1,1,0,2,2,2,1,1,1,1,2,3,3,2,3,5,3,3,2,2,2,6,3,2,5,3,4,5,5,4,7,7,7,
%T 5,4,3,5,5,8,6,2,5,4,5,3,2,5,7,6,5,4,5,8,10,8,10,4,6,6,7,8,3,4,4,9,6,
%U 4,7,8,7,5,7,7,6,9,12,6,11,8
%N a(n) = |{0 < g < prime(n): g is a primitive root modulo prime(n) and g is a product of two consecutive integers}|.
%C Conjecture: a(n) > 0 for all n > 4. In other words, any prime p > 7 has a primitive root g < p of the form k*(k+1).
%C We have verified this for all n = 5, ..., 2*10^5.
%C See also A239957 and A239963 for similar conjectures. Clearly, for any prime p > 3, one of the three numbers 1*2, 2*3, 3*4 is a quadratic residue modulo p.
%H Zhi-Wei Sun, <a href="/A241492/b241492.txt">Table of n, a(n) for n = 1..10000</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(9) = 1 since 4*5 = 20 is a primitive root modulo prime(9) = 23.
%e a(10) = 1 since 1*2 = 2 is a primitive root modulo prime(10) = 29.
%e a(11) = 1 since 3*4 = 12 is a primitive root modulo prime(11) = 31.
%t f[k_]:=f[k]=k(k+1)
%t dv[n_]:=dv[n]=Divisors[n]
%t Do[m=0;Do[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,(Sqrt[4*Prime[n]-3]-1)/2}];Print[n," ",m];Continue,{n,1,80}]
%Y Cf. A000040, A002378, A239957, A239963, A241476.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Apr 23 2014
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