Timeline for Explicit forms for the roots of Eulerian polynomials

6 events

when toggle format	what		by	license	comment
Dec 6, 2017 at 3:43	comment	added	Richard Stanley		I was mistaken when I said that it is known that $E_n(z)$ is irreducible. First, $E_{2m}(z)$ is divisible by $1+z$. It is only conjectured that $E_{2m+1}(z)$ and $E_{2m}(z)/(1+z)$ are irreducible. For some instances of irreducibility see ac.els-cdn.com/0022314X84900507/….
Dec 4, 2017 at 17:02	comment	added	Richard Stanley		As for the roots having an explicit description, if the Galois group of $E_n(z)$ were abelian then there would be a trigonometric formula for the roots. However, this is not the case. It is known that $E_n(z)$ is irreducible. By the symmetry of the coefficients, the largest possible Galois group has order $2^m m!$, where $m=\lfloor n/2\rfloor$ (and is the hyperoctahedral group $B_m$) . It has been checked for $n\leq 9$ that $B_m$ is indeed the Galois group of $E_n(z)$, so one could conjecture that this is the case for all $n$.
Dec 2, 2017 at 15:31	answer	added	Richard Stanley		timeline score: 7
Dec 2, 2017 at 13:13	history	edited	Christian Stump	CC BY-SA 3.0	typo
Dec 2, 2017 at 10:19	history	edited	Christian Stump	CC BY-SA 3.0	math typo
Dec 2, 2017 at 10:12	history	asked	Christian Stump	CC BY-SA 3.0