Int. J. Contemp. Math. Sciences, Vol. 7, 2012, no. 18, 881 - 889
Curious Congruences for Balancing Numbers Prasanta Kumar Ray International Institute of Information and Technology Bhubaneswar-751003, India
[email protected] Abstract The balancing numbers are terms of a sequence defined in a quite simple recursive fashion. However, despite its simplicity, they have some curious properties, which are worth mentioning. In this paper, we establish some fascinating congruences involving balancing and related numbers.
Mathematics Subject Classification: 11B39, 11B83 Keywords: Balancing numbers, Balancers, Lucas-balancing numbers, Recurrence relation
1
Introduction
Recently, Behera and Panda [1] introduced balancing numbers as solutions of the Diophantine equation 1 + 2 + · · · + (n − 1) = (n + 1) + (n + 2) + · · · + (n + r) calling r ∈ Z + , the balancer corresponding to the balancing number n. The numbers 6,35,204 are examples of first three balancing numbers with balancers 2,14,and 84 respectively. In [3], Panda and Ray introduced cobalancing numbers by slightly modifying the original definition. They defined the cobalancing numbers as the solution of the Diophantine equation 1 + 2 + · · · + n = (n + 1) + (n + 2) + · · · + (n + r) calling r ∈ Z + , the cobalancer corresponding to the cobalancing number n. It is shown in [1] that a positive integer n is a balancing number if and only if n2 is a triangular number or equivalently, 8n2 + 1 is a perfect square.Though the definition does not allow any balancing number to be less than 1, Panda [5, 6] and Panda and Ray [4] accepted 1 as a balancing number being the positive square root of the square triangular number 1. While the sequence {Bn }∞ n=1 of balancing numbers obey the recurrence relation Bn+1 = 6Bn − Bn−1 . If n is a balancing number, Cn is called a Lucas-balancing number, while for a cobalancing number n, cn is called a Lucas-cobalancing number [5]. The sequence of Lucas-balancing numbers satisfy the recurrence relation identical with that for balancing numbers, where as
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the recurrence relation for Lucas-cobalancing numbers are not identical with that for cobalancing numbers and surprisingly, they are identical with that for balancing numbers [5]. Liptai [2] added another interesting result to the theory of balancing numbers by proving that the only balancing number in the Fibonacci sequence is 1. Also in [3] he proved that there are no Lucas balancing numbers in the sequence of Fibonacci numbers. In [7], Ray obtain nice product formulas for balancing and Lucas-balancing numbers. √ It is proved in [1],that limn→∞ BBn+1 = 3 + 8. Throughout this paper, we n √ √ denote λ1 = 3 + 8 and its reciprocal λ2 = 3 − 8.
2
Some Basic Properties of Balancing Numbers
If Bn and Cn be the nth balancing and Lucas-balancing numbers, their closed forms, popularly known as Binet’s formulas [1,4] are respectively given by Bn =
λn1 − λn2 √ 2 8
(1)
λn1 + λn2 . (2) 2 Using (1) and (2), for all positive integer n, one can easily obtain the following important results. B−n = −Bn , C−n = Cn (3) Cn =
Cn = Bn+1 − 3Bn
(4)
Bn+1 − Bn−1 = 2Cn
(5)
Cn+1 = 3Cn + 8Bn
(6)
Cn = 3Bn − Bn−1
(7)
Keeping all these important results in mind, we suggest the proof of the following theorem. ( ) m Theorem 2.1 If denote the usual notation for combination and for j every integers k, m, n, s with m ≥ 0, we have m ( ) ∑ m m−j m Bs Bkm+n = (−1)m−j Bkj Bk−s Bjs+n (8) j j=0
and Bsm Ckm+n
=
m ( ) ∑ m j=0
j
m−j (−1)m−j Bkj Bk−s Cjs+n
(9)
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Proof. As λ1 + λ2 = 6, λ1 λ2 = 1 and by (3), applying binomial theorem, we obtain( ) ∑m m m−j (−1)m−j Bkj Bk−s Bjs+n j=0 j( ) k−s k −λk ∑ m −λk−s λjs+n −λ2js+n m−j λ1 √ 2 j λ1 √ 2 )m−j ( 1 √ = m (−1) ( ) ( ) j=0 2 8 2 8 2 8 j ( ) ∑ m m−j = (λ1 −λ12 )m+1 m (−1)m−j (λk1 − λk2 )j (λk−s − λk−s (λjs+n − λjs+n ) 1 2 ) 1 2 j=0 j ( ) ∑ m = (λ1 −λ12 )m+1 m (−1)m−j (λjs+n − λjs+n )(λk1 − λk2 )j (λs1 λk2 − λk1 λs2 )m−j 1 2 j=0 j ∑ s+k = (λ1 −λ12 )m+1 [λn1 m − λs1 λk2 )j (λs1 λk2 − λk1 λs2 )m−j j=0 (λ1 ∑ s+k j k s m−j s k k s ] −λn2 m j=0 (λ1 λ2 − λ2 ) (λ1 λ2 − λ1 λ2 ) 1 n s+k k s m n s k m = (λ1 −λ2 )m+1 {λ1 (λ1 − λ1 λ2 ) − λ2 (λ1 λ2 − λs+k 2 ) } n km s s m = (λ1 −λ12 )m+1 {(λn1 λkm 1 − λ2 λ2 )(λ1 − λ2 ) } n km s s m = (λ1 −λ12 )m+1 {(λn1 λkm 1 − λ2 λ2 )(λ1 − λ2 ) } λkm+n −λkm+n
λs −λs
)( λ11 −λ22 )m = ( 1 λ1 −λ22 m = Bs Bkm+n . In order ) prove (9), noting that Cn = Bn+1 − 3Bn and from (8), we obtain ( to ∑m m m−j (−1)m−j Bkj Bk−s Cjs+n j=0 j( ) ( ) ∑m ∑ m m m j m−j m−j = j=0 (−1)m−j Bk Bk−s Bjs+n+1 − 3 j=0 (−1)m−j Bkj Bk−s Bjs+n j j = Bsm (Bkm+n+1 − 3Bkm+n+1 ) = Bsm Ckm+n . In the special case, for s = 1 and n = 0 in (8), we have Bkm =
m ( ) ∑ m j=0
j
m−j (−1)m−j Bkj Bk−1 Bj
(10)
m−j (−1)m−j Bkj Bk−1 Bj+n .
(11)
and the general case s = 1 of (8) is Bkm+n =
m ( ) ∑ m j=0
j
Corollary 2.2 for every integers k, n, we have (i) Bk−n Bk+n = Bk2 − Bn2 (ii) B2n = 2Bn Cn , C2n = Cn2 + 8Bn2 2 (iii) B2n+1 = Bn+1 − Bn2 (iv) Bk2 − Bk−1 Bk + 1 = 1
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Proof. In particular, for m = 1, (8) and (9) become Bs Bk+n = Bk Bs+n − Bk−s Bn , Bs Ck+n = Bk Cs+n − Bk−s Cn .
(12)
Proof of (i) directly follows from (12) by setting s = k − n. Substitute s = 1, k = n in (9) and using (5), we obtain first part of (ii). One can prove the second part of (ii) by using (6) and (7) as follows: C2n = Bn Cn+1 − Bn−1 Cn = Bn (3Cn + 8Bn ) − Bn−1 Cn = Cn (3Bn − Bn−1 + 8Bn2 = Cn2 + 8Bn2 . Setting s = 1, k = n + 1 in (12), we obtain (iii). (iv) is a special case of (1) for n = 1. We observe from (iv) that, Bk−1 is prime to Bk . For m ≥ 1, it follows from (5), that m−1 m Bsm Bkm+n ≡ (−1)m Bk−s − Bn + m(−1)m−1 Bk−s − Bk Bs+n (mod Bk2 ). (13)
For s = 1 in (13), we get m−1 m Bkm+n ≡ (−1)m Bk−1 − Bn + m(−1)m−1 Bk−1 − Bk B1+n (mod Bk2 ),
(14)
and therefore for n = 0, m−1 Bkm ≡ m(−1)m−1 Bk−1 − Bk (mod Bk2 ).
(15)
If (a, b) denote the greatest common divisor of a and b, then from (14) and (15), we observe that m (Bkm+n , Bk ) = (Bk−1 , Bk ) = (Bk , Bn ).
(16)
From (16) and Euclidean algorithm, we have the following beautiful result due to Panda[8]. Theorem 2.3 Let m and n be positive integers. Then (Bm , Bn ) = B(m,n) Corollary 2.4 If m, n are positive integers, then Bm |Bn ⇔ m|n. Proof. By virtue of Theorem 2.2, we have m|n ⇔ (m, n) = m ⇔ B(m,n) = Bm ⇔ (Bm , Bn ) = Bm ⇔ Bm |Bn .
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Congruences for prime subscripted balancing numbers
Let ( ap ) be the Legendre symbol of any integer a and any prime p. If p is an odd prime, by quadratic reciprocity law, we have ( 8 ) ( p ) { 1, if p ≡ 1 (mod 8) = = −1, if p ≡ 5 (mod 8). p 8 Theorem 3.1 If p is an odd prime, then Cp ≡ 3 (mod p) (p) Bp ≡ (mod p) 8
( ) p k! = p(p − 1) · · · (p − k + 1) ≡ 0 (mod p), we observe that Proof. Since k ( ) p p| , k = 1, 2, · · · (p − 1). k From this and (4), we get √ √ 1 Cp = [(3 + 8)p + (3 − 8)p ] 2 ( ) √ 1 p p p−k √ k 3 ( 8) + (− 8)k ] = [Σk=0 k 2 ( ) p p−k k2 = Σpk=0 3 8 ≡ 3 (mod p). k Similarly using (3) and Euler’s criterion and since ( ap ) ≡ ap−1 (mod p), we get √ √ 1 Bp = √ [(3 + 8)p − (3 − 8)p ] 2 8 p ( ) √ 1 ∑ p p−k √ k = √ [ 3 {( 8) − ( 8)−k }] 2 8 k=0 k ( ) ( ) ( ) p ∑ p−1 8 p p p−k k−1 2 2 ≡8 ≡ (mod p). = = 3 8 k p 8 k=0 2|k
Theorem 3.2 Let p be an odd prime. Then (( p ) ) Bp−1 ≡ 3 − 1 (mod p) ( 8 ( p )) Bp+1 ≡ 3 1 + (mod p) 8
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Proof . From (4), we observe that Cp = Bp+1 − 3Bp = 6Bp − Bp−1 − 3Bp = 3Bp − Bp−1 Thus, Bp+1 = 3Bp + Cp and Bp−1 = 3Bp − Cp . These equations together with Theorem 3.1 yield the results. Corollary 3.3 Let p be a prime , then p|Bp−( p ) . 8 Proof .The proof of the corollary directly follows from Theorem 3.1. Corollary 3.4 Let p > 3 be a prime and q be a prime divisor of Bp , then (q ) q≡ (mod p) 8 ( ) Proof. By virtue of Corollary 3.3, since q|Bq−( q ) , we have q| Bq−( q ) , Bp . 8 ( 8 ( )) By Theorem 2.2, we get q|B(p,q−( q )) which implies p, q − 8q = p and hence, 8 ( ) q ≡ 8q (mod p). In[4], it is well known that, for any integer k, Ck2 − 8Bk2 = 1. Thus the greatest common divisor (Ck , Bk ) is either 1 or 2. Also observe that, for integers k and n with k ̸= 0, substituting s = −k in (12) and since B2k = 2Bk Ck and using (3), we find Bk+n = 2Bn Ck − Bn−k .
(17)
Theorem 3.5 Let k and n are integers with k ̸= 0, then { Bkn 2m + 1 (mod 8Bk2 ), if n=2m+1; ≡ 2mCk (mod 8Bk2 ), if n=2m. Bk Proof . Since B−kn = −Bkn , it suffices to prove the result for n ≥ 0. Clearly the theorem holds for n = 0, n = 1. Suppose n ≥ 2 and it holds for all positive integers less than n. For n = (n − 1)k, (17) becomes Bkn = 2B(n−1)k Ck − B(n−2)k and since Ck2 = 8Bk2 + 1 ≡ 1 (mod 8Bk2 ), by inductive hypothesis, one can obtain B(n−1)k B(n−2)k Bkn = 2Ck − Bk Bk Bk
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which is either 4mCk2 − (2m − 1) or 2Ck (2m − 1) − 2(m − 1)Ck according as n = 2m + 1 or n = 2m. Thus { Bkn 2m + 1 (mod 8Bk2 ), if n=2m+1; ≡ 2mCk (mod 8Bk2 ), if n=2m, Bk and therefore theorem holds for all positive integers n. Corollary 3.6 If p be an odd prime divisor of Bk and k be any nonzero integer, then Bkp ≡ p (mod 8p2 ) Bk Proof . Since p|Bk , 8p2 |8Bk2 and by virtue of Theorem 3.3, we get the desired result. Theorem 3.7 For positive integers m and n, Bmn =
n ( ) ∑ n
i
i=0
i n−i (−1)n−i Bm Bm−1 Bi ≡ 0 (mod Bm )
Proof . Proof of this theorem directly follows from (10). Corollary 3.8 For positive integers λ, m, n and prime p, if pλ ∥Bm , then p|n if and only if pλ+1 |Bmn . Where pλ ∥Bm means pλ divides Bm and pλ+1 does not divide Bm . Proof . By Theorem 3.7, n−1 2 Bmn ≡ (−1)n−1 nBm−1 Bm (mod Bm ).
Since p|Bm , m > 1, therefore (Bm−1 , Bm ) = B(m−1,m) = 1. 2 , we get Again, since pλ ∥Bm and pλ+1 |Bm n−1 pλ+1 |Bmn ⇔ pλ+1 |nBm−1 Bm ⇔ p|n.
A very interesting relation among balancing and Lucas-balancing numbers is observed through following theorem. Theorem 3.9 For any positive integer m, B2m ≡ 0 (mod Cm ), B2m−1 ≡ 1 (mod Cm )
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Proof . The first part of the theorem is obvious as B2m = 2Bm Cm . Since λ1 λ2 = 1, we observe that λ12m−1 − λ2m−1 − (λ1 − λ2 ) = λ2m−1 − λ2m−1 − (λ1 λ2 )m−1 (λ1 − λ2 ) 2 1 2 m = (λm−1 − λm−1 )(λm 1 2 1 + λ2 ) Divide (λ1 − λ2 ) both sides to get B2m−1 = 1 + 2Bm−1 Cm , which proves second part of the theorem. We conclude this section by presenting an important remark. Remark 3.10 By Corollary 3.8, any prime power divides some positive balancing numbers. Let d = pλ1 1 pλ2 2 . . . pλr r where p1 < p2 . . . pr and suppose pλi i |Bni for each i = 1, 2, . . . r. Since Bni |B[n1 ,n2 ...nk ] and pλi i |B[n1 ,n2 ...nk ] , it follows that d|B[n1 ,n2 ...nk ] . Thus any positive integer d is a divisor of some positive balancing numbers.
Acknowledgement The author wishes to thanks Prof. A Behera and Prof. G. K. Panda, N.I.T, Rourkela, INDIA for their valuable suggestion and guidance while making this paper.
References 1. Behera, A. and Panda, G.K., On the square roots of triangular numbers, The Fibonacci Quarterly, 37(2), 1999, 98 − 105. 2. Liptai, K., Fibonacci balancing numbers, The Fibonacci Quarterly, 42(4), 2004, 330 − 340. 3. Panda, G. K. and Ray, P. K., Cobalancing Numbers and Cobalancers, International Journal of Mathematics and MathematicalSciences,2005(8), 2005, 1189 - 1200. 4. Panda, G. K. and Ray, P. K., Some links of balancing and cobalancing numbers with Pell and associated Pell numbers, Bulletin of the Institute of Mathematics, Academia Sinica (New Series), 6(1), 2011, 41-72. 5. Panda, G. K., Sequence balancing and cobalancing numbers, The Fibonacci Quarterly, 45(3), 2007, 265 − 271.
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6. Panda, G. K., Some Fascinating properties of balancing Numbers, to appear in ”Applications of Fibonacci Numbers Vol.10, Kluwer Academic pub., 2006. 7. Ray, P. K., Application of Chybeshev Polynomials in Factorization of balancing and Lucas-balancing Numbers , to appear in ”Bol. Soc. Paran. Mat. Vol.30 (2), 2012. Received: October, 2011
