CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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The Eulerian numbers satisfy $A (n, k) = A (n, n - 1 - k)$

Statement

For $n \geq 1$ and $0 \leq k \leq n - 1$ ,

$A (n, k) = A (n, n - 1 - k) .$

Facts & Assumptions

Given: A natural number $n \geq 1$ and the value-complement map $R (σ) (i) : = n - 1 - σ (i)$ on $S_{n}$ .

Proof

technique · direct

1.1

For every $0 \leq i \leq n - 2$ , one has $R (σ) (i) > R (σ) (i + 1)$ exactly when $σ (i) < σ (i + 1)$ . Thus the descent set of $R (σ)$ is the complement of the descent set of $σ$ in ${0, \dots, n - 2}$ , and $des (R (σ)) = n - 1 - des (σ)$ .

givenalgebra

2.1

The map $R$ is a bijection of $S_{n}$ , since applying it twice returns the original permutation. Therefore the number of permutations with $k$ descents equals the number with $n - 1 - k$ descents, which is exactly the displayed symmetry of the Eulerian numbers.

step 1.1given∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard P. Stanley, Enumerative Combinatorics, Volume 1, second edition (standard reference, not scraped)

The Eulerian numbers satisfy A(n,k)=A(n,n−1−k)A(n,k)=A(n,n−1−k)