TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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

1.1

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.

givenL1algebra

1.2

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

given

1.3

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 $F = D + R / Q$ .

givenL2algebra

2.1

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.

step 1.3L1

3.1

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.

step 1.1step 2.1algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 34 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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)