On the Recurrence Properties of Narayana’s Cows Sequence

Abstract

In this paper, we consider the recurrence properties of two generalized forms of Narayana’s cows sequence. On the one hand, we study Narayana’s cows sequence at negative indices and express it as the linear combination of the sequence at positive indices. On the other hand, we study the convolved Narayana number and obtain a computation formula for it.

1. Introduction

Narayana’s cows sequence originated from a problem with cows proposed by the Indian mathematician Narayana in the 14th century. In this problem, we suppose that there is one cow in the beginning and that every cow produces one calf each year from the age of 4 years old. Narayana’s cow problem counts the number of calves produced every year [1,2]. This problem appears to be similar to Fibonacci’s rabbit problem. So too do the answers, known as the Narayana sequence and the Fibonacci sequence.

Narayana’s cows sequence (A000930 in [3]) satisfies a third-order recurrence relation:

$$\begin{array}{c}\hfill G_n=G_{n-1}+G_{n-3},\mathrm{for}n\ge 3.\end{array}$$

This has the initial values $G_0=0$ and $G_1=G_2=G_3=1$ [1]. Explicitly, the characteristic equation of $G_n$ is:

$$x^3-x^2-1=0,$$

and the characteristic roots are:

$$\begin{array}{ccc}\hfill \alpha & =& \frac{1}{3}\left(\sqrt[3]{\frac{1}{2}\left(29-3\sqrt{93}\right)}+\sqrt[3]{\frac{1}{2}\left(3\sqrt{93}+29\right)}+1\right),\hfill \\ \hfill \beta & =& \frac{1}{3}-\frac{1}{6}\left(1-i\sqrt{3}\right)\sqrt[3]{\frac{1}{2}\left(29-3\sqrt{93}\right)}-\frac{1}{6}\left(1+i\sqrt{3}\right)\sqrt[3]{\frac{1}{2}\left(3\sqrt{93}+29\right)},\hfill \\ \hfill \gamma & =& \frac{1}{3}-\frac{1}{6}\left(1+i\sqrt{3}\right)\sqrt[3]{\frac{1}{2}\left(29-3\sqrt{93}\right)}-\frac{1}{6}\left(1-i\sqrt{3}\right)\sqrt[3]{\frac{1}{2}\left(3\sqrt{93}+29\right)}.\hfill \end{array}$$

(1)

Then, the Narayana sequence can be obtained by Binet’s formula:

$$G_n=A{\alpha }^n+B{\beta }^n+C{\gamma }^n.$$

(2)

For $n\in {\mathbb{Z}}_{\ge 0}$, the generating function of the Narayana sequence is:

$$\begin{array}{c}\hfill g(x)=\frac{1}{1-x-x^3}=\sum _{n=0}^{\infty }{G}_{n+1}{x}^n.\end{array}$$

(3)

With the Vieta theorem, we have:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}\alpha +\beta +\gamma =1,\hfill \\ \alpha \beta +\beta \gamma +\alpha \gamma =0,\hfill \\ \alpha \beta \gamma =1.\hfill \end{array}\right.\end{array}$$

(4)

From Formula (2), we obtain:

$$\begin{array}{ccc}\hfill G_0& =& A+B+C=0,\hfill \\ \hfill G_1& =& A\alpha +B\beta +C\gamma =1,\hfill \\ \hfill G_2& =& A{\alpha }^2+B{\beta }^2+C{\gamma }^2=1,\hfill \end{array}$$

which implies:

$$\begin{array}{c}\hfill A=\frac{1-\beta -\gamma }{(\alpha -\beta )(\alpha -\gamma )},B=\frac{1-\alpha -\gamma }{(\beta -\alpha )(\beta -\gamma )},C=\frac{1-\alpha -\beta }{(\gamma -\beta )(\gamma -\alpha )}.\end{array}$$

With Formula (4), we can simplify A, B, and C and obtain:

$$\begin{array}{c}\hfill A=\frac{\alpha }{{\alpha }^2-\alpha \beta -\alpha \gamma +\beta \gamma }=\frac{\alpha }{{\alpha }^2+2\beta \gamma }=\frac{{\alpha }^2}{{\alpha }^3+2}\end{array}$$

and

$$\begin{array}{c}\hfill B=\frac{{\beta }^2}{{\beta }^3+2},C=\frac{{\gamma }^2}{{\gamma }^3+2}.\end{array}$$

(5)

The Narayana sequence was originally defined at positive indices. Actually, it can be extended to negative indices by defining:

$$\begin{array}{c}\hfill G_{-n}=\frac{A}{{\alpha }^n}+\frac{B}{{\beta }^n}+\frac{C}{{\gamma }^n},\end{array}$$

(6)

The following recurrence relation holds for all integral indices.

$$\begin{array}{c}\hfill G_n=G_{n-1}+G_{n-3},n\in \mathbb{Z}.\end{array}$$

(7)

Through a simple computation, the first few terms of $G_n$ at negative indices can be obtained from Formulas (5) and (6), so that $G_{-1}=0,G_{-2}=1,G_{-3}=0,G_{-4}=-1$, which also satisfies Relation (7).

The Narayana sequence has a close connection to some famous numbers or sequences and plays an important role in cryptography and combinatorics. For instance, it can be seen as the number of compositions of n into parts 1 and 3. For $n\ge 3$, the Narayana sequence can be expressed as the row sums of Pascal’s triangle with triplicated diagonals, while the Fibonacci number $F_n$ is the row sums of Pascal’s triangle with slope diagonals of 45 degrees [3]. Narayana’s sequence has a beautiful distribution pattern, the ratio of whose consecutive terms approximates the supergolden ratio, which is closely related to the golden ratio [2]. Moreover, the Narayana sequence satisfies good cross-correlation and autocorrelation properties, which provide wide applications in data coding, information theory, and cryptography, and especially in multiparty computation [2,4,5]. More research about the Narayana sequence and its applications can be found in [6,7,8].

In this paper, we study two natural generalized forms of the Narayana sequence $G_n$ (i.e., $\{G_n|n\in {\mathbb{Z}}_{<0}\}$ and $\{g^h(x)|h\in {\mathbb{R}}_{>0}\}$) and construct the recurrence relation between these generalized forms and $G_n$. Many mathematical efforts have been made concerning these two generalizations of the linear recurrence sequence, such as the Fibonacci sequence $F_n$ (A000035 in [3]), the Tribonacci sequence $T_n$ (A000073 in [3]), the Lucas sequence $L_n$ (A000032 in [3]), and so on.

For the first aspect, Falcon [9,10] obtained the connection between $F_n$ and $L_n$ and their values at negative indices. He proved that for $n\in {\mathbb{Z}}_{\ge 0}$, the following identities hold.

$$\begin{array}{c}\hfill F_{-n}={(-1)}^{n+1}{F}_n\mathrm{and}{L}_{-n}={(-1)}^{n+1}{L}_n.\end{array}$$

Halici and Akyuz [11] proposed a series of formulae expressing $F_n$ and $L_n$ at negative indices as the linear combination of themselves. For example, they obtained that for any integer $n\ge 0$ and $k\ge 1$, one has:

$$\begin{array}{ccc}\hfill F_{-nk}& =& F_{-n}{L}_n^{k-1}\sum _{i=0}^{[\frac{k-1}{2}]}\left(\genfrac{}{}{0pt}{}{k-1-i}{i}\right)L_n^{-2i}{(-1)}^i,n\mathrm{and}k\mathrm{are}\mathrm{even}.\hfill \\ \hfill L_{-nk}& =& 5^{\frac{k-1}{2}}{L}_{-n}\sum _{i=0}^{[\frac{k-1}{2}]}\left(\genfrac{}{}{0pt}{}{k-1-i}{i}\right)5^{-i}{F}_n^{k-1-2i}{(-1)}^i,n\mathrm{and}k\mathrm{are}\mathrm{odd},\hfill \end{array}$$

where $\left(\genfrac{}{}{0pt}{}{n}{i}\right)$ denotes the binomial coefficients. More works about the linear recurrence sequence at negative indices can be seen in [12,13,14,15].

For the second aspect, Kim et al. [16] and Chen and Qi [17] considered the convolved Fibonacci number $F_n(x)$, which is defined by the generating function:

$${\left(\frac{1}{1-t-t^2}\right)}^x=\sum _{n=0}^{\infty }{F}_n(x)\frac{t^n}{n!},x\in {\mathbb{R}}_{>0}.$$

With the combinatorial methods, they expressed the $F_n(x)$ by the linear combination of $F_n(x)$ itself [16] or $L_n$ [17].

$$\begin{array}{ccc}\hfill F_n(x)& =& \sum _{l=0}^n\left(\genfrac{}{}{0pt}{}{n}{l}\right)F_l(r)F_{n-l}(x-r).\hfill \\ \hfill F_n(x)& =& \frac{1}{2}\sum _{i=0}^n{(-1)}^i\left(\genfrac{}{}{0pt}{}{n}{i}\right){⟨x⟩}_i{⟨x⟩}_{n-i}{L}_{n-2i},\hfill \end{array}$$

where ${⟨x⟩}_0=1$ and ${⟨x⟩}_i=x(x+1)(x+2)\cdots (x+n-1)$. As a corollary of [17], one has the following identity for $k\in {\mathbb{Z}}_{\ge 0}$:

$$\begin{array}{c}\hfill \sum _{a_1+a_2+\cdots +a_k=n}{F}_{a_1}{F}_{a_2}\cdots F_{a_k}=\frac{1}{2(k-1){!}^2}\sum _{i=0}^n{(-1)}^i\frac{(k+i-1)!(k+n-i-1)!}{i!(n-i)!}{L}_{n-2i}.\end{array}$$

Zhou and Chen [18] studied the convolved Tribonacci number $T_n(x)$, defined by:

$${\left(\frac{1}{1-t-t^2-t^3}\right)}^x=\sum _{n=0}^{\infty }{T}_n(x)t^n,x\in {\mathbb{R}}_{>0}.$$

They obtained:

$$\begin{array}{ccc}\hfill T_n(x)& =& \frac{1}{6}\sum _{u+v+w=n}\frac{{⟨x⟩}_u}{u!}\frac{{⟨x⟩}_v}{v!}\frac{{⟨x⟩}_w}{w!}(3T_{w+1-u}-2T_{w-u}-T_{w-u-1})\hfill \\ & & \times (3T_{w+1-v}-2T_{w-v}-T_{w-v-1})\hfill \\ & & -\frac{1}{6}\sum _{u+v+w=n}\frac{{⟨x⟩}_u}{u!}\frac{{⟨x⟩}_v}{v!}\frac{{⟨x⟩}_w}{w!}(3T_{3w+1-n}-2T_{3w+n}-T_{3w-1-n}).\hfill \end{array}$$

In particular, for $k\in {\mathbb{Z}}_{\ge 0}$, [18] obtained the computational formula for:

$$\begin{array}{c}\hfill \sum _{a_1+a_2+\cdots +a_k=n}{T}_{a_1+1}{T}_{a_2+1}\cdots T_{a_k+1}.\end{array}$$

Other works related to convolved numbers can be found in [19,20,21,22,23,24].

Recently, Professor Tianxin Cai visited Northwest University and gave a talk about a series of linear recurrence sequences and their properties, which incited our interest in this field. There are many recursive identities concerning the Fibonacci, Tribonacci, and Lucas sequences, etc. However, few studies have been conducted regarding the Narayana sequence. Professor Cai proposed an open problem:

Whether and how can the Narayana sequence at negative indices be expressed by the sequence itself at positive indices?

Inspired by the question above, our work focuses on the new recurrence relations around the Narayana sequence. On the one hand, we study the recursive properties of the Narayana sequence at negative indices. For $n\in \mathbb{Z}$, we express $G_{-n}$ as the linear combination of $G_n$. On the other hand, we consider the convolved Narayana number $G_{n+1}(h)$ defined as:

$$g^h(x)={\left(\frac{1}{1-x-x^3}\right)}^h=\sum _{n=0}^{\infty }{G}_{n+1}(h)x^n,h\in {\mathbb{R}}_{>0}.$$

(8)

When $h=1$, $G_{n+1}(h)$ becomes the Narayana number $G_{n+1}$. We propose a computation formula for $G_{n+1}(h)$ involved with $G_{n+1}$. As a corollary, we obtain an identity between $G_{n+1}(h)$ and $G_{n+1}$.

Our main results are presented as follows.

Theorem 1.

For $n\in \mathbb{Z}$, we have:

$$\begin{array}{c}\hfill G_{-n}=G_{2n}-3G_n{G}_{n+1}+2G_n^2.\end{array}$$

Theorem 1 solves Professor Cai’s problem completely. It illustrates the connection between the Narayana sequence at the positive index and the negative index. By Theorem 1, we obtain the recurrence property of the sequence at the negative index, which deepens our knowledge of the nature of the sequence.

Theorem 2.

Let $h\in {\mathbb{R}}_{>0}$. For $n\in {\mathbb{Z}}_{\ge 0}$. We have:

$$\begin{array}{ccc}\hfill G_{n+1}(h)& =& \frac{1}{6}\sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}\hfill \\ & & \times (3G_{i-k+1}-2G_{i-k})(3G_{j-k+1}-2G_{j-k})\hfill \\ & & -\frac{1}{6}\sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}(3G_{i+j-2k+1}-2G_{i+j-2k}).\hfill \end{array}$$

Theorem 2 gives a computation formula for $G_{n+1}(h)$ involving $G_n$. If h is a positive integer, we can deduce the corollary as follows.

Corollary 1.

Let $h\in {\mathbb{Z}}_{>0}$. For $n\in {\mathbb{Z}}_{\ge 0}$. We have:

$$\begin{array}{ccc}& & \sum _{a_1+a_2+\cdots +a_h=n}{G}_{a_1+1}{G}_{a_2+1}\cdots G_{a_h+1}\hfill \\ & =& \frac{1}{6}\sum _{i+j+k=n}\left(\genfrac{}{}{0pt}{}{h+i-1}{i}\right)\left(\genfrac{}{}{0pt}{}{h+j-1}{j}\right)\left(\genfrac{}{}{0pt}{}{h+k-1}{k}\right)\hfill \\ & & \times (3G_{i-k+1}-2G_{i-k})(3G_{j-k+1}-2G_{j-k})\hfill \\ & & -\frac{1}{6}\sum _{i+j+k=n}\left(\genfrac{}{}{0pt}{}{h+i-1}{i}\right)\left(\genfrac{}{}{0pt}{}{h+j-1}{j}\right)\left(\genfrac{}{}{0pt}{}{h+k-1}{k}\right)(3G_{i+j-2k+1}-2G_{i+j-2k}).\hfill \end{array}$$

The formulae in the theorems and corollary above are recursive and computable. To better reflect our results, we calculated some values for the formula in Theorem 1 and Corollary 1; they are listed in Appendix A. The numerical results for $G_{-n}$ and $G_{n+1}(h)(h\in {\mathbb{Z}}_{>0})$ meet their definitions, which prove our results numerically. The numerical experiments were performed in Mathematica $12.0$.

2. Preliminary

To prove our theorems, we propose some lemmas.

Lemma 1.

For $n\in {\mathbb{Z}}_{\ge 0}$, denote:

$$\begin{array}{ccc}\hfill T_n& =& {\alpha }^n+{\beta }^n+{\gamma }^n,\hfill \\ \hfill S_n& =& {\alpha }^n{\beta }^n+{\alpha }^n{\gamma }^n+{\beta }^n{\gamma }^n,\hfill \end{array}$$

where α, β, and γ are as defined in Formula (1). We have $T_0=3$, $T_1=1$, $T_2=1$; $S_0=3$, $S_1=0$, $S_2=-2$; and the following recurrence relation for $n\ge 3$.

$$\begin{array}{c}\hfill T_n=T_{n-1}+T_{n-3},\mathrm{and}{S}_n=-S_{n-2}+S_{n-3}.\end{array}$$

(9)

$T_n$ and $S_n$ satisfy the identity so that:

$$\begin{array}{c}\hfill 2S_n=T_n^2-T_{2n}.\end{array}$$

(10)

Proof. 

With Formula (4), we obtain the initial value of $T_n$ and $S_n$ so that $T_0=3$, $T_1=1$, $S_0=3$, $S_1=0$, and:

$$\begin{array}{ccc}\hfill T_2& =& {(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\alpha \gamma +\beta \gamma )=1,\hfill \\ \hfill S_2& =& {(\alpha \beta +\alpha \gamma +\beta \gamma )}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )=-2.\hfill \end{array}$$

For $n\ge 3$, we have:

$$\begin{array}{ccc}\hfill T_1{T}_{n-1}& =& (\alpha +\beta +\gamma )({\alpha }^{n-1}+{\beta }^{n-1}+{\gamma }^{n-1})\hfill \\ & =& T_n+\alpha \beta \left({\alpha }^{n-2}+{\beta }^{n-2}\right)+\alpha \gamma \left({\alpha }^{n-2}+{\gamma }^{n-2}\right)+\beta \gamma \left({\beta }^{n-2}+{\gamma }^{n-2}\right)\hfill \\ & =& T_n+\alpha \beta \left(T_{n-2}-{\gamma }^{n-2}\right)+\alpha \gamma \left(T_{n-2}-{\beta }^{n-2}\right)+\beta \gamma \left(T_{n-2}-{\alpha }^{n-2}\right)\hfill \\ & =& T_n+T_{n-2}\left(\alpha \beta +\alpha \gamma +\beta \gamma \right)-\alpha \beta \gamma \left({\alpha }^{n-3}+{\beta }^{n-3}+{\gamma }^{n-3}\right)\hfill \\ & =& T_n-T_{n-3}=T_{n-1}\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill S_1{S}_{n-1}& =& (\alpha \beta +\alpha \gamma +\beta \gamma )({\alpha }^{n-1}{\beta }^{n-1}+{\alpha }^{n-1}{\gamma }^{n-1}+{\beta }^{n-1}{\gamma }^{n-1})\hfill \\ & =& S_n+\alpha \beta \gamma \left({\alpha }^{n-2}{\beta }^{n-2}(\alpha +\beta )+{\alpha }^{n-2}{\gamma }^{n-2}(\alpha +\gamma )+{\beta }^{n-2}{\gamma }^{n-2}(\beta +\gamma )\right)\hfill \\ & =& S_n+S_{n-2}-\alpha \beta \gamma ({\alpha }^{n-3}{\beta }^{n-3}+{\alpha }^{n-3}{\gamma }^{n-3}+{\beta }^{n-3}{\gamma }^{n-3})\hfill \\ & =& S_n+S_{n-2}-S_{n-3}=0.\hfill \end{array}$$

From the definition of $T_n$ and $S_n$, we obtain:

$$\begin{array}{ccc}\hfill 2S_n& =& {\alpha }^n({\beta }^n+{\gamma }^n)+{\beta }^n({\alpha }^n+{\gamma }^n)+{\gamma }^n({\alpha }^n+{\beta }^n)\hfill \\ & =& T_n({\alpha }^n+{\beta }^n+{\gamma }^n)-({\alpha }^{2n}+{\beta }^{2n}+{\gamma }^{2n})\hfill \\ & =& T_n^2-T_{2n}.\hfill \end{array}$$

The proof of Lemma is 1 completed. □

Note: In fact, the index n of sequence $T_n$ and $S_n$ in Lemma 1 can be defined with $\mathbb{Z}$. Since $\alpha \beta \gamma =1$, we have $T_n=S_{-n}$ for $n\in \mathbb{Z}$. With symmetry, it suffices to consider that $n\in {\mathbb{Z}}_{\ge 0}$.

Lemma 2.

For $n\in {\mathbb{Z}}_{\ge 0}$, we have:

$$\begin{array}{c}\hfill T_n=3G_{n+1}-2G_{n}and{G}_n=\frac{9T_{n+2}-3T_{n+1}-2T_n}{31}.\end{array}$$

(11)

Proof. 

With Formula (5), we have:

$$A({\alpha }^3+2)={\alpha }^2,B({\beta }^3+2)={\beta }^2,C({\gamma }^3+2)={\gamma }^2.$$

Then, we obtain the recurrence relation for ${\alpha }^n$, ${\beta }^n$, and ${\gamma }^n$:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\alpha }^n=A({\alpha }^3+2){\alpha }^{n-2}=A{\alpha }^{n+1}+2A{\alpha }^{n-2},\hfill \\ {\beta }^n=B({\beta }^3+2){\beta }^{n-2}=B{\beta }^{n+1}+2B{\beta }^{n-2},\hfill \\ {\gamma }^n=C({\gamma }^3+2){\gamma }^{n-2}=C{\gamma }^{n+1}+2C{\gamma }^{n-2}.\hfill \end{array}\right.\end{array}$$

(12)

Summing the formulae above, we obtain:

$$\begin{array}{ccc}\hfill T_n& =& (A{\alpha }^{n+1}+B{\beta }^{n+1}+C{\gamma }^{n+1})+2(A{\alpha }^{n-2}+B{\beta }^{n-2}+C{\gamma }^{n-2})\hfill \\ & =& G_{n+1}+2G_{n-2}=3G_{n+1}-2G_n.\hfill \end{array}$$

Now, we prove the second formula. Denote:

$$\begin{array}{c}\hfill G_n=A{\alpha }^n+B{\beta }^n+C{\gamma }^n=\frac{{\alpha }^{n+2}}{{\alpha }^3+2}+\frac{{\beta }^{n+2}}{{\beta }^3+2}+\frac{{\gamma }^{n+2}}{{\gamma }^3+2}=\frac{N_n}{D}.\end{array}$$

The nominator $N_n$ and denominator D of $G_n$ are:

$$\begin{array}{ccc}\hfill N_n& =& 4{\alpha }^{n+2}+4{\beta }^{n+2}+4{\gamma }^{n+2}+{\beta }^3{\gamma }^3{\alpha }^{n+2}+{\alpha }^3{\gamma }^3{\beta }^{n+2}+{\alpha }^3{\beta }^3{\gamma }^{n+2}\hfill \\ & & +2{\beta }^3{\alpha }^{n+2}+2{\gamma }^3{\alpha }^{n+2}+2{\alpha }^3{\beta }^{n+2}+2{\alpha }^3{\gamma }^{n+2}+2{\gamma }^3{\beta }^{n+2}+2{\beta }^3{\gamma }^{n+2}\hfill \\ & =& 4\left({\alpha }^{n+2}+{\beta }^{n+2}+{\gamma }^{n+2}\right)+{\alpha }^3{\beta }^3{\gamma }^3\left({\alpha }^{n-1}+{\beta }^{n-1}+{\gamma }^{n-1}\right)\hfill \\ & & +2{\alpha }^3{\beta }^3\left({\alpha }^{n-1}+{\beta }^{n-1}\right)+2{\alpha }^3{\gamma }^3\left({\alpha }^{n-1}+{\gamma }^{n-1}\right)+2{\beta }^3{\gamma }^3\left({\beta }^{n-1}+{\gamma }^{n-1}\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill D& =& \left({\alpha }^3+2\right)\left({\beta }^3+2\right)\left({\gamma }^3+2\right)\hfill \\ & =& {\alpha }^3{\beta }^3{\gamma }^3+4{\alpha }^3+4{\beta }^3+4{\gamma }^3+2{\alpha }^3{\beta }^3+2{\alpha }^3{\gamma }^3+2{\beta }^3{\gamma }^3+8.\hfill \end{array}$$

With Formula (4), we compute $N_n$ and D, so that:

$$\begin{array}{ccc}\hfill D& =& 1+16+6+8=31,\hfill \\ \hfill N_n& =& 4T_{n+2}+7T_{n-1}-2T_{n-4},\hfill \end{array}$$

This implies that:

$$\begin{array}{c}\hfill G_n=\frac{4T_{n+2}+7T_{n-1}-2T_{n-4}}{31}=\frac{9T_{n+2}-3T_{n+1}-2T_n}{31}.\end{array}$$

The last equality holds from Formula (9). □

Lemma 3.

For $n\in {\mathbb{Z}}_{\ge 0}$, we have:

$$\begin{array}{c}\hfill S_n=G_{-n}+3G_{-(n+2)}and{G}_{-n}=\frac{6S_{n+2}+9S_{n+1}+4S_n}{31}.\end{array}$$

Proof. 

With Formulas (4) and (5), we have:

$$A=\frac{{\alpha }^2}{{\alpha }^3+2}=\frac{\beta \gamma }{1+2{\beta }^3{\gamma }^3}\mathrm{and}A(1+2{\beta }^3{\gamma }^3)=\beta \gamma .$$

Similar to Formula (12), we obtain:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\beta }^n{\gamma }^n=A{\beta }^{n-1}{\gamma }^{n-1}+2A{\beta }^{n+2}{\gamma }^{n+2},\hfill \\ {\alpha }^n{\gamma }^n=B{\alpha }^{n-1}{\gamma }^{n-1}+2B{\alpha }^{n+2}{\gamma }^{n+2},\hfill \\ {\alpha }^n{\beta }^n=C{\alpha }^{n-1}{\beta }^{n-1}+2C{\alpha }^{n+2}{\beta }^{n+2}.\hfill \end{array}\right.\end{array}$$

Combining the formulae above, we express $S_n$ by $G_n$.

$$\begin{array}{ccc}\hfill S_n& =& (A{\beta }^{n-1}{\gamma }^{n-1}+B{\alpha }^{n-1}{\gamma }^{n-1}+C{\alpha }^{n-1}{\beta }^{n-1})\hfill \\ & & +2(A{\beta }^{n+2}{\gamma }^{n+2}+B{\alpha }^{n+2}{\gamma }^{n+2}+C{\alpha }^{n+2}{\beta }^{n+2})\hfill \\ & =& G_{-(n-1)}+2G_{-(n+2)}=G_{-n}+3G_{-(n+2)}.\hfill \end{array}$$

Now, we prove the second formula. With Formula (6), we have:

$$\begin{array}{c}\hfill G_{-n}=\frac{1}{{\alpha }^{n+1}+2{\alpha }^{n-2}}+\frac{1}{{\beta }^{n+1}+2{\beta }^{n-2}}+\frac{1}{{\gamma }^{n+1}+2{\gamma }^{n-2}}=\frac{N_{-n}}{D_{-n}},\end{array}$$

whose denominator and nominator are:

$$\begin{array}{c}\hfill D_{-n}={\alpha }^n{\beta }^n{\gamma }^n\left({\alpha }^3+2\right)\left({\beta }^3+2\right)\left({\gamma }^3+2\right)=D=31\end{array}$$

and

$$\begin{array}{ccc}\hfill N_{-n}& =& 4{\gamma }^2{\alpha }^n{\beta }^n+4{\beta }^2{\alpha }^n{\gamma }^n+4{\alpha }^2{\beta }^n{\gamma }^n+{\gamma }^2{\alpha }^{n+3}{\beta }^{n+3}+{\beta }^2{\alpha }^{n+3}{\gamma }^{n+3}\hfill \\ & & +{\alpha }^2{\beta }^{n+3}{\gamma }^{n+3}+\left(2{\gamma }^2{\alpha }^{n+3}{\beta }^n+2{\beta }^2{\alpha }^{n+3}{\gamma }^n+2{\gamma }^2{\alpha }^n{\beta }^{n+3}+2{\beta }^2{\alpha }^n{\gamma }^{n+3}\right.\hfill \\ & & \left.+2{\alpha }^2{\beta }^{n+3}{\gamma }^n+2{\alpha }^2{\beta }^n{\gamma }^{n+3}\right)\hfill \\ & =& 4{\alpha }^2{\beta }^2{\gamma }^2{S}_{n-2}+{\alpha }^2{\beta }^2{\gamma }^2{S}_{n+1}+2{\alpha }^2{\beta }^2{\gamma }^2\left({\alpha }^{n+1}{\beta }^{n-2}+{\alpha }^{n+1}{\gamma }^{n-2}\right.\hfill \\ & & \left.+{\alpha }^{n-2}{\beta }^{n+1}+{\alpha }^{n-2}{\gamma }^{n+1}+{\beta }^{n+1}{\gamma }^{n-2}+{\beta }^{n-2}{\gamma }^{n+1}\right).\hfill \end{array}$$

After a simple computation, the last term of the formula above can also be expressed by $S_n$.

$$\begin{array}{ccc}& & {\alpha }^{n+1}{\beta }^{n-2}+{\alpha }^{n+1}{\gamma }^{n-2}+{\alpha }^{n-2}{\beta }^{n+1}+{\alpha }^{n-2}{\gamma }^{n+1}+{\beta }^{n+1}{\gamma }^{n-2}+{\beta }^{n-2}{\gamma }^{n+1}\hfill \\ & =& {\alpha }^{n-2}{\beta }^{n-2}({\alpha }^3+{\beta }^3)+{\alpha }^{n-2}{\gamma }^{n-2}({\alpha }^3+{\gamma }^3)+{\beta }^{n-2}{\gamma }^{n-2}({\beta }^3+{\gamma }^3)\hfill \\ & =& 4S_{n-2}-{\alpha }^3{\beta }^3{\gamma }^3({\alpha }^{n-5}{\beta }^{n-5}+{\alpha }^{n-5}{\gamma }^{n-5}+{\beta }^{n-5}{\gamma }^{n-5})\hfill \\ & =& 4S_{n-2}-S_{n-5}.\hfill \end{array}$$

Then, we have:

$$\begin{array}{c}\hfill N_{-n}=S_{n+1}+12S_{n-2}-2S_{n-5}.\end{array}$$

Thus:

$$\begin{array}{c}\hfill G_{-n}=\frac{S_{n+1}+12S_{n-2}-2S_{n-5}}{31}=\frac{6S_{n+2}+9S_{n+1}+4S_n}{31}.\end{array}$$

□

Lemma 4.

Let $h\in {\mathbb{R}}_{>0}$. For $n\in {\mathbb{Z}}_{\ge 0}$, we have:

$$\begin{array}{ccc}& & \sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}{\alpha }^i{\beta }^j{\gamma }^k\hfill \\ & =& \frac{1}{6}\sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}(3G_{i-k+1}-2G_{i-k})(3G_{j-k+1}-2G_{j-k})\hfill \\ & & -\frac{1}{6}\sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}(3G_{i+j-2k+1}-2G_{i+j-2k}).\hfill \end{array}$$

Proof. 

For any positive integer i, j, and k, we have:

$$\begin{array}{ccc}\hfill T_{i-k}{T}_{j-k}& =& ({\alpha }^{i-k}+{\beta }^{i-k}+{\gamma }^{i-k})({\alpha }^{j-k}+{\beta }^{j-k}+{\gamma }^{j-k})\hfill \\ & =& {\alpha }^{i+j-2k}+{\beta }^{i+j-2k}+{\gamma }^{i+j-2k}+{\alpha }^{i-k}{\beta }^{j-k}+{\alpha }^{i-k}{\gamma }^{j-k}\hfill \\ & & +{\beta }^{i-k}{\alpha }^{j-k}+{\beta }^{i-k}{\gamma }^{j-k}+{\gamma }^{i-k}{\alpha }^{j-k}+{\gamma }^{i-k}{\beta }^{j-k}.\hfill \end{array}$$

Note that: ${\alpha }^{i-k}{\beta }^{j-k}=({\alpha }^{i-k}{\beta }^{j-k})({\alpha }^k{\beta }^k{\gamma }^k)={\alpha }^i{\beta }^j{\gamma }^k$. Then, the formula above can be reduced to:

$$\begin{array}{ccc}\hfill T_{i-k}{T}_{j-k}& =& {\alpha }^{i+j-2k}+{\beta }^{i+j-2k}+{\gamma }^{i+j-2k}+{\alpha }^i{\beta }^j{\gamma }^k+{\alpha }^i{\beta }^k{\gamma }^j+{\alpha }^j{\beta }^i{\gamma }^k\hfill \\ & & +{\alpha }^k{\beta }^i{\gamma }^j+{\alpha }^j{\beta }^k{\gamma }^i+{\alpha }^k{\beta }^j{\gamma }^i.\hfill \end{array}$$

(13)

From Formula (11), we have:

$$\begin{array}{ccc}\hfill & & \sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}{T}_{i-k}{T}_{j-k}\hfill \\ & =& \sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}(3G_{i-k+1}-2G_{i-k})(3G_{j-k+1}-2G_{j-k}).\hfill \end{array}$$

(14)

With Formulas (11) and (13), we obtain:

$$\begin{array}{ccc}\hfill & & \sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}{T}_{i-k}{T}_{j-k}\hfill \\ & =& \sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}(3G_{i+j-2k+1}-2G_{i+j-2k})\hfill \\ & & +6\sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}{\alpha }^i{\beta }^j{\gamma }^k.\hfill \end{array}$$

(15)

Comparing Formulas (14) and (15), we obtain:

$$\begin{array}{ccc}& & 6\sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}{\alpha }^i{\beta }^j{\gamma }^k\hfill \\ & =& \sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}(3G_{i-k+1}-2G_{i-k})(3G_{j-k+1}-2G_{j-k})\hfill \\ & & -\sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}(3G_{i+j-2k+1}-2G_{i+j-2k}).\hfill \end{array}$$

This proves Lemma 4. □

3. Proof of Theorems

Now, we complete the proof of Theorems.

Proof of Theorem 1.

For $n\ge 0$, we have:

$$\begin{array}{ccc}\hfill G_n{T}_n& =& (A{\alpha }^n+B{\beta }^n+C{\gamma }^n)({\alpha }^n+{\beta }^n+{\gamma }^n)\hfill \\ & =& A{\alpha }^{2n}+B{\beta }^{2n}+C{\gamma }^{2n}+(A+B){\alpha }^n{\beta }^n+(A+C){\alpha }^n{\gamma }^n+(B+C){\beta }^n{\gamma }^n\hfill \\ & =& G_{2n}+(A+B+C)G_{-n}-G_{-n}=G_{2n}-G_{-n}.\hfill \end{array}$$

With Lemma 2, we obtain:

$$G_{-n}=G_{2n}-3G_n{G}_{n+1}+2G_n^2.$$

Similarly, for $n<0$ we have:

$$\begin{array}{ccc}\hfill G_{-n}{T}_{-n}& =& \frac{A}{{\alpha }^{2n}}+\frac{B}{{\beta }^{2n}}+\frac{C}{{\gamma }^{2n}}+(A+B){\gamma }^n+(A+C){\beta }^n+(B+C){\alpha }^n\hfill \\ & =& G_{-2n}-G_n.\hfill \end{array}$$

With Lemma 3 and Formula (7), we obtain:

$$\begin{array}{c}\hfill G_n=G_{-2n}-G_n(3G_{-n+1}-2G_{-n})=G_{-2n}-3G_{-n}{G}_{-n+1}+2G_{-n}^2.\end{array}$$

This proves Theorem 1. □

Proof of Theorem 2.

Note that the expansion of power series:

$$\frac{1}{{(1-x)}^h}=\sum _{n=0}^{\infty }\frac{{⟨h⟩}_n}{n!}{x}^n,|x|<1.$$

Decomposing $g^h(x)$, we obtain:

$$\begin{array}{ccc}\hfill g^h(x)& =& {\left(\frac{1}{1-x-x^3}\right)}^h=\frac{1}{{(1-\alpha x)}^h{(1-\beta x)}^h{(1-\gamma x)}^h}\hfill \\ & =& \left(\sum _{n=0}^{\infty }\frac{{⟨h⟩}_n}{n!}{\alpha }^n{x}^n\right)\left(\sum _{n=0}^{\infty }\frac{{⟨h⟩}_n}{n!}{\beta }^n{x}^n\right)\left(\sum _{n=0}^{\infty }\frac{{⟨h⟩}_n}{n!}{\gamma }^n{x}^n\right)\hfill \\ & =& \sum _{n=0}^{\infty }\left(\sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}{\alpha }^i{\beta }^j{\gamma }^k\right)x^n,\hfill \end{array}$$

where $\alpha$, $\beta$, and $\gamma$ are the roots of the characteristic equation $x^3-x^2-1=0$. With Formula (8), we obtain:

$$G_{n+1}(h)=\sum _{i+j+k=n}\frac{{⟨h⟩}_i}{i!}\frac{{⟨h⟩}_j}{j!}\frac{{⟨h⟩}_k}{k!}{\alpha }^i{\beta }^j{\gamma }^k.$$

Theorem 2 follows from Lemma 4. □

Proof of Corollary 1.

Let $h\in {\mathbb{Z}}_{>0}$. With Formulas (3) and (8), we have:

$$\begin{array}{ccc}\hfill g^h(x)& =& {\left(\frac{1}{1-x-x^3}\right)}^h={\left(\sum _{n=0}^{+\infty }{G}_{n+1}{x}^n\right)}^h\hfill \\ & =& \sum _{n=0}^{+\infty }\left(\sum _{a_1+a_2+\cdots +a_h=n}{G}_{a_1+1}{G}_{a_2+1}\cdots G_{a_h+1}\right)x^n.\hfill \end{array}$$

Since $\frac{{⟨h⟩}_n}{n!}=\left(\genfrac{}{}{0pt}{}{h+n-1}{n}\right)$ for $h\in {\mathbb{Z}}_{>0}$ Corollary 1 can be obtained by Theorem 2. □

4. Discussion

The main results of this paper propose new recurrence properties of Narayana’s cow sequence in two generalized forms. First, we consider Narayana’s cows sequence $G_n$ at negative indices and construct the relationship between the sequence and itself at positive indices. This illustrates the recurrence property of the sequence at the negative index. Meanwhile, we prove that this connection holds for all integers, and not only for positive integers. Our results solve an open problem proposed by Professor Tianxin Cai completely. In addition, we obtain a computable recurrence formula for the convolved Narayana number. As a corollary, we obtain an identity related to $G_n$.