TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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$C (x) = 1 + x C (x)^{2}$

Statement

In $Q ⟦ x ⟧$ the Catalan generating function (The Catalan generating function $C (x) = \sum_{n \geq 0} C_{n} x^{n}$ in $Q ⟦ x ⟧$ ) satisfies

$C = 1 + x C^{2} .$

Facts & Assumptions

Given: the Catalan generating function $C \in Q ⟦ x ⟧$ .

[F1]

For every $n \geq 0$ , $[x^{n}] C = C_{n}$ , and a natural number written where a rational is expected denotes its image under an injective embedding preserving addition, multiplication and finite sums (The Catalan generating function $C (x) = \sum_{n \geq 0} C_{n} x^{n}$ in $Q ⟦ x ⟧$ ).

[F2]

$C_{0} = 1$ (The Catalan number $C_{n} : = ∣ D_{n} ∣$ ).

[L1]

$C_{n + 1} = \sum_{i = 0}^{n} C_{i} C_{n - i}$ in $N$ for every $n \in N$ ( $C_{n + 1} = \sum_{i = 0}^{n} C_{i} C_{n - i}$ , with $C_{0} = 1$ ).

[L2]

$[x^{n}] (f + g) = [x^{n}] f + [x^{n}] g$ ; $f = g$ if and only if $[x^{n}] f = [x^{n}] g$ for every $n$ ; $[x^{n}] (x^{k} f) = [x^{n - k}] f$ for $k \leq n$ and $0$ for $k > n$ ; and $[x^{n}] (f g) = \sum_{i = 0}^{n} [x^{i}] f [x^{n - i}] g$ (Coefficient extraction is $R$ -linear, separates formal series, shifts under multiplication by $x^{k}$ , and converts products to finite convolution).

[L3]

The coefficientwise sum and Cauchy product make $Q ⟦ x ⟧$ a commutative ring, and the constant series form an isomorphic copy of $Q$ inside it (Cauchy multiplication makes $R ⟦ x ⟧$ a commutative ring containing $R [x]$ as the finitely supported subring).

Proof

technique · direct

1.1F1F2L2L3

The constant coefficients agree: $[x^{0}] (1 + x C^{2}) = [x^{0}] 1 + [x^{0}] (x C^{2}) = 1 + 0 = 1$ , the second term vanishing by the clause of [L2] for $k = 1 > 0$ , and $[x^{0}] C = C_{0} = 1$ by [F1] and [F2].

1.2F1L1L2L3

The coefficients at every positive index agree. Let $n \in N$ . Then $[x^{n + 1}] (1 + x C^{2}) = [x^{n + 1}] (x C^{2}) = [x^{n}] (C^{2})$ by [L2], and the Cauchy product clause of [L2] evaluates $[x^{n}] (C^{2})$ as $\sum_{i = 0}^{n} C_{i} C_{n - i}$ , which is $C_{n + 1}$ by [L1], read in $Q$ through the embedding of [F1]. And $[x^{n + 1}] C = C_{n + 1}$ .

2.1L2step 1.1step 1.2∎

The two series have the same coefficient at every index by steps 1.1 and 1.2, so they are equal by the extensionality clause of [L2].

Remarks

This is the recurrence, transcribed. The equation carries exactly the content of the convolution recurrence together with the initial value $C_{0} = 1$ ; the passage between the two is the Cauchy product formula and nothing else. What the equation buys is that it can be solved, which a recurrence cannot be.
No division occurs. The equation is stated in the cleared form $C = 1 + x C^{2}$ . Solving it below produces the closed form by identifying a square root, not by dividing by $2 x$ , which is not a unit of $Q ⟦ x ⟧$ .

Depends on

Used by

Dependency tree · two levels

22 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

A. Postnikov (notes by A. Lin), MIT 18.212 Algebraic Combinatorics, Spring 2019, Proposition 6 (standard reference, not scraped)

C(x)=1+x C(x)2C(x)=1+xC(x)2