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

Table of contents
- Solution b

Solution

Using the abacus of a partition in James's convention, read runner

r

as a one-runner abacus: replace each occupied position

r + e q

by

q

. After the usual harmless shift of its beta set, the resulting partition is

λ^{(r)}

. Doing this for

r = 0, \dots, e - 1

produces

Q_{e} (λ) = (λ^{(0)}, \dots, λ^{(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} (λ^{'}) = ((λ^{(e - 1)})^{'}, \dots, (λ^{(1)})^{'}, (λ^{(0)})^{'}) .

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version