We present some variation on this MO question which is equally amusing to me.

Define the $2^{n-1}\times 2^{n-1}$ matrix $A_n$ recursively as follows: $A_1(a_1)=\begin{pmatrix} a_1\end{pmatrix}$ and $$A_n(a_1,\dots,a_n)=\begin{pmatrix} A_{n-1}(a_1,\dots,a_{n-1})& a_nJ_{n-1}\\ -a_nJ_{n-1}&A_{n-1}(a_1,\dots,a_{n-1}) \end{pmatrix}.$$ Here $J_n$ is a $2^{n-1}\times 2^{n-1}$ matrix with $1$'s on the antidiagonal and zeros elsewhere.

*Example.* For $n=2$ and $n=3$, we have
$$A_2(a_1,a_2)=\begin{pmatrix} a_1&a_2\\-a_2&a_1\end{pmatrix}
\qquad\text{and} \qquad
A_3(a_1,a_2,a_3)=\begin{pmatrix} a_1&a_2&0&a_3\\-a_2&a_1&a_3&0\\0&-a_3&a_1&a_2\\-a_3&0&-a_2&a_1\end{pmatrix}.$$

Question.Is there a closed (or interesting) formula for the determinant $\det(A_n)$?

7more comments