Symbolic method

Remarks

For unlabelled classes the basic rules are as follows. The disjoint union $\mathcal{B} + \mathcal{C}$ gives $B(z) + C(z)$. The Cartesian product $\mathcal{B} \times \mathcal{C}$, with sizes added, gives $B(z) C(z)$. The sequence construction $\operatorname{SEQ}(\mathcal{B})$, for $\mathcal{B}_0 = \emptyset$, gives $1 / (1 - B(z))$. The multiset construction $\operatorname{MSET}(\mathcal{B})$, for $\mathcal{B}_0 = \emptyset$, gives $\exp\big(\sum_{k \ge 1} B(z^k) / k\big)$. For labelled classes one uses exponential generating functions. Then the labelled product gives $B(z) C(z)$, and, for $\mathcal{B}_0 = \emptyset$, sets give $\exp(B(z))$ and cycles give $\log \frac{1}{1 - B(z)}$.

Example: binary trees satisfy $\mathcal{B} = \mathcal{E} + \mathcal{Z} \times \mathcal{B} \times \mathcal{B}$, with size the number of internal nodes. The rules give $B(z) = 1 + z B(z)^2$. Example: permutations are labelled sets of cycles, so their exponential generating function is $\exp\big(\log \frac{1}{1 - z}\big) = \frac{1}{1 - z}$, which gives $n!$.

Non-example: the operation "keep the objects of $\mathcal{B}$ that are trees of height at most $2$" is not admissible. Its counts depend on the inner structure of the objects of $\mathcal{B}$, and not only on the numbers $\abs{\mathcal{B}_n}$.