TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The Stirling numbers of the second kind satisfy $S (n, k) = k S (n - 1, k) + S (n - 1, k - 1)$

Statement

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

$S (n, k) = k S (n - 1, k) + S (n - 1, k - 1) .$

Facts & Assumptions

Given: A partition of $[n]$ into exactly $k$ nonempty blocks, with $n \geq 1$ and $k \geq 1$ , counted by The Stirling numbers of the second kind and the Bell numbers.

Proof

technique · direct

1.1given

Look at the block containing the element $n$ . If that block is the singleton ${n}$ , deleting it leaves a partition of $[n - 1]$ into exactly $k - 1$ blocks. Conversely, adjoining ${n}$ to any partition of $[n - 1]$ into $k - 1$ blocks produces such a partition of $[n]$ . So the singleton case contributes $S (n - 1, k - 1)$ .

1.2givenalgebra

If the block of $n$ is not a singleton, delete $n$ from that block. The remaining blocks form a partition of $[n - 1]$ into exactly $k$ blocks, and the original partition is recovered by choosing one of those $k$ blocks and reinserting $n$ into it. So the nonsingleton case contributes $k S (n - 1, k)$ .

2.1step 1.1step 1.2∎

The two cases are disjoint and exhaustive, so their counts add to $S (n, k)$ . This gives the displayed recurrence.

Depends on

The Stirling numbers of the second kind and the Bell numbers

Used by

S(4,2) computed by recurrence, inclusion-exclusion, and direct listing Example

Dependency tree · two levels

3 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

Andrew Lin, 18.212 Algebraic Combinatorics, Lecture 11 (standard reference, not scraped)
Herbert S. Wilf, generatingfunctionology, 2nd ed., §3.6 (standard reference, not scraped)

The Stirling numbers of the second kind satisfy S(n,k)=kS(n−1,k)+S(n−1,k−1)S(n,k)=kS(n−1,k)+S(n−1,k−1)