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$C (x)$ is not a rational formal power series, so $(C_{n})$ satisfies no eventual constant-coefficient linear recurrence

Statement

The Catalan generating function $C \in Q ⟦ x ⟧$ (The Catalan generating function $C (x) = \sum_{n \geq 0} C_{n} x^{n}$ in $Q ⟦ x ⟧$ ) is not a rational formal power series (Rational formal power series, proper presentations and reduced denominators): there are no polynomials $P, Q \in Q [x]$ with $Q (0) \neq 0$ and $Q C = P$ .

Consequently the sequence $(C_{n})_{n \geq 0}$ , read in $Q$ , satisfies no eventual constant-coefficient linear recurrence (A coefficient sequence is eventually linearly recurrent if and only if its formal generating function is rational).

Facts & Assumptions

Given: the Catalan generating function $C$ , and $Q [x]$ the polynomial ring over $Q$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[F1]

$C = 1 + x C^{2}$ in $Q ⟦ x ⟧$ ( $C (x) = 1 + x C (x)^{2}$ ).

[F2]

$Q ⟦ x ⟧$ is a commutative $Q$ -algebra and the coefficient of $C$ at the index $n$ is $C_{n}$ (The Catalan generating function $C (x) = \sum_{n \geq 0} C_{n} x^{n}$ in $Q ⟦ x ⟧$ ).

[L1]

A formal power series $F \in R ⟦ x ⟧$ is rational when there are polynomials $P, Q \in R [x]$ with $Q (0)$ a unit and $Q F = P$ (Rational formal power series, proper presentations and reduced denominators).

[L2]

For a field $K$ and a sequence $a$ in $K$ with $F = \sum_{n \geq 0} a_{n} x^{n}$ : $a$ satisfies an eventual constant-coefficient linear recurrence if and only if $F$ is a rational formal power series (A coefficient sequence is eventually linearly recurrent if and only if its formal generating function is rational).

[L3]

If $R$ is an integral domain and $f, g \in R [x]$ are nonzero, then $f g \neq 0$ and $\deg (f g) = \deg f + \deg g$ (Over an integral domain, degrees add under multiplication of nonzero polynomials).

[L5]

The degree of a nonzero polynomial is the largest index carrying a nonzero coefficient (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L6]

Every field is an integral domain (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring, clause 2).

[L7]

$Q$ is a field (The rationals form a field).

[L8]

The coefficientwise sum and Cauchy product make $Q ⟦ x ⟧$ a commutative ring, and the inclusion of $Q [x]$ into it is an injective unital ring homomorphism (Cauchy multiplication makes $R ⟦ x ⟧$ a commutative ring containing $R [x]$ as the finitely supported subring).

Proof

technique · contradiction

1.1F2L1L8assume-contra

Suppose $C$ is rational: by [L1] there are $P, Q \in Q [x]$ with $Q (0)$ a unit of $Q$ , hence $Q (0) \neq 0$ and $Q \neq 0$ , and $Q C = P$ in $Q ⟦ x ⟧$ .

2.1F1L8step 1.1

From [F1] we have $C - x C^{2} = 1$ , hence $(1 - 2 x C)^{2} = 1 - 4 x C + 4 x^{2} C^{2} = 1 - 4 x (C - x C^{2}) = 1 - 4 x$ . Multiplying by $Q^{2}$ and using $Q C = P$ gives $(Q - 2 x P)^{2} = (1 - 4 x) Q^{2}$ , an identity between polynomials, which by [L8] may be read inside $Q [x]$ . Put $R : = Q - 2 x P$ .

3.1L3L6L7step 2.1

$R \neq 0$ . Otherwise $(1 - 4 x) Q^{2} = 0$ ; but $Q$ is an integral domain by [L6] and [L7], and $1 - 4 x$ and $Q$ are nonzero, so [L3] makes the product nonzero.

4.1L3L5step 2.1step 3.1

Comparing degrees in $Q [x]$ gives a contradiction. By [L3] applied twice, $\deg (R^{2}) = 2 \deg R$ and $\deg ((1 - 4 x) Q^{2}) = \deg (1 - 4 x) + 2 \deg Q = 1 + 2 \deg Q$ , the degree of $1 - 4 x$ being $1$ by [L5]. So $2 \deg R = 1 + 2 \deg Q$ in $N$ , which is impossible: writing $r : = \deg R$ and $q : = \deg Q$ , if $r \leq q$ then $2 r \leq 2 q < 1 + 2 q$ , and if $r \geq q + 1$ then $2 r \geq 2 q + 2 > 1 + 2 q$ .

5.1F2L2L7step 1.1step 4.1discharge-contradiction∎

The assumption of step 1.1 is therefore false and $C$ is not rational; and by [L2] with $K = Q$ and $a_{n} = C_{n}$ , a sequence satisfies an eventual constant-coefficient linear recurrence exactly when its generating series is rational, so the sequence $(C_{n})$ satisfies no such recurrence.

Remarks

Why the parity argument is the whole proof. The equation $R^{2} = (1 - 4 x) Q^{2}$ says that $1 - 4 x$ is a square in the fraction field of $Q [x]$ up to squares, and the degree of a square is even while the degree of $1 - 4 x$ times a square is odd. Nothing about the specific coefficients is used, and the same argument rules out rationality for any series satisfying a quadratic equation whose discriminant has odd degree.
What the second clause does and does not say. It says no recurrence with constantly many constant coefficients holds from some index onwards. The Catalan numbers do satisfy the convolution recurrence $C_{n + 1} = \sum_{i} C_{i} C_{n - i}$ , which is not of that form, and they satisfy the two-term recurrence $(n + 2) C_{n + 1} = (4 n + 2) C_{n}$ whose coefficients depend on $n$ ; neither is excluded, and the companion page carries the false statement that conflates them.

Depends on

Used by

FALSE: the Catalan numbers satisfy a constant-coefficient linear recurrence False statement

Dependency tree · two levels

40 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 (standard reference, not scraped)
D. Guichard, An Introduction to Combinatorics and Graph Theory, §3.5 (standard reference, not scraped)

C(x)C(x) is not a rational formal power series, so (Cn)(Cn​) satisfies no eventual constant-coefficient linear recurrence