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Computations with Maple suggest the following binomial identity \begin{equation*} \forall{p,j}: \sum_{k=j+1}^{p+1} (-1)^j \dfrac{1}{k}\binom{k-1}{j} = \sum_{k=j+1}^{p+1} (-1)^{k-1} \dfrac{1} …

The EULER numbers $E_n$, $n \in \mathbb{N}$, are defined via the TAYLOR expansion of the hyperbolic secant: \begin{equation} \text{sech}(x) = \sum_{n=0}^{\infty} \frac{E_n}{n!} x^n = \sum_{n=0 …

@Sam Hopkins: The formula there reads \begin{equation*} E_{2n}=\sum_{k=1}^{2n }(-1)^k\frac{1}{2^k}\sum_{\ell=0}^{2k}(-1)^{\ell}\binom{2k}{\ell}(k-\ell)^{2n} \end{equation*} which is slightly diffe …

Come that far, one can begin to exploit the subgroups, turn one's attention to symmetric spaces, of which there are the compact and noncompact ones, the riemannian and the hermitian ones. And one can …

Let $K$ be an algebraic number field, $f(x) = 0$ an algebraic equation over $K$ of degree $n \ge 2$ with only simple roots $x_1, \dotsc, x_n$, and $L \mathrel{:=} K[x_1, \dots, x_n]$ the splitting fie …