TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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A coefficient sequence is eventually linearly recurrent if and only if its formal generating function is rational

Statement

Let $K$ be a field, let $a = (a_{n})_{n \geq 0}$ be a sequence in $K$ , and let $F (x) = \sum_{n \geq 0} a_{n} x^{n}$ . Then $a$ satisfies an eventual constant-coefficient linear recurrence if and only if $F$ is a rational formal power series.

More precisely, let $Q (x) = 1 + c_{1} x + \dots + c_{d} x^{d}$ with $d \geq 1$ and $c_{d} \neq 0$ . The sequence satisfies the corresponding recurrence from index $N$ exactly when $Q F$ has no nonzero coefficient of degree at least $N + d$ . In particular, the recurrence starts at zero exactly when $Q F$ is zero or has degree below $d$ .

Facts & Assumptions

Given: A field $K$ , a sequence $a = (a_{n})_{n \geq 0}$ , and its formal generating function $F = \sum_{n \geq 0} a_{n} x^{n}$ .

[L1]

For fixed $Q (x) = 1 + c_{1} x + \dots + c_{d} x^{d}$ with $d \geq 1$ and $c_{d} \neq 0$ , multiplication by $Q$ identifies sequences recurrent from zero with numerators of degree below $d$ (The initial-value, recurrence-sequence, numerator and fixed-denominator rational-series spaces all have dimension $d$ ).

[L2]

If $0 \neq Q \in K [x]$ and $P \in K [x]$ , there are unique $D, R \in K [x]$ with $P = D Q + R$ and either $R = 0$ or $\deg R < \deg Q$ (Division algorithm for polynomials over a field).

Proof

technique · direct

1.1givenL1algebra

Suppose first that $a$ satisfies an order- $d$ recurrence from index $N$ . For $m \geq N + d$ , coefficient extraction gives $[x^{m}] (Q F) = a_{m} + c_{1} a_{m - 1} + \dots + c_{d} a_{m - d} = 0$ , so $Q F$ is a polynomial and $F = (Q F) / Q$ is rational.

1.2given

An eventual order-zero recurrence means that $a$ is eventually zero, so $F$ is a polynomial and is rational with denominator $1$ .

1.3givenL2algebra

Conversely, suppose $F = P / Q$ with $Q (0) \neq 0$ . Rescale so that $Q (0) = 1$ , and use [L2] to write $P = D Q + R$ with $R = 0$ or $\deg R < \deg Q$ ; then $F = D + R / Q$ .

2.1step 1.3L1

If $\deg Q = d \geq 1$ , [L1] says that the coefficients of $R / Q$ satisfy the order- $d$ recurrence from zero, while the polynomial $D$ changes only finitely many coefficients; hence the coefficients of $F$ satisfy that recurrence eventually. If $Q = 1$ , then $F = P$ is eventually zero and has eventual order zero.

3.1step 1.1step 2.1algebra∎

The coefficient calculation in step 1.1 is reversible: for fixed positive-degree $Q$ , the recurrence holds at $n$ exactly when $[x^{n + d}] (Q F) = 0$ . This gives the stated starting-index clause and completes both directions.

Depends on

Used by

Dependency tree · two levels

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Sources

R. P. Stanley, Enumerative Combinatorics, vol. 1, 2nd ed., Theorem 4.1.1 and Proposition 4.2.2 (standard reference, not scraped)
B. E. Sagan, Combinatorics: The Art of Counting, Theorem 3.7.1 (standard reference, not scraped)