RECIPROCAL SUMS OF SEQUENCES INVOLVING BALANCING AND LUCAS-BALANCING NUMBERS

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1 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 reciprocal sums involving terms from Fibonacci and other related sequences The purpose of this paper is to provide bounds for reciprocal sums with terms from balancing and Lucas-balancing sequences AMS 00 Subject Classification: B39, B83 Key words: balancing and Lucas balancing numbers, recurrence relation, Binet form INTRODUCTION A positive integer x is called balancing number if x ) x ) y ) holds for some integer y x The problem of determining all balancing numbers leads to the Pell equation y 8x, whose solutions in x can be described by the recurrence B n 6B n B n n ) with B 0 0 and B see eg [,, ]) Balancing numbers have been extensively studied by many authors Karaatli et al [] expressed the positive integral solutions of a Diophantine equation in terms of balancing numbers Liptai [8] proved that there is no Fibonacci balancing number except In [3], the period rank and order of the sequence of balancing numbers are studied One of the most general extensions of the defining equation of balancing numbers is the Diophantine equation k k x ) k x ) l y ) l, where the exponents k and l are given positive integers In [9] effective and non-effective finiteness theorems for the above equation are proved In [6] a balancing problem of ordinary binomial coefficients is studied, and effective and non-effective finiteness theorems are given The numbers C n 8B n is called the n th Lucas-balancing number [], and these numbers satisfy the recurrence relation C n 6C n C n with MATH REPORTS 070), 08), 0 4

2 0 Gopal Krishna Panda, Takao Komatsu and Ravi Kumar Davala initial values C 0, C 3 The Binet forms for B n and C n are respectively B n αn β n 4, C n αn β n where α 3 and β 3 The balancing and Lucas-balancing numbers satisfy the identities B n r B nr B n B r, C n r C nr C n C r for n r respectively In particular, for n we have The identity gives B n B n B n and C n C n C n 8 B B 3 B n B n B n B n B n The proofs of the above identities are available in [4] In the subsequent sections, we shall use the above identities without further reference The intention of this paper is to develop certain interesting bounds for reciprocal sums with terms involving balancing and Lucas-balancing numbers in some combinations The reciprocal of partial infinite sums of reciprocal Fibonacci numbers has been extensively studied by many authors eg, see [4, 5, 7, 0, 5]) In [3], the following identities are shown for generalized Fibonacci numbers G n, defined by G n ag n G n n ), G 0 0, G, where a is a positive integer If a, then G n s are equal to the Fibonacci numbers Throughout this paper, integer part of a number is denoted by ) ) 3) 4) Proposition { G n G n if n is even and n ; G k G n G n if n is odd and n { ag n G n if n is even and n ; G k ag n G n if n is odd and n G n G n n ) G k G k ) G n G n 3 n )

3 3 Reciprocal sums of sequences involving balancing and Lucas-balancing numbers 03 5) G 4n G 4n 3 n ) G k G k 6) G 4n G 4n n ) G k G k 7) G G 4n G 4n 3 n ) k 8) G G 4n 3 G 4n 5 n ) k 9) G 4n G 4n 4 n ) G k G k In this paper, we shall show some analogous results for the sequences of balancing and Luca-balancing numbers ) ) 3) 4) 5) 6) Proposition B n B n n ) B k B k ) B n B n B n n ) B n B n B n B n n ) B k B k C n C n n ) C k C k ) C n C n n ) C n C n C n C n n ) C k C k

4 04 Gopal Krishna Panda, Takao Komatsu and Ravi Kumar Davala 4 RECIPROCAL SUMS INVOLVING BALANCING NUMBERS In this section, we establish bounds for several reciprocal sums involving balancing numbers By using the bounds, we obtain the results in Proposition ), ) and 3) The following theorem gives sharp bounds for reciprocal sums of balancing numbers Thus, Theorem For any positive integer n, B n B n B k B n B n Proof For any positive integer n, B n B n B n B n B n B n B n B n B n B n B n ) B n B n B n B n B n B n B n Repeating the above steps, we can show that B n B n B n B n B n Combining the above two inequalities, we get B n B n B n B n B n B n Continuing in this manner, one can arrive at the inequality ) B n B n B k On the other hand, B n B n B n B n B n B n B n B n B n B n B n B n ) B n B n B n B n B n ) B n B n B n B n B n ) B n

5 5 Reciprocal sums of sequences involving balancing and Lucas-balancing numbers 05 which gives B n B n, B n B n B n B n B n Continuing in a similar fashion, we finally obtain ) B n B n B k Combining ) and ), we get the desired inequality as stated in the theorem The following theorem provides sharp bounds for the reciprocal sum of squares of balancing numbers Theorem For any positive integer n, B n B n Proof Since for each n, we have B k B n B n B n B n B n B n 0, Bn B n B n B n Thus, for each n, B n Bn B n B n B n which yields B n BnB n Bn ) Bn B n ) Bn Bn B n ) Bn Bn, B n B n B n Bn B n

6 06 Gopal Krishna Panda, Takao Komatsu and Ravi Kumar Davala 6 This inequality is also valid for n Recursive iteration of the last inequality gives 3) B n On the other hand, B n Bn For n, Thus, we have B k Bn Bn Bn B n B nb n B n ) B n )B n ) B nb n B n ) B n B n )B n ) B nb n B n ) B nb n B n B 4 n B n B n ) 6B n B n ) B 4 n B n 34B n B n B n 0 Iterating recursively, we get 4) B n Bn B n B n B k Combining inequalities 3) and 4), we get what has been claimed The reciprocal sum of products of two consecutive balancing numbers has analogous bounds The following theorem is important in this regard Theorem 3 For any positive integer n, B n B n B n B n B k B k B n B n B n B n Proof Using the fact B n B n Bn B n B n B n B n B n B n B n B n n, we have B n B n B n ) Bn B n B n ) B n B n Bn

7 7 Reciprocal sums of sequences involving balancing and Lucas-balancing numbers 07 Thus, B n B n B n B n Iterating recursively, we get 5) On the other hand, Since B k B k B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n B n ) [ )] Bn B n B n B n B n B n B n B n B n B n B nb n B n B n we have B n )B n ) B nb n B nb n B nb n B n B n )B n B n ) B n B n B n B n B n B n The last inequality can be rearranged as 5B n B nb n B n B n B n )B n B n ) [ )] Bn B n B n B n B n B n 0, B n B n B n B n B n B n B n B n ) B n B n B n B n B n B n B n B n B n B n B n B n B n B n Repeated iteration gives 6) B k B k B n B n B n B n

8 08 Gopal Krishna Panda, Takao Komatsu and Ravi Kumar Davala 8 Combining the inequalities 5) and 6) together, we get B n B n B n B n This ends the proof B k B k B n B n B n B n The following theorem can be proved in a similar fashion However, the bounds are not as sharp as those in the previous theorems Theorem 4 For positive integers n and r 3, Bn r Bn r Bn r B n ) r B r k 3 RECIPROCAL SUMS INVOLVING LUCAS-BALANCING NUMBERS In this section, we shall establish certain bounds for the reciprocal sums involving Lucas-balancing numbers By using these bounds, we obtain the results in Proposition 4), 5) and 6) The following theorem provides sharp bounds for reciprocal sums of the Lucas-balancing numbers Theorem 3 For any positive integer n, C n C n Proof For any positive integer n, C n C n C n Thus, C n C n C n C n C k C n C n C n C n C n C n C n Repeating the above steps, we can obtain C n C n C n C n C n Combining the above two inequalities, we get C n C n C n C n ) 8 C n C n C n C n C n C n C n C n

9 9 Reciprocal sums of sequences involving balancing and Lucas-balancing numbers 09 Continuing in this manner, one can arrive at the inequality 3) C n C n C k On the other hand, since we have C n C n C n C n Cn C n C n C n C n C n C n C n ) C n 8 C n C n C n C n ) C n C n C n, C n C n C n C n C n Continuing in the same way, one can obtain 3) C n C n C k Combining inequalities 3) and 3), we get the desired inequality as stated in the theorem The following theorem provides sharp bounds for the reciprocal sum of squares of Lucas-balancing numbers we have Theorem 3 For any positive integer n, Proof For each n Cn Cn C k Cn Cn C n C n C n C n 8 C n C n 0, Cn C n C n C n Thus for each n, C n C n C n C n CnC n Cn ) ) Cn Cn Cn

10 0 Gopal Krishna Panda, Takao Komatsu and Ravi Kumar Davala 0 The above inequality yields C n C n and iterating recursively, we get 33) On the other hand, for n, Cn Cn Cn C n C n C n Cn Cn CnC n Cn C n C n Cn C n ) C n C n C k C n ) ) C n Cn If we set x n C n C n, then Hence, x n x n C n C n Cn Cn 7C n Cn 64 Cn )Cn 0 ) C Cn n Cn Thus, we have ) C Iterating recursively, we get 34) n C ) n Cn Cn CnC n C n ) Cn C n Cn Cn Cn Cn C n C n C n C k Combining inequalities 33) and 34), we get what has been claimed The reciprocal sum of products of two consecutive Lucas-balancing numbers has analogous bounds The following theorem is important in this regard

11 Reciprocal sums of sequences involving balancing and Lucas-balancing numbers Theorem 33 For any positive integer n, C n C n C n C n C k C k Proof Since for each positive integer n, C n C n C n C n C n C n )C n C n C n C n ) C n C n C n C n C n C n ) C n C n C n C n C nc n C n C n C n C n C n 8)C n 8) C nc n C n 6C n C n ) C n C n C n 8C n C n C n C n C n 63 0, we have C n C n C n C n C n C n C n C n C n C n C n C n C n C n ) C n C n C n C n Thus, C n C n C n C n C n C n C n C n C n C n Iterating recursively, we get 35) C k C k C n C n C n C n On the other hand, C n C n C n C n C n C n C n C n C n C n C n C n C n C n ) [ )] Cn C n C n C n C n C n Since C n C n C n C n C nc n C n C n we obtain C n 8)C n 8) C nc n C nc n C nc n C n C n )C n C n ) C n C n C n C n C n C n 4C n 4C nc n 8C n 64 C n C n )C n C n ) [ )] Cn C n C n C n C n C n 0,

12 Gopal Krishna Panda, Takao Komatsu and Ravi Kumar Davala The last inequality can be rearranged as C n C n C n C n C n C n C n C n ) C n C n C n C n C n C n C n C n C n C n C n C n C n C n, which on repeated iteration gives 36) C k C k C n C n C n C n Inequalities 35) and 36) combined together gives C n C n C n C n This ends the proof C k C k C n C n C n C n The following theorem can be proved in a similar fashion However, the bounds are not so sharp as compared to those in the previous theorems Theorem 34 For positive integers n and r 3 Cn r C n ) r C r k Cn r Cn r 4 ADDITIONAL RESULTS Using the techniques of last two sections, one can establish the following results for balancing and Lucas-balancing numbers ) ) 3) Proposition 4 B n B n n ) B k B k B k B k ) B n B n 3 n ) ) B n B n n )

13 3 Reciprocal sums of sequences involving balancing and Lucas-balancing numbers 3 4) Bn Bn n ) B k B k 5) B Bn Bn n ) k 6) B Bn Bn 3 n ) k ) ) 3) 4) 5) 6) Proposition 4 C n C n n ) C k C k C k C k C k C k C k C k ) C n C n 3 n ) ) C n C n 8 n ) ) C n C n 8 n ) ) C n C n n ) ) C n C n 3 n ) However, it seems difficult to get results involving higher power from Theorem 4 and Theorem 34 Acknowledgments The authors thank the anonymous referee for the careful reading of the manuscript REFERENCES [] A Behera and GK Panda, On the square roots of triangular numbers Fibonacci Quart ), [] R Finkelstein, The house problem Amer Math Monthly 7 965),

14 4 Gopal Krishna Panda, Takao Komatsu and Ravi Kumar Davala 4 [3] SH Holliday and T Komatsu, On the sum of reciprocal generalized Fibonacci numbers Integers 0), [4] T Komatsu and V Laohakosol, On the sum of reciprocals of numbers satisfying a recurrence relation of order s J Integer Seq 3 00), Article pages) [5] T Komatsu, On the nearest integer of the sum of reciprocal Fibonacci numbers Aportaciones Mat Investig 0 0), 7 84 [6] T Komatsu and L Szalay, Balancing with binomial coefficients Int J Number Theory 0 04), [7] K Kuhapatanakul, On the sums of reciprocal generalized Fibonacci numbers J Integer Seq 6 03), Article 37 8 pages) [8] K Liptai, Fibonacci balancing numbers Fibonacci Quart 4 004), [9] K Liptai, F Luca, Á Pintér and L Szalay, Generalized balancing numbers Indag Math NS) 0 009), [0] H Ohtsuka and S Nakamura, On the sum of reciprocal Fibonacci numbers Fibonacci Quart 46/47 008/009), [] O Karaatli, R Keskin and H Zhu, Infinitely many positive integer solutions of the quadratic Diophantine equation x 8B nxy y ±r Irish Math Soc Bull 73 04), 9 45 [] GK Panda, Some fascinating properties of balancing numbers Congr Numer ), [3] BK Patel and PK Ray, The period, rank and order of the sequence of balancing numbers modulo m Math Rep Bucur) 8 06), 3, [4] PK Ray, Balancing and cobalancing numbers PhD Thesis, National Institute of Technology, Rourkela, 009), 00 p Available at PhD Thesis of PK Raypdf [5] GJ Zhang, The infinite sum of reciprocal of Fibonacci numbers J Math Res Exp 3 0), Received 4 April 06 National Institute of Technology, Rourkela , India gkpanda nit@rediffmailcom Wuhan University, School of Mathematics and Statistics, Wuhan 43007, China komatsu@whueducn National Institute of Technology, Rourkela , India davalaravikumar@gmailcom