A241504 a(n) = |{0 < g < prime(n): g is not only a primitive root modulo prime(n) but also a partition number given by A000041}|.

1, 1, 2, 2, 2, 3, 4, 3, 4, 4, 3, 4, 5, 3, 5, 4, 5, 3, 3, 5, 4, 4, 6, 5, 4, 6, 4, 6, 4, 3, 4, 4, 3, 7, 8, 5, 3, 6, 5, 8, 5, 2, 5, 7, 7, 6, 4, 7, 7, 2, 7, 5, 3, 6, 6, 10, 9, 5, 8, 7, 5, 10, 5, 5, 3, 8, 5, 5, 9, 4, 5, 5, 5, 8, 7, 10, 9, 6, 7, 4

Offset: 1

Author

Zhi-Wei Sun, Apr 24 2014

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0. In other words, any prime p has a primitive root g < p which is also a partition number.

(ii) Any prime p > 3 has a primitive root g < p which is also a strict partition number (i.e., a term of A000009).

We have checked part (i) for all primes p < 2*10^7, and part (ii) for all primes p < 5*10^6. See also A241516.

Examples

			a(92) = 1 since p(13) = 101 is a primitive root modulo prime(92) = 479, where p(.) is the partition function (A000041).
a(493) = 1 since p(20) = 627 is a primitive root modulo prime(493) = 3529.
a(541) = 1 since p(20) = 627 is a primitive root modulo prime(541) = 3911.
a(1146) = 1 since p(27) = 3010 is a primitive root modulo prime(1146) = 9241.
a(1951) = 1 since p(35) = 14883 is a primitive root modulo prime(1951) = 16921.
a(2380) = 1 since p(36) = 17977 is a primitive root modulo prime(2380) = 21169.
a(5629) = 1 since p(20) = 627 is a primitive root modulo prime(5629) = 55441.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Crossrefs

Cf. A000009, A000040, A000041, A237121, A239957, A239963, A241476, A241492, A241516, A241568.

Programs

Mathematica

f[k_]:=PartitionsP[k]
dv[n_]:=Divisors[n]
Do[m=0;Do[If[f[k]>Prime[n]-1,Goto[bb]];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,Prime[n]-1}];Label[bb];Print[n," ",m];Continue,{n,1,80}]