A242242 Least positive primitive root g < prime(n) modulo prime(n) with 2^g - 1 also a primitive root modulo prime(n), or 0 if such a number g does not exist.

1, 0, 2, 5, 2, 2, 3, 2, 5, 2, 3, 2, 6, 3, 5, 2, 2, 2, 2, 7, 5, 3, 2, 3, 5, 2, 5, 2, 6, 3, 3, 2, 3, 2, 2, 6, 5, 2, 5, 2, 2, 2, 19, 5, 2, 3, 2, 3, 2, 6, 3, 7, 7, 6, 3, 5, 2, 6, 5, 3, 3, 2, 5, 17, 10, 2, 3, 10, 2, 2

Offset: 1

Author

Zhi-Wei Sun, May 09 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 except for n = 2. In other words, for any prime p > 3, there exists a primitive root 0 < g < p modulo p such that 2^g - 1 is also a primitive root modulo p.

Examples

			a(4) = 5 since both 5 and 2^5 - 1 = 31 are primitive roots modulo prime(4) = 7, but none of 1, 2, 4 and 2^3 - 1 is a primitive root modulo prime(4) = 7.
		

Links

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

Crossrefs

Cf. A000040, A000225, A234972, A236966.

Programs

Mathematica

f[n_]:=f[n]=2^n-1
dv[n_]:=Divisors[n]
Do[Do[If[Mod[f[k],Prime[n]]==0,Goto[aa]];Do[If[Mod[k^(Part[dv[Prime[n]-1],j])-1,Prime[n]]==0,Goto[aa]],{j,1,Length[dv[Prime[n]-1]]-1}];Do[If[rMod[f[k]^(Part[dv[Prime[n]-1],i])-1,Prime[n]]==0,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];
Print[n," ",k];Goto[bb];Label[aa];Continue,{k,1,Prime[n]-1}];Label[cc];Print[n," ",0];Label[bb];Continue,{n,1,70}]