The sequences of Fibonacci and Lucas for each real quadratic fields $\mathbb{Q}(\sqrt{d}\ )$

TL;DR

This paper generalizes Fibonacci and Lucas sequences to real quadratic fields using fundamental units, deriving their properties, generating functions, and identities, extending classical results to these algebraic structures.

Contribution

It introduces a construction of Fibonacci and Lucas sequences in quadratic fields using fundamental units, generalizing classical sequences and their properties.

Findings

Sequences constructed for quadratic fields with properties similar to classical Fibonacci and Lucas sequences.

Derived generating functions, Golden ratio, and Binet's formulas for these sequences.

Identified conditions under which these sequences relate to k-Fibonacci sequences.

Abstract

We construct the sequences of Fibonacci and Lucas at any quadratic field $\mathbb{Q}(\sqrt{d}\ )$ with $d>0$ square free, noting in general that the properties remain valid as those given by the classical sequences of Fibonacci and Lucas for the case $d = 5$, under the respective variants. For this construction, we use the fundamental unit of $\mathbb{Q}(\sqrt{d}\ )$ and then we observe the generalizations for any unit of $\mathbb{Q}(\sqrt{d}\ )$ where, under certain conditions, some of this constructions correspond to $k$-Fibonacci sequence for some $k\in \mathbb{N}$. Of course, for both sequences, we obtain the generating function, Golden ratio, Binet's formula and some identities that they keep.

Tables

Table 1. Table 1. Sequence of Fibonacci and Lucas of degree d 𝑑 d .

$d$	$\epsilon$	$\Delta$	Terms of the sequence	Terms of the sequence
			${\left\{F_{\epsilon ,n}\right\}}_{n\in \mathbb{N}}$	${\left\{L_{\epsilon ,n}\right\}}_{n\in \mathbb{N}}$
$2$	$1+\sqrt{2}$	$-1$	$1,2,5,12,29,70,169,408\dots$	$1,3,7,17,41,99,\dots$
			OEIS sequence A$000129$	OEIS sequence A$001333$
$3$	$2+\sqrt{3}$	$1$	$1,4,15,56,209,780,\dots$	$1,\frac{7}{2},13,\frac{97}{2},181,\frac{1351}{2},\dots$
			OEIS sequence A$001353$	OEIS sequence A$001075$
$5$	$\frac{1+\sqrt{5}}{2}$	$-1$	$1,1,2,3,5,8,\dots$	$1,3,4,7,11,18,\dots$
			OEIS sequence A$000045$	OEIS sequence A$000032$
$6$	$5+2\sqrt{6}$	$1$	$1,10,99,980,9701,96030,\dots$	$1,\frac{49}{5},97,\frac{4801}{5},9505,\frac{470449}{5},\dots$
			OEIS sequence A$004189$	OEIS sequence A$001079$
$7$	$8+3\sqrt{7}$	$1$	$1,16,255,4064,64769,\dots$	$1,\frac{127}{8},253,\frac{32257}{8},64261,\dots$
			OEIS sequence A$077412$	OEIS sequence A$001081$
$10$	$3+1\sqrt{10}$	$-1$	$1,6,37,228,1405,8658,\dots$	$1,\frac{19}{3},39,\frac{721}{3},1481,\dots$
			OEIS sequence A$005668$	OEIS sequence A$005667$
$11$	$10+3\sqrt{11}$	$1$	$1,20,399,7960,158801,\dots$	$1,\frac{199}{10},397,\frac{79201}{10},158005,\dots$
			OEIS sequence A$075843$	OEIS sequence A$001085$
$13$	$\frac{3+\sqrt{13}}{2}$	$-1$	$1,3,10,33,109,360,\dots$	$1,\frac{11}{3},12,\frac{119}{3},131,\frac{1298}{3},\dots$
			OEIS sequence A$006190$	OEIS sequence A$006497$

Table 2. Table 2. Sequence of Fibonacci of degree d 𝑑 d .

$d$	$\epsilon$	$\Delta$	Terms of the sequence	OEIS
			${\left\{F_{\epsilon ,n}\right\}}_{n\in \mathbb{N}}$	integer sequence
$37$	$6+\sqrt{37}$	$-1$	$1,12,145,1752,21169,255780,\dots$	A$041061$
$38$	$37+6\sqrt{38}$	$1$	$1,74,5475,405076,29970149,2217385950,\dots$
$39$	$25+4\sqrt{39}$	$1$	$1,50,2499,124900,6242501,312000150,\dots$
$41$	$32+5\sqrt{41}$	$-1$	$1,64,4097,262272,16789505,1074790592,\dots$
$42$	$13+2\sqrt{42}$	$1$	$1,26,675,17524,454949,11811150,\dots$	A$097309$

Equations

$$G_{\lambda}=\left\{A\in GL_{2}(\mathbb{Q})\ \left|\,A=\left[\begin{array}[]{cc}a&b\lambda\\ b&a\end{array}\right]\right\}\right.\ ,\ \ \ \ L_{\lambda}=\{A\in G_{\lambda}\ |,\det(A)=\pm 1\}\ \ \ {\rm and}$$

$$T_{d}=\left\{A\in\mathfrak{M}_{2\times 2}(\mathbb{Z})\ \left|\,A=\left[\begin{array}[]{cc}a&bd\\ b&a\end{array}\right]\right\}\right..$$

$$\left[\begin{array}[]{cc}a&bd\\ b&a\end{array}\right]\longmapsto a+b\sqrt{d}\ .$$

$$a_n$$

$$\displaystyle b_{n}=\left\{\begin{array}[]{lcc}\displaystyle{\sum_{0\leq t\leq\frac{n-2}{2}}\binom{n}{2t+1}a^{2t+1}b^{n-2t-1}d^{\frac{n-2}{2}-t}}&&if\ n\ even\\ &&\\ \displaystyle{\sum_{0\leq t\leq\frac{n-1}{2}}\binom{n}{2t}a^{2t}b^{n-2t}d^{\frac{n-1}{2}-t}}&&if\ n\ odd.\end{array}\right.$$

$$F_{\epsilon ,n}:=\frac{b_n}{b}\left(resp.L_{\epsilon ,n}:=\frac{a_n}{a}\right)(n\in \mathbb{N})$$

$$\displaystyle F_{\varepsilon,n}=\left\{\begin{array}[]{lcc}\displaystyle{\sum_{0\leq t\leq\frac{n-2}{2}}\binom{n}{2t+1}a^{2t+1}b^{n-2t-2}d^{\frac{n-2}{2}-t}}&&if\ n\ even\\ &&\\ \displaystyle{\sum_{0\leq t\leq\frac{n-1}{2}}\binom{n}{2t}a^{2t}b^{n-2t-1}d^{\frac{n-1}{2}-t}}&&if\ n\ odd\end{array}\right.$$

$$\displaystyle L_{\varepsilon,n}=\left\{\begin{array}[]{lcc}\displaystyle{\sum_{0\leq t\leq\frac{n}{2}}\binom{n}{2t}a^{2t-1}b^{n-2t}d^{\frac{n}{2}-t}}&&if\ n\ even\\ &&\\ \displaystyle{\sum_{0\leq t\leq\frac{n-1}{2}}\binom{n}{2t+1}a^{2t}b^{n-2t-1}d^{\frac{n-1}{2}-t}}&&if\ n\ odd\end{array}\right.$$

$$a_n+b_n\sqrt{d}=(a+b\sqrt{d}{)}^n=\sum _{i=0}^n(\genfrac{}{}{0ex}{}{n}{i})a^i{b}^{n-i}(\sqrt{d}{)}^{n-i}=\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)a^{n-i}{b}^i(\sqrt{d}{)}^i.$$

$$(-\Delta )F_n+2a{F}_{n+1}$$

$$F_{n+2}=r{F}_n+s{F}_{n+1}\mathrm{for}\mathrm{all}n\in \mathbb{N}.$$

$$\displaystyle\begin{array}[]{rcl}r+2as&=&4a^{2}-\Delta\\ &&\\ 2ar+(4a^{2}-\Delta)s&=&8a^{3}-4a\Delta\end{array}$$

$$F_{n+2}=F_n+F_{n+1}$$

$$(-\Delta )L_n+2a{L}_{n+1}$$

$$\begin{array}[]{ccl}L_{1}&=&1\\ L_{2}&=&2a-\displaystyle{\frac{\Delta}{a}}\\ L_{3}&=&4a^{2}-3\Delta\\ L_{4}&=&8a^{3}-8a\Delta+\displaystyle{\frac{1}{a}}\\ \vdots&&\quad\quad\quad\vdots\end{array}$$

$$L_{n+2}=r{L}_n+s{L}_{n+1}\mathrm{for}\mathrm{all}n\in \mathbb{N}.$$

$$\displaystyle\begin{array}[]{rcl}r+\left(2a-\displaystyle{\frac{\Delta}{a}}\right)s&=&4a^{2}-3\Delta\\ &&\\ \left(2a-\displaystyle{\frac{\Delta}{a}}\right)r+(4a^{2}-3\Delta)s&=&8a^{3}-8a\Delta+\displaystyle{\frac{1}{a}}\end{array}$$

$$L_{n+2}=L_n+L_{n+1}$$

$$f(x)$$

$$g(x)$$

$$\frac{2ax-\Delta y}{x}=\frac{x}{y}.$$

$$2a-\frac{\Delta }{\phi }$$

$${\phi }^2-2a\phi +\Delta =0.$$

$${\epsilon }^2=2a\epsilon -\Delta .$$

$${\epsilon }^n=F_n\epsilon -F_{n-1}\Delta .$$

$${\epsilon }^{n+1}$$

$$(\overline{\epsilon }{)}^n=F_n\overline{\epsilon }-F_{n-1}\Delta ,$$

$$F_n=\frac{{\epsilon }^n-(\overline{\epsilon }{)}^n}{\epsilon -\overline{\epsilon }}.$$

$$L_n=\frac{{\epsilon }^n+(\overline{\epsilon }{)}^n}{\epsilon +\overline{\epsilon }}.$$

Full text

1. Introduction

The Fibonacci sequence was introduced by Leonardo of Pisa in 1202 in his book Liber Abaci (Book of Calculation) [23]. Many of the properties of the Fibonacci sequence were obtained by F. Édouard Lucas who appoints such sequence by “Fibonacci” [21, Section $3{.}1{.}2$]. For more information about the history of the Fibonacci numbers, we can see [20]. But also, Lucas is who initiates the generalizations and their variants that have emerged from the Fibonacci sequence, as we can observe, for example, in [4], [7], [24], [25] and [26]. Vera W. de Spinadel introduced the Metallic Means family whose members of such a family have many wonderful and amazing properties, and applications to almost every areas of sciences and arts, such as in some areas of the physical, biology, astronomy and music (see [8], [9], [10] and [15]). On the other hand, Sergio Falcón and Ángel Plaza give properties of $k$-Fibonacci sequence in [4], [5], [6] and [7], and these are a particular case and general of metallic means families. Also in [3] M. El-Mikkawy and T. Sogabe given a new family of $k$-Fibonacci numbers. In [16], we can find hundreds of known identities, and Azarian presents in [1] some known identities as binomial sums for quick numerical calculations.

In this paper, we associate with each real quadratic field $\mathbb{Q}(\sqrt{d}\ )$, with $d>0$ square free, its owns sequences of Fibonacci and Lucas (Definition 5), which correspond to certain metallic means families (Theorem 8 and 13). These sequences of Fibonacci and Lucas are determined by their generating functions (Theorem 19) satisfying each Binet’s formula (Theorem 22 and Corollary 23). This means that each real quadratic field $\mathbb{Q}(\sqrt{d}\ )$ will have also associated its own Golden ratio (Definition 20), characteristic equation (5.2) and its Golden ratio will be the fundamental unit (Theorem 18). Finally, we will establish for each $k\in\mathbb{N}$, the $k$-Fibonacci sequence corresponds to Fibonacci sequence of the real quadratic field $\mathbb{Q}(\sqrt{d}\ )$ for a unique $d>0$ square free (Theorem 28).

At the time of submission, there is no description of the infinite family of sequences in the Online Encyclopedia of Integer Sequences, though some of the sequences do appear there, as indicated in Table 1 and Table 2.

This paper is organized as follows. In Section $2$ we collect results of quadratic fields necessary for the development of the work. In Section $3$ we construct the sequences of Fibonacci and Lucas at any real quadratic field. Also we proof that the properties remain valid as those given by the classical sequence of Fibonacci and Lucas for $d=5$. In Section $4$ the main goal is proof that Fibonacci and Lucas sequence are determined by the generating functions. In Section $5$ we give Golden ratio associated as the real quadratic field and we obtain Binet’s formula in $\mathbb{Q}(\sqrt{d}\;)$. In Section $6$ we extend our construct of the sequences of Fibonacci and Lucas over all integer number. Finally, in Section $7$ we define the sequence of Fibonacci and Lucas of degree $d$ with respect to an arbitrary unit $\eta$ of $\mathbb{Q}(\sqrt{d}\;)$ and we proof the results of the previous sections are still met.

2. Quadratic fields

In this section we collect fundamental results from quadratic fields. Throughout this paper, $d$ denotes a square free integer, $\delta$ the discriminant of the quadratic field $\mathbb{Q}(\sqrt{d}\ )$, $\mathcal{O}$ the ring of integers of $\mathbb{Q}(\sqrt{d}\ )$, and $\mathcal{O}^{*}$ the multiplicative group of all invertible elements of the ring $\mathcal{O}$. When $d>0$, we say that $\mathbb{Q}(\sqrt{d}\ )$ is a real quadratic field, while if $d<0$ then $\mathbb{Q}(\sqrt{d}\ )$ is called an imaginary quadratic field. The following results are well known.

Theorem 1. Keeping the previous notation.

1. $(i)$

If $d\equiv 1$ mod $4$, then the set $\displaystyle{\left\{\ 1\ ,\ \frac{1+\sqrt{d}}{2}\ \right\}}$ is an integral basis of $\mathbb{Q}(\sqrt{d}\ )$, $\delta=d$, $\mathcal{O}=\mathbb{Z}\displaystyle{\left[\frac{1+\sqrt{d}}{2}\right]}=\mathbb{Z}+\mathbb{Z}\ \displaystyle{\frac{1+\sqrt{d}}{2}}$ and

\displaystyle{\mathcal{O}^{*}=\left\{\ \frac{a+b\sqrt{d}}{2}\ \ \bigg{|}\ a,b\in\mathbb{Z},\ a^{2}-db^{2}=\pm 4\right\}.}** 2. $(ii)$

If $d\equiv 2$ mod $4$ or $d\equiv 3$ mod $4$, then the set $\displaystyle{\left\{\ 1\ ,\ \sqrt{d}\ \right\}}$ is an integral basis of $\mathbb{Q}(\sqrt{d}\ )$,  $\delta=4d$, $\displaystyle{\mathcal{O}=\mathbb{Z}[\sqrt{d}\ ]=\mathbb{Z}+\mathbb{Z}\ \sqrt{d}\ }$ and

\displaystyle{\mathcal{O}^{*}=\left\{\ a+b\sqrt{d}\ \ \big{|}\ a,b\in\mathbb{Z},\ a^{2}-db^{2}=\pm 1\right\}.}** 3. $(iii)$

If $d<0$, then $\mathcal{O}^{*}=\{\ -1,1\ \}$ when $d\neq-1,\ -3$,  $\mathcal{O}^{*}=\langle i\rangle=\{-1,1,i,-i\}$ when $d=-1$ and $\mathcal{O}^{*}=\langle\zeta_{6}\rangle$ if $d=-3$, where $\zeta_{6}$ is a primitive $6$-th root of unity. 4. $(iv)$

If $d>0$, then

1. $(a)$

There exists a unit $\varepsilon>1$ in $\mathcal{O}$ such that $\mathcal{O}^{*}=\langle-1\rangle\times\langle\varepsilon\rangle$. 2. $(b)$

If $u>1$ is a unit of $\mathcal{O}$, then $u=a+b\sqrt{d}$ for some $a>0$, $b>0$ in $\mathbb{Q}$. 3. $(c)$

If $N(\varepsilon)=1$, then $N(u)=1$ for all $u\in\mathcal{O}^{*}$.

Proof. See [13]. $\square$

The unit $\varepsilon$ of $\mathcal{O}$ in the Theorem 1, $(iv)$, is called the fundamental unit of $\mathcal{O}$. Hence the unit $\varepsilon$ of $\mathcal{O}$ completely determines the group $\mathcal{O}^{*}$. For example, we have for $d=5$ ($d\equiv 1$ mod $4$), $\displaystyle{\frac{1+\sqrt{5}}{2}}$ is the fundamental unit of $\mathbb{Q}(\sqrt{5})$. If $d=17$, then $\displaystyle{\frac{8+2\sqrt{17}}{2}=4+\sqrt{17}}$ is a fundamental unit of $\mathbb{Q}(\sqrt{17})$. In general, if $d\equiv 1$ mod $4$, then $\varepsilon=\displaystyle{\frac{a_{0}+b_{0}\sqrt{d}}{2}}$ where $a_{0}$ and $b_{0}$ are either both even or both odd. Of course, if $a_{0}$ and $b_{0}$ are both even, then $\varepsilon\in\mathbb{Z}[\sqrt{d}\ ]$.

On the other hand, we denote by $\mathfrak{M}_{2\times 2}(\mathbb{Z})$ the set of all matrices $2\times 2$ with integer entries. Let $GL_{2}(\mathbb{Q})$ be the multiplicative group of invertible $2\times 2$ matrices with rational entries, which is called the general lineal group of degree $2$ over $\mathbb{Q}$. The subset of all matrices of $GL_{2}(\mathbb{Q})$ with determinant $1$ is a normal subgroup of $GL_{2}(\mathbb{Q})$ called the special lineal group of degree $2$ over $\mathbb{Q}$ and denoted by $SL_{2}(\mathbb{Q})$.

For each $\lambda\in\mathbb{Q}$, let

[TABLE]

[TABLE]

We have the follows results whose proofs can be seen in [17].

Theorem 2. Keeping the previous notation we obtain

1. $(i)$

$T_{d}$* is a commutative subring with identity of $\mathfrak{M}_{2\times 2}(\mathbb{Z})$.* 2. $(ii)$

If $T^{*}_{d}$ is the multiplicative group of units of $T_{d}$, then $T^{*}_{d}=L_{d}\cap\mathfrak{M}_{2\times 2}(\mathbb{Z})$. In particular, $T^{*}_{d}$ is a subgroup of $L_{d}$. 3. $(iii)$

The rings $T_{d}$ and $\mathbb{Z}[\sqrt{d}\,]$ are isomorphic under the correspondence

[TABLE]

In particular, $T_{d}$ is an integral domain. 4. $(iv)$

The isomorphism in $(iii)$ induces an isomorphism between the multiplicative groups $T^{*}_{d}$ and $(\mathbb{Z}[\sqrt{d}\,])^{*}$. 5. $(v)$

$T_{d}/(T_{d}\cap SL_{2}(\mathbb{Q}))\cong\{-1,\ 1\}$. $\square$

Theorem 3. Let $Q_{d}$ be the set of all matrices of the form A=\left[\begin{array}[]{cc}a&bd\\ b&a\end{array}\right] with $a,b\in\mathbb{Q}$.

1. $(i)$

$Q_{d}$* is a field isomorphic $\mathbb{Q}(\sqrt{d}\,)$ under the correspondence \left[\begin{array}[]{cc}a&bd\\ b&a\end{array}\right]\longmapsto a+b\sqrt{d}\ . This is, $Q_{d}$ is the field of quotients of $T_{d}$.* 2. $(ii)$

There exists a monomorphism of the multiplicative group $\mathbb{Q}(\sqrt{d}\,)^{*}$ in the group $GL_{2}(\mathbb{Q})$. 3. $(iii)$

The group $GL_{2}(\mathbb{Q})$ contains the chain of subgroups $Q_{d}^{*}\cap SL_{2}(\mathbb{Q})<L_{m}<G_{d}=Q_{d}^{*}<GL_{2}(\mathbb{Q}).$

$\square$**

Theorem 4. Let A=\left[\begin{array}[]{cc}a&bd\\ b&a\end{array}\right]\in Q_{d} where $a$, $b$ are two rational numbers. Then the powers of $A$, A^{n}=\left[\begin{array}[]{cc}a_{n}&b_{n}d\\ b_{n}&a_{n}\end{array}\right] with $n\in\mathbb{N}$, are given as follows:

[TABLE]

and

[TABLE]

$\square$* *

3. The sequences of Fibonacci and Lucas in

$\mathbb{Q}(\sqrt{d}\ )$

In this section, we construct the sequences of Fibonacci and Lucas at any real quadratic field. We proof that the properties remain valid as those given by the classical sequence of Fibonacci and Lucas for $d=5$. Being $d>0$ a square free integer and $\varepsilon$ the fundamental unit of $\mathbb{Q}(\sqrt{d}\ )$, we will write $\varepsilon=a+b\sqrt{d}$ where $a,b\in\mathbb{Q}$ with its corresponding matrix A_{\varepsilon}=\left[\begin{array}[]{cc}a&bd\\ b&a\end{array}\right]\ and the powers $n$-th of $A_{\varepsilon}$ by A_{\varepsilon}^{n}=\left[\begin{array}[]{cc}a_{n}&b_{n}d\\ b_{n}&a_{n}\end{array}\right]\ where $a_{n}$ and $b_{n}$ are given as in the equations (2.4) and (2.8) of Theorem 4. Also, $\Delta$ will be the determinant of $A_{\varepsilon}$, that is, $\Delta=a^{2}-b^{2}d=N(\varepsilon)=\pm 1$, where $N$ is the norm function of the square field $\mathbb{Q}(\sqrt{d}\ )$.

Keeping the previous notation, we have the follows:

Definition 5. The sequence of Fibonacci $($resp. Lucas$)$ of degree d with respect to the fundamental unit $\varepsilon$ $($or simply the sequence of Fibonacci $($resp. Lucas$)$, if there is no risk of confusion with respect to $d$ and to its fundamental unit $\varepsilon)$ is the sequence $\{F_{\varepsilon,n}\}_{n\in\mathbb{N}}$ $($resp. $\{L_{\varepsilon,n}\}_{n\in\mathbb{N}})$ of positive numbers given as follows:

[TABLE]

*where the sequence $\{b_{n}\}_{n\in\mathbb{N}}$ *$($*resp. $\{a_{n}\}_{n\in\mathbb{N}})$ is given as in the equation $(\ref{eqn-2})$ *$($*resp. $(\ref{eqn-1}))$ of Theorem $4$. *

According to the equation (3.1) of the Definition 5, we have that $F_{\varepsilon,n}$ and $L_{\varepsilon,n}$ are given by the follows equations:

[TABLE]

and

[TABLE]

for each $n\in\mathbb{N}$.

In the Table 1 expresses some terms of the sequences $\{F_{\varepsilon,n}\}_{n\in\mathbb{N}}$ and $\{L_{\varepsilon,n}\}_{n\in\mathbb{N}}$ for some $d$’s square free. Unless otherwise noted, the sequences are not in the Online Encyclopedia of Integer Sequences at the time of publication, though some of the sequences do appear there, as indicated in Table 2.

Observation 6. Note that when $d=5$, we have $\{F_{\varepsilon,n}\}_{n\in\mathbb{N}}$ and $\{L_{\varepsilon,n}\}_{n\in\mathbb{N}}$ are exactly the classical sequences of Fibonacci and Lucas, respectively.

In the rest of the work, by abuse of notation, we write $F_{n}$ and $L_{n}$ instead of $F_{\varepsilon,n}$ and $L_{\varepsilon,n}$ if there is no risk of confusion with respect to the classical sequences of Fibonacci and Lucas.

Theorem 7. For each $m,n\in\mathbb{N}$,

1. $(i)$

$F_{n+1}=a(L_{n}+F_{n})$. 2. $(ii)$

$L_{n+1}=aL_{n}+\displaystyle{\frac{b^{2}d}{a}F_{n}}$. 3. $(iii)$

F_{n}=\displaystyle{\frac{a}{\Delta}\left(F_{n+1}-\frac{}{}L_{n+1}\right)}=\left\{\begin{array}[]{ccl}a(L_{n+1}-F_{n+1})&if&\Delta=-1\\ &&\\ a(F_{n+1}-L_{n+1})&if&\Delta=1\ .\end{array}\right.** 4. $(iv)$

L_{n}=\displaystyle{\frac{1}{\Delta}\left(aL_{n+1}-\frac{b^{2}d}{a}F_{n+1}\right)}=\left\{\begin{array}[]{ccl}\displaystyle{\frac{b^{2}d}{a}F_{n+1}-aL_{n+1}}&if&\Delta=-1\\ &&\\ \displaystyle{aL_{n+1}-\frac{b^{2}d}{a}F_{n+1}}&if&\Delta=1\ .\end{array}\right.** 5. $(v)$

$F_{n+1}-a^{n}F_{1}=\displaystyle{\sum_{t=0}^{n-1}a^{t+1}L_{n-t}}$* .* 6. $(vi)$

$L_{n+1}-a^{n}L_{1}=\displaystyle{b^{2}d\ \sum_{t=0}^{n-1}a^{t-1}F_{n-t}}$* .* 7. $(vii)$

$F_{m+n}=a(F_{m}L_{n}+F_{n}L_{m})$. 8. $(viii)$

$L_{m+n}=\displaystyle{\frac{b^{2}d}{a}}\cdot F_{m}F_{n}+a\ L_{m}L_{n}$. 9. $(ix)$

$b^{2}d\ F_{n}^{2}-a^{2}L_{n}^{2}=-\Delta^{n}$. 10. $(x)$

$F_{n}=\displaystyle{\sum_{t=0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n}{2t+1}a^{n-2t-1}b^{2t}d^{t}}=\displaystyle{\sum_{t=0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n}{n-2t-1}a^{n-2t-1}b^{2t}d^{t}}$. 11. $(xi)$

$L_{n}=\displaystyle{\sum_{t=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{2t}a^{n-2t-1}b^{2t}d^{t}}=\displaystyle{\sum_{t=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{n-2t}a^{n-2t-1}b^{2t}d^{t}}$.

Here $\lfloor x\rfloor$ is the integral part of $x\in\mathbb{R}$, i.e., is the greatest integer $n$ such that $n\leq x<n+1$.

Proof. $(i)$ and $(ii)$ are obtained directly from the equations (3.5) and (3.9). $(iii)$ and $(iv)$ are deducted from $(i)$ and $(ii)$. For induction, we obtain $(v)$ and $(vi)$. $(vii)$ and $(viii)$ are obtained from the relationship $A_{\varepsilon}^{m+n}=A_{\varepsilon}^{m}\cdot A_{\varepsilon}^{n}$. The relation $\det(A_{\varepsilon}^{n})=\Delta^{n}$ implies the relation $(ix)$. Finally, $(x)$ y $(xi)$ are obtained of the relationships

[TABLE]

$\square$

Theorem 8. There exist unique $r,s\in\mathbb{Q}^{*}$ such that $F_{n+2}=rF_{n}+sF_{n+1}$ for all $n\in\mathbb{N}$. More precisely, $F_{n+2}=(-\Delta)F_{n}+2aF_{n+1}$ for all $n\in\mathbb{N}$.

Proof. We have for each $n\in\mathbb{N}$,

[TABLE]

On the other hand, let $r,s\in\mathbb{Q}^{*}$ be such that

[TABLE]

As $b^{2}d=a^{2}-\Delta$, implies that $F_{1}=1$, $F_{2}=2a$, $F_{3}=4a^{2}-\Delta$ and $F_{4}=8a^{3}-4a\Delta$. In particular, by the equation (3.10) for $n=1$ and $n=2$, we obtain the system of equations

[TABLE]

which it has an unique solution, namely $r=-\Delta$ and $s=2a$. This complete the proof of theorem.

$\square$

Corollary 9. The Fibonacci sequence $\{F_{n}\}_{n\in\mathbb{N}}$ is a $k$-Fibonacci sequence for some $k\in\mathbb{N}$ $($namely, $k=2a)$ if and only if $\Delta=-1$.

Proof. It immediate by Theorem 8. $\square$

Corollary 10. The following conditions are equivalent:

1. $(i)$

$F_{n+2}=F_{n}+F_{n+1}$* for all $n\in\mathbb{N}$;* 2. $(ii)$

$F_{3}=F_{1}+F_{2}$; 3. $(iii)$

$d=5$* and $\displaystyle{\varepsilon=\frac{1+\sqrt{5}}{2}}$ .*

Proof. $(i)\Longrightarrow(ii)$: It is immediate.

$(ii)\Longrightarrow(iii)$: We have that $-\Delta+4a^{2}=(-\Delta)F_{1}+2aF_{2}=F_{3}=F_{1}+F_{2}=1+2a$, then $4a^{2}-2a-(\Delta+1)=0$. If $\Delta=1$, then $2a^{2}-a-1=0$; since $a\neq 1$, necessarily $a=-1/2$. But this implies that $4b^{2}d=-3$; contradiction. Therefore $\Delta=-1$, $a=1/2=b$ and $d=5$.

$(iii)\Longrightarrow(i)$: It is clear. $\square$

Corollary 11. If $\{F_{n}\}_{n\in\mathbb{N}}$ is the Fibonacci sequence classical, that is $d=5$, then

[TABLE]

for each $n\in\mathbb{N}$.

Proof. It is immediate. $\square$

We recall if $d\equiv 2$ or $3$ mod $4$, then $\varepsilon=a+b\sqrt{d}$ where $a,b\in\mathbb{Z}$. In this case, it is obvious that $F_{n}\in\mathbb{N}$ for all $n\in\mathbb{N}$. If $d\equiv 1$ mod $4$, then $\varepsilon=a+b\sqrt{d}=\displaystyle{\frac{a_{0}+b_{0}\sqrt{d}}{2}}$ with $a_{0},\ b_{0}\in\mathbb{N}$, where either are both even or both odd. When they are both even, we have that $a,b\in\mathbb{N}$ and, hence, $F_{n}\in\mathbb{N}$. But, in any case, $2a\in\mathbb{N}$. Therefore, we obtain the following result.

Corollary 12. $F_{n}\in\mathbb{N}$* for all $n\in\mathbb{N}$.*

Proof. By Theorem 8, we have $F_{n+2}=(-\Delta)F_{n}+2aF_{n+1}$ for all $n\in\mathbb{N}$, where $F_{1}=1$ and $F_{2}=2a\in\mathbb{N}$. Then, the show follows by induction on $n$. $\square$

Theorem 13. There exist unique $r,s\in\mathbb{Q}^{*}$ such that $L_{n+2}=rL_{n}+sL_{n+1}$ for all $n\in\mathbb{N}$. More precisely, $L_{n+2}=(-\Delta)L_{n}+2aL_{n+1}$ for all $n\in\mathbb{N}$.

Proof. We have that for each $n\in\mathbb{N}$

[TABLE]

Now we prove the uniqueness. As $b^{2}d=a^{2}-\Delta$, it follows that

[TABLE]

Let $r,s\in\mathbb{Q}^{*}$ be such that

[TABLE]

In particular, for $n=1$ and $n=2$, we have the system of equations

[TABLE]

which it has a unique solution, namely $r=-\Delta$ and $s=2a$; so that, this system of equations has the same solution that the system of equations (3.14) given in the proof of Theorem 8. Therefore, the theorem is true. $\square$

Similarly to the corollaries to Theorem 8 for Fibonacci sequence, we obtain corollaries to Theorem 13 for Lucas sequence.

Corollary 14. The Lucas sequence $\{L_{n}\}_{n\in\mathbb{N}}$ is a $k$-Lucas sequence for some $k\in\mathbb{N}$ $($namely, $k=2a)$ if and only if $\Delta=-1$.

Proof. It immediate by Theorem 13. $\square$

Corollary 15. The following conditions are equivalent:

1. $(i)$

$L_{n+2}=L_{n}+L_{n+1}$* for all $n\in\mathbb{N}$;* 2. $(ii)$

$L_{3}=L_{1}+L_{2}$; 3. $(iii)$

$d=5$* and $\displaystyle{\varepsilon=\frac{1+\sqrt{5}}{2}}$.*

Proof. $(i)\Longrightarrow(ii)$: It is immediate.

$(ii)\Longrightarrow(iii)$: Since $L_{3}=L_{1}+L_{2}$, that is, $4a^{2}-3\Delta=1+2a-\displaystyle{\frac{\Delta}{a}}$, we have that $4a^{3}-2a^{2}-a+(1-3a)\Delta=0$. If $\Delta=1$, then $4a^{3}-2a^{2}-4a+1=0$ and $a$ can not be a rational number, contradiction. Hence, $\Delta=-1$ and $(2a^{2}+1)(2a-1)=0$. This implies that $a=1/2$ and $4b^{2}d=5$. Therefore, $d=5$ and $a=1/2=b$.

$(iii)\Longrightarrow(i)$: It is clear. $\square$

Corollary 16. If $\{L_{n}\}_{n\in\mathbb{N}}$ is the Lucas sequence classical, that is $d=5$, then

[TABLE]

for each $n\in\mathbb{N}$.

Proof. It is immediate. $\square$

Corollary 17. For all $k\in\mathbb{N}$,

1. $(i)$

$L_{2k-1}\in\mathbb{N}$; 2. $(ii)$

if $a\in\mathbb{N}$, then $aL_{2k}\in\mathbb{N}$ and $(a,aL_{2k})=1$; 3. $(iii)$

if $\displaystyle{a=\frac{a_{0}}{2}}$, with $a_{0}$ odd, then $a_{0}L_{2k}\in\mathbb{N}$ and $(a_{0},a_{0}L_{2k})=1$.

Proof. Applying the Theorem 13, the proof follows by induction over all the pairs $(L_{2k-1},L_{2k})$, $k\in\mathbb{N}$. $\square$

4. Generating function

The main goal of this section is to show that the Fibonacci and Lucas sequences given in $(4)$ and $(5)$ are determined by the generating functions.

Theorem 18. We obtain

1. $(i)$

$\displaystyle{\lim_{n\rightarrow\infty}\frac{F_{n+1}}{F_{n}}}=\varepsilon=\displaystyle{\lim_{n\rightarrow\infty}\frac{L_{n+1}}{L_{n}}}$. 2. $(ii)$

The series $\displaystyle{\sum_{n=1}^{\infty}F_{n}x^{n-1}}$ and $\displaystyle{\sum_{n=1}^{\infty}L_{n}x^{n-1}}$ both have the same radius of convergence, namely $R=1/\varepsilon$.

Proof. $(i)$: By Theorem 7, we have $\displaystyle{\frac{F_{n+1}}{F_{n}}}=\displaystyle{\frac{a(L_{n}+F_{n})}{F_{n}}}=a+a\cdot\displaystyle{\frac{L_{n}}{F_{n}}}=a+b\cdot\displaystyle{\frac{a_{n}}{b_{n}}}$, and $\displaystyle{\frac{L_{n+1}}{L_{n}}}=\displaystyle{\frac{aL_{n}+\displaystyle{\frac{b^{2}d}{a}F_{n}}}{L_{n}}}=a+\displaystyle{\frac{b^{2}d}{a}}\cdot\displaystyle{\frac{F_{n}}{L_{n}}}=a+b\cdot\displaystyle{\frac{b_{n}d}{a_{n}}}$ where $\displaystyle{\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}}=\sqrt{d}=\displaystyle{\lim_{n\rightarrow\infty}\frac{b_{n}d}{a_{n}}}$, see [17, Theorem 3.1]. Thus, $\displaystyle{\lim_{n\rightarrow\infty}\frac{F_{n+1}}{F_{n}}}=\varepsilon=\displaystyle{\lim_{n\rightarrow\infty}\frac{L_{n+1}}{L_{n}}}$.

$(ii)$: For each $x\in\mathbb{R}$, $x\neq 0$, we have that $\displaystyle{\lim_{n\rightarrow\infty}\frac{F_{n+1}x^{n}}{F_{n}x^{n-1}}}=\varepsilon x=\displaystyle{\lim_{n\rightarrow\infty}\frac{L_{n+1}x^{n}}{L_{n}x^{n-1}}\ .}$ Then $\displaystyle{\lim_{n\rightarrow\infty}\frac{F_{n+1}|x|^{n}}{F_{n}|x|^{n-1}}<1}$ if and only if $\displaystyle{|x|<\frac{1}{\varepsilon}}$. Similarly, $\displaystyle{\lim_{n\rightarrow\infty}\frac{L_{n+1}|x|^{n}}{L_{n}|x|^{n-1}}<1}$ if and only if $\displaystyle{|x|<\frac{1}{\varepsilon}}$ . Therefore, both series have the same radius of convergence $R=1/\varepsilon$. This complete the proof of the theorem. $\square$

Theorem 19. (Generating function) Let $x\in\mathbb{R}$ be such that $|x|<1/\varepsilon$.

1. $(i)$

If $f(x)=\displaystyle{\sum_{n=1}^{\infty}F_{n}x^{n-1}}$, then $\displaystyle{f(x)=\frac{1}{\Delta x^{2}-2ax+1}\ .}$ 2. $(ii)$

If $g(x)=\displaystyle{\sum_{n=1}^{\infty}L_{n}x^{n-1}}$, then $\displaystyle{g(x)=\left(\frac{a-\Delta x}{a}\right)f(x)=\frac{a-\Delta x}{a(\Delta x^{2}-2ax+1)}\ .}$

Proof. $(i)$: For each $x\in\mathbb{R}$ with $|x|<1/\varepsilon$, we have that

[TABLE]

this implies that $\displaystyle{f(x)=\frac{1}{\Delta x^{2}-2ax+1}\ .}$

$(ii)$: We observe that, for each $x\in\mathbb{R}$ with $|x|<1/\varepsilon$

[TABLE]

If $x\neq 0$, then

[TABLE]

$\square$

5. Golden ratio and Binet’s formula in $\mathbb{Q}(\sqrt{d}\ )$

In this section we give Golden ratio associated as the quadratic field $\mathbb{Q}(\sqrt{d}\,)$. Also we obtain Binet’s formula in $\mathbb{Q}(\sqrt{d}\,)$. We start with

Definition 20. Let $x,y\in\mathbb{R}$ be such that $0<y<x$. We say that $x$ and $y$ are in Golden ratio with respect to the quadratic field $\mathbb{Q}(\sqrt{d}\,)$ $($or simply that they are in Golden ratio, if there is no risk of confusion with respect to the quadratic field $\mathbb{Q}(\sqrt{d}\ ))$, if

[TABLE]

Thus, if $x$ and $y$ are in Golden ratio and we write $\varphi:=\displaystyle{\frac{x}{y}}$, then we have that

[TABLE]

This is, $\varphi$ satisfies the equation

[TABLE]

But $x^{2}-2ax+\Delta$ is the irreducible polynomial of $\varepsilon$ over $\mathbb{Q}$ with $\overline{\varepsilon}$ its other root, where $\overline{\varepsilon}$ is the conjugate of $\varepsilon$. Therefore, $\varphi=\varepsilon$ or $\varphi=\overline{\varepsilon}$. As $x>y>0$ and $\overline{\varepsilon}=\Delta/\varepsilon$, necessarily $\varphi=\varepsilon$. In consequence, we have the equation

[TABLE]

Theorem 21. For each $n\in\mathbb{N}$, with $n\geq 2$,

[TABLE]

Proof. The show is by induction on $n$. It is clear for $n=2$, that is, $\varepsilon^{2}=2a\varepsilon-\Delta=F_{2}\ \varepsilon-F_{1}\ \Delta$. Hence,

[TABLE]

$\square$

Since $\overline{\varepsilon}$ also satisfies the equation (5.3), we have the equation

[TABLE]

for each $n\geq 2$.

Theorem 22. For each $n\in\mathbb{N}$,

[TABLE]

Proof. It follows to make the difference of the equations (5.4) and (5.5). $\square$

The equation (5.6) is known as the Binet’s formula

Corollary 23. For each $n\in\mathbb{N}$,

[TABLE]

Proof. It is immediate from the following

[TABLE]

$\square$

The following two theorems give us other version of the generating functions of the sequences of Fibonacci and Lucas in $\mathbb{Q}(\sqrt{d}\ )$.

Theorem 24. Let $f_{1}(x)=\displaystyle{\sum_{n=0}^{\infty}\Delta^{n}F_{n+1}x^{n}}$ and $\displaystyle{g_{1}(x)=\sum_{n=0}^{\infty}\Delta^{n}L_{n+1}x^{n}}$. Then, the series $f_{1}(x)$ and $g_{1}(x)$ are convergent for $|x|<{\rm min}\{|\varepsilon|,|\overline{\varepsilon}|\}$. Furthermore,

[TABLE]

and

[TABLE]

Proof. We have for $|x|<{\rm min}\{|\varepsilon|,|\overline{\varepsilon}|\}$

[TABLE]

This implies that

[TABLE]

or equivalently

[TABLE]

Therefore

[TABLE]

On the other hand, we have

[TABLE]

Therefore

[TABLE]

$\square$

6. Some Other Properties

Using the equations (5.6) and (5.7), we can extend the definition of the sequences of Fibonacci and Lucas over all integer number. This is, we use the Binet’s formula for all $n\in\mathbb{Z}$, Theorem 22 and Corallary 23 we obtain

[TABLE]

and

[TABLE]

Thus, it holds that for all $n\in\mathbb{Z}$

[TABLE]

and

[TABLE]

But also we can obtain, in our case, the identities established by Catalan, Cassini, D’ Ocagne, and Hosnberger which are hold for all $n\in\mathbb{Z}$, that is

Theorem 25. For all $m,n\in\mathbb{Z}$, the follows identities are holds:

1. $(i)$

$F_{n}^{2}-F_{n+m}F_{n-m}=\Delta^{n-m}F_{m}^{2}.$** 2. $(ii)$

$F_{n}^{2}-F_{n-1}F_{n+1}=\Delta^{n-1}.$** 3. $(iii)$

$L_{n}^{2}-L_{n+r}L_{n-r}=\displaystyle{\frac{\Delta^{n}}{2a^{2}}-\left(\frac{\Delta^{n-r}}{2a}\right)L_{2r}}$. 4. $(iv)$

$F_{m}F_{n+1}-F_{n}F_{m+1}=\Delta^{n}F_{m-n}.$** 5. $(v)$

F_{m-1}F_{n}+F_{m}F_{n+1}=\left\{\begin{array}[]{lcl}F_{m+n}&if&\Delta=-1\\ &&\\ \displaystyle{\frac{a}{2b^{2}d}}\cdot\bigg{(}2aL_{m+n}-L_{m-n-1}\bigg{)}&if&\Delta=1.\end{array}\right.** 6. $(vi)$

$L_{n}L_{n+r}=\displaystyle{\left(\frac{1}{2a}\right)\ L_{2n+r}+\left(\frac{\Delta^{n}}{2a}\right)\ L_{r}}$.

Proof. The show for each of the identities can be performed using the Binet’s Formula. So we will prove only $(iv)$. Hence we have

[TABLE]

$\square$

7. The sequence of Fibonacci and of Lucas of degree $d$ with

respect to an arbitrary unit

The unit group of $\mathbb{Q}(\sqrt{d}\ )$, with $d>0$, is isomorph to group $\langle-1\rangle\times\langle\varepsilon\rangle$ where $\varepsilon$ is the fundamental unit of $\mathbb{Q}(\sqrt{d}\ )$, generator of the infinity cyclic subgroup. This cyclic subgroup also is generated by $1/\varepsilon$, $-\varepsilon$ and $-1/\varepsilon$. This is, each unit of $\mathbb{Q}(\sqrt{d}\ )$ has the form $\pm\varepsilon^{l}$ for some $l\in\mathbb{Z}$. Observing the previous development, we can define the sequence of Fibonacci and Lucas of degree $d$ with respect to an arbitrary unit $\eta$ of $\mathbb{Q}(\sqrt{d}\ )$, and the results of the previous sections are still met. Essentially this is because $N(\eta)=\pm 1$. This allows us to build even more an infinity of sequences in $\mathbb{Q}(\sqrt{d}\ )$ meeting similar properties of the sequences of Fibonacci and Lucas. For example, we consider the unit $\eta=\displaystyle{\frac{-1+\sqrt{5}}{2}}$ of $\mathbb{Q}(\sqrt{5}\ )$, we have that the first terms of the sequence of Fibonacci of degree $5$ with respect to the unit $\eta$ are:

[TABLE]

where $N(\eta)=-1$. Comparing the terms of the sequence of Fibonacci with negative index, $F_{-n}$ with $n\geq 1$, we have that $F_{\eta,n}=F_{-n}$ for all $n\in\mathbb{N}$. This is, the sequence of Fibonacci with negative index of degree $5$ with respect to the fundamental unit $\displaystyle{\varepsilon=\frac{1+\sqrt{5}}{2}}$ is the sequence of Fibonacci of degree $5$ respect to the unit $\eta=\displaystyle{\frac{-1+\sqrt{5}}{2}}$ . This it is not a coincidence, that is, this fact is generalized in the following.

Theorem 26. The Fibonacci sequence of degree $d$ with respect to the unit $1/\varepsilon$ and $\Delta=-1$ is the Fibonacci sequence with negative index of degree $d$ with respect to the fundamental unit $\varepsilon$.

Proof. We write $\eta=1/\varepsilon=\Delta\overline{\varepsilon}$. Hence, $\overline{\eta}=\Delta\varepsilon$. Using the Binet’s formula, we have that for all $n\in\mathbb{N}$

[TABLE]

$\square$

Observation 27. We note that if $\Delta=1$ then the Fibonacci sequence of degree $d$ with respect to the unit $1/\varepsilon$ coincides with the Fibonacci sequence of degree $d$ with respect to the unit $\varepsilon$.

We finish our work with the following result.

Theorem 28. For each $k\in\mathbb{N}$ there exist unique $d,\ r\in\mathbb{N}$ such that $d$ is square free and $\displaystyle{\frac{k}{2}+\frac{r}{2}\sqrt{d}}$ is a unit of the quadratic field $\mathbb{Q}(\sqrt{d}\,)$ with norm $-1$. Therefore, in this case, the $k$-Fibonacci sequence is the Fibonacci sequence of degree $d$ with respect to a unit of $\mathbb{Q}(\sqrt{d}\,)$.

Proof. Let $k\in\mathbb{N}$ be arbitrary. We have that $k^{2}+4$ is not a perfect square. Hence, there exist $d,r\in\mathbb{N}$ such that $k^{2}+4=r^{2}d$ where $d$ is positive square free. This implies that $\displaystyle{\left(\frac{k}{2}\right)^{2}-\left(\frac{r}{2}\right)^{2}d=-1}$. Hence, $\displaystyle{\frac{k}{2}+\frac{r}{2}\sqrt{d}}$ is a unit of $\mathbb{Q}(\sqrt{d}\ )$ with norm $-1$. On the other hand, if $d,\ d_{1},r,\ r_{1}\in\mathbb{N}$ such that $\displaystyle{\frac{k}{2}+\frac{r}{2}\sqrt{d}}$ and $\displaystyle{\frac{k}{2}+\frac{r_{1}}{2}\sqrt{d_{1}}}$ are units of the quadratic field $\mathbb{Q}(\sqrt{d}\ )$ both with norm $-1$, then $\displaystyle{\left(\frac{k}{2}\right)^{2}-\left(\frac{r}{2}\right)^{2}d=-1=\left(\frac{k}{2}\right)^{2}-\left(\frac{r_{1}}{2}\right)^{2}d_{1}}$, thus $\displaystyle{\left(\frac{r}{2}\right)^{2}d=\left(\frac{r_{1}}{2}\right)^{2}d_{1}}$. That is, $r^{2}d=r_{1}^{2}d_{1}$, where $d$ and $d_{1}$ are square free. Therefore, $d_{1}=d$ and $r_{1}=r$. In consequence, the $k$-Fibonacci sequence is the Fibonacci sequence of degree $d$ with respect to a unit of $\mathbb{Q}(\sqrt{d}\ )$. $\square$

Corollary 29. For each $k\in\mathbb{N}$, the $k$-Fibonacci sequence is the Fibonacci sequence of degree $d$ with respect to a unit of $\mathbb{Q}(\sqrt{d}\ )$ for some $d$ square free.

Proof. It is immediately of Theorem 28 and Corollary 9. $\square$

8. Conclusions

In this work we have established that every real quadratic field $\mathbb{Q}(\sqrt{d}\ )$ has its own Fibonacci sequence and Lucas sequence, and variants of these, through the fundamental unit, being this the golden ratio. Therefore, the real quadratic field $\mathbb{Q}(\sqrt{d}\ )$ has its own gold ratio. Under these conditions, it is possible that may arise further research aimed at obtaining properties, both algebraic and geometric, related with the intrinsic properties of the real quadratic field $\mathbb{Q}(\sqrt{d}\ )$.

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