The Davenport constant $D(G)$ of a finite group $G$ is the minimal $d$ such that any sequence/multiset of length $d$ is one-product, i.e., identity can be obtained as a product (in some order) of some non-empty subsequence. The small davenport constant $d(G)$ is the maximal length of one-product free sequences, i.e., $d(G)+1=D(G)$. I am focusing on $S_n$ and $A_n$.

In another post Davenport constant $D(S_5)=10$ or $11$? for computing $D(S_5)$, it ended up showing that $d(S_5)=10$, and it is also known that $d(S_3)=3$ and $d(S_4)=6$. For these three cases $d(S_n)= {n \choose 2}=n(n-1)/2$ holds, i.e., maximal one-product free sequences have this length.

I think it might be just coincidence, however this combinatorial expression appears a lot in $S_n$. For example, it is the total number of transpositions, or the largest number of inversions an element can have. If we consider adjacent transpositions $(i, i+1)$ as generators of $S_n$, this could mean that the largest length that a permutation could have is exactly ${ n \choose 2}$. Maybe it has nothing to do but...is there any relation between ${n \choose 2}$ and $d(S_n)$??

Maybe is not equality but some lower or upper bound, which would also be very interesting. This means that either we could construct a family of maximal one-product free sequences of length ${n \choose 2}$ in $S_n$ (taking the idea of distinct tranpositions or inversions into account) or prove that any sequence with more than ${n \choose 2}$ elements must be one-product. Any other strategies to get upper and lower bounds for $S_n$ or $A_n$ are also very welcome! Thanks a lot for your help ;)