A229899 a(n) = |{0

0, 0, 0, 0, 1, 0, 2, 1, 1, 0, 1, 2, 2, 2, 4, 6, 7, 0, 1, 2, 2, 1, 10, 10, 8, 4, 4, 13, 4, 6, 3, 5, 10, 3, 20, 2, 6, 6, 19, 18, 22, 4, 11, 6, 16, 4, 3, 7, 28, 8, 28, 16, 4, 16, 32, 31, 30, 5, 8, 16, 13, 32, 7, 17, 6, 40, 7, 2, 43, 8, 36, 43, 10, 12, 8, 46, 44, 8, 30, 16, 39, 8, 24, 20, 11, 39, 30, 14, 22, 9, 58, 58, 22, 17, 22, 61, 60, 30, 21, 10

Offset: 1

Author

Zhi-Wei Sun, Oct 02 2013

Keywords

• nonn

Comments

Conjecture: a(n) > 0 except for n = 1, 2, 3, 4, 6, 10, 18. In other words, for any prime p > 7 not equal to 13 or 29 or 61, there are three consecutive integers which are primitive roots modulo p.

Let p be an odd prime. For any integer c, define S(c,p) to be the sum of the Legendre symbols ((g+c)/p) over all primitive roots g modulo p among 1, ..., p-1. If g is a primitive root modulo p, then so is the inverse g^{-1} of g modulo p, and -((g^{-1}+c)/p) = (g*(g^{-1}+c)/p) = ((1+c*g)/p). So S(1,p) = 0, and also S(-1,p) = 0 when p == 1 (mod 4). The author also showed that S(-c,p) = S(c,p) if p == 1 (mod 4), and that S(c,p) = 0 if p is a Fermat prime and c is a quadratic residue modulo p.

Zhi-Wei Sun also made the following conjectures:

(i) Let p > 13 be a prime not equal to 19 or 31, and let a,b,c be integers with a or c not divisible by p. If p does not divide b^2-4*a*c, then there is a primitive root g modulo p such that a*g^2+b*g+c is a quadratic residue modulo p, and there is also a primitive root h modulo p such that a*h^2+b*h+c is a quadratic nonresidue modulo p.

(ii) Let p be any odd prime, and let a,b,c be integers with a or c not divisible by p. If p does not divide b^2-4*a*c, then the absolute value of the sum of the Legendre symbols ((a*g^2+b*g+c)/p) over all primitive roots g modulo p among 1, ..., p-1 is smaller than 2*sqrt(p).

Examples

			a(5) = 1 since 6, 7, 8 are primitive roots modulo p_5 = 11.
a(7) = 2 since 5, 6, 7, 10, 11, 12 are primitive roots modulo p_7 = 17.
a(8) = 1 since 13, 14, 15 are primitive roots modulo p_8 = 19.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..400

Crossrefs

Cf. A001918.

Programs

Mathematica

gp[g_,p_]:=gp[g,p]=Length[Union[Table[Mod[g^k, p],{k,1,p-1}]]]==p-1
a[n_]:=Sum[If[gp[g,Prime[n]]&&gp[g-1,Prime[n]]&&gp[g+1,Prime[n]],1,0],{g,1,Prime[n]-1}]
Table[a[n],{n,1,100}]