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 numbers. Further we formulate the eigenvalues and spectral norm of the circulant 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 characteristic√ equation of them is x2 − px + q = 0 and hence the roots are α = p+ d and √ 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 √ √ −(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 √−β , b = α √−β − 1 , n 4 2 n 4 2 2 α2n+β2n α2n−1+β2n−1 √ √ 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 √ √ 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 √ 2n+1 √ 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 √−β , we easily get n 4 2

α2n+2 − β2n+2 α2n − β2n Bn+1 + Bn = √ + √ = 4 2 4 2 α2n(α2 + 1) + β2n(−β2 − 1) = √ = 4 2 √ √ α2n(1 + 2) + β2n(1 − 2) = = 2 α2n+1 + β2n+1 = = 2 = an √ √ √ √ 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 √−β and b = α √−β − 1 . So n 4 2 n 4 2 2 α4n+2 − β4n+2 α4n+1 − β4n+1 1 B2n+1 + b2n+1 = √ + √ − = 4 2 4 2 2 √ α4n(α2 + α) − β4n(β2 + β) − 2 2 = √ = 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 numbers 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. The sum of odd Pell numbers from 0 to 2n is equal to the product of (2n+1)st th Pell number and n an number, that is,

2n X P2i+1 = P2n+1an. i=0 Integer sequence 281

10. The sum of Pell–Lucas numbers from 0 to 2n is equal to the quotient of two st th times of (2n + 1) balancing number by the n an number, that is,

2n X 2B2n+1 Q = . i a i=0 n

11. The sum of even Pell–Lucas numbers from 0 to 2n is equal to the product of th th st n an number and the diﬀerence of n and (n − 1) an numbers, that is,

2n X Q2i = an(an − an−1). i=0

an+1−an−2 Proof. 1. Since a0 + a1 + ··· + an = 4 by Theorem 2.1, we easily get

n X an+1 − an − 2 a = = i 4 i=0 2n+3 2n+3 2n+1 2n+1 α +β − α +β − 2 = 2 2 = 4 α2n+1(α2 − 1) + β2n+1(β2 − 1) 1 = − = √ 8 √ 2 α2n+1(1 + 2) + β2n+1(1 − 2) 1 = − = 4 2 α2n+2(1 + α−1) + β2n+2(−1 − β−1) 1 = √ − = 4 2 2 α2n+2 − β2n+2 α2n+1 − β2n+1 1 = √ + √ − = 4 2 4 2 2

= Bn+1 + bn+1

as we claimed. The other cases can be proved similarly. For the perfect squares, we can give the following result.

th Theorem 2.5. Let an denote the n an number. Then

2 q 2 1. 1 + 8Bn+1 is a perfect square and is 1 + 8Bn+1 = 2an + Cn.

2 q 2 2. P2n+1 + P2nP2n+2 is a perfect square and is P2n+1 + P2nP2n+2 = an. 282 A. Akın and A. Tekcan q 2 2 2 2 2 3. B2n+1 +4anBnBn+1 is a perfect square and is B2n+1 + 4anBnBn+1 = an. Proof. 1. Applying Binet formulas, we get s q α2n+2 − β2n+2 2 1 + 8B2 = 1 + 8 √ = n+1 4 2 r α4n+4 + β4n+4 + 2 = = 2 α2n+2 + β2n+2 = = 2 α2n(2α + 1) + β2n(2β + 1) = = 2 α2n+1 + β2n+1 α2n + β2n = 2 + = 2 2

= 2an + Cn 2 2 since 2α + 1 = α and 2β + 1 = β . The other cases can be proved similarly. In [13], Santana and Diaz–Barrero proved that the sum of ﬁrst nonzero 4n + 1 terms of Pell numbers is a perfect square and is

4n+1 n !2 X X 2n + 1 P = 2i . i 2i i=1 i=0 2 In fact, this sum is equals to an, that is, 4n+1 X 2 Pi = an. i=1 Tekcan and Tayat ([14]) proved that the sum of ﬁrst nonzero 2n + 1 terms of Pell numbers is

 n+1 n+1 2 2n+1  α +β X  2 , for even n ≥ 0 2 Pi = αn+1−βn+1 √ i=1  2 2 , for odd n ≥ 1 √ √ where α = 1 + 2 and β = 1 − 2. By considering this result, they set two n+1 n+1 n+1 n+1 integer sequences X = α +β and Y = α √−β and proved that the n 2 n 2 sum of ﬁrst nonzero 4n + 1 terms of Pell numbers is 4n+1 X 2 n+1 2 Pi = (2Xn − 2XnYn−1 + (−1) ) . i=1 Similarly we can give the following result. Integer sequence 283

th Theorem 2.6. Let an denote the n an number. Then

1. The sum of an numbers from 0 to 2n is a perfect square and is

2n X 2 ai = an i=0 for n ≥ 1.

2. The half of the sum of odd Pell–Lucas numbers from 0 to 2n is a perfect square and is 2n P Q2i+1 i=0 = a2 2 n for n ≥ 1.

3. The sum of an numbers from 1 to n and adding 2 is a perfect square and is

n X 2 2 + ai = C n+1 2 i=1

for odd n ≥ 1, or the sum of an numbers from 1 to n and adding 1 is a perfect square and is n X 2 1 + ai = c n+2 2 i=1 for even n ≥ 2.

an+1−an−2 Proof. 1. Since a0 + a1 + ··· + an = 4 , we deduce that

2n α4n+3+β4n+3 α4n+1+β4n+1 X a2n+1 − a2n − 2 − − 2 a = = 2 2 = i 4 4 i=0 4n+1 2 4n+1 2 √ √ α (α −1)+β (β −1) − 2 α4n+1(1 + 2) + β4n+1(1 − 2) − 2 = 2 = = 4 4 α4n+2 + β4n+2 − 2 α4n+2 + 2(αβ)2n+1 + β4n+2 = = = 4 4 α2n+1 + β2n+1 2 = = 2 2 = an.

The other cases can be proved similarly. 284 A. Akın and A. Tekcan

A circulant matrix (see [3]) is a matrix M deﬁned as

  m0 m1 m2 ··· mn−2 mn−1  mn−1 m0 m1 ··· mn−3 mn−2     mn−2 mn−1 m0 ··· mn−4 mn−3     . . . ··· . .  M =   ,  . . . ··· . .     . . . ··· . .     m2 m3 m4 ··· m0 m1  m1 m2 m3 ··· mn−1 m0

where mi are constant. In this case, the eigenvalues of M are

n−1 X −ju (2.3) λj(M) = muw , u=0

2πi √ where w = e n , i = −1 and j = 0, 1, ··· , n − 1. The spectral norm of a matrix Q = [qij]n×n is deﬁned to be

p ||Q||spec = max { λj}, 0≤j≤n−1

∗ ∗ where λj are the eigenvalues of Q Q and Q denotes the conjugate transpose of Q.

Theorem 2.7. Let a denote the circulant matrix of an numbers. Then

1. The eigenvalues of a are

(a + 1)w−j + 1 − a λ (a) = n−1 n j w−2j − 6w−j + 1

for j = 0, 1, 2, ··· , n − 1.

2. The spectral norm of a is

an − an−1 − 2 ||a||spec = . 4 Integer sequence 285

Proof. 1. Applying (2.3), we deduce that

n−1 X −ju λj(a) = auw = u=0 n−1 X α2u+1 + β2u+1 = w−ju = 2 u=0 " n−1 n−1 # 1 X X = α (α2w−j)u + β (β2w−j)u = 2 u=0 u=0 1 α2n − 1 β2n − 1 = α + β = 2 α2w−j − 1 β2w−j − 1 1 (α2n+1 − α)(β2w−j − 1) + (β2n+1 − β)(α2w−j − 1) = = 2 (α2w−j − 1)(β2w−j − 1) −j h 2 α2n−1+β2n−1 −αβ2−βα2 i α+β α2n+1+β2n+1 w (αβ) 2 + 2 + 2 − 2 = = w−2j − 6w−j + 1 (a + 1)w−j + 1 − a = n−1 n w−2j − 6w−j + 1

−αβ2−βα2 α+β since αβ = −1, 2 = 1 and 2 = 1. 2. For the circulant matrix

  a0 a1 a2 ··· an−2 an−1  an−1 a0 a1 ··· an−3 an−2     . . . ··· . .    a =  . . . ··· . .  ,    . . . ··· . .     a2 a3 a4 ··· a0 a1  a1 a2 a3 ··· an−1 a0 for an numbers, we have

  a11 a12 ··· a1(n−1) a1n  a21 a22 ··· a2(n−1) a2n     . . . .  ∗   (a) a =  . . ··· . .  ,    . . . .     a(n−1)1 a(n−1)2 ··· a(n−1)(n−1) a(n−1)n  an1 an2 ··· an(n−1) ann 286 A. Akın and A. Tekcan where

2 2 2 2 a11 = a0 + an−1 + ··· + a2 + a1,

a12 = a0a1 + an−1a0 + ··· + a2a3 + a1a2, . .

a1(n−1) = a0an−2 + an−1an−3 + ··· + a2a0 + a1an−1,

a1n = a0an−1 + an−1an−2 + ··· + a2a1 + a1a0,

a21 = a1a0 + a0an−1 + ··· + a3a2 + a2a1, 2 2 2 2 a22 = a1 + a0 + ··· + a3 + a2, . .

a2(n−1) = a1an−2 + a0an−3 + ··· + a3a0 + a2an−1

a2n = a1an−1 + a0an−2 + ··· + a3a1 + a2a0, . .

an1 = an−1a0 + an−2an−1 + ··· + a1a2 + a0a1,

an2 = an−1a1 + an−2a0 + ··· + a1a3 + a0a2, . .

an(n−1) = an−1an−2 + an−2an−3 + ··· + a1a0 + a0an−1, 2 2 2 2 ann = an−1 + an−2 + ··· + a1 + a0. ∗ The eigenvalues of a a are λ0, λ1, ··· , λn−1. Here λ0 is maximum and is 2 2 2 2 λ0 = a0 + a1 + ··· + an−2 + an−1+   a0(a1 + a2 + ··· + an−2 + an−1)+  + a1(a2 + ··· + an−2 + an−1)+  + 2   =  + ··· +  + an−2an−1 2 = (a0 + a1 + ··· + an−1) . √ Thus the spectral norm of a is hence ||a||spec = λ0 = a0 + a1 + ··· + an−1. So the result is clear from Theorem 2.1. Example 2.8. Let n = 5. Then the circulant matrix is  1 7 41 239 1393   1393 1 7 41 239    a =  239 1393 1 7 41  .    41 239 1393 1 7  7 41 239 1393 1 Integer sequence 287

The eigenvalues of

 1999301 344413 68817 68817 344413   344413 1999301 344413 68817 68817  ∗   a a =  68817 344413 1999301 344413 68817     68817 68817 344413 1999301 344413  344413 68817 68817 344413 1999301 √ √ are λ0 = 2825761, λ1 = 1792686+137798 5, λ2 = 1792686√ −137798 5, λ3 = λ1 and λ4 = λ2. So the spectral norm of a is ||a||spec = λ0 = 1681. On the other hand a5−a4−2 = 1681. 4

From Theorem 2.7, we can give the following result.

Corollary 2.1. The spectral norm of a is √ 2 ( 2Pn) , for even n ≥ 2 ||a||spec = 2 (a n−1 ) , for odd n ≥ 1. 2

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A. Akın and A. Tekcan Uludag University Faculty of Science Department of Mathematics Gorukle, Bursa–Turkiye [email protected] [email protected]