Non-crossing partitions without singletons

Count the number $f(n)$ by cases of how many singletons the partition has. Let's denote by $g(n)$ the number of noncrossing partitions of $[n]$ without singletons. We have

$$f(n) = \sum_{j=0}^n {n \choose j}g(n-j)$$

because we can choose the $j$ singletons out of $n$ numbers and then the rest forms a non-crossing partition of $[n-j]$ without singletons.

We can express this as a matrix equation $f = Ag$ where $A = [{i \choose j}]_{ij}$.

Now (see for example here)

$$A^{-1} = [(-1)^{i+j}{i \choose j}]_{ij}$$

So we have a formula for the vector $g$ (since we know $f(n)=\frac{1}{n+1} { {2n} \choose n}$ are the Catalan numbers):

$$g(n) = \sum_{j=0}^n \frac{(-1)^{n+j}}{j+1} { {n} \choose j} { {2j} \choose j}. $$

EDIT:
These are the Riordan numbers (a more general OEIS-sequence was already linked by Jean Marie)

In the paper: link the following generating function equation is given and used to derive a recursion.

If we put $y = \sum g(n)x^n$, then

$$y = \frac{1}{1+x} + xy^2.$$

Implicit differentiation and arduous algebraic manipulation of the equations leads to

$$1 - x(1+x)(1-3x)y' = (1-3x^2)y.$$

From this the recursion

$$g(n) = \frac{n-1}{n+1}\left(2g(n-1)+3g(n-2)\right)$$

can be read.