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THE INTEGER SEQUENCE an and ITS RELATIONSHIP with PELL, PELL–LUCAS and BALANCING NUMBERS

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THE INTEGER SEQUENCE an and ITS RELATIONSHIP with PELL, PELL–LUCAS and BALANCING NUMBERS Annales Univ. Sci. Budapest., Sect. Comp. 46 (2017) 275{288 THE INTEGER SEQUENCE an AND ITS RELATIONSHIP WITH PELL, PELL{LUCAS AND BALANCING NUMBERS Arzu Akın and Ahmet Tekcan (Bursa, Turkiye) Communicated by Bui Minh Phong (Received October 19, 2016; accepted August 5, 2017) Abstract. In this work we deduce some algebraic relations on integer sequence an and its relationship with Pell, Pell{Lucas and balancing num- bers. Further we formulate the eigenvalues and spectral norm of the circu- lant matrix of an numbers. 1. Preliminaries Let p and q be two non{zero integers and let d = p2 − 4q 6= 0 (to exclude a degenerate case). We set the sequences Un and Vn to be Un = Un(p; q) = pUn−1 − qUn−2 Vn = Vn(p; q) = pVn−1 − qVn−2 for n ≥ 2 with U0 = 0;U1 = 1;V0 = 2;V1 = p. The characteristicp equ- ation of them is x2 − px + q = 0 and hence the roots are α = p+ d and p 2 p− d αn−βn n n β = 2 . Their Binet formulas are Un = α−β and Vn = α + β : Note Key words and phrases: Balancing numbers, Pell numbers, Pell{Lucas numbers, sums, perfect squares, circulant matrices and spectral norms. 2010 Mathematics Subject Classification: 05A19, 11B37, 11B39. 276 A. Akın and A. Tekcan that Un(1; −1) = Fn Fibonacci numbers (A000045 in OEIS), Vn(1; −1) = Ln Lucas numbers (A000032 in OEIS), Un(2; −1) = Pn Pell numbers (A000129 in OEIS) and Vn(2; −1) = Qn Pell{Lucas numbers (A002203 in OEIS) (for further details see [2, 5, 6, 7, 12]). Recently, balancing numbers have been defined by Behera and Panda in [1]. A positive integer n is called a balancing number if the Diophantine equation (1.1) 1 + 2 + ··· + (n − 1) = (n + 1) + (n + 2) + ··· + (n + r) holds for some positive integer r which is called cobalancing number (or balan- (n−1)n r(r+1) cer). From (1.1) one has 2 = rn + 2 and so p p −(2n + 1) + 8n2 + 1 2r + 1 + 8r2 + 8r + 1 (1.2) r = and n = : 2 2 th th Let Bn denote the n balancing number, and let bn denote the n cobalancing 2 number. Then from (1.2) we see that Bn is a balancing number iff 8Bn + 1 is 2 a perfect square, and bn is a cobalancing number iff 8bn + 8bn + 1 is a perfect p 2 p 2 square. So Cn = 8Bn + 1 and cn = 8bn + 8bn + 1 are integers called the nth Lucas{balancing and nth Lucas{cobalancing number, respectively. Binet 2n 2n 2n−1 2n−1 formulas for all balancing numbers are B = α p−β ; b = α p−β − 1 , n 4 2 n 4 2 2 α2n+β2n α2n−1+β2n−1 p p Cn = 2 and cn = 2 , where α = 1 + 2 and β = 1 − 2 which are the roots of the characteristic equation of Pell numbers (for further details see [8, 9, 10, 11]). 2. Main results In [13], Santana and Diaz{Barrero set a sequence (2.1) an = P2n + P2n+1 in order to determine the sum of first 4n + 1 nonzero terms of Pell numbers. They proved that an+1 = 6an − an−1 for n ≥ 1, where a0 = 1; a1 = 7. Since αn−βn p p Pn = α−β for α = 1 + 2 and β = 1 − 2, the Binet formula for an numbers is α2n+1 + β2n+1 (2.2) a = n 2 for n ≥ 0. For the sums of an numbers, we can give the following result. Integer sequence 277 th Theorem 2.1. Let an denote the n number. Then n X an+1 − an − 2 a = for n ≥ 1; i 4 i=0 n X 33a2n−1 − a2n−3 − 8 a = for n ≥ 2; 2i−1 32 i=1 n X 33a2n − a2n−2 − 8 a = for n ≥ 1: 2i 32 i=0 α2n+1+β2n+1 Proof. Since an = 2 , we easily get n X ai = a0 + a1 + ··· + an = i=0 α + β α3 + β3 α2n+1 + β2n+1 = + + ··· + = 2 2 2 α2n+2 + β2n+2 − 2 = = 4 2n+1 p 2n+1 p α [2(1+ 2)]+β [2(1− 2)] − 2 = 2 = 4 2n+1 2 2n+1 2 α (α −1)+β (β −1) − 2 = 2 = 4 2n+3 2n+3 2n+1 2n+1 α +β − α +β − 2 = 2 2 = 4 a − a − 2 = n+1 n : 4 The others can be proved similarly. For the relationship with Pell, Pell{Lucas and balancing numbers, we can give the following result. th Theorem 2.2. Let an denote the n number. Then for n ≥ 1, we have st th th 1. The sum of (n+1) and n balancing numbers is equal to the n an number, that is, Bn+1 + Bn = an: 2. The half of the difference of (n + 2)nd and nth cobalancing numbers is equal th to the n an number, that is, b − b n+2 n = a : 2 n 278 A. Akın and A. Tekcan 3. The half of the difference of (n + 1)st and nth Lucas{balancing numbers is th equal to the n an number, that is, C − C n+1 n = a : 2 n st th 4. The (n + 1) Lucas{cobalancing number is equal to the n an number, that is, cn+1 = an: st th 5. The half of (2n + 1) Pell{Lucas number is equal to the n an number, that is, Q 2n+1 = a : 2 n 2n 2n Proof. 1. Since B = α p−β , we easily get n 4 2 α2n+2 − β2n+2 α2n − β2n Bn+1 + Bn = p + p = 4 2 4 2 α2n(α2 + 1) + β2n(−β2 − 1) = p = 4 2 p p α2n(1 + 2) + β2n(1 − 2) = = 2 α2n+1 + β2n+1 = = 2 = an p p p p since α2 + 1 = 2 2(1 + 2) and −β2 − 1 = 2 2(1 − 2). The other cases can be proved similarly. We have already P2n + P2n+1 = an by (2.1). Also we can give the following theorem. th Theorem 2.3. Let an denote the n number. Then for n ≥ 0; 1. The sum of (2n + 1)st balancing and cobalancing numbers is a perfect square and is 2 B2n+1 + b2n+1 = an: 2. The sum of (2n + 1)st Lucas{balancing and Lucas{cobalancing numbers is th st equal to product of four times of n an number and (2n + 1) Pell number, that is, C2n+1 + c2n+1 = 4anP2n+1: Integer sequence 279 2n 2n 2n−1 2n−1 Proof. 1. Recall that B = α p−β and b = α p−β − 1 . So n 4 2 n 4 2 2 α4n+2 − β4n+2 α4n+1 − β4n+1 1 B2n+1 + b2n+1 = p + p − = 4 2 4 2 2 p α4n(α2 + α) − β4n(β2 + β) − 2 2 = p = 4 2 α4n+2 + 2(αβ)2n+1 + β4n+2 = = 4 α2n+1 + β2n+1 2 = = 2 2 = an since αβ = −1. The other case can be proved similarly. Panda and Ray ([10]) considered the sums of Pell numbers and proved 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 X Pi = Bn + bn: i=1 Later G¨ozeri, Ozko¸cand¨ Tekcan ([4]) proved that the sum of Pell{Lucas num- bers from 0 to 2n − 1 is equal to the sum of nth Lucas{balancing and Lucas{ cobalancing number, that is, 2n−1 X Qi = Cn + cn: i=0 Similarly, we can give the following result. th Theorem 2.4. Let an denote the n number. st 1. The sum of an numbers from 0 to n is equal to sum of the (n+1) balancing number and its balancer, that is, n X ai = Bn+1 + bn+1: i=0 st 2. The sum of even an numbers from 1 to n is equal to the product of (n + 1) th an number and n balancing number, that is, n X a2i = an+1Bn: i=1 280 A. Akın and A. Tekcan th 3. The sum of odd an numbers from 1 to n is equal to the product of n an number and nth balancing number, that is, n X a2i−1 = anBn: i=1 th 4. The sum of even an numbers from 1 to 2n + 1 is equal to the product of n nd st and (2n + 2) an numbers and (2n + 1) Pell number, that is, 2n+1 X a2i = ana2n+2P2n+1: i=1 th 5. The sum of odd an numbers from 1 to 2n + 1 is equal to the product of n st st and (2n + 1) an numbers and (2n + 1) Pell number, that is, 2n+1 X a2i−1 = ana2n+1P2n+1: i=1 6. The sum of odd balancing numbers from 0 to 2n is equal to the product of st th st (2n+1) Pell number, n an number and (2n+1) balancing number, that is, 2n X B2i+1 = P2n+1anB2n+1: i=0 7. The sum of even balancing numbers from 1 to 2n is equal to the product th th th of two times of n an number, n balancing number, n Lucas{balancing number and (2n + 1)st Pell number, that is, 2n X B2i = 2anBnCnP2n+1: i=1 8. The sum of even Pell numbers from 1 to 2n is equal to the product of two th th times of n balancing and n an number, that is, 2n X P2i = 2Bnan: i=1 9.
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