A331506 Least primitive root g < prime(n) of the n-th prime with g a product of two Fibonacci numbers, or 0 if such a number g does not exist.

1, 2, 2, 3, 2, 2, 3, 2, 5, 2, 3, 2, 6, 3, 5, 2, 2, 2, 2, 13, 5, 3, 2, 3, 5, 2, 5, 2, 6, 3, 3, 2, 3, 2, 2, 6, 5, 2, 5, 2, 2, 2, 21, 5, 2, 3, 2, 3, 2, 6, 3, 13, 13, 6, 3, 5, 2, 6, 5, 3, 3, 2, 5, 34, 10, 2, 3, 10, 2, 2, 3, 13, 6, 2, 2, 5, 2, 5, 3, 21, 2, 2, 13, 5, 15, 2, 3, 13, 2, 3, 2, 13, 3, 2, 10, 5, 2, 3, 2, 2

Offset: 1

Author

Zhi-Wei Sun, Jan 18 2020

Keywords

• nonn

Comments

Conjecture 1: a(n) > 0 for all n > 0. In other words, for each prime p there are two Fibonacci numbers F(k) and F(m) with F(k)*F(m) < p such that F(k)*F(m) is a primitive root modulo p.

This implies that for each odd prime p there exists a Fibonacci number F(k) < p which is a quadratic nonresidue modulo p.

It seems that Conjecture 1 can be strengthened as follows: For any prime p, there is a primitive root g < p modulo p such that g/F(2) = g or g/F(3) = g/2 or g/F(4) = g/3 is a Fibonacci number. We have verified this strong version for all primes p < 5*10^9.

We also have the following conjecture similar to Conjecture 1.

Conjecture 2. For any prime p, there are two Lucas numbers L(k) and L(m) with k >= m >= 0 and L(k)*L(m) < p such that L(k)*L(m) is a primitive root modulo p.

This has been verified for all primes p < 10^9.

Examples

			a(85) = 15 with 15 = 3*5 = F(4)*F(5) a primitive root modulo prime(85) = 439.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.
• Zhi-Wei Sun, Primitive roots modulo primes related to Fibonacci numbers or Lucas numbers, Question 350730 at MathOverflow, Jan. 19, 2020.

Crossrefs

Cf. A000032, A000040, A000045, A001918, A239957, A241476, A241504, A241568, A241604.

Programs

Mathematica

p[n_]:=p[n]=Prime[n];
Dv[n_]:=Dv[n]=Divisors[p[n]-1];
ls={};
Do[If[Fibonacci[k]Fibonacci[m]=p[n],tab=Append[tab,0];Goto[bb]];Do[If[PowerMod[ff[r],Dv[n][[i]],p[n]]==1,r=r+1;Goto[aa]],{i,1,Length[Dv[n]]-1}];tab=Append[tab,ff[r]];Label[bb],{n,1,100}];Print[tab]