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A235712
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Least prime p < prime(n) with 2^p + 1 a quadratic nonresidue modulo prime(n), or 0 if such a prime p does not exist.
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5
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0, 2, 0, 2, 7, 2, 2, 5, 2, 11, 11, 2, 7, 2, 2, 2, 5, 5, 2, 5, 2, 5, 2, 5, 2, 7, 2, 2, 5, 2, 2, 13, 2, 5, 13, 5, 2, 2, 2, 2, 5, 11, 5, 2, 2, 7, 5, 2, 2, 23, 2, 7, 5, 5, 2, 2, 5, 5, 2, 7, 2, 2, 2, 5, 2, 2, 7, 2, 2, 5, 2, 7, 2, 2, 11, 2, 5, 2, 5, 5, 5, 7, 7, 2, 5, 2, 5, 2, 7, 2, 2, 7, 2, 13, 7, 2, 5, 5, 2, 5
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OFFSET
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1,2
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COMMENTS
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Conjecture: a(n) > 0 for all n > 3.
Note that 2^3 + 1 = 3^2 is a quadratic residue modulo any prime p > 3. Also, there is no prime p < prime(316) = 2089 with 2^p + 1 a primitive root modulo 2089.
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LINKS
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EXAMPLE
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a(4) = 2 since 2^2 + 1 = 5 is a quadratic nonresidue modulo prime(4) = 7.
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MATHEMATICA
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Do[Do[If[JacobiSymbol[2^(Prime[k])+1, Prime[n]]==-1, 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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