Bisymmetric Hadamard matrices

Question

Definitions: An $n\times n$ Hadamard matrix is a matrix whose entries are either $1$ or $−1$ and whose rows are mutually orthogonal. A symmetric matrix is a square matrix that is equal to its own transpose. A bisymmetric matrix is a square matrix that is symmetric about both of its two longest diagonals.

Background: It is conjectured that there exists an $n\times n$ symmetric Hadamard matrix if and only if there exists an $n\times n$ Hadamard matrix.

Questions: Do you have any conjecture, conclusion or algorithm for bisymmetric Hadamard matrices? Any remark will help. Thanks!

Answer 1

There exists an $n$-by-$n$ bisymmetric Hadamard matrix, where $n=1,4,36$.

$n=1$ and $n=4$ are easy cases.

For $n=36$, let $L = (l_{ij})$ be a Latin square of order 6: that is, a 6-by-6 array with entries 1,2,3,4,5,6 such that each entry occurs exactly once in each row or column of the array. Now let H be a matrix with rows and columns indexed by the 36 cells of the array: its entry in the position corresponding to a pair (c,c') of distinct cells is defined to be 1 if c and c' lie in the same row, or in the same column, or have the same entry; all other entries (including the diagonal ones) are -1. Then H is a bisymmetric Hadamard matrix of order 36.

Answer 2

There is a theorem due to James Sylvester that says the Kronecker product of two Hadamard matrices is another Hadamard matrix. Furthermore, the Kronecker product of two bisymmetric Hadamard matrices is another bisymmetric Hadamard matrix.

So based on De Costa's answer, there exists an n-by-n bisymmetric Hadamard matrix, where $n=16,144,1296,...$

Maybe there is more than one recursive construction algorithm for bisymmetric Hadamard matrices. I will edit this answer if I retrieve other constructions.