on generalization of bi-periodic -numbers · positive integer and a;ba positive real numbers...

On generalization of bi-periodic r-numbers

N. Rosa AIT-AMRANE

University Yahia Fares, of Medea, Algeria.

USTHB, Faculty of Mathematics, RECITS Laboratory, CATI Team.

[email protected]

ALGOS 2020

26-28 August 2020

Introduction

Recently, many authors have studied generalizations of Fibonacci and

Lucas sequences. Yazlik et al.[1] introduced generalization, for r a

positive integer and a, b a positive real numbers

1. Yazlik Y, Kome C, Madhusudanan V. A new generalization of Fibonacci and

Lucas p-numbers. Journal of computational analysis and applications 25 (4), (2018),

657–669.N. Rosa AIT-AMRANE On generalization of bi-periodic r-numbers 1/1

Introduction

the bi-periodic Fibonacci r-numbers (fn), for n ≥ r + 1

fn =

{afn−1 + fn−r−1, for n ≡ 0 (mod 2),

bfn−1 + fn−r−1, for n ≡ 1 (mod 2),

and the bi-periodic Lucas r-numbers (ln), for n ≥ r + 1

ln =

{bln−1 + ln−r−1, for n ≡ 0 (mod 2),

aln−1 + ln−r−1, for n ≡ 1 (mod 2),

with the initial conditions f0 = 0, f1 = 1, f2 = a, ..., fr = abr/2cbb(r−1)/2c

and l0 = r + 1, l1 = a, l2 = ab, ..., lr = ab(r+1)/2cbbr/2c, respectively.

N. Rosa AIT-AMRANE On generalization of bi-periodic r-numbers 2/1

The bi-periodic r-Fibonacci sequence

First, we define the bi-periodic r-Fibonacci sequence (U(r)n )n

Definition

For a, b, c, d nonzero real numbers and r ∈ N, the bi-periodic

r-Fibonacci sequence (U(r)n )n is defined by, for n ≥ r + 1

U (r)n =

{aU

(r)n−1 + cU

(r)n−r−1, for n ≡ 0 (mod 2),

bU(r)n−1 + dU

(r)n−r−1, for n ≡ 1 (mod 2),

with the initial conditions

U(r)0 = 0, U

(r)1 = 1, U

(r)2 = a, . . . , U

(r)r = abr/2cbb(r−1)/2c.


The bi-periodic r-Fibonacci sequence

Which can be expressed by linear recurrence relation.

Theorem

For a, b, c, d nonzero real numbers and r ∈ N, the bi-periodic

r-Fibonacci sequence satisfies the following linear recurrence, for

n ≥ 2r + 2

U (r)n = abU

(r)n−2 +(aξ(r+1)d+ bξ(r+1)c)U

(r)n−r−1−ξ(r+1)− (−1)r+1cdU

(r)n−2r−2,

with U(r)0 = 0, U

(r)1 = 1, U

(r)2 = a, . . . , U

(r)r = abr/2cbb(r−1)/2c,

and for r + 1 ≤ n ≤ 2r + 1,

U (r)n =

ab

n2 cb

⌊(n−1)

2

⌋+(⌊n−r2

⌋d+

⌊n−r−1

2

⌋c)a

⌊(n−r−1)

2

⌋b

⌊(n−r−2)

2

⌋, for r odd,

abn2 cb

⌊(n−1)

2

⌋+⌊n−r2

⌋a

⌊(n−r−2)

2

⌋b

⌊(n−r−1)

2

⌋c+

⌊n−r−1

2

⌋a

⌊(n−r)

2

⌋bb

n−r−32 cd, for r even,

where ξ(k) = 2(k/2− bk/2c) is the parity function.


The bi-periodic r-Lucas sequence of type s

Secondly, we introduce a new family of companion sequences related to

the bi-periodic r-Fibonacci sequence, called the bi-periodic r-Lucas

sequence of type s, (V(r,s)n )n.

Definition

For any nonzero real numbers a, b, c, d and integers s, r such that

1 ≤ s ≤ r, we define for n ≥ r + 1

V (r,s)n =

{bV

(r,s)n−1 + dV

(r,s)n−r−1, for n ≡ 0 (mod 2),

aV(r,s)n−1 + cV

(r,s)n−r−1, for n ≡ 1 (mod 2),

with the initial conditions V(r,s)0 = s+ 1,

V(r,s)1 = a, V

(r,s)2 = ab, . . . , V

(r,s)r = ab(r+1)/2cbbr/2c.



The bi-periodic r-Lucas sequence of type s, 1 ≤ s ≤ r satisfy the

following linear recurrence relation.

Theorem

For a nonzero real numbers a, b, c, d and s, r such that 1 ≤ s ≤ r, the

family of the bi-periodic r-Lucas sequence of type s satisfy, for

n ≥ 2r + 2

V (r,s)n = abV

(r,s)n−2 +(aξ(r+1)d+bξ(r+1)c)V

(r,s)n−r−1−ξ(r+1)−(−1)r+1cdV

(r,s)n−2r−2,

with V(r,s)0 = s+ 1, V

(r,s)1 = a, V

(r,s)2 = ab, . . . , V

(r,s)r = ab(r+1)/2cbbr/2c,

and for r + 1 ≤ n ≤ 2r + 1,

V(r,s)n =

a

⌊(n+1)

2

⌋bb

n2 c +

((s+

⌊n−r+1

2

⌋)d+

⌊n−r2

⌋c)a

⌊(n−r)

2

⌋b

⌊(n−r−1)

2

⌋, for r odd,

a

⌊(n+1)

2

⌋bb

n2 c +

(s+

⌊n−r+1

2

⌋)a

⌊(n−r−1)

2

⌋b

⌊(n−r)

2

⌋c+

⌊n−r2

⌋a

⌊(n−r+1)

2

⌋b

⌊n−r−2

2

⌋d, for r even.

(1)



We list some particular cases

For a = b = c = d = 1 and r = s = 1, we get the classical

Fibonacci and Lucas sequences.

For a = b = 2, c = d = 1 and r = s = 1, we get the

classical Pell and Pell-Lucas sequences.

For a, b nonzero real numbers, c = d = 1 and r = s = 1,

we get the bi-periodic Fibonacci and bi-periodic Lucas

sequences.

For a, b nonzero real numbers, c = d = 2 and r = s = 1, we

get the Jacobsthal and the Jacobsthal-Lucas sequences.


The link

Now, we express the bi-periodic r-Lucas sequence of type s, V(r,s)n in

terms of U(r)n .

Theorem

Let r and s be nonnegative integers such that 1 ≤ s ≤ r, the bi-periodic

r-Fibonacci sequence and the bi-periodic r-Lucas sequence of type s

satisfy the following relationship

V (r,s)n =

U

(r)n+1 + sdU

(r)n−r, n ≥ r, for r odd,

U(r)n+1 + scbU

(r)n−r−1 + scdU

(r)n−2r−1, n ≥ 2r + 1, for r even.


The generating functions

Next, we give the generating function of the bi-periodic r-Fibonacci

sequence and the bi-periodic r-Lucas sequence of type s.

Theorem

Let r be nonnegative integer, the generating function of (U(r)n )n is

G(x) =x+ ax2 + (−1)ξ(r)cxr+2

1− abx2 − (aξ(r+1)d+ bξ(r+1)c)xr+1+ξ(r+1) − (−1)rcdx2r+2.

If r = 1 we obtain the generating function of the bi-periodic Fibonacci

sequence given by Sahin [1]

1. M. Sahin, The Gelin-Cesario identity in some conditional sequences, Hacettepe

Journal of Mathematics and statistics, vol 40 (6), (2011), 855-861.N. Rosa AIT-AMRANE On generalization of bi-periodic r-numbers 9/1

The generating functions

The following theorem express the generating function of (V(r,s)n )n

Theorem

Let r and s be nonnegative integers (1 ≤ s ≤ r), the generating

function of (V(r,s)n )n is

H(x) =(s+ 1) + ax− absx2 + (−1)ξ(r)(s+ 1)cxr+1 + (−1)ξ(r+1)adsxr+2

1− abx2 − (aξ(r+1)d+ bξ(r+1)c)xr+1+ξ(r+1) − (−1)rcdx2r+2.

If we take r = 1 and c = d = 1 we obtain the generation function of the

bi-periodic Lucas sequence given by Bilgici[1]

1. G. Bilgici, Two generalizations of Lucas sequence, Appl. Math. Comput. 245

(2014), 526-538.N. Rosa AIT-AMRANE On generalization of bi-periodic r-numbers 10/1

Explicit formulas

Next, we state an explicit formula of (U(r)n )n and (V

(r,s)n )n.

Theorem

For any integer r ≥ 1, we have

U(r)n+1 =

∑2i+(r+1)t+(r+1)k=n

(i+ t

t

)(t

k

)(ab)i(c+ d)t−k(−cd)k, for r odd,

∑2i+(r+2)t+rk=n

(i+ t

t

)(t

k

)(ab)i(ad+ bc)t−k(cd)k, for r even.


Explicit formulas

Now, we give an analogous result for the bi-periodic r-Lucas sequence

of type s.


Explicit formulas

Theorem

For any integers r and s, we have

for r odd

V (r,s)n =

∑2i+(r+1)t+(r+1)k=n

(i+ t

t

)(t

k

)(ab)i(c+ d)t−k(−cd)k

+ sd∑

2i+(r+1)t+(r+1)k=n−r−1

(i+ t

t

)(t

k

)(ab)i(c+ d)t−k(−cd)k


Explicit formulas

Theorem

for r even

V (r,s)n =

∑2i+(r+2)t+rk=n

(i+ t

t

)(t

k

)(ab)i(ad+ bc)t−k(cd)k

+ sbc∑

2i+(r+2)t+rk=n−r−2

(i+ t

t

)(t

k


+ scd∑

2i+(r+2)t+rk=n−2r−2

(i+ t

t

)(t

k



Explicit formulas

If we take r = 1 and c = d = 1, Yayenie [1] gave an explicit formula for

the bi-periodic Fibonacci sequence

pn = aξ(n−1)b(n−1)/2c∑

l=0

(n− 1− i

i

)(ab)b(n−1)/2c−i.

Tan and Ekin[2] in gave an explicit formula for the bi-periodic Lucas

sequence

qn = aξ(n)bn/2c∑l=0

n

n− i

(n− ii

)(ab)bn/2c−i.

1. O. Yayenie. A note on generalized Fibonacci sequence. Appl. Math. Comput,

(2011), 217, pp 5603-5611.

2. Tan E, Ekin A B. Bi-periodic Incomplete Lucas Sequences. Ars Combinatoria

123,(2015), 371–380.N. Rosa AIT-AMRANE On generalization of bi-periodic r-numbers 15/1

Explicit formulas

If we take r = 1 and a = b = c = d = 1 we obtain an explicit formulas

of the classical Fibonacci and Lucas numbers (Fn) and (Ln),

respectively, see Koshy [1]

Fn =

b(n−1)/2c∑l=0

(n− 1− i

i

),

and

Ln =

bn/2c∑l=0

n

n− i

(n− ii

).

1. Koshy T. Fibonacci and Lucas numbers with application. Wiley, New York,

(2001).N. Rosa AIT-AMRANE On generalization of bi-periodic r-numbers 16/1

The Binet formulas

Finally, in order to obtain the Binet formulas of the bi-periodic

r-Fibonacci sequence and the bi-periodic r-Lucas sequence of type s, we

need to express the characteristic polynomial of (U(r)n )n and (V

(r,s)n )n

xr+1 − abxr − (aξ(r+1)d+ bξ(r+1)c)xbr+12 c − (−1)ξ(r)cd.


The Binet formulas

Theorem

Let α1, α2, . . . , αr+1 be the distinct roots of the characteristic polynomial

associated with (U(r)n )n and (V

(r,s)n )n, then for n ≥ 1, we have

U (r)n =

r+1∑i=1

( r∑j=1

(−1)j σij U(r)2r−2j+ξ(n) + U

(r)2r+ξ(n)

)∏

1≤k≤r+1k 6=i

(αi − αk)αbn/2ci ,


The Binet formulas

Theorem

and

V(r,s)n =

r+1∑i=1

( r∑j=1

(−1)j σij U(r)2r−2j+ξ(n+1)

+ U(r)2r+ξ(n+1)

)∏

1≤k≤r+1k 6=i

(αi − αk)

(αb(n+1)/2ci + sdα

b(n−r)/2ci

), for r odd,

r+1∑i=1

( r∑j=1

(−1)j σij U(r)2r−2j+ξ(n+1)

+ U(r)2r+ξ(n+1)

)∏

1≤k≤r+1k 6=i

(αi − αk)

(αb(n+1)/2ci + scbα

b(n−r−1)/2ci + scdα

b(n−2r−1)/2ci

), for r even,

with the initial conditions

U(r)0 = 0, U

(r)1 = 1, U

(r)2 = a, . . . , U (r)

r = abr/2cbb(r−1)/2c,

V(r,s)0 = s+ 1, V

(r,s)1 = a, V

(r,s)2 = ab, . . . , V (r,s)

r = ab(r+1)/2cbbr/2c.


The Binet formulas

Thank you for your attention !


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