Conjecture on determinants of trimidiation matrices

Let $B_d$ be the trimidiation matrix defined by equation (3.18) in the source. That defining formula, also confirmed by the displayed matrices for $d=1,2,3$, can be written as

$$(B_d)_{r,j} = [a^r c^{d-r}] (a+2c)^{d-j}(a-c)^j, \qquad 0\le r,j\le d. \tag{1}$$

Consider the linear substitution

$$x\longmapsto a+2c,\qquad y\longmapsto a-c,$$

whose matrix is

$$T= \begin{pmatrix} 1&1\\ 2&-1 \end{pmatrix}, \qquad \det T=-3.$$

In the standard descending monomial basis

$$(a^d,a^{d-1}c,\ldots,c^d),$$

the induced substitution on homogeneous polynomials of degree $d$ is $\operatorname{Sym}^d(T)$. However, equation (1) orders its output coefficients in the reverse basis

$$(c^d,ac^{d-1},\ldots,a^d).$$

Writing $J_d$ for the coordinate-reversal matrix, we therefore have

$$B_d=J_d\operatorname{Sym}^d(T). \tag{2}$$

Put

$$h=\binom{d+1}{2}=\frac{d(d+1)}2.$$

The reversal of $d+1$ coordinates has $h$ inversions, and hence

$$\det J_d=(-1)^h.$$

The eigenvalues of $T$ are $\sqrt3$ and $-\sqrt3$. Consequently the eigenvalues of $\operatorname{Sym}^d(T)$ are

$$(\sqrt3)^{d-j}(-\sqrt3)^j, \qquad j=0,\ldots,d,$$

and their product is

$$\det\operatorname{Sym}^d(T) = (\det T)^h = (-3)^h.$$

Taking determinants in (2) yields

$$\boxed{\det B_d=(-1)^h(-3)^h = 3^{d(d+1)/2}}$$

for every $d\ge0$, exactly as conjectured.