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A321969 Distinct weight enumerators for binary self-dual codes of length 2n. 2
1, 1, 1, 2, 2, 3, 4, 6, 8, 15, 22, 45, 79, 159, 312, 696, 1264, 3504 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

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.

LINKS

Table of n, a(n) for n=1..18.

W. Cary Huffman and Vera Pless, Fundamentals of Error Correcting Codes, Cambridge University Press, 2003, Pages 7,252-282,338-393.

EXAMPLE

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.

CROSSREFS

Cf. A003179.

Sequence in context: A325832 A068598 A293165 * A163770 A035561 A068106

Adjacent sequences:  A321966 A321967 A321968 * A321970 A321971 A321972

KEYWORD

nonn,more

AUTHOR

Nathan J. Russell, Nov 22 2018

STATUS

approved

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Last modified May 28 04:02 EDT 2022. Contains 354112 sequences. (Running on oeis4.)