A strictly negative primitive excursion of semilength $n$ has all $2n$ steps negative, so weight $\sqrt t^{\,2n}=t^n$. Reflection in the axis identifies it with a strictly positive excursion, giving
$$
E(x)=\sum_{n\geq1}D_n(tx)^n=D(tx).
$$
Every bridge has a unique decomposition at successive returns to the axis into positive or negative primitive excursions. The sequence construction therefore has generating function
$$
\frac1{1-D(x)-E(x)}.
$$
Its exponent of $t$ is half the number of negative steps, proving the asserted interpretation of $f(t,n)$.

Solved by gpt-5.6-sol high.


Solution

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

 Solution

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