A242248 a(n) = |{0 < g < prime(n): g, 2^g - 1 and (g-1)! are all primitive roots modulo prime(n)}|.

1, 0, 1, 1, 1, 1, 3, 1, 1, 2, 1, 2, 2, 1, 6, 2, 10, 3, 2, 3, 5, 2, 10, 12, 3, 6, 7, 15, 3, 9, 3, 8, 18, 5, 18, 3, 7, 7, 24, 20, 26, 4, 13, 10, 15, 5, 4, 3, 35, 5, 19, 19, 3, 19, 36, 37, 38, 5, 10, 15, 16, 34, 7, 16, 6, 36, 4, 4, 44, 14

Offset: 1

Author

Zhi-Wei Sun, May 09 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there exists a positive primitive root g < p modulo p such that 2^g - 1 and (g-1)! are also primitive roots modulo p.

We have verified this for all primes p with 3 < p < 10^6.

Examples

			a(4) = 1 since 5, 2^5 - 1 = 31 and (5-1)! = 24 are all primitive roots modulo prime(4) = 7.
a(6) = 1 since 11, 2^(11) - 1 = 2047 and (11-1)! = 3628800 are all primitive roots modulo prime(6) = 13. Note that both 2047 and 3628800 are congruent to 6 modulo 13.
a(14) = 1 since 34, 2^(34) - 1 and (34-1)! are all primititive roots modulo prime(14) = 43. Note that 2^(34) - 1 == 20 (mod 43) and 33! == -14 (mod 43).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..1500
• Zhi-Wei Sun, Notes on primitive roots modulo primes, arXiv:1405.0290 [math.NT], 2014.

Crossrefs

Cf. A000040, A000142, A000225, A236966, A237112, A242242, A242250.

Programs

Mathematica

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