Combinatorial identities concerning trigonometric functions and Fibonacci/Lucas numbers

Yulei Chen; Yingming Zhu; Dongwei Guo; Yulei Chen; Yingming Zhu; Dongwei Guo

doi:10.3934/math.2024455

AIMS Mathematics

2024, Volume 9, Issue 4: 9348-9363. doi: 10.3934/math.2024455

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Research article

Combinatorial identities concerning trigonometric functions and Fibonacci/Lucas numbers

1.
School of Mathematics and Statistics, Zhoukou Normal University, Zhoukou (Henan), China
2.
School of Economics and Management, Nanjing University of Science and Technology, Nanjing (Jiangsu), China

Received: 29 December 2023 Revised: 29 February 2024 Accepted: 05 March 2024 Published: 07 March 2024
MSC : 05A19, 11B65

In this work, by means of the generating function method and the De Moivre's formula, we derive some interesting combinatorial identities concerning trigonometric functions and Fibonacci/Lucas numbers. One of them confirms the formula proposed recently by Svinin (2022).

Keywords:

Citation: Yulei Chen, Yingming Zhu, Dongwei Guo. Combinatorial identities concerning trigonometric functions and Fibonacci/Lucas numbers[J]. AIMS Mathematics, 2024, 9(4): 9348-9363. doi: 10.3934/math.2024455

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Abstract

1. Introduction and motivation

The Fibonacci and Lucas numbers are defined by the following recurrence relations ^[5,6]

$\begin{cases} F_0 = 0, F_1 = 1, \\ F_n = F_{n-1}+F_{n-2}, n\geq2 \end{cases} \quad\text{and}\quad \begin{cases} L_0 = 2, L_1 = 1, \\ L_n = L_{n-1}+L_{n-2}, n\geq2 \end{cases}$

with the explicit formulae of Binet forms

$F_n = \frac{ \alpha^n-\beta^n}{ \alpha-\beta} \quad\text{and}\quad L_n = \alpha^n+\beta^n,$

where $\alpha = \frac{1+\sqrt5}{2}$ and $\beta = \frac{1-\sqrt5}{2}$ . It is interesting that they can be expressed by binomial coefficients, which can be found in Koshy ^[8, Eqs 12.1 and 13.5] and ^[1]

$\sum\limits_{k = 0}^{\lfloor\frac{n-1}{2}\rfloor} \binom{n-k-1}{k} = F_n \quad\text{and}\quad \sum\limits_{k = 0}^{\lfloor\frac{n}{2}\rfloor} \frac{n}{n-k} \binom{n-k}{k} = L_n \quad \text{with}\quad n > 0.$

In fact, there are two general binomial identities, which can be found in Carlitz ^[2, Page 23], Comtet ^[3, §4.9] and ^[5]

$\begin{align} \sum\limits_{0\leq k < n/2}(-1)^k \binom{n-k-1}{k} (uv)^k(u+v)^{n-2k-1} & = \frac{u^n-v^n}{u-v}, \\ \sum\limits_{0\leq k < n/2}(-1)^k\frac{n}{n-k} \binom{n-k}{k}(uv)^k(u+v)^{n-2k} & = u^n+v^n. \end{align}$

When specifying $u = \alpha$ and $v = \beta$ , they reduce to the formulae mentioned above.

Recently, Svinin ^[10] proposed a problem, where including demands to show that

$\begin{align} \sum\limits_{k = 0}^{\lfloor\frac{n-1}{3}\rfloor} \frac1{2k+1} \binom{n-k-1}{2k}\bigg(\frac4{27}\bigg)^k = \frac{4^n-1}{3^{n-1}(2n+1)}, \end{align}$

(1.1)

which can be rewritten as

$\sum\limits_{k = 0}^{\lfloor\frac{n-1}{3}\rfloor} \frac{2n+1}{n-k} \binom{n-k}{2k+1} \bigg(\frac4{27}\bigg)^k = \frac{4^n-1}{3^{n-1}}.$

When making attempts to resolve (1.1), we find that a series of similar identities, concerning trigonometric function or Fibonacci/Lucas numbers, can be established. In the process of proving some of these formulae, we will use the De Moivre's formula

$(\cos\theta+\textbf{i}\sin\theta)^n = \cos n\theta+\textbf{i}\sin n\theta,$

by which we can get the following identities

$\begin{align} &(\cos\theta+\textbf{i}\sin\theta)^n +(\cos\theta-\textbf{i}\sin\theta)^n = 2\cos(n\theta), \end{align}$

(1.2)

$\begin{align} &(\cos\theta+\textbf{i}\sin\theta)^n -(\cos\theta-\textbf{i}\sin\theta)^n = 2\textbf{i}\sin(n\theta), \end{align}$

(1.3)

where $\textbf{i}$ denotes the imaginary unit.

In the following sections, we will evaluate, by means of generating functions and binomial linear relations, the following binomial sums

$\sum\limits_{k = 0}^{\lfloor\frac{n-\delta}{3}\rfloor} \frac{\tau n+\delta}{n-\upsilon k} \binom{n-\upsilon k}{\tau k+\delta}x^k, \quad\text{with}\quad \tau, \upsilon, \delta\in\{0, 1, 2\}.$

Specifically, in the next section, we shall consider the sums

$\begin{align*} \sum\limits_{k = 0}^{\lfloor\frac{n-\delta}{3}\rfloor} \frac{2n+\delta}{n-k} \binom{n-k}{2k+\delta}x^k, \end{align*}$

where $\delta\in\{0, 1\}$ . Then in Section 3, we will examine the binomial sums

$\begin{align*} \sum\limits_{k = 0}^{\lfloor\frac{n}{3}\rfloor} \frac{n}{n-2k} \binom{n-2k}{k}x^k. \end{align*}$

Finally in Section 4, the paper will end with two formulae, one of which is equivalent to (1.1).

For convenience, throughout the paper we shall make use of the notations: for a real number $x$ , the symbol $\lfloor x \rfloor$ denotes the greatest integer $\leq x$ . When $n$ is a natural number, the symbol $i\equiv_nj$ stands for " $i$ is congruent to $j$ modulo $n$ ". For a formal power series $f(x)$ , its coefficient of $x^n$ denote by $[x^n]f(x)$ . Considering two formal power series $f(x)$ and $g(x)$ , the following two equalities are well known ^[4,9]

$\begin{align} &[x^n]xf(x) = [x^{n-1}]f(x), \end{align}$

(1.4)

$\begin{align} &[x^n](\alpha f(x)+\beta g(x)) = \alpha[x^n]f(x)+\beta[x^n]g(x), \end{align}$

(1.5)

which will be used frequently in the proofs in subsequent sections.

2. Case $\upsilon=1$ and $\tau=2$

Lemma 1. The generating function of the sequence $A_n(x) = \sum\limits_{k = 0}^{n} \binom{n-k+\delta}{2k+\delta}x^k$ is

$\begin{align} \mathcal{{A}}(z, x, \delta) = \sum\limits_{n = 0}^{\infty} A_n(x)z^n = \frac{1}{(1-z)^{\delta+1}-xz^3(1-z)^{\delta-1}}. \end{align}$

(2.1)

Proof. By exchanging the summation order, we have

$\begin{align*} \mathcal{{A}}(z, x, \delta) = \sum\limits_{n = 0}^{\infty} \sum\limits_{k = 0}^{n} \binom{n-k+\delta}{2k+\delta}x^kz^n = \sum\limits_{k = 0}^{\infty}x^k \sum\limits_{n = k}^{\infty} \binom{n-k+\delta}{2k+\delta}z^n. \end{align*}$

Making the replacement $n\to n+k-\delta$ , the last equation becomes

$\begin{align*} \mathcal{{A}}(z, x, \delta) = \sum\limits_{k = 0}^{\infty}x^kz^{k-\delta} \sum\limits_{n = 0}^{\infty} \binom{n}{2k+\delta}z^n = \sum\limits_{k = 0}^{\infty}x^kz^{k-\delta} \sum\limits_{n = 2k+\delta}^{\infty} \binom{n}{2k+\delta}z^n. \end{align*}$

Keeping in mind the formula ^[11, Eq 5.57]

$\sum\limits_{k = 0}^{\infty} \binom{k}{n}z^k = \frac{z^n}{(1-z)^{n+1}},$

we get the generating function

$\begin{align*} \mathcal{{A}}(z, x, \delta) = \sum\limits_{k = 0}^{\infty} \frac{x^kz^{3k}}{(1-z)^{2k+\delta+1}} = \frac{1}{(1-z)^{\delta+1}-xz^3(1-z)^{\delta-1}}.\end{align*}$

□

2.1. Four combinatorial identities concerning trigonometric functions

Firstly, we establish four combinatorial identities concerning trigonometric functions.

Theorem 2 ( $n\in\mathbb{{N}}$ ).

$\begin{align} {\sum\limits_{k = 0}^{\lfloor\frac{n-\delta}{3}\rfloor} \frac{2n+\delta}{n-k} \binom{n-k}{2k+\delta}2^k = \begin{cases} 2^n+2\cos\frac{n\pi}{2}, &\delta = 0;\\ 2^n+\sqrt2\sin\frac{(2n-1)\pi}{4}, &\delta = 1. \end{cases}} \end{align}$

(2.2)

Proof. Letting $\delta = 1$ and $x = 2$ in (2.1), we have

$\begin{align*} \mathcal{{A}}(z, 2, 1) = \frac{1}{(1-z)^2-2z^3} = \frac1{(1-2z)(1+\textbf{i}z)(1-\textbf{i}z)}. \end{align*}$

According to partial fraction decomposition, it can be decomposed into

$\begin{align*} \mathcal{{A}}(z, 2, 1) = \frac{1}{(1-2z)(1+\textbf{i}z)(1-\textbf{i}z)} = \frac{A}{1-2z}+\frac{B}{1+\textbf{i}z}+\frac{C}{1-\textbf{i}z}, \end{align*}$

where $A, B$ and $C$ are three parameters to be determined. Multiplying both sides of the equation by $1-2z$ and taking the limit at $z\to \frac12$ yields

$\begin{align*} A = \lim\limits_{z\to \frac12}\frac{1-2z}{(1-2z)(1+\textbf{i}z)(1-\textbf{i}z)} = \lim\limits_{z\to \frac12}\frac1{(1+\textbf{i}z)(1-\textbf{i}z)} = \frac45. \end{align*}$

Similarly, multiplying both sides of the equation by $1+\textbf{i}z$ and taking the limit at $z\to \textbf{i}$ yields

$\begin{align*} B = \lim\limits_{z\to \textbf{i}}\frac{1+\textbf{i}z}{(1-2z)(1+\textbf{i}z)(1-\textbf{i}z)} = \lim\limits_{z\to \textbf{i}}\frac1{(1-2z)(1-\textbf{i}z)} = \frac{2\textbf{i}+1}{10}. \end{align*}$

Multiplying both sides of the equation by $1-\textbf{i}z$ and taking the limit at $z\to -\textbf{i}$ yields

$\begin{align*} C = \lim\limits_{z\to -\textbf{i}}\frac{1-\textbf{i}z}{(1-2z)(1+\textbf{i}z)(1-\textbf{i}z)} = \lim\limits_{z\to -\textbf{i}}\frac1{(1-2z)(1+\textbf{i}z)} = -\frac{2\textbf{i}-1}{10}. \end{align*}$

Therefore, we can decompose the rational fraction

$\begin{align*} \mathcal{{A}}(z, 2, 1) = \frac{1}{(1-z)^2-2z^3} = \frac45\frac{1}{1-2z} +\frac{2\textbf{i}+1}{10}\frac{1}{1+\textbf{i}z} -\frac{2\textbf{i}-1}{10}\frac{1}{1-\textbf{i}z}. \end{align*}$

Keeping in mind (1.4) and the fact that $\frac{1}{1-x} = \sum\limits_{n = 0}^{\infty}x^n$ , we can determine the coefficient of $z^n$

$\begin{align*} [z^n]\frac{1}{(1-z)^2-2z^3} = \frac{2^{n+2}}{5}+\frac{1}{10} \big\{[\textbf{i}^n+(-\textbf{i})^n] -2\textbf{i}[\textbf{i}^n-(-\textbf{i})^n]\big\}. \end{align*}$

By means of the relations (1.2) and (1.3), we can get the formulae

$\begin{align*} \textbf{i}^n+(-\textbf{i})^n = 2\cos\frac{n\pi}{2} \quad\text{and}\quad \textbf{i}^n-(-\textbf{i})^n = 2\textbf{i}\sin\frac{n\pi}{2}, \end{align*}$

which results in the following binomial summation formula

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k+1}2^k = \frac{2^{n+1}+\sin\frac{n\pi}{2} -2\cos\frac{n\pi}{2}}{5}. \end{align}$

(2.3)

Noting that when $\delta = 0$ and $x = 2$

$\mathcal{{A}}(z, 2, 0) = (1-z)\mathcal{{A}}(z, 2, 1),$

we have

$[z^n]\mathcal{{A}}(z, 2, 0) = [z^n]\mathcal{{A}}(z, 2, 1) -[z^{n-1}]\mathcal{{A}}(z, 2, 1),$

which yields another combinatorial identity below.

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k}2^k = \frac{2^{n+1}+3\cos\frac{n\pi}{2} +\sin\frac{n\pi}{2}}{5}. \end{align}$

(2.4)

Then by means of the binomial relation below

$\begin{align} \frac{2n+\delta}{n-k} \binom{n-k}{2k+\delta} = 3 \binom{n-k}{2k+\delta}- \binom{n-k-1}{2k+\delta}, \end{align}$

(2.5)

and after a little manipulation, we can derive, from (2.4) and (2.3), the identity in Theorem 2.□

Theorem 3 ( $n\in\mathbb{{N}}$ ).

$\begin{align} \sum\limits_{k = 0}^{\lfloor\frac{n-\delta}{3}\rfloor} \frac{2n+\delta}{n-k} \binom{n-k}{2k+\delta}(-4)^k = \begin{cases} 2^{n+1}\cos(n\theta_1)+(-1)^n, &\delta = 0;\\ 2^{n+\frac12}\sin(n\theta_1+\theta_2)-\frac{(-1)^n}{2}, &\delta = 1, \end{cases} \end{align}$

(2.6)

where $\theta_1 = \arctan\frac{\sqrt7}{3}$ and $\theta_2 = \arctan\frac{\sqrt7}{7}$ .

Proof. Letting $\delta = 1$ and $x = -4$ in (2.1), we have

$(1-z)^2+4z^3 = 4(z+1)\Bigg(z-\frac{3+\sqrt7\textbf{i}}{8}\Bigg) \Bigg(z-\frac{3-\sqrt7\textbf{i}}{8}\Bigg).$

By using partial fractional decomposition, we get the generating function

$\begin{align*} \mathcal{{A}}(z, -4, 1)& = \frac{1}{(1-z)^2+4z^3}\\ & = \frac18\frac1{1+z} +\frac{49+13\sqrt7\textbf{i}}{112} \frac1{1-\frac{3-\sqrt7\textbf{i}}{2}z} +\frac{49-13\sqrt7\textbf{i}}{112} \frac1{1-\frac{3+\sqrt7\textbf{i}}{2}z}. \end{align*}$

Computing the coefficient of $z^n$ , we obtain

$\begin{align*} [z^n]\mathcal{{A}}(z, -4, 1) = &\frac{(-1)^n}{8} +\frac{1}{112\cdot2^n} \Big\{49\big[(3+\sqrt7\textbf{i})^n +(3-\sqrt7\textbf{i})^n\big]\\ &-13\sqrt7\textbf{i}\big[(3+\sqrt7\textbf{i})^n -(3-\sqrt7\textbf{i})^n\big]\Big\}. \end{align*}$

By means of the relations (1.2) and (1.3), we have

$\begin{align*} \Big(3+\sqrt7\textbf{i}\Big)^n +\Big(3-\sqrt7\textbf{i}\Big)^n = 2\cdot4^n\cos\Bigg(n\arctan\frac{\sqrt7}{3}\Bigg), \\ \Big(3+\sqrt7\textbf{i}\Big)^n -\Big(3-\sqrt7\textbf{i}\Big)^n = 2\cdot4^n\textbf{i}\sin\Bigg(n\arctan\frac{\sqrt7}{3}\Bigg), \end{align*}$

we can establish the following combinatorial identity:

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k+1}(-4)^k = \frac{2^n}{56} \Bigg\{7\cos\Bigg(n\arctan\frac{\sqrt7}{3}\Bigg) +11\sqrt7\sin\Bigg(n\arctan\frac{\sqrt7}{3}\Bigg)\Bigg\}-\frac{(-1)^n}{8}. \end{align}$

(2.7)

Keeping in mind the relations

$\mathcal{{A}}(z, -4, 0) = (1-z)\mathcal{{A}}(z, -4, 1)$

and

$[z^n]\mathcal{{A}}(z, -4, 0) = [z^n]\mathcal{{A}}(z, -4, 1)-[z^{n-1}]\mathcal{{A}}(z, -4, 1),$

we can derive the identity below:

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k}(-4)^k = \frac{(-1)^n}{4}+\frac{2^n}{28} \Bigg\{21\cos\Bigg(n\arctan\frac{\sqrt7}{3}\Bigg) +\sqrt7\sin\Bigg(n\arctan\frac{\sqrt7}{3}\Bigg)\Bigg\}. \end{align}$

(2.8)

By employing the relation (2.5), we can get, from (2.8) and (2.7), the identity in Theorem 3.□

Theorem 4 ( $n\in\mathbb{{N}}$ ).

$\begin{align} \sum\limits_{k = 0}^{\lfloor\frac{n-\delta}{3}\rfloor} \frac{2n+\delta}{n-k} \binom{n-k}{2k+\delta}(-18)^k = \begin{cases} 2\cdot3^n\cos(n\theta_1)+(-2)^n, &\delta = 0;\\ 3^{n-1}\sqrt6\sin(n\theta_1+\theta_2)-\frac{(-2)^n}{3}, &\delta = 1, \end{cases} \end{align}$

(2.9)

where $\theta_1 = \arctan\frac{\sqrt5}{2}$ and $\theta_2 = \arctan\frac{\sqrt5}{5}$ .

Proof. By setting $t = 1$ and $x = -18$ in (2.1), we have

$\begin{align*} (1-z)^2+18z^3 = 18\bigg(z+\frac12\bigg) \bigg(z-\frac{2+\sqrt5\textbf{i}}{9}\bigg) \bigg(z-\frac{2-\sqrt5\textbf{i}}{9}\bigg). \end{align*}$

By means of the partial fractional decomposition, we can get the generating function

$\begin{align*} \mathcal{{A}}(z, -18, 1) = \frac{4}{21}\frac{1}{1+2z} +\frac{85+16\sqrt5\textbf{i}}{210} \frac1{1-\Big(2-\sqrt5\textbf{i}\Big)z} +\frac{85-16\sqrt5\textbf{i}}{210} \frac1{1-\Big(2+\sqrt5\textbf{i}\Big)z}, \end{align*}$

and determine the coefficient of $z^n$

$\begin{align*} [z^n]\mathcal{{A}}(z, -18, 1) = &\frac{4(-2)^n}{21}+\frac{3^n}{210} \Bigg\{85\bigg[\bigg(\frac23+\frac{\sqrt5}{3}\textbf{i}\bigg)^n +\bigg(\frac23-\frac{\sqrt5}{3}\textbf{i}\bigg)^n\bigg]\\ &-16\sqrt5\textbf{i}\bigg[\bigg(\frac23+\frac{\sqrt5}{3}\textbf{i}\bigg)^n -\bigg(\frac23-\frac{\sqrt5}{3}\textbf{i}\bigg)^n\bigg]\Bigg\}. \end{align*}$

Then we can derive, by utilizing the relations

$\begin{align*} &\bigg(\frac23+\frac{\sqrt5}{3}\textbf{i}\bigg)^n +\bigg(\frac23-\frac{\sqrt5}{3}\textbf{i}\bigg)^n = 2\cos\bigg(n\arctan\frac{\sqrt5}{2}\bigg), \\ &\bigg(\frac23+\frac{\sqrt5}{3}\textbf{i}\bigg)^n -\bigg(\frac23-\frac{\sqrt5}{3}\textbf{i}\bigg)^n = 2\textbf{i}\sin\bigg(n\arctan\frac{\sqrt5}{2}\bigg), \end{align*}$

the following summation formula:

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k+1}(-18)^k = \frac{4(-2)^{n-1}}{21}+\frac{3^n}{105}\chi(n), \end{align}$

(2.10)

where

$\chi(n) = 10\cos\bigg(n\arctan\frac{\sqrt5}{2}\bigg) +13\sqrt5\sin\bigg(n\arctan\frac{\sqrt5}{2}\bigg).$

According to the relation

$\mathcal{{A}}(z, -18, 0) = (1-z)\mathcal{{A}}(z, -18, 1),$

we have

$[z^n]\mathcal{{A}}(z, -18, 0) = [z^n]\mathcal{{A}}(z, -18, 1)-[z^{n-1}]\mathcal{{A}}(z, -18, 1),$

which leads to the following binomial sum:

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k}(-18)^k = \frac{2(-2)^n}{7}+\frac{3^n}{35} \Bigg\{25\cos\bigg(n\arctan\frac{\sqrt5}{2}\bigg) +\sqrt5\sin\bigg(n\arctan\frac{\sqrt5}{2}\bigg)\Bigg\}. \end{align}$

(2.11)

By means of the binomial relation (2.5), we can get, from (2.11) and (2.10), the summation formula stated in Theorem 4.□

Theorem 5 ( $n\in\mathbb{{N}}$ ).

$\begin{align} \sum\limits_{k = 0}^{\lfloor\frac{n-\delta}{3}\rfloor} \frac{2n+\delta}{n-k} \binom{n-k}{2k+\delta}36^k = \begin{cases} 4^n+2(-3)^n\cos(n\theta_1), &\delta = 0;\\ \frac{4^n}{3}+(-3)^{n-1}\sin(n\theta_1+\theta_2), &\delta = 1, \end{cases} \end{align}$

(2.12)

where $\theta_1 = \arctan2\sqrt2$ and $\theta_2 = \arctan\frac{\sqrt2}{2}$ .

Proof. Letting $\delta = 1$ and $x = 36$ in (2.1), we have

$\begin{align*} (1-z)^2-36z^3 = -36\bigg(z-\frac14\bigg) \bigg(z-\frac{2\sqrt2\textbf{i}-1}{9}\bigg) \bigg(z+\frac{2\sqrt2\textbf{i}+1}{9}\bigg), \end{align*}$

which results in, by the partial fractional decomposition, the generating function

$\begin{align*} \mathcal{{A}}(z, 36, 1) & = \frac{16}{33}\frac1{1-4z} +\frac{68+19\sqrt2\textbf{i}}{264} \frac1{1+\Big(1+2\sqrt2\textbf{i}\Big)z}\\ &+\frac{68-19\sqrt2\textbf{i}}{264} \frac1{1+\Big(1-2\sqrt2\textbf{i}\Big)z}. \end{align*}$

Computing it's coefficient of $z^n$ , we have

$\begin{align*} [z^n]\mathcal{{A}}(z, 36, 1) = &\frac{4^{n+2}}{33}+\frac{(-1)^n}{264} \bigg\{68\Big[\Big(1+2\sqrt2\textbf{i}\Big)^n +\Big(1-2\sqrt2\textbf{i}\Big)^n\Big]\\ &+19\sqrt2\textbf{i}\Big[\Big(1+2\sqrt2\textbf{i}\Big)^n -\Big(1-2\sqrt2\textbf{i}\Big)^n\Big]\bigg\}. \end{align*}$

Combining it with the relations

$\begin{align*} \Big(1+2\sqrt2\textbf{i}\Big)^n +\Big(1-2\sqrt2\textbf{i}\Big)^n = 2\cdot3^n\cos\Big(n\arctan2\sqrt2\Big), \\ \Big(1+2\sqrt2\textbf{i}\Big)^n -\Big(1-2\sqrt2\textbf{i}\Big)^n = 2\cdot3^n\textbf{i}\sin\Big(n\arctan2\sqrt2\Big), \end{align*}$

we can obtain the following identity:

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k+1}36^k = \frac{4^{n+1}}{33}-\frac{(-3)^n}{132} \eta(n), \end{align}$

(2.13)

where

$\eta(n) = 16\cos\Big(n\arctan2\sqrt2\Big) +13\sqrt2\sin\Big(n\arctan2\sqrt2\Big).$

According to the relation between the generating functions of $\delta = 1$ and $\delta = 0$

$\mathcal{{A}}(z, 36, 0) = (1-z)\mathcal{{A}}(z, 36, 1),$

we have

$[z^n]\mathcal{{A}}(z, 36, 0) = [z^n]\mathcal{{A}}(z, 36, 1)-[z^{n-1}]\mathcal{{A}}(z, 36, 1),$

which yields the summation formula below:

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k}36^k = \frac{4^{n+1}}{11}+\frac{(-3)^n}{22} \Big\{14\cos\Big(n\arctan2\sqrt2\Big) -\sqrt2\sin\Big(n\arctan2\sqrt2\Big)\Big\}. \end{align}$

(2.14)

Now, employing the binomial relation (2.5), we can get, from (2.14) and (2.13), the identity stated in Theorem 5.□

2.2. Three identities concerning Fibonacci and Lucas numbers

In this section, three identities concerning Fibonacci and Lucas numbers are derived.

For $\lambda\neq\{0, \pm1, -2, -\frac12\}$ , letting $\delta = 1$ and $x = x(\lambda) = \frac{(\lambda^2+ \lambda)^2}{(\lambda^2+ \lambda+1)^3}$ in (2.1), we have

$\begin{align*} (1-z)^2-xz^3 = -\frac{( \lambda^2+ \lambda)^2}{( \lambda^2+ \lambda+1)^3} \big(z-( \lambda^2+ \lambda+1)\big) \bigg(z-\frac{ \lambda^2+ \lambda+1}{ \lambda^2}\bigg) \bigg(z-\frac{ \lambda^2+ \lambda+1}{( \lambda+1)^2}\bigg). \end{align*}$

By means of the partial fractional decomposition, we can get the generating function

$\begin{align*} \mathcal{{A}}(z, x( \lambda), 1) & = \frac{1}{( \lambda^2-1)( \lambda^2+2 \lambda)} \frac{1}{1-\frac{z}{ \lambda^2+ \lambda+1}} -\frac{ \lambda^4}{( \lambda^2-1)(2 \lambda+1)} \frac{1}{1-\frac{ \lambda^2}{ \lambda^2+ \lambda+1}z}\\ &+\frac{( \lambda+1)^4}{( \lambda^2+ \lambda)(2 \lambda+1)} \frac{1}{1-\frac{( \lambda+1)^2}{ \lambda^2+ \lambda+1}z}. \end{align*}$

Evaluating the coefficient of $z^n$ , we get

$\begin{align*} [z^n]\mathcal{{A}}(z, x( \lambda), 1) & = \bigg\{\frac1{( \lambda^2-1)( \lambda^2+2 \lambda)} -\frac{ \lambda^{2n+4}}{( \lambda^2-1)(2 \lambda+1)} +\frac{( \lambda+1)^{2n+4}}{( \lambda^2+2 \lambda)(2 \lambda+1)}\bigg\}\\ &\quad \times\frac{1}{( \lambda^2+ \lambda+1)^n}\\ & = \frac{2 \lambda+1- \lambda^{2n+4}( \lambda^2+2 \lambda) +( \lambda+1)^{2n+4}( \lambda^2-1)} {( \lambda^2-1)( \lambda^2+2 \lambda)(2 \lambda+1)( \lambda^2+ \lambda+1)^n}, \end{align*}$

which results in the following identity:

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k+1} \bigg(\frac{( \lambda^2+ \lambda)^2}{( \lambda^2+ \lambda+1)^3}\bigg)^k = \frac{2 \lambda+1- \lambda^{2n+2}( \lambda^2+2 \lambda) +( \lambda+1)^{2n+2}( \lambda^2-1)} {( \lambda^2-1)( \lambda^2+2 \lambda)(2 \lambda+1)( \lambda^2+ \lambda+1)^{n-1}}. \end{align}$

(2.15)

Theorem 6 ( $n\in\mathbb{{N}}$ ).

$\begin{align} &\sum\limits_{k = 0}^{\lfloor\frac{n-\delta}{3}\rfloor} \frac{2n+\delta}{n-k} \binom{n-k}{2k+\delta} \bigg(\frac{5}{64}\bigg)^k = \begin{cases} \frac{5^n+L_{2n}}{4^n}, &\delta = 0;\\ \frac{5^n-F_{2n+1}}{4^{n-1}}, &\delta = 1. \end{cases} \end{align}$

(2.16)

$\begin{align} &\sum\limits_{k = 0}^{\lfloor\frac{n-\delta}{3}\rfloor} \frac{2n+\delta}{n-k} \binom{n-k}{2k+\delta} \bigg(\frac{9}{512}\bigg)^k = \begin{cases} \frac{9^n+L_{4n}}{8^n}, &\delta = 0;\\ \frac{3^{2n+1}-L_{4n+2}}{3\cdot8^{n-1}}, &\delta = 1. \end{cases} \end{align}$

(2.17)

$\begin{align} &\sum\limits_{k = 0}^{\lfloor\frac{n-\delta}{3}\rfloor} \frac{2n+\delta}{n-k} \binom{n-k}{2k+\delta} \bigg(\frac{5}{216}\bigg)^k = \begin{cases} \frac{5^n+L_{4n}}{6^n}, &\delta = 0;\\ \frac{F_{4n+2}-5^n}{6^{n-1}}, &\delta = 1. \end{cases} \end{align}$

(2.18)

Proof. Letting $\lambda = \alpha^2$ , $\lambda = \alpha^4$ and $\lambda = - \alpha^4$ in (2.15), we can get the following three combinatorial identities:

$\begin{align} & \fbox{$\lambda = \alpha^2$} \qquad\;\; \sum\limits_{k = 0}^{n} \binom{n-k}{2k+1} \bigg(\frac{5}{64}\bigg)^k = \frac{5^{n+1}-F_{2n}-F_{2n+5}}{11\cdot4^{n-1}}, \\ &\fbox{$\lambda = \alpha^4$} \;\;\qquad \sum\limits_{k = 0}^{n} \binom{n-k}{2k+1} \bigg(\frac{9}{512}\bigg)^k = \frac{3^{2n+3}-2F_{4n+4}-F_{4n+8}}{57\cdot8^{n-1}}, \\ &\fbox{$\lambda = -\alpha^4$} \qquad \sum\limits_{k = 0}^{n} \binom{n-k}{2k+1} \bigg(\frac{5}{216}\bigg)^k = \frac{F_{4n+5}-5^{n+1}}{9\cdot6^{n-1}}. \end{align}$

Observing that the generating function

$\mathcal{{A}}(z, x( \lambda), 0) = (1-z)\mathcal{{A}}(z, x( \lambda), 1),$

we can establish another three identities below about Fibonacci and Lucas numbers:

$\begin{align*} &\sum\limits_{k = 0}^{n} \binom{n-k}{2k} \bigg(\frac{5}{64}\bigg)^k = \frac{5^{n+1}+10F_{2n}+3L_{2n}}{11\cdot4^n}, \\ &\sum\limits_{k = 0}^{n} \binom{n-k}{2k} \bigg(\frac{9}{512}\bigg)^k = \frac{2\cdot3^{2n+2}+9F_{4n+4}-L_{4n+4}}{38\cdot8^n}, \\ &\sum\limits_{k = 0}^{n} \binom{n-k}{2k} \bigg(\frac{5}{216}\bigg)^k = \frac{5^{n+1}+L_{4n+3}}{9\cdot6^n}. \end{align*}$

Applying the binomial relation (2.5) to the above six identities, we can recover the three summation formulae stated in Theorem 6 □.

3. Case $\upsilon=2$ and $\tau=1$

Lemma 7. The generating function of the sequence $B_n(x) = \sum\limits_{k = 0}^{n} \binom{n-2k}{k}x^k$ is

$\begin{align} \mathcal{{B}}(z, x) = \sum\limits_{n = 0}^{\infty} \sum\limits_{k = 0}^{n} \binom{n-2k}{k}x^kz^n = \frac{1}{1-z-xz^3}. \end{align}$

(3.1)

Proof. Exchanging the summation order, we can evaluate

$\begin{align*} \mathcal{{B}}(z, x) = \sum\limits_{n = 0}^{\infty} \sum\limits_{k = 0}^{n} \binom{n-2k}{k}x^kz^n = \sum\limits_{k = 0}^{\infty}x^k \sum\limits_{n = k}^{\infty} \binom{n-2k}{k}z^n. \end{align*}$

By making the replacement $n\to n+2k$ , the last equation becomes

$\begin{align*} \mathcal{{B}}(z, x) = \sum\limits_{k = 0}^{\infty}x^kz^{2k} \sum\limits_{n = 0}^{\infty} \binom{n}{k}z^n. \end{align*}$

By means of the formula

$\sum\limits_{k = 0}^{\infty} \binom{k}{n}z^k = \frac{z^n}{(1-z)^{n+1}},$

we get the generating function

$\begin{align*} \mathcal{{B}}(z, x) = \sum\limits_{k = 0}^{\infty} \frac{x^kz^{3k}}{(1-z)^{k+1}} = \frac{1}{1-z-xz^3}. \end{align*}$

□

3.1. Three formulae concerning trigonometric functions

Analogously, in this section we establish another three formulae concerning trigonometric functions.

Theorem 8 ( $n\in\mathbb{{N}}$ ).

$\begin{align} \sum\limits_{k = 0}^{\lfloor\frac{n}{3}\rfloor} \frac{n}{n-2k} \binom{n-2k}{k}(-2)^k = 2^{\frac{n+2}{2}}\cos\frac{n\pi}{4}+(-1)^n. \end{align}$

(3.2)

Proof. Letting $x = -2$ in Lemma 7, we have the generating function

$\begin{align*} \mathcal{{B}}(z, -2) = \frac1{1-z+2z^3} = \frac15\frac{1}{1+z} +\frac{2-\textbf{i}}{5}\frac{1}{1-(1+\textbf{i})z} +\frac{2+\textbf{i}}{5}\frac{1}{1-(1-\textbf{i})z}. \end{align*}$

By extracting the coefficient of $z^n$ , we get

$\begin{align*} [z^n]\mathcal{{B}}(z, -2) = & = \frac{(-1)^n}{5} +\frac{2-\textbf{i}}{5}(1+\textbf{i})^n +\frac{2+\textbf{i}}{5}(1-\textbf{i})^n\\ & = \frac{(-1)^n}{5} +\frac15\Big\{2\big[(1+\textbf{i})^n +(1-\textbf{i})^n\big] +\textbf{i}\big[(1-\textbf{i})^n -(1+\textbf{i})^n\big]\Big\}. \end{align*}$

By means of the identities ^[7, Eqs. 1.90 and 1.96]

$\begin{align*} &\sum\limits_{k = 0}^{\lfloor\frac{n}{2}\rfloor} \binom{n}{2k}(-1)^k = \frac{(1+\textbf{i})^n+(1-\textbf{i})^n}{2} = 2^{\frac{n}{2}}\cos\frac{n\pi}{4}, \\ &\sum\limits_{k = 0}^{\lfloor\frac{n-1}{2}\rfloor} \binom{n}{2k+1}(-1)^k = \frac{(1+\textbf{i})^n-(1-\textbf{i})^n}{2\textbf{i}} = 2^{\frac{n}{2}}\sin\frac{n\pi}{4}, \end{align*}$

we obtain the following identity

$\begin{align*} \sum\limits_{k = 0}^{n} \binom{n-2k}{k}(-2)^k = \frac{1}{5}\Bigg\{ 2^{\frac{n+2}{2}} \bigg(2\cos\frac{n\pi}{4}+\sin\frac{n\pi}{4}\bigg) +(-1)^n\Bigg\}. \end{align*}$

Then the desired identity follows by using the binomial relation (3.7).□

Theorem 9 ( $n\in\mathbb{{N}}$ ).

$\begin{align} {\sum\limits_{k = 0}^{\lfloor\frac{n}{3}\rfloor} \frac{n}{n-2k} \binom{n-2k}{k}18^k = 2\Big(-\sqrt6\Big)^n\cos\Big(n\arctan\sqrt5\Big)+3^n.} \end{align}$

(3.3)

Proof. Letting $x = 18$ in Lemma 7, we have the generating function

$\begin{align*} \mathcal{{B}}(z, 18)& = \frac{1}{1-z-18z^3}\\ & = \frac{3}{7}\frac{1}{1-3z} +\frac{10+\sqrt5\textbf{i}}{35} \frac{1}{1+\Big(\sqrt5\textbf{i}+1\Big)z} +\frac{10-\sqrt5\textbf{i}}{35} \frac{1}{1-\Big(\sqrt5\textbf{i}-1\Big)z}. \end{align*}$

Evaluating the coefficient of $z^n$ , we get

$\begin{align*} [z^n]\mathcal{{B}}(z, 18) = \frac{3^{n+1}}{7} &+\frac{(-1)^n}{35} \bigg\{10\Big[\Big(1+\sqrt5\textbf{i}\Big)^n +\Big(1-\sqrt5\textbf{i}\Big)^n\Big]\\ &+\sqrt5\textbf{i}\Big[\Big(1+\sqrt5\textbf{i}\Big)^n -\Big(1-\sqrt5\textbf{i}\Big)^n\Big]\bigg\}. \end{align*}$

Keeping in mind that

$\begin{align*} &(\cos\theta+\textbf{i}\sin\theta)^n +(\cos\theta-\textbf{i}\sin\theta)^n = 2\cos(n\theta), \\ &(\cos\theta+\textbf{i}\sin\theta)^n -(\cos\theta-\textbf{i}\sin\theta)^n = 2\textbf{i}\sin(n\theta), \end{align*}$

we have

$\begin{align*} \Big(1+\sqrt5\textbf{i}\Big)^n +\Big(1-\sqrt5\textbf{i}\Big)^n & = 6^{\frac{n}{2}} \Bigg\{\bigg(\frac1{\sqrt6}+\frac{\sqrt5}{\sqrt6}\textbf{i}\bigg)^n +\bigg(\frac1{\sqrt6}-\frac{\sqrt5}{\sqrt6}\textbf{i}\bigg)^n\Bigg\}\\ & = 6^{\frac{n}{2}} \Big\{(\cos t+\textbf{i}\sin t)^n +(\cos t-\textbf{i}\sin t)^n\Big\}\\ & = 2\cdot6^{\frac{n}{2}}\cos(nt), \end{align*}$

and

$\begin{align*} \Big(1+\sqrt5\textbf{i}\Big)^n -\Big(1-\sqrt5\textbf{i}\Big)^n & = 6^{\frac{n}{2}} \Bigg\{\bigg(\frac1{\sqrt6}+\frac{\sqrt5}{\sqrt6}\textbf{i}\bigg)^n -\bigg(\frac1{\sqrt6}-\frac{\sqrt5}{\sqrt6}\textbf{i}\bigg)^n\Bigg\}\\ & = 6^{\frac{n}{2}} \big\{(\cos t+\textbf{i}\sin t)^n -(\cos t-\textbf{i}\sin t)^n\big\}\\ & = 2\cdot6^{\frac{n}{2}}\textbf{i}\sin(nt), \end{align*}$

where $\tan t = \sqrt5$ , which yield the identity

$\begin{align*} \sum\limits_{k = 0}^{n} \binom{n-2k}{k}18^k = \frac{3^{n+1}}{7}+\frac{2(-\sqrt6)^n}{35} \Big\{10\cos\Big(n\arctan\sqrt5\Big) -\sqrt5\sin\Big(n\arctan\sqrt5\Big)\Big\}. \end{align*}$

The desired identity follows by means of (3.7).□

Theorem 10 ( $n\in\mathbb{{N}}$ ).

$\begin{align} {\sum\limits_{k = 0}^{\lfloor\frac{n}{3}\rfloor} \frac{n}{n-2k} \binom{n-2k}{k}(-36)^k = 2\Big(2\sqrt3\Big)^n\cos\Big(n\arctan\sqrt2\Big)+(-3)^n.} \end{align}$

(3.4)

Proof. Letting $x = -36$ in Lemma 7, we have the generating function

$\begin{align*} &\mathcal{{B}}(z, -36) = \frac{1}{36z^3-z+1}\\ & = \frac{3}{11}\frac{1}{1+3z} +\frac{8+\sqrt2\textbf{i}}{22} \frac{1}{1-2\Big(1-\sqrt2\textbf{i}\Big)z} +\frac{8-\sqrt2\textbf{i}}{22} \frac{1}{1-2\Big(1+\sqrt2\textbf{i}\Big)z}. \end{align*}$

and its coefficient of $z^n$

$\begin{align*} [z^n]\mathcal{{B}}(z, -36) = \frac{3(-3)^n}{11} &+\frac{2^n}{22}\bigg\{ 8\Big[\Big(1-\sqrt2\textbf{i}\Big)^n +\Big(1+\sqrt2\textbf{i}\Big)^n\Big]\\ &+\sqrt2\textbf{i} \Big[\Big(1-\sqrt2\textbf{i}\Big)^n -\Big(1+\sqrt2\textbf{i}\Big)^n\Big]\bigg\}. \end{align*}$

By means of the following results

$\begin{align*} \Big(1-\sqrt2\textbf{i}\Big)^n +\Big(1+\sqrt2\textbf{i}\Big)^n = 2\cdot3^{\frac{n}{2}}\cos\Big(n\arctan\sqrt2\Big), \end{align*}$

and

$\begin{align*} \Big(1-\sqrt2\textbf{i}\Big)^n -\Big(1+\sqrt2\textbf{i}\Big)^n = 2\cdot3^{\frac{n}{2}}\textbf{i}\sin\Big(n\arctan\sqrt2\Big), \end{align*}$

we can get the identity

$\begin{align*} \sum\limits_{k = 0}^{n} \binom{n-2k}{k}(-36)^k = \frac{3(-3)^n}{11}+\frac{(2\sqrt3)^n}{11} \Big\{8\cos\Big(n\arctan\sqrt2\Big) +\sqrt2\sin\Big(n\arctan\sqrt2\Big)\Big\}. \end{align*}$

Then the proof completes by the aid of (3.7).□

3.2. Four identities concerning Fibonacci and Lucas numbers

For $x = x(\lambda) = -\frac{(\lambda^2+ \lambda)^2}{(\lambda^2+ \lambda+1)^3}$ , $\lambda\neq\{0, \pm1, -2, -\frac12\}$ , we have

$\begin{align*} 1-z-xz^3 = \frac{( \lambda^2+ \lambda)^2}{( \lambda^2+ \lambda+1)^3} \Bigg(z+\frac{ \lambda^2+ \lambda+1}{ \lambda}\Bigg) \Bigg(z-\frac{ \lambda^2+ \lambda+1}{ \lambda+1}\Bigg) \Bigg(z-\frac{ \lambda^2+ \lambda+1}{ \lambda^2+ \lambda}\Bigg). \end{align*}$

By means of the partial fractional decomposition, we can get the generating function

$\begin{align*} \mathcal{{B}}(z, x( \lambda))& = \frac{ \lambda}{( \lambda+2)(2 \lambda+1)} \frac{1}{1+\frac{ \lambda}{ \lambda^2+ \lambda+1}z} -\frac{ \lambda+1}{(2 \lambda+1)( \lambda-1)} \frac{1}{1-\frac{ \lambda+1}{ \lambda^2+ \lambda+1}z}\\ &+\frac{ \lambda( \lambda+1)}{( \lambda+2)( \lambda-1)} \frac{1}{1-\frac{ \lambda( \lambda+1)}{ \lambda^2+ \lambda+1}z}. \end{align*}$

Evaluating the coefficient of $z^n$ , we get

$\begin{align*} [z^n]\mathcal{{B}}(z, x( \lambda)) = \frac1{( \lambda^2+ \lambda+1)^n} \Bigg\{\frac{(-1)^n \lambda^{n+1}}{( \lambda+2)(2 \lambda+1)} -\frac{( \lambda+1)^{n+1}}{(2 \lambda+1)( \lambda-1)} +\frac{( \lambda^2+ \lambda)^{n+1}}{( \lambda+2)( \lambda-1)}\Bigg\}, \end{align*}$

which results in the following identity:

$\begin{align} &\sum\limits_{k = 0}^{n}(-1)^k \binom{n-2k}{k} \bigg(\frac{( \lambda^2+ \lambda)^2}{( \lambda^2+ \lambda+1)^3}\bigg)^k\\ & = \frac1{( \lambda^2+ \lambda+1)^n} \Bigg\{\frac{(-1)^n \lambda^{n+1}}{( \lambda+2)(2 \lambda+1)} -\frac{( \lambda+1)^{n+1}}{(2 \lambda+1)( \lambda-1)} +\frac{( \lambda^2+ \lambda)^{n+1}}{( \lambda+2)( \lambda-1)}\Bigg\}. \end{align}$

(3.5)

Theorem 11 ( $n\in\mathbb{{N}}$ ).

$\begin{align} {\sum\limits_{k = 0}^{\lfloor\frac{n}{3}\rfloor} \frac{n}{n-2k} \binom{n-2k}{k} \bigg(-\frac18\bigg)^k = \frac{L_n+1}{2^n}.} \end{align}$

(3.6)

Proof. Let $\lambda = \alpha$ in (3.5), we have $x(\lambda) = -\frac18$ and

$\begin{align*} \sum\limits_{k = 0}^{n} \binom{n-2k}{k} \bigg(-\frac18\bigg)^k = \frac{F_{n+3}-1}{2^n}. \end{align*}$

By means of the relation

$\begin{align} \frac{n}{n-2k} \binom{n-2k}{k} = 3 \binom{n-2k}{k}-2 \binom{n-2k-1}{k}, \end{align}$

(3.7)

we have

$\begin{align*} \sum\limits_{k = 0}^{\lfloor\frac{n}{3}\rfloor} \frac{n}{n-2k} \binom{n-2k}{k} \bigg(-\frac18\bigg)^k & = 3\sum\limits_{k = 0}^{n} \binom{n-2k}{k} \bigg(-\frac18\bigg)^k -2\sum\limits_{k = 0}^{n-1} \binom{n-2k-1}{k} \bigg(-\frac18\bigg)^k\\ & = \frac{3F_{n+3}-3}{2^n} -\frac{2F_{n+2}-2}{2^{n-1}} = \frac{3F_{n+3}-4F_{n+2}+1}{2^n}. \end{align*}$

Then the proof follows by using

$3F_{n+3}-4F_{n+2} = L_n.$

□

Theorem 12 ( $n\in\mathbb{{N}}$ ).

$\begin{align} \sum\limits_{k = 0}^{\lfloor\frac{n}{3}\rfloor} \frac{n}{n-2k} \binom{n-2k}{k} \bigg(-\frac{5}{64}\bigg)^k = \begin{cases} \frac{5^{\frac{n}{2}}L_n+1}{4^n}, &n\equiv_20;\\ \frac{5^{\frac{n+1}{2}}F_n-1}{4^n}, &n\equiv_21. \end{cases} \end{align}$

(3.8)

Proof. Let $\lambda = \alpha^2$ in (3.5), we have $x(\lambda) = -\frac{5}{64}$ and

$\begin{align*} \sum\limits_{k = 0}^{n} \binom{n-2k}{k} \bigg(-\frac{5}{64}\bigg)^k = \begin{cases} \frac{1}{11\cdot 4^n} \Big\{1+5^{\frac{n+2}{2}} (L_n+4F_n)\Big\}, &n\equiv_20;\\ \frac{1}{11\cdot 4^n} \Big\{5^{\frac{n+1}{2}} (4L_n+5F_n)-1\Big\}, &n\equiv_21. \end{cases} \end{align*}$

Then the proof follows by using the relation (3.7).□

Theorem 13 ( $n\in\mathbb{{N}}$ ).

$\begin{align} \sum\limits_{k = 0}^{\lfloor\frac{n}{3}\rfloor} \frac{n}{n-2k} \binom{n-2k}{k} \bigg(-\frac{5}{216}\bigg)^k = \begin{cases} \frac{5^{\frac{n}{2}}L_{2n}+1}{6^n}, &n\equiv_20;\\ \frac{5^{\frac{n+1}{2}}F_{2n}+1}{6^n}, &n\equiv_21. \end{cases} \end{align}$

(3.9)

Proof. Let $\lambda = - \alpha^4$ in (3.5), we have $x(\lambda) = -\frac{5}{216}$ and

$\begin{align*} \sum\limits_{k = 0}^{n} \binom{n-2k}{k} \bigg(-\frac{5}{216}\bigg)^k = \begin{cases} \frac{1}{9\cdot 6^n} \Big\{5^{\frac{n+2}{2}} F_{2n+3}-1\Big\}, &n\equiv_20;\\ \frac{1}{9\cdot 6^n} \Big\{5^{\frac{n+1}{2}} L_{2n+3}-1\Big\}, &n\equiv_21. \end{cases} \end{align*}$

By means of the relation (3.7), we can complete the proof.□

Theorem 14 ( $n\in\mathbb{{N}}$ ).

$\begin{align} \sum\limits_{k = 0}^{\lfloor\frac{n}{3}\rfloor} \frac{n}{n-2k} \binom{n-2k}{k} \bigg(-\frac{9}{512}\bigg)^k = \frac{3^nL_{2n}+(-1)^n}{8^n}. \end{align}$

(3.10)

Proof. Let $\lambda = \alpha^4$ in (3.5), we have $x(\lambda) = -\frac{9}{512}$ and

$\begin{align*} \sum\limits_{k = 0}^{n} \binom{n-2k}{k} \bigg(-\frac{9}{512}\bigg)^k = \frac{3^{n+1}(8F_{2n}+3L_{2n})+(-1)^n} {19\cdot 8^n}. \end{align*}$

By using the binomial relation (3.7), we can get the identity stated in the theorem.□

4. Concluding comments

There should be more interesting identities by choosing appropriate parameters $\lambda$ or $x$ in Lemmas 2.1 and 7. For instance, by setting $\delta = 1$ and $x = \frac{4}{27}$ in (2.1), we have the generating function

$\begin{align*} \mathcal{{A}}\bigg(z, \frac{4}{27}, 1\bigg) = -\frac{27}{4z^3-27z^2+54z-27} = \frac{16}{9}\frac1{1-\frac43z} +\frac4{27}\frac{z}{\Big(1-\frac{z}{3}\Big)^2} -\frac79\frac1{\Big(1-\frac{z}{3}\Big)^2}. \end{align*}$

Extracting the coefficient of $z^{n-1}$ , we can get the following identity:

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k+1} \bigg(\frac{4}{27}\bigg)^k = \frac{4^{n+1}-3n-4}{3^{n+1}}. \end{align}$

(4.1)

In view of (1.4) and (1.5), as well as the relation

$\mathcal{{A}}\bigg(z, \frac4{27}, 0\bigg) = (1-z)\mathcal{{A}}\bigg(z, \frac4{27}, 1\bigg),$

we have

$[z^n]\mathcal{{A}}\bigg(z, \frac4{27}, 0\bigg) = [z^n]\mathcal{{A}}\bigg(z, \frac4{27}, 1\bigg) -[z^{n-1}]\mathcal{{A}}\bigg(z, \frac4{27}, 1\bigg),$

and derive immediately another summation formula below:

$\begin{align} \sum\limits_{k = 0}^{n} \binom{n-k}{2k} \bigg(\frac{4}{27}\bigg)^k = \frac{4^{n+1}+6n+5}{3^{n+2}}. \end{align}$

(4.2)

By means of the binomial relation (2.5), we can get, from (4.2) and (4.1), the theorem below.

Theorem 15 ( $n\in\mathbb{{N}}$ ).

$\begin{align} \sum\limits_{k = 0}^{\lfloor\frac{n-\delta}{3}\rfloor} \frac{2n+\delta}{n-k} \binom{n-k}{2k+\delta} \bigg(\frac{4}{27}\bigg)^k = \begin{cases} \frac{4^n+2}{3^n}, &\delta = 0;\\ \frac{4^n-1}{3^{n-1}}, &\delta = 1. \end{cases} \end{align}$

(4.3)

In fact, the case of $\delta = 1$ is equivalent to the identity (1.1) anticipated in the introduction because of the binomial relation

$\frac{1}{n-k} \binom{n-k}{2k+1} = \frac1{2k+1} \binom{n-k-1}{2k}.$

Analogously, by letting $x = -\frac{4}{27}$ in Lemma 7, we have the generating function

$\begin{align*} \mathcal{{B}}\bigg(z, -\frac{4}{27}\bigg) = \frac{1}{9+3z}-\frac43\frac{z}{(2z-3)^2} +\frac{8}{(2z-3)^2}, \end{align*}$

which yields, by evaluating the coefficient of $z^n$ , the identity

$\begin{align*} \sum\limits_{k = 0}^{n} \binom{n-2k}{k} \bigg(-\frac{4}{27}\bigg)^k = \frac{2^{n+1}(3n+4)+(-1)^n}{3^{n+2}}. \end{align*}$

Then using the binomial relation (3.7), we can get the following theorem.

Theorem 16 ( $n\in\mathbb{{N}}$ ).

$\begin{align} \sum\limits_{k = 0}^{\lfloor\frac{n}{3}\rfloor} \frac{n}{n-2k} \binom{n-2k}{k} \bigg(-\frac{4}{27}\bigg)^k = \frac{2^{n+1}+(-1)^n}{3^n}. \end{align}$

(4.4)

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors are sincerely grateful to the anonymous referees for their careful reading and valuable suggestions, that have improved the manuscript during the revision. This work was supported by Zhoukou Normal University high-level talents start-up funds research project, China, (ZKNUC2022007).

Conflict of interest

The authors declare there is no conflicts of interest.

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[3]	L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, Revised and enlarged edition, D. Reidel Publishing Company, Dordrecht, 1974.
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[5]	D. Guo, W. Chu, Binomial Sums with Pell and Lucas Polynomials, Bull. Belg. Math. Soc. Simon Stevin, 28 (2021), 133–145. https://doi.org/10.36045/j.bbms.200525 doi: 10.36045/j.bbms.200525
[6]	D. Guo, W. Chu, Inverse Tangent Series Involving Pell and Pell-Lucas Polynomials, Math. Slovaca, 72 (2022), 869–884. https://doi.org/10.1515/ms-2022-0059 doi: 10.1515/ms-2022-0059
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Yulei Chen, Dongwei Guo, Combinatorial Identities Concerning Binomial Quotients, 2024, 16, 2073-8994, 746, 10.3390/sym16060746

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AIMS Mathematics

Combinatorial identities concerning trigonometric functions and Fibonacci/Lucas numbers

Related Papers:

Abstract

1. Introduction and motivation

2. Case $\upsilon=1$ and $\tau=2$

2.1. Four combinatorial identities concerning trigonometric functions

2.2. Three identities concerning Fibonacci and Lucas numbers

3. Case $\upsilon=2$ and $\tau=1$

3.1. Three formulae concerning trigonometric functions

3.2. Four identities concerning Fibonacci and Lucas numbers

4. Concluding comments

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

This article has been cited by:

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通讯作者: 陈斌, bchen63@163.com

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Catalog

Abstract

1. Introduction and motivation

2. Case $\upsilon=1$ and $\tau=2$

2.1. Four combinatorial identities concerning trigonometric functions

2.2. Three identities concerning Fibonacci and Lucas numbers

3. Case $\upsilon=2$ and $\tau=1$

3.1. Three formulae concerning trigonometric functions

3.2. Four identities concerning Fibonacci and Lucas numbers

4. Concluding comments

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

AIMS Mathematics

Combinatorial identities concerning trigonometric functions and Fibonacci/Lucas numbers

Related Papers:

Abstract

1. Introduction and motivation

2. Case υ=1 \upsilon=1 and τ=2 \tau=2

2.1. Four combinatorial identities concerning trigonometric functions

2.2. Three identities concerning Fibonacci and Lucas numbers

3. Case υ=2 \upsilon=2 and τ=1 \tau=1

3.1. Three formulae concerning trigonometric functions

3.2. Four identities concerning Fibonacci and Lucas numbers

4. Concluding comments

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

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Other Articles By Authors

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Content

Catalog

Abstract

1. Introduction and motivation

2. Case υ=1 \upsilon=1 and τ=2 \tau=2

2.1. Four combinatorial identities concerning trigonometric functions

2.2. Three identities concerning Fibonacci and Lucas numbers

3. Case υ=2 \upsilon=2 and τ=1 \tau=1

3.1. Three formulae concerning trigonometric functions

3.2. Four identities concerning Fibonacci and Lucas numbers

4. Concluding comments

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

2. Case $\upsilon=1$ and $\tau=2$

3. Case $\upsilon=2$ and $\tau=1$

2. Case $\upsilon=1$ and $\tau=2$

3. Case $\upsilon=2$ and $\tau=1$