Let $E_n(z)$ be the Eulerian polynomial $$E_n(z) = \sum_{\tau \in \mathfrak{S}_n} z^{\operatorname{des}(\tau)}$$ where $\mathfrak{S}_n$ denotes the set of all permutations of $\{1,\ldots,n\}$ and $\operatorname{des}(\tau)$ is the number of descents $\tau(i) > \tau(i+1)$ in the permutation $\tau$. These are studied in great detail, see its OEIS wiki page for other definitions and several properties. In particular, it is known the Eulerian polynomial has only negative and simple roots, $$E_n(z) = \prod_{i=1}^n (1+q^{(n)}_i)$$ for different positive numbers $q^{(n)}_i$.

My question now is

What is known about the $q^{(n)}_i$'s? Do they have an explicit description?

(It is known that the roots of $E_n$ separate the roots of $E_{n+1}$. This is, $$q_1^{(n+1)} < q_1^{(n)} < \cdots < q_{n}^{(n+1)}< q_n^{(n)} < q_{n+1}^{(n+1)}$$ when they come in sorted order. That's not the type of property I am looking for, but only properties towards their explicit values.)

Here are the first two examples and the type of property I would like to have answered from a description I search for:

$$ \begin{align*} E_2(z) &= z^2 + 4z + 1 = (z+2-\sqrt{3})(z+2+\sqrt{3}) \\ E_3(z) &= z^3 + 11z^2 + 11z + 1 = (z+5+2\sqrt{6})(z+5-2\sqrt{6})(z+1) \end{align*} $$

(The roots become more complicated than $a\pm b\sqrt{c}$ for bigger $n$'s.)

It is known that the mean value of the discrete distribution given by $E_n$ is $n/2$ and the variance is $(n+2)/12$. This can be used to show that $$ \begin{align*} \sum_i \frac{1}{1+q^{(n)}_i} &= \frac{n}{2} \\ \sum_i \frac{q^{(n)}_i}{\big(1+q^{(n)}_i\big)^2} &= \frac{n+2}{12} \end{align*} $$

Doing this computation in the first example yields $$ \frac{1}{1+2+\sqrt{3}} + \frac{1}{1+2-\sqrt{3}} = 1 $$ and $$ \frac{2+\sqrt{3}}{(1+2+\sqrt{3})^2} + \frac{2-\sqrt{3}}{(1+2-\sqrt{3})^2} = \frac{1}{3} $$

Is there any known immediate property of the roots that can be used to get these identities?

Let $E_n(z)$ be the Eulerian polynomial $$E_n(z) = \sum_{\tau \in \mathfrak{S}_n} z^{\operatorname{des}(\tau)}$$ where $\mathfrak{S}_n$ denotes the set of all permutations of $\{1,\ldots,n\}$ and $\operatorname{des}(\tau)$ is the number of descents $\tau(i) > \tau(i+1)$ in the permutation $\tau$. These are studied in great detail, see its OEIS wiki page for other definitions and several properties. In particular, it is known the Eulerian polynomial has only negative and simple roots, $$E_n(z) = \prod_{i=1}^n (z+q^{(n)}_i)$$ for different positive numbers $q^{(n)}_i$.

My question now is

What is known about the $q^{(n)}_i$'s? Do they have an explicit description?

(It is known that the roots of $E_n$ separate the roots of $E_{n+1}$. This is, $$q_1^{(n+1)} < q_1^{(n)} < \cdots < q_{n}^{(n+1)}< q_n^{(n)} < q_{n+1}^{(n+1)}$$ when they come in sorted order. That's not the type of property I am looking for, but only properties towards their explicit values.)

Here are the first two examples and the type of property I would like to have answered from a description I search for:

$$ \begin{align*} E_2(z) &= z^2 + 4z + 1 = (z+2-\sqrt{3})(z+2+\sqrt{3}) \\ E_3(z) &= z^3 + 11z^2 + 11z + 1 = (z+5+2\sqrt{6})(z+5-2\sqrt{6})(z+1) \end{align*} $$

(The roots become more complicated than $a\pm b\sqrt{c}$ for bigger $n$'s.)

It is known that the mean value of the discrete distribution given by $E_n$ is $n/2$ and the variance is $(n+2)/12$. This can be used to show that $$ \begin{align*} \sum_i \frac{1}{1+q^{(n)}_i} &= \frac{n}{2} \\ \sum_i \frac{q^{(n)}_i}{\big(1+q^{(n)}_i\big)^2} &= \frac{n+2}{12} \end{align*} $$

Doing this computation in the first example yields $$ \frac{1}{1+2+\sqrt{3}} + \frac{1}{1+2-\sqrt{3}} = 1 $$ and $$ \frac{2+\sqrt{3}}{(1+2+\sqrt{3})^2} + \frac{2-\sqrt{3}}{(1+2-\sqrt{3})^2} = \frac{1}{3} $$

Is there any known immediate property of the roots that can be used to get these identities?

Let $E_n(z)$ be the Eulerian polynomial $$E_n(z) = \sum_{\tau \in \mathfrak{S}_n} z^{\operatorname{des}(\tau)}$$ where $\mathfrak{S}_n$ denotes the set of all permutations of $\{1,\ldots,n\}$ and $\operatorname{des}(\tau)$ is the number of descents $\tau(i) > \tau(i+1)$ in the permutation $\tau$. These are studied in great detail, see its OEIS wiki page for other definitions and several properties. In particular, it is known the Eulerian polynomial has only negative and simple roots, $$E_n(z) = \prod_{i=1}^n (1+q^{(n)}_i)$$ for different positive numbers $q^{(n)}_i$.

My question now is

What is known about the $q^{(n)}_i$'s? Do they have an explicit description?

(It is known that the roots of $E_n$ separate the roots of $E_{n+1}$. This is, $$q_1^{(n+1)} < q_1^{(n)} < \cdots < q_{n}^{(n+1)}< q_n^{(n)} < q_{n+1}^{(n+1)}$$ when they come in sorted order. That's not the type of property I am looking for, but only properties towards their explicit values.)

Here are the first two examples and the type of property I would like to have answered from a description I search for:

$$ \begin{align*} E_2(z) &= z^2 + 4z + 1 = (z+2-\sqrt{3})(z+2+\sqrt{3}) \\ E_3(z) &= z^3 + 11z^2 + 11z + 1 = (z+5+2\sqrt{6})(z+5-2\sqrt{6})(z+1) \end{align*} $$

(The roots become more complicated than $a\pm b\sqrt{c}$ for bigger $n$'s.)

It is known that the mean value of the discrete distribution given by $E_n$ is $n/2$ and the variance is $(n+2)/12$. This can be used to show that $$ \begin{align*} \sum_i \frac{1}{1+q^{(n)}_i} &= \frac{n}{2} \\ \sum_i \frac{q^{(n)}_i}{\big(1+q^{(n)}_i\big)^2} &= \frac{n+2}{12} \end{align*} $$

Doing this computation in the first example yields $$ \frac{1}{1+2+\sqrt{3}} + \frac{1}{1+2-\sqrt{3}} = 1 $$ and $$ \frac{1+2+\sqrt{3}}{(1+2+\sqrt{3})^2} + \frac{1+2-\sqrt{3}}{(1+2-\sqrt{3})^2} = \frac{1}{3} $$

Is there any immediate property of the roots that can be used to get these identities?

Let $E_n(z)$ be the Eulerian polynomial $$E_n(z) = \sum_{\tau \in \mathfrak{S}_n} z^{\operatorname{des}(\tau)}$$ where $\mathfrak{S}_n$ denotes the set of all permutations of $\{1,\ldots,n\}$ and $\operatorname{des}(\tau)$ is the number of descents $\tau(i) > \tau(i+1)$ in the permutation $\tau$. These are studied in great detail, see its OEIS wiki page for other definitions and several properties. In particular, it is known the Eulerian polynomial has only negative and simple roots, $$E_n(z) = \prod_{i=1}^n (1+q^{(n)}_i)$$ for different positive numbers $q^{(n)}_i$.

My question now is

What is known about the $q^{(n)}_i$'s? Do they have an explicit description?

(It is known that the roots of $E_n$ separate the roots of $E_{n+1}$. This is, $$q_1^{(n+1)} < q_1^{(n)} < \cdots < q_{n}^{(n+1)}< q_n^{(n)} < q_{n+1}^{(n+1)}$$ when they come in sorted order. That's not the type of property I am looking for, but only properties towards their explicit values.)

Here are the first two examples and the type of property I would like to have answered from a description I search for:

$$ \begin{align*} E_2(z) &= z^2 + 4z + 1 = (z+2-\sqrt{3})(z+2+\sqrt{3}) \\ E_3(z) &= z^3 + 11z^2 + 11z + 1 = (z+5+2\sqrt{6})(z+5-2\sqrt{6})(z+1) \end{align*} $$

(The roots become more complicated than $a\pm b\sqrt{c}$ for bigger $n$'s.)

It is known that the mean value of the discrete distribution given by $E_n$ is $n/2$ and the variance is $(n+2)/12$. This can be used to show that $$ \begin{align*} \sum_i \frac{1}{1+q^{(n)}_i} &= \frac{n}{2} \\ \sum_i \frac{q^{(n)}_i}{\big(1+q^{(n)}_i\big)^2} &= \frac{n+2}{12} \end{align*} $$

Doing this computation in the first example yields $$ \frac{1}{1+2+\sqrt{3}} + \frac{1}{1+2-\sqrt{3}} = 1 $$ and $$ \frac{2+\sqrt{3}}{(1+2+\sqrt{3})^2} + \frac{2-\sqrt{3}}{(1+2-\sqrt{3})^2} = \frac{1}{3} $$

Is there any known immediate property of the roots that can be used to get these identities?