A241605 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}|.

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, 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, 5, 7, 5, 3, 4, 4, 6, 3, 6, 7, 4, 3

Offset: 1

Author

Zhi-Wei Sun, Apr 26 2014

Keywords

• nonn

Comments

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.

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).

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.

Examples

			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).
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).
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).
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).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A000045, A241568, A241604.

Programs

Mathematica

f[k_]:=f[k]=Fibonacci[k]
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}]