Hi everybody, for my research I am dealing with the following function:

$$\alpha_n(x):=\left.\frac{\partial^{2n+1}}{\partial z^{2n+1}}\frac{\sinh(z)}{\cosh(z)-1+x}\right|_{z=0},\quad n\in \mathbb{N},$$

It is possible to show that $$\alpha_n(x)=\frac{P_n(x)}{x^{n+1}},$$ where $P_n(\cdot)$ is a polynomial of order $n$ in $x$, having integer coeffients.

To make few concrete examples $$\alpha_0(x)=\frac{1}{x}$$ $$\alpha_1(x)=\frac{-3+x}{x^2}$$ $$\alpha_2(x)=\frac{30-15 x+x^2}{x^3}$$ $$\alpha_3(x)=\frac{-630+420 x-63 x^2+x^3}{x^4}$$ $$\alpha_4(x)=\frac{22680-18900 x+4410 x^2-255 x^3+x^4}{x^5}$$ and so on.

What I would need to show (and it is veryfied for all the special cases I was able to compute, like those above) is that all the roots of $P_n(x)$ (and therefore those of $\alpha_n(x)$) are real and strictly greater than 2.

An explicit albeit complicated expression for $\alpha_n(x)$ can be obtained, namely:

```
\begin{equation*}
\begin{array}{ll}
\alpha_n(x)=&x^{-n-1}\sum_{j=0}^{n} x^{n-j}\sum_{k=j+1}^{n} (2k)!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{2k}\sum_{i=j}^{k}\left[ {i+1\choose j+1}\binom{2k}{2 i+1}-{i\choose j+1}\binom{2k}{2 i}\right](-2)^{j+1-2k}+\\
&x^{-n-1}\sum_{j=0}^{n} x^{n-j}\sum_{k=j}^{n}{(2k+1)!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{2k+1}}\sum_{i=j}^{k} \left[ {i+1\choose j+1}\binom{2k+1}{2 i+1}-{i\choose j+1}\binom{2k+1}{2 i}\right](-2)^{j-2k},
\end{array}
\end{equation*}
```

where the number between the curly brakets are the Stirling number of the second kind; Moreover,

$$\alpha_n(2)=-\sum_{k=1}^{2n+1}{k!\genfrac{\lbrace}{\rbrace}{0pt}{}{2n+1}{k}}(-2)^{-k}\neq0.$$

If someone is interested, I can post more on how I got these expressions. Thanks in advance to everybody that will try to help me.

Best Regards

Enzo