Coefficient-matrix determinant conjecture for lattice-path tables

The proposed determinant has an incorrect sign. In fact, writing $h=\lceil m/2\rceil$, the correct general formula is

$$\det\mathcal C(T)=(-1)^{h(h-1)/2}2^{\lfloor m/2\rfloor}.$$

Here is an explicit counterexample with $m=7$ and $n=5$. Write the symmetric population vector in column $n-4$ as

$$(x_1,x_2,x_3,x_4,x_3,x_2,x_1)^{\mathsf T}.$$

One right/up/down step induces the half-vector transition

$$B=\begin{pmatrix}1&1&0&0\\1&1&1&0\\0&1&1&1\\0&0&2&1\end{pmatrix}, \qquad \lambda=(2,2,2,1)^{\mathsf T}.$$

Consequently the columns of the coefficient matrix, in exactly the stated order corresponding to $\mathcal I_7(n-1),\mathcal I_7(n-2),\mathcal I_7(n-3),\mathcal I_7(n-4)$, are

$$(B^{\mathsf T})^3\lambda,\quad (B^{\mathsf T})^2\lambda,\quad B^{\mathsf T}\lambda,\quad\lambda.$$

Therefore

$$\mathcal C(T)= \begin{pmatrix} 26&10&4&2\\ 44&16&6&2\\ 52&18&6&2\\ 27&9&3&1 \end{pmatrix}, \qquad \det\mathcal C(T)=8\ne-8=-2^{\lfloor7/2\rfloor}.$$

For completeness, the corrected formula holds for every $m$. Let $A$ be the $m\times m$ tridiagonal matrix with diagonal and adjacent diagonals equal to $1$. Let $S$ map an $h$-vector to its symmetric $m$-vector, and define $B$ by $AS=SB$. Put $W=S^{\mathsf T}S$ and $u=(1,\ldots,1)^{\mathsf T}$. Then

$$W=\operatorname{diag}(2,\ldots,2,1)\quad(m\text{ odd}), \qquad W=2I_h\quad(m\text{ even}),$$

so $\det W=2^{\lfloor m/2\rfloor}$. Symmetry of $A$ implies $B^{\mathsf T}W=WB$, and the population functional is $\lambda=Wu$. Thus

$$\mathcal C(T)=W[B^{h-1}u,B^{h-2}u,\ldots,Bu,u].$$

We claim $\det[u,Bu,\ldots,B^{h-1}u]=1$. The case $h=1$ is immediate. For $h\ge2$, put $L=3I_h-B$. The first row sum of $B$ is $2$, and all remaining row sums are $3$, so $Lu=e_1$. Therefore

$$[u,Lu,\ldots,L^{h-1}u] =[u,e_1,Le_1,\ldots,L^{h-2}e_1].$$

The first $h-2$ subdiagonal entries of $L$ are $-1$. Expanding along the last row, the remaining triangular minor has diagonal $1,-1,(-1)^2,\ldots,(-1)^{h-2}$, giving

$$\det[u,Lu,\ldots,L^{h-1}u] =(-1)^{h+1+(h-1)(h-2)/2} =(-1)^{h(h-1)/2}.$$

On the other hand, replacing $B^ju$ by $(3I-B)^ju$ is an upper-triangular change of columns with diagonal $(-1)^j$, whose determinant is also $(-1)^{h(h-1)/2}$. Hence

$$\det[u,Bu,\ldots,B^{h-1}u]=1.$$

Finally, reversing the $h$ columns contributes $(-1)^{h(h-1)/2}$, proving the corrected formula. The proposed constant negative sign therefore fails whenever $h\equiv0,1\pmod4$.