Alphabeta MathHow statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
- 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.
- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
The Stirling numbers of the second kind and the Bell numbers
Definition
For , let be the set of partitions of into exactly blocks. By The set partitions of a finite set form a finite set, this set is finite.
The Stirling number of the second kind is
The Bell number is
the total number of set partitions of .
The definition includes the standard boundary values:
- because the empty family is the unique partition of ;
- for because a nonempty set cannot be partitioned into zero nonempty blocks; and
- for because distinct elements cannot be split into more than nonempty blocks.
Depends on
Used by
- Dobinski's formula expresses the Bell numbers as Bₙ=e⁻¹∑_ℓ≥0ℓⁿ/ℓ! Corollary
- The Bell numbers satisfy Bₙ₊₁=∑ₖ₌₀ⁿC(n, k)Bₖ Corollary
- Ordered Bell numbers already differ from Bell numbers at n=3 Example
- S(4,2) computed by recurrence, inclusion-exclusion, and direct listing Example
- The set partitions of [4] grouped by their number of blocks Example
- FALSE: Bell numbers count ordered set partitions False statement
- Ordinary powers expand in the falling-factorial basis by the second-kind Stirling numbers Theorem
- The exponential formula gives the Bell-number generating function Theorem
- The Stirling numbers of the second kind are given by S(n,k)=1/k!∑ᵢ₌₀ᵏ(-1)ⁱC(k, i)(k-i)ⁿ Theorem
- The Stirling numbers of the second kind satisfy S(n,k)=kS(n-1,k)+S(n-1,k-1) Theorem
- The two Stirling triangles have the expected vertical exponential generating functions Theorem
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
- Herbert S. Wilf, generatingfunctionology, 2nd ed., §3.6 (standard reference, not scraped)
- Andrew Lin, 18.212 Algebraic Combinatorics, Lecture 11 (standard reference, not scraped)