A239963 Number of triangular numbers below prime(n) which are also primitive roots modulo prime(n).

1, 0, 1, 1, 1, 1, 3, 3, 3, 4, 2, 1, 3, 2, 3, 3, 3, 3, 1, 3, 3, 4, 5, 5, 3, 5, 4, 9, 3, 7, 6, 4, 7, 3, 9, 3, 7, 5, 10, 9, 10, 9, 5, 10, 7, 7, 2, 5, 8, 6, 8, 7, 6, 6, 12, 10, 8, 9, 7, 10, 8, 11, 6, 6, 12, 14, 8, 7, 16, 5, 11, 10, 9, 6, 14, 14, 11, 8, 14, 7

Offset: 1

Author

Zhi-Wei Sun, Apr 23 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there is a primitive root 0 < g < p of the form k*(k+1)/2, where k is a positive integer.

Examples

			a(5) = 1 since the triangular number 3*4/2 = 6 is a primitive root modulo prime(5) = 11.
a(12) = 1 since the triangular number 5*6/2 = 15 is a primitive root modulo prime(12) = 37.
a(19) = 1 since the triangular number 7*8/2 = 28 is a primitive root modulo prime(19) = 67.
		

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, A000217, A236308, A236966, A237112, A237121, A237594, A239957.

Programs

Mathematica

f[k_]:=f[k]=k(k+1)/2
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[8Prime[n]-7]-1)/2}];Print[n," ",m];Continue,{n,1,80}]