Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-16
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The Lucas generating function and its two-root closed form

Example

With ϕ=(1+5)/2 and ϕ^=(1−5)/2, the Lucas sequence satisfies

∑n≥0Lnxn=2−x1−x−x2

in Q⟦x⟧, and

Ln=ϕn+ϕ^n

in Q(5) for every n≥0.

Facts & Assumptions

Given: The Lucas initial values and recurrence.

[L1]

The Lucas sequence satisfies L0=2, L1=1, and Ln+2=Ln+1+Ln (The Fibonacci sequence F0=0,F1=1 and Lucas sequence L0=2,L1=1).

[L2]

A recurrence beginning at zero has a rational generating function whose numerator is obtained by multiplying by its reciprocal denominator (A coefficient sequence is eventually linearly recurrent if and only if its formal generating function is rational).

[L3]

Distinct roots of a characteristic polynomial give a unique pure-exponential closed form over a characteristic-zero splitting field (Over a named splitting field in characteristic zero, repeated characteristic roots give polynomial-times-exponential closed forms).

Verification

technique · direct
1.1givenL1L2algebra

For L(x)=∑n≥0Lnxn, [L1] gives (1−x−x2)L(x)=L0+(L1−L0)x=2−x, so [L2] proves the generating-function formula.

1.2L3algebra

Since t2−t−1=(t−ϕ)(t−ϕ^), [L3] gives Ln=Aϕn+Bϕ^n.

2.1step 1.2L1algebra

The initial equations A+B=2 and Aϕ+Bϕ^=1 have the solution A=B=1, because ϕ+ϕ^=1.

3.1step 2.1∎

Substitution in step 2.1 proves the displayed closed form for all n≥0.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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