A237112 Number of primes p < prime(n)/2 with p! a primitive root modulo prime(n).

0, 0, 1, 0, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 6, 4, 2, 3, 3, 4, 7, 9, 5, 7, 5, 8, 4, 7, 6, 7, 8, 7, 11, 8, 9, 7, 13, 10, 16, 4, 7, 8, 13, 9, 8, 12, 17, 10, 14, 12, 4, 10, 14, 15, 18, 8, 9, 8, 8, 18, 6, 8, 7, 16, 9, 11, 21, 15

Offset: 1

Author

Zhi-Wei Sun, Apr 22 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 4. In other words, for any prime p > 7, there exists a prime q < p/2 such that q! is a primitive root modulo p.

See also A236306 for a similar conjecture.

Examples

			a(7) = 1 since 3 is a prime smaller than prime(7)/2 = 17/2 and 3! = 6 is a primitive root modulo prime(7) = 17.
a(9) = 1 since  5 is a prime smaller than prime(9)/2 = 23/2 and 5! = 120 is a primitive root modulo prime(9) = 23.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..600
• Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

Crossrefs

Cf. A000040, A000142, A234972, A235709, A235712, A236306, A236308, A236966, A237121.

Programs

Mathematica

f[k_]:=(Prime[k])!
dv[n_]:=Divisors[n]
Do[m=0;Do[Do[If[Mod[f[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];m=m+1;Label[aa];Continue,{k,1,PrimePi[(Prime[n]-1)/2]}];Print[n," ",m];Continue,{n,1,70}]