Tri–Periodic Fibonacci Numbers and Tri–Periodic Leonardo Numbers

1. Introduction

The number sequences have a crucial place in the literature. Because these sequences have wide-ranging applications, i.e, cryptology, computer science,art, architecture, finance and algorithm analysis to natural phenomena [1,2,3,4,5,6,7,8,9,10]. As a special cases of number sequences, named Fibonacci and Leonardo sequences have attracted the authors in the literature and have celebrated their recurrence relations. Also, they have connections to different mathematical problems.

Over the years, interest has grown in generalized forms of these sequences, particularly those governed by periodic recurrence relations. In this study, we focus on tri-periodic variants of the Fibonacci and Leonardo sequences, where the recurrence relation alternates cyclically every three terms. First, we remember some essential properties of Fibonacci and Leonardo numbers.

Fibonacci sequence is defined as reccursevily by

$$F_n=F_{n-1}+F_{n-2},F_0=0,F_1=1$$

respectively. The first few Fibonacci sequence numbers are given below

$$0,1,1,2,3,5,8,13,21,34,55,89,144,...$$

Then the characteristic equation of these sequences is given by

$$x^2-x-1=0$$

that the roots are

$$\alpha ={\displaystyle \frac{1+\sqrt{5}}{2}}\mathrm{and}\beta ={\displaystyle \frac{1+\sqrt{5}}{2}}$$

Hence, the Binet’s formula of Fibonacci sequence is

$$F_n={\displaystyle \frac{{\alpha }^n-{\beta }^n}{\alpha -\beta }}$$

and generating function of Fibonacci sequence is

$$\sum _{n=0}^{\infty }{F}_n{z}^n={\displaystyle \frac{z}{1-z-z^2}}.$$

There are many study in literature related to the Fibonacci sequence and so there are many properties found in the literature [10,11,12,13,14,15,16].

Now, we give the recurrance relation of Leonardo sequence as

$$L_n=L_{n-1}+L_{n-2}+1,L_0=1,L_1=1$$

The first few Leonardo numbers are given below

$$1,1,3,5,9,15,25,41,67,109,177,287,465,753,1219,1973,3193,5167,8361,...$$

The Leonardo numbers are related to the Fibonacci numbers given by

$$L_n=2F_{n+1}-1.$$

Hence, Binet’s formula of Leonardo numbers can be given by

$$L_n={\displaystyle \frac{{\alpha }^n-{\beta }^n}{\alpha -\beta }}-1.$$

Also there are many study in literature corresponding to the Leonardo sequence. So there are many relation between Leonardo numbers and Fibonacci numbers [17,18,19,20,21,22,23].

The extensions of number sequences studied in literature are named bi-periodic number sequences, where the terms of bi-periodic sequences are generated based on odd and even indices. Unlike classical definition of number sequences, the structure of bi-periodic sequences makes them suitable for modeling systems with changing conditions for odd and even indices. Also many authors studied bi periodic sequences related to the special numbers and its properties, i.e , the generating function, Binet’s formula, Catalan’s and D’ocgane’s identities. Now we present some examples from the literature raleted to the study of bi-periodic sequences. Next, we present some study that found in literature.

• Edson Yayenie whose first studied bi-periodic sequences based on Fibonacci numbers [24], defined bi-periodic Fibonacci sequences indicated by ${\left\{q_n\right\}}_{n\ge 0}$ which is defined as

  $$q_n=\left\{\begin{array}{c}a{q}_{n-1}+q_{n-2}\\ b{q}_{n-1}+q_{n-2}\end{array}\right.\begin{array}{c}\mathrm{if}n\mathrm{is}\mathrm{even},\\ \mathrm{if}n\mathrm{is}\mathrm{odd},\end{array}n\ge 2$$

  (1)

  with the initial conditions $q_0=0$ and $q_1=1,$ where a and b are nonzero real numbers.
• Catarino and Spreafico [25] defined bi-periodic Leonardo sequences denoted by ${\left\{GL{e}_n\right\}}_{n\ge 0}$ which is defined as

  $$GL{e}_n=\left\{\begin{array}{c}aGL{e}_{n-1}+GL{e}_{n-2}+a\\ bGL{e}_{n-1}+GL{e}_{n-2}+b\end{array}\right.\begin{array}{c}\mathrm{if}n\mathrm{is}\mathrm{even},\\ \mathrm{if}n\mathrm{is}\mathrm{odd},\end{array}n\ge 2$$

  with the initial conditions $GL{e}_0=2a-1$ and $GL{e}_1=2ab-1,$ where a and b are nonzero real numbers.
• Costa et al. [26] defined bi-periodic Eduard and Eduard-Lucas sequences denoted by ${\left\{E_n^{(a,b)}\right\}}_{n\ge 0}$ and ${\left\{K_n^{(a,b)}\right\}}_{n\ge 0},$ respecitively, which are defined as

  $$E_n^{(a,b)}=\left\{\begin{array}{c}6a{E}_{n-1}^{(a,b)}+E_{n-2}^{(a,b)}\\ 6b{E}_{n-1}^{(a,b)}+E_{n-2}^{(a,b)}\end{array}\right.\begin{array}{c}\mathrm{if}n\mathrm{is}\mathrm{even},\\ \mathrm{if}n\mathrm{is}\mathrm{odd},\end{array}n\ge 2$$

  (2)

  with the initial conditions $E_0^{(a,b)}=0$ and $E_1^{(a,b)}=1,$ and

  $$K_n^{(a,b)}=\left\{\begin{array}{c}6a{K}_{n-1}^{(a,b)}+K_{n-2}^{(a,b)}\\ 6b{K}_{n-1}^{(a,b)}+K_{n-2}^{(a,b)}\end{array}\right.\begin{array}{c}\mathrm{if}n\mathrm{is}\mathrm{even},\\ \mathrm{if}n\mathrm{is}\mathrm{odd},\end{array}n\ge 2$$

  (3)

  with the initial conditions $K_0^{(a,b)}=3$ and $K_1^{(a,b)}=7$ where a and b are nonzero real numbers.
• Edson et al. [27] defined k-Periodic Fibonacci Sequence as

  $$q_n=\left\{\begin{array}{c}{x}_1{q}_{n-1}+q_{n-2},\\ x_2{q}_{n-1}+q_{n-2},\\ .\\ .\\ .\\ x_{k-1}{q}_{n-1}+q_{n-2},\\ x_k{q}_{n-1}+q_{n-2},\end{array}\right.\begin{array}{c}\mathrm{if}n\equiv 2\left(\mathrm{mod}k\right)\\ \mathrm{if}n\equiv 3\left(\mathrm{mod}k\right)\\ .\\ .\\ .\\ \mathrm{if}n\equiv 0\left(\mathrm{mod}k\right)\\ \mathrm{if}n\equiv 1\left(\mathrm{mod}k\right)\end{array}n\ge 2$$

  with the initial condition $q_0=0,$$q_1=1.$

In this study we present the different expension of Fibonacci and Leonardo sequences named tri-periodic Fibonacci and tri-periodic Leonardo sequences where the terms of these sequences based on mod 3 indices. Then we investigate generating function of tri-periodic Fibonacci and tri-periodic Leonardo sequences, Binet’s formula of tri-periodic Fibonacci and tri-periodic Leonardo sequences and also Cassani’s identity of tri-periodic Fibonacci numbers.

2. Tri–Periodic Fibonacci Numbers and Tri–periodic Leonardo Numbers

In this section, we define generalization of Fibonacci numbers denoted $F_n^{(a,b,c)}$ called tri-periodic Fibonacci numbers and generalization of Leonardo numbers denoted $L_n^{(a,b,c)}$ called tri-periodic Leonardo numbers.

Definition 1. 

For any real nonzero numbers a, b and c the tri-periodic Fibonacci sequence is defined as reccursevily by,

$$F_n^{(a,b,c)}=\left[\begin{array}{c}a{F}_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}\\ b{F}_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}\\ c{F}_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}\end{array}\right.\begin{array}{c}n\equiv 0(\mathrm{mod}3)\\ n\equiv 1(\mathrm{mod}3)\\ n\equiv 2(\mathrm{mod}3)\end{array}n\ge 3,$$

(4)

with the initial condition $F_0^{(a,b,c)}=0,$$F_1^{(a,b,c)}=1,$and $F_2^{(a,b,c)}=c.$

Definition 2. 

For any real nonzero numbers a, b and c the tri-periodic Leonardo sequence is defined as recurrsively by,

$$l_n^{(a,b,c)}=\left[\begin{array}{c}a{l}_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+a\\ b{l}_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+b\\ c{l}_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+c\end{array}\right.\begin{array}{c}n\equiv 0(\mathrm{mod}3)\\ n\equiv 1(\mathrm{mod}3)\\ n\equiv 2(\mathrm{mod}3)\end{array}n\ge 3,$$

(5)

with the initial condition $l_0^{(a,b,c)}=1,$$l_1^{(a,b,c)}=1,$and $l_2^{(a,b,c)}=2c+1.$

Note that, if $a=b=1$, (4) gives Fibonacci numbers and if $a=b=1$, (4) gives k-Fibonacci numbers. Similarly, if $a=b=1$, (5) gives Leonardo numbers and if $a=b=1$, (5) gives k-Leonardo numbers. To write (4) and (5) in simple form, for all integer n, we define moduler Kronecker delta function that we denote ${\delta }_k\left(n\right)$ as

$${\delta }_k\left(n\right)=\left[\begin{array}{c}1\\ 0\end{array}\right.\begin{array}{c}\mathrm{if}n=k(\mathrm{mod}3)\\ \mathrm{otherwise}\end{array}$$

where $k=0,1,2.$ Thus the recurrence relation (4) and (5) can be written as

$$\begin{array}{ccc}\hfill F_n^{(a,b,c)}& =& a^{{\delta }_0\left(n\right)}{b}^{{\delta }_1\left(n\right)}{c}^{{\delta }_2\left(n\right)}{F}_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)},n\ge 3\hfill \end{array}$$

(6)

$$\begin{array}{ccc}\hfill l_n^{(a,b,c)}& =& a^{{\delta }_0\left(n\right)}{b}^{{\delta }_1\left(n\right)}{c}^{{\delta }_2\left(n\right)}{l}_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+a^{{\delta }_0\left(n\right)}{b}^{{\delta }_1\left(n\right)}{c}^{{\delta }_2\left(n\right)},n\ge 3\hfill \end{array}$$

(7)

The (4) and (5) gives the following nonlinear quadratic equation for the tri-periodic Fibonacci sequence.

$$x^2-abcx-abc=0$$

with roots $\phi$ and $\psi$ defined by

$$\phi ={\displaystyle \frac{abc+\sqrt{abc(abc+4)}}{2}},\psi ={\displaystyle \frac{abc-\sqrt{abc(abc+4)}}{2}}.$$

(8)

Note that, the following equalities are true,

$$\begin{array}{ccc}\hfill \phi +\psi & =& abc,\phi \psi =-abc\hfill \\ \hfill {\displaystyle \frac{{\psi }^2}{abc}}& =& \psi +1,{\displaystyle \frac{{\phi }^2}{abc}}=\phi +1,\hfill \\ \hfill \phi \psi & =& -abc,\phi +\psi =abc,\hfill \\ \hfill \left(\phi +1\right)\left(\psi +1\right)& =& 1,\left(\phi +1\right)\psi =-\phi ,\left(\psi +1\right)\phi =-\psi .\hfill \end{array}$$

In the Table 1, some terms of the tri-periodic Fibonacci sequence is given.

In the Table 2, some terms of the tri-periodic Leonardo sequence is given.

Theorem 1. 

The tri-periodic Fibonacci sequence $F_n^{(a,b,c)}$ and tri-periodic Leonardo sequence are hold the following equality.

(a)

$F_{n+2}^{(a,b,c)}=(\omega -{\delta }_2\left(n\right)a-{\delta }_0\left(n\right)b-{\delta }_1\left(n\right)c)F_{n-1}^{(a,b,c)}+(a^{1-{\delta }_0\left(n\right)}{b}^{1-{\delta }_1\left(n\right)}{c}^{1-{\delta }_2\left(n\right)}+1)F_{n-2}^{(a,b,c)}.$

(b)

$l_{n+2}^{(a,b,c)}=(\omega -{\delta }_2\left(n\right)a-{\delta }_0\left(n\right)b-{\delta }_1\left(n\right)c)l_{n-1}^{(a,b,c)}+(a^{1-{\delta }_0\left(n\right)}{b}^{1-{\delta }_1\left(n\right)}{c}^{1-{\delta }_2\left(n\right)}+1)l_{n-2}^{(a,b,c)}+\omega -{\delta }_2\left(n\right)a-{\delta }_0\left(n\right)b-{\delta }_1\left(n\right)c+a^{1-{\delta }_0\left(n\right)}{b}^{1-{\delta }_1\left(n\right)}{c}^{1-{\delta }_2\left(n\right)}.$

where $\omega =(abc+a+b+c).$

Proof. (a) Let $n\equiv 0$ (mod 3) then using (4), we obtain

$$\begin{array}{ccc}\hfill F_n^{(a,b,c)}& =& a{F}_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)},\hfill \\ \hfill F_{n+1}^{(a,b,c)}& =& b{F}_n^{(a,b,c)}+F_{n-1}^{(a,b,c)},\hfill \\ \hfill F_{n+2}^{(a,b,c)}& =& c{F}_{n+1}^{(a,b,c)}+F_n^{(a,b,c)},\hfill \end{array}$$

and summing side by side and reorganizing these equalities we have

$$F_{n+2}^{(a,b,c)}=(c-1)F_{n+1}^{(a,b,c)}+b{F}_n^{(a,b,c)}+(a+1)F_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}.$$

Hence, we obtain

$$\begin{array}{ccc}\hfill F_{n+2}^{(a,b,c)}& =& (c-1)(b{F}_n^{(a,b,c)}+F_{n-1}^{(a,b,c)})+b{F}_n^{(a,b,c)}+(a+1)F_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}\hfill \\ & =& cb(a{F}_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)})+(c+a)F_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}\hfill \\ & =& (abc+a+c)F_{n-1}^{(a,b,c)}+(bc+1)F_{n-2}^{(a,b,c)}.\hfill \end{array}$$

Let $n\equiv 1$ (mod 3) then using (4), we obtain

$$\begin{array}{ccc}\hfill F_n^{(a,b,c)}& =& b{F}_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)},\hfill \\ \hfill F_{n+1}^{(a,b,c)}& =& c{F}_n^{(a,b,c)}+F_{n-1}^{(a,b,c)},\hfill \\ \hfill F_{n+2}^{(a,b,c)}& =& a{F}_{n+1}^{(a,b,c)}+F_n^{(a,b,c)},\hfill \end{array}$$

and summing side by side and reorganizing these equalities we have

$$F_{n+2}^{(a,b,c)}=(a-1)F_{n+1}^{(a,b,c)}+c{F}_n^{(a,b,c)}+(b+1)F_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}.$$

Hence, we obtain

$$\begin{array}{ccc}\hfill F_{n+2}^{(a,b,c)}& =& (a-1)(c{F}_n^{(a,b,c)}+F_{n-1}^{(a,b,c)})+c{F}_n^{(a,b,c)}+(b+1)F_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}\hfill \\ & =& ac(b{F}_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)})+(a+b)F_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}\hfill \\ & =& (abc+a+b)F_{n-1}^{(a,b,c)}+(ac+1)F_{n-2}^{(a,b,c)}.\hfill \end{array}$$

Similarly, Let $n\equiv 2$ (mod 3) then using (4), we obtain

$$\begin{array}{ccc}\hfill F_n^{(a,b,c)}& =& c{F}_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)},\hfill \\ \hfill F_{n+1}^{(a,b,c)}& =& a{F}_n^{(a,b,c)}+F_{n-1}^{(a,b,c)},\hfill \\ \hfill F_{n+2}^{(a,b,c)}& =& b{F}_{n+1}^{(a,b,c)}+F_n^{(a,b,c)},\hfill \end{array}$$

and summing side by side and reorganizing these equalities we have

$$F_{n+2}^{(a,b,c)}=(b-1)F_{n+1}^{(a,b,c)}+a{F}_n^{(a,b,c)}+(c+1)F_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}.$$

Hence, we obtain

$$\begin{array}{ccc}\hfill F_{n+2}^{(a,b,c)}& =& (b-1)(a{F}_n^{(a,b,c)}+F_{n-1}^{(a,b,c)})+a{F}_n^{(a,b,c)}+(c+1)F_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}\hfill \\ & =& ab(a{F}_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)})+(b+c)F_{n-1}^{(a,b,c)}+F_{n-2}^{(a,b,c)}\hfill \\ & =& (abc+b+c)F_{n-1}^{(a,b,c)}+(ab+1)F_{n-2}^{(a,b,c)}.\hfill \end{array}$$

(b) Let $n\equiv 0$ (mod 3) then using (4), we obtain

$$\begin{array}{ccc}\hfill l_n^{(a,b,c)}& =& a{l}_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+a,\hfill \\ \hfill l_{n+1}^{(a,b,c)}& =& b{l}_n^{(a,b,c)}+l_{n-1}^{(a,b,c)}+b,\hfill \\ \hfill l_{n+2}^{(a,b,c)}& =& c{l}_{n+1}^{(a,b,c)}+l_n^{(a,b,c)}+c,\hfill \end{array}$$

and summing side by side and reorganizing these equalities we have

$$l_{n+2}^{(a,b,c)}=(c-1)l_{n+1}^{(a,b,c)}+b{l}_n^{(a,b,c)}+(a+1)l_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+a+b+c.$$

Hence, we obtain

$$\begin{array}{ccc}\hfill l_{n+2}^{(a,b,c)}& =& (c-1)(b{l}_n^{(a,b,c)}+l_{n-1}^{(a,b,c)}+b)+b{l}_n^{(a,b,c)}+(a+1)l_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+a+b+c\hfill \\ & =& bc(a{l}_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+a)+(a+c)l_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+bc+a+c\hfill \\ & =& (abc+a+c)l_{n-1}^{(a,b,c)}+(bc+1)l_{n-2}^{(a,b,c)}+abc+bc+a+c.\hfill \end{array}$$

Let $n\equiv 1$ (mod 3) then using (4), we obtain

$$\begin{array}{ccc}\hfill l_n^{(a,b,c)}& =& b{l}_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+b,\hfill \\ \hfill l_{n+1}^{(a,b,c)}& =& c{l}_n^{(a,b,c)}+l_{n-1}^{(a,b,c)}+c,\hfill \\ \hfill l_{n+2}^{(a,b,c)}& =& a{l}_{n+1}^{(a,b,c)}+l_n^{(a,b,c)}+a,\hfill \end{array}$$

and summing side by side and reorganizing these equalities we have

$$l_{n+2}^{(a,b,c)}=(a-1)l_{n+1}^{(a,b,c)}+c{l}_n^{(a,b,c)}+(b+1)l_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+a+b+c.$$

Hence, we obtain

$$\begin{array}{ccc}\hfill l_{n+2}^{(a,b,c)}& =& (a-1)(c{l}_n^{(a,b,c)}+l_{n-1}^{(a,b,c)}+c)+c{l}_n^{(a,b,c)}+(b+1)l_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+a+b+c\hfill \\ & =& ac(b{l}_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+b)+(a+b)l_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+ac+a+b\hfill \\ & =& (abc+a+b)l_{n-1}^{(a,b,c)}+(ac+1)l_{n-2}^{(a,b,c)}+abc+ac+a+b.\hfill \end{array}$$

Similarly, Let $n\equiv 2$ (mod 3) then using (4), we obtain

$$\begin{array}{ccc}\hfill l_n^{(a,b,c)}& =& c{l}_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+c,\hfill \\ \hfill l_{n+1}^{(a,b,c)}& =& a{l}_n^{(a,b,c)}+l_{n-1}^{(a,b,c)}+a,\hfill \\ \hfill l_{n+2}^{(a,b,c)}& =& b{l}_{n+1}^{(a,b,c)}+l_n^{(a,b,c)}+b,\hfill \end{array}$$

and summing side by side and reorganizing these equalities we have

$$l_{n+2}^{(a,b,c)}=(b-1)l_{n+1}^{(a,b,c)}+a{l}_n^{(a,b,c)}+(c+1)l_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+a+b+c.$$

Hence, we obtain

$$\begin{array}{ccc}\hfill l_{n+2}^{(a,b,c)}& =& (b-1)(a{l}_n^{(a,b,c)}+l_{n-1}^{(a,b,c)}+a)+a{l}_n^{(a,b,c)}+(c+1)l_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+a+b+c\hfill \\ & =& ab(c{l}_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+c)+(b+c)l_{n-1}^{(a,b,c)}+l_{n-2}^{(a,b,c)}+ab+b+c\hfill \\ & =& (abc+b+c)l_{n-1}^{(a,b,c)}+(ab+1)l_{n-2}^{(a,b,c)}+abc+ab+b+c.\square \hfill \end{array}$$

Theorem 2. 

The tri-periodic Fibonacci sequence $F_n^{(a,b,c)}$ and tri-periodic Leonardo sequence hold the following equality.

$$\begin{array}{ccc}\hfill F_n^{(a,b,c)}& =& \omega F_{n-3}^{(a,b,c)}+F_{n-6}^{(a,b,c)},\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill l_n^{(a,b,c)}& =& \omega l_{n-3}^{(a,b,c)}+l_{n-6}^{(a,b,c)}+\omega ,\hfill \end{array}$$

(10)

where $\omega =(abc+a+b+c).$

Proof. (a) For the case $n\equiv 0$ (mod 3), using (4), we have

$$\begin{array}{ccc}\hfill F_{3n}^{(a,b,c)}& =& a{F}_{3n-1}^{(a,b,c)}+F_{3n-2}^{(a,b,c)}\hfill \\ & =& ac{F}_{3n-2}^{(a,b,c)}+a{F}_{3n-3}^{(a,b,c)}+F_{3n-2}^{(a,b,c)}\hfill \\ & =& acb{F}_{3n-3}^{(a,b,c)}+ac{F}_{3n-4}^{(a,b,c)}+a{F}_{3n-3}^{(a,b,c)}+F_{3n-2}^{(a,b,c)}\hfill \\ & =& acb{F}_{3n-3}^{(a,b,c)}+c{F}_{3n-3}^{(a,b,c)}-c{F}_{3n-5}^{(a,b,c)}+a{F}_{3n-3}^{(a,b,c)}+F_{3n-2}^{(a,b,c)}\hfill \\ & =& (abc+a+b+c)F_{3n-3}^{(a,b,c)}-(F_{3n-4}^{(a,b,c)}-F_{3n-6}^{(a,b,c)})+F_{3n-4}^{(a,b,c)}\hfill \\ & =& (abc+a+b+c)F_{3n-3}^{(a,b,c)}+F_{3n-6}^{(a,b,c)}.\hfill \end{array}$$

For the case $n\equiv 1$ (mod 3), using (4), we have

$$\begin{array}{ccc}\hfill F_{3n+1}^{(a,b,c)}& =& b{F}_{3n}^{(a,b,c)}+F_{3n-1}^{(a,b,c)}\hfill \\ & =& ab{F}_{3n-1}^{(a,b,c)}+b{F}_{3n-2}^{(a,b,c)}+F_{3n-1}^{(a,b,c)}\hfill \\ & =& abc{F}_{3n-2}^{(a,b,c)}+ab{F}_{3n-3}^{(a,b,c)}+b{F}_{3n-2}^{(a,b,c)}+F_{3n-1}^{(a,b,c)}\hfill \\ & =& abc{F}_{3n-2}^{(a,b,c)}+a{F}_{3n-2}^{(a,b,c)}-a{F}_{3n-4}^{(a,b,c)}+b{F}_{3n-2}^{(a,b,c)}+F_{3n-1}^{(a,b,c)}\hfill \\ & =& (abc+a+b+c)F_{3n-2}^{(a,b,c)}-(F_{3n-3}^{(a,b,c)}-F_{3n-5}^{(a,b,c)})+F_{3n-3}^{(a,b,c)}\hfill \\ & =& (abc+a+b+c)F_{3n-2}^{(a,b,c)}+F_{3n-5}^{(a,b,c)}.\hfill \end{array}$$

For the case $n\equiv 2$ (mod 3), using (4), we have

$$\begin{array}{ccc}\hfill F_{3n+2}^{(a,b,c)}& =& c{F}_{3n+1}^{(a,b,c)}+F_{3n}^{(a,b,c)}\hfill \\ & =& cb{F}_{3n}^{(a,b,c)}+c{F}_{3n-1}^{(a,b,c)}+F_{3n}^{(a,b,c)}\hfill \\ & =& abc{F}_{3n-1}^{(a,b,c)}+cb{F}_{3n-2}^{(a,b,c)}+c{F}_{3n-1}^{(a,b,c)}+F_{3n}^{(a,b,c)}\hfill \\ & =& abc{F}_{3n-1}^{(a,b,c)}+b{F}_{3n-1}^{(a,b,c)}-b{F}_{3n-3}^{(a,b,c)}+c{F}_{3n-1}^{(a,b,c)}+F_{3n}^{(a,b,c)}\hfill \\ & =& (abc+a+b+c)F_{3n-1}^{(a,b,c)}-(F_{3n-2}^{(a,b,c)}-F_{3n-5}^{(a,b,c)})+F_{3n-2}^{(a,b,c)}\hfill \\ & =& (abc+a+b+c)F_{3n-1}^{(a,b,c)}+F_{3n-5}^{(a,b,c)}.\square \hfill \end{array}$$

(b) For the case $n\equiv 0$ (mod 3), using (4), we have

$$\begin{array}{ccc}\hfill l_{3n}^{(a,b,c)}& =& a{l}_{3n-1}^{(a,b,c)}+l_{3n-2}^{(a,b,c)}+a\hfill \\ & =& ac{l}_{3n-2}^{(a,b,c)}+a{l}_{3n-3}^{(a,b,c)}+l_{3n-2}^{(a,b,c)}+ac+a\hfill \\ & =& acb{l}_{3n-3}^{(a,b,c)}+ac{l}_{3n-4}^{(a,b,c)}+a{l}_{3n-3}^{(a,b,c)}+l_{3n-2}^{(a,b,c)}+abc+ac+a\hfill \\ & =& acb{l}_{3n-3}^{(a,b,c)}+c{l}_{3n-3}^{(a,b,c)}-c{l}_{3n-5}^{(a,b,c)}+a{l}_{3n-3}^{(a,b,c)}+l_{3n-2}^{(a,b,c)}+abc+a\hfill \\ & =& (abc+a+b+c)l_{3n-3}^{(a,b,c)}-(l_{3n-4}^{(a,b,c)}-l_{3n-6}^{(a,b,c)}-c)+l_{3n-4}^{(a,b,c)}+abc+a+b\hfill \\ & =& (abc+a+b+c)l_{3n-3}^{(a,b,c)}+l_{3n-6}^{(a,b,c)}+abc+a+b+c.\hfill \end{array}$$

For the case $n\equiv 1$ (mod 3), using (4), we have

$$\begin{array}{ccc}\hfill l_{3n+1}^{(a,b,c)}& =& b{l}_{3n}^{(a,b,c)}+l_{3n-1}^{(a,b,c)}+b\hfill \\ & =& ab{l}_{3n-1}^{(a,b,c)}+b{l}_{3n-2}^{(a,b,c)}+l_{3n-1}^{(a,b,c)}+ab+b\hfill \\ & =& abc{l}_{3n-2}^{(a,b,c)}+ab{l}_{3n-3}^{(a,b,c)}+b{l}_{3n-2}^{(a,b,c)}+l_{3n-1}^{(a,b,c)}+abc+ab+b\hfill \\ & =& abc{l}_{3n-2}^{(a,b,c)}+a{l}_{3n-2}^{(a,b,c)}-a{l}_{3n-4}^{(a,b,c)}+b{l}_{3n-2}^{(a,b,c)}+l_{3n-1}^{(a,b,c)}+abc+b\hfill \\ & =& (abc+a+b+c)l_{3n-2}^{(a,b,c)}-a{l}_{3n-4}^{(a,b,c)}+l_{3n-3}^{(a,b,c)}+abc+b+c\hfill \\ & =& (abc+a+b+c)l_{3n-2}^{(a,b,c)}+l_{3n-5}^{(a,b,c)}+abc+a+b+c.\hfill \end{array}$$

For the case $n\equiv 2$ (mod 3), using (4), we have

$$\begin{array}{ccc}\hfill l_{3n+2}^{(a,b,c)}& =& c{l}_{3n+1}^{(a,b,c)}+l_{3n}^{(a,b,c)}+c\hfill \\ & =& cb{l}_{3n}^{(a,b,c)}+c{l}_{3n-1}^{(a,b,c)}+l_{3n}^{(a,b,c)}+bc+c\hfill \\ & =& abc{l}_{3n-1}^{(a,b,c)}+cb{l}_{3n-2}^{(a,b,c)}+c{l}_{3n-1}^{(a,b,c)}+l_{3n}^{(a,b,c)}+abc+bc+c\hfill \\ & =& abc{l}_{3n-1}^{(a,b,c)}+b{l}_{3n-1}^{(a,b,c)}-b{l}_{3n-3}^{(a,b,c)}+c{l}_{3n-1}^{(a,b,c)}+l_{3n}^{(a,b,c)}+abc+c\hfill \\ & =& (abc+a+b+c)l_{3n-1}^{(a,b,c)}-(l_{3n-2}^{(a,b,c)}-l_{3n-5}^{(a,b,c)}-b)+l_{3n-2}^{(a,b,c)}+abc+a+c\hfill \\ & =& (abc+a+b+c)l_{3n-1}^{(a,b,c)}+l_{3n-5}^{(a,b,c)}+abc+a+b+c.\square \hfill \end{array}$$

The recurrence (9) and (10) hold the following characteristic equation:

$$x^6-\omega x^3-1=\left(x^3-\alpha \right)\left(x^3-\beta \right)=0$$

(11)

where $\alpha ={\displaystyle \frac{\omega +\sqrt{{\omega }^2+4}}{2}}$ and $\beta ={\displaystyle \frac{\omega -\sqrt{{\omega }^2+4}}{2}}.$

Theorem 3. 

Suppose that $F\left(z\right)={\sum }_{n=0}^{\infty }{F}_n^{(a,b,c)}{z}^n$ is the ordinary generating function of the tri-periodic Fibonacci sequence. Then $F\left(z\right)$ is given by

$$F\left(z\right)={\displaystyle \frac{F_0^{(a,b,c)}+F_1^{(a,b,c)}z+F_2^{(a,b,c)}{z}^2+(F_3^{(a,b,c)}-\omega F_0^{(a,b,c)})z^3+(F_4^{(a,b,c)}-\omega F_1^{(a,b,c)})z^4+(F_5^{(a,b,c)}-\omega F_2^{(a,b,c)})z^5}{1-\omega z^3-z^6}}.$$

Proof. Consider the generating function as

$$\begin{array}{ccc}\hfill F\left(z\right)& =& \sum _{n=0}^{\infty }{F}_n^{(a,b,c)}{z}^n\hfill \\ & =& \sum _{n=0}^{\infty }{F}_{3n}^{(a,b,c)}{z}^{3n}+\sum _{n=0}^{\infty }{F}_{3n+1}^{(a,b,c)}{z}^{3n+1}+\sum _{n=0}^{\infty }{F}_{3n+2}^{(a,b,c)}{z}^{3n+2}\hfill \\ & =& G\left(z\right)+H\left(z\right)+T\left(z\right).\hfill \end{array}$$

(12)

Hence, if we multiply $G\left(z\right)={\sum }_{n=0}^{\infty }{F}_{3n}^{(a,b,c)}{z}^{3n}$ with $\left(1-\omega z^3-z^6\right)$ and using (9), we obtain

$$\begin{array}{ccc}\hfill \left(1-\omega z^3-z^6\right)G\left(z\right)& =& \left(1-\omega z^3-z^6\right)\sum _{n=0}^{\infty }{F}_{3n}^{(a,b,c)}{z}^{3n}\hfill \\ & =& \sum _{n=0}^{\infty }{F}_{3n}^{(a,b,c)}{z}^{3n}-\omega \sum _{n=1}^{\infty }{F}_{3n-3}^{(a,b,c)}{z}^{3n}-\sum _{n=2}^{\infty }{F}_{3n-6}^{(a,b,c)}{z}^{3n}\hfill \\ & =& F_0^{(a,b,c)}+(F_3^{(a,b,c)}-\omega F_0^{(a,b,c)})z^3\hfill \\ & & +\sum _{n=2}^{\infty }\underset{\mathrm{equal}\mathrm{to}\mathrm{zero}}{\underbrace{(F_{3n}^{(a,b,c)}-\omega F_{3n-3}^{(a,b,c)}-F_{3n-6}^{(a,b,c)})}}{z}^{3n}.\hfill \end{array}$$

That means

$$G\left(z\right)={\displaystyle \frac{F_0^{(a,b,c)}+(F_3^{(a,b,c)}-\omega F_0^{(a,b,c)})z^3}{\left(1-\omega z^3-z^6\right)}}$$

Next, if we multiply $H\left(z\right)={\sum }_{n=0}^{\infty }{F}_{3n+1}^{(a,b,c)}{z}^{3n+1}$ with $\left(1-\omega z^3-z^6\right)$ and using (9), we obtain

$$\begin{array}{ccc}\hfill \left(1-\omega z^3-z^6\right)H\left(z\right)& =& \left(1-\omega z^3-z^6\right)\sum _{n=0}^{\infty }{F}_{3n+1}^{(a,b,c)}{z}^{3n+1}\hfill \\ & =& \sum _{n=0}^{\infty }{F}_{3n+1}^{(a,b,c)}{z}^{3n+1}-\omega \sum _{n=1}^{\infty }{F}_{3n-2}^{(a,b,c)}{z}^{3n+1}-\sum _{n=2}^{\infty }{F}_{3n-5}^{(a,b,c)}{z}^{3n+1}\hfill \\ & =& F_1^{(a,b,c)}z+(F_4^{(a,b,c)}-\omega F_1^{(a,b,c)})z^4\hfill \\ & & +\sum _{n=2}^{\infty }\underset{\mathrm{equal}\mathrm{to}\mathrm{zero}}{\underbrace{(F_{3n+1}^{(a,b,c)}-\omega F_{3n-2}^{(a,b,c)}-F_{3n-5}^{(a,b,c)})}}{z}^{3n+1}.\hfill \end{array}$$

That means

$$H\left(z\right)={\displaystyle \frac{F_1^{(a,b,c)}z+(F_4^{(a,b,c)}-\omega F_1^{(a,b,c)})z^4}{\left(1-\omega z^3-z^6\right)}}$$

Eventually, if we multiply $T\left(z\right)={\sum }_{n=0}^{\infty }{F}_{3n+2}^{(a,b,c)}{z}^{3n+2}$ with $\left(1-\omega z^3-z^6\right)$ and using (9), we obtain

$$\begin{array}{ccc}\hfill \left(1-\omega z^3-z^6\right)T\left(z\right)& =& \left(1-\omega z^3-z^6\right)\sum _{n=0}^{\infty }{F}_{3n+2}^{(a,b,c)}{z}^{3n+2}\hfill \\ & =& \sum _{n=0}^{\infty }{F}_{3n+2}^{(a,b,c)}{z}^{3n+2}-\omega \sum _{n=1}^{\infty }{F}_{3n-1}^{(a,b,c)}{z}^{3n+2}-\sum _{n=2}^{\infty }{F}_{3n-4}^{(a,b,c)}{z}^{3n+2}\hfill \\ & =& F_2^{(a,b,c)}{z}^2+(F_5^{(a,b,c)}-\omega F_2^{(a,b,c)})z^5\hfill \\ & & +\sum _{n=2}^{\infty }\underset{\mathrm{equal}\mathrm{to}\mathrm{zero}}{\underbrace{(F_{3n+2}^{(a,b,c)}-\omega F_{3n-1}^{(a,b,c)}-F_{3n-4}^{(a,b,c)})}}{z}^{3n+2}.\hfill \end{array}$$

That means

$$T\left(z\right)={\displaystyle \frac{F_2^{(a,b,c)}{z}^2+(F_5^{(a,b,c)}-\omega F_2^{(a,b,c)})z^5}{\left(1-\omega z^3-z^6\right)}}$$

So, if we substitute the expressions $G\left(z\right),$$H\left(z\right)$ and $T\left(z\right)$ into (12) we get the result.□

Theorem 4. 

Suppose that $l\left(z\right)={\sum }_{n=0}^{\infty }{l}_n^{(a,b,c)}{z}^n$ is the ordinary generating function of the tri-periodic Leonardo sequence. Then $l\left(z\right)$ is given by

$$\begin{array}{ccc}\hfill l\left(z\right)& =& {\displaystyle \frac{1}{(1-z)\left(1-\omega z^3-z^6\right)}}(l_0^{(a,b,c)}(1-z)+l_1^{(a,b,c)}z(1-z)+l_2^{(a,b,c)}{z}^2(1-z)\hfill \\ & & +(l_3^{(a,b,c)}-\omega l_0^{(a,b,c)})z^3(1-z)+(l_4^{(a,b,c)}-\omega l_1^{(a,b,c)})z^4(1-z)\hfill \\ & & +(l_5^{(a,b,c)}-\omega l_2^{(a,b,c)})z^5(1-z)+\omega z^6).\square \hfill \end{array}$$

Proof. Consider the generating function as

$$\begin{array}{ccc}\hfill l\left(z\right)& =& \sum _{n=0}^{\infty }{l}_n^{(a,b,c)}{z}^n\hfill \\ & =& \sum _{n=0}^{\infty }{l}_{3n}^{(a,b,c)}{z}^{3n}+\sum _{n=0}^{\infty }{l}_{3n+1}^{(a,b,c)}{z}^{3n+1}+\sum _{n=0}^{\infty }{l}_{3n+2}^{(a,b,c)}{z}^{3n+2}\hfill \\ & =& K\left(z\right)+M\left(z\right)+N\left(z\right).\hfill \end{array}$$

(13)

Hence, if we multiply $K\left(z\right)={\sum }_{n=0}^{\infty }{l}_{3n}^{(a,b,c)}{z}^{3n}$ with $\left(1-\omega z^3-z^6\right)$ and using (9), we obtain

$$\begin{array}{ccc}\hfill \left(1-\omega z^3-z^6\right)K\left(z\right)& =& \left(1-\omega z^3-z^6\right)\sum _{n=0}^{\infty }{l}_{3n}^{(a,b,c)}{z}^{3n}\hfill \\ & =& \sum _{n=0}^{\infty }{l}_{3n}^{(a,b,c)}{z}^{3n}-\omega \sum _{n=1}^{\infty }{l}_{3n-3}^{(a,b,c)}{z}^{3n}-\sum _{n=2}^{\infty }{l}_{3n-6}^{(a,b,c)}{z}^{3n}\hfill \\ & =& l_0^{(a,b,c)}+(l_3^{(a,b,c)}-\omega l_0^{(a,b,c)})z^3+\sum _{n=2}^{\infty }\underset{\mathrm{equal}\mathrm{to}\omega }{\underbrace{(F_{3n}^{(a,b,c)}-\omega F_{3n-3}^{(a,b,c)}-F_{3n-6}^{(a,b,c)})}}{z}^{3n}\hfill \\ & =& l_0^{(a,b,c)}+(l_3^{(a,b,c)}-\omega l_0^{(a,b,c)})z^3+\omega ({\displaystyle \frac{1}{1-z^3}}-1-z^3)\hfill \\ & =& {\displaystyle \frac{l_0^{(a,b,c)}(1-z^3)+(l_3^{(a,b,c)}-\omega l_0^{(a,b,c)})z^3(1-z^3)+\omega z^6}{1-z^3}}.\hfill \end{array}$$

That means

$$K\left(z\right)={\displaystyle \frac{l_0^{(a,b,c)}(1-z^3)+(l_3^{(a,b,c)}-\omega l_0^{(a,b,c)})z^3(1-z^3)+\omega z^6}{\left(1-\omega z^3-z^6\right)(1-z^3)}}$$

Next, if we multiply $M\left(z\right)={\sum }_{n=0}^{\infty }{F}_{3n+1}^{(a,b,c)}{z}^{3n+1}$ with $\left(1-\omega z^3-z^6\right)$ and using (9), we obtain

$$\begin{array}{ccc}\hfill \left(1-\omega z^3-z^6\right)M\left(z\right)& =& \left(1-\omega z^3-z^6\right)\sum _{n=0}^{\infty }{l}_{3n+1}^{(a,b,c)}{z}^{3n+1}\hfill \\ & =& \sum _{n=0}^{\infty }{l}_{3n+1}^{(a,b,c)}{z}^{3n+1}-\omega \sum _{n=1}^{\infty }{l}_{3n-2}^{(a,b,c)}{z}^{3n+1}-\sum _{n=2}^{\infty }{l}_{3n-5}^{(a,b,c)}{z}^{3n+1}\hfill \\ & =& l_1^{(a,b,c)}z+(l_4^{(a,b,c)}-\omega l_1^{(a,b,c)})z^4+\sum _{n=2}^{\infty }\underset{\mathrm{equal}\mathrm{to}\omega }{\underbrace{(l_{3n+1}^{(a,b,c)}-\omega l_{3n-2}^{(a,b,c)}-l_{3n-5}^{(a,b,c)})}}{z}^{3n+1}\hfill \\ & =& l_1^{(a,b,c)}z+(l_4^{(a,b,c)}-\omega l_1^{(a,b,c)})z^4+\omega z({\displaystyle \frac{1}{1-z^3}}-1-z^3)\hfill \\ & =& {\displaystyle \frac{l_1^{(a,b,c)}z(1-z^3)+(l_4^{(a,b,c)}-\omega l_1^{(a,b,c)})z^4(1-z^3)+\omega z^7}{1-z^3}}.\hfill \end{array}$$

That means

$$M\left(z\right)={\displaystyle \frac{l_1^{(a,b,c)}z(1-z^3)+(l_4^{(a,b,c)}-\omega l_1^{(a,b,c)})z^4(1-z^3)+\omega z^7}{(1-z^3)\left(1-\omega z^3-z^6\right)}}$$

Eventually, if we multiply $N\left(z\right)={\sum }_{n=0}^{\infty }{F}_{3n+2}^{(a,b,c)}{z}^{3n+2}$ with $\left(1-\omega z^3-z^6\right)$ and using (9), we obtain

$$\begin{array}{ccc}\hfill \left(1-\omega z^3-z^6\right)N\left(z\right)& =& \left(1-\omega z^3-z^6\right)\sum _{n=0}^{\infty }{l}_{3n+2}^{(a,b,c)}{z}^{3n+2}\hfill \\ & =& \sum _{n=0}^{\infty }{l}_{3n+2}^{(a,b,c)}{z}^{3n+2}-\omega \sum _{n=1}^{\infty }{l}_{3n-1}^{(a,b,c)}{z}^{3n+2}-\sum _{n=2}^{\infty }{l}_{3n-4}^{(a,b,c)}{z}^{3n+2}\hfill \\ & =& l_2^{(a,b,c)}{z}^2+(l_5^{(a,b,c)}-\omega l_2^{(a,b,c)})z^5+\sum _{n=2}^{\infty }\underset{\mathrm{equal}\mathrm{to}\omega }{\underbrace{(l_{3n+2}^{(a,b,c)}-\omega l_{3n-1}^{(a,b,c)}-l_{3n-4}^{(a,b,c)})}}{z}^{3n+2}\hfill \\ & =& l_2^{(a,b,c)}{z}^2+(l_5^{(a,b,c)}-\omega l_2^{(a,b,c)})z^5+\omega z^2({\displaystyle \frac{1}{1-z^3}}-1-z^3)\hfill \\ & =& {\displaystyle \frac{l_2^{(a,b,c)}{z}^2(1-z^3)+(l_5^{(a,b,c)}-\omega l_2^{(a,b,c)})z^5(1-z^3)+\omega z^8}{1-z^3}}.\hfill \end{array}$$

That means

$$T\left(z\right)={\displaystyle \frac{l_2^{(a,b,c)}{z}^2(1-z^3)+(l_5^{(a,b,c)}-\omega l_2^{(a,b,c)})z^5(1-z^3)+\omega z^8}{(1-z^3)\left(1-\omega z^3-z^6\right)}}$$

So, if we substitute the expressions $K\left(z\right),$$M\left(z\right)$ and $N\left(z\right)$ into (12), we obtain

$$\begin{array}{ccc}\hfill l\left(z\right)& =& {\displaystyle \frac{1}{(1-z)\left(1-\omega z^3-z^6\right)}}(l_0^{(a,b,c)}(1-z)+l_1^{(a,b,c)}z(1-z)+l_2^{(a,b,c)}{z}^2(1-z)\hfill \\ & & +(l_3^{(a,b,c)}-\omega l_0^{(a,b,c)})z^3(1-z)+(l_4^{(a,b,c)}-\omega l_1^{(a,b,c)})z^4(1-z)\hfill \\ & & +(l_5^{(a,b,c)}-\omega l_2^{(a,b,c)})z^5(1-z)+\omega z^6).\square \hfill \end{array}$$

Note that the following equality is true by using Maclaurin series expansion

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{1-d{z}^3}}& =& \sum _{n=0}^{\infty }{d}^n{z}^{3n},\left|d{z}^3\le 1\right|,\hfill \end{array}$$

(14)

$$\begin{array}{ccc}\hfill {\displaystyle \frac{z}{1-d{z}^3}}& =& \sum _{n=0}^{\infty }{d}^n{z}^{3n+1},\left|d{z}^3\le 1\right|,\hfill \end{array}$$

(15)

$$\begin{array}{ccc}\hfill {\displaystyle \frac{z^2}{1-d{z}^3}}& =& \sum _{n=0}^{\infty }{d}^n{z}^{3n+2},\left|d{z}^3\le 1\right|.\hfill \end{array}$$

(16)

Theorem 5. 

The Binet’s formula of the tri-periodic Fibonacci sequence is given by

$$F_n^{(a,b,c)}={\tilde{F}}_0{\delta }_0\left(n\right)+{\tilde{F}}_1{\delta }_1\left(n\right)+{\tilde{F}}_2{\delta }_2\left(n\right).$$

(17)

where

$$\begin{array}{ccc}\hfill {\tilde{F}}_0& =& {\displaystyle \frac{(1+ac)({\alpha }^{\frac{n}{3}}-{\beta }^{\frac{n}{3}})}{\alpha -\beta }},\hfill \\ \hfill {\tilde{F}}_1& =& {\displaystyle \frac{({\alpha }^{\frac{n-1}{3}+1}-{\beta }^{\frac{n-1}{3}+1})-a({\alpha }^{\frac{n-1}{3}}-{\beta }^{\frac{n-1}{3}})}{\alpha -\beta }},\hfill \\ \hfill {\tilde{F}}_2& =& {\displaystyle \frac{c({\alpha }^{\frac{n-2}{3}+1}-{\beta }^{\frac{n-2}{3}+1})+({\alpha }^{\frac{n-2}{3}}-{\beta }^{\frac{n-2}{3}})}{\alpha -\beta }}.\hfill \end{array}$$

Proof. Using Theorem [28], (14)-(16) and (11), we obtain

$$\begin{array}{ccc}\hfill F\left(z\right)& =& \frac{F_0^{(a,b,c)}+F_1^{(a,b,c)}z+F_2^{(a,b,c)}{z}^2+(F_3^{(a,b,c)}-\omega F_0^{(a,b,c)})z^3+(F_4^{(a,b,c)}-\omega F_1^{(a,b,c)})z^4+(F_5^{(a,b,c)}-\omega F_2^{(a,b,c)})z^5}{1-\omega z^3-z^6}\hfill \\ & =& {\displaystyle \frac{r_1{z}^2+r_{2}z+r_3}{\left(1-\alpha z^3\right)}}+{\displaystyle \frac{r_4{z}^2+r_{5}z+r_6}{\left(1-\beta z^3\right)}}\hfill \\ & =& \sum _{n=0}^{\infty }{r}_1{\alpha }^n{z}^{3n+2}+\sum _{n=0}^{\infty }{r}_2{\alpha }^n{z}^{3n+1}+\sum _{n=0}^{\infty }{r}_3{\alpha }^n{z}^{3n}+\sum _{n=0}^{\infty }{r}_4{\beta }^n{z}^{3n+2}+\sum _{n=0}^{\infty }{r}_5{\beta }^n{z}^{3n+1}+\sum _{n=0}^{\infty }{r}_6{\beta }^n{z}^{3n}.\hfill \end{array}$$

(18)

Hence, we obtain

$F_0^{(a,b,c)}+F_1^{(a,b,c)}z+F_2^{(a,b,c)}{z}^2+(F_3^{(a,b,c)}-\omega F_0^{(a,b,c)})z^3+(F_4^{(a,b,c)}-\omega F_1^{(a,b,c)})z^4+(F_5^{(a,b,c)}-\omega F_3^{(a,b,c)})z^5=\left(r_3+r_6\right)+$$z\left(r_2+r_5\right)+$$z^2\left(r_1+r_4\right)-z^3\left(\beta r_3+\alpha r_6\right)-z^4\left(\beta r_2+\alpha r_5\right)-z^5\left(\beta r_1+\alpha r_4\right).$

From this, we have the following equalities:

$$\begin{array}{ccc}\hfill r_3+r_6& =& F_0^{(a,b,c)}\hfill \\ \hfill r_2+r_5& =& F_1^{(a,b,c)}\hfill \\ \hfill r_1+r_4& =& F_2^{(a,b,c)}\hfill \\ \hfill -\beta r_3-\alpha r_6& =& F_3^{(a,b,c)}-\omega F_0^{(a,b,c)}\hfill \\ \hfill -\beta r_2-\alpha r_5& =& F_4^{(a,b,c)}-\omega F_1^{(a,b,c)}\hfill \\ \hfill -\beta r_1-\alpha r_4& =& F_5^{(a,b,c)}-\omega F_2^{(a,b,c)}\hfill \end{array}$$

(19)

Hence, we have

$$\left(\begin{array}{cccccc}0& 0& 1& 0& 0& 1\\ 0& 1& 0& 0& 1& 0\\ 1& 0& 0& 1& 0& 0\\ 0& 0& -\beta & 0& 0& -\alpha \\ 0& -\beta & 0& 0& -\alpha & 0\\ -\beta & 0& 0& -\alpha & 0& 0\end{array}\right)\left(\begin{array}{c}{r}_1\\ r_2\\ r_3\\ r_4\\ r_5\\ r_6\end{array}\right)=\left(\begin{array}{c}{F}_0^{(a,b,c)}\\ F_1^{(a,b,c)}\\ F_2^{(a,b,c)}\\ F_3^{(a,b,c)}-\omega F_0^{(a,b,c)}\\ F_4^{(a,b,c)}-\omega F_1^{(a,b,c)}\\ F_5^{(a,b,c)}-\omega F_3^{(a,b,c)}\end{array}\right).$$

and

$$\begin{array}{ccc}\hfill r_1& =& {\displaystyle \frac{1}{\alpha -\beta }}\left(F_2^{(a,b,c)}\alpha +F_5^{(a,b,c)}-\omega F_2^{(a,b,c)}\right),\hfill \\ \hfill r_2& =& {\displaystyle \frac{1}{\alpha -\beta }}\left(F_1^{(a,b,c)}\alpha +F_4^{(a,b,c)}-\omega F_1^{(a,b,c)}\right),\hfill \\ \hfill r_3& =& {\displaystyle \frac{1}{\alpha -\beta }}\left(F_0^{(a,b,c)}\alpha +F_3^{(a,b,c)}-\omega F_0^{(a,b,c)}\right),\hfill \\ \hfill r_4& =& -{\displaystyle \frac{1}{\alpha -\beta }}\left(F_2^{(a,b,c)}\beta +F_5^{(a,b,c)}-\omega F_2^{(a,b,c)}\right),\hfill \\ \hfill r_5& =& -{\displaystyle \frac{1}{\alpha -\beta }}\left(F_1^{(a,b,c)}\beta +F_4^{(a,b,c)}-\omega F_1^{(a,b,c)}\right),\hfill \\ \hfill r_6& =& -{\displaystyle \frac{1}{\alpha -\beta }}\left(F_0^{(a,b,c)}\beta +F_3^{(a,b,c)}-\omega F_0^{(a,b,c)}\right).\hfill \end{array}$$

Thus, (18) can be written as

$$\begin{array}{ccc}\hfill F\left(z\right)& =& \sum _{n=0}^{\infty }{\displaystyle \frac{1}{\alpha -\beta }}\left(F_0^{(a,b,c)}\alpha +F_3^{(a,b,c)}-\omega F_0^{(a,b,c)}\right){\alpha }^n{z}^{3n}-\sum _{n=0}^{\infty }{\displaystyle \frac{1}{\alpha -\beta }}\left(F_0^{(a,b,c)}\beta +F_3^{(a,b,c)}-\omega F_0^{(a,b,c)}\right){\beta }^n{z}^{3n}\hfill \\ & & +\sum _{n=0}^{\infty }{\displaystyle \frac{1}{\alpha -\beta }}\left(F_1^{(a,b,c)}\alpha +F_4^{(a,b,c)}-\omega F_1^{(a,b,c)}\right){\alpha }^n{z}^{3n+1}-\sum _{n=0}^{\infty }{\displaystyle \frac{1}{\alpha -\beta }}\left(F_1^{(a,b,c)}\beta +F_4^{(a,b,c)}-\omega F_1^{(a,b,c)}\right){\beta }^n{z}^{3n+1}\hfill \\ & & +\sum _{n=0}^{\infty }{\displaystyle \frac{1}{\alpha -\beta }}\left(F_2^{(a,b,c)}\alpha +F_5^{(a,b,c)}-\omega F_2^{(a,b,c)}\right){\alpha }^n{z}^{3n+2}-\sum _{n=0}^{\infty }{\displaystyle \frac{1}{\alpha -\beta }}\left(F_2^{(a,b,c)}\beta +F_5^{(a,b,c)}-\omega F_2^{(a,b,c)}\right){\beta }^n{z}^{3n+2}.\hfill \end{array}$$

So, using (9), we have

$$\begin{array}{ccc}\hfill F\left(z\right)& =& \sum _{n=0}^{\infty }{\displaystyle \frac{1}{\alpha -\beta }}\left(F_0^{(a,b,c)}({\alpha }^{n+1}-{\beta }^{n+1})+(F_3^{(a,b,c)}-\omega F_0^{(a,b,c)})({\alpha }^n-{\beta }^n)\right)z^{3n}\hfill \\ & & +\sum _{n=0}^{\infty }{\displaystyle \frac{1}{\alpha -\beta }}\left(F_1^{(a,b,c)}({\alpha }^{n+1}-{\beta }^{n+1})+(F_4^{(a,b,c)}-\omega F_1^{(a,b,c)})({\alpha }^n-{\beta }^n)\right)z^{3n+1}\hfill \\ & & +\sum _{n=0}^{\infty }{\displaystyle \frac{1}{\alpha -\beta }}\left(F_2^{(a,b,c)}({\alpha }^{n+1}-{\beta }^{n+1})+(F_5^{(a,b,c)}-\omega F_2^{(a,b,c)})({\alpha }^n-{\beta }^n)\right)z^{3n+2}.\hfill \end{array}$$

This implies that

$$F_n^{(a,b,c)}=\left\{\begin{array}{c}{\displaystyle \frac{F_0^{(a,b,c)}({\alpha }^{k+1}-{\beta }^{k+1})+(F_3^{(a,b,c)}-\omega F_0^{(a,b,c)})({\alpha }^k-{\beta }^k)}{\alpha -\beta }}\\ {\displaystyle \frac{F_1^{(a,b,c)}({\alpha }^{k+1}-{\beta }^{k+1})+(F_4^{(a,b,c)}-\omega F_1^{(a,b,c)})({\alpha }^k-{\beta }^k)}{\alpha -\beta }}\\ {\displaystyle \frac{F_2^{(a,b,c)}({\alpha }^{k+1}-{\beta }^{k+1})+(F_5^{(a,b,c)}-\omega F_2^{(a,b,c)})({\alpha }^k-{\beta }^k)}{\alpha -\beta }}\end{array}\right.\begin{array}{c}n=3k\\ n\equiv 3k+1\\ n\equiv 3k+2\end{array},k\in \mathbb{Z}$$

Using (9), the above equality can be written as

$$\begin{array}{ccc}\hfill F_n^{(a,b,c)}& =& \left\{\begin{array}{c}{\displaystyle \frac{F_0^{(a,b,c)}({\alpha }^{\frac{n}{3}+1}-{\beta }^{\frac{n}{3}+1})+F_{-3}^{(a,b,c)}({\alpha }^{\frac{n}{3}}-{\beta }^{\frac{n}{3}})}{\alpha -\beta }}\\ {\displaystyle \frac{F_1^{(a,b,c)}({\alpha }^{\frac{n-1}{3}+1}-{\beta }^{\frac{n-1}{3}+1})+F_{-2}^{(a,b,c)}({\alpha }^{\frac{n-1}{3}}-{\beta }^{\frac{n-1}{3}})}{\alpha -\beta }}\\ {\displaystyle \frac{F_2^{(a,b,c)}({\alpha }^{\frac{n-2}{3}+1}-{\beta }^{\frac{n-2}{3}+1})+F_{-1}^{(a,b,c)}({\alpha }^{\frac{n-2}{3}}-{\beta }^{\frac{n-2}{3}})}{\alpha -\beta }}\end{array}\right.\begin{array}{c}n\equiv 0(\mathrm{mod}3)\\ n\equiv 1(\mathrm{mod}3)\\ n\equiv 2(\mathrm{mod}3)\end{array}n\ge 3,\hfill \\ & =& {\tilde{F}}_0{\delta }_0\left(n\right)+{\tilde{F}}_1{\delta }_1\left(n\right)+{\tilde{F}}_2{\delta }_2\left(n\right).\square \hfill \end{array}$$

Theorem 6 

(Cassani’s identity). For $n\ge 0$ we have following equalities.

(a)

For $n\equiv 0$ (mod 3)

$$F_{n-1}^{(a,b,c)}{F}_{n+1}^{(a,b,c)}-{\left(F_n^{(a,b,c)}\right)}^2={\displaystyle \frac{1}{{\omega }^2+4}}(A_0\sqrt[3]{{\alpha }^{2n}}+B_0\sqrt[3]{{\beta }^{2n}}+C_0{(-1)}^n)$$

where

$$\begin{array}{ccc}\hfill A_0& =& {\displaystyle \frac{(1+\alpha F_2^{(a,b,c)})(\alpha +F_{-2}^{(a,b,c)})}{\alpha }}-{\left(F_3^{(a,b,c)}\right)}^2\hfill \\ \hfill B_0& =& {\displaystyle \frac{(1+\beta F_2^{(a,b,c)})(\beta +F_{-2}^{(a,b,c)})}{\beta }}-{\left(F_3^{(a,b,c)}\right)}^2\hfill \\ \hfill C_0& =& {\omega }^2+2\omega -2{\left(F_3^{(a,b,c)}\right)}^2.\hfill \end{array}$$

(b)

For $n\equiv 1$ (mod 3)

$$F_{n-1}^{(a,b,c)}{F}_{n+1}^{(a,b,c)}-{\left(F_n^{(a,b,c)}\right)}^2={\displaystyle \frac{1}{{\omega }^2+4}}(A_1\sqrt[3]{{\alpha }^{2n-2}}+B_1\sqrt[3]{{\beta }^{2n-2}}+C_1{(-1)}^n)$$

where

$$\begin{array}{ccc}\hfill A_1& =& F_3^{(a,b,c)}(1+F_2^{(a,b,c)}\alpha )-{(\alpha -F_2^{(a,b,c)})}^2\hfill \\ \hfill B_1& =& F_3^{(a,b,c)}(1+F_2^{(a,b,c)}\beta )-{(\beta -F_2^{(a,b,c)})}^2\hfill \\ \hfill C_1& =& .F_3^{(a,b,c)}(-F_2^{(a,b,c)}\omega -2)+2F_{-2}^{(a,b,c)}-2{\left(F_{-2}^{(a,b,c)}\right)}^2.\hfill \end{array}$$

(c)

For $n\equiv 2$ (mod 3)

$$F_{n-1}^{(a,b,c)}{F}_{n+1}^{(a,b,c)}-{\left(F_n^{(a,b,c)}\right)}^2={\displaystyle \frac{1}{{\omega }^2+4}}(A_2\sqrt[3]{{\alpha }^{2n-4}}+B_2\sqrt[3]{{\beta }^{2n-4}}+C_2{(-1)}^n$$

where

$$\begin{array}{ccc}\hfill A_2& =& F_3^{(a,b,c)}({\alpha }^2+F_{-2}^{(a,b,c)}\alpha )-{(F_2^{(a,b,c)}+\alpha )}^2\hfill \\ \hfill B_1& =& F_3^{(a,b,c)}({\beta }^2+F_{-2}^{(a,b,c)}\beta )-{(F_2^{(a,b,c)}+\beta )}^2\hfill \\ \hfill C_1& =& F_3^{(a,b,c)}(2+F_{-2}^{(a,b,c)}\omega )-2F_2^{(a,b,c)}\left(F_2^{(a,b,c)}+\omega \right)-2.\hfill \end{array}$$

Proof. (a) For the case, $n=3k,$$k\in \mathbb{Z}$ and using Theorem (5)

$$F_{n-1}^{(a,b,c)}{F}_{n+1}^{(a,b,c)}-{\left(F_n^{(a,b,c)}\right)}^2=F_{3k-1}^{(a,b,c)}{F}_{3k+1}^{(a,b,c)}-{\left(F_{3k}^{(a,b,c)}\right)}^2.$$

Hence, we have

$$\begin{array}{ccc}\hfill F_{3k-1}^{(a,b,c)}{F}_{3k+1}^{(a,b,c)}& =& {\displaystyle \frac{c({\alpha }^k-{\beta }^k)+({\alpha }^{k-1}-{\beta }^{k-1})}{\alpha -\beta }}{\displaystyle \frac{({\alpha }^{k+1}-{\beta }^{k+1})-a({\alpha }^k-{\beta }^k)}{\alpha -\beta }}\hfill \\ & =& {\displaystyle \frac{1}{{\left(\alpha -\beta \right)}^2}}({\alpha }^{2k-1}(1+c\alpha )(\alpha -a)+{\beta }^{2k-1}(1+c\beta )(\alpha -\beta )\hfill \\ & & +{(-1)}^k({\omega }^2+2\omega ).\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill {\left(F_{3k}^{(a,b,c)}\right)}^2& =& {\left({\displaystyle \frac{(1+ac)({\alpha }^k-{\beta }^k)}{\alpha -\beta }}\right)}^2\hfill \\ & =& {(1+ac)}^2{\alpha }^{2k}+(1+ac){\beta }^{2k}-2{(-1)}^k{(1+ac)}^2.\hfill \end{array}$$

Therefore, we obtain

$$F_{n-1}^{(a,b,c)}{F}_{n+1}^{(a,b,c)}-{\left(F_n^{(a,b,c)}\right)}^2={\displaystyle \frac{1}{{\omega }^2+4}}(A_0\sqrt[3]{{\alpha }^{2n}}+B_0\sqrt[3]{{\beta }^{2n}}+C_0)$$

(b)For the case, $n=3k+1,$$k\in \mathbb{Z}$ and using Theorem (5)

$$F_{n-1}^{(a,b,c)}{F}_{n+1}^{(a,b,c)}-{\left(F_n^{(a,b,c)}\right)}^2=F_{3k}^{(a,b,c)}{F}_{3k+2}^{(a,b,c)}-{\left(F_{3k+1}^{(a,b,c)}\right)}^2.$$

Hence, we have

$$\begin{array}{ccc}\hfill F_{3k}^{(a,b,c)}{F}_{3k+2}^{(a,b,c)}& =& {\displaystyle \frac{(1+ac)({\alpha }^k-{\beta }^k)}{\alpha -\beta }}{\displaystyle \frac{c({\alpha }^{k+1}-{\beta }^{k+1})+({\alpha }^k-{\beta }^k)}{\alpha -\beta }}\hfill \\ & =& {\displaystyle \frac{1}{{\left(\alpha -\beta \right)}^2}}({\alpha }^{2k}(1+ac)(1+c\alpha )+{\beta }^{2k}(1+ac)(1+c\beta )\hfill \\ & & +{(-1)}^k\left((1+ac)(-c\omega -2)\right))\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill {\left(F_{3k+1}^{(a,b,c)}\right)}^2& =& {\left({\displaystyle \frac{({\alpha }^{k+1}-{\beta }^{k+1})-a({\alpha }^k-{\beta }^k)}{\alpha -\beta }}\right)}^2\hfill \\ & =& {\displaystyle \frac{1}{{\left(\alpha -\beta \right)}^2}}({\alpha }^{2k}{(\alpha -a)}^2+{\beta }^{2k}{(\beta -a)}^2-{(-1)}^k(2+2a\omega +2a^2))\hfill \end{array}$$

Therefore, we obtain

$$F_{3k}^{(a,b,c)}{F}_{3k+2}^{(a,b,c)}-{\left(F_{3k+1}^{(a,b,c)}\right)}^2={\displaystyle \frac{1}{{\omega }^2+4}}(A_1\sqrt[3]{{\alpha }^{2n-2}}+B_1\sqrt[3]{{\beta }^{2n-2}}+C_1{(-1)}^n)$$

(c)For the case, $n=3k+2,$$k\in \mathbb{Z}$ and using Theorem (5)

$$F_{n-1}^{(a,b,c)}{F}_{n+1}^{(a,b,c)}-{\left(F_n^{(a,b,c)}\right)}^2=F_{3k+1}^{(a,b,c)}{F}_{3k+3}^{(a,b,c)}-{\left(F_{3k+2}^{(a,b,c)}\right)}^2.$$

Hence, we have

$$\begin{array}{ccc}\hfill F_{3k+1}^{(a,b,c)}{F}_{3k+3}^{(a,b,c)}& =& {\displaystyle \frac{({\alpha }^{k+1}-{\beta }^{k+1})-a({\alpha }^k-{\beta }^k)}{\alpha -\beta }}{\displaystyle \frac{(1+ac)({\alpha }^{k+1}-{\beta }^{k+1})}{\alpha -\beta }}\hfill \\ & =& {\displaystyle \frac{1}{{\left(\alpha -\beta \right)}^2}}({\alpha }^{2k}((1+ac)({\alpha }^2-a\alpha )+{\beta }^{2k}((1+ac)({\beta }^2-a\beta )\hfill \\ & & +{(-1)}^k\left((1+ac)(2-a\omega )\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill {\left(F_{3k+2}^{(a,b,c)}\right)}^2& =& {\displaystyle \frac{1}{{\left(\alpha -\beta \right)}^2}}({\alpha }^{2k}{(c+\alpha )}^2+{\beta }^{2k}{(c+\beta )}^2\hfill \\ & & +{(-1)}^k(2c^2-2c\omega -2))\hfill \end{array}$$

Hence, we have

$$F_{3k+1}^{(a,b,c)}{F}_{3k+3}^{(a,b,c)}-{\left(F_{3k+2}^{(a,b,c)}\right)}^2={\displaystyle \frac{1}{{\omega }^2+4}}(A_3\sqrt[3]{{\alpha }^{2n-4}}+B_3\sqrt[3]{{\beta }^{2n-4}}+C_3{(-1)}^n.$$