Eulerian

The Eulerian number $A(n, k)$, the number of permutations of $\{1, …, n\}$ with exactly $k$ descents (equally, $k$ ascents).

WikipediaEulerian number↗MathWorldEulerianNumber↗OEISA008292↗

Eulerian(n, k)the number of permutations of $\{1, …, n\}$ with exactly $k$ descents

CombinatoricaEulerian[n, k]SageMathenumeratio_eulerian(n, k)

Typed signatures

Type	Package	Applies
(integer, integer) -> integer	compute-engine	always

Domain: Combinatorics

Details

• Counts from $k = 0$: $A(n, 0) = 1$ is the identity, and $A(n, k) = 0$ for $k \ge n$ (except $A(0, 0) = 1$). Combinatorica's Eulerian[n, k] has the same convention; the Function Repository's ResourceFunction["EulerianNumber"][n, k] counts $k$ from $1$, so it equals Eulerian(n, k - 1). Wolfram 15 has no core Eulerian.
• Satisfies $A(n, k) = (k + 1)\,A(n - 1, k) + (n - k)\,A(n - 1, k - 1)$ and the symmetry $A(n, k) = A(n, n - 1 - k)$, which is why descents and ascents agree.
• Equals $Count(KDescentPermutations(n, k))$, and the number of permutations $p$ with $Descents(p) = k$; compute-engine's native, exact in bigints.
• A negative argument is left unevaluated (as is MahonianNumber); Combinatorica answers $0$ for a negative $k$.

Plot

Examples

2 more kept as data, checked against the oracles

Basic #

row $4$: $1, 11, 11, 1$

live ↗

In$\mathrm{Map}(k↦\mathrm{Eulerian}(4, k), 0..1..3)$Out$\bigl\lbrack1, 11, 11, 1\bigr\rbrack$

Known value · OEIS A008292

row $5$: $1, 26, 66, 26, 1$

live ↗

In$\mathrm{Map}(k↦\mathrm{Eulerian}(5, k), 0..1..4)$Out$\bigl\lbrack1, 26, 66, 26, 1\bigr\rbrack$

Known value · wolframscript 15.0.0, Combinatorica Eulerian[n, k]

the $11$ permutations of $4$ with one descent

live ↗

In$\mathrm{Eulerian}(4, 1)$Out$11$

Known value · wolframscript 15.0.0, Combinatorica Eulerian[n, k]

the empty permutation has no descents

live ↗

In$\mathrm{Eulerian}(0, 0)$Out$1$

Known value · wolframscript 15.0.0, Combinatorica Eulerian[n, k]

a permutation of $3$ has at most $2$ descents

live ↗

In$\mathrm{Eulerian}(3, 3)$Out$0$

Known value · wolframscript 15.0.0, Combinatorica Eulerian[n, k]

$A(6, 2) = 302$ both ways

live ↗

In$\mathrm{Eulerian}(6, 2)=\mathrm{Count}(\mathrm{KDescentPermutations}(6, 2))$Out$\mathrm{True}$

About

Examples	8
Checked against	Combinatorica 7 agreeSageMath 5 agree
Crosswalk	WikipediaMathWorldOEISCombinatoricaSageMath

Implementation

mappedcombinatoricapackages/oracle/src/mappings.ts

mappedsagepackages/oracle/src/mappings.ts

See also: Descents, Ascents, KDescentPermutations, MahonianNumber