The tensor rank of a three dimensional array $M[i,j,k], i,j,k\in [1,\ldots,n]$ is the minimal number of vectors $x\_i,y\_i,z\_i$, such that $M=\sum\_{i=1}^d x\_i\otimes y\_i\otimes z\_i$. 
From dimension argument it easily follows that there exists a tensor of tensor rank at least $\frac{1}{3}n^2$. One can also easily show that every tensor is of tensor rank at most $n^2$. 


So I know that the maximal tensor rank is between $\frac{1}{3}n^2$ and $n^2$. Does any one know what is the maximal tensor rank?


p.s. As far as I understand maximal border rank is $\frac{1}{3}n^2$.



What is the largest tensor rank of $n \times n \times n$ tensor?
The lower bound can be improved slightly to $n^3/(3n-2)$ by noting that in $x\otimes y\otimes z$ one can assume $|y|=|z|=1$. See also Chapter 20, "Typical Tensorial Rank", in the book *Algebraic Complexity Theory*, by Peter Bürgisser, Michael Clausen, and Mohammad Amin Shokrollahi.


An upper bound of $n^2−n−1$ is shown in <http://arxiv.org/abs/0909.4262v4> so that is probably the best known upper bound.


