The question is inspired by G. Rzadkowski and M. Urlinska's examples in their paper A Generalization of the Eulerian Numbers. They refer to the discussion Expressions involving Eulerian numbers of the second kind, with Pietro Majer's answer being particularly relevant.

There are slightly different definitions of the Eulerian numbers, we follow the definitions in GKP, Concrete Mathematics, (6.35) and (6.41). In the OEIS the numbers are listed as A173018 and A201637.

The first-order Eulerian numbers are defined as $$ \left\langle n\atop k \right\rangle = (k+1) \left\langle n-1\atop k \right\rangle + (n-k) \left\langle n-1\atop k-1 \right\rangle, $$ with boundary conditions $\left\langle 0\atop 0 \right\rangle=1$, $\left\langle n\atop k \right\rangle =0$ for $k<0$ or $k > n$.

The second-order Eulerian numbers are defined by the recurrence

$$ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle = (k+1) \left\langle\!\!\left\langle n-1\atop k\right\rangle\!\!\right\rangle + (2n-k-1) \left\langle\!\!\left\langle n-1\atop k-1\right\rangle\!\!\right\rangle . $$

The same boundary conditions as for the first-order numbers apply.

Based on a hypothesis about the representation of the Bernoulli numbers (formula 29), one can derive from Rzadkowski and Urlinska's formula 20 and the last formula in their paper:

$$ \frac{1}{n+1}\, \sum_{k=0}^{n-1} (-1)^{k} \frac{ \left\langle n\atop k \right\rangle }{ \binom{n}{k} } = \frac{1}{2}\sum\limits_{k=0}^{n-1}(-1)^k \frac{\left\langle\!\!\left\langle n-1\atop k \right\rangle\!\!\right\rangle} { \binom {2n-1}{k+1}} \quad(n \ge 0). $$

Apparently, both sides have for $n \ge 1$ the value $B_{n}(1)$, where $B_{n}(x)$ denotes the Bernoulli polynomials. Can this equation be proved, or can the relation between the two kinds of Eulerian numbers be expressed more succinctly?

**Edit**: Definitions adapted to GKP, Concrete Mathematics and adjusted the conclusion to them.