Do you know if there is at least some characterisation or properties that maximal zero-sum free sequences must satisfy for n=3? I will write him an email maybe regarding this "easy" case.

Thank you very much! I was trying with $n=3$ and it was already difficult to find some characterisation...

@PeterTaylor Thanks a lot! I am trying Visual Code using Python and Sage environment. As you said, it takes some time but it works! I am now looking for one-free sequences in $A_6$ of length 12 and 4 distinct elements, it is taking long time (maybe there are not). By the way, I've just came up with an interesting question: do you think that proving there is NO one-product free sequence with $k$ distinct elements and length $n$, implies there is also NO one-free sequence of length $n$ and $l$ distinct elements, for any $l>k$? Intuitively, the less distinct elements, more likely is but...

@PeterTaylor Okay, thanks a lot! It seems I have to wait bit more then! The point is that the symbol of "loading" disappears after some time, which makes me think the computation is over. It seems it is not, or my computer does not wanna work ;)

@PeterTaylor I am literally taking the Sage code you posted in the answer of the other question for $D(S_5)$, and changing the last parameters instead of oneFreeProduct(S5, 10, 3) now I write oneFreeProduct(S6, 12. 4). I just change that line.

@PeterTaylor Yes, I agree! I asked you this since there is some contradiction I do not understand why it fails, maybe you could help me or my computer does not compute it correctly. As you previously mentioned in one comment above there is some one-free sequence of length 11 in $A_6$ with three distict elements, thus adding any odd permutation in $S_6$ it should be one-free sequence of length 12 in $S_6$ with 4 distinct elements. However, in the Sage code it seems not to exist such a sequence, when I look for it. Do you know why could that be?

@PeterTaylor If I understood well, the code runs thrugh all possible subsequence products, with all possible ordering right? Because that is so important point, for a fixed subsequence taking the different orders for the product into account!

@PeterTaylor Thanks a lot once again! I am definitely going to use your Sage code from now on since it is very useful for the calculations I need! I really appreciate your help ;)

Thanks a lot! That makes sense. Do you think there is also such an easy upper bound for $A_n$ or $S_n$? The unique idea I come up with is that $D(S_n) \leq n*D(S_{n-1})$ and of course $D(S_n) \leq 2*D(A_n)$, but still is need to know previous values...

Thank you so much for the contribution! It is weird I dind't realise to check first such a natural lower bound of $D(S_n)$. By the way, do you know if there is an equivalent function as the Landau function for alternating groups? Thanks :)