What is the number of ways to parenthesize $n$ elements using applications of operators of *arbitrary* arities larger than or equal to $2$? For example, for $n=3$, there are $3$ ways:
$$
abc, a(bc),(ab)c
$$
and for $n=4$ there are 11 ways:
$$
abcd,\ ab(cd),\ a(bc)d,\ (ab)cd ,\ a(bcd), (abc)d,$$
$$
a(b(cd)),\ a((bc)d),\ (ab)(cd),\ (a(bc))d,\ ((ab)c)d
$$
Note that, if we restrict the operators to have arity $2$ (i.e. binary operators), then the answer would be given by the Catalan number $C\_{n-1}$. (More generally, if we restrict the operators to have arity $p$, the answer would be given by generalized Catalan numbers. So the point here is that the arity is arbitrary, corresponding to a situation where I can select between operators of arities 2,3,4,...


An aymptotic formula for $n\to\infty$ would also be highly appreciated.



Even more generalized Catalan numbers
A little bit of programming and a look up in the OEIS tells me that this is the sequence A001003, the solution to [Schroeder's second problem](http://oeis.org/A001003), see also [Wikipedia](http://en.wikipedia.org/wiki/Schr%25C3%25B6der%25E2%2580%2593Hipparchus_number).


According to the page in OEIS, the asymptotic form is
$$ \frac{n^{-3/2}}{4}\sqrt{\frac{\sqrt{18}-4}{\pi}}(3+\sqrt{8})^n. $$


