Deleting the first up-step and last down-step of a strictly positive walk of semilength $n$ and lowering the remainder by one gives a Dyck path of semilength $n-1$, bijectively. Hence $D_n=C_{n-1}$ and
$$
D(x)=\sum_{n\geq1}D_nx^n=xC(x).
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 145 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-145/3/b/solution