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A235709
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Least prime p < prime(n) with 2^p - 1 a quadratic residue modulo prime(n), or 0 if such a number does not exist.
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6
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0, 0, 0, 0, 2, 2, 7, 3, 2, 3, 3, 2, 5, 5, 2, 3, 2, 2, 7, 2, 2, 5, 2, 23, 2, 5, 3, 2, 2, 3, 5, 2, 3, 3, 3, 5, 2, 11, 2, 5, 2, 2, 2, 2, 3, 3, 11, 3, 2, 2, 3, 2, 2, 2, 5, 2, 7, 3, 2, 3, 3, 5, 3, 2, 2, 3, 5, 2, 2, 2, 7, 2, 3, 2, 7, 2, 3, 2, 3, 2, 2, 2, 2, 2, 3, 2, 3, 2, 11, 5, 2, 2, 5, 2, 5, 2, 7, 5, 3, 2
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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.
We have verified this for all n = 5, ..., 10^8.
Note that the conjecture in A234972 implies that for any prime p > 3 there is a prime q < p with 2^q - 1 a quadratic nonresidue modulo p.
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LINKS
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EXAMPLE
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a(8) = 3 since 2^3 - 1 = 7 is a quadratic residue modulo prime(8) = 19, but 2^2 - 1 = 3 is not.
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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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