On the half-skew-centrosymmetric Hadamard matrices

Question

Definition 1: A Hadamard matrix is an $n\times n$ matrix $H$ whose entries are either $1$ or $-1$ and whose rows are mutually orthogonal.

Definition 2: A matrix $A$ is half-skew-centrosymmetric if there exist two square matrices $B$ and $C$ of order $n$ such that

\begin{equation} A = \begin{bmatrix} B & R C R \\ C & -R B R \end{bmatrix}. \end{equation}

where $R$ is the reverse identity matrix.

One day I found that these two definitions can be considered simultaneously. For example, \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} is the simplest half-skew-centrosymmetric Hadamard matrix.

Here comes my questions:

First, can you give some references about half-skew-centrosymmetric Hadamard matrices? Are there other names for these matrices?

Second one is about construction methods. It is easy to figure out how to construct a half-skew-centrosymmetric Hadamard matrix of order $2^k\cdot n$ based on a Hadamard matrix of order $n$ by using a variant of Sylvester's construction. Can you propose more methods to construct half-skew-centrosymmetric Hadamard matrices?

Third, can you prove the following conjecture or give a counter-example?

Conjecture: A half-skew-centrosymmetric Hadamard matrix exists for $n=2$ or $n$ is a multiple of $4$.

Answer 1

Let $H_n$ be an $n×n$ Hadamard matrix and $R_n$ the $n×n$ reverse identity matrix.

The matrix $X= \begin{pmatrix} H_n & R_nH_n \\ H_n & -R_nH_n \end{pmatrix}$ has entries of length $1$ and $$XX^* = 2nI_{2n} + ((nI_n - R_nH_nH_n^*R_n) \otimes R_2)$$ which is simply $2nI_{2n}$ so it is a Hadamard matrix. Permute the last $n$ columns with $R_n$ and you have it in the form you give. This gives half-skew-centrosymmetric Hadamard matrices of twice the size of a Hadamard matrix. (It also works for complex Hadamard matrices or even Hadamard matrices over *-rings)

Another construction (edit): suppose we apply the Paley construction II to a finite field with $q=4k+1$ elements giving an $2n×2n$ Hadamard matrix $H= \begin{pmatrix} H_1 & H_2 \\ H_3 & H_4 \end{pmatrix} $ after permuting the odd columns and rows to the first $n$ rows/columns, the even as last $n$ AND put the first even row/column at the last position. Define $X := X_{i,j} = (i-j)^{0.5(q-1)}$ (in the finite field this is $\pm 1$ with zero diagonal and symmetric because $-1$ is a square). Note that $X_{i,j+k}=X_{j+k,i}=X_{n-i+k,n-j}$. Then for Paley construction II, using $j$ as all-ones vector $$H_2^T = H_3 = \begin{pmatrix} j & X-I \\ -1 & j^T \end{pmatrix}$$ $$H_1 = \begin{pmatrix} 1 & j^T \\ j & X+I \end{pmatrix}$$ $$H_4 = \begin{pmatrix} -X-I & -j \\ -j^T & -1 \end{pmatrix}$$The $H_i$ now have the properties $R_nH_2R_n = H_3$ and $-R_nH_1R_n = H_4$ due to our note above. Thus $H$ is a half-skew-centrosymmetric Hadamard matrix!