EasyChair Preprint № 4342
Balcobalancing Numbers
Ahmet Tekcan and Meryem Yi̇ldi̇z
EasyChair preprints are intended for rapid dissemination of research results and are integrated with the rest of EasyChair.
October 10, 2020 Notes on Number Theory and Discrete Mathematics Print ISSN 1310–5132, Online ISSN 2367–8275 Vol. 25, 2019, No. 1, xx–xxx DOI: 10.7546/nntdm.2019.25.1.xx-xxx
Balcobalancing Numbers
Ahmet Tekcan and Meryem Yıldız
Bursa Uludag University, Faculty of Science Department of Mathematics, Bursa, Turkey e-mail: [email protected],[email protected]
Received: 05 October 2020 Revised: 1 November 2018 Accepted: 23 February 2019
Abstract: In this work we determine the general terms of balcobalancing numbers, Lucas-bal- cobalancing numbers and balcobalancers. Further we formulate the sums of these numbers and derive relationship with square triangular numbers. Keywords: Balancing numbers, cobalancing numbers, square triangular numbers, Pell equations. 2010 Mathematics Subject Classification: 11B37, 11B39, 11D09, 11D79.
1 Introduction
A positive integer n is called a balancing number ([2]) if the Diophantine equation
1 + 2 + ··· + (n − 1) = (n + 1) + (n + 2) + ··· + (n + r) (1) holds for some positive integer r which is called balancer corresponding to n. If n is a balancing number with balancer r, then from (1) √ −2n − 1 + 8n2 + 1 r = . (2) 2 From (2) we note that n is a balancing number if and only if 8n2 + 1 is a perfect square. Though the definition of balancing numbers suggests that no balancing number should be less than 2. But from (2) we note that 8(0)2 + 1 = 1 and 8(1)2 + 1 = 32 are perfect squares. So we accept 0 th and 1 to be balancing numbers. Let Bn denote the n balancing number. Then B0 = 0,B1 = 1,
B2 = 6 and Bn+1 = 6Bn − Bn−1 for n ≥ 2.
1 Later Panda and Ray ([14]) defined that a positive integer n is called a cobalancing number if the Diophantine equation
1 + 2 + ··· + n = (n + 1) + (n + 2) + ··· + (n + r) (3)
holds for some positive integer r which is called cobalancer corresponding to n. If n is a cobal- ancing number with cobalancer r, then from (3) √ −2n − 1 + 8n2 + 8n + 1 r = . (4) 2 From (4) we note that n is a cobalancing number if and only if 8n2 + 8n + 1 is a perfect square. Since 8(0)2 + 8(0) + 1 = 1 is a perfect square, we accept 0 to be a cobalancing number, just like
Behera and Panda accepted 0, 1 balancing numbers. Cobalancing number is denoted by bn. Then b0 = b1 = 0, b2 = 2 and bn+1 = 6bn − bn−1 + 2 for n ≥ 2. It is clear from (1) and (3) that every balancing number is a cobalancer and every cobalancing th number is a balancer, that is, Bn = rn+1 and Rn = bn for n ≥ 1, where Rn is the n the balancer th and rn is the n cobalancer. Since Rn = bn, we get from (1) that
−2B − 1 + p8B2 + 1 2b + 1 + p8b2 + 8b + 1 b = n n and B = n n n . (5) n 2 n 2
p 2 p 2 Thus from (5), we see that Cn = 8Bn + 1 and cn = 8bn + 8bn + 1 are integers which are called the nth Lucas-balancing number and nth Lucas-cobalancing number, respectively. √ √ Let α = 1 + 2 and β = 1 − 2 be the roots of the characteristic equation for Pell and
Pell-Lucas numbers which are the numbers defined by P0 = 0,P1 = 1,Pn = 2Pn−1 + Pn−2 and
Q0 = Q1 = 2,Qn = 2Qn−1 + Qn−2 for n ≥ 2. Then Ray ([17]) derived some nice results on balancing numbers and Pell numbers his Phd thesis. Since x is a balancing number if and only if 8x2 + 1 is a perfect square, he set 8x2 + 1 = y2 for some integer y ≥ 1. Then he get the Pell equation ([1, 3, 9]) y2 − 8x2 = 1. (6) √ √ The fundamental solution of (6) is (y , x ) = (3, 1). So y + x 8 = (3 + 8)n for n ≥ 1 and √ √ 1 1 √ n n √ n γn−δn similarly yn − xn 8 = (3 − 8) . Let γ = 3 + 8 and δ = 3 − 8. Then he get xn = γ−δ γn−δn 2 which is the Binet formula for balancing numbers, that is, Bn = γ−δ . Since α = γ and 2n 2n β2 = δ, he conclude that the Binet formula for balancing numbers is B = α √−β . Similarly he n 4 2 2n−1 2n−1 2n 2n 2n−1 2n−1 get b = α √−β − 1 ,C = α +β and c = α +β for n ≥ 1 (see also [10, 13, 16]). n 4 2 2 n 2 n 2 Balancing numbers and their generalizations have been investigated by several authors from many aspects. In [7], Liptai proved that there is no Fibonacci balancing number except 1 and in [8] he proved that there is no Lucas balancing number. In [20], Szalay considered the same problem and obtained some nice results by a different method. In [5], Kovacs,´ Liptai and Olajos extended the concept of balancing numbers to the (a, b)-balancing numbers defined as follows: Let a > 0 and b ≥ 0 be coprime integers. If
(a + b) + ··· + (a(n − 1) + b) = (a(n + 1) + b) + ··· + (a(n + r) + b)
2 for some positive integers n and r, then an + b is an (a, b)-balancing number. The sequence of (a,b) (a, b)-balancing numbers is denoted by Bm for m ≥ 1. In [6], Liptai, Luca, Pinter´ and Szalay generalized the notion of balancing numbers to numbers defined as follows: Let y, k, l ∈ Z+ such that y ≥ 4. Then a positive integer x with x ≤ y − 2 is called a (k, l)-power numerical center for y if 1k + ··· + (x − 1)k = (x + 1)l + ··· + (y − 1)l. They studied the number of solutions of the equation above and proved several effective and ineffective finiteness results for (k, l)-power
numerical centers. For positive integers k, x, let Πk(x) = x(x + 1) ... (x + k − 1). Then it was
proved in [5] that the equation Bm = Πk(x) for fixed integer k ≥ 2 has only infinitely many solutions and for k ∈ {2, 3, 4} all solutions were determined. In [23] Tengely, considered the
case k = 5, that is, Bm = x(x + 1)(x + 2)(x + 3)(x + 4) and proved that this Diophantine equation has no solution for m ≥ 0 and x ∈ Z. In [15], Panda, Komatsu and Davala considered the reciprocal sums of sequences involving balancing and Lucas-balancing numbers and in [18], Ray considered the sums of balancing and Lucas-balancing numbers by matrix methods. In [12], Panda and Panda defined the almost balancing number and its balancer. In [21], the first author considered almost balancing numbers, triangular numbers and square triangular numbers and in [22], he considered the sums and spectral norms of almost balancing numbers.
2 Balcobalancing Numbers.
In this work we define a new balancing number called balcobalancing number, Lucas-balcoba- lancing number and balcobalancer and determine the general terms of them. If we sum of both sides of (1) and (3), then we get the Diophantine equation 1 + 2 + ··· + (n − 1) + 1 + 2 + ··· + (n − 1) + n = 2[(n + 1) + (n + 2) + ··· + (n + r)]. (7) So a positive integer n is called a balcobalancing number if the Diophantine equation (7) holds for some positive integer r which is called balcobalancer. For example, 10, 348, 11830, 401880, ··· are balcobalancing numbers with balcobalancers 4, 144, 4900, 166464, ··· . From (7), we get √ −2n − 1 + 8n2 + 4n + 1 r = . (8) 2 bc th bc th Let Bn denote the n balcobalancing number and let Rn denote the n balcobalancer. Then bc bc 2 bc from (8), we get Bn is a balcobalancing number if and only if 8(Bn ) + 4Bn + 1 is a perfect square. Thus bc p bc 2 bc Cn = 8(Bn ) + 4Bn + 1 (9) is an integer which are called the nth Lucas-balcobancing number. For example 29, 985, 33461, ··· are Lucas-balcobancing numbers (Here we notice that balcobalancing numbers should be grater that 0. But 8(0)2 + 4(0) + 1 = 1 is a perfect square, so we assume that 0 is a balcobalancing bc bc bc number, that is, B0 = 0. In this case, R0 = 0 and C0 = 1). In order to determine the general terms of balcobalancing numbers, Lucas-balcobancing num- bers and balcobalancers, we have to determine the set of all (positive) integer solutions of the Pell equation x2 − 2y2 = −1. (10)
3 bc bc 2 bc We see from (8) that Bn is a balcobalancing number if and only if 8(Bn ) + 4Bn + 1 is a perfect bc 2 bc 2 square. So we set 8(Bn ) + 4Bn + 1 = y for some integer y ≥ 1. If we multiply both sides of bc 2 bc 2 bc 2 2 the last equation by 2, then we get 16(Bn ) + 8Bn + 2 = 2y and hence (4Bn + 1) + 1 = 2y . bc Taking x = 4Bn + 1, we get the Pell equation in (10). For the set of all integer solutions of (10), we can give the following theorem.
Theorem 2.1. The set of all integer solutions of (10) is {(cn, 2bn + 1) : n ≥ 1}.
Proof. For the Pell equation x2 − 2y2 = −1, the set of representatives is Rep = {[±1 1]} and " # 3 2 M = . In this case [−1 1]M n generates all integer solutions (x , y ) for n ≥ 1. It can 4 3 n n " # C 2B be easily seen that the nth power of M is M n = n n for n ≥ 1. So 4Bn Cn " # Cn 2Bn [xn yn] = [−1 1] = [−Cn + 4Bn − 2Bn + Cn]. 4Bn Cn
Thus the set of all integer solutions is {(−Cn + 4Bn, −2Bn + Cn): n ≥ 1}. But we notice that
α2n + β2n α2n − β2n α2n−1 + β2n−1 −Cn + 4Bn = −( ) + 4( √ ) = = cn 2 4 2 2 and similarly −2Bn + Cn = 2bn + 1. So the result is clear.
From Theorem 2.1, we can give the following result. Theorem 2.2. The general terms of balcobalancing numbers, Lucas-balcobalancing numbers and balcobalancers are c − 1 4b − c + 1 Bbc = 2n+1 ,Cbc = 2b + 1 and Rbc = 2n+1 2n+1 n 4 n 2n+1 n 4 for n ≥ 1.
bc 2 bc 2 bc Proof. Notice that we multiply the equation 8(Bn ) +4Bn +1 = y by 2 and since x = 4Bn +1, we get x − 1 c − 1 Bbc = 2n+1 = 2n+1 n 4 4 for n ≥ 1. Thus from (9),
bc p bc 2 bc Cn = 8(Bn ) + 4Bn + 1 r c − 1 c − 1 = 8( 2n+1 )2 + 4( 2n+1 ) + 1 4 4 r c2 + 1 = 2n+1 2 s ( α4n+1+β4n+1 )2 + 1 = 2 2
4 s α4n+1 − β4n+1 1 2 = 2( √ − ) + 1 2 2 2
= 2b2n+1 + 1 and from (8), we conclude that
−2( c2n+1−1 ) − 1 + 2b + 1 4b − c + 1 Rbc = 4 2n+1 = 2n+1 2n+1 . n 2 4 This completes the proof.
We can also give the general terms of balcobalancing numbers, Lucas-balcobalancing num- bers and balcobalancers in terms of balancing and cobalancing numbers as follows. Theorem 2.3. The general terms of balcobalancing numbers, Lucas-balcobalancing numbers and balcobalancers are B + b −B + b Bbc = 2n 2n+1 ,Cbc = 2b + 1 and Rbc = 2n 2n+1 n 2 n 2n+1 n 2 for n ≥ 1.
bc c2n+1−1 Proof. We proved in Theorem 2.2 that Bn = 4 . So we easily deduce that c − 1 Bbc = 2n+1 n 4 α4n+1 + β4n+1 1 = − 8 4 α−1+1 −β−1−1 α4n+1( √ ) + β4n+1( √ ) 1 = 4 2 4 2 − 2 4 4n 4n 4n+1 4n+1 α √−β + α √−β − 1 = 4 2 4 2 2 2 B + b = 2n 2n+1 . 2
bc bc −B2n+b2n+1 Cn = 2b2n+1 + 1 is already proved in Theorem 2.2. Similarly we get Rn = 2 . Recall that the general terms of all balancing numbers can be given in terms of Pell numbers, namely, P P − 1 B = 2n , b = 2n−1 ,C = P + P , c = P + P . n 2 n 2 n 2n 2n−1 n 2n−1 2n−2 Similarly we can give the following theorem. Theorem 2.4. The general terms of balcobalancing numbers, Lucas–balcobalancing numbers and balcobalancers are P + P − 1 P − P − 1 Bbc = 4n+1 4n ,Cbc = P and Rbc = 4n+1 4n n 4 n 4n+1 n 4 for n ≥ 1.
5 Proof. Notice that
α4n+1 + β4n+1 α4n+1(1 + α−1) − β4n+1(1 + β−1) c2n+1 = = √ = P4n+1 + P4n. 2 2 2 So from Theorem 2.2, we get c − 1 P + P − 1 Bbc = 2n+1 = 4n+1 4n . n 4 4 The others can be proved similarly.
We notice that n is a a balancing number if and only if n2 is a triangular number (triangular n(n+1) numbers denoted by Tn are the numbers of the form Tn = 2 for n ≥ 0). Indeed from (1) we (n+r)(n+r+1) 2 get 2 = n , that is, 2 TBn+Rn = Bn. Similarly we can give the following theorem.
bc bc bc 2 Bn Theorem 2.5. Bn is a balcobalancing number if and only if (Bn ) + 2 is a triangular number, that is, bc bc 2 Bn T bc bc = (B ) + . Bn +Rn n 2 Proof. From (7), we get n2 = 2nr + r(r + 1) and hence
(n + r)(n + r + 1) n = n2 + . 2 2 So the result is obvious.
As in Theorem 2.5, we can give the following result.
bc bc bc 2 bc bc bc Bn Theorem 2.6. Bn is a balcobalancing number if and only if (Rn ) + 2Bn Rn + Rn + 2 is a triangular number, that is,
bc bc 2 bc bc bc Bn T bc bc = (R ) + 2B R + R + . Bn +Rn n n n n 2
2 (n+r)(n+r+1) 2 n Proof. Since n = 2nr + r(r + 1) ⇔ 2 = r + 2nr + r + 2 , the result is clear. We notice that the sum of nth balancing number and it is balancer is equal to the half of the th Cn−1 n Lucas-balancing number −1, that is, Bn + Rn = 2 . Similarly we can give the following result. Theorem 2.7. The sum of nth balcobalancing number and it is balancer is equals to the (2n+1)st bc bc th cobalancing number, that is, Bn + Rn = b2n+1, and the difference of n balcobalancing number nd bc bc and it is balancer is equals to the (2n) balancing number, that is, Bn − Rn = B2n. Proof. From Theorem 2.3, we get the desired result.
bc Theorem 2.8. Rn is a perfect square for every n ≥ 1.
6 bc −B2n+b2n+1 Proof. Notice that Rn = 2 by Theorem 2.3. So we get r p −B2n + b2n+1 Rbc = n 2
s 4n 4n 4n+1 4n+1 − α √−β + α √−β − 1 = 4 2 4 2 2 2 r α4n + β4n − 2 = 8 s α2n − β2n 2 = 2( √ ) 4 2
= 2Bn as we wanted.
Further we can give the following theorem. Theorem 2.9. The ration of the nth balcobalancing number to the nth balancing number is
bc Bn = 4bn+1 + 2 Bn and the ration of the nth Lucas-balcobalancing number to the nth Lucas-balancing number is
Cbc 2b + 1 n = 2n+1 Cn 2Bn + 2bn + 1 for n ≥ 1. The ration of the nth balcobalancer to the nth balancer is
2 8C n B n 2 2 for even n ≥ 2 c n Rbc 2 n = R 2 n 2(2b n+1 +1) c n+1 2 2 for odd n ≥ 3. B n−1 2
Proof. It can be easily derived from Theorem 2.3.
3 Binet Formulas, Recurrence Relations and Companion Mat- rix.
Theorem 3.1. The Binet formulas for balcobalancing numbers, Lucas-balcobalancing numbers and balcobalancers are α4n+1 + β4n+1 1 α4n+1 − β4n+1 α4n + β4n 1 Bbc = − ,Cbc = √ and Rbc = − n 8 4 n 2 2 n 8 4 for n ≥ 1.
7 2n 2n 2n−1 2n−1 Proof. Note that B = α √−β and b = α √−β − 1 . So we get from Theorem 2.3 that n 4 2 n 4 2 2 B + b Bbc = 2n 2n+1 n 2 4n 4n 4n+1 4n+1 α √−β + α √−β − 1 = 4 2 4 2 2 2 1+α −1−β α4n( √ ) + β4n( √ ) 1 = 4 2 4 2 − √ √2 √ 4√ 2(1+ 2) 2(1− 2) α4n( √ ) + β4n( √ ) 1 = 4 2 4 2 − 2 4 α4n+1 + β4n+1 1 = − . 8 4 The others can be proved similarly.
bc bc bc Theorem 3.2. Bn ,Cn and Rn satisfy the recurrence relations
bc bc bc bc Bn = 35(Bn−1 − Bn−2) + Bn−3 bc bc bc bc Rn = 35(Rn−1 − Rn−2) + Rn−3 for n ≥ 3 and
bc bc bc Cn = 34Cn−1 − Cn−2 for n ≥ 2.
bc α4n+1+β4n+1 1 −3 −7 −11 Proof. Recall that Bn = 8 − 4 by Theorem 3.1. Since 35α − 35α + α = α and 35β−3 − 35β−7 + β−11 = β, we conclude that
bc bc bc 35(Bn−1 − Bn−2) + Bn−3 α4n−3 + β4n−3 1 α4n−7 + β4n−7 1 α4n−11 + β4n−11 1 = 35 ( − ) − ( − ) + − 8 4 8 4 8 4 α4n(35α−3 − 35α−7 + α−11) + β4n(35β−3 − 35β−7 + β−11) 1 = − 8 4 α4n+1 + β4n+1 1 = − 8 4 bc = Bn The others can be proved similarly.
Recall that the companion matrix for balancing numbers is " # 6 −1 M = . 1 0
It can be easily seen that the nth power of M is " # B −B M n = n+1 n Bn −Bn−1
8 bc bc bc bc bc bc bc bc for n ≥ 1. Since Bn = 35(Bn−1 − Bn−2) + Bn−3 and Rn = 35(Rn−1 − Rn−2) + Rn−3 by Theorem 3.2, the companion matrix for balcobalancing numbers and balcobalancers are same and is 35 −35 1 bc M = 1 0 0 0 1 0
bc bc bc and since Cn = 34Cn−1 − Cn−2, the companion matrix for Lucas-balcobalancing numbers is " # 34 −1 N bc = . 1 0
Hence we can give the following theorem. Theorem 3.3. The nth power of M bc is
n n−2 2 n 2 PB −PB P B 4i+1 2i+1 4i+3 i=0 i=1 i=0 n−2 n−2 n−1 bc n 2 2 (M ) = P B − P B P B 4i+3 2i+1 4i+1 i=0 i=1 i=0 n−2 n−4 n−2 P2 P P2 B4i+1 − B2i+1 B4i+3 i=0 i=1 i=0 for even n ≥ 4 or n−1 n−1 2 n 2 P B −PB P B 4i+3 2i+1 4i+1 i=0 i=1 i=0 n−1 n−3 n−1 bc n 2 2 (M ) = P B − P B P B 4i+1 2i+1 4i+3 i=0 i=1 i=0 n−3 n−3 n−2 P2 P P2 B4i+3 − B2i+1 B4i+1 i=0 i=1 i=0 for odd n ≥ 3, and the nth power of N bc is
n+1 n P i+1 P i+1 (−1) B2i−1 (−1) B2i−1 i=1 i=1 bc n n (N ) = (−1) n n−1 P i+1 P i+1 − (−1) B2i−1 − (−1) B2i−1 i=1 i=1 for every n ≥ 1.
Proof. It can be proved by induction on n.
9 We can rewrite the nth power of M bc and N bc in terms of balancing and Lucas-balancing numbers instead of sums of balancing numbers. For this purpose, we set the integer sequences −8B + 3C − 3 −288B − 102C + 102 k = 2n 2n and l = 2n 2n n 96 n 96 for n ≥ 0. Then we can give the following theorem. Theorem 3.4. The nth power of M bc is kn+2 ln kn+1 bc n (M ) = kn+1 ln−1 kn kn ln−2 kn−1
for every n ≥ 2, and the nth power of N is kn+2 − kn+1 kn − kn+1 for even n ≥ 2 −k + k −k + k n n+1 n n−1 (N bc)n = (−1)n k − k k − k n+1 n+2 n+1 n for odd n ≥ 1. −kn+1 + kn −kn−1 + kn
Proof. It can be proved by induction on n.
4 Sums of Balcobalancing Numbers.
bc bc bc Theorem 4.1. The sum of first n−terms of Bn ,Cn and Rn is
n X b2n+2 − 2n − 2 Bbc = i 8 i=1 n X c2n+2 − 7 Cbc = i 8 i=1 n X B2n+1 − 2n − 1 Rbc = i 8 i=1 for n ≥ 1.
bc α4n+1+β4n+1 1 Proof. Recall that Bn = 8 − 4 by Theorem 3.1. So
n n X X α4i+1 + β4i+1 1 Bbc = ( − ). (11) i 8 4 i=1 i=1
10 n n −α3(1−α4n) β3(1−β4n) Here we notice that Pα4i+1 = √ and Pβ4i+1 = √ . So (11) becomes 4 2 4 2 i=1 i=1
n n X X α4i+1 + β4i+1 1 Bbc = ( − ) i 8 4 i=1 i=1 −α3(1−α4n) β3(1−β4n) √ + √ n = 4 2 4 2 − 8 4 α4n+3 − β4n+3 − α3 + β3 n = √ − 32 2 4 α4n+3 − β4n+3 5 n = √ − − 32 2 16 4 α4n+3−β4n+3 1 1 ( √ − ) + 5 n = 4 2 2 2 − − 8 16 4 b − 2n − 2 = 2n+2 . 8 The others can be proved similarly.
In [13], Panda and Ray proved that the sum of first 2n − 1 Pell numbers is equals to the sum of nth balancing number and its balancer, that is,
2n−1 X Pi = Bn + Rn. (12) i=1
Later in [4], Gozeri,¨ Ozkoc¸and¨ Tekcan proved that the sum of Pell-Lucas numbers from 0 to 2n − 1 is equals to the sum of the nth Lucas-balancing and the nth Lucas-cobalancing number, that is, 2n−1 X Qi = Cn + cn. i=0 As in (12), we can give the following theorem. Theorem 4.2. The sum of even ordered Pell numbers from 1 to (2n) is equals to the sum of the nth balcobalancing numbers and its balancer, that is,
2n X bc bc P2i = Bn + Rn . i=1
2n 2n n n −α(1−α4n) −β(1−β4n) Proof. Recall that P = α √−β . Since Pα2i = and Pβ2i = , we observe n 2 2 2 2 i=1 i=1 that
2n 2n X X α2i − β2i P2i = ( √ ) i=1 i=1 2 2 4n 4n −α(1−α ) − −β(1−β ) = 2 √ 2 2 2
11 α4n+1 − β4n+1 1 = √ − 4 2 2 α4n+1(1 + α−1) + β4n+1(1 + β−1) 1 = − 8 2 α4n+1 + β4n+1 1 α4n + β4n 1 = − + − 8 4 8 4 bc bc = Bn + Rn as we claimed.
Similarly we can give the following theorem which can be proved similarly. Theorem 4.3. For the sums of Pell, Pell-Lucas and balancing numbers, we have
1. the sum of odd ordered Pell numbers from 1 to (2n) is equals to the difference of the nth balcobalancing number and its balancer, that is,
2n X bc bc P2i−1 = Bn − Rn . i=1
2. the half of the sum of Pell numbers from 1 to (4n) is equals to the nth balcobalancing number, that is, 4n P Pi i=1 = Bbc. 2 n 3. the sum of Pell-Lucas numbers from 0 to (4n + 1) is equals to the sum of the twelve times of the nth balcobalancing number, four times of the its balancer plus 4, that is,
4n+1 X bc bc Qi = 12Bn + 4Rn + 4. i=0
4. the sum of Pell-Lucas numbers from 1 to (4n) is equals to the two times of the nth Lucas- balcobalancing number mines 1, that is,
4n X bc Qi = 2(Cn − 1). i=1
5. the sum of balancing numbers from 1 to (4n + 1) is equals to the product of the three times of the nth balcobalancing number, its balancer plus 1 and the four times of the nth balcobalancing number plus 1, that is,
4n+1 X bc bc bc Bi = (3Bn + Rn + 1)(4Bn + 1). i=1
12 In [19], Santana and Diaz-Barrero proved that the sum of first nonzero 4n + 1 terms of Pell numbers is a perfect square, that is,
4n+1 " n ! #2 X X 2n + 1 P = 2i . i 2i i=1 i=0 In fact this sum equals to the square of the (n + 1)st Lucas-cobalancing number, that is,
4n+1 X 2 Pi = cn+1. i=1 Similarly we can give the following result. Theorem 4.4. The sum of Pell numbers from 1 to (8n + 1) is a perfect square and is
8n+1 X bc 2 Pi = (4Bn + 1) . i=1
Proof. It can be proved as in the same way that Theorems 4.1 and 4.2 were proved.
Also they proved that
2n 2n X X P2n+1 P2i+1 and P2n P2i−1 . i=0 i=1 Similarly we can give the following result. 4n bc P Theorem 4.5. Cn P2i+1 . i=0 Proof. It can be easily derived that
4n X bc bc P2i+1 = Cn (4Bn + 1). i=0 So the result is obvious.
Apart from Theorem 4.4, we can give the following theorem which can be proved similarly. Theorem 4.6. For the sums of Pell, Pell-Lucas, balancing and Lucas-cobalancing numbers, we have
1. the sum of Pell numbers from 1 to (8n + 3) plus 1 is a perfect square and is
8n+3 X bc bc 2 1 + Pi = (4Bn + 2Cn + 1) . i=1
2. the sum of odd ordered Pell-Lucas numbers from 1 to (4n + 2) is a perfect square and is
4n+2 X bc bc 2 Q2i−1 = (8Bn + 2Cn + 2) . i=1
13 3. the half of the sum of odd Pell-Lucas numbers from 0 to (4n) is a perfect square and is
4n P Q2i+1 i=0 = (4Bbc + 1)2. 2 n 4. the sum of odd ordered balancing numbers from 1 to (2n + 1) is a perfect square and is
2n+1 X bc bc 2 B2i−1 = (3Bn + Rn + 1) i=1 and the four times of the sum of odd ordered balancing numbers from 1 to n is a perfect square and is n X bc 4 B2i−1 = Rn (by Theorem 2.8) i=1 5. the sum of Lucas-cobalancing numbers from 1 to (4n + 2) plus 1 is a perfect square and is
4n+2 X bc bc 2 1 + ci = (8Bn + 4Rn + 3) . i=1
5 Relationship with Square Triangular Numbers.
Recall that there are infinitely many triangular numbers that are also square numbers which are
called square triangular numbers and are denoted by Sn. For example, 1, 36, 1225, 41616,... are square triangular numbers. th For the n square triangular number Sn, we can write t (t + 1) S = s2 = n n , n n 2
where sn and tn are the sides of the corresponding square and triangle. Their Binet formulas are α4n + β4n − 2 α2n − β2n α2n + β2n − 2 Sn = , sn = √ and tn = (13) 32 4 2 4 for n ≥ 1 (see [2, 11]). In [21], the first author gave the general terms of almost balancing numbers in terms of square triangular numbers. Similarly, we can give the general terms of balcobalancing numbers, Lucas- balcobalancing numbers and balcobalancers in terms of squares and triangles as follows. Theorem 5.1. The general terms of balcobalancing numbers, Lucas-balcobalancing numbers and balcobalancers are 2s − t − 1 Bbc = 2n+1 2n+1 n 2 bc Cn = −2s2n+1 + 2t2n+1 + 1 −4s + 3t + 1 Rbc = 2n+1 2n+1 n 2 for n ≥ 1.
14 bc B2n+b2n+1 Proof. Since Bn = 2 by Theorem 2.3, we find that B + b Bbc = 2n 2n+1 n 2 4n 2n 4n+1 4n+1 α √−β + α √−β − 1 = 4 2 4 2 2 2 α4n+1 + β4n+1 − 2 = 8 α4n+2( √1 − 1 ) + β4n+2(− √1 − 1 ) − 1 = 2 2 4 2 2 4 2 2 4n+2 4n+2 4n+2 4n+2 2( α √−β ) − ( α +β −2 ) − 1 = 4 2 4 2 2s − t − 1 = 2n+1 2n+1 2 by (13). The others can be proved similarly.
Finally we can give the following result.
bc Rn Theorem 5.2. Sn = 4 for n ≥ 1. Proof. Appyling (13), we get
4n 4n α4n + β4n − 2 α +β − 1 Rbc S = = 8 4 = n n 32 4 4 by Theorem 3.1.
References
[1] Barbeau, E.J. (2003). Pell’s Equation. Springer-Verlag New York, Inc.
[2] Behera, A. & Panda, G.K. (1999). On the Square Roots of Triangular Numbers. The Fi- bonacci Quart., 37(2), 98-105.
[3] Flath, D.E. (1989). Introduction to Number Theory. Wiley.
[4] Gozeri,¨ G.K., Ozkoc¸,¨ A. & Tekcan, A. (2017). Some Algebraic Relations on Balancing Numbers. Utilitas Mathematica, 103, 217-236.
[5] Kovacs, T., Liptai, K. & Olajos, P. (2010). On (a, b)-Balancing Numbers. Publ. Math. Deb., 77(3-4), 485-498.
[6] Liptai, K., Luca, F., Pinter, A. & Szalay, L. (2009). Generalized Balancing Numbers. Indag. Mathem. N.S., 20(1), 87-100.
[7] Liptai, K. (2004). Fibonacci Balancing Numbers. The Fibonacci Quarterly, 42(4), 330-340.
[8] Liptai, K. (2006). Lucas Balancing Numbers. Acta Math. Univ. Ostrav., 14, 43-47.
15 [9] Mollin, R.A. (1996). Quadratics. CRS Press, Boca Raton, New York, London, Tokyo.
[10] Olajos, P. (2010). Properties of Balancing, Cobalancing and Generalized Balancing Num- bers. Annales Mathematicae et Informaticae, 37, 125-138.
[11] Ozkoc¸,¨ A., Tekcan, A. & Gozeri,¨ G.K. (2017). Triangular and Square Triangular Numbers Involving Generalized Pell Numbers. Utilitas Mathematica, 102, 231-254.
[12] Panda, G.K. & Panda, A.K. (2015). Almost Balancing Numbers. Journal of the Indian Math. Soc., 82 (3-4), 147-156.
[13] Panda, G.K. & Ray, P.K. (2011). Some Links of Balancing and Cobalancing Numbers with Pell and Associated Pell Numbers. Bul. of Inst. of Math. Acad. Sinica, 6(1), 41-72.
[14] Panda, G.K. & Ray, P.K. (2005). Cobalancing Numbers and Cobalancers. Int. Jour. Math. Math. Sci., 8, 1189-1200.
[15] Panda, G.K., Komatsu, T. & Davala, R.K. (2018). Reciprocal Sums of Sequences Involving Balancing and Lucas-balancing Numbers. Math. Reports., 20(70)(2018), 201-214.
[16] Panda, A.K. (2017). Some Variants of the Balancing Sequences. Ph.D. thesis, National In- stitute of Technology Rourkela, India.
[17] Ray, P.K. (2009). Balancing and Cobalancing Numbers. Ph.D. thesis, Department of Math- ematics, National Institute of Technology, Rourkela, India.
[18] Ray, P.K. (2015). Balancing and Lucas-balancing Sums by Matrix Methods. Math. Reports., 17(67), 225-233.
[19] Santana, S.F. & Diaz-Barrero, J.L. (2006). Some Properties of Sums Involving Pell Num- bers. Missouri Journal of Mathematical Science., 18(1), 33-40.
[20] Szalay, L. (2007). On the Resolution of Simultaneous Pell Equations. Ann. Math. Inform., 34, 77-87.
[21] Tekcan, A. (2019). Almost Balancing, Triangular and Square Triangular Numbers. Notes on Number Theory and Discrete Maths., 25(1), 108-121.
[22] Tekcan, A. (2019). Sums and Spectral Norms of all Almost Balancing Numbers. Creative Mathematics and Informatics, 28(2), 203-214.
[23] Tengely, S. (2013). Balancing Numbers which are Products of Consecutive Integers. Publ. Math. Deb., 83(1-2), 197-205.
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