Binet's formula for generalized tribonacci numbers

Abstract

In this note, we derive Binet's formula for the general term ${\mathcal{T}}_n$ of the generalized tribonacci sequence. This formula gives ${\mathcal{T}}_n$ explicitly as a function of the index n, the roots of the associated characteristic equation, and the initial terms ${\mathcal{T}}_0$, ${\mathcal{T}}_1$, and ${\mathcal{T}}_2$. By way of illustration, we obtain Binet's formula for the Cordonnier, Perrin, and Van der Laan numbers. In addition, we establish a double identity that can be regarded as a parent of Binet's formula for generalized tribonacci numbers.

Keywords:

• tribonacci numbers
• generalized tribonacci sequence
• characteristic equation
• Binet's formula
• symmetric function

Acknowledgements

The author thanks an anonymous reviewer for detailed instructive suggestions that led to improvements in the presentation of this note.

Disclosure statement

No potential conflict of interest was reported by the author.

Notes

1. Unfortunately, Mark Feinberg died in a motorcycle accident four years later.

2. The closed-form formula for the nth Fibonacci number, $F_n=({\varphi }^n-{\phi }^n)/\sqrt{5}$ (where $\varphi =(1+\sqrt{5})/2$ and $\phi =(1-\sqrt{5})/2$), is usually attributed to the French mathematician Jacques Philippe Marie Binet (1786–1856) who published it in 1843, although it was known in the eighteenth century, to such famous mathematicians as Abraham de Moivre (1667–1754), Daniel Bernoulli (1700–1782), and Leonhard Euler (1707–1783). Amazingly enough, the exponential growth of the Fibonacci numbers played a prominent role in the resolution of Hilbert's 10th problem in 1970 by the combined efforts of Yuri Matiyasevich, Martin Davis, Hilary Putnam, and Julia Robinson.

3. Throughout this note, we will assume that α ≠ β, α ≠ γ, and β ≠ γ. This is tantamount to assuming that Δ, the discriminant, is not zero. That is, Δ² = (α − β)²(α − γ)²(β − γ)² = r²s² − 27t² + 4s³ − 4r³t − 18rst > 0.

4. Cordonnier numbers are more commonly known in the literature as Padovan numbers.

5. The proof of Theorem 2.1 is analogous to that given in [8] for the case r = s = t = 1.

6. From Equations (10a(10a) $$P_n(\alpha ,\beta ,\gamma )=\sum _{i=0}^{n-1}\sum _{j=0}^{n-i}\left({\alpha }^{i+2}{\beta }^{j+1}{\gamma }^{n-i-j}+{\beta }^{i+2}{\alpha }^j{\gamma }^{n+1-i-j}+{\gamma }^{i+2}{\beta }^j{\alpha }^{n+1-i-j}\right),$$(10a) ) and (10b(10b) $$\begin{array}{c}\hfill Q_n(\alpha ,\beta ,\gamma )=\sum _{i=0}^{n-1}\sum _{j=0}^{n-i}\left({\alpha }^{i+2}{\beta }^j{\gamma }^{n+1-i-j}+{\beta }^{i+2}{\alpha }^{j+1}{\gamma }^{n-i-j}+{\gamma }^{i+2}{\beta }^{j+1}{\alpha }^{n-i-j}\right).\end{array}$$(10b) ), it is not difficult to show that P_n(α, β, γ) and Q_n(α, β, γ) are symmetric functions of α, β, and γ. We omit the proof here for the sake of brevity (although the reader might want to prove that, for example, P_n(α, β, γ) = P_n(β, α, γ)). Moreover, from the expressions (10a(10a) $$P_n(\alpha ,\beta ,\gamma )=\sum _{i=0}^{n-1}\sum _{j=0}^{n-i}\left({\alpha }^{i+2}{\beta }^{j+1}{\gamma }^{n-i-j}+{\beta }^{i+2}{\alpha }^j{\gamma }^{n+1-i-j}+{\gamma }^{i+2}{\beta }^j{\alpha }^{n+1-i-j}\right),$$(10a) ) and (10b(10b) $$\begin{array}{c}\hfill Q_n(\alpha ,\beta ,\gamma )=\sum _{i=0}^{n-1}\sum _{j=0}^{n-i}\left({\alpha }^{i+2}{\beta }^j{\gamma }^{n+1-i-j}+{\beta }^{i+2}{\alpha }^{j+1}{\gamma }^{n-i-j}+{\gamma }^{i+2}{\beta }^{j+1}{\alpha }^{n-i-j}\right).\end{array}$$(10b) ), it is readily seen that P_n(β, α, γ) = Q_n(α, β, γ), from which we conclude that P_n(α, β, γ) = Q_n(α, β, γ). Specifically, for n = 1, 2, and 3, we have $$\begin{array}{ccc}\hfill P_1(\alpha ,\beta ,\gamma )& =& Q_1(\alpha ,\beta ,\gamma )\hfill \\ & =& \left(\alpha +\beta +\gamma \right)\left(\alpha \beta \gamma \right)+{\left(\alpha \beta \right)}^2+{\left(\alpha \gamma \right)}^2+{\left(\beta \gamma \right)}^2,\hfill \\ \hfill P_2(\alpha ,\beta ,\gamma )& =& Q_2(\alpha ,\beta ,\gamma )\hfill \\ & =& \left({\alpha }^2+{\beta }^2+{\gamma }^2\right)\left(\alpha \beta \gamma \right)+{\left(\alpha \beta \right)}^2\left(\alpha +\beta +2\gamma \right)\hfill \\ & & +{\left(\alpha \gamma \right)}^2\left(\alpha +2\beta +\gamma \right)+{\left(\beta \gamma \right)}^2\left(2\alpha +\beta +\gamma \right),\hfill \\ \hfill P_3(\alpha ,\beta ,\gamma )& =& Q_3(\alpha ,\beta ,\gamma )\hfill \\ & =& \left({\alpha }^3+{\beta }^3+{\gamma }^3\right)\left(\alpha \beta \gamma \right)+3{\left(\alpha \beta \gamma \right)}^2+{\left(\alpha \beta \right)}^2\left[\left(\alpha +\beta \right)\left(\alpha +\beta +2\gamma \right)-\alpha \beta \right]\hfill \\ & & +{\left(\alpha \gamma \right)}^2\left[\left(\alpha +\gamma \right)\left(\alpha +2\beta +\gamma \right)-\alpha \gamma \right]+{\left(\beta \gamma \right)}^2\left[\left(\beta +\gamma \right)\left(2\alpha +\beta +\gamma \right)-\beta \gamma \right]\hfill \\ & =& \left({\alpha }^2+\alpha \beta +{\beta }^2\right)\left({\alpha }^2+\alpha \gamma +{\gamma }^2\right)\left({\beta }^2+\beta \gamma +{\gamma }^2\right),\hfill \end{array}$$which are clearly symmetric functions of α, β, and γ.