Positivity problems in enumerative and algebraic combinatorics

Theme 2: Total positivity of combinatorial matrices

Open
P2.1 Brenti's Eulerian triangle

The clean Eulerian numbers \( \cleaneuler{n}{k} \) count the number of permutations of \([n+1]\) with \(k\) excedances/descents. These numbers satisfy the recurrence: \[ \cleaneuler{n}{k} = (n-k+1)\cleaneuler{n-1}{k-1} + (k+1)\cleaneuler{n-1}{k}. \]

Conjecture: The matrix \(\left(\cleaneuler{n}{k}\right)_{n,k\geq 0}\) is totally positive.

Origin: Posed by Francesco Brenti in [Brenti (1996), Conjecture 6.10]
Open
P2.2 \(acdefg\)-triangle

Let the lower-triangular matrix \( T = (T(n,k))_{n,k \geq 0} \) be defined by the recurrence \[ T(n,k) = [a(n-k)+c]\,T(n-1,k-1) + (dk+e)\,T(n-1,k) + [f(n-2)+g]\,T(n-2,k-1) \] for \( n \geq 1 \), with initial conditions \( T(0,k) = \delta_{k0} \) and \( T(-1,k) = 0 \). Here \( a, c, d, e, f, g \) are treated as algebraic indeterminates, so that the matrix entries belong to the polynomial ring \( \mathbb{Z}[a,c,d,e,f,g] \).


Conjecture: The matrix \( T = (T(n,k))_{n,k \geq 0} \) is coefficientwise totally positive in the indeterminates \( a, c, d, e, f, g \).

Origin: Posed by Xi Chen, Bishal Deb, Alexander Dyachenko, Tomack Gilmore, and Alan D. Sokal in [Chen et. al., Conjecture 1.4]
Remarks:
Open
P2.3 \(r\)-th order Eulerian triangle

The \(r\)-th order Eulerian numbers \( \euler{n}{k}^{(r)} \) counts the number of Stirling permutations of the multiset \(\{1^r,\ldots,n^r\}\) with \(k\) ascents. They satisfy the recurrence \[ \euler{n}{k}^{(r)} = \big[rn - k - (r-1)\big] \euler{n-1}{k-1}^{(r)} + (k+1)\euler{n-1}{k}^{(r)} \] for \( n \geq 1 \), with initial condition \( \euler{n}{k}^{(r)} = \delta_{k0} \). Here \( r \geq 1 \) is a fixed integer.


(P.2.3a) Conjecture: For any \( r \geq 1 \), the \( r \)-th order Eulerian triangle \( E^{(r)} = \left( \euler{n}{k}^{(r)} \right)_{n,k \geq 0} \) is totally positive.

(P.2.3b) Conjecture: For any \( r \geq 1 \), the shifted-reversed \( r \)-th order Eulerian triangle \( E^{(r)} = \left( \euler{n}{n-k-1}^{(r)} \right)_{n,k \geq 0} \) is totally positive.

Origin: Posed by Bishal Deb, Alexander Dyachenko, and Alan Sokal in [Deb--Sokal (2025), Conjecture 1.7] and also by Ming-Jian Ding, Lili Mu, and Bao-Xuan Zhu in [Ding--Mu--Zhu (2026), Conjecture 5.1].
Remarks:
Open
P2.4 \((r,s)\)-generalised Eulerian triangle

The \((r,s)\)-generalised Eulerian numbers \( \euler{n}{k}^{(r,s)} \) are defined by the recurrence \[ \euler{n}{k}^{(r,s)} = \big[r(n-1) - k +1\big] \euler{n-1}{k-1}^{(r,s)} + [s(n-1)+k+1]\euler{n-1}{k}^{(r,s)} \] for \( n \geq 1 \), with initial condition \( \euler{n}{k}^{(r,s)} = \delta_{k0} \).


Conjecture: For any \(r,s \in \mathbb{N}\), the triangle \(\left( \euler{n}{k}^{(r,s)} \right)_{n,k \geq 0} \) is totally positive.

Origin: Posed by Ming-Jian Ding, Lili Mu, and Bao-Xuan Zhu in [Ding--Mu--Zhu (2026), Conjecture 5.1].
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