CREAT. MATH. INFORM. Volume 30 (2021), No. 2, Pages 203–222 Online version at https://creative-mathematics.cunbm.utcluj.ro/ Print Edition: ISSN 1584 - 286X; Online Edition: ISSN 1843 - 441X DOI: https://doi.org/10.37193/CMI.2021.02.11 Special issue “30 years of publication of CMI Balcobalancing numbers and balcobalancers AHMET TEKCAN and MERYEM YILDIZ ABSTRACT. In this work, we determine the general terms of balcobalancing numbers, balcobalancers and also Lucas–balcobalancing numbers in terms of balancing numbers. Further we formulate the sums of these numbers and derive some relations associated with Pell, Pell–Lucas and square triangular numbers. 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) (1.1) holds for some positive integer r which is called balancer corresponding to n. If n is a balancing number with balancer r, then from (1.1) r = −2n− 1 + √ 8n2 + 1 2 . (1.2) From (1.2) 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 (1.2) we note that 8(0)2 + 1 = 1 and 8(1)2 + 1 = 32 are perfect squares. So we accept 0 and 1 to be balancing numbers. Let Bn denote the nth balancing number. Then B0 = 0, B1 = 1, B2 = 6 and Bn+1 = 6Bn −Bn−1 for n ≥ 2. Later Panda and Ray ([12]) defined that a positive integer n is called a cobalancing number if the Diophantine equation 1 + 2 + · · ·+ n = (n+ 1) + (n+ 2) + · · ·+ (n+ r) (1.3) holds for some positive integer r which is called cobalancer corresponding to n. If n is a cobalancing number with cobalancer r, then from (1.3) r = −2n− 1 + √ 8n2 + 8n+ 1 2 . (1.4) From (1.4) 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 to be balancing numbers. Cobalancing number is denoted by bn, and b0 = b1 = 0, b2 = 2 and bn+1 = 6bn − bn−1 + 2 for n ≥ 2. It is clear from (1.1) and (1.3) that every balancing number is a cobalancer and every cobalancing number is a balancer, that is, Bn = rn+1 and Rn = bn for n ≥ 1, where Rn is Received: 03.05.2021. In revised form: 25.07.2021. Accepted: 02.08.2021 2010 Mathematics Subject Classification. 11B37, 11B39, 11D09, 11D79. Key words and phrases. Balancing numbers, cobalancing numbers, square triangular numbers, Pell equations. Corresponding author: Ahmet Tekcan; tekcan@uludag.edu.tr 203 204 Ahmet Tekcan and Meryem Yıldız the nth the balancer and rn is the nth cobalancer. Since Rn = bn, we get from (1.1) that bn = −2Bn − 1 + √ 8B2 n + 1 2 and Bn = 2bn + 1 + √ 8b2n + 8bn + 1 2 . (1.5) Thus from (1.5), we see that Bn is a balancing number if and only if 8B2 n + 1 is a perfect square and bn is a cobalancing number if and only if 8b2n + 8bn + 1 is a perfect square. So Cn = √ 8B2 n + 1 and cn = √ 8b2n + 8bn + 1 (1.6) are integers which are called the Lucas–balancing number and Lucas–cobalancing num- ber, 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. Ray ([16]) 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 y2−8x2 = 1 which is a Pell equation ([1, 3, 9]). The fundamental solution is (y1, x1) = (3, 1). So yn+xn √ 8 = (3+ √ 8)n for n ≥ 1 and similarly yn−xn √ 8 = (3− √ 8)n. Let γ = 3+ √ 8 and δ = 3 − √ 8. Then he get xn = γn−δn γ−δ which is the Binet formula for balancing numbers, that is, Bn = γn−δn γ−δ . Since α2 = γ and β2 = δ, he conclude that the Binet formula for balancing numbers is Bn = α2n−β2n 4 √ 2 . Similarly he get bn = α2n−1−β2n−1 4 √ 2 − 1 2 , Cn = α2n+β2n 2 and cn = α2n−1+β2n−1 2 for n ≥ 1 (see also [10, 11, 15]). 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 [19], Szalay considered the same problem and obtained some nice results by a different method. In [5], Kovács, Liptai and Olajos extended the concept of balancing numbers to the (a, b)−ba- lancing 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) for some positive integers n and r, then an+b is an (a, b)−balancing number. The sequence of (a, b)−balancing numbers is denoted byB(a,b) m form ≥ 1. In [6], Liptai, Luca, Pintér and Szalay generalized the notion of balancing numbers to numbers defined as follows: Let y, k, l ∈ Z+ with 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 and proved that this Diophan- tine equation has no solution for m ≥ 0 and x ∈ Z. In [14], Panda, Komatsu and Davala considered the reciprocal sums of sequences involving balancing and Lucas–balancing numbers and in [17], Ray considered the sums of balancing and Lucas–balancing num- bers by matrix methods. In [13], Panda and Panda defined the almost balancing number Balcobalancing numbers 205 and its balancer. In [21], the first author considered amost balancing numbers, triangular numbers and square triangular numbers and in [22], he considered the sums and spectral norms of all almost balancing numbers. 2. RESULTS. In this work, we set three new integer sequences called balcobalancing number, bal- cobalancer and Lucas–balcobalancing number and try to determine the general terms of them in terms of balancing numbers. We also want to derive some relations with Pell, Pell–Lucas and square triangular numbers. If we sum both sides of (1.1) and (1.3), then we get the Diophantine equation 1 + 2 + · · ·+ (n− 1) + 1 + 2 + · · ·+ (n− 1) + n = 2[(n+ 1) + (n+ 2) + · · ·+ (n+ r)]. (2.7) Thus a positive integer n is called a balcobalancing number if the Diophantine equation in (2.7) verified for some positive integer r which is called balcobalancer. For example, 10, 348, 11830, · · · are balcobalancing numbers with balcobalancers 4, 144, 4900, · · · . (Here we want to use name “balcobalancing” since it comes from balancing and cobalancing numbers). From (2.7), we get r = −2n− 1 + √ 8n2 + 4n+ 1 2 . (2.8) Let Bbcn denote the balcobalancing number and let Rbcn denote the balcobalancer. Then from (2.8), Bbcn is a balcobalancing number if and only if 8(Bbcn )2 + 4Bbcn + 1 is a perfect square. Thus Cbcn = √ 8(Bbcn )2 + 4Bbcn + 1 (2.9) is an integer which are called the Lucas–balcobancing number. (Here we notice that bal- cobalancing numbers should be greater that 0. But in (2.9), 8(0)2 + 4(0) + 1 = 1 is a perfect square, so we assume that 0 is a balcobalancing number, that is, Bbc0 = 0. In this case, Rbc0 = 0 and Cbc0 = 1). In order to determine the general terms of balcobalancing numbers, balcobalancers and Lucas–balcobalancing numbers we have to determine the set of all (positive) integer solutions of the Pell equation x2 − 2y2 = −1. (2.10) We see from (2.8) that Bbcn is a balcobalancing number if and only if 8(Bbcn )2 + 4Bbcn + 1 is a perfect square. So we set 8(Bbcn )2 + 4Bbcn + 1 = y2 for some integer y ≥ 1. If we multiply both sides of the last equation by 2, then we get 16(Bbcn )2 + 8Bbcn + 2 = 2y2 and hence (4Bbcn + 1)2 + 1 = 2y2. Taking x = 4Bbcn + 1, we get the Pell equation in (2.10). Let Ω denotes the set of all integer solutions of (2.10), that is, Ω = {(x, y) : x2 − 2y2 = −1}. Then we can give the following theorem. Theorem 2.1. The set of all integer solutions of (2.10) is Ω = {(cn, 2bn + 1) : n ≥ 1}. Proof. For the Pell equation x2−2y2 = −1, the set of representatives Rep = {[±1 1]} and M = [ 3 2 4 3 ] . In this case [−1 1]Mn generates all integer solutions (xn, yn) for n ≥ 1. It can be easily seen that the nth power of M is Mn = [ Cn 2Bn 4Bn Cn ] 206 Ahmet Tekcan and Meryem Yıldız for n ≥ 1. So [xn yn] = [−1 1] [ Cn 2Bn 4Bn Cn ] = [−Cn + 4Bn − 2Bn + Cn]. Thus the set of all integer solutions is Ω = {(−Cn + 4Bn,−2Bn + Cn) : n ≥ 1}. But it can be easily seen that −Cn + 4Bn = cn and −2Bn + Cn = 2bn + 1. So we conclude that the set of all integer solutions of (2.10) is Ω = {(cn, 2bn + 1) : n ≥ 1}. � 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 Bbcn = c2n+1 − 1 4 , Cbcn = 2b2n+1 + 1 and Rbcn = 4b2n+1 − c2n+1 + 1 4 for n ≥ 1. Proof. We proved in Theorem 2.1 that Ω = {(cn, 2bn + 1) : n ≥ 1}. Since x = 4Bbcn + 1, we get Bbcn = x2n+1 − 1 4 = c2n+1 − 1 4 for n ≥ 1. Thus from (2.9), we obtain Cbcn = √ 8(Bbcn )2 + 4Bbcn + 1 = √ 8( c2n+1 − 1 4 )2 + 4( c2n+1 − 1 4 ) + 1 = √ c22n+1 + 1 2 = √ (α 4n+1+β4n+1 2 )2 + 1 2 = √[ 2( α4n+1 − β4n+1 2 √ 2 − 1 2 ) + 1 ]2 = 2b2n+1 + 1. Finally from (2.8), we deduce that Rbcn = 4b2n+1 − c2n+1 + 1 4 as we wanted. � We can also give the general terms of balcobalancing numbers, Lucas–balcobalancing numbers 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 Bbcn = B2n + b2n+1 2 , Cbcn = 2b2n+1 + 1 and Rbcn = −B2n + b2n+1 2 for n ≥ 1. Balcobalancing numbers 207 Proof. We proved in Theorem 2.2 that Bbcn = c2n+1−1 4 . So we easily deduce that Bbcn = c2n+1 − 1 4 = α4n+1 + β4n+1 8 − 1 4 = α4n+1(α −1+1 4 √ 2 ) + β4n+1(−β −1−1 4 √ 2 ) 2 − 1 4 = α4n−β4n 4 √ 2 + α4n+1−β4n+1 4 √ 2 − 1 2 2 = B2n + b2n+1 2 . Cbcn = 2b2n+1 + 1 is already proved in Theorem 2.2. Similarly it can be proved that Rbcn = −B2n+b2n+1 2 . � As in Theorem 2.3, we can give the general terms of balcobalancing numbers, Lucas– balcobalancing numbers and balcobalancers in terms of only balancing numbers or only Lucas–balancing numbers as follows. Theorem 2.4. The general terms of balcobalancing numbers, Lucas–balcobalancing numbers and balcobalancers are Bbcn = 2Bn(Bn+1 −Bn), Cbcn = B2n+1 −B2n, R bc n = 4B2 n or Bbcn = C2n+1 − C2n − 2 8 , Cbcn = C2n+1 + C2n 4 , Rbcn = C2n − 1 4 for n ≥ 1. Proof. From Theorem 2.3, we get Bbcn = B2n + b2n+1 2 = α4n−β4n 4 √ 2 + α4n+1−β4n+1 4 √ 2 − 1 2 2 = α4n+1 + β4n+1 8 − 1 4 = α4n+1 − α2nβ2n+1 − β2nα2n+1 + β4n+1 8 = 2 ( α2n − β2n 4 √ 2 )( α2n+1 − β2n+1 2 √ 2 ) = 2 ( α2n − β2n 4 √ 2 )( α2n(α2 − 1)− β2n(β2 − 1) 4 √ 2 ) = 2 ( α2n − β2n 4 √ 2 )( α2n+2 − β2n+2 4 √ 2 − α2n − β2n 4 √ 2 ) = 2Bn(Bn+1 −Bn). The others can be proved similarly. � In Theorems 2.2, 2.3 and 2.4, we can give the general terms of balcobalancing numbers, Lucas–balcobalancing numbers and balcobalancers in terms of balancing, cobalancing, 208 Ahmet Tekcan and Meryem Yıldız Lucas–balancing and Lucas–cobalancing numbers. Conversely, we can give the gene- ral terms of balancing, cobalancing, Lucas–balancing and Lucas–cobalancing numbers in terms of balcobalancing numbers, Lucas–balcobalancing numbers and balcobalancers as follows. Theorem 2.5. The general terms of balancing, cobalancing, Lucas–balancing and Lucas–cobalan- cing numbers are Bn = { Bbcn 2 −Rbcn 2 n ≥ 2 even (Bbcn+1 2 +Bbcn−1 2 − 2Rbcn+1 2 )/2 n ≥ 1 odd bn = { −Bbcn 2 + 3Rbcn 2 n ≥ 2 even Bbcn−1 2 +Rbcn−1 2 n ≥ 1 odd Cn = { −4Bbcn 2 + 2Cbcn 2 − 1 n ≥ 2 even 4Bbcn−1 2 + 2Cbcn−1 2 + 1 n ≥ 1 odd cn = { 12Bbcn 2 − 4Cbcn 2 + 3 n ≥ 2 even 4Bbcn−1 2 + 1 n ≥ 1 odd. Proof. From Theorem 2.3, we get Bbcn = B2n+b2n+1 2 and Rbcn = −B2n+b2n+1 2 . Thus we get B2n = Bbcn −Rbcn and hence Bn = Bbcn 2 −Rbcn 2 for even n ≥ 2. The others can be proved similarly. � Thus we construct one-to-one correspondence between all balcobalancing numbers and all balancing numbers. 3. BINET FORMULAS, RECURRENCE RELATIONS AND COMPANION MATRIX. Theorem 3.6. Binet formulas for balcobalancing numbers, Lucas–balcobalancing numbers and balcobalancers are Bbcn = α4n+1 + β4n+1 8 − 1 4 , Cbcn = α4n+1 − β4n+1 2 √ 2 and Rbcn = α4n + β4n 8 − 1 4 for n ≥ 1. Proof. Since Bn = α2n−β2n 4 √ 2 and bn = α2n−1−β2n−1 4 √ 2 − 1 2 , we get from Theorem 2.3 that Bbcn = B2n + b2n+1 2 = α4n−β4n 4 √ 2 + α4n+1−β4n+1 4 √ 2 − 1 2 2 = α4n( 1+α 4 √ 2 ) + β4n(−1−β 4 √ 2 ) 2 − 1 4 = α4n+1 + β4n+1 8 − 1 4 . The others can be proved similarly. � Recall that balancing numbers satisfy recurrence relationBn = 6Bn−1−Bn−2 for n ≥ 2. Similarly we can give the following result. Balcobalancing numbers 209 Theorem 3.7. Bbcn , Cbcn and Rbcn satisfy the recurrence relations Bbcn = 35(Bbcn−1 −Bbcn−2) +Bbcn−3 Rbcn = 35(Rbcn−1 −Rbcn−2) +Rbcn−3 for n ≥ 3 and Cbcn = 34Cbcn−1 − Cbcn−2 for n ≥ 2. Proof. Recall that Bbcn = α4n+1+β4n+1 8 − 1 4 by Theorem 3.6. Since 35α−3−35α−7 +α−11 = α and 35β−3 − 35β−7 + β−11 = β, we get 35(Bbcn−1 −Bbcn−2) +Bbcn−3 = 35 [ ( α4n−3 + β4n−3 8 − 1 4 )− ( α4n−7 + β4n−7 8 − 1 4 ) ] + α4n−11 + β4n−11 8 − 1 4 = α4n(35α−3 − 35α−7 + α−11) + β4n(35β−3 − 35β−7 + β−11) 8 − 1 4 = α4n+1 + β4n+1 8 − 1 4 = Bbcn The others can be proved similarly. � Recall that the companion matrix for balancing numbers is M = [ 6 −1 1 0 ] . It can be easily seen that the nth power of M is Mn = [ Bn+1 −Bn Bn −Bn−1 ] (3.11) for n ≥ 1. Since Bbcn = 35(Bbcn−1 − Bbcn−2) + Bbcn−3 and Rbcn = 35(Rbcn−1 − Rbcn−2) + Rbcn−3 by Theorem 3.7, the companion matrix for balcobalancing numbers and balcobalancers are same and is M bc =  35 −35 1 1 0 0 0 1 0  and sinceCbcn = 34Cbcn−1−Cbcn−2, the companion matrix for Lucas–balcobalancing numbers is N bc = [ 34 −1 1 0 ] . As in (3.11), we can give the following theorem. 210 Ahmet Tekcan and Meryem Yıldız Theorem 3.8. The nth power of M bc is (M bc)n =  n 2∑ i=0 B4i+1 − n∑ i=1 B2i+1 n−2 2∑ i=0 B4i+3 n−2 2∑ i=0 B4i+3 − n−1∑ i=1 B2i+1 n−2 2∑ i=0 B4i+1 n−2 2∑ i=0 B4i+1 − n−2∑ i=1 B2i+1 n−4 2∑ i=0 B4i+3  for even n ≥ 4 or (M bc)n =  n−1 2∑ i=0 B4i+3 − n∑ i=1 B2i+1 n−1 2∑ i=0 B4i+1 n−1 2∑ i=0 B4i+1 − n−1∑ i=1 B2i+1 n−3 2∑ i=0 B4i+3 n−3 2∑ i=0 B4i+3 − n−2∑ i=1 B2i+1 n−3 2∑ i=0 B4i+1  for odd n ≥ 3, and the nth power of N bc is (N bc)n = (−1)n  n+1∑ i=1 (−1)i+1B2i−1 n∑ i=1 (−1)i+1B2i−1 − n∑ i=1 (−1)i+1B2i−1 − n−1∑ i=1 (−1)i+1B2i−1  for every n ≥ 1. Proof. It can be proved by induction on n. � We can rewrite the nth power ofM bc andN bc in terms of balancing and Lucas–balancing numbers instead of sums of balancing numbers. For this purpose, we set two integer se- quences kn and ln to be kn = −8B2n + 3C2n − 3 96 and ln = −288B2n − 102C2n + 102 96 for n ≥ 0. Then we can give the following theorem. Theorem 3.9. The nth power of M bc is (M bc)n =  kn+2 ln kn+1 kn+1 ln−1 kn kn ln−2 kn−1  for every n ≥ 2, and the nth power of N bc is Balcobalancing numbers 211 (N bc)n = (−1)n   kn+2 − kn+1 kn − kn+1 −kn + kn+1 −kn + kn−1  for even n ≥ 2  kn+1 − kn+2 kn+1 − kn −kn+1 + kn −kn−1 + kn  for odd n ≥ 1. Proof. It can be proved by induction on n. � 4. RELATIONSHIP WITH PELL AND PELL–LUCAS NUMBERS. Recall that general terms of all balancing numbers can be given in terms of Pell num- bers Bn = P2n 2 , bn = P2n−1 − 1 2 , Cn = P2n + P2n−1 and cn = P2n−1 + P2n−2 and also in terms of Pell–Lucas numbers Bn = Q2n +Q2n−1 8 , bn = Q2n −Q2n−1 − 4 8 , Cn = Q2n 2 and cn = Q2n−1 2 . Similarly we can give the following theorem. Theorem 4.10. The general terms of balcobalancing numbers, Lucas–balcobalancing numbers and balcobalancers are Bbcn = P2n+1P2n, C bc n = P 2 2n+1 + P 2 2n, R bc n = P 2 2n or Bbcn = Q2n+1Q2n − 4 8 , Cbcn = Q2 2n+1 +Q2 2n 8 , Rbcn = ( Q2n+1 −Q2n 4 )2 for n ≥ 1. Proof. We deduce from Theorem 2.3 that Bbcn = B2n + b2n+1 2 = α4n−β4n 4 √ 2 + α4n+1−β4n+1 4 √ 2 − 1 2 2 = α4n+1(α−1 + 1) + β4n+1(−1− β−1) 8 √ 2 − 1 4 = α4n+1 + β4n+1 8 − 1 4 = α4n+1 + β4n+1 − (αβ)2n(α+ β) 8 = α4n+1 − α2n+1β2n − β2n+1α2n + β4n+1 8 = ( α2n+1 − β2n+1 2 √ 2 )( α2n − β2n 2 √ 2 ) = P2n+1P2n. The others can be proved similarly. � Conversely, we can give the general terms of Pell and Pell–Lucas numbers in terms of balcobalancing numbers, Lucas–balcobalancing numbers and balcobalancers as follows. 212 Ahmet Tekcan and Meryem Yıldız Theorem 4.11. The general terms of Pell and Pell–Lucas numbers are Pn =  2(Bbcn 4 −Rbcn 4 ) n ≡ 0(mod4) Cbcn−1 4 n ≡ 1(mod4) 4Bbcn−2 4 + Cbcn−2 4 + 1 n ≡ 2(mod4) 8Bbcn−3 4 + 3Cbcn−3 4 + 2 n ≡ 3(mod4) and Qn =  8Rbcn 4 + 2 n ≡ 0(mod4) 8Bbcn−1 4 + 2 n ≡ 1(mod4) 8Bbcn−2 4 + 4Cbcn−2 4 + 2 n ≡ 2(mod4) (Cbcn+1 4 − Cbcn−3 4 )/2 n ≡ 3(mod4). Proof. It can be proved as in the same way that Theorem 2.3 was proved. � Thus we construct one-to-one correspondence between all balcobalancing numbers and Pell and Pell–Lucas numbers. 5. RELATIONSHIP WITH TRIANGULAR AND SQUARE TRIANGULAR NUMBERS. Recall that triangular numbers denoted by Tn are the numbers of the form Tn = n(n+ 1) 2 . It is known that there is a correspondence between balancing (and also cobalancing) num- bers and triangular numbers. Indeed from (1.1), we note that n is a balancing number if and only if n2 is a triangular number since (n+ r)(n+ r + 1) 2 = n2. So TBn+Rn = B2 n, (5.12) Similarly from (1.3), n is a cobalancing number if and only if n2 +n is a triangular number since (n+ r)(n+ r + 1) 2 = n2 + n. So Tbn+rn = b2n + bn. As in (5.12), we can give the following theorem. Theorem 5.12. Bbcn is a balcobalancing number if and only if (Bbcn )2+ Bbc n 2 is a triangular number, that is, TBbc n +Rbc n = (Bbcn )2 + Bbcn 2 . Proof. From (2.7), we get n2 = 2nr + r2 + r and hence (n+ r)(n+ r + 1) 2 = n2 + n 2 . So the result is obvious. � Balcobalancing numbers 213 There are infinitely many triangular numbers that are also square numbers which are called square triangular numbers and is denoted by Sn. Notice that Sn = s2n = tn(tn + 1) 2 , where sn and tn are the sides of the corresponding square and triangle. We can give the general terms of Sn, sn and tn in terms of balancing and cobalancing numbers, namely, Sn = B2 n, sn = Bn and tn = Bn + bn. Their Binet formulas are Sn = α4n + β4n − 2 32 , sn = α2n − β2n 4 √ 2 and tn = α2n + β2n − 2 4 (5.13) for n ≥ 1. We can give the general terms of balcobalancing numbers, Lucas–balcobalan- cing numbers and balcobalancers in terms of sn and tn as follows. Theorem 5.13. The general terms of balcobalancing numbers, Lucas–balcobalancing numbers and balcobalancers are Bbcn = 2s2n+1 − t2n+1 − 1 2 Cbcn = −2s2n+1 + 2t2n+1 + 1 Rbcn = −4s2n+1 + 3t2n+1 + 1 2 for n ≥ 1. Proof. From Theorem 2.3, we get Bbcn = B2n + b2n+1 2 = α4n−β4n 4 √ 2 + α4n+1−β4n+1 4 √ 2 − 1 2 2 = α4n+2( 1 2 √ 2 − 1 4 ) + β4n+2(− 1 2 √ 2 − 1 4 )− 1 2 2 = 2(α 4n+2−β4n+2 4 √ 2 )− (α 4n+2+β4n+2−2 4 )− 1 2 = 2s2n+1 − t2n+1 − 1 2 by (5.13). The others can be proved similarly. � Conversely, we can give the following theorem. Theorem 5.14. The general terms of Sn, sn and tn are Sn = Rbcn 4 sn = { Bbcn 2 −Rbcn 2 n ≥ 2 even (4Bbcn−1 2 + Cbcn−1 2 + 1)/2 n ≥ 1 odd tn = { 2Rbcn 2 n ≥ 2 even 2Bbcn−1 2 + Cbcn−1 2 n ≥ 1 odd. 214 Ahmet Tekcan and Meryem Yıldız Proof. From Theorem 2.3, we get Rbcn = −B2n + b2n+1 2 = −α 4n−β4n 4 √ 2 + α4n+1−β4n+1 4 √ 2 − 1 2 2 = α4n(−1 + α) + β4n(1− β) 8 √ 2 − 1 4 = α4n + β4n 8 − 1 4 . So from (5.13), we observe that Sn = α4n + β4n − 2 32 = α4n+β4n 8 − 1 4 4 = Rbcn 4 as we wanted. The others can be proved similarly. � Thus we construct one-to-one correspondence between all balcobalancing numbers and square triangular numbers. Finally, we want to construct a correspondence between triangular and square trian- gular numbers via balcobalancing numbers, that is, we want to find out that for which balcobalancing numbers m, the equation Tm = Sn holds. The answer is given below. Theorem 5.15. For triangular numbers Tn and square triangular numbers Sn, we have (1) if n ≥ 1 is odd, then T2Bbc n−1 2 +Cbc n−1 2 = Sn. (2) if n ≥ 2 is even, then T−2Bbc n 2 +Cbc n 2 −1 = Sn. Proof. (1) Let n ≥ 1 be odd. Then T2Bbc n−1 2 +Cbc n−1 2 = (2Bbcn−1 2 + Cbcn−1 2 )(2Bbcn−1 2 + Cbcn−1 2 + 1) 2 =  [ 2 ( α2n−1+β2n−1 8 − 1 4 ) + α2n−1−β2n−1 2 √ 2 ] ×[ 2 ( α2n−1+β2n−1 8 − 1 4 ) + α2n−1−β2n−1 2 √ 2 + 1 ]  /2 = ( α2n + β2n − 2 ) (α2n + β2n + 2) 32 = α4n + β4n − 2 32 = Sn by (5.13). The other case can be proved similarly. � Balcobalancing numbers 215 6. SUMS OF BALCOBALANCING NUMBERS. Theorem 6.16. The sum of first n−terms of Bbcn , Cbcn and Rbcn is n∑ i=1 Bbci = b2n+2 − 2n− 2 8 n∑ i=1 Cbci = c2n+2 − 7 8 n∑ i=1 Rbci = B2n+1 − 2n− 1 8 for n ≥ 1. Proof. Recall that Bbcn = α4n+1+β4n+1 8 − 1 4 by Theorem 3.6. Since n∑ i=1 α4i+1 = −α3(1− α4n) 4 √ 2 and n∑ i=1 β4i+1 = β3(1− β4n) 4 √ 2 , we get n∑ i=1 Bbci = n∑ i=1 ( α4i+1 + β4i+1 8 − 1 4 ) = −α3(1−α4n) 4 √ 2 + β3(1−β4n) 4 √ 2 8 − n 4 = α4n+3 − β4n+3 − α3 + β3 32 √ 2 − n 4 = α4n+3 − β4n+3 − 10 √ 2 32 √ 2 − n 4 = α4n+3 − β4n+3 32 √ 2 − 5 16 − n 4 = α4n+3−β4n+3 4 √ 2 − 1 2 + 1 2 8 − 5 16 − n 4 = α4n+3−β4n+3 4 √ 2 − 1 2 8 − n+ 1 4 = b2n+2 − 2n− 2 8 . The others can be proved similarly. � We can give the sums of first n−terms of balcobalancing numbers in terms of balancing numbers, sums of first n−terms of Lucas–balcobalancing numbers in terms of Lucas– balancing numbers and sums of first n−terms of balcobalancers in terms of balancers as follows. 216 Ahmet Tekcan and Meryem Yıldız Theorem 6.17. The sum of first n−terms of Bbcn , Cbcn and Rbcn is n∑ i=1 Bbci = B2n+2 −B2n+1 − 4n− 5 16 n∑ i=1 Cbci = 5C2n+1 − C2n − 14 16 n∑ i=1 Rbci = R2n+2 −R2n+1 − 4n− 2 16 for n ≥ 1. Proof. It can be easily seen that B2n+2 − B2n+1 = 2b2n+2 + 1. So from Theorem 6.16, we get n∑ i=1 Bbci = b2n+2 − 2n− 2 8 = B2n+2−B2n+1−1 2 − 2n− 2 8 = B2n+2 −B2n+1 − 4n− 5 16 . The others can be proved similarly. � Recall that the sum of first n−terms of all balancing numbers can be given in terms of same balancing numbers, that is, n∑ i=1 Bi = 5Bn −Bn−1 − 1 4 , n∑ i=1 bi = 5bn − bn−1 + 2− 2n 4 n∑ i=1 Ci = 5Cn − Cn−1 − 2 4 , n∑ i=1 ci = 5cn − cn−1 − 2 4 . Similarly we can give the sums of first n−terms of balcobalancing numbers in terms of bal- cobalancing numbers, sums of first n−terms of Lucas–balcobalancing numbers in terms of Lucas–balcobalancing numbers and sums of first n−terms of balcobalancers in terms of balcobalancers as follows. Theorem 6.18. The sum of first n−terms of Bbcn , Cbcn and Rbcn is n∑ i=1 Bbci = 33Bbcn −Bbcn−1 − 8n− 2 32 n∑ i=1 Cbci = 33Cbcn − Cbcn−1 − 28 32 n∑ i=1 Rbci = 33Rbcn −Rbcn−1 − 8n+ 4 32 for n ≥ 1. Proof. It can be proved similarly. � Balcobalancing numbers 217 We also note that n∑ i=1 (−1)iBi = { 2B2 n 2 +Bn 2 Cn 2 n ≥ 2 even −2Bn+1 2 (bn+1 2 + 1 2 ) n ≥ 1 odd n∑ i=1 (−1)ibi = { 2B2 n 2 n ≥ 2 even −2b2n+1 2 − 2bn+1 2 n ≥ 1 odd n∑ i=1 (−1)iCi = { Bn + 8B2 n 2 n ≥ 2 even −Bn − 8(bn+1 2 + 1 2 )2 n ≥ 1 odd n∑ i=1 (−1)ici = { Bn n ≥ 2 even −Bn n ≥ 1 odd. Similarly we can give the following theorem. Theorem 6.19. For Bbcn , Cbcn and Rbcn , we get n∑ i=1 (−1)iBbci = { (35Bbcn −Bbcn−1 − 2)/36 n ≥ 2 even (−35Bbcn +Bbcn−1 − 10)/36 n ≥ 1 odd n∑ i=1 (−1)iCbci = { (35Cbcn − Cbcn−1 − 30)/36 n ≥ 2 even (−35Cbcn + Cbcn−1 − 30)/36 n ≥ 1 odd n∑ i=1 (−1)iRbci = { (35Rbcn −Rbcn−1 + 4)/36 n ≥ 2 even (−35Rbcn +Rbcn−1 − 4)/36 n ≥ 1 odd. Proof. It can be proved similarly. � In [20], Tekcan and Tayat set two integer sequences Xn = αn+1 + βn+1 2 and Yn = αn+1 − βn+1 √ 2 for n ≥ 0 and proved that n∑ i=1 BiCi = XnXn−1YnYn−1 8 . It can be easily seen that n∑ i=1 BiCi = C2n+1 − 3 32 . (6.14) As in (6.14), we can give the following theorem. Theorem 6.20. For Bbcn and Cbcn , we get n∑ i=1 Bbci C bc i = (3Bbcn + Cbcn )2 − 1 12 . Proof. From Theorem 3.6, we find that n∑ i=1 Bbci C bc i = Bbc1 C bc 1 +Bbc2 C bc 2 + · · ·+Bbcn C bc n 218 Ahmet Tekcan and Meryem Yıldız = ( α5 + β5 8 − 1 4 )( α5 − β5 2 √ 2 ) + ( α9 + β9 8 − 1 4 )( α9 − β9 2 √ 2 ) + · · ·+ ( α4n+1 + β4n+1 8 − 1 4 )( α4n+1 − β4n+1 2 √ 2 ) = (α10 + α18 + · · ·+ α8n+2)− (β10 + β18 + · · ·+ β8n+2) 16 √ 2 − (α5 + α9 + · · ·+ α4n+1)− (β5 + β9 + · · ·+ β4n+1) 8 √ 2 = 1 16 √ 2 [ α10(α8n − 1) α8 − 1 − β10(β8n − 1) β8 − 1 ] − 1 8 √ 2 [ α5(α4n − 1) α4 − 1 − β5(β4n − 1) β4 − 1 ] = 1 32 [ α8n+6 + β8n+6 − 198 24 ] − 1 16 [ α4n+3 + β4n+3 − 14 4 ] = 1 32.24 [ (α4n+3 + β4n+3 − 6)2 − 64 ] = 1 12 [ 3 ( α4n+1 + β4n+1 8 − 1 4 ) + ( α4n+1 − β4n+1 2 √ 2 )]2 − 1 12 = (3Bbcn + Cbcn )2 − 1 12 . This completes the proof. � In [18], Santana and Diaz–Barrero proved that P2n+1 ∣∣∣∣∣ 2n∑ i=0 P2i+1 and P2n ∣∣∣∣∣ 2n∑ i=1 P2i−1. Similarly we can give the following theorem. Theorem 6.21. Cbcn ∣∣∣∣ 4n∑ i=0 P2i+1. Proof. As in Theorem 6.16, we find that 4n∑ i=0 P2i+1 = Cbcn (4Bbcn + 1). So the result is obvious. � 7. SUMS OF PELL AND BALANCING NUMBERS. Panda and Ray proved in [11] that the sum of first 2n − 1 Pell numbers is equal to the sum of nth balancing number and its balancer, that is, 2n−1∑ i=1 Pi = Bn + bn. (7.15) Later Gözeri, Özkoç and Tekcan proved in [4] that the sum of Pell–Lucas numbers from 0 to 2n − 1 is equal to the the sum of nth Lucas–balancing and Lucas–cobalancing number, that is, 2n−1∑ i=0 Qi = Cn + cn. Balcobalancing numbers 219 Since Rn = bn, (7.15) becomes 2n−1∑ i=1 Pi = Bn +Rn. (7.16) As in (7.16), we can give the following result. Theorem 7.22. The sum of even ordered Pell numbers from 1 to 2n is equal to the sum of the nth balcobalancing number and its balcobalancer, that is, 2n∑ i=1 P2i = Bbcn +Rbcn . Proof. Since 2n∑ i=1 α2i = −α(1−α4n) 2 and 2n∑ i=1 β2i = −β(1−β4n) 2 , we deduce that 2n∑ i=1 P2i = 2n∑ i=1 ( α2i − β2i 2 √ 2 ) = −α(1−α4n) 2 − −β(1−β 4n) 2 2 √ 2 = α4n+1 − β4n+1 4 √ 2 − 1 2 = α4n+1(1 + α−1) + β4n+1(1 + β−1) 8 − 1 2 = ( α4n+1 + β4n+1 8 − 1 4 ) + ( α4n + β4n 8 − 1 4 ) = Bbcn +Rbcn by Theorem 3.6. � Similarly we can give the following theorem which can be proved similarly. Theorem 7.23. 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 equal to the difference of the nth balcobalancing number and its balcobalancer, that is, 2n∑ i=1 P2i−1 = Bbcn −Rbcn . (2) the half of the sum of Pell numbers from 1 to 4n is equal to the nth balcobalancing number, that is, 4n∑ i=1 Pi 2 = Bbcn . (3) the sum of Pell–Lucas numbers from 0 to 4n+ 1 is equal to the sum of the twelve times of the nth balcobalancing number, four times of the its balcobalancer plus 4, that is, 4n+1∑ i=0 Qi = 12Bbcn + 4Rbcn + 4. 220 Ahmet Tekcan and Meryem Yıldız (4) the sum of Pell–Lucas numbers from 1 to 4n is equal to the two times of the nth Lucas– balcobalancing number minus 1, that is, 4n∑ i=1 Qi = 2(Cbcn − 1). (5) the sum of balancing numbers from 1 to 4n+ 1 is equal to the product of the sum of three times of the nth balcobalancing number, its balcobalancer plus 1 and the four times of the nth balcobalancing number plus 1, that is, 4n+1∑ i=1 Bi = (3Bbcn +Rbcn + 1)(4Bbcn + 1). In [18], 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∑ i=1 Pi = [ n∑ i=0 ( 2n+ 1 2i ) 2i ]2 . In fact this sum is equals to c2n+1, that is, 4n+1∑ i=1 Pi = c2n+1. Similarly we can give the following result. Theorem 7.24. The sum of Pell numbers from 1 to 8n+ 1 is a perfect square and is 8n+1∑ i=1 Pi = (4Bbcn + 1)2. Proof. Since n∑ i=1 Pi = Pn+1+Pn−1 2 , we get 8n+1∑ i=1 Pi = P8n+2 + P8n+1 − 1 2 = α8n+2−β8n+2 2 √ 2 + α8n+1−β8n+1 2 √ 2 − 1 2 = α8n+2(1+α−1)+β8n+2(−1−β−1) 2 √ 2 2 − 1 2 = α8n+2 + β8n+2 4 − 1 2 = α8n+2 + 2α4n+1β4n+1 + β8n+2 4 Balcobalancing numbers 221 = 16 [( α4n+1 + β4n+1 8 )2 − 2 ( α4n+1 + β4n+1 8 ) ( 1 4 ) + 1 16 ] + 8 ( α4n+1 + β4n+1 8 ) − 2 + 1 = 16 [ α4n+1 + β4n+1 8 − 1 4 ]2 + 8 [ α4n+1 + β4n+1 8 − 1 4 ] + 1 = 16(Bbcn )2 + 8Bbcn + 1 = (4Bbcn + 1)2 by Theorem 3.6. � Apart from Theorem 7.24, we can give the following theorem which can be proved similarly. Theorem 7.25. 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 1 + 8n+3∑ i=1 Pi = (4Bbcn + 2Cbcn + 1)2. 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Generalized Pell numbers, balancing numbers and binary quadratic forms. Creat. Math. Inform. 23 (2014), no. 1, 115–122. [21] Tekcan, A. Almost Balancing, Triangular and Square Triangular Numbers. Notes on Number Theory and Discrete Math. 25 (2019), no. 1, 108–121. [22] Tekcan, A. Sums and Spectral Norms of all Almost Balancing Numbers. Creat. Math. Inform. 28 (2019), no. 2, 203–214. [23] Tengely, S. Balancing Numbers which are Products of Consecutive Integers. Publ. Math. Deb. 83 (2013), no. (1-2), 197–205. BURSA ULUDAG UNIVERSITY FACULTY OF SCIENCE DEPARTMENT OF MATHEMATICS BURSA, TURKIYE Email address: tekcan@uludag.edu.tr, 502011002@ogr.uludag.edu.tr Copyright of Creative Mathematics & Informatics is the property of Creative Mathematics & Informatics and its content may not be copied or emailed to multiple sites or posted to a listserv without the copyright holder's express written permission. However, users may print, download, or email articles for individual use.