A321969 Distinct weight enumerators for binary self-dual codes of length 2n.

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

Offset: 1

Author

Nathan J. Russell, Nov 22 2018

Keywords

• nonn
• more

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A003179.