A241492 a(n) = |{0 < g < prime(n): g is a primitive root modulo prime(n) and g is a product of two consecutive integers}|.

0, 1, 1, 0, 2, 2, 2, 1, 1, 1, 1, 2, 3, 3, 2, 3, 5, 3, 3, 2, 2, 2, 6, 3, 2, 5, 3, 4, 5, 5, 4, 7, 7, 7, 5, 4, 3, 5, 5, 8, 6, 2, 5, 4, 5, 3, 2, 5, 7, 6, 5, 4, 5, 8, 10, 8, 10, 4, 6, 6, 7, 8, 3, 4, 4, 9, 6, 4, 7, 8, 7, 5, 7, 7, 6, 9, 12, 6, 11, 8

Offset: 1

Author

Zhi-Wei Sun, Apr 23 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 4. In other words, any prime p > 7 has a primitive root g < p of the form k*(k+1).

We have verified this for all n = 5, ..., 2*10^5.

See also A239957 and A239963 for similar conjectures. Clearly, for any prime p > 3, one of the three numbers 1*2, 2*3, 3*4 is a quadratic residue modulo p.

Examples

			a(9) = 1 since 4*5 = 20 is a primitive root modulo prime(9) = 23.
a(10) = 1 since 1*2 = 2 is a primitive root modulo prime(10) = 29.
a(11) = 1 since 3*4 = 12 is a primitive root modulo prime(11) = 31.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

Crossrefs

Cf. A000040, A002378, A239957, A239963, A241476.

Programs

Mathematica

f[k_]:=f[k]=k(k+1)
dv[n_]:=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,(Sqrt[4*Prime[n]-3]-1)/2}];Print[n," ",m];Continue,{n,1,80}]