TITLE: de Bruijn sequence in which order of subsquences doesn't matter
QUESTION [10 upvotes]: 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?

REPLY [2 votes]: When size of the alphabet (k) equals 1 and the length of the subsequences (n) is 1 there is a valid cycle, for higher n the additional repetitions from overlap invalidate the cycle (00"0" for instance has it's two 0's and then the loop adds a second instance of two 0's, in other words, it repeats.) When n is 0 you must have a valid cycle for any alphabet but this is trivial. When n=2 there is a valid cycle for k=3,5 (001122"0", 002113224330441"0") I have not found valid cycles for other k when n=2. I can confirm that if they exist they are of odd k. 
When k is even each digit must touch an odd number of other digits, which is impossible since it must be a cycle, so each digit must touch 2 other digits, one on each side. there is the exception of the digit that touches itself, but those form pairs and do not solve the problem. I do not believe this holds for n other than 2, but it may, or it may hold on all even n.
While standard De Bruijn cycles are very simple to compute (for how complicated they are), but these are very difficult it seems. though that may just be because I have not yet found a method of doing so.