# Higher order balancing numbers: new sequences, recurrence relations, generating functions and identities

Kalika Prasad¹, Indubala², Munesh Kumari¹*

¹Government Engineering College Bhojpur, Bihar, India

² Indian Institute of Science, Bengaluru, Karnataka, India

Email(s): klkaprsd@gmail.com, indubala@iisc.ac.in, muneshnasir94@gmail.com

## Abstract

In this article, we introduce a new extension of the balancing numbers referred as the higher order balancing numbers which is analogous to the balancing sequence. Here we present various algebraic properties of this newly introduced sequence such as recurrence relation, generating functions, partial and binomial sums, well-known name identities, etc. Moreover we show interrelations with the existing integer sequences.

**Keywords:** Higher order balancing numbers, partial and binomial sums, binomial transform, generating functions, recurrence relations

**2020 MS Classifications:** 11B37, 11B39, 11B83

## 1 Introduction

In number theory, the Fibonacci numbers emerges as celebrity kind numbers. Beside Fibonacci numbers, in number theory Lucas, Pell, Jacobsthal, Mersenne, Leonardo, Perrin, Padovan, etc. also follows the same pattern of study and opens an area of research for further investigation. One of such fascinating number sequences is the balancing sequence. The concept of balancing numbers (and balancers) was originally introduced in 1999 by Behera et al. [1]. Let us recall some important properties of it. A natural number $n$ is said to be balancing number with balancer $r$ if it satisfy the Diophantine equation

$$1 + 2 + 3 + ... + (n - 1) = (n + 1) + (n + 2) + ... + (n + r).$$

Thus the balancing numbers {$B_n$}$_{n \ge 0}$ are defined recursively as

$$B_{n+2} = 6B_{n+1} - B_n \text{ with } B_0 = 0, B_1 = 1.$$

Another sequence close to balancing numbers is 'Lucas-balancing numbers {$C_n$}' which is defined as $C_n = \sqrt{8B_n^2 + 1}$ and that satisfy the same recurrence relation but with initial condition $C_0 = 1, C_1 = 3$. The first few values of these sequence are:

<table><thead><tr><th>n</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th><th>...</th></tr></thead><tbody><tr><th>B<sub>n</sub></th><td>0</td><td>1</td><td>6</td><td>35</td><td>204</td><td>1189</td><td>6930</td><td>40391</td><td>235416</td><td>...</td></tr><tr><th>C<sub>n</sub></th><td>1</td><td>3</td><td>17</td><td>99</td><td>577</td><td>3363</td><td>19601</td><td>114243</td><td>665857</td><td>...</td></tr></tbody></table>
The closed form formula for the above sequences are given as

$$B_n = \frac{\lambda_1^n - \lambda_2^n}{4\sqrt{2}} \quad \text{and} \quad C_n = \frac{\lambda_1^n + \lambda_2^n}{2}, \qquad (1.1)$$

where $\lambda_1 = 3 + \sqrt{8}$ and $\lambda_2 = 3 - \sqrt{8}$ are the roots of the characteristic equation $x^2 - 6x + 1 = 0$ and hold the following relations:

$$\lambda_1 + \lambda_2 = 6, \quad \lambda_1^2 + \lambda_2^2 = 34, \quad \lambda_1 - \lambda_2 = 2\sqrt{8} \quad \text{and} \quad \lambda_1\lambda_2 = 1.$$

Some identities involving the balancing and Lucas-balancing numbers which we need later are given in the following lemma.

**Lemma 1.1.** For integers $m$ and $n$, we have

$$1. \quad B_{n-m}B_{n+m} - B_n^2 = -B_m^2,$$

$$2. \quad C_{n-m}C_{n+m} - C_n^2 = \frac{1}{2}(C_{2m} - 1),$$

$$3. \quad C_{2n} = 2C_n^2 - 1 \quad \text{and} \quad C_n^2 = 8B_n^2 + 1.$$

**Theorem 1.2.** For positive integer $k$ and $n \ge 0$, we have

$$B_{k(n+1)} = \lambda_1^k B_{kn} + \lambda_2^{kn} B_k \qquad (1.2)$$

*Proof.* From Binet's formula (1.1), the LHS can be written as

$$\begin{aligned} B_{k(n+1)} &= \frac{\lambda_1^{k(n+1)} - \lambda_2^{k(n+1)}}{4\sqrt{2}} \\ &= \frac{\lambda_1^{k(n+1)} - \lambda_2^{k(n+1)} + \lambda_1^k \lambda_2^{kn} - \lambda_1^k \lambda_2^{kn}}{4\sqrt{2}} \\ &= \frac{\lambda_1^k (\lambda_1^{kn} - \lambda_2^{kn}) + \lambda_2^{kn} (\lambda_1^k - \lambda_2^k)}{4\sqrt{2}} \\ &= \lambda_1^k B_{kn} + \lambda_2^{kn} B_k. \quad \square \end{aligned}$$

In next section, first we define the higher order balancing numbers then we study their algebraic properties. In recent years, several articles appeared on higher order sequences associated with the famous number sequence, where the authors presented study on the algebraic properties. The study begins with the earlier work of Randić et al. (1996) [15] where the authors proposed higher order Fibonacci numbers and investigated their properties in context of application in chemistry.

Recently, Uysal and Özkan [9, 20] studied the quaternion algebra of higher order Jacobsthal–Lucas numbers. Özimamoğlu [8] studied the hyper complex numbers whose components are higher order Pell numbers. Prasad et. al. [11] studied the higher order Mersenne numbers and Kumari et al. [5, 6] used these numbers to study the algebra of quaternions and octonions and Gaussian forms. Similar concept was applied in [13] to study the new sequences balancing numbers. A polynomial version of the balancing numbers and their algebraic properties were studied by Frontczak [2] and Ray [16]. Some interesting properties and recent developments on balancing numbers are due to Ray [17–19], Prasad et al. [12, 14], Frontczak [2–4], Panda [10], Liptai [7], etc.

Similar to these work, many algebraic properties of our proposed new sequences can be investigated after this study.
## 2 Higher order balancing numbers

Throughout the article, we adopt the symbols {$B_n$}, {$C_n$} and {$B_n^{(k)}$} for the nth balancing, Lucas-balancing and higher-order balancing numbers, respectively.

**Definition 2.1.** Let $k \in \mathbb{N}$, then the higher-order balancing numbers {$B_n^{(k)}$} is defined as

$$B_n^{(k)} = \frac{B_{kn}}{B_k}, \quad n = 0, 1, 2, \dots \qquad (2.1)$$

From Definition 2.1, note that the following identities hold obviously:

1. $B_0^{(k)} = 0$ and $B_1^{(k)} = 1$.

2. $B_2^{(k)} = \lambda_1^k + \lambda_2^k = 2C_k$.

For different values of $k$, some higher-order balancing numbers are listed in Table 1:

<table><caption>Table 1: List of some higher order balancing numbers ($B_n^{(k)}$)</caption><thead><tr><th>Numbers</th><th>$k=1$</th><th>$k=2$</th><th>$k=3$</th><th>$k=4$</th><th>$k=5$</th></tr></thead><tbody><tr><td>$B_0^{(k)}$</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td></tr><tr><td>$B_1^{(k)}$</td><td>1</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>$B_2^{(k)}$</td><td>6</td><td>34</td><td>198</td><td>1154</td><td>6726</td></tr><tr><td>$B_3^{(k)}$</td><td>35</td><td>1155</td><td>39203</td><td>1331715</td><td>45239075</td></tr><tr><td>$B_4^{(k)}$</td><td>204</td><td>39236</td><td>7761996</td><td>1536797956</td><td>304278011724</td></tr><tr><td>$B_5^{(k)}$</td><td>1189</td><td>1332869</td><td>1536836005</td><td>1773463509509</td><td>2046573861616547</td></tr></tbody></table>

**Theorem 2.1.** The sequence {$B_n^{(k)}$} satisfy the following recurrence relation

$$B_{n+2}^{(k)} - (\lambda_1^k + \lambda_2^k)B_{n+1}^{(k)} + B_n^{(k)} = 0, \quad \text{with } B_0^{(k)} = 0, B_1^{(k)} = 1. \qquad (2.2)$$

*Proof.* Dividing both sides of (1.2) by $B_k$ and then using Definition 2.1, we get

$$B_{n+1}^{(k)} = \lambda_1^k B_n^{(k)} + \lambda_2^{kn}. \qquad (2.3)$$

Now replacing $n$ by $n + 1$ in (2.3) yields

$$B_{n+2}^{(k)} = \lambda_1^k B_{n+1}^{(k)} + \lambda_2^{k(n+1)}. \qquad (2.4)$$

Multiplying both sides of (2.4) by $\lambda_2^{-k}$, we get

$$B_{n+2}^{(k)} \lambda_2^{-k} = \lambda_1^k B_{n+1}^{(k)} \lambda_2^{-k} + \lambda_2^{kn}. \qquad (2.5)$$

Thus, subtracting (2.3) from (2.5) gives the required result. $\square$

Now we give the Binet's formula for higher order balancing numbers and using the Binet's formula we prove some algebraic identities for this sequence.
## 2.1 Binet's formula and some identities

For fixed positive integer $k$, the characteristic equation for (2.2) is $x^2 - (\lambda_1^k + \lambda_2^k)x + 1 = 0$, whose roots are $\lambda_1^k$ and $\lambda_2^k$. Thus, the Binet's formula for the higher order balancing numbers $B_n^{(k)}$ is given by

$$B_n^{(k)} = \frac{\lambda_1^{kn} - \lambda_2^{kn}}{\lambda_1^k - \lambda_2^k} \qquad (2.6)$$

**Theorem 2.2.** The limiting ratio of higher order balancing numbers is $\lambda_1^k$ i.e.

$$\lim_{n \to \infty} \frac{B_{n+1}^{(k)}}{B_n^{(k)}} = \lambda_1^k.$$

*Proof.* From Binet's formula (2.6), we can write

$$\begin{aligned}
\lim_{n \to \infty} \frac{B_{n+1}^{(k)}}{B_n^{(k)}} &= \lim_{n \to \infty} \left( \frac{\frac{(\lambda_1^k)^{n+1} - (\lambda_2^k)^{n+1}}{\lambda_1^k - \lambda_2^k}}{\frac{(\lambda_1^k)^n - (\lambda_2^k)^n}{\lambda_1^k - \lambda_2^k}} \right) \\
&= \lim_{n \to \infty} \left( \frac{(\lambda_1^k)^{n+1} - (\lambda_2^k)^{n+1}}{(\lambda_1^k)^n - (\lambda_2^k)^n} \right) \\
&= \lim_{n \to \infty} \left( \frac{(\lambda_1^k)^n \lambda_1^k - (\lambda_2^k)^n \lambda_2^k}{(\lambda_1^k)^n - (\lambda_2^k)^n} \right) \\
&= \lim_{n \to \infty} \left( \frac{(\lambda_1^k) \left( \frac{(\lambda_2^k)^{n+1}}{(\lambda_1^k)^{n+1}} - 1 \right)}{\left( \frac{(\lambda_2^k)^n}{(\lambda_1^k)^n} - 1 \right)} \right).
\end{aligned}$$

Now taking into account that $\lim_{n \to \infty} \left( \frac{(\lambda_1^k)^{n+1}}{(\lambda_1^k)^{n+1}} \right) = 0$ and $\lim_{n \to \infty} \left( \frac{(\lambda_2^k)^n}{(\lambda_1^k)^n} \right) = 0$ as $|\lambda_2|/|\lambda_1| < 1$ for $k \in \mathbb{N}$, we have

$$\lim_{n \to \infty} \left( \frac{(\lambda_2^k)^{n+1}}{(\lambda_1^k)^{n+1}} - 1 \right) = -1$$

So the original limit simplify to

$$\lim_{n \to \infty} \frac{B_{n+1}^{(k)}}{B_n^{(k)}} = \lambda_1^k. \qquad \square$$

**Theorem 2.3 (Catalan's identity).** For $n \ge r$, we have

$$B_{n-r}^{(k)} B_{n+r}^{(k)} - (B_n^{(k)})^2 = -((\lambda_1 \lambda_2)^k)^{n-r} (B_r^{(k)})^2 = -(B_r^{(k)})^2. \qquad (2.7)$$

*Proof.* From Binet's formula (2.6), we have

$$\begin{aligned}
& B_{n-r}^{(k)} B_{n+r}^{(k)} - (B_n^{(k)})^2 \\
&= \left( \frac{(\lambda_1^k)^{n-r} - (\lambda_2^k)^{n-r}}{\lambda_1^k - \lambda_2^k} \right) \left( \frac{(\lambda_1^k)^{n+r} - (\lambda_2^k)^{n+r}}{\lambda_1^k - \lambda_2^k} \right) - \left( \frac{(\lambda_1^k)^n - (\lambda_2^k)^n}{\lambda_1^k - \lambda_2^k} \right)^2 \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( (\lambda_1^k)^{n-r} (\lambda_1^k)^{n+r} - (\lambda_1^k)^{n-r} (\lambda_2^k)^{n+r} - (\lambda_2^k)^{n-r} (\lambda_1^k)^{n+r} + (\lambda_2^k)^{n-r} (\lambda_2^k)^{n+r} \right) \\
&\quad - \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( (\lambda_1^k)^{2n} - 2(\lambda_1^k)^n (\lambda_2^k)^n + (\lambda_2^k)^{2n} \right)
\end{aligned}$$
$$
\begin{align*}
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( (\lambda_1^k)^{2n} - (\lambda_1^k)^{n-r} (\lambda_2^k)^{n+r} - (\lambda_2^k)^{n-r} (\lambda_1^k)^{n+r} + (\lambda_2^k)^{2n} \right. \\
&\qquad \left. - (\lambda_1^k)^{2n} + 2(\lambda_1^k)^n (\lambda_2^k)^n - (\lambda_2^k)^{2n} \right) \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( 2(\lambda_1^k \lambda_2^k)^n - (\lambda_1^k \lambda_2^k)^n \frac{(\lambda_2^k)^r}{(\lambda_1^k)^r} - (\lambda_1^k \lambda_2^k)^n \frac{(\lambda_1^k)^r}{(\lambda_2^k)^r} \right) \\
&= \frac{(\lambda_1 \lambda_2)^{kn}}{(\lambda_1^k - \lambda_2^k)^2} \left( 2 - \frac{(\lambda_2^k)^r}{(\lambda_1^k)^r} - \frac{(\lambda_1^k)^r}{(\lambda_2^k)^r} \right) \\
&= \frac{(\lambda_1 \lambda_2)^{kn}}{(\lambda_1^k - \lambda_2^k)^2 (\lambda_1 \lambda_2)^{kr}} \left( 2(\lambda_1 \lambda_2)^{kr} - \lambda_1^{2kr} - \lambda_2^{2kr} \right) \\
&= - \frac{(\lambda_1 \lambda_2)^{kn-kr}}{(\lambda_1^k - \lambda_2^k)^2} \left( (\lambda_1^{kr} - \lambda_2^{kr})^2 \right) \\
&= -((\lambda_1 \lambda_2)^k)^{n-r} (B_r^{(k)})^2 \\
&= -(B_r^{(k)})^2.
\end{align*}
$$

**Corollary 2.4 (Cassini's identity).** For positive integer $n$, we have

$$B_{n-1}^{(k)} B_{n+1}^{(k)} - (B_n^{(k)})^2 = -(B_1^{(k)})^2 = -1.$$

**Theorem 2.5 (d'Ocagane identity).** For $m > n$, we have

$$B_{n+1}^{(k)} B_{m}^{(k)} - B_{n}^{(k)} B_{m+1}^{(k)} = B_{m-n}^{k}$$

*Proof.* From Binet's formula (2.6), we have

$$
\begin{align*}
B_{n+1}^{(k)} B_{m}^{(k)} - B_{n}^{(k)} B_{m+1}^{(k)} &= \left( \frac{\lambda_1^{k(n+1)} - \lambda_2^{k(n+1)}}{\lambda_1^k - \lambda_2^k} \right) \left( \frac{\lambda_1^{mk} - \lambda_2^{mk}}{\lambda_1^k - \lambda_2^k} \right) - \left( \frac{\lambda_1^{nk} - \lambda_2^{nk}}{\lambda_1^k - \lambda_2^k} \right) \left( \frac{\lambda_1^{k(m+1)} - \lambda_2^{k(m+1)}}{\lambda_1^k - \lambda_2^k} \right) \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( (\lambda_1^{k(n+m+1)} + \lambda_2^{k(n+m+1)} - \lambda_1^{k(n+1)} \lambda_2^{mk} - \lambda_2^{k(n+1)} \lambda_1^{mk}) \right. \\
&\qquad \left. - (\lambda_1^{k(n+m+1)} + \lambda_2^{k(n+m+1)} - \lambda_1^{k(m+1)} \lambda_2^{nk} - \lambda_2^{k(m+1)} \lambda_1^{nk}) \right) \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( \lambda_1^{k(m+1)} \lambda_2^{nk} - \lambda_2^{k(n+1)} \lambda_1^{mk} + \lambda_2^{k(m+1)} \lambda_1^{nk} - \lambda_1^{k(n+1)} \lambda_2^{mk} \right) \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( \lambda_1^{km} \lambda_2^{nk} (\lambda_1^k - \lambda_2^k) - \lambda_1^{kn} \lambda_2^{mk} (\lambda_1^k - \lambda_2^k) \right) \\
&= \frac{\lambda_1^{km} \lambda_2^{nk} - \lambda_2^{km} \lambda_1^{nk}}{\lambda_1^k - \lambda_2^k} \\
&= \frac{\lambda_1^{km} \lambda_1^{kn} \lambda_2^{nk} \lambda_1^{-kn} - \lambda_2^{km} \lambda_2^{kn} \lambda_1^{nk} \lambda_2^{-kn}}{\lambda_1^k - \lambda_2^k} \\
&= \frac{\lambda_1^{k(m-n)} - \lambda_2^{k(m-n)}}{\lambda_1^k - \lambda_2^k} \quad (\text{as } \lambda_1 \lambda_2 = 1) \\
&= B_{m-n}^{(k)}
\end{align*}
$$

**Theorem 2.6 (Vajda's identity).** For every $n, m, r \ge 0$, we have

$$B_{n+m}^{(k)} B_{n+r}^{(k)} - B_{n}^{(k)} B_{n+m+r}^{(k)} = B_{m}^{k} B_{r}^{k}.$$
Proof. We start by using the definition of $B_n^{(k)}$ in terms of the roots $\lambda_1$ and $\lambda_2$:

$$B_n^{(k)} = \frac{\lambda_1^{kn} - \lambda_2^{kn}}{\lambda_1^k - \lambda_2^k}.$$

So,

$$B_{n+m}^{(k)} B_{n+r}^{(k)} - B_{n}^{(k)} B_{n+m+r}^{(k)} = \left( \frac{\lambda_{1}^{k(n+m)} - \lambda_{2}^{k(n+m)}}{\lambda_{1}^{k} - \lambda_{2}^{k}} \right) \left( \frac{\lambda_{1}^{k(n+r)} - \lambda_{2}^{k(n+r)}}{\lambda_{1}^{k} - \lambda_{2}^{k}} \right) - \left( \frac{\lambda_{1}^{kn} - \lambda_{2}^{kn}}{\lambda_{1}^{k} - \lambda_{2}^{k}} \right) \left( \frac{\lambda_{1}^{k(n+m+r)} - \lambda_{2}^{k(n+m+r)}}{\lambda_{1}^{k} - \lambda_{2}^{k}} \right).$$

Expanding the products:

$$\begin{aligned} & = \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( (\lambda_1^{k(n+m)} - \lambda_2^{k(n+m)})(\lambda_1^{k(n+r)} - \lambda_2^{k(n+r)}) \right. \\ & \qquad \left. - (\lambda_1^{kn} - \lambda_2^{kn})(\lambda_1^{k(n+m+r)} - \lambda_2^{k(n+m+r)}) \right) \\ & = \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( (\lambda_1^{k(n+m+n+r)} - \lambda_1^{k(n+m)}\lambda_2^{k(n+r)} - \lambda_2^{k(n+m)}\lambda_1^{k(n+r)} + \lambda_2^{k(n+m+n+r)}) \right. \\ & \qquad \left. - (\lambda_1^{k(n+n+m+r)} - \lambda_1^{kn}\lambda_2^{k(n+m+r)} - \lambda_2^{kn}\lambda_1^{k(n+m+r)} + \lambda_2^{k(n+n+m+r)}) \right) \\ & = \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( (\lambda_1^{k(2n+m+r)} - \lambda_1^{k(n+m)}\lambda_2^{k(n+r)} - \lambda_2^{k(n+m)}\lambda_1^{k(n+r)} + \lambda_2^{k(2n+m+r)}) \right. \\ & \qquad \left. - (\lambda_1^{k(2n+m+r)} - \lambda_1^{kn}\lambda_2^{k(n+m+r)} - \lambda_2^{kn}\lambda_1^{k(n+m+r)} + \lambda_2^{k(2n+m+r)}) \right). \end{aligned}$$

Since $\lambda_1\lambda_2 = 1$, so

$$\begin{aligned} & = \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( (\lambda_1^{k(m+r)} + \lambda_2^{k(m+r)} - \lambda_1^{km}\lambda_2^{kr} - \lambda_2^{km}\lambda_1^{kr}) \right) \\ & = \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( (\lambda_1^{km}(\lambda_1^{kr} - \lambda_2^{kr}) - \lambda_2^{km}(\lambda_1^{kr} - \lambda_2^{kr})) \right) \\ & = \left( \frac{\lambda_1^{km} - \lambda_2^{km}}{\lambda_1^k - \lambda_2^k} \right) \left( \frac{\lambda_1^{kr} - \lambda_2^{kr}}{\lambda_1^k - \lambda_2^k} \right) \\ & = B_m^{(k)} B_r^{(k)}. \end{aligned}$$

Thus, we conclude that

$$B_{n+m}^{(k)} B_{n+r}^{(k)} - B_{n}^{(k)} B_{n+m+r}^{(k)} = B_{m}^{(k)} B_{r}^{(k)}. \quad \square$$

**Theorem 2.7 (Honsberger's identity).** For any integer $p > 0$, we have

$$B_{p-1}^{(k)} B_{n}^{(k)} + B_{p}^{(k)} B_{n+1}^{(k)} = \frac{(C_{k(p+n-1)} + C_{k(p+n+1)} - 2C_{k(p+n-1)})}{16(B_k)^2}.$$

Proof. Taking into account $\lambda_1\lambda_2 = 1$, we obtain

$$\begin{aligned} & B_{p-1}^{(k)} B_{n}^{(k)} + B_{p}^{(k)} B_{n+1}^{(k)} \\ & = \left( \frac{\lambda_{1}^{k(p-1)} - \lambda_{2}^{k(p-1)}}{\lambda_{1}^{k} - \lambda_{2}^{k}} \right) \left( \frac{\lambda_{1}^{kn} - \lambda_{2}^{kn}}{\lambda_{1}^{k} - \lambda_{2}^{k}} \right) + \left( \frac{\lambda_{1}^{kp} - \lambda_{2}^{kp}}{\lambda_{1}^{k} - \lambda_{2}^{k}} \right) \left( \frac{\lambda_{1}^{k(n+1)} - \lambda_{2}^{k(n+1)}}{\lambda_{1}^{k} - \lambda_{2}^{k}} \right) \\ & = \frac{1}{(\lambda_{1}^{k} - \lambda_{2}^{k})^2} \left[ (\lambda_{1}^{k(p-1+n)} - \lambda_{1}^{k(p-1)}\lambda_{2}^{kn} - \lambda_{2}^{k(p-1)}\lambda_{1}^{kn} + \lambda_{2}^{k(p-1+n)}) \right] \\ & \quad + \frac{1}{(\lambda_{1}^{k} - \lambda_{2}^{k})^2} \left[ (\lambda_{1}^{k(p+1+n)} - \lambda_{1}^{kp}\lambda_{2}^{k(n+1)} - \lambda_{2}^{kp}\lambda_{1}^{k(n+1)} + \lambda_{2}^{k(p+1+n)}) \right] \end{aligned}$$
$$
\begin{aligned}
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left[ \lambda_1^{k(p+n)} [\lambda_1^{-kn} + \lambda_1^{kn}] + \lambda_2^{k(p+n)} [\lambda_2^{-kn} + \lambda_2^{kn}] \right. \\
&\qquad \left. - \lambda_1^{k(p-1-n)} - \lambda_2^{k(p-n-1)} - \lambda_1^{k(p-n-1)} - \lambda_2^{k(p-n-1)} \right] \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left[ (\lambda_1^{k(p+n)} + \lambda_2^{k(p+n)})(\lambda_1^{kn} + \lambda_2^{kn}) - 2(\lambda_1^{k(p-1-n)} + \lambda_2^{k(p-n-1)}) \right] \\
&= \frac{C_{kn}C_{k(p+n)} - C_k(p-n-1)}{16B_k^2}.
\end{aligned}
\quad \square
$$

3 Some identities involving HOBN

**Theorem 3.1 (Generating function).** For higher order balancing sequence, we have

$$
\sum_{n=0}^{\infty} B_{n}^{(k)} x^{n} = \frac{x}{x^{2} - 2C_{k}x + 1}.
$$

Proof. Let $G(x, k) = \sum_{n=0}^{\infty} B_n^{(k)} x^n$ be the generating function for the higher-order balancing sequence.
Now multiplying both sides of (2.2) by $x^{n+2}$ and then taking the summation over 0 to $\infty$, we get

$$
\sum_{n=0}^{\infty} B_{n+2}^{(k)} x^{n+2} - (\lambda_1^k + \lambda_2^k) \sum_{n=0}^{\infty} B_{n+1}^{(k)} x^{n+2} + \sum_{n=0}^{\infty} B_n^{(k)} x^{n+2} = 0.
$$

Now substituting

$$
\begin{gather*}
\sum_{n=0}^{\infty} B_{n+2}^{(k)} x^{n+2} = G(x, k) - B_1^{(k)} x - B_0^{(k)}, \quad \sum_{n=0}^{\infty} B_{n+1}^{(k)} x^{n+2} = x[G(x, k) - B_0^{(k)}] \\
\text{and} \quad \sum_{n=0}^{\infty} B_n^{(k)} x^{n+2} = x^2 G(x, k).
\end{gather*}
$$

We get

$$
\begin{gather*}
(G(x, k) - B_1^{(k)} x - B_0^{(k)}) - (\lambda_1^k + \lambda_2^k) x [G(x, k) - B_0^{(k)}] + x^2 G(x, k) = 0 \\
G(x, k) [1 - (\lambda_1^k + \lambda_2^k) x + x^2] + (\lambda_1^k + \lambda_2^k) x B_0^{(k)} - B_1^{(k)} x - B_0^{(k)} = 0.
\end{gather*}
$$

Since $B_0^{(k)} = 0$ and $B_1^{(k)} = 1$, so after simplifications it becomes

$$
G(x, k) = \frac{x}{x^2 - 2C_k x + 1}. \quad \square
$$

For instance, setting $k = 1$ in the above theorem gives the generating function for balancing se-
quence i.e.

$$
G(x, 1) = \frac{x}{x^2 - 6x + 1}.
$$

**Theorem 3.2.** The generating function for odd and even indexed sequences of higher-order balancing numbers is:

$$
\sum_{r=0}^{\infty} B_{2r}^{(k)} x^r = \frac{2xC_k}{x^2 - 2C_2kx + 1} \quad \text{and} \quad \sum_{r=0}^{\infty} B_{2r+1}^{(k)} x^r = \frac{x+1}{x^2 - 2C_2kx + 1}.
$$
Proof. We have

$$
\begin{align*}
\sum_{r=0}^{\infty} B_{2r}^{(k)} x^r &= \sum_{r=0}^{\infty} \frac{\lambda_1^{2kr} - \lambda_2^{2kr}}{\lambda_1^k - \lambda_2^k} x^r \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left[ \sum_{r=0}^{\infty} \lambda_1^{2kr} - \lambda_2^{2kr} \right] \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left[ \frac{1}{1 - \lambda_1^{2k}x} - \frac{1}{1 - \lambda_2^{2k}x} \right] \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left[ \frac{x(\lambda_1^{2k} - \lambda_2^{2k})}{x^2 - x(\lambda_1^{2k} + \lambda_2^{2k}) + 1} \right] \\
&= \frac{x B_2^{(k)}}{x^2 - 2C_{2k}x + 1} = \frac{2xC_k}{x^2 - 2C_{2k}x + 1}.
\end{align*}
$$

In a similar fashion,

$$
\begin{align*}
\sum_{r=0}^{\infty} B_{2r+1}^{(k)} &= \sum_{r=0}^{\infty} \frac{\lambda_1^{(2r+1)k} - \lambda_2^{(2r+1)k}}{\lambda_1^k - \lambda_2^k} \\
&= \sum_{r=0}^{\infty} \frac{\lambda_1^{(2r+1)k}}{\lambda_1^k - \lambda_2^k} x^r - \sum_{r=0}^{\infty} \frac{\lambda_2^{(2r+1)k}}{\lambda_1^k - \lambda_2^k} x^r \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left[ \frac{\lambda_1^k}{1 - \lambda_1^{2k}x} - \frac{\lambda_2^k}{1 - \lambda_2^{2k}x} \right] \\
&= \frac{x+1}{x^2 - 2C_{2k}x + 1}.
\end{align*}
$$

**Theorem 3.3.** The exponential generating function $E(x, k)$ for higher order balancing numbers is

$$
E(x, k) = \frac{e^{\lambda_1^k x} - e^{\lambda_2^k x}}{4\sqrt{2}B_k}.
$$

Proof. Using Binet’s formula 2.6, we have

$$
\begin{align*}
E(x, k) &= \sum_{n=0}^{\infty} \frac{B_n^{(k)} x^n}{n!} \\
&= \sum_{n=0}^{\infty} \left( \frac{(\lambda_1^{kn} - \lambda_2^{kn})}{\lambda_1^k - \lambda_2^k} \right) \frac{x^n}{n!} \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( \sum_{n=0}^{\infty} \frac{(\lambda_1^k x)^n}{n!} - \sum_{n=0}^{\infty} \frac{(\lambda_2^k x)^n}{n!} \right) \\
&= \frac{e^{\lambda_1^k x} - e^{\lambda_2^k x}}{\lambda_1^k - \lambda_2^k} \\
&= \frac{e^{\lambda_1^k x} - e^{\lambda_2^k x}}{4\sqrt{2}B_k}.
\end{align*}
$$

**Theorem 3.4.** The Poisson generating function for the higher-order balancing numbers is:

$$
\sum_{r=0}^{\infty} B_{r}^{(k)} \frac{x^{-r}}{r!} = \frac{e^{\frac{\lambda_{1}^{k}}{x}} - e^{\frac{\lambda_{2}^{k}}{x}}}{4\sqrt{2}B_{k}}.
$$
*Proof.* In previous theorem replacing x by $x^{-1}$, we get the required result

$$ \sum_{r=0}^{\infty} B_r^{(k)} \frac{x^{-r}}{r!} = \frac{e^{\frac{\lambda_1^k}{x}} - e^{\frac{\lambda_2^k}{x}}}{4\sqrt{2}B_k}. \quad \square $$

## 3.1 Partial sum and Binomial sum

**Theorem 3.5 (Partial sum).** For $k, n \in \mathbb{N}$, we have

$$ \sum_{i=1}^{n} B_{i}^{(k)} = \frac{B_{n}^{(k)} + B_{n+1}^{(k)} k + 1}{2(1 - C_{k})}. $$

*Proof.* Using the Binet's formula of higher order Balancing numbers, we have

$$ \begin{aligned} \sum_{i=1}^{n} B_{i}^{(k)} &= \sum_{i=1}^{n} \frac{1}{\lambda_{1}^{k} - \lambda_{2}^{k}} (\lambda_{1}^{ki} - \lambda_{2}^{ki}) \\ &= \frac{1}{\lambda_{1}^{k} - \lambda_{2}^{k}} \left( \lambda_{1}^{k} \frac{(\lambda_{1}^{k})^{n} - 1}{\lambda_{1}^{k} - 1} - \lambda_{2}^{k} \frac{(\lambda_{2}^{k})^{n} - 1}{\lambda_{2}^{k} - 1} \right). \\ &= \frac{1}{\lambda_{1}^{k} - \lambda_{2}^{k}} \left[ \frac{((\lambda_{1}^{k})^{n+1} \lambda_{2}^{k} - \lambda_{1}^{k} \lambda_{2}^{k} - (\lambda_{1}^{k})^{n+1} + \lambda_{1}^{k}) - ((\lambda_{2}^{k})^{n+1} \lambda_{1}^{k} - \lambda_{1}^{k} \lambda_{2}^{k} - (\lambda_{2}^{k})^{n+1} + \lambda_{2}^{k})}{(\lambda_{1}^{k} - 1)(\lambda_{2}^{k} - 1)} \right] \\ &= \frac{1}{\lambda_{1}^{k} - \lambda_{2}^{k}} \left[ \frac{((\lambda_{1}^{k})^{n} - (\lambda_{2}^{k})^{n}) - ((\lambda_{1}^{k})^{n+1} - (\lambda_{2}^{k})^{n+1}) + (\lambda_{1}^{k} - \lambda_{2}^{k})}{(\lambda_{1} \lambda_{2})^{k} - \lambda_{1}^{k} - \lambda_{2}^{k} + 1} \right] \\ &= \frac{B_{n}^{(k)} - B_{n+1}^{(k)} + 1}{2(1 - C_{k})}. \quad \square \end{aligned} $$

**Theorem 3.6 (Partial sum with even indexes).** For $k, n \in \mathbb{N}$, we have

$$ \sum_{i=1}^{n} B_{2i}^{(k)} = \frac{B_{2}^{(k)} + B_{2n}^{(k)} - B_{2(n+1)}^{(k)}}{2 - 2C_{2k}}. $$

*Proof.* We write

$$ \begin{aligned} \sum_{i=1}^{n} B_{2i}^{(k)} &= \sum_{i=1}^{n} \frac{1}{\lambda_{1}^{k} - \lambda_{2}^{k}} (\lambda_{1}^{2ki} - \lambda_{2}^{2ki}) \\ &= \frac{1}{\lambda_{1}^{k} - \lambda_{2}^{k}} \left( \lambda_{1}^{2k} \frac{(\lambda_{1}^{2k})^{n} - 1}{\lambda_{1}^{2k} - 1} - \lambda_{2}^{2k} \frac{(\lambda_{2}^{2k})^{n} - 1}{\lambda_{2}^{2k} - 1} \right). \end{aligned} $$

After simplifying, we get

$$ \sum_{i=1}^{n} B_{2i}^{(k)} = \frac{B_{2}^{(k)} + B_{2n}^{(k)} - B_{2(n+1)}^{(k)}}{2 - 2C_{2k}}. \quad \square $$

**Theorem 3.7 (Partial sum with odd indexes).** For $k, n \in \mathbb{N}$, we have

$$ \sum_{i=0}^{n} B_{2i+1}^{(k)} = \frac{C_{2k(n+1)-1}}{C_{2k}-1}. $$
*Proof.* Using the Binet's formula, we have

$$
\begin{align*}
\sum_{i=0}^{n} B_{2i+1}^{(k)} &= \sum_{i=0}^{n} \frac{1}{\lambda_1^k - \lambda_2^k} (\lambda_1^{k(2i+1)} - \lambda_2^{k(2i+1)}) \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( \lambda_1^k \frac{(\lambda_1^{2k})^{n+1} - 1}{\lambda_1^{2k} - 1} - \lambda_2^k \frac{(\lambda_2^{2k})^{n+1} - 1}{\lambda_2^{2k} - 1} \right) \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left[ \frac{(\lambda_1^k (\lambda_1^{2k})^{n+1} \lambda_2^{2k} - \lambda_1^k \lambda_2^{2k} - \lambda_1^k (\lambda_1^{2k})^{n+1} + \lambda_1^k) - (\lambda_2^k (\lambda_2^{2k})^{n+1} \lambda_1^{2k} - \lambda_1^{2k} \lambda_2^k - \lambda_2^k (\lambda_2^{2k})^{n+1} + \lambda_2^k)}{(\lambda_1^k - 1)(\lambda_2^k - 1)} \right] \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left[ \frac{(\lambda_1^k (\lambda_1^{2k})^n - \lambda_1^k (\lambda_1^{2k})^{n+1} - \lambda_2^k (\lambda_2^{2k})^n + \lambda_2^k (\lambda_2^{2k})^{n+1} + 2(\lambda_1^k - \lambda_2^k))}{(\lambda_1^k - 1)(\lambda_2^k - 1)} \right] \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left[ \frac{((\lambda_1^{2k})^{n+1} (\lambda_2^k - \lambda_1^k) + (\lambda_1^{2k})^{n+1} (\lambda_2^k - \lambda_1^k) + 2(\lambda_1^k - \lambda_2^k))}{(\lambda_1 \lambda_2)^k - \lambda_1^k - \lambda_2^k + 1} \right] \\
&= \frac{C_{2k(n+1)-1}}{C_{2k}-1}.
\end{align*}
$$

**Theorem 3.8 (Binomial sum).** For the higher order balancing numbers {$B_n^{(k)}$}, we have

$$
1. \sum_{n=0}^{s-1} \binom{s-1}{n} B_n^{(k)} = \frac{4\sqrt{2}}{B_k} \left( (1 + \lambda_1^k)^{s-1} - (1 + \lambda_2^k)^{s-1} \right)
$$

$$
2. \sum_{n=0}^{s-1} (-1)^n \binom{s-1}{n} B_n^{(k)} = \frac{4\sqrt{2}}{B_k} \left( (1 - \lambda_1^k)^{s-1} - (1 - \lambda_2^k)^{s-1} \right).
$$

*Proof.* 1. From Binet's formula (2.6), we write

$$
\begin{align*}
\sum_{n=0}^{s-1} \binom{s-1}{n} B_n^{(k)} &= \sum_{n=0}^{s-1} \binom{s-1}{n} \frac{\lambda_1^{kn} - \lambda_2^{kn}}{\lambda_1^k - \lambda_2^k} \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( \sum_{n=0}^{s-1} \binom{s-1}{n} \lambda_1^{kn} - \sum_{n=0}^{s-1} \binom{s-1}{n} \lambda_2^{kn} \right) \\
&= \frac{4\sqrt{2}}{B_k} \left( (1 + \lambda_1^k)^{s-1} - (1 + \lambda_2^k)^{s-1} \right) \quad \text{(using the Binomial theorem)}
\end{align*}
$$

2. From Binet's formula (2.6), we write

$$
\begin{align*}
\sum_{n=0}^{s-1} (-1)^n \binom{s-1}{n} B_n^{(k)} &= \sum_{n=0}^{s-1} \binom{s-1}{n} \frac{(-\lambda_1^k)^n - (-\lambda_2^k)^n}{\lambda_1^k - \lambda_2^k} \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( \sum_{n=0}^{s-1} \binom{s-1}{n} (-\lambda_1^k)^n - \sum_{n=0}^{s-1} \binom{s-1}{n} (-\lambda_2^k)^n \right) \\
&= \frac{4\sqrt{2}}{B_k} \left( (1 - \lambda_1^k)^{s-1} - (1 - \lambda_2^k)^{s-1} \right) \quad \text{(using the Binomial theorem)}.
\end{align*}
$$

**Theorem 3.9.** If *m* and *n* are positive integers, then

$$
1. \sum_{r=0}^{2n} \binom{2n}{r} B_{2r}^{(k)} = \frac{4\sqrt{2}}{B_k} \left( (1 + \lambda_1^{2k})^{2n} - (1 + \lambda_2^{2k})^{2n} \right)
$$

$$
2. \sum_{r=0}^{2n+1} \binom{2n+1}{r} B_{2r}^{(k)} = \frac{4\sqrt{2}}{B_k} \left( (1 + \lambda_1^{2k})^{2n+1} - (1 + \lambda_2^{2k})^{2n+1} \right).
$$
Proof. 1. From Binet's formula (2.6), we write

$$
\begin{align*}
\sum_{r=0}^{2n} \binom{2n}{r} B_{2r}^{(k)} &= \sum_{r=0}^{2n} \binom{2n}{r} \frac{\lambda_1^{2kr} - \lambda_2^{2kr}}{\lambda_1^k - \lambda_2^k} \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( \sum_{r=0}^{2n} \binom{2n}{r} \lambda_1^{2kr} - \sum_{r=0}^{2n} \binom{2n}{r} \lambda_2^{2kr} \right) \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( (1 + \lambda_1^{2k})^{2n} - (1 + \lambda_2^{2k})^{2n} \right) \quad \text{(using the Binomial theorem)} \\
&= \frac{4\sqrt{2}}{B_k} \left( (1 + \lambda_1^{2k})^{2n} - (1 + \lambda_2^{2k})^{2n} \right).
\end{align*}
$$

2. From Binet's formula (2.6), we write

$$
\begin{align*}
\sum_{r=0}^{2n+1} \binom{2n+1}{r} B_{2r}^{(k)} &= \sum_{r=0}^{2n+1} \binom{2n+1}{r} \frac{\lambda_1^{2kr} - \lambda_2^{2kr}}{\lambda_1^k - \lambda_2^k} \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( \sum_{r=0}^{2n+1} \binom{2n+1}{r} \lambda_1^{2kr} - \sum_{r=0}^{2n+1} \binom{2n+1}{r} \lambda_2^{2kr} \right) \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( (1 + \lambda_1^{2k})^{2n+1} - (1 + \lambda_2^{2k})^{2n+1} \right) \quad \text{(using the Binomial theorem)} \\
&= \frac{4\sqrt{2}}{B_k} \left( (1 + \lambda_1^{2k})^{2n+1} - (1 + \lambda_2^{2k})^{2n+1} \right). \quad \square
\end{align*}
$$

With the help of Binet's formula, the following identities can be easily established.

**Theorem 3.10.** For $n \ge 0$, the following identities are verified:

$$
1. B_{-n}^{(k)} = -B_{n}^{(k)}
$$

$$
2. B_{2n}^{(k)} = 2C_{kn}B_{n}^{k}
$$

$$
3. B_{3n}^{(k)} = 32B_{n}^{(k)}(B_{kn}^{2} + 3)
$$

$$
4. B_{n \pm r}^{(k)} = B_{n}^{(k)} C_{kr} \pm B_{r}^{(k)} C_{kn}
$$

$$
5. B_{n+r}^{(k)} + B_{n-r}^{(k)} = 2C_{kr}B_{n}^{(k)}
$$

$$
6. B_{n+r}^{(k)} - B_{n-r}^{(k)} = 2C_{kn}B_{r}^{(k)}
$$

$$
7. (B_{n+r}^{(k)})^2 - (B_{n-r}^{(k)})^2 = 4C_{kn}C_{kr}B_{n}^{(k)}B_{r}^{(k)}
$$

$$
8. B_{n+r}^{(k)} B_{n-r}^{(k)} = (B_{n}^{(k)})^{2} - (B_{r}^{(k)})^{2}
$$

Proof. 1. Since $B_{-nk} = -B_{nk}$, so

$$
B_{-n}^{(k)} = \frac{B_{-nk}}{B_k} = -\frac{B_{nk}}{B_k} = -B_n^{(k)}.
$$

2. Using $B_{2k} = 2B_k C_k$, we can write $B_{2n}^{(k)} = \frac{B_{2kn}}{B_k} = 2C_{kn} \frac{B_{kn}}{B_k} = 2C_{kn} B_n^k$.

3. Taking into account the identity $B_{3k} = 32B_k^3 + 3B_k$, we have

$$
B_{3n}^{(k)} = \frac{B_{3nk}}{B_k} = \frac{32B_{kn}^3 + 3B_{kn}}{B_k} = 32B_n^{(k)}(B_{kn}^2 + 3).
$$

Along with Definition 2.1, proof of identity (4), (5) and (6) use $B_{m \pm n} = B_m C_n \pm C_m B_n$), $B_{a+b} + B_{a-b} = 2B_a C_b$ and $B_{a+b} - B_{a-b} = 2C_a B_b$, respectively.
7. Using $B_{a+b} + B_{a-b} = 2B_a C_b$ and $B_{a+b} - B_{a-b} = 2C_a B_b$, we have

$$ 
\begin{aligned}
(B_{n+r}^{(k)})^2 - (B_{n-r}^{(k)})^2 &= \frac{(B_{k(n+r)})^2 - (B_{k(n-r)})^2}{(B_k)^2} \\
&= \frac{(B_{kn+kr} + B_{kn-kr})(B_{kn+kr} - B_{kn-kr})}{(B_k)^2} \\
&= \frac{4B_{kn}B_{kr}C_{kn}C_{kr}}{(B_k)^2} \\
&= 4C_{kn}C_{kr}B_{n}^{(k)}B_{r}^{(k)}.
\end{aligned}
$$

8. Since $B_{a+b}B_{a-b} = B_a^2 - B_b^2$, so

$$ B_{n+r}^{(k)} B_{n-r}^{(k)} = \frac{B_{kn+kr} B_{kn-kr}}{B_k^2} = \frac{B_{kn}^2 - B_{kr}^2}{B_k^2} = (B_n^{(k)})^2 - (B_r^{(k)})^2. \quad \square $$

**Theorem 3.11.** If *m* and *n* are positive integers, then

$$ 1. \sum_{r=1}^{n} B_{2mr}^{(k)} = \frac{B_{2mn}^{(k)} - B_{2m(n+1)}^{(k)} + B_{2m}^{(k)}}{2 - 2C_{2mk}}. $$

$$ 2. \sum_{r=1}^{n} (-1)^r B_{2mr}^{(k)} = \frac{(-1)^n (B_{2mn}^{(k)} + B_{2m(n+1)}^{(k)}) - B_{2m}^{(k)}}{2 + 2C_{2mk}}. $$

$$ 3. \sum_{r=1}^{n} (B_{mr}^{(k)})^2 = \frac{1}{C_{2k}-1} \left( \frac{C_{2mk} + C_{2mkn} - C_{2mk(1+n)} - 1}{2(1-C_{2mk})} - n \right). $$

$$ 4. \sum_{r=1}^{n} (-1)^r (B_{mr}^{(k)})^2 = \begin{cases} \frac{1}{2(C_{2k}-1)} \left[ \frac{(-1)^n (C_{2mkn} + C_{2mk(n+1)}) - C_{2mk}}{C_{2mk} + 1} \right]; & n = \text{even} \\ \frac{1}{2(C_{2k}-1)} \left[ \frac{(-1)^n (C_{2mkn} + C_{2mk(n+1)}) - C_{2mk}}{C_{2mk} + 1} + 2 \right]; & n = \text{odd}. \end{cases} $$

*Proof.* 1. Using the Binet's formula (2.6), we have

$$ 
\begin{aligned}
\sum_{r=1}^{n} B_{2mr}^{(k)} &= \sum_{r=1}^{n} \frac{1}{\lambda_1^k - \lambda_2^k} (\lambda_1^{2mrk} - \lambda_2^{2mrk}) \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( \lambda_1^{2mk} \frac{(\lambda_1^{2mk})^n - 1}{\lambda_1^{2mk} - 1} - \lambda_2^{2mk} \frac{(\lambda_2^{2mk})^n - 1}{\lambda_2^{2mk} - 1} \right) \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left[ \frac{((\lambda_1^{2mk})^{n+1} \lambda_2^{2mk} - (\lambda_1 \lambda_2)^{2mk} - (\lambda_1^{2mk})^{n+1} + \lambda_1^{2mk})}{(\lambda_1^k - 1)(\lambda_2^k - 1)} \right] \\
&\quad - \frac{1}{\lambda_1^k - \lambda_2^k} \left[ \frac{((\lambda_2^{2mk})^{n+1} \lambda_1^{2mk} - (\lambda_1 \lambda_2)^{2mk} - (\lambda_2^{2mk})^{n+1} + \lambda_2^{2mk})}{(\lambda_1^k - 1)(\lambda_2^k - 1)} \right] \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left[ \frac{((\lambda_1^{2mk})^n - (\lambda_2^{2mk})^n) - ((\lambda_1^{2mk})^{n+1} - (\lambda_2^{2mk})^{n+1}) + (\lambda_1^{2mk} - \lambda_2^{2mk})}{(\lambda_1 \lambda_2)^k - \lambda_1^k - \lambda_2^k + 1} \right] \\
&= \frac{B_{2mn}^{(k)} - B_{2m(n+1)}^{(k)} + B_{2m}^{(k)}}{2 - 2C_{2mk}}.
\end{aligned}
$$
2. Using the Binet's formula (2.6), we have

$$
\begin{aligned}
\sum_{r=1}^{n} (-1)^r B_{2mr}^{(k)} &= \sum_{r=1}^{n} \frac{(-1)^r}{\lambda_1^k - \lambda_2^k} (\lambda_1^{2mrk} - \lambda_2^{2mrk}) \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( \sum_{r=1}^{n} (-\lambda_1^{2mk})^r - \sum_{r=1}^{n} (-\lambda_2^{2mk})^r \right) \\
&= \frac{1}{\lambda_1^k - \lambda_2^k} \left( -\lambda_1^{2mk} \frac{1 - (-\lambda_1^{2mk})^n}{1 - (-\lambda_1^{2mk})} - (-\lambda_2^{2mk}) \frac{1 - (-\lambda_2^{2mk})^n}{1 - (-\lambda_2^{2mk})} \right).
\end{aligned}
$$

After simplifying, we get

$$
\sum_{r=1}^{n} (-1)^{r} B_{2mr}^{k} = \frac{(-1)^{n} \left( B_{2mn}^{(k)} + B_{2m(n+1)}^{(k)} \right) - B_{2m}^{(k)}}{2 + 2C_{2mk}}.
$$

3. Using the Binet's formula (2.6), we have

$$
\begin{aligned}
\sum_{r=1}^{n} (B_{mr}^{(k)})^2 &= \sum_{r=1}^{n} \left( \frac{\lambda_1^{mrk} - \lambda_2^{mrk}}{\lambda_1^k - \lambda_2^k} \right)^2 \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \sum_{r=1}^{n} \left( \lambda_1^{2mkr} + \lambda_2^{2mkr} - 2(\lambda_1 \lambda_2)^{mkr} \right) \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left( \lambda_1^{2mk} \frac{(\lambda_1^{2mk})^n - 1}{\lambda_1^{2mk} - 1} + \lambda_2^{2mk} \frac{(\lambda_2^{2mk})^n - 1}{\lambda_2^{2mk} - 1} - 2n \right).
\end{aligned}
$$

After simplifying, we get

$$
\sum_{r=1}^{n} (B_{mr}^{(k)})^2 = \frac{1}{C_{2k}-1} \left( \frac{C_{2mk} + C_{2mkn} - C_{2mk(1+n)} - 1}{2(1-C_{2mk})} - n \right).
$$

4. Using the Binet's formula of higher order Balancing number, we have

$$
\begin{aligned}
\sum_{r=1}^{n} (-1)^r (B_{mr}^{(k)})^2 &= \sum_{r=1}^{n} (-1)^r \left( \frac{\lambda_1^{mrk} - \lambda_2^{mrk}}{\lambda_1^k - \lambda_2^k} \right)^2 \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \sum_{r=1}^{n} (-1)^r \left( \lambda_1^{2mkr} + \lambda_2^{2mkr} - 2(\lambda_1 \lambda_2)^{mkr} \right) \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left[ \sum_{r=1}^{n} (-\lambda_1^{2mk})^r + \sum_{r=1}^{n} (-\lambda_2^{2mk})^r - 2 \sum_{r=1}^{n} (-1)^r \right] \\
&= \frac{1}{(\lambda_1^k - \lambda_2^k)^2} \left[ (-\lambda_1^{2mk}) \frac{(-\lambda_1^{2mk})^n - 1}{(-\lambda_1^{2mk}) - 1} + (-\lambda_2^{2mk}) \frac{(-\lambda_2^{2mk})^n - 1}{(-\lambda_2^{2mk}) - 1} - 2 \sum_{r=1}^{n} (-1)^r \right] \\
&= \frac{1}{2(C_{2k} - 1)} \left[ \frac{(-1)^n (C_{2mkn} + C_{2mk(n+1)}) - C_{2mk}}{C_{2mk} + 1} - 2 \sum_{r=1}^{n} (-1)^r \right].
\end{aligned}
$$

Since $\sum_{r=1}^{n} (-1)^r = 0$ and $-1$ for $n$ = even and odd, respectively, so

$$
\sum_{r=1}^{n} (-1)^r (B_{mr}^{(k)})^2 = \begin{cases} \frac{1}{2(C_{2k}-1)} \left[ \frac{(-1)^n (C_{2mkn} + C_{2mk(n+1)}) - C_{2mk}}{C_{2mk} + 1} \right] & : n = \text{even} \\ \frac{1}{2(C_{2k}-1)} \left[ \frac{(-1)^n (C_{2mkn} + C_{2mk(n+1)}) - C_{2mk}}{C_{2mk} + 1} + 2 \right] & : n = \text{odd}. \end{cases}
$$

□
# 4 Conclusion

Through this article we introduced a new extension of the balancing numbers, which generalizes the balancing numbers in a different way. For this integer sequence, we presented Binet's formula, algebraic identities, various generating functions, various partial and binomial sums, etc. In addition, we presented several identities in connections with the classic balancing numbers.

In future study, this sequence can be extended in many directions like hypercomplex numbers, polynomials, quaternions, octonions, matrix theory, etc. We are working on hypercomplex/2<sup>m</sup>-ions numbers using this new sequence.

# References

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