

A321969


Distinct weight enumerators for binary selfdual codes of length 2n.


2



1, 1, 1, 2, 2, 3, 4, 6, 8, 15, 22, 45, 79, 159, 312, 696, 1264, 3504
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OFFSET

1,4


COMMENTS

All binary selfdual codes must have even length. This is due to the fact that all binary selfdual codes contain the "all 1's codeword". If a binary selfdual 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 selfdual.


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 7,252282,338393.


EXAMPLE

For n=1 there is only a(1) = 1 distinct weight enumerator for the length 2 binary selfdual codes. Up to permutation equivalence there is only 1 length 2 binary selfdual code.
For n=18 there are a(18)=3504 distinct weight enumerators. There are 519492 binary selfdual codes of length 36. However there are only 3504 distinct weight enumerators among the 519492 binary selfdual 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



