Fibonacci-Like Sequence and its Properties - Hikari Ltd

Int. J. Contemp. Math. Sciences, Vol. 5, 2010, no. 18, 859 - 868

Fibonacci-Like Sequence and its Properties Bijendra Singh 1 , Omprakash Sikhwal 2 and Shikha Bhatnagar 3 1

School of Studies in Mathematics Vikram University Ujjain, India [email protected] 2

Department of Mathematics Mandsaur Institute of Technology, Mandsaur, India [email protected] 3

Bhanpura, District, Mandsaur (M. P.) India [email protected]

Abstract The Fibonacci sequence is a source of many nice and interesting identities. A similar interpretation exists for Lucas sequence. In this paper, we study Fibonacci-Like sequence that is defined by the recurrence S n = Sn −1 + Sn − 2 , for all n ≥ 2, S0 = 2, S1 = 2 . The associated initial conditions are the sum of initial conditions of Fibonacci and Lucas sequences respectively. We shall define Binet’s formula and generating function of Fibonacci-Like sequence. Mainly, Inducion method and Binet’s formula will be used to establish properties of Fibonacci-Like sequence. Mathematics Subject classification: 11B39, 11B37, 11B99 Keywords: Fibonacci sequence, Lucas sequence, Fibonacci-like sequence

1. INTRODUCTION As illustrated in the tome by Koshy [11], the Fibonacci and Lucas numbers are arguably two of the most interesting sequences in all of mathematics. Many identities have been documented in an extensive list that appears in the work of Vajda [10], where they are proved by algebraic means, even though combinatorial proofs of many of these interesting identities are also given [3]. The sequence of Fibonacci numbers Fn is defined by

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Fn = Fn-1 + F n-2 , n ≥ 2, F0 = 0, F1 = 1.

(1.1)

The sequence of Lucas numbers Ln is defined by Ln = L n-1 + L n-2 , n ≥ 2, L0 = 2, L1 = 1.

(1.2)

The Binet’s formula for Fibonacci sequence is given by n n φn −φ n 1 ⎧⎪⎛ 1 + 5 ⎞ ⎛ 1 − 5 ⎞ ⎫⎪ (1.3) Fn = = ⎟ −⎜ ⎟ ⎬ , ⎨⎜ φ −φ 5 ⎪⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎪ ⎩ ⎭ 1+ 5 1− 5 where φ = = Golden ratio ≈ 1.618 and φ = ≈ −0.618. 2 2 In this paper, we present properties of Fibonacci-like sequence that is defined by S n = Sn −1 + Sn − 2 , for all n ≥ 2, S0 = 2 and S1 = 2. (1.4) Here initial conditions S 0 and S1 are the sum of initial conditions of Fibonacci and Lucas sequences respectively, i.e. S0 = F0 + L0 , S1 = F1 + L1. The few terms of the sequence S n are 2, 2, 4, 6, 10, 16, 26 and so on.

2. PRELIMINARY RESULTS OF FIBONACCI-LIKE SEQUENCE

We will need to introduce some basic results of Fibonacci-Like sequence and Fibonacci sequence. The Fibonacci-Like sequence may also be written as n [ ] ⎛n ⎞ 2 ⎛n−k ⎞ ⎛ n ⎞ ⎛ n − 1⎞ ⎛ n − 2 ⎞ 2⎟ ⎜ Sn = ∑ 2 ⎜ (2.1) ⎟ = 2⎜ ⎟ + 2⎜ ⎟ + 2⎜ ⎟ +L + 2⎜ ⎟, n k =0 ⎝ k ⎠ ⎝0⎠ ⎝ 1 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ where [x] is defined as the greatest integer less than or equal to x. The relation between Fibonacci sequence and Fibonacci-Like sequence be written as Sn = 2 Fn . (2.2) The recurrence relation (1.1) has the characteristic equation x 2 -x-1 = 0 , which produces two roots as 1+ 5 1− 5 ≈ −0.618. (2.3) α= = Golden ratio ≈ 1.618 = φ , β = 2 2 Now notice a few things about α and β: α + β = 1, α − β = 5 and αβ = −1. (2.4) Using these two roots, we obtain Binet’s formula of recurrence relation (1.4) α n +1 − β n +1 2 = Sn = 2 (α n+1 − β n+1 ) , α −β 5 (2.5) n +1 n +1 ⎛ 1 − 5 ⎞ ⎫⎪ 2 ⎧⎪⎛ 1 + 5 ⎞ = ⎟ − ⎜⎜ ⎟ ⎬. ⎨⎜ 2 ⎟⎠ ⎪ 5 ⎪⎜⎝ 2 ⎟⎠ ⎝ ⎩ ⎭

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861

The generating function of S n is defined by ∞

S = ∑ Sn x n = n=0

2 . (1 − x − x 2 )

(2.6)

This can be written as 2 2 S= = ( x + α )( x + β ) (1 − α x )(1 − β x ) =

⎞ 2 ⎛ 1 1 − ⎜⎜ ⎟ 5 ⎝ (1 − α x ) (1 − β x ) ⎟⎠

=

2 ⎛ ∞ k +1 k ∞ k +1 k ⎞ 2 ∞ α k +1 − β k +1 ) x k . ( ∑ ⎜ ∑α x − ∑ β x ⎟ = 5 ⎝ k =0 5 k =0 k =0 ⎠

(2.7)

3. PROPERTIES OF FIBONACCI-LIKE SEQUENCE

Despite its simple appearance the Fibonacci-Like sequence contains a wealth of subtle and fascinating properties [8], [10]. Sums of Fibonacci –Like numbers: Theorem 3.1: Sum of first n terms of the Fibonacci –Like sequence is defined by n

S1 + S 2 + s3 + L + S n = ∑ S k = S n + 2 − 4.

(3.1)

k =1

This identity becomes 2n

S1 + S 2 + s3 + L + S 2 n = ∑ S k = S 2 n + 2 − 4.

(3.2)

k =1

Theorem 3.2: Sum of first n terms with odd indices is defined by n

S1 + S3 + s5 + L + S 2 n −1 = ∑ S 2 k −1 = S 2 n − 2.

(3.3)

k =1

Theorem 3.3: Sum of first n terms with even indices is defined by n

S 2 + S 4 + s6 + L + S 2 n −1 = ∑ S 2 k = S 2 n +1 − 2.

(3.4)

k =0

The identities from 3.1 to 3.3 can be derived by induction method. If we subtract 3.4 termwise from 3.3, we get alternating sum of the first n numbers (3.5) S1 − S 2 + S 3 − S 4 + L + S 2 n −1 − S 2 n = S 2 n − 2 − S 2 n +1 + 2 = − S 2 n −1 Adding S2 n +1 to both sides of 2.5, we get (3.6) S1 − S 2 + S 3 − S 4 + L + S 2 n −1 − S 2 n + S 2 n +1 = − S 2 n −1 + S 2 n +1 = S 2 n Combining (3.5) and (3.6), we get

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S1 − S 2 + S 3 − S 4 + L + ( −1) n +1 S n = ( −1) n +1 S n −1

(3.7)

Theorem 3.4: Sum of square of first n terms of the Fibonacci –Like sequence is n

S12 + S 22 + S 32 + L + S n2 = ∑ S k2 = S n S n +1 − 4. k =1

(3.8)

Now we state and prove some nice identities similar to those obtained for Fibonacci and Lucas sequences [1], [2], [3] and [11]. Theorem 3.5: For positive integer n, prove that S − n = ( −1) n S n − 2 , n ≠ 1.

(3.9)

Theorem 3.6: For positive integer n, prove that S n2 + S n2+1 = 2 S 2 n + 2

(3.10)

Proof: By Binet’s formula, first we calculate 4 4 S n2 + S n2+1 = {(α n +1 − β n +1 ) 2 + (α n + 2 − β n + 2 ) 2 } = {α 2 n + 2 (1 + α 2 ) + β 2 n + 2 (1 + β 2 )} . 5 5 2 2 Because 1 + α = 5α and 1 + β = − 5β , we get 4 S n2 + S n2+1 = α 2 n +3 − β 2 n +3 } = 2 S 2 n + 2 . { 5

Theorem 3.7: For positive integer n, prove that S n2+1 − S n2−1 = 2 S 2 n +1 .

(3.11)

Proof: By Binet’s formula, first we calculate 4 4 S n2+1 − S n2−1 = {(α n + 2 − β n + 2 ) 2 − (α n − β n ) 2 } = {α 2 n (α 4 − 1) + β 2 n ( β 4 − 1)} . 5 5 4 2 4 Because (α − 1) = 5α and ( β − 1) = − 5β 2 , we get 4 S n2+1 − S n2−1 = α 2 n + 2 − β 2 n + 2 )} = 2 S 2 n +1. { 5

Theorem 3.8: For positive integer n, prove that S S5 + S8 + S11 + L + S3 n + 2 = 3 n + 4 − 5. 2 Proof: By Binet’s formula, we have

(3.12)

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S5 + S8 + S11 + L + S3n + 2 2 (α 6 − β 6 ) + (α 9 − β 9 ) + (α 12 − β 12 ) + L + (α 3n +3 − β 3n +3 )} { 5 2 = (α 6 + α 9 + α 12 + L + α 3n +3 ) − ( β 6 + β 9 + β 12 + L + β 3n +3 )} { 5 2 ⎧ (α 3 n + 6 − α 6 ) ( β 6 − β 3n + 6 ) ⎫ = + ⎨ ⎬ 3 β 3 −1 ⎭ 5 ⎩ α −1 =

=

S3n + 4 − 5. 2

(since α 3 − 1 = α , β 3 − 1 = β )

Theorem 3.9: For positive integer n, prove that 4 S n3 = ⎡⎣ S3 n + 2 − 3( −1) n +1 S n ⎤⎦ . 5 This can be derived same as theorem 2.3.

(3.13)

Theorem 3.10: For positive integer n, prove that S13 + S 23 + S33 + L + S n3 4 = ⎡ ( S5 + S8 + S11 + L + S3 n + 2 ) − 3 ( S1 − S 2 + S3 − S 4 + L + (−1) n +1 S n ) ⎤ ⎦ 5⎣ 1 n +1 = 2 S3 n + 4 − 20 − ( −1) 12 S n −1 . (By 3.8 and 3.12) 5 This can be derived same as theorem 2.3.

{

{

}

(3.14)

}

Theorem 3.11: For positive integer n, prove that 2S m + n +1 = Sm Sn +1 + Sm −1Sn , m ≥ 1, n ≥ 0. Proof: Let m be fixed and we proceed by inducting on n. When n = 0, then 2 S m +1 = Sm S1 + S m −1

= 2Sm + 2 S m −1S0 = 2 ( Sm + S m −1 ) = 2 S m +1 , which is true.

(3.15)

,

When n = 1, then 2 S m + 2 = Sm S 2 + S m −1S1 = 4Sm + 2 S m −1 = 2Sm + 2Sm + 2 S m −1

= 2Sm + 2Sm +1 = 2 ( Sm + S m +1 ) = 2 S m + 2 , which is also true.

Now assume that identity is true for n = k and by assumption, 2 Sm + k = Sm Sk + Sm −1S k −1 and 2S m + k +1 = Sm Sk +1 + Sm −1Sk . Adding these two relations, we get

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2 Sm + k +1 + 2 Sm + k = Sm ( S k +1 + S k ) + S m −1 ( Sk + Sk −1 ) = 2 ( Sm + k +1 + Sm + k ) = Sm Sk + 2 + S m −1Sk +1 = Sm S k + 2 + Sm −1Sk +1 = Sm + k + 2 , which is true for n = k + 1. Hence, 2 Sm + n +1 = Sm Sn +1 + Sm −1S n .

Corollary 3.11.1: If m=n, then we get 2 S 2 n +1 = Sn S n +1 + S n −1S n = Sn ( S n +1 + S n −1 ) = S n2+1 − S n2−1. Corollary 3.11.2: If m=n-1, then we get 2 S2 n = Sn-1S n +1 + Sn − 2 S n = Sn-1 ( Sn + Sn −1 ) + S n − 2 Sn

= Sn ( Sn −1 + Sn − 2 ) + Sn2−1 = S2n + Sn2−1. Corollary 3.11.3: If m=2n-1, then we get 2 S3n = S2n-1S n +1 + S 2 n − 2 S n = S2n-1 ( S n + S n −1 ) + S 2 n − 2 S n = S nS2n + S 2 n −1S n −1.

Theorem 3.12: Prove that S n2 -Sn + r S n − r = (−1) n − r +1 Sr2−1 , r ≥ 1, n ≥ 1.

(3.16)

This can be derived same as theorem 2.3. Theorem 3.13: Prove that Sm +1S n − Sm Sn +1 = 2(−1) m +1 S n − m−1 , n ≥ 1, m ≥ 1.

(3.17)

This can be derived same as theorem 2.3. Theorem 3.14: Prove that Sn +1S n −1 − S2n = 4(−1) n +1 , n ≥ 0.

(3.18)

This can be derived same as theorem 2.3. We derive the identity related to binomial coefficients. Theorem 3.15: Prove that n ⎛n⎞ S2n = ∑ ⎜ ⎟ Sn−k . k =0 ⎝ k ⎠ Proof: By recurrence relation, we have

(3.19)

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S 2 n = S2 n −1 + S2 n − 2 = ( S2 n − 2 + S2 n −3 ) + ( S 2 n −3 + S 2 n − 4 ) = S 2 n − 2 + 2 S 2 n −3 + S 2 n − 4 . Again S 2 n = S 2 n − 2 + 2 S 2 n −3 + S 2 n − 4 = ( S 2 n −3 + S 2 n − 4 ) + 2 ( S 2 n − 4 + S 2 n − 5 ) + ( S 2 n −5 + S 2 n − 6 ) = S2 n −3 + 3S 2 n − 4 + 3S 2 n −5 + S 2 n −6 . Going on n times, we get n n ( n − 1) n ( n − 1) ⎛n⎞ S 2 n = Sn + nSn −1 + Sn − 2 + ... + S2 + nS1 + S0 = ∑ ⎜ ⎟ Sn − k . 2 2 k =0 ⎝ k ⎠

4. CONNECTION FORMULAE Theorem 4.1: Prove that 2 Ln +1 = S n +1 + S n −1 , n ≥ 1. Proof: We shall prove this identity by induction. It is easy to see that for n = 1 2 L2 = 2 × 3 = 6 = 4 + 2 = S 2 + S0 . Now suppose that the identity holds for n = k − 2 and n = k − 1. Then, 2 Lk −1 = S k −1 + S k −3 . 2 Lk = S k + S k − 2 . Adding equation (4.2) and equation (4.3), we get 2 Lk −1 + 2 Lk = ( S k −1 + S k ) + ( S k − 2 + S k −3 )

(4.1)

(4.2) (4.3)

i.e. 2 Lk +1 = S k +1 + S k −1. Which is precisely our identity when n = k . Hence 2 Ln +1 = S n +1 + S n −1 , n ≥ 1.

Theorem 4.2: Prove that 2 Fn +1 = S n +1 − Sn −1 , n ≥ 1. Proof: We shall prove this identity by induction over n. For n = 0 , we see that 2 F1 = 2 × 1 = 2 = 4 − 2 = S2 − S0 , which is true. Now suppose that the identity holds for n = k − 1 and n = k − 2. Then, 2 Fk = Sk − Sk − 2 . 2 Fk −1 = Sk −1 − Sk −3 . Adding equation (4.4) and equation (4.5), we get

(4.4)

(4.5) (4.6)

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2 Fk + 2 Fk −1 = ( S k + S k −1 ) − ( S k − 2 + S k −3 ) i.e. 2 Fk +1 = S k +1 − S k −1 , which is true for n = k . Hence 2 Fn +1 = Sn +1 − Sn −1 , n ≥ 1.

Theorem 4.3: Prove that S n +1 = Ln +1 + Fn +1.

(4.7)

Proof: By (4.1) and (4.4), we have 2 Fn +1 = Sn +1 − Sn −1 , n ≥ 1. 2 Ln +1 = S n +1 + S n −1 , n ≥ 1. Adding both above results, we get 2 Fn +1 + 2 Ln +1 = 2S n +1 , i.e. Sn +1 = Ln +1 + Fn +1.

Theorem 4.4: Prove that S n −1 = Ln +1 − Fn +1 , n ≥ 1.

(4.8)

Proof: Subtracting (4.1) from (4.4), we get 2 Ln +1 − 2 Fn +1 = 2S n −1

i.e. S n −1 = Ln +1 − Fn +1.

Theorem 4.5: Prove that L2 n + F2 n = S2 n .

(4.9)

Proof: we shall use mathematical induction method over n . For n = 0, L0 + F0 = 2 + 0 = 2 = S0 = S 2×0 . , which is true for n=0. For n = 1, L2 + F2 = 3 + 1 = 4 = S2 = S2×1. , which is also true for n=1 Assume that the result is true for n = k , then L2 k + F2 k = S 2 k . Now L2( k +1) + F2( k +1) = ( L2( k +1) −1 + L2( k +1)− 2 ) + ( F2( k +1)−1 + F2( k +1)− 2 )

= ( L2( k +1) −1 + F2( k +1) −1 ) + ( L2( k +1)− 2 + F2( k +1)− 2 ) = S2( k +1) −1 + S 2( k +1) − 2

(By induction hypothesis)

= S 2( k +1) . Therefore, L2( k +1) + F2( k +1) = S2( k +1) . , which is also true for n=k+1. Hence, the result is true for all n.

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5. CONCLUSION

There are many known identities established for Fibonacci and Lucas sequences. This paper describes comparable identities of Fibonacci-Like sequence. We have also developed connection formulas for Fibonacci-Like sequence, Fibonacci sequence and Lucas sequence respectively. It is easy to discover new identities simply by varying the pattern of known identities and using inductive reasoning to guess new results. Of course, the ideas can be extended to more general recurrent sequences in obvious way.

REFERENCES

[1]

A. F. Horadam, Basic Properties of Certain Generalized Sequence of Numbers, The Fibonacci Quarterly, 3 (1965), 161-176.

[2]

A. F. Horadam, The Generalized Fibonacci Sequences, The American Math. Monthly, 68, No. 5 (1961), 455-459.

[3]

A. T. Benjamin and J. J. Quinn, Recounting Fibonacci and Lucas identities, College Math. J., 30, No. 5 (1999), 359-366.

[4]

D. M. Burton, Elementary Number Theory, Tata McGraw-Hill Publishing Company Ltd., New Delhi, 2006.

[5]

E. Weisstein et al., Fibonacci number, from Math World- a Wolfram Web Resource, http://mathworld.wolfram.com/FibonacciNumber.html.

[6]

J. M. Patel, Problem H-635, The Fibonacci Quarterly, 44, No.1 (2006), 91.

[7]

M. E. Waddill and Louis Sacks, Another Generalized Fibonacci Sequence, The Fibonacci Quarterly, Vol. 5 No. 3, (1967), 209-222.

[8]

N. N. Vorobyov, The Fibonacci Numbers, D. C. Health and company, Boston, 1963.

[9]

S. Basir and V. Hoggatt, Jr., A Primer on the Fibonacci Sequence, Part II, The Fibonacci Quarterly, 1 (1963), 61-68.

[10]

S. Vajda, Fibonacci & Lucas numbers and the Golden section, Ellis Horwood Limited, Chichester, England, 1989.

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[11]

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T. Koshy, Fibonacci and Lucas numbers with Applications, Wiley, 2001.

Received: June, 2009