Higher-order Eulerian triangle total positivity conjecture

For r1r\geq 1, let E(r)=(langlenkrangle ⁣(r))n,k0E^{(r)} = \left(\genfrac{\langle}{\rangle}{0pt}{}{n}{k}^{\! (r)}\right)_{n,k\geq 0} be the rrth-order Eulerian triangle, and let Eˇ(r)=(langlennk1rangle ⁣(r))n,k0\widecheck{E}^{(r)} = \left(\genfrac{\langle}{\rangle}{0pt}{}{n}{n-k-1}^{\! (r)}\right)_{n,k\geq 0} be its shifted reversal. Higher-order Eulerian conjecture. Both E(r)E^{(r)} and Eˇ(r)\widecheck{E}^{(r)} are totally positive for every r1r\geq 1. The cases concerning real-rootedness and coefficientwise Hankel-total positivity are already known, but the supplied status evidence does not establish that the two total-positivity assertions themselves are solved.

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Primary source

Bishal Deb and Alan D. Sokal, “Higher-order Stirling cycle and subset triangles: Total positivity, continued fractions and real-rootedness”, arXiv:2507.18959 (2025).

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