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Search: id:A111931
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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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+0 2
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OFFSET
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2,1
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COMMENT
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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
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a(2)=11 since A000236(2)=3 and 1/2=6, 2/3=8 are non-squares modulo 11 and there no smaller prime modulo which 1/2 and 2/3 are non-squares.
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CROSSREFS
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Cf. A000236, A000445.
Sequence in context: A165673 A120792 A145833 this_sequence A066433 A038741 A099336
Adjacent sequences: A111928 A111929 A111930 this_sequence A111932 A111933 A111934
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KEYWORD
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hard,nonn
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AUTHOR
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Max Alekseyev (maxale(AT)gmail.com), Aug 21 2005
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