A241605 - OEIS

%I #30 Apr 16 2018 03:02:39

%S 0,0,2,2,1,3,2,3,1,3,2,4,2,4,2,5,3,3,6,3,4,2,5,2,4,4,3,4,3,4,2,2,5,4,

%T 7,2,6,5,4,4,5,3,2,3,6,4,3,4,5,5,4,2,4,4,3,3,3,3,3,5,6,7,8,2,5,7,6,3,

%U 5,7,5,3,4,4,6,3,6,7,4,3

%N a(n) = |{0 < k < sqrt(prime(n))*log(prime(n)) : k is not only a quadratic nonresidue modulo prime(n) but also a Fibonacci number}|.

%C It might seem that a(n) > 0 for all n > 2, but this is not true. In fact, for the prime p = prime(139570751) = 2893610639, the integer 1346269 is the least Fibonacci number which is a quadratic nonresidue modulo p, and 1346269/(sqrt(p)*log(p)) is approximately equal to 1.1488.

%C Conjecture: For any odd prime p, let f(p) be the least Fibonacci number which is a quadratic nonresidue modulo p. Then f(p) = o(p^(0.7)) as p tends to infinity. Moreover,f(p) = O(p^c) for any constant c > c_0 = log_2((1+sqrt(5))/2).

%C This refinement of the conjecture in A241568 seems reasonable in view of the heuristic arguments described below. Let c be any constant greater than c_0 (which has the approximating value 0.694). Roughly speaking, there are about (log_2(p^c))/c_0 positive Fibonacci numbers below p^c. Suppose that a positive Fibonacci number is a quadratic residue modulo a fixed odd prime p with "probability" 1/2. Then we might expect that the "probability" for all positive Fibonacci numbers below p^c being quadratic residues modulo p is about (1/2)^(c*(log_2 p)/c_0) = 1/p^(c/c_0). As sum_p 1/p^(c/c_0) converges, it seems reasonabale to think that there are only finitely many odd primes p for which all positive Fibonacci numbers below p^c are quadratic residues mod p.

%H Zhi-Wei Sun, <a href="/A241605/b241605.txt">Table of n, a(n) for n = 1..10000</a>

%e a(3) = 2 since the Fibonacci numbers F(3) = 2 and F(4) = 3 are quadratic nonresidues modulo prime(3) = 5 which are also smaller than sqrt(5)*log(5).

%e a(4) = 2 since the Fibonacci numbers F(4) = 3 and F(5) = 5 are quadratic nonresidues modulo prime(4) = 7 which are also smaller than sqrt(7)*log(7).

%e a(5) = 1 since the Fibonacci number F(3) = 2 is a quadratic nonresidue modulo prime(5) = 11 which is also smaller than sqrt(11)*log(11).

%e a(9) = 1 since the Fibonacci number F(5) = 5 is a quadratic nonresidue modulo prime(9) = 23 which is also smaller than sqrt(23)*log(23).

%t f[k_]:=f[k]=Fibonacci[k]

%t Do[m=0;Do[If[f[k]>=Sqrt[Prime[n]]*Log[Prime[n]],Goto[aa]];If[JacobiSymbol[f[k],Prime[n]]==-1,m=m+1];Continue,{k,2,Prime[n]}];Label[aa];Print[n," ",m];Continue,{n,1,80}]

%Y Cf. A000040, A000045, A241568, A241604.

%K nonn

%O 1,3

%A _Zhi-Wei Sun_, Apr 26 2014