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.

1 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 [1–6]).

2 Quadra Fibona-Pell sequence

In [7], 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 [8], 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) $W_{n}=P_{n}-F_{n}$ for $n\geq0$.

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

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

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

• (5) $2(W_{n+1}-W_{n}+F_{n-1})=\gamma^{n}+\delta^{n}$ for $n\geq1$.

• (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 $W_{2}=[0]_{2\times2}$. 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.