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%I #18 Dec 04 2018 04:13:24
%S 1,1,1,2,2,3,4,6,8,15,22,45,79,159,312,696,1264,3504
%N Distinct weight enumerators for binary self-dual codes of length 2n.
%C 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.
%H W. Cary Huffman and Vera Pless, <a href="https://doi.org/10.1017/CBO9780511807077">Fundamentals of Error Correcting Codes</a>, Cambridge University Press, 2003, Pages 7,252-282,338-393.
%e 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.
%e 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.
%Y Cf. A003179.
%K nonn,more
%O 1,4
%A _Nathan J. Russell_, Nov 22 2018
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