TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

How 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.

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The two Stirling triangles have the expected vertical exponential generating functions

Statement

For every fixed $k \in N$ ,

$\sum_{n \geq 0} S (n, k) \frac{x^{n}}{n!} = \frac{(e^{x} - 1)^{k}}{k!},$

and

$\sum_{n \geq 0} c (n, k) \frac{x^{n}}{n!} = \frac{1}{k!} {(\log \frac{1}{1 - x})}^{k} .$

Facts & Assumptions

Given: The second-kind and first-kind definitions and the labelled symbolic rules of The labelled constructions translate into the usual exponential-generating-function rules.

Proof

technique · direct

1.1

A partition of $[n]$ into exactly $k$ blocks is a labelled set of exactly $k$ nonempty labelled sets. The EGF of a nonempty labelled set of atoms is $e^{x} - 1$ , and taking an unordered set of exactly $k$ such blocks contributes the factor $1 / k!$ . Therefore $\sum_{n \geq 0} S (n, k) \frac{x^{n}}{n!} = \frac{(e^{x} - 1)^{k}}{k!} .$

given

1.2

A permutation with exactly $k$ cycles is a labelled set of exactly $k$ labelled cycles of atoms. The EGF of one labelled cycle is $\log (1 / (1 - x))$ , so the same labelled-set rule gives $\sum_{n \geq 0} c (n, k) \frac{x^{n}}{n!} = \frac{1}{k!} {(\log \frac{1}{1 - x})}^{k} .$

given

2.1

Steps 1.1 and 1.2 are exactly the two claimed vertical exponential generating functions.

step 1.1step 1.2∎

Depends on

Used by

Standard labelled specializations give involutions, ordered Bell numbers, and partitions without singletons Corollary

Dependency tree · two levels

12 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

Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics — Symbolic Combinatorics (standard reference, not scraped)
Herbert S. Wilf, generatingfunctionology, 2nd ed., §§3.5-3.6 (standard reference, not scraped)