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A237112
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Number of primes p < prime(n)/2 with p! a primitive root modulo prime(n).
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7
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0, 0, 1, 0, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 6, 4, 2, 3, 3, 4, 7, 9, 5, 7, 5, 8, 4, 7, 6, 7, 8, 7, 11, 8, 9, 7, 13, 10, 16, 4, 7, 8, 13, 9, 8, 12, 17, 10, 14, 12, 4, 10, 14, 15, 18, 8, 9, 8, 8, 18, 6, 8, 7, 16, 9, 11, 21, 15
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OFFSET
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1,5
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COMMENTS
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Conjecture: a(n) > 0 for all n > 4. In other words, for any prime p > 7, there exists a prime q < p/2 such that q! is a primitive root modulo p.
See also A236306 for a similar conjecture.
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LINKS
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EXAMPLE
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a(7) = 1 since 3 is a prime smaller than prime(7)/2 = 17/2 and 3! = 6 is a primitive root modulo prime(7) = 17.
a(9) = 1 since 5 is a prime smaller than prime(9)/2 = 23/2 and 5! = 120 is a primitive root modulo prime(9) = 23.
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MATHEMATICA
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f[k_]:=(Prime[k])!
dv[n_]:=Divisors[n]
Do[m=0; Do[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}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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