Theory and Modern Applications

# Some algebraic identities on quadra Fibona-Pell integer sequence

## Abstract

In this work, we define a quadra Fibona-Pell integer sequence $$W_{n}=3W_{n-1}-3W_{n-3}-W_{n-4}$$ for $$n\geq4$$ with initial values $$W_{0}=W_{1}=0$$, $$W_{2}=1$$, $$W_{3}=3$$, and we derive some algebraic identities on it including its relationship with Fibonacci and Pell numbers.

## Preliminaries

Let p and q be non-zero integers such that $$D=p^{2}-4q\neq0$$ (to exclude a degenerate case). We set the sequences $$U_{n}$$ and $$V_{n}$$ to be

\begin{aligned} &U_{n}=U_{n}(p,q)=pU_{n-1}-qU_{n-2}, \\ &V_{n}=V_{n}(p,q)=pV_{n-1}-qV_{n-2} \end{aligned}
(1)

for $$n\geq2$$ with initial values $$U_{0}=0$$, $$U_{1}=1$$, $$V_{0}=2$$, and $$V_{1}=p$$. The sequences $$U_{n}$$ and $$V_{n}$$ are called the (first and second) Lucas sequences with parameters p and q. $$V_{n}$$ is also called the companion Lucas sequence with parameters p and q.

The characteristic equation of $$U_{n}$$ and $$V_{n}$$ is $$x^{2}-px+q=0$$ and hence the roots of it are $$x_{1}=\frac{p+\sqrt{D}}{2}$$ and $$x_{2}=\frac{p-\sqrt{D}}{2}$$. So their Binet formulas are

$$U_{n}=\frac{x_{1}^{n}-x_{2}^{n}}{x_{1}-x_{2}} \quad \mbox{and}\quad V_{n}=x_{1}^{n}+x_{2}^{n}$$

for $$n\geq0$$. For the companion matrix $$M=\bigl[ {\scriptsize\begin{matrix} p & -q \cr 1 & 0\end{matrix}} \bigr]$$, one has

$$\left [ \begin{array}{@{}c@{}} U_{n} \\ U_{n-1} \end{array} \right ] =M^{n-1}\left [ \begin{array}{@{}c@{}} 1 \\ 0 \end{array} \right ]\quad \mbox{and} \quad \left [ \begin{array}{@{}c@{}} V_{n} \\ V_{n-1}\end{array} \right ] =M^{n-1}\left [ \begin{array}{@{}c@{}} p \\ 2\end{array} \right ]$$

for $$n\geq1$$. The generating functions of $$U_{n}$$ and $$V_{n}$$ are

$$U(x)=\frac{x}{1-px+qx^{2}}\quad \mbox{and} \quad V(x)=\frac{2-px}{1-px+qx^{2}}.$$
(2)

Fibonacci, Lucas, Pell, and Pell-Lucas numbers can be derived from (1). Indeed for $$p=1$$ and $$q=-1$$, the numbers $$U_{n}=U_{n}(1,-1)$$ are called the Fibonacci numbers (A000045 in OEIS), while the numbers $$V_{n}=V_{n}(1,-1)$$ are called the Lucas numbers (A000032 in OEIS). Similarly, for $$p=2$$ and $$q=-1$$, the numbers $$U_{n}=U_{n}(2,-1)$$ are called the Pell numbers (A000129 in OEIS), while the numbers $$V_{n}=V_{n}(2,-1)$$ are called the Pell-Lucas (A002203 in OEIS) (companion Pell) numbers (for further details see ).

In , the author considered the quadra Pell numbers $$D(n)$$, which are the numbers of the form $$D(n)=D(n-2)+2D(n-3)+D(n-4)$$ for $$n\geq4$$ with initial values $$D(0)=D(1)=D(2)=1$$, $$D(3)=2$$, and the author derived some algebraic relations on it.

In , the authors considered the integer sequence (with four parameters) $$T_{n}=-5T_{n-1}-5T_{n-2}+2T_{n-3}+2T_{n-4}$$ with initial values $$T_{0}=0$$, $$T_{1}=0$$, $$T_{2}=-3$$, $$T_{3}=12$$, and they derived some algebraic relations on it.

In the present paper, we want to define a similar sequence related to Fibonacci and Pell numbers and derive some algebraic relations on it. For this reason, we set the integer sequence $$W_{n}$$ to be

$$W_{n}=3W_{n-1}-3W_{n-3}-W_{n-4}$$
(3)

for $$n\geq4$$ with initial values $$W_{0}=W_{1}=0$$, $$W_{2}=1$$, $$W_{3}=3$$ and call it a quadra Fibona-Pell sequence. Here one may wonder why we choose this equation and call it a quadra Fibona-Pell sequence. Let us explain: We will see below that the roots of the characteristic equation of $$W_{n}$$ are the roots of the characteristic equations of both Fibonacci and Pell sequences. Indeed, the characteristic equation of (3) is $$x^{4}-3x^{3}+3x+1=0$$ and hence the roots of it are

$$\alpha=\frac{1+\sqrt{5}}{2},\qquad \beta=\frac{1-\sqrt{5}}{2},\qquad \gamma =1+ \sqrt{2} \quad \mbox{and}\quad \delta=1-\sqrt{2}.$$
(4)

(Here α, β are the roots of the characteristic equation of Fibonacci numbers and γ, δ are the roots of the characteristic equation of Pell numbers.) Then we can give the following results for $$W_{n}$$.

### Theorem 1

The generating function for $$W_{n}$$ is

$$W(x)=\frac{x^{2}}{x^{4}+3x^{3}-3x+1}.$$

### Proof

The generating function $$W(x)$$ is a function whose formal power series expansion at $$x=0$$ has the form

$$W(x)=\sum_{n=0}^{\infty}W_{n}x^{n}=W_{0}+W_{1}x+W_{2}x^{2}+ \cdots +W_{n}x^{n}+\cdots.$$

Since the characteristic equation of (3) is $$x^{4}-3x^{3}+3x+1=0$$, we get

\begin{aligned} \bigl(1-3x+3x^{3}+x^{4}\bigr)W(x) =&\bigl(1-3x+3x^{3}+x^{4} \bigr) \bigl(W_{0}+W_{1}x+\cdots +W_{n}x^{n}+ \cdots\bigr) \\ =&W_{0}+(W_{1}-3W_{0})x+(W_{2}-3W_{1})x^{2} \\ &{}+(W_{3}-3W_{2}+3W_{0})x^{3}+\cdots \\ &{}+(W_{n}-3W_{n-1}+3W_{n-3}+W_{n-4})x^{n}+ \cdots. \end{aligned}

Notice that $$W_{0}=W_{1}=0$$, $$W_{2}=1$$, $$W_{3}=3$$, and $$W_{n}=3W_{n-1}-3W_{n-3}-W_{n-4}$$. So $$(1-3x+3x^{3}+x^{4})W(x)=x^{2}$$ and hence the result is obvious. □

### Theorem 2

The Binet formula for $$W_{n}$$ is

$$W_{n}= \biggl( \frac{\gamma^{n}-\delta^{n}}{\gamma-\delta } \biggr) - \biggl( \frac{\alpha^{n}-\beta^{n}}{\alpha-\beta } \biggr)$$

for $$n\geq0$$.

### Proof

Note that the generating function is $$W(x)=\frac{x^{2}}{x^{4}+3x^{3}-3x+1}$$. It is easily seen that $$x^{4}+3x^{3}-3x+1=(1-x-x^{2})(1-2x-x^{2})$$. So we can rewrite $$W(x)$$ as

$$W(x)=\frac{x}{1-2x-x^{2}}-\frac{x}{1-x-x^{2}}.$$
(5)

From (2), we see that the generating function for Pell numbers is

$$P(x)=\frac{x}{1-2x-x^{2}}$$
(6)

and the generating function for the Fibonacci numbers is

$$F(x)=\frac{x}{1-x-x^{2}}.$$
(7)

From (5), (6), (7), we get $$W(x)=P(x)-F(x)$$. So $$W_{n}= ( \frac{\gamma^{n}-\delta^{n}}{\gamma-\delta } ) - ( \frac{\alpha^{n}-\beta^{n}}{\alpha-\beta } )$$ as we wanted. □

The relationship with Fibonacci, Lucas, and Pell numbers is given below.

### Theorem 3

For the sequences $$W_{n}$$, $$F_{n}$$, $$L_{n}$$, and $$P_{n}$$, we have:

1. (1)

$$W_{n}=P_{n}-F_{n}$$ for $$n\geq0$$.

2. (2)

$$W_{n+1}+W_{n-1}=(\gamma^{n}+\delta^{n})-(\alpha^{n}+\beta^{n})$$ for $$n\geq1$$.

3. (3)

$$\sqrt{5}F_{n}+2\sqrt{2}P_{n}=(\gamma^{n}-\delta^{n})+(\alpha ^{n}-\beta^{n})$$ for $$n\geq1$$.

4. (4)

$$L_{n}+P_{n+1}+P_{n-1}=\alpha^{n}+\beta^{n}+\gamma^{n}+\delta^{n}$$ for $$n\geq1$$.

5. (5)

$$2(W_{n+1}-W_{n}+F_{n-1})=\gamma^{n}+\delta^{n}$$ for $$n\geq1$$.

6. (6)

$$\lim_{n\rightarrow\infty}\frac{W_{n}}{W_{n-1}}=\gamma$$.

### Proof

(1) It is clear from the above theorem, since $$W(x)=P(x)-F(x)$$.

(2) Since $$6W_{n-1}+W_{n+2}=3W_{n+1}-3W_{n-1}-W_{n-2}+6W_{n-1}$$, we get

\begin{aligned} W_{n+1}+W_{n-1} =&2W_{n-1}+\frac{1}{3}W_{n-2}+ \frac{1}{3}W_{n+2} \\ =&\frac{6}{3} \biggl( \frac{\gamma^{n-1}-\delta^{n-1}}{\gamma-\delta }-\frac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta} \biggr) \\ &{}+\frac{1}{3} \biggl( \frac{\gamma^{n-2}-\delta^{n-2}}{\gamma-\delta }-\frac{\alpha^{n-2}-\beta^{n-2}}{\alpha-\beta} \biggr) \\ &{}+\frac{1}{3} \biggl( \frac{\gamma^{n+2}-\delta^{n+2}}{\gamma-\delta }-\frac{\alpha^{n+2}-\beta^{n+2}}{\alpha-\beta} \biggr) \\ =&\frac{1}{3(\gamma-\delta)} \biggl[ \gamma^{n} \biggl( \frac{6}{\gamma }+\frac{1}{\gamma^{2}}+\gamma^{2} \biggr) +\delta^{n} \biggl( \frac{-6}{\delta }-\frac{1}{\delta^{2}}-\delta^{2} \biggr) \biggr] \\ &{}+\frac{1}{3(\alpha-\beta)} \biggl[ \alpha^{n} \biggl( \frac{-6}{\alpha }-\frac{1}{\alpha^{2}}-\alpha^{2} \biggr) + \beta^{n} \biggl( \frac {6}{\beta}+\frac{1}{\beta^{2}}+ \beta^{2} \biggr) \biggr] \\ =&\bigl(\gamma^{n}+\delta^{n}\bigr)-\bigl( \alpha^{n}+\beta^{n}\bigr), \end{aligned}

since $$\frac{6}{\gamma}+\frac{1}{\gamma^{2}}+\gamma ^{2}=\frac{-6}{\delta}-\frac{1}{\delta^{2}}-\delta ^{2}=6\sqrt{2}$$ and $$\frac{-6}{\alpha}-\frac{1}{\alpha ^{2}}-\alpha^{2}=\frac{6}{\beta}+ \frac{1}{\beta^{2}}+\beta ^{2}=-3\sqrt{5}$$.

(3) Notice that $$F_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}$$ and $$P_{n}=\frac{\gamma^{n}-\delta^{n}}{\gamma-\delta}$$. So we get $$\sqrt{5}F_{n}=\alpha^{n}-\beta^{n}$$ and $$2\sqrt{2} P_{n}=\gamma ^{n}-\delta^{n}$$. Thus clearly, $$\sqrt{5}F_{n}+2\sqrt{2}P_{n}=(\gamma^{n}-\delta^{n})+(\alpha ^{n}-\beta^{n})$$.

(4) It is easily seen that $$P_{n+1}+P_{n-1}=\gamma^{n}+\delta^{n}$$. Also $$L_{n}=\alpha^{n}+\beta^{n}$$. So $$L_{n}+P_{n+1}+P_{n-1}=\alpha ^{n}+\beta^{n}+\gamma^{n}+\delta^{n}$$.

(5) Since $$W_{n+1}=3W_{n}-3W_{n-2}-W_{n-3}$$, we easily get

\begin{aligned} W_{n+1}-W_{n} =&2W_{n}-3W_{n-2}-W_{n-3} \\ =&2 \biggl( \frac{\gamma^{n}-\delta^{n}}{\gamma-\delta }-\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta} \biggr) -3 \biggl( \frac{\gamma^{n-2}-\delta ^{n-2}}{\gamma-\delta}-\frac{\alpha^{n-2}-\beta^{n-2}}{\alpha -\beta} \biggr) \\ &{}- \biggl( \frac{\gamma^{n-3}-\delta^{n-3}}{\gamma-\delta }-\frac{\alpha ^{n-3}-\beta^{n-3}}{\alpha-\beta} \biggr) \\ =&\frac{1}{\gamma-\delta} \biggl[ \gamma^{n} \biggl( 2- \frac{3}{\gamma ^{2}}-\frac{1}{\gamma^{3}} \biggr) +\delta^{n} \biggl( -2+\frac{3}{\delta ^{2}}+\frac{1}{\delta^{3}} \biggr) \biggr] \\ &{}+\frac{1}{\alpha-\beta} \biggl[ \alpha^{n-1} \biggl( 2\alpha- \frac{3}{ \alpha}-\frac{1}{\alpha^{2}} \biggr) -\beta^{n-1} \biggl( 2\beta- \frac {3}{\beta}-\frac{1}{\beta^{2}} \biggr) \biggr] \end{aligned}

and hence

\begin{aligned}& 2W_{n+1}-2W_{n} = \frac{2}{2\sqrt{2}} \biggl[ \gamma^{n} \biggl( \frac{2\gamma^{3}-3\gamma-1}{\gamma^{3}} \biggr) +\delta ^{n} \biggl( \frac{-2\delta ^{2}+3\delta+1}{\delta^{3}} \biggr) \biggr] \\& \hphantom{2W_{n+1}-2W_{n} ={}}{} -\frac{2}{\alpha-\beta} \biggl[ \alpha^{n-1} \biggl( \frac{2\alpha^{3}-3\alpha-1}{\alpha^{2}} \biggr) -\beta ^{n-1} \biggl( \frac{2\beta ^{3}-3\beta-1}{\beta^{2}} \biggr) \biggr] \\& \quad \Leftrightarrow \quad 2W_{n+1}-2W_{n} + \frac{2}{\alpha-\beta} \biggl[ \alpha^{n-1} \biggl( \frac{2\alpha^{3}-3\alpha-1}{\alpha^{2}} \biggr) -\beta ^{n-1} \biggl( \frac{2\beta ^{3}-3\beta-1}{\beta^{2}} \biggr) \biggr] \\& \hphantom{\quad \Leftrightarrow \quad}\quad = \frac{1}{\sqrt{2}} \biggl[ \gamma^{n} \biggl( \frac{2\gamma ^{3}-3\gamma-1}{\gamma^{3}} \biggr) +\delta^{n} \biggl( \frac{-2\delta^{3}+3\delta+1}{\delta^{3}} \biggr) \biggr] \\& \quad \Leftrightarrow\quad 2 ( W_{n+1}-W_{n}+F_{n-1} ) =\gamma ^{n}+\delta^{n}, \end{aligned}

since $$\frac{2\gamma^{3}-3\gamma-1}{\gamma^{3}}=\frac{-2\delta ^{3}+3\delta+1}{\delta^{3}}=\sqrt{2}$$ and $$\frac{2\alpha ^{3}-3\alpha-1}{\alpha^{2}}=\frac{2\beta^{3}-3\beta-1}{\beta ^{2}}=1$$.

(6) It is just an algebraic computation, since $$W_{n}= ( \frac{\gamma^{n}-\delta^{n}}{\gamma-\delta} ) - ( \frac{\alpha^{n}-\beta^{n}}{\alpha-\beta} )$$. □

### Theorem 4

The sum of the first n terms of $$W_{n}$$ is

$$\sum_{i=1}^{n}W_{i}= \frac{W_{n}+4W_{n-1}+4W_{n-2}+W_{n-3}+1}{2}$$
(8)

for $$n\geq3$$.

### Proof

Recall that $$W_{n}=3W_{n-1}-3W_{n-3}-W_{n-4}$$. So

$$W_{n-3}+W_{n-4}=3W_{n-1}-2W_{n-3}-W_{n}.$$
(9)

Applying (9), we deduce that

\begin{aligned} &W_{1}+W_{0} = 3W_{3}-2W_{1}-W_{4} , \\ &W_{2}+W_{1} = 3W_{4}-2W_{2}-W_{5} , \\ &W_{3}+W_{2} = 3W_{5}-2W_{3}-W_{6} , \\ & \ldots , \\ &W_{n-4}+W_{n-5} = 3W_{n-2}-2W_{n-4}-W_{n-1} , \\ &W_{n-3}+W_{n-4} = 3W_{n-1}-2W_{n-3}-W_{n}. \end{aligned}
(10)

If we sum of both sides of (10), then we obtain $$W_{n-3}+W_{0}+2(W_{1}+\cdots+W_{n-4})=3(W_{3}+W_{4}+\cdots +W_{n-1})-2(W_{1}+W_{2}+\cdots+W_{n-3})-(W_{4}+W_{5}+\cdots +W_{n})$$. So we get $$W_{n-3}+2(W_{1}+W_{2}+\cdots +W_{n-4})=1-W_{n-2}-W_{n-1}-W_{n}+3W_{n-2}+3W_{n-1}$$ and hence we get the desired result. □

### Theorem 5

The recurrence relations are

\begin{aligned}& W_{2n} = 9W_{2n-2}-20W_{2n-4}+9W_{2n-6}-W_{2n-8} , \\& W_{2n+1} = 9W_{2n-1}-20W_{2n-3}+9W_{2n-5}-W_{2n-7} \end{aligned}

for $$n\geq4$$.

### Proof

Recall that $$W_{n}=3W_{n-1}-3W_{n-3}-W_{n-4}$$. So $$W_{2n}=3W_{2n-1}-3W_{2n-3}-W_{2n-4}$$ and hence

\begin{aligned} W_{2n} =&3W_{2n-1}-3W_{2n-3}-W_{2n-4} \\ =&9W_{2n-2}-9W_{2n-4}-3W_{2n-5}-9W_{2n-4}+9W_{2n-6}+3W_{2n-7} \\ &{}+W_{2n-8}-W_{2n-8}-W_{2n-4} \\ =&-(3W_{2n-5}-3W_{2n-7}-W_{2n-8})+9W_{2n-2}-18W_{2n-4}+9W_{2n-6} \\ &{}-W_{2n-8}-W_{2n-4} \\ =&-W_{2n-4}+9W_{2n-2}-9W_{2n-4}-9W_{2n-4}+9W_{2n-6}-W_{2n-8}-W_{2n-4} \\ =&9W_{2n-2}-20W_{2n-4}+9W_{2n-6}-W_{2n-8}. \end{aligned}

The other assertion can be proved similarly. □

The rank of an integer N is defined to be

$$\rho(N)=\left \{ \begin{array}{l@{\quad}l} p & \mbox{if }p\mbox{ is the smallest prime with }p|N, \\ \infty& \mbox{if }N\mbox{ is prime}. \end{array} \right .$$

Thus we can give the following theorem.

### Theorem 6

The rank of $$W_{n}$$ is

$$\rho(W_{n})=\left \{ \begin{array}{l@{\quad}l} 2 & \textit{if }n=5+6k,6+6k,7+6k, \\ 3 & \textit{if }n=8+12k,9+12k,15+12k,16+12k, \\ 5 & \textit{if }n=14+60k,46+60k \end{array} \right .$$

for an integer $$k\geq0$$.

### Proof

Let $$n=5+6k$$. We prove it by induction on k. Let $$k=0$$. Then we get $$W_{5}=24=2^{3}\cdot3$$. So $$\rho(W_{5})=2$$. Let us assume that the rank of $$W_{n}$$ is 2 for $$n=k-1$$, that is, $$\rho(W_{6k-1})=2$$, so $$W_{5+6(k-1)}=W_{6k-1}=2^{a}\cdot B$$ for some integers $$a\geq1$$ and $$B>0$$. For $$n=k$$, we get

\begin{aligned} W_{6k+5} =&3W_{6k+4}-3W_{6k+2}-W_{6k+1} \\ =&3(3W_{6k+3}-3W_{6k+1}-W_{6k})-3W_{6k+2}-W_{6k+1} \\ =&9W_{6k+3}-9W_{6k+1}-3W_{6k}-3W_{6k+2}-W_{6k+1} \\ =&9(3W_{6k+2}-3W_{6k}-W_{6k-1})-9W_{6k+1}-3W_{6k}-3W_{6k+2}-W_{6k+1} \\ =&27W_{6k+2}-27W_{6k}-9W_{6k-1}-9W_{6k+1}-3W_{6k}-3W_{6k+2}-W_{6k+1} \\ =&24W_{6k+2}-30W_{6k}-10W_{6k+1}-9W_{6k-1} \\ =&24W_{6k+2}-30W_{6k}-10W_{6k+1}-9 \cdot2^{a}B \\ =&2\bigl[12W_{6k+2}-15W_{6k}-5W_{6k+1}-9 \cdot2^{a-1}B\bigr]. \end{aligned}

Therefore $$\rho(W_{5+6k})=2$$. Similarly it can be shown that $$\rho(W_{6+6k})=\rho(W_{7+6k})=2$$.

Now let $$n=8+12k$$. For $$k=0$$, we get $$W_{8}=387=3^{2}\cdot43$$. So $$\rho(W_{8})=3$$. Let us assume that for $$n=k-1$$ the rank of $$W_{n}$$ is 3, that is, $$\rho(W_{8+12(k-1)})=\rho (W_{12k-4})=3^{b}\cdot H$$ for some integers $$b\geq1$$ and $$H>0$$ which is not even integer. For $$n=k$$, we get

\begin{aligned} W_{12k+8} =&3W_{12k+7}-3W_{12k+5}-W_{12k+4} \\ =&3W_{12k+7}-3W_{12k+5}-(3W_{12k+3}-3W_{12k+1}-W_{12k}) \\ =&3W_{12k+7}-3W_{12k+5}-3W_{12k+3}+3W_{12k+1}+W_{12k} \\ =&3W_{12k+7}-3W_{12k+5}-3W_{12k+3}+3W_{12k+1} \\ &{}+(3W_{12k-1}-3W_{12k-3}-W_{12k-4}) \\ =&3W_{12k+7}-3W_{12k+5}-3W_{12k+3}+3W_{12k+1}+3W_{12k-1} \\ &{}-3W_{12k-3}-W_{12k-4} \\ =&3W_{12k+7}-3W_{12k+5}-3W_{12k+3}+3W_{12k+1}+3W_{12k-1} \\ &{}-3W_{12k-3}-3^{b}\cdot H \\ =&3\bigl(W_{12k+7}-W_{12k+5}-W_{12k+3}+W_{12k+1}+W_{12k-1} \\ &{}-W_{12k-3}-3^{b-1}\cdot H\bigr). \end{aligned}

So $$\rho(W_{12k+8})=3$$. The others can be proved similarly. □

### Remark 1

Apart from the above theorem, we see that $$\rho(W_{22})=\rho (W_{26})=\infty$$, while $$\rho(W_{70})=\rho(W_{98})=13$$ and $$\rho (W_{10})=\rho(W_{34})=\rho(W_{50})=23$$. But there is no general formula.

The companion matrix for $$W_{n}$$ is

$$M=\left [ \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} 3 & 0 & -3 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\end{array} \right ].$$

Set

$$N=\left [ \begin{array}{@{}c@{}} 1 \\ 0 \\ 0 \\ 0 \end{array} \right ]$$

and

$$R=[ \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} 3 & 1 & 0 & 0\end{array} ] .$$

Then we can give the following theorem, which can be proved by induction on n.

### Theorem 7

For the sequence $$W_{n}$$, we have:

(1) $$RM^{n}N=W_{n+3}+P_{n}+2(W_{n+1}-F_{n})$$ for $$n\geq1$$.

(2) $$R(M^{T})^{n-3}N=W_{n}$$ for $$n\geq3$$.

(3) If $$n\geq7$$ is odd, then

$$M^{n}=\left [ \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} m_{11} & m_{12} & m_{13} & m_{14} \\ m_{21} & m_{22} & m_{23} & m_{24} \\ m_{31} & m_{32} & m_{33} & m_{34} \\ m_{41} & m_{42} & m_{43} & m_{44}\end{array} \right ] ,$$

where

\begin{aligned}& m_{11} =W_{n+2},\qquad m_{21}=W_{n+1}, \qquad m_{31}=W_{n},\qquad m_{41}=W_{n-1} , \\& m_{14} =-W_{n+1},\qquad m_{24}=-W_{n}, \qquad m_{34}=-W_{n-1},\qquad m_{44}=-W_{n-2}, \\& m_{12} =-1-W_{n+1}-2\sum_{i=0}^{\frac {n-5}{2}}W_{n-1-2i}, \qquad m_{13}=-W_{n+2}-2\sum _{i=0}^{\frac{n-3}{2}}W_{n-2i} , \\& m_{22} =-W_{n}-2\sum_{i=0}^{\frac{n-5}{2}}W_{n-2-2i}, \qquad m_{23}=-1-W_{n+1}-2\sum _{i=0}^{\frac{n-5}{2}}W_{n-1-2i}, \\& m_{32} =-1-W_{n-1}-2\sum_{i=0}^{\frac{n-7}{2}}W_{n-3-2i}, \qquad m_{33}=-W_{n}-2\sum _{i=0}^{\frac{n-5}{2}}W_{n-2-2i}, \\& m_{42} =-W_{n-2}-2\sum_{i=0}^{\frac{n-7}{2}}W_{n-4-2i}, \qquad m_{43}=-1-W_{n-1}-2\sum _{i=0}^{\frac{n-7}{2}}W_{n-3-2i}, \end{aligned}

and if $$n\geq8$$ is even, then

$$M^{n}=\left [ \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} m_{11} & m_{12} & m_{13} & m_{14} \\ m_{21} & m_{22} & m_{23} & m_{24} \\ m_{31} & m_{32} & m_{33} & m_{34} \\ m_{41} & m_{42} & m_{43} & m_{44}\end{array} \right ] ,$$

where

\begin{aligned}& m_{11} =W_{n+2},\qquad m_{21}=W_{n+1}, \qquad m_{31}=W_{n},\qquad m_{41}=W_{n-1}, \\& m_{14} =-W_{n+1},\qquad m_{24}=-W_{n}, \qquad m_{34}=-W_{n-1},\qquad m_{44}=-W_{n-2}, \\& m_{12} =-W_{n+1}-2\sum_{i=0}^{\frac {n-4}{2}}W_{n-1-2i}, \qquad m_{13}=-1-W_{n+2}-2\sum _{i=0}^{\frac{n-4}{2}}W_{n-2i}, \\& m_{22} =-1-W_{n}-2\sum_{i=0}^{\frac{n-6}{2}}W_{n-2-2i}, \qquad m_{23}=-W_{n+1}-2\sum _{i=0}^{\frac{n-4}{2}}W_{n-1-2i}, \\& m_{32} =-W_{n-1}-2\sum_{i=0}^{\frac{n-6}{2}}W_{n-3-2i}, \qquad m_{33}=-1-W_{n}-2\sum _{i=0}^{\frac{n-6}{2}}W_{n-2-2i}, \\& m_{42} =-1-W_{n-2}-2\sum_{i=0}^{\frac{n-8}{2}}W_{n-4-2i}, \qquad m_{43}=-W_{n-1}-2\sum _{i=0}^{\frac{n-6}{2}}W_{n-3-2i}. \end{aligned}

A circulant matrix is a matrix $$A=[a_{ij}]_{n\times n}$$ defined to be

$$A=\left [ \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} a_{0} & a_{1} & a_{2} & \cdots& a_{n-1} \\ a_{n-1} & a_{0} & a_{1} & \cdots& a_{n-2} \\ a_{n-2} & a_{n-1} & a_{0} & \cdots& a_{n-3} \\ \cdot&\cdot&\cdot&\cdots&\cdot \\ \cdot&\cdot&\cdot&\cdots&\cdot \\ a_{1} & a_{2} & a_{3} & \cdots& a_{0}\end{array} \right ] ,$$

where $$a_{i}$$ are constants. The eigenvalues of A are

$$\lambda_{j}(A)=\sum_{k=0}^{n-1}a_{k}w^{-jk},$$
(11)

where $$w=e^{\frac{2\pi i}{n}}$$, $$i=\sqrt{-1}$$, and $$j=0,1,\ldots ,n-1$$. The spectral norm for a matrix $$B=[b_{ij}]_{n\times m}$$ is defined to be $$\|B\|_{\mathrm{spec}}=\max\{\sqrt{\lambda_{i}}\}$$, where $$\lambda_{i}$$ are the eigenvalues of $$B^{H}B$$ for $$0\leq j\leq n-1$$ and $$B^{H}$$ denotes the conjugate transpose of B.

For the circulant matrix

$$W=W(W_{n})=\left [ \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} W_{0} & W_{1} & W_{2} & \cdots& W_{n-1} \\ W_{n-1} & W_{0} & W_{1} & \cdots& W_{n-2} \\ W_{n-2} & W_{n-1} & W_{0} & \cdots& W_{n-3} \\ \cdot&\cdot&\cdot&\cdots&\cdot \\ \cdot&\cdot&\cdot&\cdots&\cdot \\ W_{1} & W_{2} & W_{3} & \cdots& W_{0}\end{array} \right ]$$

for $$W_{n}$$, we can give the following theorem.

### Theorem 8

The eigenvalues of W are

$$\lambda_{j}(W)=\frac{\left \{ \begin{array}{@{}l@{}} W_{n-1}w^{-3j}+(W_{n}+P_{n-1}-2F_{n-1}+1)w^{-2j} \\ \quad {}+(P_{n}-2F_{n}-W_{n-1})w^{-j}-W_{n}\end{array} \right \} } {w^{-4j}+3w^{-3j}-3w^{-j}+1}$$

for $$j=0,1,2,\ldots,n-1$$.

### Proof

Applying (11) we easily get

\begin{aligned} \lambda_{j}(W) =&\sum_{k=0}^{n-1}W_{k}w^{-jk} =\sum_{k=0}^{n-1} \biggl( \frac{\gamma^{k}-\delta^{k}}{\gamma-\delta }-\frac{\alpha^{k}-\beta^{k}}{\alpha-\beta} \biggr) w^{-jk} \\ =&\frac{1}{\gamma-\delta} \biggl[ \frac{\gamma^{n}-1}{\gamma w^{-j}-1}-\frac{\delta^{n}-1}{\delta w^{-j}-1} \biggr] -\frac{1}{\alpha -\beta} \biggl[ \frac{\alpha^{n}-1}{\alpha w^{-j}-1}-\frac{\beta^{n}-1}{\beta w^{-j}-1} \biggr] \\ =&\frac{1}{\gamma-\delta} \biggl[ \frac{(\gamma^{n}-1)(\delta w^{-j}-1)-(\delta^{n}-1)(\gamma w^{-j}-1)}{(\gamma w^{-j}-1)(\delta w^{-j}-1)} \biggr] \\ &{}-\frac{1}{\alpha-\beta} \biggl[ \frac{(\alpha^{n}-1)(\beta w^{-j}-1)-(\beta^{n}-1)(\alpha w^{-j}-1)}{(\alpha w^{-j}-1)(\beta w^{-j}-1)} \biggr] \\ =&\frac{1}{\gamma-\delta} \biggl[ \frac{w^{-j}(\gamma^{n}\delta -\delta^{n}\gamma+\gamma-\delta)+\delta^{n}-\gamma ^{n}}{\delta\gamma w^{-2j}-w^{-j}(\delta+\gamma)+1} \biggr] \\ &{}-\frac{1}{\alpha-\beta} \biggl[ \frac{w^{-j}(\alpha^{n}\beta -\beta^{n}\alpha+\alpha-\beta)+\beta^{n}-\alpha^{n}}{\beta \alpha w^{-2j}-w^{-j}(\beta+\alpha)+1} \biggr] \\ =&\frac{\left \{ \begin{array}{@{}l@{}} w^{-3j}[\sqrt{5}(\delta-\gamma+\gamma\delta^{n}-\delta\gamma ^{n})+2\sqrt{2}(\alpha-\beta+\alpha^{n}\beta-\alpha\beta^{n})] \\ \quad {}+w^{-2j}[\sqrt{5}(\gamma^{n}-\delta^{n}+\delta-\gamma+\gamma \delta ^{n}-\gamma^{n}\delta)+2\sqrt{2}(\beta^{n}-\alpha^{n}) \\ \quad {}+4\sqrt{2}(\alpha-\beta+\alpha^{n}\beta-\alpha\beta ^{n})]+w^{-j}[\sqrt{5}(\gamma^{n}-\delta^{n}+\gamma-\delta \\ \quad {}+\gamma^{n}\delta-\gamma\delta^{n})+2\sqrt{2}(\beta-\alpha +\beta^{n}\alpha-\alpha^{n}\beta)+4\sqrt{2}(\beta^{n}-\alpha^{n})] \\ \quad {}+[\sqrt{5}(\delta^{n}-\gamma^{n})+2\sqrt{2}(\alpha^{n}-\beta^{n})]\end{array} \right \} } {2\sqrt{10}(w^{-4j}+3w^{-3j}-3w^{-j}+1)} \\ =&\frac{\left \{ \begin{array}{@{}l@{}} W_{n-1}w^{-3j}+(W_{n}+P_{n-1}-2F_{n-1}+1)w^{-2j} \\ \quad {}+(P_{n}-2F_{n}-W_{n-1})w^{-j}-W_{n}\end{array} \right \} } {w^{-4j}+3w^{-3j}-3w^{-j}+1}, \end{aligned}

since $$\alpha\beta=-1$$, $$\gamma\delta=-1$$, $$\alpha+\beta=1$$, $$\alpha-\beta=\sqrt{5}$$, $$\gamma+\delta=2$$, and $$\gamma-\delta =2\sqrt{2}$$. □

After all, we consider the spectral norm of W. Let $$n=2$$. Then $$W_{2}=_{2\times2}$$. So $$\|W_{2}\|_{\mathrm{spec}}=0$$. Similarly for $$n=3$$, we get

$$W_{3}=\left [ \begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array} \right ]$$

and hence $$W_{3}^{H}W_{3}=I_{3}$$. So $$\|W_{3}\|_{\mathrm{spec}}=1$$. For $$n\geq4$$, the spectral norm of $$W_{n}$$ is given by the following theorem, which can be proved by induction on n.

### Theorem 9

The spectral norm of $$W_{n}$$ is

$$\|W_{n}\|_{\mathrm{spec}}=\frac{W_{n-1}+4W_{n-2}+4W_{n-3}+W_{n-4}+1}{2}$$

for $$n\geq4$$.

For example, let $$n=6$$. Then the eigenvalues of $$W_{6}^{H}W_{6}$$ are

$$\lambda_{0}=1\text{,}369,\qquad \lambda_{1}=289, \qquad \lambda_{2}=\lambda _{4}=784 \quad \mbox{and} \quad \lambda_{3}=\lambda_{5}=388.$$

So the spectral norm is $$\|W_{6}\|_{\mathrm{spec}}=\sqrt{\lambda_{0}}=37$$. Also $$\frac{W_{5}+4W_{4}+4W_{3}+W_{2}+1}{2}=37$$. Consequently,

$$\|W_{6}\|_{\mathrm{spec}}=\frac{W_{5}+4W_{4}+4W_{3}+W_{2}+1}{2}=37$$

as we claimed.

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## Acknowledgements

The author wishes to thank Professor Ahmet Tekcan of Uludag University for constructive suggestions.

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Correspondence to Arzu Özkoç. 