So here is the problem (I.1.3 in Grillet's Abstract Algebra)
Let $X$ be a set with a binary operation $\cdot:X \times X \to X,$ where $\cdot(x,y):= xy, \text{ for all } x,y\in X$.
A product $x\in X $ of $\,x_1, \ldots, x_n \in X$ (in that order) is defined in the following way.
(1) If $n=1,$ then the only possibility is $x=x_1.$
(2) If $n\geq 2,$ then $x$ is a product of $x_1, \ldots, x_n$(in that order) iff for some $1\leq k < n$, $\, x=yz,$ where $y$ is a product of $x_1, \ldots, x_k$(in that order) and $z$ is a product of $x_{k+1}, \ldots, x_n$ (in that order).
Count all possible products of $x_1,\ldots, x_n$(in that order) for $n=6,7,8.$
For example $x=x_1(x_2x_3)\lor x=(x_1x_2)x_3$ are all the possible products for $n=3.$
I constructed all the products for $n=4$ and $n=5$ (previous exercises) and came up with an answer.
Let $|\Pi_n|$ denote the number all possible products of $x_1,\ldots, x_n$(in that order). Then
$$\big\vert \,\Pi_n \big\vert= \sum_{k=1}^{n-1}\big\vert \,\Pi_k \big\vert \cdot\big\vert \,\Pi_{n-k} \big\vert \,\,\,(n\geq 2)$$
I haven't proved it, but by going through the motions with $n=4$ and $n=5,$ I feel very comfortable with it. So I got $$\big\vert \,\Pi_6 \big\vert =42, \big\vert \,\Pi_7 \big\vert =132, \big\vert \,\Pi_8 \big\vert=429 $$
The problem with it is that in order to count, for example, $\big\vert \,\Pi_{100} \big\vert$ you need to know the result for all $n\leq 99.$
Is there a formula for doing the calculation without needing all the previous ones?
Thanks for your help in advance. If my formula is not correct, please say so.