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I am looking for all $k$-dimensional subspaces of $(\mathbb{Z}/2\mathbb{Z})^n$ up to permutational equivalence.

Is there a database of all $[n,k]$-codes up to equivalence for reasonable values of $(n,k)$? For instance, $1 \leq k \leq 6$ and $6 \leq n \leq 12$ (giving around 35,000 codes).

When $k=1$, there are $n$ such subspaces, each uniquely determined by the Hamming weight of the nonzero element.

When $k=2$, there are 0, 1, 3, 6, 10, 16, 23, 32 … (OEIS:A034198) such subspaces.

When $k=3$, there are 0, 0, 1, 4, 10, 22, 43, 77, … (OEIS:A034357) such subspaces.

When $k=4$, there are 0, 0, 0, 1, 5, 16, 43, 106, … (OEIS:A034358) such subspaces.

For each $k\geq 2$, I have trouble finding representatives of the codes for $n \geq 8$, but surely someone has recorded them somewhere. For $n=8$, there are just 8, 32, 77, 106 such codes for $k=1,2,3,4$, so it seems reasonable to me that someone wrote them down.

I have found several databases of "good" codes, but I actually want the bad codes too.

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For $k=2$ the code is determined up to equivalence by the Hamming weights $n\ge w_1\ge w_2\ge w_3>0$ of the non-zero words subject to the obvious constraints: $2\mid s\le 2n$, where $s=w_1+w_2+w_3$, and $w_2+w_3\ge w_1$. The sequence $(w_i)_{i=1}^3$ of numbers is an invariant of the code. It is also easy see that all codes sharing the same sequence are equivalent by permuting the coordinates in such a way that the 1s get stacked into the beginning. The weight enumerator is always an invariant, but there are known examples of non-equivalent codes with the same weight enumerator. –  Jyrki Lahtonen Jul 1 '11 at 12:16
    
Rats. The last sentence was meant to say that there are examples of non-equivalent codes sharing the same weight enumerator, but only when $k$ is larger. –  Jyrki Lahtonen Jul 3 '11 at 9:32

1 Answer 1

A more comprehensive list of those numbers can be found here. The numbers are growing extremely fast, so that in all but the smallest cases, it is hopeless to store a complete list of representatives.

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