TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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If $A$ has no size-zero objects then $MSET (A)$ has generating function $\prod_{n \geq 1} (1 - x^{n})^{- a_{n}}$

Statement

Let $A$ be a combinatorial class with no size-zero objects, and write

$A (x) = \sum_{n \geq 1} a_{n} x^{n}$

for its ordinary generating function. Then $MSET (A)$ is a combinatorial class and

$OGF (MSET (A)) = \prod_{n \geq 1} (1 - x^{n})^{- a_{n}} .$

Facts & Assumptions

Given: A combinatorial class $A$ with no size-zero objects and counting sequence $(a_{n})_{n \geq 1}$ .

[L1]

A formal power series is a unit exactly when its constant coefficient is a unit (A formal power series is a unit exactly when its constant coefficient is a unit).

[L2]

Summable formal families may be regrouped and rearranged, distribute over multiplication, and have well-defined locally finite products (Summable formal families may be regrouped and rearranged, distribute over multiplication, and have well-defined locally finite products).

Proof

technique · direct

1.1

For one fixed object $u \in A$ of size $n \geq 1$ , the possible multiplicities $0, 1, 2, \dots$ contribute the formal series $1 + x^{n} + x^{2 n} + \dots$ , which is $(1 - x^{n})^{- 1}$ because its product with $1 - x^{n}$ is $1$ coefficientwise. The factor exists by [L1], since $1 - x^{n}$ has constant coefficient $1$ .

L1algebra

2.1

A multiset of $A$ is exactly a choice of one multiplicity for each object of $A$ . Because every object has positive size, only finitely many objects can contribute to any fixed degree $< N$ , so the product of the per-object series of step 1.1 is locally finite and may be regrouped by [L2]. This also shows that each size layer of $MSET (A)$ is finite.

step 1.1L2

3.1

Regroup the factors of step 2.1 by object size. For each $n \geq 1$ there are exactly $a_{n}$ objects of size $n$ , and each contributes one factor $(1 - x^{n})^{- 1}$ , so the total contribution of size $n$ objects is $(1 - x^{n})^{- a_{n}}$ . Multiplying over all sizes gives the displayed product formula.

step 2.1L2∎

Depends on

Used by

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 (standard reference, not scraped)
Stephen Melczer, An Invitation to Enumeration, Chapter 5: Combinatorial Constructions (standard reference, not scraped)

If AA has no size-zero objects then MSET⁡(A)MSET(A) has generating function ∏n≥1(1−xn)−an∏n≥1​(1−xn)−an​