ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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$A (4, 2) = 11$ by the recurrence, by Worpitzky's identity and by excedances

Example

The Eulerian number $A (4, 2)$ equals $11$ , and three different routes produce the same value.

Facts & Assumptions

Given: The Eulerian recurrence, Worpitzky's identity, and the equidistribution of descents and excedances (The Eulerian numbers satisfy $A (n, k) = (k + 1) A (n - 1, k) + (n - k) A (n - 1, k - 1)$ , Worpitzky's identity for $n \geq 1$ , Descents and excedances are equidistributed on $S_{n}$ ).

Verification

technique · direct

1.1

The recurrence gives $A (4, 2) = 3 A (3, 2) + 2 A (3, 1) = 3 \cdot 1 + 2 \cdot 4 = 11$ .

givenalgebra

1.2

Worpitzky's identity with $n = 4$ and $m = 2$ gives $2^{4} = A (4, 2) (\binom{4}{4}) + A (4, 3) (\binom{5}{4})$ , because the $k = 0, 1$ terms vanish. Since $A (4, 3) = 1$ , this reads $16 = A (4, 2) + 5$ , so again $A (4, 2) = 11$ .

givenalgebra

2.1

By descents/excedances equidistribution, $A (4, 2)$ also counts permutations of $S_{4}$ with exactly two excedances, and the $S_{4}$ table on this companion page contains exactly $11$ such permutations.

step 1.1given∎

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A(4,2)=11A(4,2)=11 by the recurrence, by Worpitzky's identity and by excedances