A236306 Least prime p < prime(n) for which both p and p! are primitive roots modulo prime(n), or 0 if such a prime does not exist.

0, 2, 2, 0, 2, 2, 3, 2, 5, 2, 11, 2, 17, 5, 5, 2, 2, 2, 2, 13, 5, 3, 2, 3, 5, 2, 11, 2, 67, 3, 3, 2, 3, 2, 2, 13, 53, 2, 5, 2, 2, 2, 47, 5, 2, 3, 2, 3, 2, 29, 3, 7, 137, 11, 3, 5, 2, 59, 31, 13, 17, 2, 5, 23, 47, 2, 101, 23, 2, 2, 13, 7, 43, 2, 2, 5, 2, 109, 3, 127

Offset: 1

Author

Zhi-Wei Sun, Apr 21 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 4. In other words, for any prime p > 7, there exists a prime q < p such that both q and q! are primitive roots modulo p.

P. Erdős asked whether for any sufficiently large prime p there exists a prime q < p which is a primitive root modulo p (see page 377 of Guy's book in the reference).

Examples

			a(7) = 3 since both 3 and 3! = 6 are primititive roots modulo prime(7) = 17, but 2 is not a primitive root modulo 17.
		

References

• R. K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, New York, 2004.

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A000142, A234972, A235709, A235712.

Programs

Mathematica

f[k_]:=Prime[k]!
dv[n_]:=Divisors[n]
Do[Do[Do[If[Mod[Prime[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1||Mod[f[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];Print[n," ",Prime[k]];Goto[bb];Label[aa];Continue,{k,1,n-1}];Print[n," ",0];Label[bb];Continue,{n,1,80}]