[Paper Review] A simplified Binet formula for k-generalized Fibonacci numbers

Gregory P. Dresden|arXiv (Cornell University)|May 4, 2009

Advanced Combinatorial MathematicsMathematics5 references20 citations

TL;DR

This paper presents a simplified Binet-type formula for k-generalized Fibonacci numbers (e.g., Tribonacci, Tetranacci) that expresses each term as the nearest integer to the first term of the formula, using only the dominant real root of the characteristic equation. The key contribution is proving that rounding the leading term of the formula yields exact k-generalized Fibonacci numbers for all n ≥ 2−k, offering a computationally efficient and exact closed-form expression.

ABSTRACT

We present a particularly nice Binet-style formula that can be used to produce the k-generalized Fibonacci numbers (that is, the Tribonaccis, Tetranaccis, etc). Furthermore, we show that in fact one needs only take the integer closest to the first term of this Binet-style formula to generate the desired sequence. These results were also found (independently) at about the same time by Zhaohui Du of Singapore, China. We are working on a joint paper.

Motivation & Objective

• To derive a simplified, closed-form expression for k-generalized Fibonacci numbers (e.g., Tribonacci, Tetranacci) using a Binet-style formula.
• To show that the integer closest to the first term of this formula exactly generates the k-generalized Fibonacci sequence.
• To establish the validity of this rounding method across all n ≥ 2−k, including negative indices.
• To provide a unified and analytically tractable representation that simplifies prior formulas and enables error analysis.

Proposed method

• Derive a Binet-style formula for Fₙ⁽ᵏ⁾ using the roots α₁,…,αₖ of the characteristic polynomial xᵏ − xᵏ⁻¹ − ⋯ − 1 = 0.
• Express Fₙ⁽ᵏ⁾ as a sum over these roots: Fₙ⁽ᵏ⁾ = Σᵢ₌₁ᵏ [ (αᵢ − 1) / (2 + (k+1)(αᵢ − 2)) ] αᵢⁿ⁻¹.
• Identify the dominant real root α > 1 and show that the contributions from other roots decay exponentially.
• Define the error term Eₙ = Fₙ⁽ᵏ⁾ − m⁽ᵏ⁾(α)αⁿ⁻¹, where m⁽ᵏ⁾(α) is the coefficient of the dominant term.
• Prove that |Eₙ| < 1/2 for all n ≥ 2−k by induction and recurrence properties, ensuring rounding yields the correct integer.
• Use the recurrence of Eₙ to show that if |Eₙ| ≥ 1/2 for some n ≥ 2, a contradiction arises due to unbounded growth, contradicting asymptotic decay.

Experimental results

Research questions

• RQ1 Can a simplified Binet-type formula be derived for k-generalized Fibonacci numbers that avoids complex root contributions?
• RQ2 Is the integer closest to the dominant term of the Binet formula exactly equal to Fₙ⁽ᵏ⁾ for all n ≥ 2−k?
• RQ3 What is the behavior of the error term Eₙ = Fₙ⁽ᵏ⁾ − m⁽ᵏ⁾(α)αⁿ⁻¹, and does it remain bounded within 1/2 for all relevant n?
• RQ4 Why does the rounding property fail for some sequences (e.g., scaled Fibonacci), and what distinguishes k-generalized Fibonacci sequences that allow this?
• RQ5 How does the coefficient m⁽ᵏ⁾(α) = (α−1)/(2+(k+1)(α−2)) relate to alternative known formulas, such as Spickerman’s for Tribonacci numbers?

Key findings

• The paper proves that Fₙ⁽ᵏ⁾ = Round[ m⁽ᵏ⁾(α)αⁿ⁻¹ ] for all n ≥ 2−k, where α is the unique positive real root of xᵏ − xᵏ⁻¹ − ⋯ − 1 = 0.
• The error term Eₙ = Fₙ⁽ᵏ⁾ − m⁽ᵏ⁾(α)αⁿ⁻¹ satisfies the same k-th order linear recurrence as Fₙ⁽ᵏ⁾, enabling inductive error control.
• For n = 0, the error |E₀| = |m⁽ᵏ⁾(α)α⁻¹| < 1/2 holds for all k ≥ 2, with explicit bounds derived using α > 2 − 1/(3k) and m⁽ᵏ⁾(α) < (3k−1)/(5k−1).
• For n = 1, |E₁| < 1/2 because m⁽ᵏ⁾(α) ∈ (1/2, 1), and F₁⁽ᵏ⁾ = 1.
• The assumption that |Eₙ| ≥ 1/2 for some n ≥ 2 leads to a contradiction via unbounded growth of |Eₙ|, violating the asymptotic decay of the error.
• The formula is equivalent to Spickerman’s earlier result for k=3 (Tribonacci), confirming consistency and generalizing it to all k ≥ 2.