OFFSET
1,7
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.
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.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 23 2014
STATUS
approved