A Note on Catalan’s Identity for the k-Fibonacci Quaternions

Emrah Polatlı and Seyhun Kesim Department of Mathematics Bülent Ecevit University 67100 Zonguldak Turkey

                    emrahpolatli@gmail.com
                   seyhun.kesim@beun.edu.tr

Abstract Ramírez recently conjectured a version of Catalan’s identity for the k-Fibonacci quaternions. In this note we give a proof of (a suitably reformulated version of) this identity.

1 Introduction

For any positive real number k, define the k-Fibonacci and k-Lucas sequences, (F_{k,n})_{n∈ℕ} and (L_{k,n})_{n∈ℕ}, as follows:

F_{k,0} = 0, F_{k,1} = 1, and F_{k,n} = kF_{k,n−1} + F_{k,n−2}, n ≥ 2,

and

L_{k,0} = 2, L_{k,1} = k, and L_{k,n} = kL_{k,n−1} + L_{k,n−2}, n ≥ 2,

respectively.

Let α and β be the roots of the characteristic equation x^{2} − kx − 1 = 0. Then the Binet formulas for the k-Fibonacci and k-Lucas sequences are

F_{k,n} = \frac{α^{n} − β^{n}}{α − β}

and

L_{k,n} = α^{n} + β^{n},

where α = (k + \sqrt{k^{2} + 4}) /2 and β = (k − \sqrt{k^{2} + 4}) /2.

A quaternion p, with real components a_{0}, a_{1}, a_{2}, a_{3} and basis 1, i, j, k, is an element of the form

p = a_{0} + a_{1}i + a_{2}j + a_{3}k = (a_{0}, a_{1}, a_{2}, a_{3}) , (a_{0}1 = a_{0}) ,

where

i^{2} = j^{2} = k^{2} = −1,
ij = k = −ji, jk = i = −kj, ki = j = −ik.

Ramírez [1] defined the nth k-Fibonacci quaternion, D_{k,n}, as follows:

D_{k,n} = F_{k,n} + F_{k,n+1}i + F_{k,n+2}j + F_{k,n+3}k, n ≥ 0,

where F_{k,n} is the nth k-Fibonacci number. Ramírez [1] also gave the Binet formula for the k-Fibonacci quaternion as follows:

D_{k,n} = \frac{α̂α^{n} − β̂β^{n}}{α − β} ,

where α̂ = 1 + αi + α^{2}j + α^{3}k and β̂ = 1 + βi + β^{2}j + β^{3}k.

Ramírez [1] conjectured that the Catalan identity for the k-Fibonacci quaternions is

D_{k,n−r}D_{k,n+r} − D_{k,n}^{2} = (−1)^{n−r} (2F_{k,r}D_{k,r} − G_{k,r}k) ,

for n ≥ r ≥ 1, where G_{k,r} is the sequence satisfying the following recurrence:

G_{k,0} = 0, G_{k,1} = k^{2} + 2k, and G_{k,n} = (k^{2} + 2) G_{k,n−1} − G_{k,n−2}, n ≥ 2.

However, in this short paper, we show that this conjecture is incorrect, by giving the correct Catalan identity and proving it.

2 Catalan identity for the k-Fibonacci quaternions

We need the following lemma.

Lemma 1. For r ≥ 1, we have

\frac{α̂β̂β^{r} − β̂α̂α^{r}}{α − β} = −2D_{k,r} + L_{k,2}L_{k,r}k.

Proof. Since

α̂β̂ = 2 + 2βi + 2β^{2}j + (α^{3} + β^{3} + α − β) k

and

β̂α̂ = 2 + 2αi + 2α^{2}j + (α^{3} + β^{3} + β − α) k,

we get

\frac{α̂β̂β^{r} − β̂α̂α^{r}}{α − β} = −2F_{k,r} − 2F_{k,r+1}i − 2F_{k,r+2}j + (−2F_{k,r+3} + L_{k,2}L_{k,r}) k
= −2D_{k,r} + L_{k,2}L_{k,r}k.

Theorem 2. For n ≥ r ≥ 1, Catalan identity for the k-Fibonacci quaternions is

D_{k,n−r}D_{k,n+r} − D_{k,n}^{2} = (−1)^{n−r+1} (2F_{k,r}D_{k,r} − L_{k,2}F_{k,2r}k) .

Proof. By considering the Binet formula for the k-Fibonacci quaternions, quaternion multiplication and Lemma 1, we obtain

D_{k,n−r}D_{k,n+r} − D_{k,n}^{2} = (\frac{α̂α^{n−r} − β̂β^{n−r}}{α − β} ) (\frac{α̂α^{n+r} − β̂β^{n+r}}{α − β} ) − (\frac{α̂α^{n} − β̂β^{n}}{α − β} )^{2}

= \begin{matrix}(αβ) & n \\ (α − β) & 2 \end{matrix} (α̂β̂(1 − \frac{β^{r}}{α^{r}} ) + β̂α̂ (1 − \frac{α^{r}}{β^{r}} ))

= (αβ)^{n} \frac{α^{r} − β^{r}}{(α − β)^{2}} (\frac{α̂β̂}{α^{r}} − \frac{β̂α̂}{β^{r}})

= (αβ)^{n−r} \frac{α^{r} − β^{r}}{α − β} (\frac{α̂β̂β^{r} − β̂α̂α^{r}}{α − β} )

= (αβ)^{n−r} F_{k,r} (−2D_{k,r} + L_{k,2}L_{k,r}k)

= (−1)^{n−r+1} (2F_{k,r}D_{k,r} − L_{k,2}F_{k,2r}k) .

References

[1] J. L. Ramírez, Some combinatorial properties of the k -Fibonacci and the k-Lucas quaternions, An. Ştiinţ. Univ. Ovidius Constanţa Ser. Mat. 23 (2015), 201–212.

2010 Mathematics Subject Classification: Primary 11B39. Keywords: k-Fibonacci quaternion, Catalan’s identity.

Received May 18 2015; revised version received May 20 2015; June 18 2015; June 24 2015. Published in Journal of Integer Sequences, July 16 2015.

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