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Article

On the Reciprocal Sums of Products of Balancing and Lucas-Balancing Numbers

Department of Electronic and Electrical Convergence Engineering, Hongik University, Sejong-Ro 2639, Sejong 30016, Korea
Mathematics 2021, 9(4), 350; https://doi.org/10.3390/math9040350
Submission received: 23 January 2021 / Revised: 7 February 2021 / Accepted: 9 February 2021 / Published: 10 February 2021
(This article belongs to the Section Computational and Applied Mathematics)

Abstract

:
Recently Panda et al. obtained some identities for the reciprocal sums of balancing and Lucas-balancing numbers. In this paper, we derive general identities related to reciprocal sums of products of two balancing numbers, products of two Lucas-balancing numbers and products of balancing and Lucas-balancing numbers. The method of this paper can also be applied to even-indexed and odd-indexed Fibonacci, Lucas, Pell and Pell–Lucas numbers.

1. Introduction

The classical Fibonacci numbers { F n } n = 0 are generated from the recurrence relation F n = F n 1 + F n 2 ( n 2 ) with the initial conditions F 0 = 0 and F 1 = 1 . As is well known, the Fibonacci numbers possess many interesting properties and appear in a variety of application fields [1].
Recently Ohtsuka and Nakamura [2] reported an interesting property of the Fibonacci numbers and proved the following identities:
k = n 1 F k 1 = F n F n 1 , if   n     2   and n   is   even ; F n F n 1 1 , if   n     1   and   n   is   odd ,
k = n 1 F k 2 1 = F n 1 F n 1 , if   n     2   and   n   is   even ; F n 1 F n , if   n     1   and   n   is   odd ,
where · is the floor function.
Following the work of Ohtsuka and Nakamura, diverse results in the same direction have been reported in the literature [3,4,5,6,7,8,9,10,11,12,13,14,15].
A positive integer n is called the balancing number if [16]
1 + 2 + + ( n 1 ) = ( n + 1 ) + ( n + 2 ) + + ( n + r ) ,
for some positive integer r. As shown in [16], the balancing numbers { B n } n = 0 satisfy the recurrence relation B n = 6 B n 1 B n 2 ( n 2 ) with the initial conditions B 0 = 0 and B 1 = 1 . The balancing numbers are useful in studying the Diophantine equations [17,18]. The numbers { C n } n = 0 with C n = 8 B n 2 + 1 are called the Lucas-balancing numbers [19] and obtained from the recurrence relation C n = 6 C n 1 C n 2 ( n 2 ) with the initial conditions C 0 = 1 and C 1 = 3 .
Panda et al. [20] recently studied the reciprocal sums of balancing and Lucas-balancing numbers, and obtained various identities. For example, they showed that
k = n 1 B k 1 = B n B n 1 1 ( n 1 ) ,
k = n 1 B k 2 1 = B n 2 B n 1 2 1 ( n 1 ) ,
k = n 1 B 2 k 1 = B 2 n B 2 n 2 1 ( n 1 ) ,
k = n 1 B 2 k 2 1 = B 2 n 2 B 2 n 2 2 1 ( n 1 ) ,
k = n 1 B k B k + 1 1 = B n B n + 1 B n 1 B n 1 ( n 1 ) ,
k = n 1 B 2 k B 2 k + 2 1 = B 2 n + 1 2 B 2 n 1 2 2 ( n 1 ) ,
and
k = n 1 C k 1 = C n C n 1 ( n 2 ) ,
k = n 1 C k 2 1 = C n 2 C n 1 2 ( n 1 ) ,
k = n 1 C 2 k 1 = C 2 n C 2 n 2 ( n 1 ) ,
k = n 1 C 2 k 2 1 = C 2 n 2 C 2 n 2 2 ( n 1 ) ,
k = n 1 C k C k + 1 1 = C n C n + 1 C n 1 C n + 1 ( n 1 ) ,
k = n 1 C 2 k C 2 k + 2 1 = C 2 n + 1 2 C 2 n 1 2 + 8 ( n 1 ) ,
etc.
We note that (3), (4) and (9), (10) also can be obtained, respectively from ([12] [Theorem 2.1]) and ([12] [Theorem 2.2]).
In this paper, we derive general identities related to reciprocal sums of products of two balancing numbers, products of two Lucas-balancing numbers and products of balancing and Lucas-balancing numbers. The results obtained here not only include most identities in [20] as special cases but also can be used to derive similar identities for even-indexed and odd-indexed Fibonacci, Lucas, Pell and Pell–Lucas numbers.

2. Results

For the ease of presentation, we use the notation G n = S ( G 0 , G 1 , a , b ) to denote the numbers { G n } n = 0 generated from the recurrence relation
G n = a G n 1 + b G n 2 ( n 2 ) ,
with the initial conditions G 0 and G 1 .
To deal with the balancing numbers B n = S ( 0 , 1 , 6 , 1 ) and Lucas-balancing numbers C n = S ( 1 , 3 , 6 , 1 ) in a unified manner, we consider the numbers G n = S ( G 0 , G 1 , a , 1 ) , where G 0 is a nonnegative integer, G 1 and a are positive integers. As in [12], we assume that
a max { 3 , 1 + G 0 / G 1 } .
Firstly we present two lemmas which will be used to prove our main results. For G n = S ( G 0 , G 1 , a , 1 ) and H n = S ( H 0 , H 1 , a , 1 ) , define
Φ G : = a G 0 G 1 G 0 2 G 1 2 , Φ H : = a H 0 H 1 H 0 2 H 1 2 , Δ n : = G n 1 G n + 1 H n + m 1 H n + m + 1 G n 2 H n + m 2 G n + 1 H n + m + 1 G n 1 H n + m = 1 , Δ m : = lim n Δ n .
Lemma 1
(See [21]). For G n = S ( G 0 , G 1 , a , 1 ) , we have
G n 2 G n r G n + r = ( G 1 G r G 0 G r + 1 ) Q r ,
where Q n = S ( 0 , 1 , a , 1 ) .
Lemma 2.
For G n = S ( G 0 , G 1 , a , 1 ) and H n = S ( H 0 , H 1 , a , 1 ) , we have
Δ m = ( Φ H G 1 2 + Φ G H m + 1 2 ) α 4 2 ( Φ H G 0 G 1 + Φ G H m H m + 1 ) α 3 + ( Φ H G 0 2 + Φ G H m 2 ) α 2 G 1 H m + 1 α 6 ( G 1 H m + G 0 H m + 1 ) α 5 + G 0 H 0 α 4 G 1 H m + 1 α 2 + ( G 1 H m + G 0 H m + 1 ) α G 0 H m ,
where
α = a + a 2 4 2 .
Proof. 
From Lemma 1, we have
G n 1 G n + 1 H n + m 1 H n + m + 1 G n 2 H n + m 2 = Φ H G n 2 + Φ G H n + m 2 + Φ G Φ H .
G n and H n + m can be expressed as [21]
G n = G 1 ( α n β n ) G 0 ( α n 1 β n 1 ) α β , H n + m = H m + 1 ( α n β n ) H m ( α n 1 β n 1 ) α β ,
where α > 1 and 0 < β < 1 are solutions of the equation x 2 a x + 1 = 0 , i.e.,
α = a + a 2 4 2 and β = a a 2 4 2 .
Since
( α β ) 2 ( Φ H G n 2 + Φ G H n + m 2 + Φ G Φ H ) = α 2 n 2 ( Φ H G 1 2 + Φ G H m + 1 2 ) α 2 2 ( Φ H G 0 G 1 + Φ G H m H m + 1 ) α + ( Φ H G 0 2 + Φ G H m 2 ) + γ n ,
and
( α β ) 2 ( G n + 1 H n + m + 1 G n 1 H n + m 1 ) = α 2 n 4 G 1 H m + 1 α 6 ( G 1 H m + G 0 H m + 1 ) α 5 + G 0 H 0 α 4 G 1 H m + 1 α 2 + ( G 1 H m + G 0 H m + 1 ) α G 0 H m + δ n ,
where lim n γ n = 0 and lim n δ n = 0 , then the proof is completed. □
Now we state our main results.
Theorem 1.
For G n = S ( G 0 , G 1 , a , 1 ) and H n = S ( H 0 , H 1 , a , 1 ) , there exists a positive integer N such that
k = n 1 G k H k + m 1 = G n H n + m G n 1 H n + m 1 + g m , i f   n N ,
where
g m = Δ m .
Proof. 
Consider
1 G n H n + m G n 1 H n + m 1 + g m 1 G n + 1 H n + m + 1 G n H n + m + g m 1 G n H n + m = X 1 ( G n H n + m G n 1 H n + m 1 + g m ) ( G n + 1 H n + m + 1 G n H n + m + g m ) G n H n + m ,
where
X 1 = G n 1 G n + 1 H n + m 1 H n + m + 1 G n 2 H n + m 2 g m ( G n + 1 H n + m + 1 G n 1 H n + m 1 ) g m 2 = ( G n + 1 H n + m + 1 G n 1 H n + m 1 ) ( Δ n g m ) g m 2 .
Since Δ n converges to Δ m and Δ m g m > 0 , there exists a positive integer n 0 such that X 1 > 0 if n n 0 or
1 G n H n + m < 1 G n H n + m G n 1 H n + m 1 + g m 1 G n + 1 H n + m + 1 G n H n + m + g m , if   n     n 0 .
Repeatedly applying the above inequality, we have
k = n 1 G k H k + m < 1 G n H n + m G n 1 H n + m 1 + g m , if   n     n 0 .
Similarly,
1 G n H n + m G n 1 H n + m 1 + g m + 1 1 G n + 1 H n + m + 1 G n H n + m + g m + 1 1 G n H n + m = X 2 ( G n H n + m G n 1 H n + m 1 + g m + 1 ) ( G n + 1 H n + m + 1 G n H n + m + g m + 1 ) G n H n + m ,
where
X 2 = G n 1 G n + 1 H n + m 1 H n + m + 1 G n 2 H n + m 2 G n + 1 H n + m + 1 + G n 1 H n + m 1 g m ( G n + 1 H n + m + 1 G n 1 H n + m 1 ) ( g m + 1 ) 2 = ( G n + 1 H n + m + 1 G n 1 H n + m 1 ) ( Δ n g m ) G n + 1 H n + m + 1 + G n 1 H n + m 1 ( g m + 1 ) 2 .
Since Δ n converges to Δ m and 0 < Δ m g m < 1 , then there exists a positive integer n 1 such that X 2 < 0 if n n 1 or
1 G n H n + m G n 1 H n + m 1 + g m 1 G n + 1 H n + m + 1 G n H n + m + g m < 1 G n H n + m , if   n     n 1 ,
from which we obtain
1 G n H n + m G n 1 H n + m 1 + g m + 1 < k = n 1 G k H k + m , if   n     n 1 .
Then (15) follows from (16) and (17). □
Setting G n = H n = S ( 0 , 1 , 6 , 1 ) in Theorem 1, we obtain Corollary 1 below.
Corollary 1.
For balancing numbers B n = S ( 0 , 1 , 6 , 1 ) , there exists a positive integer N such that
k = n 1 B k B k + m 1 = B n B n + m B n 1 B n + m 1 + g m , i f   n N ,
where
g m = ( B m + 1 2 + 1 ) α 3 + 2 B m B m + 1 α 2 B m 2 α B m + 1 α 5 B m α 4 B m + 1 α + B m ,
with α = 3 + 2 2 .
For balancing numbers, g 0 = g 1 = 1 and we obtain (4) and (7) from (18). In addition we have
k = n 1 B k B k + 2 1 = B n B n + 2 B n 1 B n + 1 2 , if   n     1 , k = n 1 B k B k + 3 1 = B n B n + 3 B n 1 B n + 2 6 , if   n     1 ,
etc.,
Setting G n = H n = S ( 1 , 3 , 6 , 1 ) in Theorem 1, we obtain Corollary 2 below.
Corollary 2.
For Lucas-balancing numbers C n = S ( 0 , 1 , 6 , 1 ) , there exists a positive integer N such that
k = n 1 C k C k + m 1 = C n C n + m C n 1 C n + m 1 + g m , i f   n N ,
where
g m = ( 8 C m + 1 2 + 72 ) α 4 ( 16 C m C m + 1 + 48 ) α 3 + ( 8 C m 2 + 8 ) α 2 3 C m + 1 α 6 ( 3 C m + C m + 1 ) α 5 + α 4 3 C m + 1 α 2 + ( 3 C m + C m + 1 ) α C m ,
with α = 3 + 2 2 .
For Lucas-balancing numbers, g 0 = 0 and g 1 = 1 and we obtain (10) and (13) from (19). In addition we have
k = n 1 C k C k + 2 1 = C n C n + 2 C n 1 C n + 1 + 8 , if   n     1 , k = n 1 C k C k + 3 1 = C n C n + 3 C n 1 C n + 2 + 46 , if   n     1 ,
etc.
Setting G n = S ( 0 , 1 , 6 , 1 ) and H n = S ( 1 , 3 , 6 , 1 ) in Theorem 1, we obtain Corollary 3 below.
Corollary 3.
For balancing numbers B n = S ( 0 , 1 , 6 , 1 ) and Lucas-balancing numbers C n = S ( 0 , 1 , 6 , 1 ) , there exists a positive integer N such that
k = n 1 B k C k + m 1 = B n C n + m B n 1 C n + m 1 + g m , i f   n N ,
where
g m = ( 8 C m + 1 2 ) α 3 + 2 C m C m + 1 α 2 C m 2 α C m + 1 α 5 C m α 4 C m + 1 α + C m ,
with α = 3 + 2 2 .
From (20), we have
k = n 1 B k C k 1 = B n C n B n 1 C n 1 1 , if   n     1 , k = n 1 B k C k + 1 1 = B n C n + 1 B n 1 C n 1 , if   n     1 , k = n 1 B k C k + 2 1 = B n C n + 2 B n 1 C n + 1 3 , if   n     1 ,
etc.
Setting G n = S ( 1 , 3 , 6 , 1 ) and H n = S ( 0 , 1 , 6 , 1 ) in Theorem 1, we obtain Corollary 4 below.
Corollary 4.
For balancing numbers B n = S ( 0 , 1 , 6 , 1 ) and Lucas-balancing numbers C n = S ( 0 , 1 , 6 , 1 ) , there exists a positive integer N such that
k = n 1 B k + m C k 1 = B n + m C n B n + m 1 C n 1 + g m , i f   n N ,
where
g m = ( 8 B m + 1 2 9 ) α 4 ( 16 B m B m + 1 6 ) α 3 + ( 8 B m 2 1 ) α 2 3 B m + 1 α 6 ( 3 B m + B m + 1 ) α 5 3 B m + 1 α 2 + ( 3 B m + B m + 1 ) α B m ,
with α = 3 + 2 2 .
From (21), we have
k = n 1 B k + 1 C k 1 = B n + 1 C n B n C n 1 , if   n     1 , k = n 1 B k + 2 C k 1 = B n + 2 C n B n + 1 C n 1 + 2 , if   n     1 ,
etc.
We can obtain similar results for the even-indexed and odd-indexed numbers of G n = S ( G 0 , G 1 , a , 1 ) and H n = S ( H 0 , H 1 , a , 1 ) . It is easily seen that
G 2 n = ( a 2 2 ) G 2 n 2 G 2 n 4 , G 2 n + 1 = ( a 2 2 ) G 2 n 1 G 2 n 3 .
Let G n e = G 2 n and G n o = G 2 n + 1 . Then
G n e = S ( G 0 e , G 1 e , a 2 2 , 1 ) , G n o = S ( G 0 o , G 1 o , a 2 2 , 1 ) ,
where G 0 e = G 0 , G 1 e = a G 1 G 0 , G 0 o = G 1 and G 1 o = ( a 2 1 ) G 1 a G 0 . Similarly
H n e = S ( H 0 e , H 1 e , a 2 2 , 1 ) , H n o = S ( H 0 o , H 1 o , a 2 2 , 1 ) ,
where H 0 e = H 0 , H 1 e = a H 1 H 0 , H 0 o = H 1 and H 1 o = ( a 2 1 ) H 1 a H 0 .
As before, for U n { G n e , G n o , H n e , H n o } and V n { G n e , G n o , H n e , H n o } , let
Φ U : = ( a 2 2 ) U 0 U 1 U 0 2 U 1 2 , Φ V : = ( a 2 2 ) V 0 V 1 V 0 2 V 1 2 , Δ ^ n : = U n 1 U n + 1 V n + m 1 V n + m + 1 U n 2 V n + m 2 U n + 1 V n + m + 1 U n 1 V n + m = 1 , Δ ^ m : = lim n Δ ^ n .
Then
Δ ^ m = ( Φ V U 1 2 + Φ U V m + 1 2 ) α ^ 4 2 ( Φ V U 0 U 1 + Φ U V m V m + 1 ) α ^ 3 + ( Φ V U 0 2 + Φ U V m 2 ) α ^ 2 U 1 V m + 1 α ^ 6 ( U 1 V m + U 0 V m + 1 ) α ^ 5 + U 0 V 0 α ^ 4 U 1 V m + 1 α ^ 2 + ( U 1 V m + U 0 V m + 1 ) α ^ U 0 V m ,
where
α ^ = a 2 2 + ( a 2 2 ) 2 4 2 ,
and we obtain the following results.
Theorem 2.
For G n = S ( G 0 , G 1 , a , 1 ) and H n = S ( H 0 , H 1 , a , 1 ) , let U n { G n e , G n o , H n e , H n o } and V n { G n e , G n o , H n e , H n o } . Then, for each pair ( U n , V n ) , there exists a positive integer N such that
k = n 1 U k V k + m 1 = U n V n + m U n 1 V n + m 1 + g ^ m , i f   n N ,
where
g ^ m = Δ ^ m .
For balancing numbers B n = S ( 0 , 1 , 6 , 1 ) , setting U n = V n = B n e in Theorem 2, we obtain Corollary 5 below.
Corollary 5.
For balancing numbers B n = S ( 0 , 1 , 6 , 1 ) , there exists a positive integer N such that
k = n 1 B 2 k B 2 k + 2 m 1 = B 2 n B 2 n + 2 m B 2 n 2 B 2 n + 2 m 2 + g ^ m = B 2 n + m 2 B 2 n + m 2 2 + g ^ m , i f   n N ,
where
g ^ m = ( 6 B 2 m + 2 2 + 216 ) α ^ 3 + 12 B 2 m B 2 m + 2 α ^ 2 6 B 2 m 2 α ^ B 2 m + 2 α ^ 5 B 2 m α ^ 4 B 2 m + 2 α ^ + B 2 m ,
with α ^ = 17 + 12 2 .
For balancing numbers, g ^ 0 = 1 and g ^ 1 = 2 , and (6) and (8) are obtained from (23). In addition we have
k = n 1 B 2 k B 2 k + 4 1 = B 2 n + 2 2 B 2 n 2 37 , if   n     1 , k = n 1 B 2 k B 2 k + 6 1 = B 2 n + 3 2 B 2 n + 1 2 1223 , if   n     1 ,
etc.
For Lucas-balancing numbers C n = S ( 1 , 3 , 6 , 1 ) , setting U n = V n = C n e in Theorem 2, we obtain Corollary 6 below.
Corollary 6.
For Lucas-balancing numbers C n = S ( 1 , 3 , 6 , 1 ) , there exists a positive integer N such that
k = n 1 C 2 k C 2 k + 2 m 1 = C 2 n C 2 n + 2 m C 2 n 2 C 2 n + 2 m 2 + g ^ m = C 2 n + m 2 C 2 n + m 2 2 + g ^ m , i f   n N ,
where
g ^ m = ( 288 C 2 m + 2 2 + 83232 ) α ^ 4 ( 576 C 2 m C 2 m + 2 + 9792 ) α ^ 3 + ( 288 C 2 m 2 + 288 ) α ^ 2 17 C 2 m + 2 α ^ 6 ( 17 C 2 m + C 2 m + 2 ) α ^ 5 + α ^ 4 17 C 2 m + 2 α ^ 2 + ( 17 C 2 m + C 2 m + 2 ) α ^ C 2 m ,
with α ^ = 17 + 12 2 .
For Lucas-balancing numbers, g ^ 0 = 0 and g ^ 1 = 8 , and (12) and (14) are obtained from (24). In addition we have
k = n 1 C 2 k C 2 k + 4 1 = C 2 n + 2 2 C 2 n 2 + 288 , if   n     1 , k = n 1 C 2 k C 2 k + 6 1 = C 2 n + 3 2 C 2 n + 1 2 + 9783 , if   n     1 ,
etc.
For U n { G n e , G n o , H n e , H n o } and V n { G n e , G n o , H n e , H n o } , we have sixteen pairs of ( U n , V n ) , and we can obtain more identities from Theorem 2. Other identities are left to the interested readers.

3. Discussion

In this paper, we derived general identities related to reciprocal sums of products of two balancing numbers, products of two Lucas-balancing numbers and products of balancing and Lucas-balancing numbers. Repeatedly applying Theorem 2, we can obtain similar results for ( B 4 n , C 4 n ) , ( B 8 n , C 8 n ) , etc.
The method of this paper can also be applied to even-indexed and odd-indexed numbers of G n = S ( G 0 , G 1 , a , 1 ) . In fact, for the numbers of the form G n = S ( G 0 , G 1 , a , 1 ) , we have
G 2 n = ( a 2 + 2 ) G 2 n 2 G 2 n 4 , G 2 n + 1 = ( a 2 + 2 ) G 2 n 1 G 2 n 3 .
Hence Theorem 2 can be used to obtain various identities for even-indexed and odd-indexed Fibonacci, Lucas, Pell and Pell–Lucas numbers.

Funding

This research received no external funding.

Conflicts of Interest

The author declares no conflict of interest.

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Choo, Y. On the Reciprocal Sums of Products of Balancing and Lucas-Balancing Numbers. Mathematics 2021, 9, 350. https://doi.org/10.3390/math9040350

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Choo Y. On the Reciprocal Sums of Products of Balancing and Lucas-Balancing Numbers. Mathematics. 2021; 9(4):350. https://doi.org/10.3390/math9040350

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Choo, Younseok. 2021. "On the Reciprocal Sums of Products of Balancing and Lucas-Balancing Numbers" Mathematics 9, no. 4: 350. https://doi.org/10.3390/math9040350

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