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Number of primes p < prime(n)/2 such that P(p) is a primitive root modulo prime(n), where P(.) is the partition function given by A000041.
7

%I #8 Apr 22 2014 10:56:31

%S 0,0,1,1,2,2,2,3,3,5,1,3,4,1,4,5,5,5,3,4,6,6,5,7,6,8,5,8,5,8,10,9,9,9,

%T 11,7,6,9,11,9,14,5,6,4,10,4,6,7,12,9,14,9,8,11,11,17,23,11,15,6,13,

%U 22,14,14,11,19,11,7,22,13

%N Number of primes p < prime(n)/2 such that P(p) is a primitive root modulo prime(n), where P(.) is the partition function given by A000041.

%C Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there is a prime q < p/2 with P(q) = A000041(q) a primitive root modulo p.

%H Zhi-Wei Sun, <a href="/A237121/b237121.txt">Table of n, a(n) for n = 1..1000</a>

%e a(14) = 1 since 3 is a prime smaller than prime(14)/2 = 43/2 and P(3) = 3 is a primitive root modulo prime(14) = 43.

%t f[k_]:=PartitionsP[Prime[k]]

%t dv[n_]:=Divisors[n]

%t 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,1,70}]

%Y Cf. A000040, A000041, A236308, A236966, A237112.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Apr 22 2014