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A237121
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
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, 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, 22, 14, 14, 11, 19, 11, 7, 22, 13
OFFSET
1,5
COMMENTS
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.
EXAMPLE
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.
MATHEMATICA
f[k_]:=PartitionsP[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, 70}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 22 2014
STATUS
approved