How many entries in a bisymmetric matrix can be chosen independently?

Question

A bisymmetric matrix is symmetric along both diagonals (it is both symmetric and persymmetric). A simple geometric argument (filling in by rows, in which for every new iteration, the outer two entries are non-unique) shows that for a bysymmetric n x n matrix, the answer is n + (n-2) + (n-4) + ... + 4 + 2 (for n even) or n + (n-2) + (n-4) + ... + 3 + 1 (for n odd). In other words, like n!! (but with addition rather than multiplication). Is there a way to simplify this?

Answer 1

Thanks to Joppy's comment I figured it out: $n(n+2)/4$ for even n, and ($n+1)^2/4$ for odd $n$.