The Bernoulli numbers $B_n$ are the backbone of calculus, and according to B. Mazur, they "act as a unifying force, holding together seemingly disparate fields of mathematics."
The Catalan numbers $C_n$ are the delight of the combinatorialist. The entry of these numbers is the longest in the OEIS, and according to N. Sloane, "rightly so."
But what is the relationship between these extraordinary numbers?
I found a formula that is amazingly simple and perhaps worthy of closer consideration:
$$ \frac{B_n}{C_n} \, = \, \sum_{k=0}^{n} (-1)^{k} \frac{E(n, k)}{F(n,k)} $$
The $E(n,k)$ are the Eulerian numbers, GKP CM 6.35 and A173018. The triangle $F(n, k)$ is the factorial counterpart of Leibniz's harmonic triangle. They are defined for $n \ge 0$ and $ 0 \le k \le n$ as
$$ F(n,k) \, = \, \frac{ (n+1)^{ \overline{n} } }{ k! \, (n-k)! },$$
where $ n^{\overline{k} } $ denotes the rising factorial. (The Leibniz triangle is obtained by choosing the falling factorial instead).
Notice: We set $B(1)=1/2$, following Donald Knuth's new definition of the Bernoulli numbers. The sign has reversed, in Concrete Mathematics since the 34-th printing, Jan. 2022.
Now the question arises, is there a simple proof of this identity?
Postscript: The identity is only a particular case of
$$\operatorname{\beta}_{n}(x) = \frac{1}{n+1}\, \sum_{k=0}^{n} (-1)^{k}\frac{ \left\langle n\atop k \right\rangle}{\binom{n}{k}}\, x^{n-k} \quad(n \ge 0).$$
I dubbed these polynomials Eulberian polynomials because I don't know any other name, but I had the impression that they should have one. They provide only one of several representations of the Bernoulli numbers which go back to Julius Worpitzky's 1883 'Studien'. The paper suggested by Jean Marie in the comments provides a proof in a modern presentation. See also OEIS A342321 and A356601.
Best Answer
We seek to show with Bernoulli numbers, Catalan numbers, Eulerian numbers and the Factorial counterpart of Leibniz's harmonic triangle:
$$\frac{B_n}{C_n} = \sum_{k=0}^n (-1)^k \frac{k! (n-k)!}{(n+1)^\overline{n}} \left\langle {n \atop k} \right\rangle.$$
Seeing that $(n+1)^\overline{n} = (2n)^\underline{n}$ we get the equivalent statement
$$B_n = \frac{1}{(n+1)!} \sum_{k=0}^n (-1)^k k! (n-k)! \left\langle {n \atop k} \right\rangle \\ = \sum_{k=0}^{n} (-1)^k \mathrm{B}(k+1, n-k+1) \left\langle {n \atop k} \right\rangle \\ = \int_0^1 u^n \sum_{k=0}^{n} (-1)^k \frac{(1-u)^k}{u^k} \left\langle {n \atop k} \right\rangle \; du.$$
We suppose as given that
$$\sum_{k=0}^n \left\langle {n \atop k} \right\rangle t^k = n! [x^n] \frac{t-1}{t-\exp((t-1)x)}.$$
This yields
$$\int_0^1 u^n n! [x^n] \frac{(u-1)/u-1}{(u-1)/u-\exp(((u-1)/u-1)x)} \; du \\ = n! [x^n] \int_0^1 \frac{(u-1)-u}{(u-1)-u\exp((u-1-u)x)} \; du \\ = n! [x^n] \int_0^1 \frac{1}{1-u+u\exp(-x)} \; du \\ = n! [x^n] \left. \frac{\log(1+(\exp(-x)-1)u)}{\exp(-x)-1} \right|_0^1 \\ = n! [x^n] \frac{(-x)}{\exp(-x)-1} = n! [x^n] \frac{x \exp(x)}{\exp(x)-1}.$$
Now the EGF of the Bernoulli numbers $\{B^{-{}}_n\}$ is
$$\frac{x}{\exp(x)-1}$$
so that we get for the revised EGF of $\{B^{+{}}_n\}$ by Luschny, Knuth et al. correcting the value at $n=1$ from $-\frac{1}{2}x$ to $\frac{1}{2} x$
$$\frac{x}{\exp(x)-1}+x = \frac{x+x(\exp(x)-1)}{\exp(x)-1} = \frac{x\exp(x)}{\exp(x)-1}$$
as claimed.
What we have here is another match score in favor of $\{B^{+{}}_n\}$ as defined in Luschny's Bernoulli Manifesto.