INCOMPLETE BALANCING AND LUCAS-BALANCING NUMBERS

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1 INCOMPLETE BALANCING AND LUCAS-BALANCING NUMBERS BIJAN KUMAR PATEL, NURETTIN IRMAK and PRASANTA KUMAR RAY Communicated by Alexandru Zaharescu The aim of this article is to establish some combinatorial expressions of balancing and Lucas-balancing numbers and investigate some of their properties. AMS 010 Subect Classification: 11B37, 11B39. Key words: balancing numbers, Lucas-balancing numbers, incomplete balancing numbers, incomplete Lucas-balancing numbers. 1. INTRODUCTION The terms balancing numbers and Lucas-balancing numbers are used to describe the series of numbers generated by the recursive formulas B n = 6B n 1 B n ; B 0 = 0, B 1 = 1 with n and C n = 6C n 1 C n ; C 0 = 1, C 1 = 3, with n respectively [1,10]. The roots λ 1 = 3+ 8 and λ = 3 8 for both these sequences, the respective Binet formulas are B n = λn 1 λn λ 1 λ and C n = λ 1+λ [1, 10]. Many interesting results of balancing numbers and their related sequences can be found in [5, 10 1]. In [4], Filipponi established two interesting classes of integers namely, incomplete Fibonacci numbers and incomplete Lucas numbers which were obtained from some of the well-known combinatorial forms of Fibonacci and Lucas numbers. He has also studied some of the congruence properties for incomplete Lucas numbers in [4]. Filipponi dreamt a glimpse of possible generalizations of incomplete Fibonacci and Lucas numbers which were fulfilled by some authors later [, 7 9, 13]. In this article we establish some combinatorial expressions for balancing and Lucas-balancing numbers and introduce incomplete balancing and incomplete Lucas-balancing numbers. Balancing sequence has been generalized in many ways. One important generalization of balancing numbers is the k-balancing numbers introduced in [6, 1]. k-balancing numbers are defined recursively by B k,n+1 = 6kB k,n B k,n 1 n, MATH. REPORTS 0(70, 1 (018, 59 7

2 60 Bian Kumar Patel, Nurettin Irmak and Prasanta Kumar Ray with initials B k,0 = 0 and B k,1 = 1. In [1], Ray has introduced sequence of balancing polynomials {B n (x} n=0that are natural extension of k-balancing numbers and is defined recursively by 1, if n = 0 B n (x = 6x, if n = 1 6xB n 1 (x B n (x, if n > 1. Further, he has also established the Binet formula as where B n (x = λn 1 (x λn (x, λ 1 (x = 3x +, λ (x = 3x and = 9x 1. The first few balancing polynomials are B (x = 6x, B 3 (x = 36x 1, B 4 (x = 16x 3 1x, B 5 (x = 196x 4 108x +1. The derivatives of balancing polynomials in the form of convolution of these polynomials are also presented in [1]. In a similar manner, the n th Lucas-balancing polynomial C n (x is defined as C n (x = and its Binet formula is 1, if n = 0 3x, if n = 1 6xC n 1 (x C n (x, if n > 1, C n (x = λn 1 (x + λn (x. The first few Lucas-balancing polynomials are C (x = 18x 1, C 3 (x = 108x 3 9x, C 4 (x = 648x 4 7x + 1, C 5 (x = 3888x 5 540x x. In this article, authors aim is to establish some combinatorial expressions of balancing and Lucas-balancing numbers and investigate some of their properties.. SOME COMBINATORIAL EXPRESSIONS OF BALANCING AND LUCAS BALANCING NUMBERS In this section, we establish some combinatorial expressions for balancing and Lucas-balancing numbers. These expressions lead to form some combinatorial expressions for balancing and Lucas-balancing polynomials.

3 3 Incomplete balancing and Lucas-balancing numbers 61 Theorem.1. Let B n and C n denote n th balancing and Lucas-balancing numbers respectively, then (.1 (. B n = (n 1/ n/ C n = 3 ( 1 where. denotes the floor function. ( 1 ( n 1 n n 6 n 1, ( n 6 n 1, Proof. From the Binet formulas for both balancing and Lucas-balancing numbers and by well-known Waring formulas [3], B n = λn 1 λn (n 1/ = λ 1 λ C n = λ n 1 + λ n = n/ ( 1 n n ( 1 ( n 1 and the result follows as λ 1 λ = 1 and λ 1 + λ = 6. (λ 1 λ (λ 1 + λ n 1, ( n (λ 1 λ (λ 1 + λ n, Indeed, the combinatorial expression for polynomial B n (x leads to B n (x = (n 1/ ( 1 ( n 1 (6x n 1 for n 1, which is shown in [1]. Similarly the combinatorial expression for polynomial C n (x will be n/ C n (x = 3 ( 1 n n ( n 6 n 1 x n for n 1. Further, their derivatives are respectively given by (n 1/ (.3 B n(x= ( 1 (n 1 n/ (.4 C n(x n(n = 3 ( 1 n ( n 1 6 n 1 x n for n 1, ( n (6x n 1 for n 1.

4 6 Bian Kumar Patel, Nurettin Irmak and Prasanta Kumar Ray 4 Clearly, B 0 (x = C 0 (x = 0. Some simple properties of the polynomials B n(x and C n(x can be derived from the Binet formulas. In fact, letting λ 1(x = d dx (λ 1(x = 3λ 1(x, λ (x = d dx (λ (x = 3λ (x, which follows (λ n 1 (x = d dx (λn 1 (x = 3nλn 1 (x, (λn (x = d dx (λn (x = 3nλn (x, we can write B n(x = d ( λ n 1 (x λ n (x = 3nC n(x 9xB n (x (.5 dx, (.6 C n(x = d dx (λn 1 (x + λ n (x = 3nB n (x. 3. INCOMPLETE BALANCING AND LUCAS-BALANCING NUMBERS The combinatorial expressions (.1 and (. give rise to two interesting classes of integers B n (k and C n (k, for integral values n and k. We call these integers as incomplete balancing numbers and incomplete Lucasbalancing numbers respectively which will be defined subsequently. In this section, authors aim to establish certain properties of incomplete balancing and incomplete Lucas-balancing numbers. Definition 3.1. The incomplete balancing numbers B n (k be defined as for any natural number n, k ( B n (k = ( 1 n 1 (3.7 6 n 1, 0 k ñ, where ñ = n 1. The numbers B n (k are shown in the Table 1 for the first few values of n and the corresponding admissible values of k. Observation of Table 1 gives rise to, We define the incomplete Lucas-balancing numbers in the following manner: B n (0 = 6 n 1 for n 1, B n (ñ = B n for n 1, { Bn ± 3n, if n is even B n (ñ 1 = B n ± 1, if n is odd for n 3.

5 5 Incomplete balancing and Lucas-balancing numbers 63 TABLE 1 Incomplete balancing numbers n/k Definition 3.. The incomplete Lucas-balancing numbers C n (k be defined as for any natural number n, k ( C n (k = 3 ( 1 n n (3.8 6 n 1, 0 k ˆn, n where ˆn = n. The numbers C n (k are shown in Table for the first few values of n and the corresponding admissible values of k. TABLE Incomplete Lucas-balancing numbers n/k Observation of Table gives rise to, C n (0 = 3.6 n 1 for n 1, C n (ˆn = C n for n 1, { Cn ± 1, if n is even C n (ˆn 1 = C n ± n, if n is odd for n 3.

6 64 Bian Kumar Patel, Nurettin Irmak and Prasanta Kumar Ray SOME IDENTITIES CONCERNING THE INCOMPLETE BALANCING NUMBERS B n (k Like balancing and Lucas-balancing numbers, incomplete balancing numbers also satisfy the second order recurrence relation. The following result demonstrates this fact. Proposition 3.3. The numbers B n (k obey the second order recurrence relation (3.9 B n+ (k + 1 = 6B n+1 (k + 1 B n (k for 0 k (n /. Proof. By virtue of the Definition 3.7, k+1 6B n+1 (k+1 B n (k = 6 and the result follows. k+1 = 6 ( 1 ( n ( 1 [( n 6 n + k ( n 1 k+1 ( = ( 1 n n+1, ( 1 ( n 1 6 n 1 ( ] n + 6 n 6 n 1 Observe that, the relation (3.9 can be transformed into the non-homogeneous recurrence relation ( (3.10 B n+ (k = 6B n+1 (k B n (k ( 1 k+1 6 n 1 k n 1 k. k The relation (3.10 can be generalized as follows. Proposition 3.4. Let 0 k n h 1. Then the identity h ( (3.11 ( h B n+ (k + = ( 1 h+1 B n+h (k + h holds. Proof. Using induction on h, the result (3.11 clearly holds for h = 0 and h = 1. Assume that the result holds for some h > 1. For the inductive step, we have h+1 ( ( h + 1 B n+ (k +

7 7 Incomplete balancing and Lucas-balancing numbers 65 h+1 [( ( ] = ( h h + B 1 n+ (k + = ( 1 h 1 B n+h (k + h + = ( 1 h 1 B n+h (k + h + = ( 1 h 1 B n+h (k + h 6 h+1 ( ( 1 i+ 6 i+1 h B i i+n+1 (k i i= 1 h ( ( 1 i+ 6 i+1 h B i i+n+1 (k i i=0 h ( ( 1 i+1 6 i h B i i+n+1 (k i i=0 = ( 1 h 1 B n+h (k + h 6( 1 h 1 B n+h+1 (k h = ( 1 h 1 [B n+h (k + h 6B n+h+1 (k h] = ( 1 h B n+(h+1 (k h, which completes the proof. Using induction on m, the following identity can be proved analogously. h m ( ( m h (3.1 B m + n+ ( = ( 1 h+1 B n+h m (h m, where h m and n h m + 1. Notice that, the identity obtained from (3.1 by setting m = 0 is identical to the identity obtained from (3.11 for k = 0. The following is an interesting relation concerning incomplete balancing numbers. The sum of all elements of the n th row of the array of Table 1 is expressed in terms of balancing and Lucas-balancing numbers. Proposition 3.5. For incomplete balancing numbers B n (k, the following identity holds. ñ { (3nCn B B n (k = n /16, if n is even (3nC n + 7B n /16 if n is odd. k=0 Proof. Recall that ñ = n 1. Then, B n (0 + B n (1 + + B n (ñ ( n 1 0 = 6 n [ ( ( 1 0 n [ ( ( 1 0 n n ( 1ñ 0 ( 6 n 1 0 +( 1 1 n ( n 1 ñ 6 n 1 ñ] ñ 6 n 1 ]

8 66 Bian Kumar Patel, Nurettin Irmak and Prasanta Kumar Ray 8 ( n 1 0 = (ñ ( n 1 ñ + + ( 1ñ ñ ñ ( = (ñ + 1 ( 1 n 1 = (ñ + 1 ñ = (ñ + 1B n ( 6 n (ñ + 1 1( 1 1 n ( 1 ( n 1 6 n 1 ñ 6 n 1 6 n 1 ñ ( ( 1 n 1 6 n 1. 6 n 1 ñ ( ( 1 n 1 6 n 1 The second term of the right hand side expression of the above equation ñ leads to Bn(8n+1 3nCn 16 for x = 1 in (.3 and (.5. For n is even, B n (k is nb n Bn(8n+1 3nCn 16 and for n is odd, This completes the proof. ñ k=0 k=0 B n (k is (n+1bn Bn(8n+1 3nCn 16. In [4], Filipponi has shown that for n = m, where m is a non-negative integer, the incomplete Fibonacci numbers F n (k is odd for all admissible values of k. The following result shows that for n = m, the incomplete balancing numbers B n (k is even for all admissible values of k. Proposition 3.6. If n = m for any non-negative integer m, then B n (k is even for all admissible values of k. Proof. We prove this result by induction on m. The basic step is true for m = 1. In the inductive step, assume that the result is valid for all m n. Now, using the (3.9; we have B m+ (k + 1 = 6B m+1 (k + 1 B m (k. Clearly the first term is divisible by and by the hypothesis the second term is also even and hence the result follows. Filipponi has also shown that for any prime p, the incomplete Lucas numbers satisfy the identity L p (k 1 (mod p for all admissible values of k in [4]. Whereas no such type of identity exists in incomplete Fibonacci numbers. To demonstrate this fact, consider the following example.

9 9 Incomplete balancing and Lucas-balancing numbers 67 Example 3.7. Let p = 7. Consider k = 1,, 3 and observe that F 7 (1 = 6 1 (mod 7, F 7 ( = 1 (mod 7 and F 7 (3 = 13 1 (mod 7, respectively. However, the similar type of identities do exist in both incomplete balancing and Lucas-balancing numbers. The following result demonstrates this fact. Proposition 3.8. If n = m for any non-negative integer m, then for all admissible values of k, B n (k 0 (mod n. Proof. By virtue of Definition (3.7, (3.13 B n (k = 6 m 1 + k =1 ( 1 ( m 1 6 m 1. The first term on the right hand side (3.13 is an integer when B m(k is divisible by m for all m 0. The second term is also an integer due to Filipponi [3], and thus the congruence follows. 3.. SOME IDENTITIES INVOLVING THE INCOMPLETE LUCAS-BALANCING NUMBERS C n (k Balancing numbers and Lucas-balancing numbers are related by an identity B n+1 B n 1 = C n. Similar properties are valid for incomplete balancing and Lucas-balancing numbers. Proposition 3.9. The identity (3.14 B n+1 (k B n 1 (k 1 = C n (k, holds for 0 k ˆn, where ˆn = n. Proof. By virtue of Definition 3.7, k ( B n+1 (k B n 1 (k 1 = ( 1 n k 1 6 n = = k k ( 1 ( n ( 1 ( n n + k =1 ( 1 ( n ( 1 1 ( n 1 1 ( n 6 n 6 n 6 n

10 68 Bian Kumar Patel, Nurettin Irmak and Prasanta Kumar Ray 10 which ends the proof. = 3 k ( 1 n n ( n n 1 Proposition The numbers C n (k obey the second order recurrence relation (3.15 C n+ (k + 1 = 6C n+1 (k + 1 C n (k. Proof. Using (3.14 and (3.9, C n+ (k + 1 = 1 {B n+3(k + 1 B n+1 (k} and the result follows. = 1 [{6B n+(k + 1 B n+1 (k} {6B n (k B n 1 (k 1}] = 3(B n+ (k + 1 B n (k + 1 (B n 1(k 1 B n+1 (k = 6C n+1 (k + 1 C n (k, Notice that, the relation (3.15 can be transformed into the non-homogeneous recurrence relation (3.16 ( C n+ (k = 6C n+1 (k C n (k + ( 1 k+ 3n n k 6 n k 1 for n k. n k k Proposition For 0 k ñ, then the following identity holds. ( C n (k = B n+ (k B n (k. Proof. In view of (3.14 and (3.15, from which the result follows. B n+ (k = C n+1 (k + B n (k 1, B n (k = B n (k 1 C n 1 (k 1, Using the relations (3.11 and (3.14, the identity (3.15 can be generalized as follows. holds. Proposition 3.1. For 0 k n h 1, then h ( ( h C n+ (k + = ( 1 h+1 C n+h (k + h,

11 11 Incomplete balancing and Lucas-balancing numbers 69 The following is an interesting relation concerning incomplete Lucasbalancing numbers. The sum of all elements of the n th row of the array of Table is expressed in terms of balancing and Lucas-balancing numbers. Proposition For incomplete Lucas-balancing numbers C n (k, ˆn k=0 C n (k = { Cn + n(3b n C n /, if n is even {C n + n(3b n C n }/, if n is odd. Proof. Recall that ˆn = n. Therefore, we have C n (0 + C n (1 + + C n (ˆn = (ˆn + 1C n 3 ˆn ( 1 n n ( n 6 n 1. Putting x = 1 in (.4 and (.6 and adopting the same procedure described earlier, we obtain B n = ˆn ( 1 n n ( n 6 n 1. Therefore, 3 ˆn ( 1 n n ( n 6 n 1 = 3n(C n B n /. For n is even and odd, ˆn k=0 C n (k is ( n + 1C n 3n(C n B n / and ( n+1 C n 3n(C n B n / respectively. Therefore, the proof is completed. Proposition If n = 3 m for any non-negative integer m, then for all admissible values of k, C n (k 0 (mod n. Proof. By virtue of Definition 3.16, k ( (3.18 C n (k = 3m 1 3 3m + 3 m+1 ( m 3 m 6 3m 1. =1 The first term on the right hand side of (3.18 is an integer when C 3 m(k is divisible by 3 m for all m 0. The second term is also an integer again due to Filipponi [4] and thus the congruence follows.

12 70 Bian Kumar Patel, Nurettin Irmak and Prasanta Kumar Ray 1 4. GENERATING FUNCTIONS OF THE INCOMPLETE BALANCING AND LUCAS-BALANCING NUMBERS The generating functions for balancing and Lucas-balancing numbers play a vital role to find out many important identities for these numbers. In this section, we derive generating functions for both incomplete balancing and Lucas-balancing numbers. The following result [7] is useful to prove the subsequent theorem. Lemma 4.1. Let {s n } n N be a complex sequence satisfying the following non-homogeneous second-order recurrence relation: s n = as n 1 + bs n + r n, n > 1, where a, b C and r n : N C is a given sequence. Then the generating function U(t of s n is U(t = G(t + s 0 r 0 + (s 1 s 0 a r 1 t 1 at bt, where G(t denotes generating function of r n. The following are the generating functions for incomplete balancing and incomplete Lucas-balancing numbers. Theorem 4.. Let k be a fixed positive integer. Then B k (t = t k+1 (1 6tk+1 (B k+1 B k t ( 1 k+1 t (4.19 (1 6t + t (1 6t k+1, (4.0 C k (t = t k (1 6tk+1 (C k C k 1 t ( 1 k+1 t (1 6t + t (1 6t k+1. Proof. By virtue of the identity (3.7, for 0 n < k + 1, B n (k = 0 and for other values of n and r 0 integers, then B k+1+r (k = B k+r. It follows from (3.10 that, for n k + 3, B n (k = 6B n 1 (k B n (k ( 1 k+1 ( n 3 k n 3 k 6 n 3 k. Letting s 0 = B k+1 (k, s 1 = B k+ (k, (..., s n = B n+k+1 and suppose that n + k r 0 = r 1 = 0 and r n = ( 1 k+1 6 n. It is easy to deduce the n generating function of the sequence {r n } as G(t = ( 1k+1 t. Therefore by (1 6t k+1 Lemma 4.1, the generating function of the incomplete balancing numbers satisfies the following: U k (t(1 6t + t + ( 1k+1 t (1 6t k+1 = B k+1 + (B k+ 6B k+1 t,

13 13 Incomplete balancing and Lucas-balancing numbers 71 where U k (t is the generating function of the sequence {s n }. It follows that B k (t = t k+1 U k (t. In the proof of the identity (4.0, we use the following facts: For 0 n < k, C n (k = 0, and for other values of n and r 0 integers then, C k+1+r (k = C k+r. It follows from (3.16 that, for n k +, C n (k = 6C n 1 (k C n (k + ( 1 k+ 3(n (n k ( n k n k 6 n k 3. Letting s 0 = C k (k, s 1 = C k+1 (k,..., s n = C n+k and suppose that r 0 = r 1 = 0 and ( r n = 3.6 n 3 k+ n + k n + k ( 1. n + k n The generating function of the sequence {r n } is G 1 (t = ( 1k+ 3t ( t. Again (1 6t k+1 from Lemma 4.1, the generating function of the incomplete Lucas-balancing numbers satisfies the equation: U k (t(1 6t + t + ( 1k+ 3t ( t (1 6t k+1 = C k + (C k+1 6C k t. Finally, we conclude that C k (t = t k U k (t. Acknowledgments. The authors would like to express their gratitude to the anonymous referee for his valuable comments and suggestions which improved the presentation of the article to a great extent. REFERENCES [1] A. Behera and G.K. Panda, On the square roots of triangular numbers. Fibonacci Quart. 37 (1999,, [] G.B. Dordevic and H.M. Srivastava, Incomplete generalized Jacobsthal and Jacobsthal- Lucas numbers. Math. Comput. Modelling 4 (005, [3] P. Filipponi, Waring s formula, the binomial formula, and generalized Fibonacci matrices. Fibonacci Quart. 30 (199, 3, [4] P. Filipponi, Incomplete Fibonacci and Lucas numbers. Rend. Circ. Mat. Palermo. 45 (1996,, [5] R. Keskin and O. Karaath, Some new properties of balancing numbers and square triangular numbers. J. Integer Seq. 15 (01, Article [6] A. Özkoc, Tridigonal Matrices via k-balancing Number. British J. of Math. Comput. Sci. 10 (015, 4, [7] Á. Pinter and H.M. Srivastava, Generating functions of the incomplete Fibonacci and Lucas numbers. Rend. Circ. Mat. Palermo ( 48 (1999,,

14 7 Bian Kumar Patel, Nurettin Irmak and Prasanta Kumar Ray 14 [8] J. Ramirez, Bi-periodic incomplete Fibonacci sequence. Ann. Math. Inform. 4 (013, [9] J. Ramirez, Incomplete generalized Fibonacci and Lucas polynomials. arxiv: , 013. [10] P.K. Ray, Balancing and cobalancing numbers. Ph.D. thesis, Department of Mathematics, National Institute of Technology, Rourkela, India, 009. [11] P.K. Ray, Some congruences for balancing and Lucas balancing numbers and their applications. Integers 14 (014, #A8. [1] P.K. Ray, On the properties of k-balancing numbers. Ain Shams Eng. J., 016. In press. [13] D. Tasci and M. Cetin Firengiz, Incomplete Fibonacci and Lucas p-numbers. Math. Comput. Modelling 5(9-10 (010, Received 3 February 016 IIIT Bhubaneswar, Department of Mathematics, Bhubaneswar iiit.bian@gmail.com Niǧde University, Department of Mathematics, Niǧde 5140, Turkey nirmak@nigde.edu.tr Sambalpur University, School of Mathematical Sciences, Sambalpur , India prasantamath@suniv.ac.in

RECIPROCAL SUMS OF SEQUENCES INVOLVING BALANCING AND LUCAS-BALANCING NUMBERS

RECIPROCAL SUMS OF SEQUENCES INVOLVING BALANCING AND LUCAS-BALANCING NUMBERS GOPAL KRISHNA PANDA, TAKAO KOMATSU and RAVI KUMAR DAVALA Communicated by Alexandru Zaharescu Many authors studied bounds for