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

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

Statement

Let $I_{n}$ be the number of involutions of $[n]$ , let $F_{n}$ be the number of ordered set partitions of $[n]$ , and let $N_{n}$ be the number of set partitions of $[n]$ having no singleton block. The labelled symbolic method gives the following exponential generating functions:

involutions: $\sum_{n \geq 0} I_{n} \frac{x^{n}}{n!} = \exp ⁣ (x + \frac{x^{2}}{2});$
ordered Bell numbers: $\sum_{n \geq 0} F_{n} \frac{x^{n}}{n!} = \frac{1}{2 - e^{x}};$
set partitions with no singleton blocks: $\sum_{n \geq 0} N_{n} \frac{x^{n}}{n!} = \exp (e^{x} - 1 - x) .$

Facts & Assumptions

Given: The labelled symbolic rules and the set-partition formulas already proved on this page.

Proof

technique · direct

1.1

An involution is a labelled set of fixed points and transpositions, so its EGF is $\exp (x + x^{2} / 2)$ .

given

1.2

An ordered Bell structure is a sequence of nonempty labelled sets. The EGF of one nonempty labelled set is $e^{x} - 1$ , so the sequence rule gives $1 / (1 - (e^{x} - 1)) = 1 / (2 - e^{x})$ .

given

1.3

A partition with no singleton blocks is a labelled set of blocks of size at least $2$ . Subtracting the singleton contribution $x$ from the basic block EGF $e^{x} - 1$ leaves $e^{x} - 1 - x$ , and the set rule gives $\exp (e^{x} - 1 - x)$ .

given

2.1

These are exactly the three claimed specializations.

step 1.1step 1.2step 1.3∎

Depends on

Used by

Ordered Bell numbers already differ from Bell numbers at n=3 Example
The involutions of [4] and their cycle decomposition Example
The set partitions of [4] with no singleton blocks Example
FALSE: Bell numbers count ordered set partitions False statement

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10 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., §3.8 (standard reference, not scraped)
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics — Symbolic Combinatorics (standard reference, not scraped)