Some algebraic identities on quadra Fibona-Pell integer sequence

Transcription

1 Özkoç Advances in Difference Equations ( :148 DOI /s R E S E A R C H Open Access Some algebraic identities on quadra Fibona-Pell integer sequence Arzu Özkoç * * Correspondence: arzuozkoc@duzce.edu.tr Department of Mathematics, Faculty of Arts and Science, Düzce University,Düzce,Turkey Abstract In this work, we define a quadra Fibona-Pell integer sequence W n 3W n 1 3W n 3 W n 4 for n 4 with initial values W 0 W 1 0,W 1,W 3 3, and we derive some algebraic identities on it including its relationship with Fibonacci and Pell numbers. Keywords: Fibonacci numbers; Lucas numbers; Pell numbers; Binet s formula; binary linear recurrences 1 Preliminaries Let p and q be non-zero integers such that D p 4q 0 (to exclude a degenerate case. We set the sequences U n and V n to be U n U n (p, qpu n 1 qu n, (1 V n V n (p, qpv n 1 qv n for n with initial values U 0 0,U 1 1,V 0,andV 1 p. ThesequencesU n and V n are called the (first and second Lucas sequences with parameters p and q. V n is also called the companion Lucas sequence with parameters p and q. The characteristic equation of U n and V n is x px + q 0 and hence the roots of it are x 1 p+ D and x p D.SotheirBinetformulasare U n xn 1 xn x 1 x and V n x n 1 + xn for n 0. For the companion matrix M p q] 1 0,onehas ] ] ] ] M n 1 M n 1 U n U n and V n V n 1 for n 1. The generating functions of U n and V n are U(x p x px and V(x 1 px + qx 1 px + qx. ( Fibonacci, Lucas, Pell, and Pell-Lucas numbers can be derived from (1. Indeed for p 1 and q 1,thenumbersU n U n (1, 1 are called the Fibonacci numbers (A in 015 Özkoç; licensee Springer. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Özkoç Advances in Difference Equations ( :148 Page of 10 OEIS, while the numbers V n V n (1, 1 are called the Lucas numbers (A00003 in OEIS. Similarly, for p andq 1,thenumbersU n U n (, 1 are called the Pell numbers (A00019 in OEIS, while the numbers V n V n (, 1 are called the Pell-Lucas (A0003 in OEIS (companion Pell numbers (for further details see 1 6]. Quadra Fibona-Pell sequence In 7], the author considered the quadra Pell numbers D(n, which are the numbers of the form D(n D(n +D(n 3+D(n 4forn 4 with initial values D(0 D(1 D( 1, D(3, and the author derived some algebraic relations on it. In 8], the authors considered the integer sequence (with four parameters T n 5T n 1 5T n +T n 3 +T n 4 with initial values T 0 0,T 1 0,T 3,T 3 1,andtheyderived some algebraic relations on it. In the present paper, we want to define a similar sequence related to Fibonacci and Pell numbers and derive some algebraic relations on it. For this reason, we set the integer sequence W n to be W n 3W n 1 3W n 3 W n 4 (3 for n 4 with initial values W 0 W 1 0,W 1,W 3 3 and call it a quadra Fibona- Pell sequence. Here one may wonder why we choose this equation and call it a quadra Fibona-Pell sequence. Let us explain: We will see below that the roots of the characteristic equation of W n are the roots of the characteristic equations of both Fibonacci and Pell sequences. Indeed, the characteristic equation of (3isx 4 3x 3 +3x +10andhencethe roots of it are α 1+ 5, β 1 5, γ 1+ and δ 1. (4 (Here α, β are the roots of the characteristic equation of Fibonacci numbers and γ, δ are the roots of the characteristic equation of Pell numbers. Then we can give the following results for W n. Theorem 1 The generating function for W n is W(x x x 4 +3x 3 3x +1. Proof The generating function W(x is a function whose formal power series expansion at x 0 has the form W(x W n x n W 0 + W 1 x + W x + + W n x n +. n0 Since the characteristic equation of (3isx 4 3x 3 +3x +10,weget ( 1 3x +3x 3 + x 4 W(x ( 1 3x +3x 3 + x 4( W 0 + W 1 x + + W n x n + W 0 +(W 1 3W 0 x +(W 3W 1 x

3 Özkoç Advances in Difference Equations ( :148 Page 3 of 10 +(W 3 3W +3W 0 x 3 + +(W n 3W n 1 +3W n 3 + W n 4 x n +. Notice that W 0 W 1 0,W 1,W 3 3,andW n 3W n 1 3W n 3 W n 4.So(1 3x + 3x 3 + x 4 W(xx and hence the result is obvious. Theorem The Binet formula for W n is ( γ n δ n ( α n β n W n γ δ for n 0. x Proof Note that the generating function is W(x x 4 +3x 3 3x+1.Itiseasilyseenthatx4 + 3x 3 3x +1(1 x x (1 x x. So we can rewrite W(xas W(x x 1 x x x 1 x x. (5 From (, we see that the generating function for Pell numbers is P(x x 1 x x (6 and the generating function for the Fibonacci numbers is F(x x 1 x x. (7 From (5, (6, (7, we get W(xP(x F(x. So W n ( γ n δ n γ δ (αn β n α β aswewanted. The relationship with Fibonacci, Lucas, and Pell numbers is given below. Theorem 3 For the sequences W n, F n, L n, and P n, we have: (1 W n P n F n for n 0. ( W n+1 + W n 1 (γ n + δ n (α n + β n for n 1. (3 5F n + P n (γ n δ n +(α n β n for n 1. (4 L n + P n+1 + P n 1 α n + β n + γ n + δ n for n 1. (5 (W n+1 W n + F n 1 γ n + δ n for n 1. (6 lim n W n W n 1 γ. Proof (1 It is clear from the above theorem, since W(xP(x F(x. ( Since 6W n 1 + W n+ 3W n+1 3W n 1 W n +6W n 1,weget W n+1 + W n 1 W n W n W n+ 6 3 ( γ n 1 δ n γ δ ( γ n δ n γ δ αn 1 β n 1 αn β n

4 Özkoç Advances in Difference Equations ( :148 Page 4 of ( γ n+ δ n+ αn+ β n+ 3 γ δ ( 1 6 γ n 3(γ δ γ + 1 ( 6 γ + γ + δ n δ 1 ] δ δ ( α n 3( α 1 ( 6 α α + β n β + 1 ] β + β ( γ n + δ n ( α n + β n, since 6 γ + 1 γ + γ 6 δ 1 δ δ 6 and 6 α 1 α α 6 β + 1 (3 Notice that F n αn β n α β + β 3 5. β and P n γ n δ n γ δ.soweget 5F n α n β n and P n γ n δ n. Thus clearly, 5F n + P n (γ n δ n +(α n β n. (4 It is easily seen that P n+1 + P n 1 γ n + δ n.alsol n α n + β n.sol n + P n+1 + P n 1 α n + β n + γ n + δ n. (5 Since W n+1 3W n 3W n W n 3,weeasilyget W n+1 W n W n 3W n W n 3 ( γ n δ n γ δ αn β n ( γ n δ n 3 γ δ ( γ n 3 δ n 3 αn 3 β n 3 γ δ ( γ n 3 ( + δ n + 3 δ + 1 and hence 1 γ δ + 1 γ 1 γ 3 ( α n 1 α 3 α 1 α W n+1 W n ( γ γ n 3 3γ 1 γ 3 W n+1 W n + α n 1 ( α 3 3α 1 1 γ n ( γ 3 3γ 1 γ 3 (W n+1 W n + F n 1 γ n + δ n, α αn β n δ 3 ] β n 1 ( β 3 β 1 β + δ n ( δ +3δ +1 α n 1 ( α 3 3α 1 α δ 3 ] ] β n 1 ( β 3 3β 1 + δ n ( δ 3 +3δ +1 δ 3 since γ 3 3γ 1 δ3 +3δ+1 and α3 3α 1 γ 3 δ 3 α (6Itisjustanalgebraiccomputation,sinceW n ( γ n δ n Theorem 4 ThesumofthefirstntermsofW n is β3 3β 1 β 1. β ] β n 1 ( β 3 3β 1 γ δ ] β ] (αn β n α β. n i1 W i W n +4W n 1 +4W n + W n 3 +1 (8 for n 3.

5 Özkoç Advances in Difference Equations ( :148 Page 5 of 10 Proof Recall that W n 3W n 1 3W n 3 W n 4.So W n 3 + W n 4 3W n 1 W n 3 W n. (9 Applying (9, we deduce that W 1 + W 0 3W 3 W 1 W 4, W + W 1 3W 4 W W 5, W 3 + W 3W 5 W 3 W 6,..., (10 W n 4 + W n 5 3W n W n 4 W n 1, W n 3 + W n 4 3W n 1 W n 3 W n. Ifwesumofbothsidesof(10, then we obtain W n 3 + W 0 +(W W n 4 3(W 3 + W W n 1 (W 1 + W + + W n 3 (W 4 + W W n. So we get W n 3 +(W 1 + W + + W n 4 1 W n W n 1 W n +3W n +3W n 1 and hence we get the desired result. Theorem 5 The recurrence relations are W n 9W n 0W n 4 +9W n 6 W n 8, W n+1 9W n 1 0W n 3 +9W n 5 W n 7 for n 4. Proof Recall that W n 3W n 1 3W n 3 W n 4.SoW n 3W n 1 3W n 3 W n 4 and hence W n 3W n 1 3W n 3 W n 4 9W n 9W n 4 3W n 5 9W n 4 +9W n 6 +3W n 7 + W n 8 W n 8 W n 4 (3W n 5 3W n 7 W n 8 +9W n 18W n 4 +9W n 6 W n 8 W n 4 W n 4 +9W n 9W n 4 9W n 4 +9W n 6 W n 8 W n 4 9W n 0W n 4 +9W n 6 W n 8. The other assertion can be proved similarly. The rank of an integer N is defined to be ρ(n { p if p isthesmallestprimewithp N, if N is prime. Thus we can give the following theorem.

6 Özkoç Advances in Difference Equations ( :148 Page 6 of 10 Theorem 6 The rank of W n is if n 5+6k,6+6k,7+6k, ρ(w n 3 if n 8+1k,9+1k,15+1k,16+1k, 5 if n 14+60k,46+60k for an integer k 0. Proof Let n 5+6k. We prove it by induction on k.letk 0.ThenwegetW So ρ(w 5. Let us assume that the rank of W n is for n k 1,thatis,ρ(W 6k 1,so W 5+6(k 1 W 6k 1 a B for some integers a 1andB >0.Forn k,weget W 6k+5 3W 6k+4 3W 6k+ W 6k+1 3(3W 6k+3 3W 6k+1 W 6k 3W 6k+ W 6k+1 9W 6k+3 9W 6k+1 3W 6k 3W 6k+ W 6k+1 9(3W 6k+ 3W 6k W 6k 1 9W 6k+1 3W 6k 3W 6k+ W 6k+1 7W 6k+ 7W 6k 9W 6k 1 9W 6k+1 3W 6k 3W 6k+ W 6k+1 4W 6k+ 30W 6k 10W 6k+1 9W 6k 1 4W 6k+ 30W 6k 10W 6k+1 9 a B 1W 6k+ 15W 6k 5W 6k+1 9 a 1 B ]. Therefore ρ(w 5+6k. Similarly it can be shown that ρ(w 6+6k ρ(w 7+6k. Now let n 8+1k. Fork 0,wegetW So ρ(w 8 3.Letusassume that for n k 1therankofW n is 3, that is, ρ(w 8+1(k 1 ρ(w 1k 4 3 b H for some integers b 1andH > 0 which is not even integer. For n k,weget W 1k+8 3W 1k+7 3W 1k+5 W 1k+4 3W 1k+7 3W 1k+5 (3W 1k+3 3W 1k+1 W 1k 3W 1k+7 3W 1k+5 3W 1k+3 +3W 1k+1 + W 1k 3W 1k+7 3W 1k+5 3W 1k+3 +3W 1k+1 +(3W 1k 1 3W 1k 3 W 1k 4 3W 1k+7 3W 1k+5 3W 1k+3 +3W 1k+1 +3W 1k 1 3W 1k 3 W 1k 4 3W 1k+7 3W 1k+5 3W 1k+3 +3W 1k+1 +3W 1k 1 3W 1k 3 3 b H 3 ( W 1k+7 W 1k+5 W 1k+3 + W 1k+1 + W 1k 1 W 1k 3 3 b 1 H. So ρ(w 1k+8 3. The others can be proved similarly.

7 Özkoç Advances in Difference Equations ( :148 Page 7 of 10 Remark 1 Apart from the above theorem, we see that ρ(w ρ(w 6, while ρ(w 70 ρ(w and ρ(w 10 ρ(w 34 ρ(w But there is no general formula. The companion matrix for W n is M Set 1 0 N 0 0 and R ]. Then we can give the following theorem, which can be proved by induction on n. Theorem 7 For the sequence W n, we have: (1 RM n N W n+3 + P n +(W n+1 F n for n 1. ( R(M T n 3 N W n for n 3. (3 If n 7 is odd, then where m 11 m 1 m 13 m 14 M n m 1 m m 3 m 4 m 31 m 3 m 33 m 34, m 41 m 4 m 43 m 44 m 11 W n+, m 1 W n+1, m 31 W n, m 41 W n 1, m 14 W n+1, m 4 W n, m 34 W n 1, m 44 W n, m 1 1 W n+1 m W n n 5 m 3 1 W n 1 n 5 W n 1 i, m 13 W n+ n 3 W n i, m 3 1 W n+1 n 5 n 7 W n 3 i, m 33 W n W n i, W n 1 i, n 5 W n i,

8 Özkoç Advances in Difference Equations ( :148 Page 8 of 10 m 4 W n n 7 and if n 8 is even, then m 11 m 1 m 13 m 14 M n m 1 m m 3 m 4 m 31 m 3 m 33 m 34, m 41 m 4 m 43 m 44 W n 4 i, m 43 1 W n 1 n 7 W n 3 i, where m 11 W n+, m 1 W n+1, m 31 W n, m 41 W n 1, m 14 W n+1, m 4 W n, m 34 W n 1, m 44 W n, m 1 W n+1 m 1 W n m 3 W n 1 n 4 W n 1 i, m 13 1 W n+ n 6 n 6 W n i, m 3 W n+1 W n 3 i, m 33 1 W n m 4 1 W n n 8 n 4 n 4 W n i, W n 1 i, n 6 W n i, W n 4 i, m 43 W n 1 AcirculantmatrixisamatrixA a ij ] n n defined to be a 0 a 1 a a n 1 a n 1 a 0 a 1 a n A a n a n 1 a 0 a n 3, a 1 a a 3 a 0 n 6 W n 3 i. where a i are constants. The eigenvalues of A are n 1 λ j (A a k w jk, (11 k0 where w e πi n, i 1, and j 0,1,...,n 1. The spectral norm for a matrix B bij ] n m is defined to be B spec max{ λ i },whereλ i are the eigenvalues of B H B for 0 j n 1 and B H denotes the conjugate transpose of B.

9 Özkoç Advances in Difference Equations ( :148 Page 9 of 10 For the circulant matrix W 0 W 1 W W n 1 W n 1 W 0 W 1 W n W W(W n W n W n 1 W 0 W n 3 W 1 W W 3 W 0 for W n, we can give the following theorem. Theorem 8 The eigenvalues of W are { } W n 1 w 3j +(W n + P n 1 F n 1 +1w j λ j (W +(P n F n W n 1 w j W n w 4j +3w 3j 3w j +1 for j 0,1,,...,n 1. Proof Applying (11weeasilyget n 1 n 1 ( γ λ j (W W k w jk k δ k γ δ αk β k w jk k0 k0 1 γ n ] 1 γ δ γ w j 1 δn 1 1 α n ] 1 δw j 1 αw j 1 βn 1 βw j 1 1 (γ n 1(δw j 1 (δ n 1(γw j ] 1 γ δ (γ w j 1(δw j 1 1 (α n 1(βw j 1 (β n 1(αw j ] 1 (αw j 1(βw j 1 1 w j (γ n δ δ n γ + γ δ+δ n γ n ] γ δ δγw j w j (δ + γ +1 1 w j (α n β β n α + +β n α n ] βαw j w j (β + α+1 w 3j 5(δ γ + γδ n δγ n + ( + α n β αβ n ] + w j 5(γ n δ n + δ γ + γδ n γ n δ+ (β n α n +4 ( + α n β αβ n ] + w j 5(γ n δ n + γ δ + γ n δ γδ n + (β α + β n α α n β+4 (β n α n ] + 5(δ n γ n + (α n β n ] 10(w 4j +3w 3j 3w j +1 { } W n 1 w 3j +(W n + P n 1 F n 1 +1w j +(P n F n W n 1 w j W n w 4j +3w 3j 3w j +1, since αβ 1,γδ 1,α + β 1, 5, γ + δ,andγ δ.

10 Özkoç Advances in Difference Equations ( :148 Page 10 of 10 After all, we consider the spectral norm of W.Letn.ThenW 0].So W spec 0. Similarly for n 3,weget W and hence W H 3 W 3 I 3.So W 3 spec 1.Forn 4, the spectral norm of W n is given by the following theorem, which can be proved by induction on n. Theorem 9 The spectral norm of W n is for n 4. W n spec W n 1 +4W n +4W n 3 + W n 4 +1 For example, let n 6. Then the eigenvalues of W H 6 W 6 are λ 0 1,369, λ 1 89, λ λ and λ 3 λ So the spectral norm is W 6 spec λ 0 37.Also W 5+4W 4 +4W 3 +W Consequently, W 6 spec W 5 +4W 4 +4W 3 + W +1 as we claimed. 37 Competing interests The author declares that they have no competing interests. Acknowledgements The author wishes to thank Professor Ahmet Tekcan of Uludag University for constructive suggestions. Received: 5 March 015 Accepted: 6 April 015 References 1. Conway, JH, Guy, RK: Fibonacci numbers. In: The Book of Numbers. Springer, New York (1996. Hilton, P, Holton, D, Pedersen, J: Fibonacci and Lucas numbers. In: Mathematical Reflections in a Room with Many Mirrors, Chapter 3. Springer, New York ( Koshy, T: Fibonacci and Lucas Numbers with Applications. Wiley, New York ( Niven, I, Zuckerman, HS, Montgomery, HL: An Introduction to the Theory of Numbers, 5th edn. Wiley, New York ( Ogilvy, CS, Anderson, JT: Fibonacci numbers. In: Excursions in Number Theory, Chapter 11. Dover, New York ( Ribenboim, P: My Numbers, My Friends: Popular Lectures on Number Theory. Springer, New York ( Taşcı, D: On quadrapell numbers and quadrapell polynomials. Hacet. J. Math. Stat. 38(3, ( Tekcan, A, Özkoç, A, Engür, M, Özbek, ME: On algebraic identities on a new integer sequence with four parameters. Ars Comb. (accepted