Advances in Difference Equations

, 2018:126 | Cite as

On the Lucas polynomials and some of their new identities

Open Access


The main purpose of this paper is, using the elementary and combination methods, to study the arithmetical properties of the Lucas polynomials and to obtain some new and interesting identities for them.


Fibonacci polynomials Lucas polynomials Identity Elementary method Combination method 



1 Introduction

For any non-negative integer n, the Fibonacci polynomials \(\{F_{n}(x)\}\) and Lucas polynomials \(\{L_{n}(x)\}\) are defined by the second order linear recursive formulas \(F_{n+2}(x)=xF_{n+1}(x)+F_{n}(x)\) and \(L_{n+2}(x)=xL_{n+1}(x)+L_{n}(x)\) with \(F_{0}(x)=0\), \(F_{1}(x)=1\), \(L_{0}(x)=2\), and \(L_{1}(x)=x\). The general terms of \(F_{n}(x)\) and \(L_{n}(x)\) are given by
$$ F_{n+1}(x)=\sum_{k=0}^{ [ \frac{n}{2} ] } \binom{n-k}{k}x^{n-2k} $$
$$\begin{aligned} L_{n}(x)= \sum_{k=0}^{ [ \frac{n}{2} ] } \frac{n}{n-k} \binom{n-k}{k}x^{n-2k}, \end{aligned}$$
where \(\binom{m}{n}=\frac{m!}{n!(m-n)!}\), and \([x]\) denotes the greatest integer ≤x.
It is easy to prove the identities
$$ F_{n}(x)=\frac{1}{\sqrt{x^{2}+4}} \biggl[ \biggl( \frac{x+ \sqrt{x^{2}+4}}{2} \biggr) ^{n}- \biggl( \frac{x-\sqrt{x^{2}+4}}{2} \biggr) ^{n} \biggr] $$
$$\begin{aligned} L_{n}(x)= \biggl( \frac{x+\sqrt{x^{2}+4}}{2} \biggr) ^{n}+ \biggl( \frac{x-\sqrt{x ^{2}+4}}{2} \biggr) ^{n}. \end{aligned}$$

If \(x=1\), then \(\{F_{n}(x)\}\) becomes the famous Fibonacci sequences \(\{F_{n}\}\) and \(\{L_{n}(x)\}\) becomes the Lucas sequences \(\{L_{n}\}\).

These sequences and polynomials occupy very important positions in the theory and application of mathematics, so many scholars have studied their various arithmetical properties and obtained a series of important results. For example, Ozeki [1] proved the identity
$$\begin{aligned} \sum_{k=1}^{n}F_{2k}^{2m+1}=\frac{1}{5^{m}}\sum_{j=0}^{m}\frac{(-1)^{j}}{L _{2m+1-2j}} \bigl( ^{2m+1}_{ j} \bigr) ( F_{(2m+1-2j)(2n+1)}- F _{2m+1-2j} ) . \end{aligned}$$
Prodinger [2] studied the more general summation \(\sum_{k=0}^{n}F_{2k+ \delta }^{2m+1+\epsilon }\), where \(\delta , \epsilon \in \{0, 1\}\), and obtained many interesting results.

Ma and Zhang [3] used the properties of Chebyshev polynomials to obtain some identities involving Fibonacci numbers and Lucas numbers. Wang and Zhang [4] proved some divisible properties involving Fibonacci numbers and Lucas numbers. Some of other related papers can also be found in references [5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15], here we are not going to list them all.

In this paper, we shall use the elementary and combination methods to study the arithmetical properties of Lucas polynomials, and give some new identities for them. That is, we shall prove the following results.

Theorem 1

For any positive integerhand integer\(k \geq 0\), we have
$$\begin{aligned} \sum_{n=1}^{h}\frac{L_{2n(2k+1)}(x)-2}{n}= \sum _{n=1}^{h} \frac{ \binom{h+n}{h-n}}{n}L_{2k+1}^{2n}(x). \end{aligned}$$

Theorem 2

For any integershand\(k\geq 0\), we have
$$\begin{aligned} \sum_{n=0}^{h}\frac{L_{(2n+1)(2k+1)}(x)}{2n+1}= \sum _{n=0}^{h} \frac{ \binom{h+n+1}{h-n}}{2n+1}L_{2k+1}^{2n+1}(x). \end{aligned}$$

Theorem 3

For any integers\(n\geq 1\)and\(h\geq 0\), we have the identity
$$\begin{aligned} \int_{0}^{x} L^{2n}_{2h+1}(y)\,dy = \sum_{k=0}^{n}\frac{(-1)^{n-k} \binom{2n}{n-k}}{4kh+2k+1}L_{4kh+2k+1}(x)+ \sum_{k=1}^{n} \frac{(-1)^{n-k} \binom{2n}{n-k}}{4kh+2k-1}L_{4kh+2k-1}(x). \end{aligned}$$

Theorem 4

For any integers\(n\geq 1\)and\(h\geq 0\), we have the identity
$$\begin{aligned} \int_{0}^{x} L^{2n+1}_{2h+1}(y)\,dy &= \frac{1}{2}\sum_{k=0}^{n}(-1)^{n-k} \binom{2n+1}{n-k} \biggl( \frac{L_{4kh+2k+2h+2}(x)}{2kh+k+h+1}+\frac{L _{4kh+2k+2h}(x)}{2kh+k+h} \biggr) \\ &\quad {}- \sum_{k=0}^{n}(-1)^{n-k} \binom{2n+1}{n-k} \biggl( \frac{1}{2hk+k+h+1}+\frac{1}{2hk+k+h} \biggr) . \end{aligned}$$

Taking \(k=0\), from Theorems 1 and 2 we can deduce the following:

Corollary 1

For any positive integerh, we have the identities
$$\begin{aligned} \sum_{n=1}^{h}\frac{L_{2n}(x)-2}{n}= \sum _{n=1}^{h} \frac{ \binom{h+n}{h-n}}{n}x^{2n} \end{aligned}$$
$$\begin{aligned} \sum_{n=0}^{h}\frac{L_{2n+1}(x)}{2n+1}= \sum _{n=0}^{h} \frac{ \binom{h+n+1}{h-n}}{2n+1}x^{2n+1}. \end{aligned}$$

If \(x=1\) and \(k=0\), then we also have the following:

Corollary 2

For any positive integerh, we have the identities
$$\begin{aligned} \sum_{n=1}^{h}\frac{L_{2n}}{n}= \sum _{n=1}^{h} \frac{\binom{h+n}{h-n}+2}{n}\quad \textit{and}\quad \sum_{n=0}^{h} \frac{L_{2n+1}}{2n+1}= \sum_{n=0}^{h} \frac{ \binom{h+n+1}{h-n}}{2n+1}. \end{aligned}$$

From Theorems 3 and 4 we can deduce the following corollaries.

Corollary 3

For any integers\(n\geq 1\)and\(h\geq 0\), we have
$$\begin{aligned} \int_{0}^{1} L^{2n}_{2h+1}(y)\,dy = \sum_{k=0}^{n}\frac{(-1)^{n-k} \binom{2n}{n-k}}{4kh+2k+1}L_{4kh+2k+1}+ \sum_{k=1}^{n} \frac{(-1)^{n-k} \binom{2n}{n-k}}{4kh+2k-1}L_{4kh+2k-1}. \end{aligned}$$

Corollary 4

For any integers\(n\geq 1\)and\(h\geq 0\), we have
$$\begin{aligned} \int_{0}^{1} L^{2n+1}_{2h+1}(y)\,dy = \sum_{k=0}^{n} \frac{(-1)^{n-k}}{2} \binom{2n+1}{n-k} \biggl( \frac{L_{4kh+2k+2h+2}-2}{2kh+k+h+1}+\frac{L_{4kh+2k+2h}-2}{2kh+k+h} \biggr) . \end{aligned}$$

2 Several simple lemmas

Lemma 1

For any positive integersn, we have the identities
$$\begin{aligned}& \int_{0}^{x}L_{2n}(y)\,dy= \frac{L_{2n+1}(x)}{2n+1}+ \frac{L_{2n-1}(x)}{2n-1}; \\& \int_{0}^{x}L_{2n+1}(y)\,dy= \frac{L_{2n+2}(x)}{2n+2}+ \frac{L_{2n}(x)}{2n} -\frac{2n+1}{n(n+1)}. \end{aligned}$$


Note the identities \(( x+\sqrt{x^{2}+4} ) '= 1+\frac{x}{\sqrt{x ^{2}+4}}=\frac{x+\sqrt{x^{2}+4}}{\sqrt{x^{2}+4}}\) and \(( x-\sqrt{x ^{2}+4} ) '= 1-\frac{x}{\sqrt{x^{2}+4}}=-\frac{x- \sqrt{x^{2}+4}}{\sqrt{x^{2}+4}}\). From the definitions of the polynomials \(F_{n}(x)\) and \(L_{n}(x)\), we have
$$\begin{aligned} L'_{n}(x)&= \frac{n}{\sqrt{x^{2}+4}} \biggl( \frac{x+\sqrt{x^{2}+4}}{2} \biggr) ^{n}-\frac{n}{\sqrt{x^{2}+4}} \biggl( \frac{x-\sqrt{x^{2}+4}}{2} \biggr) ^{n} \\ &=n\cdot F_{n-1}(x). \end{aligned}$$
Applying (3), the integration by parts, and the recursive formulae of \(L_{n}(x)\) and \(F_{n}(x)\), we have
$$\begin{aligned} \int_{0}^{x}L_{n}(y)\,dy &= xL_{n}(x)- \int_{0}^{x} y L'_{n}(y)\,dy \\ &= xL _{n}(x)- n \int_{0}^{x}y F_{n-1}(y)\,dy \\ &=L_{n+1}(x)-L_{n-1}(x)- n \int_{0}^{x} \bigl( F_{n}(y)-F_{n-2}(y) \bigr)\,dy \\ &=L_{n+1}(x)-L_{n-1}(x)-\frac{n}{n+1} \bigl( L_{n+1}(x)-L_{n+1}(0) \bigr) + \frac{n}{n-1} \bigl( L_{n-1}(x)-L_{n-1}(0) \bigr) \\ &=\frac{L_{n+1}(x)}{n+1}+ \frac{L_{n-1}(x)}{n-1}+ \frac{nL_{n+1}(0)}{n+1} - \frac{nL_{n-1}(0)}{n-1}. \end{aligned}$$
If \(n=2k\), then note that \(L_{2k+1}(0)=L_{2k-1}(0)=0\). From (4) we have
$$\begin{aligned} \int_{0}^{x}L_{2k}(y)\,dy = \frac{L_{2k+1}(x)}{2k+1}+ \frac{L_{2k-1}(x)}{2k-1}. \end{aligned}$$
If \(n=2k+1\), then by (2) we have \(L_{2k+2}(0)=L_{2k}(0)=2\). From (4) we have
$$\begin{aligned} \int_{0}^{x}L_{2k+1}(y)\,dy = \frac{L_{2k+2}(x)}{2k+1}+ \frac{L_{2k}(x)}{2k}- \frac{2k+1}{k(k+1)}. \end{aligned}$$
Now Lemma 1 follows from (5) and (6). □

Lemma 2

For any positive integernand non-negative integerk, we have the identity
$$\begin{aligned} L_{n} \bigl( L_{2k+1}(x) \bigr) =L_{n(2k+1)}(x). \end{aligned}$$


Let \(\alpha =\frac{x+\sqrt{x^{2}+4}}{2}\) and \(\beta = \frac{x-\sqrt{x^{2}+4}}{2}\). Then replace x by \(L_{2k+1}(x)\) in (2) and note that \(\alpha^{2k+1}\beta^{2k+1}=-1\), we have
$$\begin{aligned} L_{2k+1}(x)+\sqrt{L^{2}_{2k+1}(x)+4}&= \alpha^{2k+1}+\beta^{2k+1}+\sqrt{ \bigl( \alpha^{2k+1}+ \beta^{2k+1} \bigr) ^{2}+4} \\ &=\alpha^{2k+1}+\beta^{2k+1}+ \sqrt{\alpha^{2(2k+1)}+ \beta^{2(2k+1)}-2+4} \\ &=\alpha^{2k+1}+\beta^{2k+1}+\sqrt{ \bigl( \alpha^{2k+1}-\beta^{2k+1} \bigr) ^{2}}=2 \alpha^{2k+1} \end{aligned}$$
$$\begin{aligned} L_{2k+1}(x)-\sqrt{L^{2}_{2k+1}(x)+4}&=\alpha^{2k+1}+\beta^{2k+1}-\sqrt{ \bigl( \alpha^{2k+1}+ \beta^{2k+1} \bigr) ^{2}+4} \\ &=\alpha^{2k+1}+\beta^{2k+1}-\sqrt{ \bigl( \alpha^{2k+1}-\beta^{2k+1} \bigr) ^{2}}=2 \beta^{2k+1}. \end{aligned}$$
From (2) we have the identity
$$\begin{aligned} L_{n} \bigl( L_{2k+1}(x) \bigr) =& \biggl( \frac{L_{2k+1}+\sqrt{L^{2} _{2k+1}+4}}{2} \biggr) ^{n}+ \biggl( \frac{L_{2k+1}- \sqrt{L^{2}_{2k+1}+4}}{2} \biggr) ^{n} \\ =&\alpha^{n(2k+1)}+ \beta^{n(2k+1)}=L_{n(2k+1)}(x). \end{aligned}$$
This proves Lemma 2. □

Lemma 3

For any non-negative integern, we have the identities
$$\begin{aligned} x^{2n}= \frac{(-1)^{n}}{2}\binom{2n}{n}\cdot L_{0}(x)+ \sum_{k=1}^{n}(-1)^{n-k} \binom{2n}{n-k}\cdot L_{2k}(x) \end{aligned}$$
$$\begin{aligned} x^{2n+1}= \sum_{k=0}^{n}(-1)^{n-k} \binom{2n+1}{n-k}\cdot L_{2k+1}(x). \end{aligned}$$


From the definition of \(L_{n}(x)\) we know that \(L_{2k}(x)\) is an even function. So we may suppose that
$$\begin{aligned} x^{2n}= \sum_{k=0}^{n} a_{k} \cdot L_{2k}(x). \end{aligned}$$
Taking \(x=2i\cos \theta \) in (7) and noting that \(x^{2}+4=4-4\cos^{2} \theta =4\sin^{2}\theta \), from Euler’s formula we have
$$\begin{aligned} L_{2k} ( 2i\cos \theta ) &= \biggl( \frac{2i\cos \theta + \sqrt{4 \sin^{2}\theta }}{2} \biggr) ^{2k}+ \biggl( \frac{2i\cos \theta - \sqrt{4 \sin^{2}\theta }}{2} \biggr) ^{2k} \\ &= ( i\cos \theta +\sin \theta ) ^{2k}+ ( i\cos \theta - \sin \theta ) ^{2k} \\ &=(-1)^{k} ( \cos \theta -i\sin \theta ) ^{2k}+(-1)^{k} ( \cos \theta +i\sin \theta ) ^{2k} \\ &=(-1)^{k} \cdot 2\cdot \cos (2k\theta ). \end{aligned}$$
Then from (7) and (8) we have
$$\begin{aligned} (-1)^{n}4^{n}\cos^{2n} \theta = \sum _{k=0}^{n} a_{k}\cdot L_{2k} ( 2i \cos \theta ) =2\sum_{k=0}^{n} a_{k}\cdot (-1)^{k}\cos ( 2k \theta ) . \end{aligned}$$
Note the identities
$$ \int_{0}^{\pi }2\cos (m\theta )\cos (n\theta )\,d\theta = \textstyle\begin{cases} \pi , &\mbox{if } m=n\neq 0, \\ 0, &\mbox{if } m\neq n, \\ 2\pi , &\mbox{if } m= n=0 \end{cases} $$
$$ \int_{0}^{\pi }\cos^{2n} (\theta )\cos ( 2k \theta )\,d\theta =\pi \cdot \frac{(2n)!}{(2n-2k)!!(2n+2k)!!}=\frac{\pi }{4^{n}}\cdot \binom{2n}{n-k}. $$
From (9) and (10) we have
$$\begin{aligned} a_{k}\cdot (-1)^{k} \pi =(-1)^{n} 4^{n} \int_{0}^{\pi }\cos^{2n} ( \theta )\cos ( 2k \theta )\,d\theta =(-1)^{n}\pi \cdot \binom{2n}{n-k} \end{aligned}$$
$$\begin{aligned} a_{0}=\frac{(-1)^{n}}{2} \cdot \binom{2n}{n}\quad \mbox{and}\quad a_{k}=(-1)^{n-k} \cdot \binom{2n}{n-k}, \quad 1\leq k\leq n. \end{aligned}$$
Combining (7), (9), and (11), we can deduce the first identity of Lemma 3.
Similarly, since \(L_{2k+1}(x)\) is an odd function, we can suppose that
$$\begin{aligned} x^{2n+1}= \sum_{k=0}^{n} b_{k} \cdot L_{2k+1}(x). \end{aligned}$$
Taking \(x=2i\cos \theta \) in (12) and noting that
$$\begin{aligned} L_{2k} ( 2i\sin \theta ) &= \biggl( \frac{2i\sin \theta + \sqrt{4 \cos^{2}\theta }}{2} \biggr) ^{2k}+ \biggl( \frac{2i\sin \theta - \sqrt{4 \cos^{2}\theta }}{2} \biggr) ^{2k} \\ &= ( i\sin \theta +\cos \theta ) ^{2k}+ ( i\sin \theta - \cos \theta ) ^{2k} \\ &= ( \cos \theta +i\sin \theta ) ^{2k}+ ( \cos \theta -i \sin \theta ) ^{2k} \\ &=\cos (2k\theta )+i\sin (2k\theta )+\cos (2k\theta )-i\sin (2k \theta ) \\ &=2 \cdot \cos (2k\theta ), \end{aligned}$$
we have
$$\begin{aligned} (-1)^{n} 4^{n} \cos^{2n+1} \theta = \sum _{k=0}^{n} b_{k} \cdot (-1)^{k} \cos \bigl( (2k+1)\theta \bigr) . \end{aligned}$$
From (10) and (13) we may immediately deduce that
$$\begin{aligned} b_{k}= \frac{2(-1)^{n-k}}{\pi }\cdot 4^{n}\cdot \int_{0}^{\pi }\cos^{2n+1} (\theta ) \cos \bigl( (2k+1)\theta \bigr)\,d\theta =(-1)^{n-k} \binom{2n+1}{n-k}. \end{aligned}$$
Now the second identity of Lemma 3 follows from (12) and (14). □

3 Proofs of the theorems

Using the lemmas in Sect. 2, we can prove our theorems easily. First we prove Theorem 2. Similarly, we can also deduce Theorem 1 and then omit its proving process here. From Lemma 1 and the definition of \(L_{n}(x)\), we have
$$\begin{aligned} \sum_{n=0}^{h} L_{2n}(x)&=\sum _{n=0}^{h} \bigl( \alpha^{2n}+ \beta^{2n} \bigr) \\ &=\frac{ \alpha^{2h+2}-1}{\alpha^{2}-1}+\frac{\beta^{2h+2}-1}{\beta^{2}-1} \\ &=\frac{L_{2h+2}(x)-L_{2h}(x)}{L_{2}(x)-2}+1 \\ &=\frac{L_{2h+1}(x)}{x}+1. \end{aligned}$$
Applying (15) and Lemma 1, we have
$$\begin{aligned} & \int_{0}^{x}\sum_{n=1}^{h} L_{2n}(y)\,dy= \int_{0}^{x} \biggl( \frac{L _{2h+1}(y)}{y}-1 \biggr)\,dy= \sum_{n=1}^{h} \biggl( \frac{L_{2n+1}(x)}{2n+1}+ \frac{L_{2n-1}(x)}{2n-1} \biggr) \end{aligned}$$
$$\begin{aligned} 2\sum_{n=0}^{h}\frac{L_{2n+1}(x)}{2n+1}&= \int_{0}^{x} \frac{L_{2h+1}(y)}{y}\,dy + \frac{L_{2h+1}(x)}{2h+1} \\ &= \sum_{k=0}^{h}\frac{2h+1}{2h+1-k} \binom{2h+1-k}{k} \int_{0}^{x}y ^{2h-2k}\,dy \\ & \quad {}+ \frac{1}{2h+1}\sum_{k=0}^{h} \frac{2h+1}{2h+1-k}\binom{2h+1-k}{k}x ^{2h+1-2k} \\ &=\sum_{k=0}^{h}\frac{2}{2h+1-2k} \binom{2h+1-k}{k}\cdot x^{2h+1-2k} \\ &=2 \sum_{n=0}^{h} \frac{\binom{h+n+1}{h-n}}{2n+1}x^{2n+1}. \end{aligned}$$
Now Theorem 2 follows from (16) and Lemma 2 with \(x=L_{2k+1}(y)\).
To prove Theorem 3, taking \(x=L_{2h+1}(y)\) in Lemma 3, we have
$$\begin{aligned} L_{2h+1}^{2n}(y)= (-1)^{n}\binom{2n}{n}+ \sum _{k=1}^{n}(-1)^{n-k} \binom{2n}{n-k}\cdot L_{2k(2h+1)}(y). \end{aligned}$$
Integrating for y from 0 to x in (17), then applying Lemma 1, we may immediately deduce
$$\begin{aligned} \int_{0}^{x} L^{2n}_{2h+1}(y)\,dy = \sum_{k=0}^{n}\frac{(-1)^{n-k} \binom{2n}{n-k}}{4kh+2k+1}L_{4kh+2k+1}(x)+ \sum_{k=1}^{n} \frac{(-1)^{n-k} \binom{2n}{n-k}}{4kh+2k-1}L_{4kh+2k-1}(x). \end{aligned}$$
This proves Theorem 3.

Similarly, we can also deduce Theorem 4.



The author would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the N.S.F. (Grant No. 11771351) of P.R. China.

Authors’ contributions

All authors read and approved the final manuscript.

Competing interests

The author declares that she has no competing interests.


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Authors and Affiliations

  1. 1.School of Information EngineeringXi’an UniversityXi’anP.R. China

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