TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The Eulerian numbers satisfy $A (n, k) = (k + 1) A (n - 1, k) + (n - k) A (n - 1, k - 1)$

Statement

For $n \geq 1$ and every natural number $k$ ,

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

where the Eulerian numbers are extended by $A (m, j) = 0$ for $j < 0$ or $j \geq m$ , except for the defining value $A (0, 0) = 1$ .

Facts & Assumptions

Given: A natural number $n \geq 1$ .

Proof

technique · direct

1.1given

Take a permutation $π \in S_{n - 1}$ and insert the new largest letter $n - 1$ into one of the $n$ slots of its one-line notation. If a slot lies after a descent of $π$ , or is the final slot, then the insertion preserves the number of descents: one old descent is replaced by one new descent, or no descent is created at the end. Every other slot creates one new descent.

2.1step 1.1

If $π$ has exactly $k$ descents, step 1.1 gives exactly $k + 1$ insertion slots producing a permutation of $S_{n}$ with $k$ descents. If $π$ has exactly $k - 1$ descents, the remaining $n - k$ slots produce a permutation with $k$ descents. These two cases are disjoint and exhaust the permutations of $S_{n}$ with $k$ descents.

3.1step 2.1∎

Counting the two cases of step 2.1 gives $A (n, k) = (k + 1) A (n - 1, k) + (n - k) A (n - 1, k - 1)$ . The out-of-range convention makes the same formula correct at the boundary values of $k$ .

Depends on

Eulerian numbers and Eulerian polynomials

Used by

Dependency tree · two levels

2 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

Felix Gotti, MIT 18.211 Combinatorial Analysis, Lecture 13 (standard reference, not scraped)
Andrew Lin, MIT 18.212 Algebraic Combinatorics, Lecture 12 (standard reference, not scraped)

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