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A234972
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Least prime p < prime(n) such that 2^p - 1 is a primitive root modulo prime(n), or 0 if such a prime p does not exist.
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9
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0, 0, 2, 2, 3, 3, 2, 2, 3, 2, 2, 17, 3, 2, 5, 2, 5, 3, 3, 3, 5, 2, 11, 2, 3, 2, 13, 3, 7, 2, 2, 5, 2, 2, 2, 3, 11, 2, 11, 2, 3, 7, 7, 7, 2, 2, 2, 2, 5, 3, 2, 3, 3, 7, 2, 3, 2, 11, 5, 2, 2, 2, 5, 5, 5, 2, 2, 5, 3, 3, 2, 3, 7, 7, 2, 7, 2, 3, 2, 7, 5, 31, 3, 3, 5, 3, 2, 5, 2, 2, 5, 5, 2, 3, 3, 5, 2, 2, 7, 7
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OFFSET
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1,3
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COMMENTS
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Conjecture: a(n) > 0 for all n > 2.
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LINKS
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EXAMPLE
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a(3) = 2 since 2 is a prime smaller than prime(3) = 5 with 2^2 - 1 = 3 a primitive root modulo prime(3) = 5.
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MATHEMATICA
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gp[g_, p_]:=Mod[g, p]>0&&(Length[Union[Table[Mod[g^k, p], {k, 1, p-1}]]]==p-1)
Do[Do[If[gp[2^(Prime[k])-1, Prime[n]], Print[n, " ", Prime[k]]; Goto[aa]], {k, 1, n-1}]; Print[n, " ", 0]; Label[aa]; Continue, {n, 1, 100}]
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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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