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The first two determinants are $\det T(\emptyset)=1$ and $\det T(a_1)=1-a_1$. …

Yes. It's real when $N\equiv 0 \text{ mod } 4$ and imaginary when $N\equiv 2\text{ mod } 4$. The square of the determinant is $\det(A+iB)^2=\det(1-1+i(AB+BA))=i^N\det(AB+BA)$, so for either parity of …

Johann Cigler and I have posted a solution on arXiv: "An interesting class of Hankel determinants", arXiv:1807.08330. Let $d_r(N)=\det\left({2i+2j+r\choose i+j}\right)_{i,j=0}^{N-1}$. …

Flip the order of the Kronecker products to get $M'=A_n(I_n,I_n)+B_n\otimes T_n$, where $T_n=A_n(1,0)$. Note that $\det M=\det M'$. Since all blocks are polynomial in $A$, they commute, and therefore …

To complete Mahdi's answer, it suffices to show $$(1-\sum_i\frac{x_i}{1+2x_i})\prod_i(1+2x_i)=\sum_{j\ge 0} (2-j)2^{j-1}e_j.$$ Clearly $\prod_i(1+ax_i)=\sum_{j\ge 0} a^je_j$, which explains the $2^je_ …

Let $M$ be the matrix in question. The entry $M_{ij}$ is of the form $\frac{2x_i^2}{(2j+2)!}p_j(x_i)$ for some even polynomial $p_j$ of degree $2j$. After factoring out the $2x_i^2$ terms from each ro …

Johann Cigler and I have posted a proof of many of these observations on arXiv: "An interesting class of Hankel determinants", arXiv:1807.08330. …