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A321969
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Distinct weight enumerators for binary self-dual codes of length 2n.
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2
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1, 1, 1, 2, 2, 3, 4, 6, 8, 15, 22, 45, 79, 159, 312, 696, 1264, 3504
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OFFSET
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1,4
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COMMENTS
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All binary self-dual codes must have even length. This is due to the fact that all binary self-dual codes contain the "all 1's codeword". If a binary self-dual code could have odd length, the dot product of the "all 1's codeword" with itself would be equal to 1. A dot product of 1 would imply it is not orthogonal to itself and so the code could not be self-dual.
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LINKS
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EXAMPLE
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For n=1 there is only a(1) = 1 distinct weight enumerator for the length 2 binary self-dual codes. Up to permutation equivalence there is only 1 length 2 binary self-dual code.
For n=18 there are a(18)=3504 distinct weight enumerators. There are 519492 binary self-dual codes of length 36. However there are only 3504 distinct weight enumerators among the 519492 binary self-dual codes.
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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STATUS
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approved
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