On the Shifted Product of Binary Recurrences

Transcription

1 Journal of Integer Sequences, Vol. 13 (2010), rticle On the Shifted Product of Binary Recurrences Omar Khadir epartment of Mathematics University of Hassan II Mohammedia, Morocco khadir@hotmail.com Kálmán Liptai Institute of Mathematics and Informatics Eszterházy Károly College Eger, Hungary liptaik@gemini.ektf.hu László Szalay 1 Institute of Mathematics and Statistics University of West Hungary Sopron, Hungary laszalay@ktk.nyme.hu bstract The present paper studies the diophantine equation G n H n + c = x 2n and related questions, where the integer binary recurrence sequences {G}, {H} and {x} satisfy the same recurrence relation, and c is a given integer. We prove necessary and sufficient conditions for the solubility of G n H n + c = x 2n. Finally, a few relevant examples are provided. 1 Research supported by the János Bolyai Scholarship of HS, by the Hungarian National Foundation for Scientific Research Grant No. T FT, and by the Mexican-Hungarian bilateral TeT. 1

2 1 Introduction Let the binary recurrences {a} n=0, {b} n=0, {c} n=0 and {d} n=0 satisfy the recurrence relation with the initial values X n+2 = 6X n+1 X n, n N, (1) a 0 = 1, a 1 = 1; b 0 = 1, b 1 = 5; c 0 = 0, c 1 = 2; d 0 = 3, d 1 = 17; respectively (Sloane: {c} is , {d} is ). Here, and in the sequel, N denotes the set of non-negative integers. By the recurrences above, we can define a fifth sequence {x} n=0 via x 2n = a n b n + 1, x 2n+1 = c n d n + 1, n N. (2) The sequence {x} (the first few terms are: 0, 1, 6, 35, 204, 1189,..., Sloane: ) also satisfies (1), in spite of the unusual method of its composition, which we discovered while studying the terms of {x}, called balancing numbers. positive integer x 2 is called a balancing number for the integer y if (x 1) = (x + 1) + + (y 1) (3) holds. This definition was introduced by Finkelstein [3] (he called balancing numbers numerical centers) when he solved a puzzle from the book [1]. For other properties of balancing numbers see, for example [2]. Clearly, (3) leads to the Pell equation (2y 1) 2 2(2x) 2 = 1, and the possible values of x can be described by a suitable recurrence relation of type (1). It is known that all real binary recurrences satisfying (1) form a vector space of dimension 2 over R. efinition (2) (and later (6)) does not fit the vector space structure. The question arises which other integer binary recurrences can possess the interesting property described by (2) or its generalization. The phenomena appearing in (2) is a specific case of a more general property. Let t N + and put α = t + t 2 + 1, β = t t efine the sequences {T } n=0 and {U} n=0 by α n = T n + U n t It is easy to see that the sequences {T } and {U} both satisfy the recurrence relation X n = 2tX n 1 + X n 2 with the initial values T 0 = 1, T 1 = t and U 0 = 0, U 1 = 1, respectively. The corresponding explicit formulae T n = αn + β n 2 and U n = αn β n 2 t show that U n ( 1) nt n±( 1) n t = 1 2 U 2n holds. Subsequently, the choice a n = T 2n 1, b n = U 2n+1, c n = T 2n+2, d n = U 2n implies a n b n + t = U 4n /2 N and c n d n + t = U 4n+2 /2 N. That is, the example given at the start of the introduction is a particular case of this construction with t = 1. Observe, that X n = (4t 2 + 2)X n 1 X n 2 holds for all the sequences {a}, {b}, {c} and {d}, and also for {x n } = {U 2n /2}. Thus, we can provide infinitely many, but not all examples similar to (2) 2

3 (see in Section 4) since the coefficients = 4t and B = 1 do not represent the most general case. Now consider a slightly different property having a similar flavor. ssume that > 0 and B 0 are integers with non-vanishing = 2 + 4B, further let B denote the set of all integer binary recurrences {X} n=0 satisfying the recurrence relation X n+2 = X n+1 + BX n, n N. (4) If {u} n=0 and {v} n=0 are elements of B with the initial values u 0 = 0, u 1 = 1 and v 0 = 2, v 1 =, respectively, then the equality u n v n = u 2n (5) holds for any non-negative integer n. Identity (5) is well-known, and all books or papers dealing, for instance, with the Fibonacci sequence mention the formula F n L n = F 2n, where F n and L n are the nth term of the Fibonacci (Sloane: ) and Lucas sequences (Sloane: ), respectively. Note, that the sequences {F } and {L} both satisfy (4) with = B = 1. In this paper, we examine how we can construct a binary recurrence {x} as a shifted product of two binary recurrences satisfying the same recurrence relation. More precisely, given the integers c and l, we consider the equation G n H n + c = x 2n+l, (6) where {G} n=0, {H} n=0 and {x} n=0 belong to the class of binary recurrences given by (4). There is no restriction in assuming that l = 0, since l causes only a translation in the subscript. Moreover, suppose that c is a fixed integer, and the binary recursive sequences {G} and {J} n=0 also belong to B. nalogously to (2), we investigate and determine the binary recurrences {H} and {K} n=0 of B, which generate a sequence {x} B satisfying x 2n = G n H n + c, x 2n+1 = J n K n + c, n N. It is worth noting that if one forgets about the integrality conditions, then obviously c can be considered to be either 0 or 1. Indeed, if c 0, then we can replace {x n } by {x n /c}, further {G n }, {H n }, {J n } and {K n } by {G n / c}, {H n / c}, {J n / c} and {K n / c}, respectively, which results in assuming that c = 1. 2 Preliminaries For any complex numbers α, β,... and for any sequence {X} n=0 B put α X = X 1 αx 0, β X = X 1 βx 0, etc. Recall, that > 0 and B 0 are integers and = 2 + 4B 0. Lemma 1. ssume that {G} B. Then the zeros α = ( + )/2 and β = ( )/2 of the companion polynomial p(x) = x 2 x B of {G} are distinct and nonzero. Further 3

4 αβ = B, α + β =, α β =, > 0 implies β < α and 1 < α. Moreover, G n = β Gα n α G β n. (7) Proof. ll formulae and statements of Lemma 1 are known. Nevertheless, the first two conditions are immediate from the determination of α and β, while (7) can be derived from the basic theorem concerning the linear recurrences (see, for instance, [4]). Note that α and β are conjugate zeros of p(x) if they are not integers. Remark 2. Later we will use the results of Lemma 1 without any comment. In the present and next sections we assume that c Z, l = 0. Further {G} B, {H} B and {x} B satisfy (6). Lemma 3. x 1 = G 1H 1 BG 0 H 0 + c(1 B). Proof. Combining x 0 = G 0 H 0 + c, x 2 = G 1 H 1 + c and x 2 = x 1 + Bx 0, we immediately get the desired statement. Now we define some important constants which will be useful in studying our problems. Put E α = β Gβ H β x, + α x, Further, let E β = α Gα H E αβ = β Gα H + α G β H. (8) = (2G 1 G 0 )H 1 (2BG 0 + G 1 )H 0 Z. (9) Remark 4. We call { G} n=0 the companion sequence of {G} B if { G} is also in B and G 0 = 2G 1 G 0, G 1 = 2BG 0 + G 1. Thus, can be simplified as G 0 H 1 G 1 H 0. Sometimes it is more convenient to write (9) in the form = 2G 1 H 1 2BG 0 H 0 (G 1 H 0 + G 0 H 1 ). Lemma 5. The quantities from (8) can also be given in the forms E α = β ( 1 ) β2 c, E β = α ( + 1 ) α2 c, E αβ = β α. (10) Proof. Recall the definition of E α and β G = G 1 βg 0, β H = H 1 βh 0, β x = x 1 βx 0, x 0 = G 0 H 0 + c. These formulae, together with Lemma 3, yield E α = β (2G 1 H 1 2BG 0 H 0 (G 1 H 0 + G 0 H 1 ) 1 ) β2 c. β In order to complete the proof on E α, observe that the first three terms in the paranthesis above make up (see Remark 4). If α Z, then E β is a conjugate of E α ; hence, the second part of the lemma is obvious. Otherwise, we can use the same procedure we applied to E α. Finally, one can easily get E αβ = / from the parts of definition (8) concerned with E αβ. 4

5 3 The equation G n H n + c = x 2n Now we proceed to the solution of the first problem. The following result provides necessary conditions for the solubility of equation (6). Theorem 6. Let c Z. If there exist sequences {G} B, {H} B and {x} B such that G n H n + c = x 2n then one of the following cases holds. If c = 0 then = 0; if c 0 then either B = 1, = c; or β = ±1, = 0. Remark 7. The assertion B = 1, together with β R implies 0 < β < 1. Then the two instances in the second part of Theorem 6 are not-overlapping. Proof. G n H n + c = x 2n is equivalent to E α α 2n + E β β 2n E αβ (αβ) n + c = 0, n N. (11) Since (11) is true for all n, therefore it is true for n = 0, 1, 2, 3. Consequently, (11) at n = 0, 1, 2, 3 is a homogeneous linear system of four equations in the unknowns E α, E β, E αβ and c. The determinant = α 2 β 2 αβ 1 α 4 β 4 (αβ) 2 1 α 6 β 6 (αβ) 3 1 of the coefficient matrix is the negative of the Vandermonde of α 2, β 2, αβ and 1. Hence, = V (α 2,β 2,αβ, 1) = αβ(α β) 3 (α + β)(1 α 2 )(1 β 2 )(1 αβ). (12) If the homogeneous system has only the trivial solution c = 0 and E α = E β = E αβ = 0, then, by Lemma 5, we obtain E α = β = 0, E β = α = 0, E αβ = 1 = 0. The coefficient of is nonzero in any of the three equalities. Subsequently, = 0 follows. If the system has infinitely many solutions, then = 0. Recalling (12) and the conditions αβ 0, α β (see Lemma 1), we distinguish four cases. 1. α + β = 0. We deduce a degenerate recursion characterized by α/β = 1. Thus = 0, and we have arrived at a contradiction α 2 = 0. Now α = ±1 R. This is impossible since α > 1 holds β 2 = 0. Both cases β = ±1 lead to the solutions E α = E αβ = 0 and E β = c, where c is a free variable. Clearly, by (10), E αβ implies = 0. 5

6 4. 1 αβ = 0. Obviously, B = 1. Moreover = yields 2. The infinitely many solutions of the system can be described by E α = E β = 0 and αβ(α β)2 c = (α 2 1)(β 2 1) E αβ. (13) Since αβ = B and (α β) 2 = 2 4, relation (13) together with (α 2 1)(β 2 1) = (α 2)(β 2) = gives E αβ = c, where c is a free variable. Then, again by (10), = c. Theorem 6 provides necessary conditions for the sequences {G}, {H} and {x} satisfying G n H n + c = x 2n. Now we show that the conditions are, essentially, sufficient as well. More precisely, we prove the following theorem. Theorem 8. Suppose that the integers c, G 0, G 1, H 0 and H 1 are fixed. Put x 0 = G 0 H 0 + c. If hold, as well as c = 0, = 0; or c 0, B = 1, = c; or c 0, β = ±1, = 0, x 1 = G 1H 1 BG 0 H 0 x 1 x 1 = G 1H 1 + G 0 H 0 + 2c Z, Z, or = G 1H 1 BG 0 H 0 + (1 B)c or Z, respectively, then the terms of the sequences {G} B, {H} B and {x} B satisfy G n H n + c = x 2n, n N. Remark 9. Theorem 8 asserts necessarily that x 1 is an integer, otherwise {x} would not be an integer sequence although it satisfies G n H n +c = x 2n. For example, let c = 0, = 14, B = 5, G 0 = 2, G 1 = 3, H 0 = 1, H 1 = 1. Then = 0, x 1 = 1/2, and {G} : 2, 3, 32, 433,... ; {H} : 1, 1, 19, 261,... ; {x} : 2 = G 0 H 0, 1 2, 3 = G 1H 1, 89 2, 608 = G 2H 2, 16579, = G 3 H 3, Proof. We will use the notation of the previous sections. By (9), it is easy to verify that G 1 H 0 + G 0 H 1 = (2G 1 H 1 2BG 0 H 0 )/. Combining it with β G β H = G 1 H 1 +β 2 G 0 H 0 β(g 1 H 0 +G 0 H 1 ) and α G α H = G 1 H 1 +α 2 G 0 H 0 α(g 1 H 0 +G 0 H 1 ), 6

7 respectively, we obtain (G1 H 1 β 2 G 0 H 0 ) + β β G β H = By x 0 = G 0 H 0 + c and Lemma 3, we also have β x = x 1 βx 0 = G 1H 1 β 2 G 0 H 0 + (1 β 2 )c Thus, by (14) and (15), G n H n + c = β Gα n α G β n β H α n α H β n which is equivalent to β x α 2n α x β 2n + 1 ( β 1 β2 and α G α H = ( G1 H 1 + α 2 G 0 H 0 ) + α. (14) and α x = x 1 αx 0 = G 1H 1 α 2 G 0 H 0 + (1 α 2 )c. (15) + c = β Gβ H α 2n + α G α H β 2n (αβ) n + c, ) c α 2n + 1 ( α + 1 ) ( ) α2 c β 2n (αβ)n c. }{{} W (16) Here, the first summand is just x 2n. Now we show that the remaining part of (16), denoted by W, vanishes under certain conditions. First, suppose that c = 0 and = 0. Then, obviously W = 0. ssume now c 0, and consider the case B = 1 (i.e., αβ = 1) with = c. It follows that W = c (β 1 ) β2 α 2n + c (α + 1 ) α2 β 2n = c ( α 2n β 2n) (αβ 1). But the last factor vanishes since αβ = 1. Consequently, W is zero again. In the third case, we have β = ±1 and = 0. Thus, W = c ( 1 ) β2 α 2n + c ( ) 1 α 2 β 2n + c = c ( ) 1 α 2 + c = c (1 β2 ) = 0. The proof is therefore complete. 4 The system x 2n = G n H n + c, x 2n+1 = J n K n + c The investigation here is based on the results of the previous sections. We have to handle only two more problems, namely the question of the translation by 1 in the subscript, and the question of putting two subsequences of {x} together. ccording to the definition of, we introduce the notation 2 = (2J 1 J 0 )K 1 (2BJ 0 + J 1 )K 0 Z. The next statement is an immediate consequence of Theorem 6. Theorem 10. Suppose that c Z and the sequences {G}, {H}, {J}, {K} and {x}, all are in B, satisfy x 2n = G n H n + c and x 2n+1 = J n K n + c. Suppose furthermore that the following two conditions 7

8 hold. If c = 0 then = 2 = 0, if c 0 then either B = 1, = 2 = c or β = ±1, = 2 = 0 Recall, that > 0, B 0 with 0. Now we are going to formulate the reciprocal of Theorem 10. Theorem 11. Let c Z be given. Suppose that {G n }, {H n }, {J n }, {K n } all are elements of the set B, and the integers G i, H i, J i, K i (i = 0, 1) fulfill one of the following conditions = 2 = 0 if c = 0; B = 1 and = 2 = c if c 0; β = ±1 and = 2 = 0 if c 0. If x 0 = G 0 H 0 + c, x 1 = J 0 K 0 + c, x 2 = G 1 H 1 + c and x 3 = J 1 K 1 + c satisfy x 2 = x 1 + Bx 0 and x 3 = x 2 + Bx 1, then the sequence {x n } given by belongs to B. x 2n = G n H n + c, x 2n+1 = J n K n + c, n N Proof. First we prove that, under the given conditions, the relation x 2n+2 = x 2n+1 + Bx 2n holds, i.e., G n+1 H n+1 + c = (J n K n + c) + B(G n H n + c). By (7), it is equivalent to β G α n+1 α G β n+1 β H α n+1 α H β n+1 + c = ( ) βj α n α J β n β = K α n α K β n + c + B ( βg α n α G β n β H α n α H β n ) + c, which in turn is equivalent to α ( ) 2n (α 2 B)β G β H β J β K +β ( ) 2n (β 2 B)α G α H α J α K + }{{}}{{} W 1 W 2 ( B) (2B + 2 ) + (1 B)c = 0. (17) }{{} W 3 The value of W 3 seems to be the simplest to determine. Obviously, = 2 = 0, and c = 0 provide immediately W 3 = 0. If c 0 and = 2 = c, B = 1, then again W 3 = 0. Otherwise, when c 0 and = 2 = 0, β = ±1, we must distinguish two cases. When β = 1, we obtain W 3 = 0, while β = 1 we get W 3 = 2(1 α)c. 8

9 We next compute W 1. Here condition α 2 B = α, together with the definition of β G, β H, β J and β K, provides W 1 = (αg 1 H 1 βbg 0 H 0 + B(G 1 H 0 + G 0 H 1 )) α (αj 1K 1 βbj 0 K 0 + B(J 1 K 0 + J 0 K 1 )). We can replace the terms G 1 H 0 + G 0 H 1 and J 1 K 0 + J 0 K 1 in (18) by 2G 1 H 1 2BG 0 H 0 respectively. Thus, W 1 and 2J 1K 1 2BJ 0 K 0 2, = (α + 2B)G 1 H 1 B(β + 2B)G 0 H 0 B 1 α ((α + 2B)J 1K 1 B(β + 2B)J 0 K 0 B 2 ) ( ) = 1 α α G 1 H 1 + βb G 0 H 0 B (α J 1 K 1 + βb ) J 0 K 0 B 2, since α + 2B = α and B(β + 2B) = βb. Using G 0 H 0 = x 0 c, etc., it follows that W 1 = ( (αx 2 x 3 ) + βb ( α (αx 0 x 1 ) + 1 α βb + βb ) ) ( ) 2 c + α α B. We use x 3 = x 2 +Bx 1 and x 2 = x 1 +Bx 0 to get αx 2 x 3 = β 2 (αx 0 x 1 ). Consequently, we get (αx 2 x 3 ) + βb (αx α 0 x 1 ) = 0. Since 1 α βb + βb/α = (α 1)(β 2 1), we conclude that W 1 = ( ) (α 1)(β 2 2 1)c + B α. (19) Similar arguments applied to W 2 give W 2 = ( ) (β 1)(α 2 2 1)c + B β. (20) Note that if α Z then W 1 and W 2 are conjugates, therefore (20) comes directly from (19). Clearly, W 1 = W 2 = 0 if c = 0 and = 2 = 0 hold. Now consider the second possibility, when c 0, = 2 = c and B = 1. The last condition shows that β is the reciprocal of α. Thus, the coefficients (α 1)(β 2 1) and (β 1)(α 2 1) in (19) and (20), respectively, coincide (β 1) and (α 1), respectively. Hence, W 1 = W 2 = 0 holds again. Finally, c 0 and = 2 = 0 yield W 1 = c(1 α)(1 β 2 ) and W 2 = c(1 β)(1 α 2 ). This leads to β = ±1. Subsequently, W 1 = 0, while W 2 = 0, or W 2 = 2 (1 α 2 ) depending on the sign of β. Consider now W 1, W 2 and W 3 again. Ignoring the trivial cases, it is sufficient to look at only the case β = 1. Now W 1 = 0, W 2 = 2 c(1 α 2 ) and W 3 = 2(1 α)c. Therefore, x 2n+2 x 2n+1 Bx 2n = 0 α 2n 2 c(1 α 2 )( 1) 2n +2(1 α)c = 2 c(1 α)(1+β) = 0. So, we have showed that x 2n+2 = x 2n+1 +Bx 2n. The argument for x 2n+3 = x 2n+2 +Bx 2n is entirely similar to the procedure we have just applied. 9 (18)

10 5 Examples Example 12. This example indicates some difficulties arising if c Z, the sequence {G} B is fixed, and one intends to determine a sequence {H} B (and {x} B) such that G n H n + c = x 2n. Let = 3, B = 1 and c = 7, G 0 = 1, G 1 = 2 (Sloane: ). Thus, = 5, = c = 35. To find {H}, it is necessary to solve the linear diophantine equation H 1 4H 0 = 35 (see Remark 4). Clearly, it is solvable. For example, put H 0 = t Z and H 1 = 4t Now, by Theorem 8, we must verify that x 1 is an integer. Here, x 1 = 3t+28. The sequences are given by {G} : 1, 2, 5, 13,... ; {H} : t, 4t + 35, 11t + 105, 29t + 280,... ; {x} : t + 7, 3t + 28, 8t + 77, 21t + 203, 55t + 532, 144t , 377t ,.... It is easy to see that for any G 1 Z there exist suitable sequences {H}. Indeed, for an arbitrary G 1 we obtain the diophantine equation (2G 1 3)H 1 + (2 3G 1 )H 0 = 35. (21) But (21) is solvable in H 1 and H 0 since gcd(2g 1 3, 2 3G 1 ) divides 5. Note, that there are fewer problems in the other two cases of Theorem 8, since, instead of c, the right hand side of the corresponding linear diophantine equations is zero. Therefore, we only have to guarantee x 1 Z. Example 13. We give a class of recurrences {G} having no pair {H} with the desired property. Suppose that B = 1, c = 1, and to facilitate the calculations, let G 1 = G 0. Thus gcd(2g 1 G 0, 2G 0 + G 1 ) = G 0 gcd(2, 2 + ) = G 0 ( 2). But = excludes = 2, therefore G 0 ( 2) ( 2 4) G 0 ( + 2). Hence, if we choose an integer G 0 which does not divide + 2, then the linear diophantine equation G 0 (2 )H 1 + G 0 ( 2)H 0 = 2 4 is not solvable in integers H 1 and H 0. Example 14. Consider now the first example in the Introduction, where the composition of four sequences was defined. Let = 6, B = 1 and c = 1. Let also G 0 = 1, G 1 = 1, J 0 = 0, and J 1 = 2. We have = 32, = 32. Both of the diophantine equations 8H 1 8H 0 = 32 4K 1 12K 0 = 32 10

11 are solvable. Take the parametrizations H 0 = t Z, H 1 = t+4 and K 0 = s Z, K 1 = 3s+8. To join the subsequences with even and odd subscripts, we must check the recurrence relation x n+2 = 6x n+1 x n. Since the four sequences generate {G} : 1, 1, 7, 41,... ; {H} : t, t + 4, 5t + 24, 29t + 140,... ; {J} : 0, 2, 12, 70,... ; {K} : s, 3s + 8, 17s + 48, 99s + 280,... {x} : t + 1, 1, t + 5, 6s + 17, 35t + 169, 204s + 577,..., therefore, by the given recurrence rule, 6(t+5) 1 = 6s +17, or equivalently s = t+2 must hold. So we obtain {x} : t + 1, 1, t + 5, 6t + 29, 35t + 169, 204t + 985,... where the particular case t = 1 gives the sequence of balancing numbers. {x} : 0, 1, 6, 35, 204, 1189, cknowledgements The authors thank the anonymous referee for valuable remarks, especially for the idea to shorten the proof of Theorem 6. References [1] J. P. dams, Puzzles for Everybody, von Publications, [2]. Behera and G. Panda, On the square roots of triangular numbers, Fibonacci Quart., 37 (1999), [3] R. P. Finkelstein, The house problem, merican Math. Monthly, 72 (1965), [4] T. N. Shorey and R. Tijdeman, Exponential iophantine Equations, Cambridge University Press, Mathematics Subject Classification: Primary 11B39; Secondary Keywords: binary recurrence. 11

12 (Concerned with sequences , , , , , and ) Received ugust ; revised version received pril ; May Published in Journal of Integer Sequences, June Return to Journal of Integer Sequences home page. 12