The (generalized) Clausen numbers A160014 are defined as
$$\operatorname{C}_{n, k} = \prod_{ p\, -\, k\, |\, n} p \quad (p \in \mathbb{P})$$ where $\mathbb{P}$ denotes the primes. The classical Clausen numbers are $\operatorname{C}_{n} = \prod_{ p-1\, |\,n} p$. They are by the von Staudt-Clausen theorem the denominators of the Bernoulli numbers. Define
$$ b_{n} \, = \, \frac{1}{2}\sum\limits_{k=0}^{n-1}(-1)^k \frac{\left\langle\!\!\left\langle n-1\atop k+1\right\rangle\!\!\right\rangle} {\binom {2n-1}{k+1}}$$
where $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$ are the second-order Eulerian numbers A340556. Apparently the product $\operatorname{C}_{n}b_{n}$ are integers. $$\operatorname{C}_{n}b_{n} \in \mathbb{N} \quad (n \ge 0) $$
If this is true, it needs proof. Who can give it?