Theorem: a(6*n-1) >= n + 3*n*(6*n-1).

Proof:

Let D(n,k) denote the max no of diagonals that can be put in the array, and L(n,k) denotes the number of diagonals that the L-arrangement admits (which is the triangular number for 2k by 2k arrays as above), we consider how large D(n,k)-L(n,k) can get.

For D(6k-1,6k-1), consider the middle 2k+1 by 2k+1 square and leaving its corners empty, fill in diagonals 'checkerboard style'. Then continue with 'branches' out of this middle square turning as needed. The 11 by 11 picture is shown in a separate link - think of the middle 5 by 5 square and the branches coming out of it.

Counting up the diagonals in this construction versus the L-arrangement proves that D(6k-1,6k-1)-L(6k-1,6k-1)>=k. This was Steve Wasielewski's idea, one of the students who worked on this project.  QED