CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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The least eventual recurrence order is the degree of the reduced denominator

Statement

Let $F = \sum_{n \geq 0} a_{n} x^{n} \in K ⟦ x ⟧$ be rational over a field, and let $P / Q$ be a normalised reduced presentation. The least order of an eventual constant-coefficient recurrence satisfied by $(a_{n})$ is $\deg Q$ , with the convention that $\deg 1 = 0$ . Thus polynomial series have minimal eventual order zero.

Facts & Assumptions

Given: A rational series $F = P / Q$ over a field $K$ , where $Q (0) = 1$ and $P, Q$ are coprime.

[L1]

A sequence is eventually linearly recurrent exactly when its generating function is rational, and a denominator $D$ of degree $e$ supplies an eventual recurrence of order $e$ (A coefficient sequence is eventually linearly recurrent if and only if its formal generating function is rational).

[L2]

The polynomial ring over a field is a unique factorisation domain (For every field $F$ , $F [x]$ is a unique factorisation domain).

Proof

technique · direct

1.1

If $(a_{n})$ satisfies an eventual recurrence of order $e$ , [L1] gives a polynomial $A$ and a normalised denominator $D$ of degree $e$ with $F = A / D$ , so $P D = A Q$ .

givenL1

2.1

Since $P$ and $Q$ are coprime in the UFD $K [x]$ , the identity $P D = A Q$ forces $Q ∣ D$ ; therefore $\deg Q \leq e$ .

step 1.1L2algebra

3.1

If $\deg Q > 0$ , the presentation $P / Q$ itself gives by [L1] an eventual recurrence of order $\deg Q$ , so step 2.1 proves minimality. If $Q = 1$ , then $F = P$ is a polynomial and its coefficients are eventually zero, giving minimal order zero.

step 2.1L1∎

Depends on

Used by

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Sources

R. P. Stanley, Enumerative Combinatorics, vol. 1, 2nd ed., Corollary 4.2.1 (standard reference, not scraped)
M. Waldschmidt, Linear Recurrence Sequences VI, Order of a linear recurrence sequence (standard reference, not scraped)