Identities concerning k-balancing and k-Lucas-balancing numbers of arithmetic indexes

1.   Introduction

Balancing numbers are originally obtained from a simple Diophantine equation. They are the solutions of the Diophantine equation $1+2+3+\ldots+ (n-1) = (n+1)+(n+2)+\ldots +(n+r),$ where $r$ is a balancer corresponds to a balancing number $n$ ^[1,3]. Balancing numbers satisfy the recurrence relation

with $B_{0} = 0$ and $B_{1} = 1$, where $B_{n}$ denotes the $n^{th}$ balancing number ^[1]. On the other hand, Lucas-balancing numbers $C_{n}$ are obtained from the formula $C_{n} = \sqrt{8B_{n}^{2}+1}$ and are the terms of the sequence $\{1, 3, 17, 99,577, \ldots\}$ ^[7]. They are recursively defined same as that of balancing numbers but with different initial values, that is,

with $C_{0} = 1$ and $C_{1} = 3$ ^[7].

Balancing numbers are generalized in many ways. For details review of some recent works, one can go through ^{[2,4,5,6,8,9,10]}. One of the generalization of balancing numbers called as $k$-balancing numbers depending on one real parameter $k$, are recently introduced by Ray in ^[9]. The $n^{th}$ $k$-balancing numbers $B_{k, n}$ are terms of the sequence $\{0, 1, 6k, 36k^{2}-1,216k^{3}-12k, \ldots\}$ and are recursively defined by

Notice that, for $k = 1$, balancing numbers $0, 1, 6, 35,204, \ldots$ are obtained.

On the other hand, $k$-Lucas-balancing numbers that are the natural extension of Lucas-balancing numbers extensively studied in ^[9]. The sequence of $k$-Lucas-balancing numbers $\{C_{k, n}\} = \{1, 3k, 18k^{2}-1,108k^{3}-9k, \ldots\}$ satisfies the same recurrence relation as that of $k$-balancing numbers with different initial conditions, i.e.,

Few properties that the $k$-balancing numbers satisfy are summarized below.

● Binet formula for $k$-balancing numbers: $B_{k, n} = \frac{\lambda_{k}^{n}-\lambda_{k}^{-n}}{\lambda_{k}-\lambda_{k}^{-1}}$, where $\lambda_{k} = 3k+\sqrt{9k^2-1}$.

● Catalan identity for $k$-balancing numbers: $B_{k, n}^{2}-B_{k, n-r}B_{k, n+r} = B_{k, r}^{2}$.

● Simson's identity for $k$-balancing numbers: $B_{k, n}^{2}-B_{k, n-1}B_{k, n+1} = 1$.

● D' Ocagne identity for $k$-balancing numbers: $B_{k, m}B_{k, n+1}-B_{k, m+1}B_{k, n} = B_{k, m-n}^{2}.$

● For odd $k$-balancing numbers, $B_{k, 2n+1} = B_{k, n+1}^{2}-B_{k, n}^{2}.$

● For even $k$-balancing numbers, $B_{k, 2n} = \frac{1}{6k}[B_{k, n+1}^{2}-B_{k, n-1}^{2}].$

● Generating function for $k$-balancing numbers: $f_{k}(x) = \frac{x}{1-6kx-x^{2}}$.

● First combinatorial formula for $k$-balancing numbers:

● Second combinatorial formula for $k$-balancing numbers:

2.   Identities concerning $k$-balancing numbers of arithmetic indexes

In this section, we study different sums of $k$-balancing numbers of arithmetic indexes, say $an+p$ for fixed integers $a$ and $p$ with $0\leq p\leq a-1$. Several identities concerning such numbers are established straightforwardly.

The following lemmas are useful while proving the subsequent results.

Lemma 2.1. For all integers $n\geq 1$, $\lambda_{k_{1}}^{n} +\lambda_{k_{2}}^{n} = B_{k, n+1}-B_{k, n-1}$.

Proof. The proof of this result can be easily shown by using Binet formula for $k$-balancing numbers and the fact $\lambda_{k_{1}}^{n} \lambda_{k_{2}}^{n} = 1$.

Lemma 2.2. $B_{k, a(n+2)+p} = (B_{k, a+1}-B_{k, a-1})B_{k, a(n+1)+p}-B_{k, an+p}$.

Proof. Using Binet formula for $k$-balancing numbers and the result from Lemma 2.1, the first term of the right hand side expression reduces

and the result follows. Since $B_{k, n+1}-B_{k, n-1} = 2C_{k, n}$ ^[9], the previous formula reduces to an identity

This identity gives the general term of the sequence of $k$-balancing numbers $\{B_{k, an+p}\}$ as a linear combination of two preceding terms. Iterative application of this result gives the general term as a combination of first two terms as follows:

In particular, for $a = 1$, then $r = 0$ and we have the corresponding identity for $k$-balancing numbers,

Now we will find the generating function for the sequence $\{B_{k, an+p}\}$. Let $G(k, a, p, x)$ be the generating function for the sequence $\{B_{k, an+p}\}$, where $0\leq p\leq a-1$, then

Multiplying $2C_{k, a}x$ and $x^{2}$ in (2.1) by turns and subtracting the first one from (2.1) and then adding the second one, we obtain

The expression within the summation vanishes in view of Lemma 2.2. On the other hand, using the convolution identity $B_{k, a+p} = B_{k, p}B_{k, a+1}-B_{k, p-1}B_{k, a}$ and the fact $B_{k, n+1}-B_{k, n-1} = 2C_{k, n}$, the expression $B_{k, p}+(B_{k, a+p}-2B_{k, p}C_{k, a})$ reduces to

Therefore, (2.2) gives

For $a = 1$, then $p = 0$ and $G(k, 1, 0, x) = \frac{x}{1-6kx+x^{2}}$ which is indeed the generating function for $k$-balancing numbers. While Choosing $a = 2$, $p$ will be $0$ and $1$ and we have $G(k, 2, 0, x) = \frac{6kx}{1-6kx+x^{2}}$ and $G(k, 2, 1, x) = \frac{1+x}{1-6kx+x^{2}}.$

The following theorem establishes the sum for $k$-balancing numbers of the type $an+p$.

Theorem 2.3. Let $a$ be any integer and $0\leq p\leq a-1$, then

Proof. Using Binet formula, the formula for geometric series and the fact $\lambda_{k_{1}}^{n} \lambda_{k_{2}}^{n} = 1$, we get

By virtue of Lemma 2.1 and by Binet formula, we get the desired result.

The following results are immediate consequence of Theorem 2.3 by setting $a = 2t+1$ and $a = 2t$ respectively.

Corollary 2.4. The sum of odd $k$-balancing numbers of the kind $an+p$ is

Corollary 2.5. The sum of even $k$-balancing numbers of the kind $an+p$ is

Obsevation 2.6. Let $t = 0$, then $a = 1$ and $p = 0$. Therefore from Corollary 2.4, we obtain

For $k = 1$, the sum of balancing numbers is $\sum\limits_{i = 0}^{n}B_{k, i} = \frac{B_{n+1}-B_{n}-1}{4}.$ Similarly, for $t = 1$, $a = 3$, we have

For $p = 0$,

and for $k = 0,$$\sum\limits_{i = 0}^{n}B_{3i} = \frac{B_{3n+3}-B_{3n}-35}{196}$. For $p = 1$,

For $k = 0,$ the sum formula for balancing numbers is given by $\sum\limits_{i = 0}^{n}B_{3i+1} = \frac{B_{3n+4}-B_{3n+1}-7}{196}$.

Finally, for $p = 2$,

Again, $k = 0$ gives $\sum\limits_{i = 0}^{n}B_{3i+2} = \frac{B_{3n+5}-B_{3n+2}-7}{196}$. Similarly, by virtue of Corollary 2.5, for $t = 1$, then $a = 2$ implies that for $p = 0$, $\sum\limits_{i = 0}^{n}B_{k, 2i} = \frac{B_{k, 2n+2}-B_{k, 2n}-6k}{36k^{2}-4}.$ For $p = 1$, we have $\sum\limits_{i = 0}^{n}B_{k, 2i+1} = \frac{B_{k, 2n+3}-B_{k, 2n+1}-2}{36k^{2}-4}$ and so on.

Now we will find the recurrence relation for the sequence $\{B_{k, an+p}\}$. For that, let us denote $\sum\limits_{i = 0}^{n}B_{k, ai+p}$ as $S_{k, an+p}$. In view of Lemma 2.2, we have

It follows that,

Consequently,

which is the desired recurrence relation for the sequence $\{B_{k, an+p}\}$.

3.   Identities concerning $k$-Lucas-balancing numbers of arithmetic indexes

In this section, we study the $k$-Lucas-balancing numbers of arithmetic indexes of the form $an+p$.

A repeated application of the formula in Lemma 2.2 gives an identity that relates $k$-Lucas-balancing numbers with $k$-balancing numbers, that is, for natural numbers $n$ and $l$,

But for $l = -n$, $C_{k, n} = C_{k, 2n}B_{k, n+1}-C_{k, 2n+1}B_{k, n}.$ Also it is observed that $C_{k, -n} = C_{k, n}$.

Lemma 3.1. Let $a \neq 0$ and $0\leq p\leq a-1$, then $C_{k, a(n+1)+p} = 2C_{k, a}C_{k, an+p}-C_{k, a(n-1)+p}.$

Proof. Clearly $2C_{k, a(n+1)+p} = B_{k, a(n+1)+p+1}-B_{k, a(n+1)+p-1}.$ Therefore, using Lemma 2.2, we get

and we obtain the desired result.

In particular, for $p = 0$ the above identity reduces to $C_{k, an+a} = 2C_{k, a}C_{k, an}-C_{k, an-a}.$ Applying iteratively the identity in Lemma 3.1 gives rise to the following sum formula.

Further, for $m = n$, this formula reduces to $C_{k, an+p} = \sum\limits_{i = 0}^{n} (-1)^{i}\begin{pmatrix} n \\ i \\ \end{pmatrix}(2C_{k, a})^{n-i} C_{k, p-ai}.$

In order to find the generating function of the sequence $\{C_{k, an+p}\}$, we proceed in the following way. Let $g(k, a, p, x)$ be the generating function for the sequence $\{C_{k, an+p}\}$, where $0\leq p\leq a-1$, then

Multiply (3.1) by $2C_{k, a}x$ and $x^{2}$ and proceed as in case of the sequence $\{B_{k, an+p}\}$, we get the generating function for the sequence $\{C_{k, an+p}\}$ as

It is observed that for $a = 1$, then $r = 0$ and we have the generating function for $k$-Lucas-balancing numbers, $g(k, x) = \frac{1-3kx}{1-6kx+x^{2}}.$ Further, for $k = 1$, the generating function for Lucas-balancing numbers $g(x) = \frac{1-3x}{1-6x+x^{2}}$ is obtained.

Theorem 3.2. Let $a$ be any integer and $0\leq p\leq a-1$, then

Proof. The proof of this theorem is analogous to Theorem 2.3.

As an observation, one can see that, for $a = 1$ then $r = 0$ gives the identity $\sum\limits_{i = 0}^{n}C_{i} = \frac{C_{n+1}-C_{n}+2}{4} = \frac{c_{n}+1}{2},$ where $c_{n}$ is the $n^{th}$ Lucas-cobalancing number with $C_{n+1}-C_{n} = 2c_{n}$ ^[2].

We end this section by establishing an important relation between $k$-balancing and $k$-Lucas-balancing numbers.

Theorem 3.3. For $t\geq 1$,

Proof. Using Binet formula and Lemma 2.1, the left side expression becomes

Continuing in this way, we finally get

This completes the proof.

In particular, for $t = 1$, the above identity reduces to $B_{k, 2n} = 2B_{k, n}C_{k, n}$, a known identity for $k$-balancing numbers.

Conflict of interest

The author declares no conflict of interest.