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A237594
Number of primes p < prime(n)/2 such that the Bell number B(p) is a primitive root modulo prime(n).
3
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
1,5
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.
LINKS
Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 22 2014
STATUS
approved