Palestine Journal of Mathematics Vol. 5(2) (2016) , 122–129
© Palestine Polytechnic University-PPU 2016
GENERATING FUNCTIONS FOR CERTAIN BALANCING AND LUCAS-BALANCING NUMBERS Prasanta Kumar Ray and Juli Sahu Communicated by Ayman Badawi MSC 2010 Classifications: 11B 39, 11B 83 Keywords and phrases: Balancing numbers; Lucas-balancing numbers; Generating functions
Abstract It is well known that the generating function for any sequence {an }, denoted by the ∞ X function g (x) and defined by g (x) = an xn . This function is used to solve both homogenous n=0
and non-homogenous recurrence relations. In this study, we find generating function of certain balancing and Lucas-balancing numbers.
1 Introduction Balancing numbers n and the balancers r are the solutions of the Diophantine equation 1 + 2 + . . . + (n − 1) = (n + 1) + (n + 2) + . . . + (n + r). The square roots of 8n2 + 1 also generate a sequence of numbers called as Lucas-balancing numbers. The balancing numbers and the Lucas-balancing numbers satisfy the same recurrence relation with different initial values, that is, Bn+1 = 6Bn − Bn−1 ; B0 = 0, B1 = 1 and Cn+1 = 6Cn − Cn−1 ; C0 = 1, C1 = 3, where n ≥ 1 and Bn and Cn are the nth balancing and Lucas-balancing numbers respectively [1, 3]. The details of balancing and Lucas-balancing numbers are available in [1–19]. It is well known that for the sequence a0 , a1 , . . . of real numbers, the function g (x) = a0 + a1 x + a2 x2 + . . . + an xn + . . . is called the generating function for the sequence {an }. n X Also, by letting ai = 0 for i > n; g (x) = an xn represents the generating function for the i=0
finite sequence {an }. In this study, authors main aim is to establish some generating functions of certain balancing and Lucas-balancing numbers. Generating functions are used to solve both homogenous and non-homogenous recurrence relations. The following example show how generating function is used to solve recurrence relation for balancing numbers and derive the famous Binet’s formula for these numbers. Example 1.1. Use generating functions to solve the balancing recurrence relation Bn+1 = 6Bn − Bn−1 , where B1 = 1, B2 = 6. Solution. Let g (x) = B0 + B1 x + B2 x2 + . . . + Bn xn + . . .. be the generating function of the balancing sequence. Using the recurrence relation Bn+1 = 6Bn − Bn−1 , we can find 6xg (x) and x2 g (x) as follows: 6xg (x) = 6B0 x + 6B1 x2 + 6B2 x3 + . . . + 6Bn−1 xn + . . .. x2 g (x) = B0 x2 + B1 x3 + B2 x4 + . . . + Bn−2 xn + . . .,
which follows that g (x) − 6xg (x) + x2 g (x) = B0 + (B1 − 6B0 )x + (B2 − 6B1 + B0 )x2 + . . . = x,
and therefore, we have g (x) =
1 1 1 x √ = − . (1 − 6x + x2 ) 2 8 1 − λ1 x 1 − λ2 x
Generating Functions for Certain Balancing...
123
Which implies that g (x) =
∞ X
Bn xn =
n=0
∞ X (λn1 − λn2 )xn √ . 2 8 n=0
From this expression, it follows that Bn =
λn1 − λn2 , λ1 − λ2
which is nothing but the Binet’s formula for balancing numbers.
2 Generating functions for certain balancing and Lucas-balancing numbers In this section, we establish generating functions of certain balancing and Lucas-balancing numbers. 3 2.1 Generating functions for B3n and Bn
Let g1 (x) be the generating function for B3n . Then g1 (x) = B0 + B3 x + B6 x2 + . . . + B3n xn + . . .,
4xg1 (x) = 4xB0 + 4B3 x2 + 4B6 x3 + . . . + 4B3n−3 xn + . . ., x2 g1 (x) = B0 x2 + B3 x3 + B6 x4 + . . . + B3n−6 xn + . . ..
Therefore, we have (1 − 4x − x2 )g1 (x) = B0 + (B3 − 4B0 )x + (B6 − 4B3 − B0 )x2 + . . . , which follows that g1 (x) =
35x + 6790x2 + . . . . 1 − 4x − x2
Similarly, if g2 (x) be the generating function for Bn3 , then we have g2 (x) = B03 + B13 x + B23 x2 + . . . + Bn3 xn + . . ., 3 n 3xg2 (x) = 3xB03 + 3B13 x2 + 3B23 x3 + . . . + 3Bn− 1 x + . . ., 3 n 6x2 g2 (x) = 6B03 x2 + 6B13 x3 + 6B23 x4 + . . . + 6Bn− 2 x + . . ., 3 n 3x3 g2 (x) = 3B03 x3 + 3B13 x4 + . . . + 3Bn− 3 x + . . ., 3 n x4 g2 (x) = B03 x4 + B13 x5 + B23 x6 + . . . + Bn− 4 x + . . ..
Therefore, we have (1 − 3x − 6x2 + 3x3 + x4 )g2 (x) = x + 213x2 + 42221x3 + . . . . Which follows that g2 (x) =
x + 213x2 + 42221x3 + . . . . (1 − 3x − 6x2 + 3x3 + x4 )
2.2 Generating function for B2n+1 and C2n+2 Let g3 (x) be the generating function for B2n+1 , then g3 (x) = B1 + B3 x + B5 x2 + . . . + B2n+1 xn + . . .,
3xg3 (x) = 3xB1 + 3B3 x2 + 3B5 x3 + . . . + 3B2n−1 xn + . . ., x2 g3 (x) = B1 x2 + B3 x3 + B5 x4 + . . . + B2n−3 xn + . . ..
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Prasanta Kumar Ray and Juli Sahu
Therefore, we have (1 − 3x − x2 )g3 (x) = B1 + (B3 − 3B1 )x + (B5 − 3B3 − B1 )x2 + . . . = 1 + 32x + 1083x2 + . . . , which follows that 1 + 32x + 1083x2 + . . . . (1 − 3x − x2 )
g3 (x) =
In a similar manner, we can find the generating function g4 (x) of C2n+2 as C2 + (C4 − 3C2 )x + . . . . (1 − 3x + x2 )
g4 (x) =
2.3 Generating functions for Bm+n and Cm+n The generating function of Bm+n is given by; ∞ X
Bm+n xn =
∞ X (λ1 m+n − λ2 m+n )xn
λ1 − λ2
n=0
n=0
∞ ∞ X X 1 m n n m λ1 λ1 x − λ2 λn2 xn = λ1 − λ2 n=0 n=0 m λm λ 1 1 2 = − λ1 − λ2 1 − λ1 x 1 − λ2 x Bm − Bm−1 x = 2 . x − 6x + 1
"
#
Likewise, it can be shown that ∞ X
Cm+n xn =
n=0
∞ X (λ1 m+n + λ2 m+n )xn
2
n=0
∞ ∞ X 1 mX n n = λ1 λ1 x + λm λn2 xn 2 2 n=0 n=0 m m λ1 λ2 1 = + 2 1 − λ1 x 1 − λ2 x Cm − Cm−1 x = 2 . x − 6x + 1
"
#
These two generating functions can be applied to derive identities. For example, ∞ X n=0
∞ ∞ X 1 X 6x − 1 1 − 3x n Bn+1 x = , Bn−1 x = and C n xn = , D D D n
n=0
where D = x2 − 6x + 1. Since 2
2
∞ X n=0
n
Cn x =
∞ X n=0
1−3x D
n=0
n
=
Bn+1 x −
1 D
−
∞ X n=0
6x−1 D ,
we have
∞ X Bn−1 x = (Bn+1 − Bn−1 )xn , n
n=0
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Generating Functions for Certain Balancing...
which implies that Bn+1 − Bn−1 = 2Cn . To prove the identity Bm Cn − Bm−1 Cn−1 = Cm+n−1 , we proceed as follows: ∞ X
(Bm Cn − Bm−1 Cn−1 )xm = Cn
m=0
∞ X
∞ X
Bm xm − Cn−1
m=0
Bm−1 xm
m=0
6x − 1 x − Cn−1 D D Cn−1 + (Cn − 6Cn−1 )x = D Cn−1 − Cn−2 x = D ∞ X = Cm+n−1 xn , = Cn
m=0
which follows that Bm Cn − Bm−1 Cn−1 = Cm+n−1 .
3 Exponential generating functions Bn n!
In this section, we develop the generating functions for
Cn n! .
and
As et =
∞ n X t , it follows n! n=0
that eλ 1 x =
∞ X λ1 n xn
n!
n=0
and eλ2 x =
∞ X λ2 n xn
n!
n=0
.
Therefore, we have ∞
∞
n=0
n=0
X (λ1 n − λ2 n ) xn X eλ1 x − eλ2 x xn Bn . = = λ1 − λ2 λ1 − λ2 n! n! eλ1 x − eλ2 x Bn generates the numbers . More generally, we can λ1 − λ2 n! ∞ n X k λ kx x −eλ2 x = Bkn . show that e 1 λ1 −λ 2 n! n=0 ∞ X √ xn Likewise, the generating function eλ1 x +eλ2 x = 2Cn can derive the formula e3x cosh( 8x) = n! n=0 ∞ ∞ X X xn tn Cn . Let us consider the functions A(t) and B (t) defined by A(t) = an and B (t) = n! n! n=0 n=0 ∞ n X t bn , so that their products A(t)B (t) and A(t)B (−t) are given by n!
Thus, the exponential function
n=0
A(t).B (t) =
" n ∞ X X n=0
and
n k
k=0
" n ∞ X X A(t).B (−t) = (−1)n−k n=0
In particular for A(t) =
e
−6λ1 t
−6λ2 t
−e λ1 − λ2
k=0
!
# ak bn−k
n k
tn , n!
!
(3.1)
# ak bn−k
and B (t) = et , we have ∞
∞
n=0
n=0
et (e−6λ1 t − e−6λ2 t ) X tn X (−6t)n = Bn . λ1 − λ2 n! n!
tn . n!
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Prasanta Kumar Ray and Juli Sahu
Which follows that " n ∞ X X e(1−6λ1 )t − e(1−6λ2 )t = λ1 − λ2 n=0
n k
k=0
!
# Bk (−6)k
tn . n!
Using the characteristic equation λ2 = 6λ − 1, we obtain ! # " n 2 2 ∞ n X X n e−λ1 t − e−λ2 t k t = Bk (−6) , λ1 − λ2 n! k n=0 k=0 which implies that "∞ # " n ∞ ∞ X (−λ1 2 t)n X X X 1 (−λ2 2 t)n − = λ1 − λ2 n! n! n=0
n=0
n=0
k=0
n k
!
# k
Bk (−6)
tn . n!
That is, ∞ X n=0
n nt
B2n (−1)
n!
=
" n ∞ X X n=0
k=0
n k
#
! k
Bk (−6)
tn . n!
Equating the coefficients of tn /n!, we have B2n (−1)n =
n X k=0
n k
! Bk (−6)k .
Replacing B (t) by e−t and proceeding similarly, we get " n ! # "∞ # ∞ ∞ ∞ X X X n (−2t)n X (−λ1 2 t)n X (−λ2 2 t)n tn 1 k Bk (−6) (−1)n − = λ1 − λ2 n! n! n! n! k n=0 n=0 n=0 k=0 n=0 " ! # ∞ n X X n tn (−2)k (−1)n B2n n! k n=0 k=0 " ! # ∞ n X X n tn = Bk (−6)k (−1)n , n! k n=0 k=0
which follows a new combinatorial identity ! n n X X n (−2)k B2n = k k=0 k=0
n k
! Bk (−6)k .
4 Some hybrid identities containing both balancing and Lucas-balancing numbers In this section, once again we consider the functions A(t) and B (t) to develop some hybrid eλ1 t − eλ2 t identities that contain both balancing and Lucas- balancing numbers. Let A(t) = λ1 − λ2 √ √ and B (t) = eλ1 t + eλ2 t , where λ1 = 3 + 8 and λ2 = 3 − 8. Then we have the following results. Lemma 4.1. If!Bn and Cn respectively denote the nth balancing and Lucas-balancing numbers, n X n then Bk Cn−k = 2n−1 Bn . k k=0
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Generating Functions for Certain Balancing...
Proof. Using (3.1), we have " n ∞ X X n=0
k=0
n k
!
# Bk Cn−k
∞
∞
X tn tn X tn = Cn Bn n! n! n! n=0
n=0
e2λ1 t
− 2(λ1 − λ2 )
= =
e2λ2 t
∞ X
2n−1 Bn
n=0
tn , n!
and hence the result follows. Lemma 4.2. If!Bn and Cn respectively denote the nth balancing and Lucas-balancing numbers, ! n n X X n n then Bk Bn−k = 2n−4 (Cn − 3n ) and Ck Cn−k = 2n−1 (Cn + 3n ). k k k=0 k=0 Proof. Again using (3.1), we have " n ! # ∞ ∞ ∞ X X X n tn tn X tn = Bn Bn Bk Bn−k n! n! n! k n=0 n=0 n=0 k=0 = =
(eλ1 t − eλ2 t )2 (λ1 − λ2 )2 ∞ X
2n−4 (Cn − 3n )
n=0
tn , n!
and the first result follows. The second result follows analogously. Lemma 4.3. For nth balancing number Bn and nth Lucas-balancing number Cn , the following identities are valid: ! n X n Bmk Cmn−mk = 2n−1 Bmn , k k=0 ! n X n n Bmk Bmn−mk = 2n−4 (Cmn − Cm ), k k=0 ! n X n n Cmk Cmn−mk = 2n−1 (Cmn + Cm ). k k=0 Proof. In view of (3.1), we have " n ! # ∞ ∞ ∞ X X n X tn tn X tn Bmk Cmn−mk = Bmn Cmn n! n! n! k n=0 k=0 n=0 n=0 m
= =
m
(e2λ1 t − e2λ2 t ) 2(λ1 − λ2 ) ∞ X n=0
2n−1 Bmn
tn , n!
which follows the first result. The other results can be proved similarly. The following result can be obtained by using the differential operator d/dt. Since A(t) = ∞ X tn (−6)n an , we have n! n=0 ∞ X dr (−6t)n r A ( t ) = ( − 6) a . n + r dtr n! n=0
128
Setting A(t) =
REFERENCES
eλ1 t − eλ2 t and B (t) = eλ1 t , then by using (3.1), we get λ1 − λ2 ! " n # "∞ # ∞ ∞ n X X X n X tn (−6t)n t k = Bn+r (−6) Bk+r n! n! n! k n=0 k=0 n=0 n=0 " # ∞ et dr X (−6t)n = Bn (−6)r dtr n! n=0
d e−6λ1 t − e−6λ2 t e ( ) (−6)r dtr λ1 − λ2 t
= =
∞ X
r
(−1)n B2n+r
n=0
tn , n!
which yields the identity n X k=0
n k
!
(−6)k Bk+r = (−1)n B2n+r .
References [1] A. Behera and G.K. Panda, On the square roots of triangular numbers, The Fibonacci Quarterly, 37(1999), 98-105. [2] A. Berczes, K. Liptai, I. Pink, On generalized balancing numbers, Fibonacci Quarterly, 48(2010), 121-128. [3] R. Keskin and O. Karaatly, Some new properties of balancing numbers and square triangular numbers, Journal of Integer Sequences, 15(2012). [4] K. Liptai, Fibonacci balancing numbers, The Fibonacci Quarterly, 42(2004), 330-340. [5] K. Liptai, Lucas balancing numbers, Acta Mathematica Universitatis Ostraviensis, 14(2006), 43-47. [6] K. Liptai, F. Luca, A. Pinter and L. Szalay, Generalized balancing numbers, Indagationes Math. N. S., 20(2009), 87-100. [7] P. Olajos, Properties of balancing, cobalancing and generalized balancing numbers, Annales Mathematicae et Informaticae, 37(2010), 125-138. [8] G.K. Panda and P.K. Ray, Cobalancing numbers and cobalancers, International Journal of Mathematics and Mathematical Sciences, 2005(2005), 1189-1200. [9] G.K. Panda and P.K. Ray, Some links of balancing and cobalancing numbers with Pell and associated Pell numbers, Bulletin of the Institute of Mathematics, Academia Sinica (New Series), 6(2011), 41-72. [10] G.K. Panda, Some fascinating properties of balancing numbers, Proceeding Eleventh International Conference on Fibonacci Numbers and Their Applications, Cong. Numerantium, 194(2009), 185-189. [11] G.K. Panda, Arithmetic progression of squares and solvability of the Diophantine equation 8x4 + 1 = y 2 , East-West Journal of Mathematics, 1(2012), 131-137 [12] P.K. Ray, Application of Chybeshev polynomials in factorization of balancing and Lucasbalancing numbers, Boletim da Sociedade Paranaense de Matemática 30 (2012), 49-56. [13] P.K. Ray, Curious congruences for balancing numbers, International Journal of Contemporary Mathematical Sciences, 7(2012), 881-889. [14] P. K. Ray, Certain matrices associated with balancing and Lucas-balancing numbers, Matematika, 28 (2012), 15-22.
REFERENCES
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[15] P.K. Ray, Factorization of negatively subscripted balancing and Lucas-balancing numbers, Boletim da Sociedade Paranaense de Matemática, 31(2013), 161-173. [16] P. K. Ray, New identities for the common factors of balancing and Lucas-balancing numbers, International Journal of Pure and Applied Mathematics, 85(2013), 487-494. [17] P.K. Ray, Some congruences for balancing and Lucas-balancing numbers and their applications, Integers, 14(2014), #A8. [18] P.K. Ray, On the properties of Lucas-balancing numbers by matrix method, Sigmae, Alfenas, 3(2014), 1-6. [19] P.K. Ray, K. Parida, Generalization of Cassini formulas for balancing and lucas-balancing numbers, Matematychni Studii, 42(2014), 9-14. Author information Prasanta Kumar Ray and Juli Sahu, Department of Mathematics, Veer Surendra Sai University of Technology, Burla-768018, Sambalpur and Department of Mathematics, National Institute of Technology, Rourkela769008, India. E-mail:
[email protected] and
[email protected]
Received:
February 7, 2015.
Accepted:
July 28, 2015.
