Coefficient-matrix determinant conjecture for lattice-path tables

About 10 years old · traced to 2016

Let $T$ be an $m\times n$ table with $2n\geqslant m$ . Let $\mathcal{C}(T)$ be the coefficient matrix whose entries are the coefficients arising when the relevant quantities $\mathcal{I}_m(n-s)$ are expressed in terms of the lattice-path counts $x_t$ . Coefficient-matrix determinant conjecture. One has

\det(\mathcal{C}(T))=-2^{\left\lfloor\frac{m}{2}\right\rfloor}.

This is stated at the end of the section as a conjecture about the coefficient matrix. The supplied text gives no resolution or proof, so its status remains open.

References

Primary source

Daniel Yaqubi, Mohammad Farrokhi Derakhshandeh Ghouchan and Hamed Ghasemian Zoeram, “Lattice paths inside a table, I”, arXiv:1612.08697 (2019).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A posted calculation claims the conjecture is false and gives a corrected sign formula, but nobody has independently verified it.

Yaqubi, Derakhshandeh Ghouchan, and Ghasemian Zoeram stated the conjecture in 2016 for the coefficient matrix associated with an $m\times n$ lattice-path table. It asserts a constant determinant $-2^{\left\lfloor m/2\right\rfloor}$ whenever $2n\geqslant m$ .

Known results

The source gives the $5\times n$ coefficient matrix and computes determinant $-4$ , matching the conjecture for $m=5$ ; it supplies no general proof.

Posted attempt

A posted calculation claims an explicit $m=7$ , $n=5$ matrix with determinant $8$ , contradicting the conjectured value $-8$ , and proposes the general formula $(-1)^{h(h-1)/2}2^{\left\lfloor m/2\right\rfloor}$ , where $h=\lceil m/2\rceil$ . This is a claimed complete disproof, not independently verified.

Current status (as of August 2026): The published conjecture has only the displayed low-dimensional check; a posted counterexample and corrected formula are unverified, so the original statement is not mathematically settled.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

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 $h\equiv0,1\pmod4$ .