A Grammatical Interpretation of Horadam Sequences

Abstract

The Horadam sequence ${\left\{H_n(a,b;p,q)\right\}}_{n⩾0}$ has been widely studied in combinatorics and number theory. In this paper, we find that the context-free grammar $G=\{x\to px+y,y\to qx\}$ can be used to generate Horadam sequences. Using this grammar, we deduce several identities, including Cassini-like identities. Moreover, we investigate the relationship between two distinct Horadam sequences $H_n(a,b;p,q)$ and $H_n(c,d;p,q)$ with $(a,b)\ne (c,d)$ and provide an approach to derive identities, which can be illustrated by the Fibonacci and Lucas sequences as well as the two kinds of Chebyshev polynomials.

1. Introduction

Setting $H_0=a$ and $H_1=b$, the linear recurrence relation

$$H_n=p{H}_{n-1}+q{H}_{n-2}$$

(1)

generates a sequence ${\left\{H_n(a,b;p,q)\right\}}_{n⩾0}$, which can be called the $Horadam sequence$ since Horadam initiated the study in 1961 [1]. It unifies a number of well-known sequences, see Table 1 for special cases. Recently, there have been many extensions and variations of the Horadam sequence, see [2,3,4,5,6,7,8,9,10,11]. For the sake of brevity, we shall denote $H_n(a,b;p,q)$ by $H_n$.

Table 1. Some special cases of Horadam sequence.

Name	Parameter Representation
Fibonacci number	$H_n(0,1;1,1)$
Lucas number	$H_n(2,1;1,1)$
Pell number	$H_n(0,1;2,1)$
Pell–Lucas number	$H_n(2,2;2,1)$
Jacobsthal number	$H_n(0,1;1,2)$
Jacobsthal–Lucas number	$H_n(2,1;1,2)$
Fermat number	$H_n(0,1;3,-2)$
Fermat–Lucas number	$H_n(2,3;3,-2)$

In this paper, we find that the context-free grammar can be used to generate the Horadam sequence. The grammatical method was introduced by Chen [12] when he investigated exponential structures in combinatorics. The context-free grammar is a formal language that can be used to discover new combinatorial models [13,14], new combinatorial algorithms [7], and new convolution formulas [14,15]. Moreover, using the changing of the grammars, different expansions and novel relationships can also be discovered [7,14,15]. Recent studies have demonstrated broad applications of context-free grammars in combinatorics, see [16,17,18,19,20] for instance.

The paper is organized as follows: In Section 2, we present a grammatical description of the Horadam sequence. In Section 3, as applications of the Horadam grammar, we establish some identities associated with Horadam numbers and investigate the relationship between two distinct Horadam numbers $H_n(a,b;p,q)$ and $H_n(c,d;p,q)$ with $(a,b)\ne (c,d)$.

2. The Horadam Grammar

Let A be an alphabet with letters as independent commutative indeterminates. Following Chen [12], a context-free grammar G over A consists of substitution rules replacing letters in A by formal functions over A. We can define a formal derivative $D_G$ associated with G as a differential operator on formal functions over A. More precisely, $D_G$ is a linear operator satisfying the relations

$$D_G(x+y)=D_G\left(x\right)+D_G\left(y\right),D_G\left(xy\right)=D_G\left(x\right)y+x{D}_G\left(y\right),$$

and the Leibnitz formula

$$D_G^n\left(xy\right)=\sum _{k=0}^{\infty }\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)D_G^k\left(x\right)D_G^{n-k}\left(y\right).$$

In the following, we give an illustration of the grammar:

$$G_0=\{x\to x+y,y\to x\}.$$

(2)

Consider the Fibonacci sequence defined by $F_n=F_{n-1}+F_{n-2}$, $F_0=0,F_1=1$. Suppose we are given a row of squares, with two types of tiles: dominos, which can cover two squares, and monominos, which can cover one square. A tiling of the row is a set of tiles that covers each square exactly once. It is well known that for $n\ge 2$, $F_n$ is the number of tilings of a row of $n-1$ squares [19]. We distinguish two cases: $\left(i\right)$ if the tiling ends with a monomino, then we mark it by a x; $\left(ii\right)$ if the tiling ends with a domino, then we mark it by a y. It is easy to verify that $D_{G_0}^n\left(x\right)=x{F}_{n+1}+y{F}_n$. For example, $D_{G_0}^2\left(x\right)=2x+y$ and $D_{G_0}^3\left(x\right)=3x+2y$. Motivated by the above discussion, we can now present the main result of the paper.

Theorem 1. 

We call $G=\{x\to px+y,y\to qx\}$ the Horadam grammar. We have

$$D_G^n(bx+ay)=H_{n+1}x+H_{n}y.$$

In particular, $D_G^n{(bx+ay)|}_{x=0,y=1}=H_n$.

Proof. 

Note that

$$D_G(bx+ay)=b{D}_G\left(x\right)+a{D}_G\left(y\right)=b(px+y)+aqx=(bp+aq)x+by=H_{2}x+H_{1}y.$$

So, the result holds for $n=1$. By induction, assume that the result holds for $n=m-1$. Then,

$$\begin{array}{cc}\hfill D_G^m(bx+ay)& =D_G\left(D_G^{m-1}(bx+ay)\right)\hfill \\ & =D_G(H_{m}x+H_{m-1}y)\hfill \\ & =H_m{D}_G\left(x\right)+H_{m-1}{D}_G\left(y\right)\hfill \\ & =H_m(px+y)+H_{m-1}\left(qx\right)\hfill \\ & =(p{H}_m+q{H}_{m-1})x+H_{m}y\hfill \\ & =H_{m+1}x+H_{m}y,\hfill \end{array}$$

as desired. This completes the proof. □

3. Applications of the Horadam Grammar

3.1. Explicit Formulas and Identities

As an application of Horadam grammar, we first deduce a known formula (see (Theorem 37) in [9]) of the Horadam sequence. We adopt a grammar-based proof, which is more concise.

Theorem 2. 

For $n>1$,

$$H_n=b\sum _{i=0}^{\lfloor {\displaystyle \frac{n-1}{2}}\rfloor }\left({\displaystyle \genfrac{}{}{0pt}{}{n-1-i}{i}}\right)p^{n-1-2i}{q}^i+a\sum _{i=0}^{\lfloor {\displaystyle \frac{n-2}{2}}\rfloor }\left({\displaystyle \genfrac{}{}{0pt}{}{n-2-i}{i}}\right)p^{n-2-2i}{q}^{i+1}.$$

Proof. 

Note that

$$\begin{array}{cc}\hfill & D_G\left(x\right)=px+y,\hfill \\ \hfill & D_G^2\left(x\right)=(p^2+q)x+py,\hfill \\ \hfill & D_G^3\left(x\right)=(p^3+2pq)x+(p^2+q)y.\hfill \end{array}$$

By introduction, it is routine to verify that

$$D_G^n\left(x\right)=\sum _{i=0}^{\lfloor {\displaystyle \frac{n}{2}}\rfloor }\left({\displaystyle \genfrac{}{}{0pt}{}{n-i}{i}}\right)p^{n-2i}{q}^{i}x+\sum _{i=0}^{\lfloor {\displaystyle \frac{n-1}{2}}\rfloor }\left({\displaystyle \genfrac{}{}{0pt}{}{n-1-i}{i}}\right)p^{n-1-2i}{q}^{i}y,$$

and, similarly,

$$\begin{array}{cc}\hfill D_G^n\left(y\right)& =q{D}_G^{n-1}\left(x\right)\hfill \\ \hfill & =\sum _{i=0}^{\lfloor {\displaystyle \frac{n-1}{2}}\rfloor }\left({\displaystyle \genfrac{}{}{0pt}{}{n-1-i}{i}}\right)p^{n-1-2i}{q}^{i+1}x+\sum _{i=0}^{\lfloor {\displaystyle \frac{n-2}{2}}\rfloor }\left({\displaystyle \genfrac{}{}{0pt}{}{n-2-i}{i}}\right)p^{n-2-2i}{q}^{i+1}y.\hfill \end{array}$$

Taking $x=0,y=1$ on both sides of $D_G^n(bx+ay)=b{D}_G^n\left(x\right)+a{D}_G^n\left(y\right)$, we can obtain

$$H_n=b\sum _{i=0}^{\lfloor {\displaystyle \frac{n-1}{2}}\rfloor }\left({\displaystyle \genfrac{}{}{0pt}{}{n-1-i}{i}}\right)p^{n-1-2i}{q}^i+a\sum _{i=0}^{\lfloor {\displaystyle \frac{n-2}{2}}\rfloor }\left({\displaystyle \genfrac{}{}{0pt}{}{n-2-i}{i}}\right)p^{n-2-2i}{q}^{i+1},$$

as desired. □

Theorem 2 can also be written in the equivalent form

$$H_n=\left\{\begin{array}{cc}{\displaystyle \sum _{i=0}^{k-1}\left(pb\left({\displaystyle \genfrac{}{}{0pt}{}{2k-1-i}{i}}\right)+qa\left({\displaystyle \genfrac{}{}{0pt}{}{2k-2-i}{i}}\right)\right)p^{2k-2-2i}{q}^i}\hfill & \mathrm{if}n=2k,\hfill \\ {\displaystyle b{q}^k+\sum _{i=0}^{k-1}\left(pb\left({\displaystyle \genfrac{}{}{0pt}{}{2k-i}{i}}\right)+qa\left({\displaystyle \genfrac{}{}{0pt}{}{2k-1-i}{i}}\right)\right)p^{2k-1-2i}{q}^i}\hfill & \mathrm{if}n=2k+1.\hfill \end{array}\right.$$

Siar and Keskin [8] of (Theorem 6) showed

$$H_{n+m-1}=H_n{H}_m(0,1;p,q)-H_{n-1}{H}_{m-1}(0,1;p,q).$$

Next, we present a new expression of $H_{n+m-1}$ using solely the terms from the $H_n$ sequence via the grammatical approach, which is unavailable in the literature.

Theorem 3. 

For $n,m⩾1$,

$$H_{n+m-1}={\displaystyle \frac{a(H_n{H}_{m+1}+q{H}_{n-1}{H}_m)-b(H_n{H}_m+q{H}_{n-1}{H}_{m-1})}{q{a}^2+pab-b^2}}.$$

(3)

Taking $m=n$ and $m=n+1$, we can, respectively, obtain that

$$\begin{array}{cc}\hfill H_{2n-1}& ={\displaystyle \frac{a(H_n{H}_{n+1}+q{H}_{n-1}{H}_n)-b(H_n{H}_n+q{H}_{n-1}{H}_{n-1})}{q{a}^2+pab-b^2}},\hfill \\ \hfill H_{2n}& ={\displaystyle \frac{a(H_n{H}_{n+2}+q{H}_{n-1}{H}_{n+1})-b(H_n{H}_{n+1}+q{H}_{n-1}{H}_n)}{q{a}^2+pab-b^2}}.\hfill \end{array}$$

For Fibonacci numbers $F_n=H_n(0,1;1,1)$ and Lucas numbers $L_n=H_n(2,1;1,1)$, we obtain

$$F_{n+m-1}=F_n{F}_m+F_{n-1}{F}_{m-1};$$

$$L_{n+m-1}=L_n{F}_m+L_{n-1}{F}_{m-1}.$$

(4)

Proof. 

We only need to prove (3) and (4). From Theorem 1, we see that $D_G^n(bx+ay)=H_{n+1}x+H_{n}y$. So, we obtain

$$D_G^n\left((pb+qa)x+by\right)=D_G^n(H_{2}x+H_{1}y)=D_G^{n+1}(bx+ay)=H_{n+2}x+H_{n+1}y.$$

Combining these two equations, we obtain a system of linear equations

$$\left\{\begin{array}{c}b{D}_G^n\left(x\right)+a{D}_G^n\left(y\right)=H_{n+1}x+H_{n}y\hfill \\ (pb+qa)D_G^n\left(x\right)+b{D}_G^n\left(y\right)=H_{n+2}x+H_{n+1}y.\hfill \end{array}\right.$$

Solving the equations yields $D_G^n\left(x\right)$ and $D_G^n\left(y\right)$ as follows:

$$\left\{\begin{array}{c}{D}_G^n\left(x\right)={\displaystyle \frac{1}{q{a}^2+pab-b^2}}\left((a{H}_{n+2}-b{H}_{n+1})x+(a{H}_{n+1}-b{H}_n)y\right)\hfill \\ D_G^n\left(y\right)={\displaystyle \frac{q}{q{a}^2+pab-b^2}}\left((a{H}_{n+1}-b{H}_n)x+(a{H}_n-b{H}_{n-1})y\right).\hfill \end{array}\right.$$

Since

$$\begin{array}{cc}\hfill D_G^{n+m-1}(bx+ay)& =D_G^m\left(D_G^{n-1}(bx+ay)\right)\hfill \\ \hfill & =D_G^m(H_{n}x+H_{n-1}y)\hfill \\ \hfill & =H_n{D}_G^m\left(x\right)+H_{n-1}{D}_G^m\left(y\right),\hfill \end{array}$$

by taking $x=0$ and $y=1$, we can obtain

$$\begin{array}{cc}\hfill & D_G^{n+m-1}(bx+ay){|}_{x=0,y=1}\hfill \\ & =H_n{D}_G^m\left(x\right){|}_{x=0,y=1}+H_{n-1}{D}_G^m\left(y\right){|}_{x=0,y=1}\hfill \\ \hfill & =H_n·{\displaystyle \frac{a{H}_{m+1}-b{H}_m}{q{a}^2+pab-b^2}}+q{H}_{n-1}·{\displaystyle \frac{a{H}_m-b{H}_{m-1}}{q{a}^2+pab-b^2}}\hfill \\ \hfill & ={\displaystyle \frac{a(H_n{H}_{m+1}+q{H}_{n-1}{H}_m)-b(H_n{H}_m+q{H}_{n-1}{H}_{m-1})}{q{a}^2+pab-b^2}},\hfill \end{array}$$

which yields the desired identity (3).

Next, we prove the last identity (4). From (3) and the identity $2L_{m+1}-L_m=5F_m$ in [21], we have

$$\begin{array}{cc}\hfill L_{n+m-1}& ={\displaystyle \frac{2(L_n{L}_{m+1}+L_{n-1}{L}_m)-(L_n{L}_m+L_{n-1}{L}_{m-1})}{5}}\hfill \\ \hfill & ={\displaystyle \frac{L_n(2L_{m+1}-L_m)+L_{n-1}(2L_m-L_{m-1})}{5}}\hfill \\ \hfill & ={\displaystyle \frac{L_n\left(5F_m\right)+L_{n-1}\left(5F_{m-1}\right)}{5}}\hfill \\ \hfill & =L_n{F}_m+L_{n-1}{F}_{m-1},\hfill \end{array}$$

as desired. This completes the proof. □

Next, we will present Vajda-like identities. The new result differs in form from the usual expressions. For example, Soykan [9] of (Theorem 11) showed

$$H_n{H}_m-H_{n-k}{H}_{m+k}={(-q)}^{n-k}(b^2-pab-q{a}^2)H_{m+k-n}(0,1;p,q)H_k(0,1;p,q);$$

Yazlik and Taskara [11] of (Theorem 7) presented

$$H_n{H}_m-H_{n-k}{H}_{m+k}={(-q)}^{n-k}{\displaystyle \frac{(b{H}_k-a{H}_{k+1})(b{H}_{m+k-n}-a{H}_{m+k-n+1})}{b^2-pab-q{a}^2}}.$$

However, when $m=n$ and $k=1$, our result degenerates into a Cassini-like identity, coinciding with [9].

Theorem 4. 

For $m⩾n⩾k⩾0$,

$$\left|\begin{array}{cc}{H}_n& H_{m+k}\\ H_{n-k}& H_m\end{array}\right|={(-q)}^{n-k}\left|\begin{array}{cc}{H}_k& H_{m-n+2k}\\ H_0& H_{m-n+k}\end{array}\right|.$$

(5)

In particular, we have

(i) 

$\left|\begin{array}{cc}{H}_n& H_{n+k}\\ H_{n-k}& H_n\end{array}\right|={(-q)}^{n-k}\left|\begin{array}{cc}{H}_k& H_{2k}\\ H_0& H_k\end{array}\right|$ when $m=n$;

(ii) 

$\left|\begin{array}{cc}{H}_n& H_{m+1}\\ H_{n-1}& H_m\end{array}\right|={(-q)}^{n-1}\left|\begin{array}{cc}{H}_1& H_{m-n+2}\\ H_0& H_{m-n+1}\end{array}\right|$ when $k=1$;

(iii) 

$\left|\begin{array}{cc}{H}_n& H_{n+1}\\ H_{n-1}& H_n\end{array}\right|={(-q)}^{n-1}(b^2-pab-q{a}^2)$ when $m=n$ and $k=1$.

Proof. 

For the Horadam grammar $G=\{x\to px+y,y\to qx\}$, in order to compute more efficiently, we consider a change of it. Setting $u={\displaystyle \frac{1}{q}}y$ and $v=x$, we see that $u\to v,v\to pv+qu$. For this reason, we obtain a new grammar:

$$G_1=\{u\to v,v\to pv+qu\}.$$

In the same way as in the proof of Theorem 1, it is easy to deduce that

$$D_{G_1}^n\left(u\right){|}_{u=a,v=b}=H_n.$$

Therefore, to study $H_m{H}_n-H_{m+k}{H}_{n-k}$, we need to compute

$$D_{G_1}^m\left(u\right)D_{G_1}^n\left(u\right)-D_{G_1}^{m+k}\left(u\right)D_{G_1}^{n-k}\left(u\right).$$

Let us first examine how to calculate it for the case of $k=1$.

$$\begin{array}{cc}\hfill & D_{G_1}^m\left(u\right)D_{G_1}^n\left(u\right)-D_{G_1}^{m+1}\left(u\right)D_{G_1}^{n-1}\left(u\right)\hfill \\ \hfill & =D_{G_1}^{m-1}\left(v\right)D_{G_1}^{n-2}(pv+qu)-D_{G_1}^{m-1}(pv+qu)D_{G_1}^{n-2}\left(v\right)\hfill \\ \hfill & =p{D}_{G_1}^{m-1}\left(v\right)D_{G_1}^{n-2}\left(v\right)+q{D}_{G_1}^{m-1}\left(v\right)D_{G_1}^{n-2}\left(u\right)-p{D}_{G_1}^{m-1}\left(v\right)D_{G_1}^{n-2}\left(v\right)-q{D}_{G_1}^{m-1}\left(u\right)D_{G_1}^{n-2}\left(v\right)\hfill \\ \hfill & =q{D}_{G_1}^{m-1}\left(v\right)D_{G_1}^{n-2}\left(u\right)-q{D}_{G_1}^{m-1}\left(u\right)D_{G_1}^{n-2}\left(v\right)\hfill \\ \hfill & =q{D}_{G_1}^m\left(u\right)D_{G_1}^{n-2}\left(u\right)-q{D}_{G_1}^{m-1}\left(u\right)D_{G_1}^{n-1}\left(u\right)\hfill \\ \hfill & =(-q)\left(D_{G_1}^{m-1}\left(u\right)D_{G_1}^{n-1}\left(u\right)-D_{G_1}^m\left(u\right)D_{G_1}^{n-2}\left(u\right)\right)\hfill \\ \hfill & \dots \hfill \\ \hfill & ={(-q)}^{n-k}\left(D_{G_1}^{m-n+k}\left(u\right)D_{G_1}^k\left(u\right)-D_{G_1}^{m-n+k+1}\left(u\right)D_{G_1}^{k-1}\left(u\right)\right).\hfill \end{array}$$

To compute $D_{G_1}^m\left(u\right)D_{G_1}^n\left(u\right)-D_{G_1}^{m+k}\left(u\right)D_{G_1}^{n-k}\left(u\right)$, we decompose the difference into a cascading sequence of intermediate terms. Through systematic cancellation of consecutive paired terms, the entire chain of equalities collapses into a concise expression involving only the boundary terms. That is,

$$\begin{array}{cc}\hfill & D_{G_1}^m\left(u\right)D_{G_1}^n\left(u\right)-D_{G_1}^{m+k}\left(u\right)D_{G_1}^{n-k}\left(u\right)\hfill \\ \hfill & =\left(D_{G_1}^m\left(u\right)D_{G_1}^n\left(u\right)-D_{G_1}^{m+1}\left(u\right)D_{G_1}^{n-1}\left(u\right)\right)+\left(D_{G_1}^{m+1}\left(u\right)D_{G_1}^{n-1}\left(u\right)-D_{G_1}^{m+2}\left(u\right)D_{G_1}^{n-2}\left(u\right)\right)\hfill \\ \hfill & +\dots +\left(D_{G_1}^{m+k-1}\left(u\right)D_{G_1}^{n-k+1}\left(u\right)-D_{G_1}^{m+k}\left(u\right)D_{G_1}^{n-k}\left(u\right)\right)\hfill \\ \hfill & ={(-q)}^{n-k}\left(D_{G_1}^{m-n+k}\left(u\right)D_{G_1}^k\left(u\right)-D_{G_1}^{m-n+k+1}\left(u\right)D_{G_1}^{k-1}\left(u\right)\right)\hfill \\ \hfill & +{(-q)}^{n-k}\left(D_{G_1}^{m-n+k+1}\left(u\right)D_{G_1}^{k-1}\left(u\right)-D_{G_1}^{m-n+k+2}\left(u\right)D_{G_1}^{k-2}\left(u\right)\right)\hfill \\ \hfill & +\dots +{(-q)}^{n-k}\left(D_{G_1}^{m-n+2k-1}\left(u\right)D_{G_1}^1\left(u\right)-D_{G_1}^{m-n+2k}\left(u\right)D_{G_1}^0\left(u\right)\right)\hfill \\ \hfill & ={(-q)}^{n-k}\left(D_{G_1}^{m-n+k}\left(u\right)D_{G_1}^k\left(u\right)-D_{G_1}^{m-n+2k}\left(u\right)D_{G_1}^0\left(u\right)\right).\hfill \end{array}$$

Since $D_{G_1}^n\left(u\right){|}_{u=a,v=b}=H_n$, we obtain

$$H_m{H}_n-H_{m+k}{H}_{n-k}={(-q)}^{n-k}(H_{m-n+k}{H}_k-H_{m-n+2k}{H}_0),$$

which leads to (5). □

3.2. The Relationship Between $H_n(a,b;p,q)$ and $H_n(c,d;p,q)$ with $(a,b)\ne (c,d)$

We will provide a new method to establish the relationship between two Horadam-type sequences that have different initial values.

Theorem 5. 

Let $H_n$ be the Horadam numbers $H_n(a,b;p,q)$. For arbitrary numbers s and r and non-negative integer k, we have

$$s{H}_{n+k}-r{H}_n=H_n(s{H}_k-ra,s{H}_{k+1}-rb;p,q).$$

(6)

In particular, if for some $s,r,$ and k, there exists $c,d,$ and e and non-negative integer l, such that

$$\left\{\begin{array}{cc}s{H}_k-ra\hfill & =e{H}_l(c,d;p,q),\hfill \\ s{H}_{k+1}-rb\hfill & =e{H}_{l+1}(c,d;p,q),\hfill \end{array}\right.$$

then

$$s{H}_{n+k}-r{H}_n=e{H}_{n+l}(c,d;p,q).$$

(7)

If $s=a$ and $r=H_k$, then

$$a{H}_{n+k}-H_k{H}_n=(a{H}_{k+1}-b{H}_k)H_n(0,1;p,q).$$

(8)

Proof. 

By Theorem 1, we have

$$\begin{array}{cc}\hfill s{D}_G^{n+k}(bx+ay)-r{D}_G^n(bx+ay)& =s{D}_G^n\left(H_{k+1}x+H_{k}y\right)-r{D}_G^n(bx+ay)\hfill \\ \hfill & =D_G^n\left((s{H}_{k+1}-rb)x+(s{H}_k-ra)y\right).\hfill \end{array}$$

(9)

Taking $x=0$ and $y=1$ on both sides of the above equation, we obtain (6).

If there exists $c,d,$ and e and non-negative integer l, such that

$$\left\{\begin{array}{cc}s{H}_k-ra\hfill & =e{H}_l(c,d;p,q),\hfill \\ s{H}_{k+1}-rb\hfill & =e{H}_{l+1}(c,d;p,q),\hfill \end{array}\right.$$

then we can follow from (9) that

$$\begin{array}{cc}\hfill s{D}_G^{n+k}(bx+ay)-r{D}_G^n(bx+ay)& =D_G^n\left(e{H}_{l+1}(c,d;p,q)x+e{H}_l(c,d;p,q)y\right)\hfill \\ \hfill & =e{D}_G^n\left(H_{l+1}(c,d;p,q)x+H_l(c,d;p,q)y\right)\hfill \\ \hfill & =e{D}_G^{n+l}\left(H_1(c,d;p,q)x+H_0(c,d;p,q)y\right)\hfill \\ \hfill & =e{D}_G^{n+l}(dx+cy).\hfill \end{array}$$

Taking $x=0$ and $y=1$ on both sides, we obtain (7).

If $s=a$ and $r=H_k$ in (9), then

$$a{D}_G^{n+k}(bx+ay)-H_k{D}_G^n(bx+ay)=D^n\left((a{H}_{k+1}-H_{k}b)x+0y\right)=(a{H}_{k+1}-b{H}_k)D_G^n\left(1x+0y\right),$$

which yields the desired result (8). □

Remark 1. 

For a given $H_n(a,b;p,q)$ and $H_n(c,d;p,q)$ with $(a,b)\ne (c,d)$, an effective way to deduce their relationship is as follows:

(i) 

Fix the value of $k,l,$ and e.

(ii) 

Solve for s and r in the following equations

$$\left\{\begin{array}{cc}s{H}_k-ra\hfill & =e{H}_l(c,d;p,q),\hfill \\ s{H}_{k+1}-rb\hfill & =e{H}_{l+1}(c,d;p,q).\hfill \end{array}\right.$$

(iii) 

Substitute s and r from (7) and simplify.

An illustration of Theorem 5 is given as follows:

Corollary 1 

([21]). For the Fibonacci and Lucas numbers, we have

$$\begin{array}{cc}\hfill 2L_{n+1}-1L_n& =5F_n;\hfill \\ \hfill 2L_{n+2}-3L_n& =5F_n;\hfill \\ \hfill L_{n+3}-2L_n& =5F_n;\hfill \\ \hfill & \dots \hfill \\ \hfill 2L_{n+k}-L_k{L}_n& =5F_k{F}_n.\hfill \end{array}$$

Proof. 

Since $p=1$ and $q=1$, we have $H_2=a+b$, $H_3=a+2b$, and $H_4=2a+3b$. Thus, letting $p=1,q=1,s=a,$ and $r=H_k$ and, respectively, taking $k=1$, $k=2$, $k=3$, …, in (8), we can obtain

$$\begin{array}{cc}\hfill a{H}_{n+1}(a,b;1,1)-b{H}_n(a,b;1,1)& =(ab+a^2-b^2)F_n;\hfill \\ \hfill a{H}_{n+2}(a,b;1,1)-(a+b)H_n(a,b;1,1)& =(ab+a^2-b^2)F_n;\hfill \\ \hfill a{H}_{n+3}(a,b;1,1)-(a+2b)H_n(a,b;1,1)& =2(ab+a^2-b^2)F_n;\hfill \\ \hfill & \dots \hfill \\ \hfill a{H}_{n+k}(a,b;1,1)-H_k(a,b;1,1)H_n(a,b;1,1)& =(ab+a^2-b^2)F_k{F}_n.\hfill \end{array}$$

Since the Lucas number $L_n=H_n(2,1;1,1)$ (i.e., $a=2$ and $b=1$), we obtain

$$\begin{array}{cc}\hfill 2L_{n+1}-1L_n& =5F_n;\hfill \\ \hfill 2L_{n+2}-3L_n& =5F_n;\hfill \\ \hfill L_{n+3}-2L_n& =5F_n;\hfill \\ \hfill & \dots \hfill \\ \hfill 2L_{n+k}-L_k{L}_n& =5F_k{F}_n.\hfill \end{array}$$

□

We now consider the Chebyshev polynomials as another application of Theorem 5. The Chebyshev polynomial $T_n\left(t\right)$ of the first kind is a polynomial in t of degree n, satisfying the recurrence relation

$$T_{n+1}\left(t\right)=2t{T}_n\left(t\right)-T_{n-1}\left(t\right),$$

with the initial conditions $T_0\left(t\right)=1$ and $T_1\left(t\right)=t$. The first few Chebyshev polynomials are listed as follows:

$$T_2\left(t\right)=2t^2-1,T_3\left(t\right)=4t^3-3t,T_4\left(t\right)=8t^4-8t^2+1.$$

The Chebyshev polynomial $U_n\left(t\right)$ of the second kind satisfies the same recurrence relation

$$U_{n+1}\left(t\right)=2t{U}_n\left(t\right)-U_{n-1}\left(t\right),$$

together with the initial conditions $U_0\left(t\right)=1$ and $U_1\left(t\right)=2t$. The first few terms are

$$U_2\left(t\right)=4t^2-1,U_3\left(t\right)=8t^3-4t,U_4\left(t\right)=16t^4-12t^2+1.$$

The polynomial sequences $T_n\left(t\right)$ and $U_n\left(t\right)$ can be regarded as two different Horadam sequences with the same $p=2t$ and $q=-1$. Therefore, according Theorem 1, for the grammar $G_2=\{x\to 2tx+y,y\to -x\}$, we have

$$\begin{array}{cc}\hfill D_{G_2}^n{(tx+y)|}_{x=0,y=1}& =T_n\left(t\right)\hfill \\ \hfill D_{G_2}^n{(2tx+y)|}_{x=0,y=1}& =U_n\left(t\right).\hfill \end{array}$$

Corollary 2 

([22]). We have

$$t{U}_{n+1}\left(t\right)-U_n\left(t\right)=T_{n+2}\left(t\right),U_{n+2}\left(t\right)-U_n\left(t\right)=2T_{n+2}\left(t\right).$$

Proof. 

In (7), by taking $H_n=U_n\left(t\right)$ (i.e., $a=1,b=2t,p=2t,$ and $q=-1$), $s=t,r=1,$ and $k=1$, we arrive at

$$\begin{array}{cc}\hfill s{H}_k-ra& =2t^2-1=T_2\left(t\right),\hfill \\ \hfill s{H}_{k+1}-rb& =4t^3-3t=T_3\left(t\right),\hfill \end{array}$$

and so

$$t{U}_{n+1}\left(t\right)-U_n\left(t\right)=T_{n+2}\left(t\right).$$

Similarly, we can obtain the second formula by taking $H_n=U_n\left(t\right)$, $s=1,r=1,$ and $k=2$. □

By taking appropriate values, many such relationships can be obtained. In (7), by taking $H_n=T_n\left(t\right)$, $s=t,$ and $r=T_{k-1}\left(t\right)$, we can obtain a new result.

Corollary 3. 

We have

$$t{T}_{n+k}\left(t\right)-T_{k-1}\left(t\right)T_n\left(t\right)=\left(t{T}_k\left(t\right)-T_{k-1}\left(t\right)\right)U_n\left(t\right).$$

In particular, we have

$$t{T}_{n+1}\left(t\right)-T_n\left(t\right)=(t^2-1)U_n\left(t\right),$$

$$T_{n+2}\left(t\right)-T_n\left(t\right)=2(t^2-1)U_n\left(t\right),$$

$$t{T}_{n+3}\left(t\right)-(2t^2-1)T_n\left(t\right)=(4t^4-5t^2+1)U_n\left(t\right),$$

$$T_{n+4}\left(t\right)-(4t^2-3)T_n\left(t\right)=4(2t^4-3t^2+1)U_n\left(t\right).$$

Equivalently, we have

$$t{T}_{n+k}\left(t\right)-T_{k-1}\left(t\right)T_n\left(t\right)=(t^2-1)U_{k-1}\left(t\right)U_n\left(t\right).$$

In (7), by taking

$$H_n=U_n\left(t\right),s=t,r=t{U}_k\left(t\right)-U_{k-1}\left(t\right)=T_{k+1}\left(t\right),$$

we obtain another new result.

Corollary 4. 

We have

$$t{U}_{n+k}\left(t\right)-T_{k+1}{U}_n\left(t\right)=U_{k-1}\left(t\right)T_n\left(t\right).$$

It should be noted that the Corollaries 1–4 merely present a subset of all possible identities. In fact, by Remark 1, we can obtain additional identities. For example, if we take $H_n=T_n\left(t\right)$, $e=1$, $k=4$, and $l=3$, then we can obtain

$$T_4\left(t\right)T_{n+4}\left(t\right)-T_n\left(t\right)=(T_5\left(t\right)-t{T}_4\left(t\right))U_{n+3}\left(t\right).$$

Since $T_5\left(t\right)-t{T}_4\left(t\right)=t{T}_4\left(t\right)-T_3\left(t\right)$, by Corollary 3, we obtain a new identity:

$$T_4\left(t\right)T_{n+4}\left(t\right)-T_n\left(t\right)=(t^2-1)U_3\left(t\right)U_{n+3}\left(t\right).$$

4. Conclusions

This paper uses context-free grammars as a new tool to investigate the properties of the Horadam sequence. By using a simple context-free grammar $G=\{x\to px+y,y\to qx\}$, we successfully generated the Horadam sequence. Our grammars not only derive several new or known identities but also uncover meaningful connections between Horadam sequences with different initial values $(a,b)\ne (c,d)$ and the same recurrence coefficients $(p,q)$. Furthermore, we demonstrate how this grammatical method can be extended to derive identities for special cases, including the Fibonacci and Lucas numbers, as well as Chebyshev polynomials of the first and second kinds.

To the best of our knowledge, Theorems 1, 3, 4, and 5 and Corollaries 3 and 4 introduce new identities or relationships, whereas Theorem 2 and Corollaries 1 and 2 provide new proofs of known results, illustrating the versatility of the grammars. Our result highlights its capacity to both generate new knowledge and deepen the understanding of classical identities. The grammars can also be flexibly changed to derive different relational expressions. This is demonstrated in the paper by the transformation from grammar $\{x\to px+y,y\to qx\}$ to $\{u\to v,v\to pv+qu\}$, which is used to obtain the identity in Theorem 4.

Indeed, many combinatorial objects admit grammatical interpretations, making this method an important tool in combinatorics. For example, if the polynomials $M_n\left(t\right)$ satisfy

$$M_{n+1}\left(t\right)=p\left(n\right)M_n\left(t\right)+q\left(n\right)M_{n-1}\left(t\right),$$

we can always use a series of grammars, such as

$$G_i=\{x\to p\left(i\right)x+y,y\to q\left(i\right)x\},i=1,2,\dots ,n,$$

to obtain the relation

$$D_n\dots D_2{D}_1(M_1\left(t\right)x+M_0\left(t\right)y)=M_{n+1}\left(t\right)x+M_n\left(t\right)y,$$

where $D_i$ denotes the formal derivative associated with $G_i$. Interested readers can use the above model to study the Hermite polynomials, Legendre polynomials, and other similar polynomials [23]. We next give another application. Let $G_i=\{x\to x-ity,y\to (i-1)x\}$; we can study the Meixner polynomials through $D_n\dots D_2{D}_1\left(x\right)$. Beyond combinatorics, the grammars can also find applications in number theory, algorithm analysis, and the mathematical modeling of recursive processes in fields such as computer science and physics.