Using the <abacus of a partition> in James's convention, read runner $r$ as a one-runner abacus: replace each occupied position $r+eq$ by $q$. After the usual harmless shift of its beta set, the resulting partition is $\lambda^{(r)}$. Doing this for $r=0,\ldots,e-1$ produces
$$
Q_e(\lambda)=(\lambda^{(0)},\ldots,\lambda^{(e-1)}).
$$

Conjugating a partition complements beads and gaps and reflects the abacus. Reflection sends runner $r$ to runner $e-1-r$, reverses bead-gap order, and conjugates the runner partition. Hence
$$
Q_e(\lambda')
=\bigl((\lambda^{(e-1)})',\ldots,
(\lambda^{(1)})',(\lambda^{(0)})'\bigr).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-160/3/b/solution