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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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If $A$ has no size-zero objects then $SEQ (A)$ has generating function $1 / (1 - A (x))$

Statement

Let $A$ be a combinatorial class with ordinary generating function

$A (x) = \sum_{n \geq 0} a_{n} x^{n},$

and suppose $A$ has no size-zero objects, so $a_{0} = 0$ . Then $SEQ (A)$ is a combinatorial class and

$OGF (SEQ (A)) = \frac{1}{1 - A (x)} .$

Consequently,

$OGF ({SEQ}_{\geq 1} (A)) = \frac{A (x)}{1 - A (x)} .$

Facts & Assumptions

Given: A combinatorial class $A$ with ordinary generating function $A (x)$ and no size-zero objects.

[L1]

Disjoint union and Cartesian product translate to addition and multiplication of ordinary generating functions (Disjoint union and Cartesian product translate to addition and multiplication of ordinary generating functions).

[L2]

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).

Proof

technique · direct

1.1

Every sequence in $SEQ (A)$ is either empty or has the form $(a, σ)$ with $a \in A$ and $σ \in SEQ (A)$ , so $SEQ (A) = E + A \times SEQ (A)$ as combinatorial classes. Also, because every object of $A$ has positive size, a sequence of total size $n$ has length at most $n$ , so the size- $n$ layer of $SEQ (A)$ is finite.

construct

2.1

Let $F (x)$ be the ordinary generating function of $SEQ (A)$ . Step 1.1 and [L1] give $F (x) = 1 + A (x) F (x)$ .

step 1.1L1

3.1

Since $A$ has no size-zero objects, the constant coefficient of $A (x)$ is $0$ , so the constant coefficient of $1 - A (x)$ is $1$ , which is a unit. By [L2], $1 - A (x)$ is invertible, and solving the equation of step 2.1 gives $F (x) = 1 / (1 - A (x))$ .

step 2.1L2

4.1

The class ${SEQ}_{\geq 1} (A)$ is $SEQ (A) - E$ , so [L1] and step 3.1 give $OGF ({SEQ}_{\geq 1} (A)) = F (x) - 1 = A (x) / (1 - A (x))$ .

step 3.1L1∎

Depends on

Used by

Binary words have generating function 1/(1-2x) Corollary
Positive-integer compositions have generating function x/(1-2x) Corollary
Rooted plane trees satisfy T(x)=x/(1-T(x)) Corollary
FALSE: SEQ(A) is a combinatorial class even when A has a size-zero object False statement

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)
Robert Sedgewick and Kevin Wayne, Analysis of Algorithms, Section 3.9 (standard reference, not scraped)

If AA has no size-zero objects then SEQ⁡(A)SEQ(A) has generating function 1/(1−A(x))1/(1−A(x))