A242250 Least positive integer g < prime(n) such that g, 2^g - 1 and (g-1)! are all primitive roots modulo prime(n), or 0 if such a number g does not exist.

1, 0, 3, 5, 8, 11, 5, 13, 21, 10, 12, 22, 24, 34, 13, 31, 18, 6, 41, 11, 14, 53, 8, 6, 26, 3, 12, 5, 47, 10, 45, 10, 5, 32, 12, 6, 24, 3, 15, 3, 6, 41, 19, 10, 8, 30, 3, 67, 5, 35, 20, 13, 99, 19, 7, 7, 3, 118, 5, 15, 22, 3, 73, 59, 91, 8, 137, 46, 20, 55

Offset: 1

Author

Zhi-Wei Sun, May 09 2014

Keywords

• nonn

Comments

According to the conjecture in A242248, a(n) should be positive for all n > 2.

Examples

			a(5) = 8 since 8, 2^8 - 1 = 255 and (8-1)! = 5040 are all primitive roots modulo prime(5) = 11 with 255 == 5040 == 2 (mod 11), but none of 1, 2^2 - 1, 3, 4, 5, (6-1)! and (7-1)!
is a primitive root modulo 11.
		

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, A000142, A000225, A236966, A237112, A242242, A242248.

Programs

Mathematica

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