Iterated Binomial Transforms of the k-Fibonacci ... - Journal Repository

British Journal of Mathematics & Computer Science 4(22): 3135-3145, 2014

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Iterated Binomial Transforms of the k-Fibonacci Sequence Sergio Falcon1∗ 1 Department of Mathematics, University of Las Palmas de Gran Canaria, 35017-Las Palmas de

Gran Canaria, Spain. Article Information DOI: 10.9734/BJMCS/2014/11783 Editor(s): (1) Jaime Rangel-Mondragon, Faculty of Informatics, Queretaros Institute of Technology, Mexico and Faculty of Computer Science, Autonomous University of Quertaro, Mexico. Reviewers: (1) Kikawa Cliff Ricahrd, Mathematics and Statistics, Tshwane University of Technology. Republic of South Africa. (2) Gerhard-Wilhelm Weber, IAM, METU, Ankara, Turkey. (3) E. M. Elsayed, King Abdulaziz University, Faculty of Science, Mathematics Department, P. O. Box 80203, Jeddah 21589, Saudi Arabia. Peer review History: http://www.sciencedomain.org/review-history.php?iid=636id=6aid=5918

Original Research Article Received: 04 June 2014 Accepted: 04 July 2014 Published: 28 August 2014

Abstract In this paper, we study the iterated application of the Binomial transform, the k–Binomial transform, the Rising k–Binomial transform, and the Falling k–Binomial transform to the k–Fibonacci sequence. In particular, we obtain the recurrence relation between the terms of the sequences obtained from these transforms and prove that they are all generalized Fibonacci sequences. As a consequence of this result, we obtain the Generating Function of these sequences, and Binnet Identity and Combinatorial Formula for the general term of each of them. We can consider the iterated application of the Binomial transforms as a new way to get integer sequences. But the way we have done, we have also found the recurrence relation between the terms of these sequences and how to find the general term of the same, either by the Binet Identity and the Combinatorial formula. Keywords: Generalized Fibonacci sequence, k–Fibonacci numbers, Recurrence formula, Generating function, Binomial transforms. 2010 Mathematics Subject Classification: 11B39 *Corresponding author: E-mail: [email protected]

British Journal of Mathematics and Computer Science 4(22), 3135-3145, 2014

1

Introduction

We begin this Introduction recalling both the definition of the k–Fibonacci sequence [1] and the generalized Fibonacci sequences [2]. Then we give the generating function of these last sequence and a way to obtain it. We finish this section finding both the Binet Identity and the Combinatorial formula for these numbers. For a positive real number k, the k–Fibonacci sequence, say Fk = {Fk,n } is defined recurrently by Fk,0 = 0, Fk,1 = 1 and Fk,n+1 = k Fk,n + Fk,n−1 for n ≥ 1. So, the terms of k–Fibonacci sequence are Fk = {Fk,n } = {0, 1, k, k2 + 1, k3 + 2k, k4 + 3k2 + 1, . . .}. For the case k = 1, we obtain the classical Fibonacci sequence F1 = F = {0, 1, 1, 2, 3, 5, 8, . . .}; while if k = 2, the Pell sequence F2 = P = {0, 1, 2, 5, 12, 29, 70, . . .} is generated. Classical Fibonacci sequence had been studied for several authors Hoggat [3]; Horadam [2], and applied to other sciences, as El Naschie M.S. [4,5,6]. Let the integer sequence {Vn }n∈N be with the initial conditions V0 = 0 and V1 . Let us suppose the terms of this sequence verify the recurrence relation Vn+1 = aVn − bVn−1 , a, b > 0

(1.1)

for n ≥ 1. The sequence {Vn } is called a generalized Fibonacci sequence. We say f (x) is the generating function of the sequence {Vn } if f (x) can be expanded in Taylor series as f (x) = V0 + V1 x + V2 x2 + V3 x3 + · · · Next we will prove three theorems related to this sequence. Theorem 1.1 (Generating function). The generating function of the sequence {Vn } is f (x) =

V1 x 1 − ax + bx2

(1.2)

Proof. If f (x) = V0 + V1 x + V2 x2 + V3 x3 + · · · is the generating function of {Vn }, multiplying it successively by ax and bx2 , we have f (x)

=

axf (x)

=

2

bx f (x)

V0 + V1 x + V2 x2 + V3 x3 + · · · aV0 x + aV1 x2 + aV2 x3 + · · · bV0 x2 + bV1 x3 + · · ·

=

Then f (x)(1 − ax + bx2 ) = V0 + (V1 − aV0 )x + (V2 − aV1 + bV0 )x2 + (V3 − aV2 + bV1 )x3 + · · · And taking into account both the recurrence relation (1.1) and V0 = 0, it is f (x)(1 − ax + bx2 ) = V1 x from where the generating function of {Vn } is f (x) =

V1 x 1 − ax + bx2

(1.3)

V1 Theorem 1.2 (Binet Identity). The general term of the sequence {Vn } is Vn = (r1n −r2n ) being r − r 1 2 √ √ a + a2 − 4b a − a2 − 4b and r2 = the solutions of the recurrence equation r2 − ar + b = 0. r1 = 2 2 Proof. Characteristic equation √ of the recurrence relation (1.1) is r2 − ar + b = 0 which solutions √ 2 a+ a −4b a− a2 −4b are r1 = and r2 = . Then, the general term of the sequence {Vn } is Vn = 2 2 C1 r1n + C2 r2n . For n = 0 it is C1 + C2 = V0 = 0 → C2 = −C1 , and for n = 1 it is V1 V1 = −C2 . Finally C1 r1 + C2 r2 = V 1 → C1 = = √ 2 r1 − r2 a − 4b √ √  n  n  V1 a + a2 − 4b a − a2 − 4b Vn = √ − (1.4) 2 2 a2 − 4b

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√ Characteristic roots, verify r1 − r2 = a2 − 4b, r1 + r2 = a , r1 · r2 = b. If a = = 1, we obtain the k–Fibonacci sequence which characteristic roots are √ k, b = −1 and V1 √ k+

k2 +4

k−

k2 +4

σ1 = and σ2 = . The positive values σ1 are known as metallic ratios Spinadel [7]: 2 2 Golden ratio (or Divine proportion), Silver ratio, and Bronze ratio, for k = 1, 2, 3, respectively. Theorem 1.3 (Combinatorial formula for Vn ). For n ≥ 1,

Vn =

V1 2n−1

b n−1 c 2

X j=0

! n an−2j−1 (a2 − 4b)j 2j + 1

(1.5)

Proof. If we expand the expression between the bracket of Formula (1.4), the respective odd terms vanish among √ themselves while the even terms are added. Consequently,

Vn = √

V1 a2 − 4b

a2 − 4b 2n−1

       n n−1 n n−3 2 n n−5 2 a + a (a − 4b) + a (a − 4b)2 + · · · and 1 3 5

from here, the combinatorial Formula (1.5). As particular case, if we apply these three theorems to the k-Fibonacci sequence in which a = k, b = −1 and V1 = 1, then Falcon and Plaza [1,8] x 1. Generating function: f (x) = 1 − kx − x2 √ √  n  n  k + k2 + 4 k − k2 + 4 1 2. Binnet Identity: Vn = √ − 2 2 k2 + 4 n−1 ! b 2 c X n 1 kn−2j−1 (k2 + 4)j 3. Combinatorial formula: Vn = n−1 2 2j + 1 j=0

2

Iterated Binomial Transforms

In this section we apply iteratly the four Binomial transforms defined in Falcon and Plaza [8], studying each separately. Binomial transforms are defined and studied in Chen [9]; Prodinger [10]; Spivey and Steil [11], and they have been applied to the k–Fibonacci sequence in Falcon and Plaza [8]: ! n X n • Binomial transform: bk,n = Fk,i . i i=0 ! n X n n−i k Fk,i . • Falling k–Binomial transform: fk,n = i i=0 ! n X n i k Fk,i . • Rising k–Binomial transform: zk,n = i i=0 ! n X n n • k–Binomial transform: wk,n = k Fk,i . i i=0 For k = 1, these four transforms coincide and therefore when applied to the classical Fibonacci sequence, produce the bisection sequence of the classical Fibonacci sequence {0, 1, 3, 8, 21, . . .}, A001906 in Sloane [12], OEIS from now on.

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2.1

Iterated Binomial transforms of the classical Fibonacci sequence

Let us suppose we apply ”r” times each of these transforms to the k–Fibonacci sequence. If k = 1 the four Binomial transforms coincide and, consequently, if T is one of these transforms, the sequence T (r) (Fk ) is independent of the transform. If we represent the classical Fibonacci sequence as F1 , then: T (1) (F1 ) = A001906, T (2) (F1 ) = A093131, T (3) (F1 ) = A099453∪{0}. T (4) (F1 ) = A081574, T (5) (F1 ) = A081575 and none more of the sequences n so o obtained appears in OEIS. (r)

The terms of the sequence T (r) = tn

verify the recurrence relation

(r)

(r)

(r)

(r)

(r) tn+1 = (2r + 1)tn − (r2 + r − 1)tn−1 , n ≥ 1, with t0 = 0, t1 = 1

This formula is a particular case of Formula √ (2.1) that we will prove in the next subsection. 1± 5 Characteristic roots are st,1,2 = + r → st,1 = Φ + r. 2 (r) So, sequence T is a generalized Fibonacci sequence that verifies (1.1) with a = 2r + 1, b = r2 + r − 1, and V1 = 1 so a2 − 4b = 5. If we apply Theorems 1, 2, and 3, for the sequence T (r) it is: 1. Generating function: t(x, r) = (r)

(r)

2. Binnet Identity: tn+1 =

x 1 − (2r + 1)x + (r2 + r − 1)x2 (r)

st,1 − st,2 √ 5

3. Combinatorial formula: Vn =

1

X

2n−1

j=0

! n (2r + 1)n−2j−1 · 5j 2j + 1

Next, we will study these transforms in a general case.

2.2

Iterated Binomial transform of the k–Fibonacci sequence

If we apply n ”r”o times the Binomial transform to the k–Fibonacci sequence, then obtain the sequence (r) (r) Bk = bk,n . In the sequel we will try to find a recurrence formula for the terms of this sequence. It is obvious (r) (r) that bk,0 = 0 and bk,1 = 1. Before finding a formula for the general term of this sequence, we need to prove the following Lemma.

2.2.1

Lemma (r)

For r ≥ 1 and n ≥ 0, the terms of the sequence Bk verify the recurrence relation ! n X n (r−1) (r) (r) bk,n+1 = bk,n + b j k,j+1 j=0 (r)

Proof. By definition of Binomial transform, bk,n+1 =

n+1 X j=0

Now, we apply the Pascal Identity

n+1 j




=

n j




+

n j−1




! n + 1 (r−1) bk,j . j

proved in Graham, Knuth and Patashnik [13],

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so, (r) bk,n+1

=

=

! !# ! ! n+1 X n (r−1) n+1 X n n n (r−1) (r−1) + bk,j = bk,j + bk,j j j − 1 j j − 1 j=1 j=0 j=0 ! ! ! n n n X X X n (r−1) n (r−1) n (r−1) (r) bk,j + bk,j+1 = bk,n + b j j j k,j+1 j=0 j=0 j=0

n+1 X

"

We are ready to prove the following Theorem. (r)

Theorem 2.1. The terms of the sequence Bk verify the recurrence relation (r)

(r)

(r)

bk,n+1 = (2r + k)bk,n − (r2 + k r − 1)bk,n−1 (r)

(2.1)

(r)

with the initial conditions bk,0 = 0 and bk,1 = 1. Proof. We will prove this formula by double induction of the two variables, r and n. We begin with the proof for r = 1, then n = 1, and continue for general values of r and n, jointly. 1. For r = 1. Let the k–Fibonacci sequence Fk = {0, 1, Fk,2 , Fk,3 , . . .} be. We repeat the proof of this transform if r = 1 as we have proved in Falcon and Plaza [8]. Previously to this proof, we need to demonstrate the following Lemma. Lemma 2.2. Binomial transform {bk,n } of the k–Fibonacci sequence verifies the relation bk,n+1

! n X n (Fk,i+1 + Fk,i ) = i i=0

Proof. Taking into account the Pascal Identity Graham, Knuth and Patashnik [13], ! ! n n + , it is i i−1

bk,n+1

=

n+1 X i=0

=

n+1 X i=0

(2.2)

n+1 i

! =

! " ! !# n+1 X n+1 n n Fk,i = + Fk,i i i i−1 i=1 ! ! ! n n X X n n n Fk,i + Fk,i+1 = (Fk,i+1 + Fk,i ) i i i i=0 i=0

Note that Equation (2.2) can also be written as bk,n+1

! n X n = Fk,i+1 + bk,n . Now, we i i=0

proceed to demonstrate this theorem. Theorem 2.3. Binomial transform of the k-Fibonacci sequence, Bk = {bk,n }, verifies the recurrence relation bk,n+1 = (k + 2)bk,n − k bk,n−1 for n ≥ 1, (2.3) with initial conditions bk,0 = 0 and bk,1 = 1

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Proof. With Lemma 2.2 in mind and formula (??),

bk,n+1

=

=

=

! n X n (Fk,i+1 + Fk,i ) + (Fk,1 + Fk,0 ) i i=1 ! n X n (k Fk,i + Fk,i + Fk,i−1 ) + (Fk,1 + Fk,0 ) i i=1 ! ! n n X X n n (k + 1) Fk,i + Fk,i−1 + Fk,1 i i i=1 i=1

And therefore, bk,n+1

! n X n = (k + 1)bk,n + Fk,i−1 + 1 i i=1

On the other hand, taking into account that

bk,n

=

=

=

=

=

=

n−1 n




(2.4)

= 0, it is

! n X n−1 (k + 1)bk,n−1 + Fk,i−1 + 1 i i=1 ! ! n−1 n−1 X n−1 X n−1 k bk,n−1 + Fk,i + Fk,i−1 + 1 i i i=1 i=0 ! ! n−1 n−1 X n−1 X n−1 k bk,n−1 + Fk,i + Fk,i + 1 i i+1 i=0 i=0 " ! !# n−1 X n−1 n−1 + Fk,i+1 k bk,n−1 + i i+1 i=0 ! n−1 X n k bk,n−1 + Fk,i + 1 i+1 i=0 ! n X n k bk,n−1 + Fk,i−1 + 1 i i=1

And, hence, ! n X n Fk,i−1 + 1 = bk,n − k bk,n−1 i i=1

(2.5)

Therefore, by substituting this expression in Equation (2.4), the looked formula is obtained. Then, this formula is true for r = 1. 2. If we apply iteratly the Binomial transform to the sequence Fk , then we find that the three first (r) (r) (r) elements of the transformed sequence are bk,0 = 0, bk,1 = 1 and bk,2 = 2r + k = (2r + k)F1 . So this formula is true for n = 1. 3. Let us suppose Formula (2.1) is true for (r − 1, n) and (r, n − 1): (r−1) (r−1) (r−1) bk,n+1 = (2(r − 1) + k)bk,n − (r2 − 2r + (r − 1)k)bk,n−1 (r)

(r)

(r)

bk,n = (2r + k)bk,n−1 − (r2 + r k − 1)bk,n−2

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Applying Lemma (2.2.1) (r) bk,n+1

=

=

=

! ! n n X n (r−1) X n (r−1) b + b j k,j j k,j+1 j=0 j=0 ! n  X n  (r−1) (r−1) (r−1) bk,j + bk,j+1 + bk,1 j 1 ! n i X n h (r) (r−1) (r−1) bk,n + (2r − 2 + k)bk,j − (r2 − 2r + rk − k)bk,j−1 + 1 j 1

So (r) bk,n+1

= (2r + k −

(r) 1)bk,n

! n X n (r−1) − (r − 2r + rk − k) b +1 j k,j−1 1 2

(2.6)

From this equation we can deduce ! n − 1 (r−1) bk,j−1 + 1 j 1 ! n X n − 1 (r−1) (r) 2 (2r + k − 1)bk,n−1 − (r − 2r + rk − k) bk,j−1 + 1 j 1 !# " ! n X n−1 n (r−1) (r) 2 − bk,j−1 + 1 (2r + k − 1)bk,n−1 − (r − 2r + rk − k) j j−1 1 ! n X n (r−1) (r) 2 (2r + k − 1)bk,n−1 − (r − 2r + rk − k) b j k,j−1 1 ! n−1 X n − 1 (r−1) +(r2 − 2r + rk − k) bk,j + 1 j j=0 ! n X n (r−1) (r) 2 (2r + k − 1)bk,n−1 − (r − 2r + rk − k) b j k,j−1 1 (r)

(r)

bk,n = (2r + k − 1)bk,n−1 − (r2 − 2r + rk − k)

=

=

=

=

n−1 X

(r)

+(r2 − 2r + rk − k)bk,n−1 + 1 =

→

! n X n (r−1) (r + rk − − (r − 2r + rk − k) b +1 j k,j 0 ! n X n (r−1) (r) (r) −(r2 − 2r + rk − k) bk,j−1 + 1 = bk,n − (r2 + rk − 1)bk,n−1 j 1 2

(r) 1)bk,n−1

2

Finally, we substitute this expression in Formula (2.6) and this Theorem is proved. (r)

Sequence Bk is a generalized Fibonacci sequence (1.1) with a = 2r + k, b = r2 + k r − 1, and V1 = 1, so a2 − 4b = k2 + 4. Now, we will apply Theorems 1, 2, and 3, to the iterated Binomial transform of the k—Fibonacci sequence and obtain the following results: x 1. Generating function: b(k, x, r) = . 1 − (2r + k)x + (r2 + rk − 1)x2 n sn (σ1 + r)n − (σ2 + r)n b,1 − sb,2 (r) 2. Binnet Identity: bk,n = = sb,1 − sb,2 σ1 − σ2 ! b n−1 c 2 X 1 n (r) 3. Combinatorial formula: bk,n = n−1 (2r + k)n−2j−1 (k2 + 4)j 2 2j + 1 0

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Unique Binomial transforms of the k–Fibonacci sequence listed in OEIS (for k > 1) are the following:

Binomial transforms of the k–Fibonacci sequence k\ r 2 3 4

1 A007070 A116415 A084326

2 A081179 A099459 A091870

3 A081180 – A093145

(k)

4 A081182 – A153882

5–6–7 A081183–4–5 – A153884 – –

8 A153593 – –

(1)

In particular, if r = k, then Bk = Fk : the k–th binomial transform of the k–Fibonacci sequence equalizes the Falling Binomial transform.

2.3

Iterated Falling k–Binomial transform

Applying ”r” times the Falling Binomial transform to the k–Fibonacci sequence {Fk,n }, we find the (r) (r) (r) (r) (r) (r) (r) sequence Fk = {fk,n } where fk,0 = 0, fk,1 = 1, and fk,n+1 = (2r +1)kfk,n −((r2 +r)k2 −1)fk,n−1 for n ≥ 1, proved in a similar form to the used in the Iterated Binomial transform. (r) As before, the sequence Fk is a generalized Fibonacci sequence of the type (1.1) with  a = 2rk + k → a2 − 4b = k2 + 4 b = r2 k2 + rk2 − 1 Applying Theorems 1, 2, and 3, for the iterated Falling k—Fibonacci transform, we have: x 1. Generating function: f (k, x, r) = 1 − (2r + 1)kx + ((r2 + r)k2 − 1)x2 2. Binnet Identity: Solutions of the recurrence equation 2 2 r2 − (2r + 1)k · r + (r √ + r)k − 1 = 0 are (2r + 1)k ± k2 + 4 = σ1,2 + rk and the Binnet Identity is sf,1,2 = 2 n n n − s s (σ + rk) − (σ2 + rk)n 1 f,2 f,1 (r) fk,n = = sf,1 − rf,2 σ1 − σ2 3. Combinatorial formula: (r) fk,n

=

b n−1 c 2

1

X

2n−1

j=0

! n ((2r + 1)k)n−2j−1 (k2 + 4)j 2j + 1 (1)

(3) F2

Sh =

h X j=1

2.4

(2)

The unique Falling k–Binomial transforms listed in OEIS are F2 = A081179, F2 = A081182, (4) (2) = A081184, F2 = A153953, and F4 = A153882. In the sequel, we will take into account (Graham, Knuth and Patashnik [13], Formula (2.26)) j·x

a+j

xa+1 = 1−x



 xh − 1 h − h · x , if x 6= 1 x−1

(2.7)

Iterated Rising k–Binomial transform

If we apply ”r” times the Rising k–Binomial transform to the k–Fibonacci sequence, we find the (r) (r) (r) sequence {zk,n }n∈N with zk,0 = 0, zk,1 = kr and, for n ≥ 1,

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(r) zk,n+1

=

k

r+1

+2

r−1 X

! k

j

(r) rk,n

r−1 X

−

j=0

jk

2r−j

r−1 X + (r − 1)k + (j + 1)kj

!

r

j=1

(r)

rk,n−1

j=0

Applying Formula (2.7) to these coefficients, the relation between three consecutive terms of the sequence is written as     r (k − 1)(kr+1 − 1) kr − 1 (r) (r) (r) r rk,n−1 zk,n+1 = kr+1 + 2 rk,n − − k k−1 (k − 1)2 Then, V1 = kr , and r

)

−1 a = kr+1 + 2 kk−1

b=

(kr −1)(kr+1 −1) (k−1)2

→ a2 − 4b = (k2 + 4)k2r

− kr

Solutions of the recurrence equation   r r+1 r −1) −1 r + (k −1)(k r2 − kr+1 + 2 kk−1 − kr = 0 are sz,1,2 = (k−1)2 kr −1 k−1

√ kr −1 k−1

+ kr

k±

k2 +4 , 2

and then sz,1,2 =

r

+ k σ1,2 . Then it holds for the iterated Rising k–Binomial transform:

1. Generating function: z(k, x, r) = (kr − 1)(kr+1 − 1) − kr (k − 1)2 2. Binnet Identity

kr x kr − 1 r+1 , with a = k + 2 , 1 − ax + bx2 k−1

b=

kr z(k, x, r) = √ a2 − 4b



√ √ n  n  a + kr k2 + 4 a − kr k2 + 4 − 2 2

3. Combinatorial formula z(k, x, r) =

kr 2n−1

c b n−1 2

X j=0

! n an−2j−1 (kr (k2 + 4))j 2j + 1

The unique sequences obtained from the Rising k–Binomial transforms that appear in OEIS are for r = 1 Falcon and Plaza [8].

2.5

Iterated k–Binomial transform

If we apply ”r” times the k–Binomial transform to the k–Fibonacci sequence, we find the sequence (r) (r) (r) {wk,n }n∈N with wk,0 = 0, wk,1 = kr and, for n ≥ 1, ! ! r r r X X X (r) (r) (r) r+1 j 2r+2−j j+1 2r wk,n+1 = k +2 k wk,n − jk + jk −k wk,n−1 j=1

j=1

j=1

Applying Formula (2.7), the relation between consecutive terms   r three r+1  is  (k − 1)(k − 1) 2 kr − 1 (r) (r) (r) r+1 2r k wk,n − k −k wk,n−1 wk,n+1 = k +2 )(k − 1)2 kr+1 − 1 kr −1 a=k + 2 k−1 k Then, V1 = kr , and (kr −1)(kr+1 → a2 − 4b = (k2 + 4)k2r −1) 2 2r b= k − k 2 (k−1) √ r+1 (k+1)−2k Solutions of the recurrence equation are sw,1,2 = k 2(k−1) ± 12 kr k2 + 4 so sw,1,2 = kr σ1,2 . Then it holds for the iterated Rising k–Binomial transform: 1. Generating function: r+1 kr x w(k, x, r) = 1−ax+bx + 2 with a = k b=

k2 (kr −1)(kr+1 −1) (k−1)2

2k(kr −1) k−1

kr −1 k+ k−1

and

2r

−k .

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2. Binnet Identity w(k, x, r) =

kr √ kr k2 + 4



√ √ n  n  a + kr k2 + 4 a − kr k2 + 4 − 2 2

3. Combinatorial formula

z(k, x, r) =

kr 2n−1

c b n−1 2

X j=0

! n an−2j−1 (k2r (k2 + 4))j 2j + 1

The only sequence obtained from the Iterated k–Binomial transform for k 6= 1 and cited in OEIS (1) is W2 = 2 · A057084 (see Falcon and Plaza [8]).

3

Conclusions

So we applied the four different iterated Binomial transforms and we have obtained different sequences according to the iteration performed. We have found different ways to each of the generating function, Binet Identity and the binomial formula for each of its general terms. We propose to apply a similar study to other sequences as ”the generalized k–Fibonacci sequences”, ”the k–Lucas sequences”, ”the k–Jacobsthal sequences”, etc.

Competing interests The author declares that no competing interests exist.

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2014 Falcon; This is an Open Access article distributed under the terms of the Creative Commons Attribution License http://creativecommons.org/licenses/by/3.0, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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