de Bruijn sequence in which order of subsquences doesn't matter A de Bruijn sequence of of size $k$ will contain every single subsequence of length $n$ exactly once within a single cycle. Is there another similar type of sequences in which the order of the subsequence elements is not considered? 
For instance, consider this de Bruijn sequence of alphabet size 3 and subsequence length 3:
$0\ 0\ 0\ 1\ 0\ 0\ 2\ 0\ 1\ 1\ 0\ 1\ 2\ 0\ 2\ 1\ 0\ 2\ 2\ 1\ 1\ 1\ 2\ 1\ 2\ 2\ 2\ (0\ 0)$
In the case above, certain subsequences will have the same elements but in different order (e.g. $0\ 0\ 1$, $0\ 1\ 0$, $1\ 0\ 0$). 
For the case above, I would like to find a sequence which would list all subsequences below exactly once before looping:
$0\ 0\ 0$
$0\ 0\ 1$
$0\ 0\ 2$
$0\ 1\ 1$
$0\ 1\ 2$
$0\ 2\ 2$
$1\ 1\ 1$
$1\ 1\ 2$
$1\ 2\ 2$
$2\ 2\ 2$
Does such a sequence exist?

For the purposes of this post, I will call unordered de Bruijn sequences the sequences satisfying the definition given in the question.
For any alphabet size $k$ and $n=1$, the sequence $1 \ 2 \ \ldots \ k$ is an unordered de Bruijn sequence.
For alphabet size $k=2$, such sequences exist only if $n=1$. Indeed, there are $n+1$ sequences of length $n$ on a two-letter alphabet (up to reordering), so an unordered de Bruijn sequence has to contain exaclty $n+1$ letters.  However, it must contain $0\ 0 \ \ldots \ 0$ and $1\ 1 \ \ldots \ 1$ as subwords, so it must have at least $2n$ letters.  Thus $n=1$.
In general, we can say this:


*

*An unordered de Bruijn sequence contains exactly $\binom{n+k-1}{k-1}$ letters, since this is the number of words of length $n$ in $k$ letters (up to reordering).

*In an unordered de Bruijn sequence, the number of occurences of a given letter is the same for all letters.  In particular, $k$ divides $\binom{n+k-1}{k-1}$.
This last point rules out your example where $n=k=3$.  In fact, for $k=3$, it rules out the cases where $n$ is divisible by $3$.
Beyond this, I don't know what can be said.  I haven't seen an unordered de Bruijn sequence yet (except when $n=1$).
