A237594 Number of primes p < prime(n)/2 such that the Bell number B(p) is a primitive root modulo prime(n).

0, 0, 1, 1, 2, 1, 2, 3, 1, 1, 2, 4, 2, 4, 6, 5, 7, 3, 4, 3, 4, 2, 12, 7, 3, 5, 4, 9, 5, 6, 4, 5, 12, 6, 7, 5, 9, 6, 12, 11, 13, 7, 7, 7, 14, 5, 5, 14, 14, 8, 13, 11, 7, 10, 19, 17, 16, 8, 11, 7, 7, 23, 11, 12, 10, 22, 14, 8, 22, 11, 20, 22, 13, 13, 15, 24, 27, 14, 18, 18

Offset: 1

Author

Zhi-Wei Sun, Apr 22 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there exists a prime q < p/2 such that the Bell number B(q) is a primitive root modulo p.

Examples

			a(9) = 1 since 3 is a prime smaller than prime(9)/2 = 23/2 and B(3) = 5 is a primitive root modulo prime(9) = 23.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..500
• Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

Crossrefs

Cf. A000040, A000110, A236308, A236966, A237112, A237121.

Programs

Mathematica

f[k_]:=BellB[Prime[k]]
dv[n_]:=Divisors[n]
Do[m=0;Do[If[Mod[f[k],Prime[n]]==0,Goto[aa],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,PrimePi[(Prime[n]-1)/2]}];Print[n," ",m];Continue,{n,1,80}]