TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26

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Over a commutative $Q$ -algebra, $MSET (A)$ has generating function $\exp (\sum_{k \geq 1} A (x^{k}) / k)$

Statement

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

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

be its ordinary generating function. Over a commutative $Q$ -algebra,

$OGF (MSET (A)) = \exp (\sum_{k \geq 1} \frac{A (x^{k})}{k}) .$

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[L1]

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

[L2]

Formal $\exp$ and $\log$ are inverse homomorphisms, and $\log ((1 + u) (1 + v)) = \log (1 + u) + \log (1 + v)$ (Formal $\exp$ and $\log$ are inverse homomorphisms and formal binomial powers obey the expected addition laws).

[L3]

The formal logarithm is $\log (1 + u) = \sum_{j \geq 1} (- 1)^{j - 1} u^{j} / j$ (Formal exponential, logarithm, and binomial powers over a commutative $Q$ -algebra).

Proof

technique · direct

1.1L1L2

Let $M (x)$ denote the multiset generating function. By [L1], $M (x) = \prod_{n \geq 1} (1 - x^{n})^{- a_{n}}$ , so applying $\log$ and using [L2] gives $\log M (x) = \sum_{n \geq 1} - a_{n} \log (1 - x^{n})$ .

2.1step 1.1L3algebra

By [L3], $- \log (1 - x^{n}) = \sum_{k \geq 1} x^{n k} / k$ , so $\log M (x) = \sum_{n \geq 1} \sum_{k \geq 1} a_{n} x^{n k} / k = \sum_{k \geq 1} (1 / k) \sum_{n \geq 1} a_{n} (x^{k})^{n} = \sum_{k \geq 1} A (x^{k}) / k$ . For each fixed degree, only finitely many pairs $(n, k)$ contribute, so the rearrangement is coefficientwise finite.

3.1step 2.1L2∎

Exponentiating the identity of step 2.1 and using that $\exp$ and $\log$ are inverse maps by [L2] gives $M (x) = \exp (\sum_{k \geq 1} A (x^{k}) / k)$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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

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)

Over a commutative QQ-algebra, MSET⁡(A)MSET(A) has generating function exp⁡(∑k≥1A(xk)/k)exp(∑k≥1​A(xk)/k)