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A111931
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Smallest prime p such that 1/2, 2/3, 3/4, ..., (m-1)/m are n-th power non-residues modulo p for maximum possible m (=A000236(n)).
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2
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OFFSET
| 2,1
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COMMENTS
| A000236(n) is the maximum length of a run of consecutive residues modulo prime p starting with 1 where no two adjacent elements belong to the same n-th power residue class (i.e., there is no n-th power residue modulo p in the sequence 1/2, 2/3, ..., A000236(n-1)/A000236(n)). A111931(n) equals the smallest p admitting a run of maximum length A000236(n).
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EXAMPLE
| a(2)=11 since A000236(2)=3 and 1/2=6, 2/3=8 are nonsquares modulo 11 and there no smaller prime modulo which 1/2 and 2/3 are nonsquares.
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CROSSREFS
| Cf. A000236, A000445.
Sequence in context: A165673 A120792 A145833 * A066433 A038741 A169731
Adjacent sequences: A111928 A111929 A111930 * A111932 A111933 A111934
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KEYWORD
| hard,nonn,more
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AUTHOR
| Max Alekseyev (maxale(AT)gmail.com), Aug 21 2005
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