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%I #29 Aug 05 2019 02:49:35
%S 0,0,1,1,1,2,2,3,1,3,2,4,2,4,2,5,3,3,5,3,4,2,5,2,4,4,3,4,3,5,3,2,5,4,
%T 7,2,6,5,4,4,5,4,3,4,7,4,4,4,5,6,4,3,5,4,3,3,3,3,3,5,6,7,8,2,5,7,6,3,
%U 5,7,5,3,4,4,6,3,6,7,4,3
%N a(n) = |{0 < k < prime(n)/2: k is not only a quadratic nonresidue modulo prime(n) but also a Fibonacci number}|.
%C Conjecture: (i) a(n) > 0 for all n > 2. In other words, for any prime p > 3, there is a Fibonacci number among 1, ..., (p-1)/2 which is a quadratic nonresidue modulo p.
%C (ii) For any n > 2, there is a prime q < prime(n) such that the q-th Fibonacci number is a quadratic nonresidue modulo prime(n).
%C (iii) For any odd prime p, there is a Lucas number (i.e., a term of A000032) smaller than p which is a quadratic nonresidue modulo p.
%C We have checked part (i) for all primes p < 3*10^9, part (ii) for n up to 10^8, and part (iii) for the first 10^7 primes.
%C See also A241604 for a sequence related to part (i) of the conjecture.
%H Zhi-Wei Sun, <a href="/A241568/b241568.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(3) = 1 since F(3) = 2 is a quadratic nonresidue modulo prime(3) = 5, where F(n) denotes the n-th Fibonacci number.
%e a(4) = 1 since F(4) = 3 is a quadratic nonresidue modulo prime(4) = 7.
%e a(5) = 1 since F(3) = 2 is a quadratic nonresidue modulo prime(5) = 11.
%e a(9) = 1 since F(5) = 5 is a quadratic nonresidue modulo prime(9) = 23.
%t f[k_]:=Fibonacci[k]
%t Do[m=0;Do[If[f[k]>Prime[n]/2,Goto[aa]];If[JacobiSymbol[f[k],Prime[n]]==-1,m=m+1];Continue,{k,2,(Prime[n]+1)/2}]; Label[aa];Print[n," ",m];Continue,{n,1,80}]
%Y Cf. A000040, A000032, A000045, A236966, A239957, A239963, A241492, A241504, A241516, A241604.
%K nonn
%O 1,6
%A _Zhi-Wei Sun_, Apr 25 2014
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