The k-Fibonacci group and periods of the k-step Fibonacci sequences

MUNESH KUMARI1, KALIKA PRASAD1, BAHAR KULOĞLU2, ENGIN ÖZKAN3

1Department of Mathematics,Central University of Jharkhand, 835205, INDIA

2Graduate School of Natural and Applied Sciences, Erzincan Binali Yıldırım University, TURKEY

3Department of Mathematics, Erzincan Binali Yıldırım University, TURKEY

Abstract: - In this paper, we have introduced a new infinite cyclic group called the k-Fibonacci group and studied

its algebraic properties. Further, we have obtained periods for k-step Fibonacci sequences in the 2-generator

groups such as S3,D3,A3,Q8and in the 3-generator group, Q8×Z2m.

Key-Words: -k-step Fibonacci sequence; Fibonacci Group; Period and Basic Period.

Received: October 5, 2022. Revised: November 9, 2022. Accepted: November 21, 2022. Published: December 12, 2022.

1 Introduction

Wall [1] studied Fibonacci sequences in groups first.

Wall showed that the Fibonacci sequence formed in

the cyclic group is periodic according to a prime num-

ber and gave the relationship between the period and

p. Wilcox [2] carried this problem to abelian groups.

Campell et al [3] introduced the Fibonacci length and

Fibonacci orbit concepts and moved this problem to

some finite simple groups. Özkan et al [11] reported

that 3-step Fibonacci sequences are periodic to an m

number and the relationship between the period and

m. Lu and Wang generalized similar work to k-step

Fibonacci sequences[4]. Some recent work in this di-

rection can be seen in[5, 6, 7].

Similar studies have been done on some fi-

nite groups, nilpotent groups, and binary polyhedral

groups [8, 9, 10, 12, 13, 14].

Gwang-Yeon Lee, et. al.[15] defined the k-

generalized Fibonacci matrix Qkand gave some of

its properties. Later, Gwang-Yeon Lee and Jin-Soo

Kim [16] defined the k-Fibonacci and the symmetric

k-Fibonacci matrix from the k-Fibonacci sequences

and discussed its algebra.

Let {fk,n}n≥0denote the generalized Fibonacci

sequence of order k(≥2) ∈Ngiven by

fk,k+n=fk,k+n−1+fk,k+n−2+fk,k+n−3+

... +fk,n+1 +fk,n, n ≥0,

with fk,0=fk,1=... =fk,k−2= 0 and fk,k−1= 1.

The generalized Fibonacci sequence {fk,n}n≥0is also

known as the k-step Fibonacci sequence.

Let the matrix Qkof order k be given as

Qk=Q1







111 ... 1 1

100 ... 0 0

010 ... 0 0

.....

000 ... 1 0







On the usual multiplication of Qk-matrix to ntimes,

we get Qn

k, where Qn

kis the generalized Fibonacci ma-

trix[17]. The generalized Fibonacci matrix Qn

kis de-

fined as







fk,n+k−1fk,n+k−2+fk,n+k−3+... +fk,n ... fk,n+k−2

fk,n+k−2fk,n+k−3+fk,n+k−4+... +fk,n−1... fk,n+k−3

fk,n fk,n−1+fk,n−2+... +fk,−k+n+1 ... fk,n−1





,

where

k=Ik,(Q1

k)n=Qn

k, Qm

kQn

k=Qm+n

kand

det(Qn

k) = (−1)(k−1)nfor m, n ∈Z.(1)

2 The k-Fibonacci Group

We will now show that the set of Qn

k,n∈Zforms a

commutative group according to the matrix product.

We will also show that this group is isomorphic to the

group of integers Zand hence is cyclic.

Theorem 2.1. The collection of all generalized Fi-

bonacci matrices forms a group with respect to matrix

multiplication.

That is, given the following set G,

G={Qn

k:k(≥2) ∈N, n ∈Z}.(2)

Then G forms an abelian group with respect to the

matrix multiplication.

Proof. For n= 0, since Ik∈G, so G is non-empty. It

can easily be seen that the set Gsatisfies the grouping

conditions, so we omit it.

We refer to the group Gdefined in Theorem 2.1 as

the k-Fibonacci group.

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Example 1. The following is a 3-Fibonacci group

with entries from a combination of tribonacci num-

bers:

G=





3=



f3,n+2 f3,n+1 +f3,n f3,n+1

f3,n+1 f3,n +f3,n−1f3,n

f3,n f3,n−1+f3,n−2f3,n−1



:n∈Z





Clearly, Gsatisfies all the axioms of the multiplica-

tive abelian group.

Also, Gis an infinite group.

Theorem 2.2. The k-Fibonacci group Gis isomor-

phic to Z.

Proof. Let us consider the set G={Qn

k:k(≥2) ∈

N, n ∈Z}.

Define a mapping φ:Z→Gsuch that φ(n) = Qn

clearly φis well defined.

Let n1, n2∈Z, then we have

φ(n1) = φ(n2) =⇒Qn1

k=Qn2

k=⇒Qn1−n2

k=Q0

=⇒n1−n2= 0.

Thus the mapping φ(n)is one-one.

Now, let Qm

k∈G, then there exists m∈Zsuch that

φ(m) = Qm

kimplies φis onto.

Since, for n1, n2∈Z, we have

φ(n1+n2) = Qn1+n2

k=Qn1

k∗Qn2

k=φ(n1)∗φ(n2)

=⇒φis homomorphism.

Thus φis one-one, onto and homomorphism implies

G∼

=Z.

Hence, the following corollaries are consequences

of the above theorem.

Corollary 2.3. The k-Fibonacci group is an infinite

cyclic group and its generators are Q1

kand Q−1

Corollary 2.4. The center of the k-Fibonacci group

Gis the group itself i.e Z(G) = G.

Theorem 2.5. For a positive integer k≥2, let G=

{Qn

k:n∈Z}and H={(Qn

k)m:m, n ∈Z}. Then

Hforms a subgroup of G. Moreover, only possible

subgroups of G are of the form < QnZ

k>.

Proof. Since (Qn

k)0=Ik∈H, thus His non-empty.

Let two elements of Hare (Qn

k)m1and (Qn

k)m2where

m1, m2∈Z, then from equation (1), we have

(Qn

k)m1[(Qn

k)m2]−1= (Qn

k)m1(Qn

k)−m2

= (Qn

k)m1−m2∈H

as m1−m2∈Z.

So, His a subgroup of G.

Now, to prove other parts of the theorem, let Qn

kbe

the element of Hsuch that nis the smallest positive

integer. Let Qm

k, m ∈Zbe any arbitrary element of

Hthen by division algorithm, we have m=nq +r

where q, r ∈Zand 0≤r < n.

Since r=m−nq ∈Z, so Qn

k=Qm−nq

k∈H

and m, n ∈Zimplies Qm

k, Qn

k∈Hhence Qr

Qm−nq

k=Q0

k=I.

This implies r= 0, which gives

m=nq

=⇒Qm

k=Qnq

k∈H, q ∈Z

and Qm

kis an arbitrary element.

Corollary 2.6. All the subgroups of the k-Fibonacci

group are normal.

Proof. Since the k-Fibonacci group Gis abelian it im-

plies that all the subgroups of Gare normal.

3 Periods for the k-step Fibonacci

Sequence

Throughout, we use BP erk(G;x0, x1, ..., xj−1)and

P erk(G;x0, x1, ..., xj−1)notation to denote the basic

period and period of the k-step Fibonacci sequence

Fk,k+n(G;x0, x1, ..., xj−1), respectively.

From [18], we note the following definition.

Definition 3.1. In a finite group, a k-nacci

sequence is a sequence of group elements

{x0, x1, ..., xn, xn+1, ...}where initial set

{x0, x1, ..., xj−1}is provided and

xn=x0x1...xn..., : j ≤n<k

xn−kxn−k+1...xn−1..., : n ≥k.

It is denoted by Fk(G;x0, x1, ..., xj−1)where Gis

a group generated by the initial set.

A sequence is simply periodic with period kif the

first kelements in the sequence form a repeating sub-

sequence. Let k(p)denote the fundamental period of

the sequence and call it the Wall number.

Asequence of group elements is said to be periodic

if, after a particular point, it contains only a fixed sub-

sequence repeatedly.

Definition 3.2. The action of automorphism group

AutGof Gon Xand on the k-nacci sequences

Fk(G;x0, x1, ..., xj−1),(x0, x1, ..., xj−1)∈X,

AutGconsists of all isomorphism θ:G→G

and if θ∈AutGand (x0, x1, ..., xj−1)∈Xthen

(x0θ, x1θ, ..., xj−1θ)∈X.

Let Abe a subset of Gand θ∈AutG, then the

image of Ais Aθ ={aθ :a∈A}under θ.

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Theorem 3.1. Let S3=hx, y :x3=y2= (xy)2=

eibe the presentation for the group S3. Then in S3,

the periods of the k-step Fibonacci sequences and the

basic periods of the basic k-step Fibonacci sequences

are given as:

(i) For k= 2,P er2(S3;x, y)=6 and

BP er2(S3;x, y)=6.

(ii) For k≥3,P erk(S3;x, y) = 2k+ 2 and

BP erk(S3;x, y) = 2k+ 2.

Proof. (i). For k= 2, we have the following se-

quence,

x, y, xy, x2, yx, xy, x, y, ...,

which repeats after 6 terms hence its period is 6.

Since, we have xθ =x,yθ =yx,xyθ =y, for

the inner automorphism θinduced by conjugation by

x, so the basic period is 6.

(ii). For k≥3, the first kterms of sequence are

x, y, x2= (xy)20, x3= (xy)21, x4= (xy)22,

..., xk−1= (xy)2k−3.

So, from the above, the following sequence is ob-

tained:

x0=x, x1=y, x2=xy and xj=efor 3≤j≤k−1.

Hence, we have

xk−1=e, xk=e, xk+1 =x2, xk+2 =yx,

xk+3 =xy, xk+4 =e, xk+5 =e, ...

x2k+2 =x, x2k+3 =y, x2k+4 =xy, x2k+5 =e,

x2k+6 =e, ...

x3k+3 =x2, x3k+4 =yx, x3k+5 =xy, x3k+6 =e,

x3k+7 =e, ..., x4k+4 =x, ...,

and whenever nk + (n+ 3) ≤j≤(n+ 1)k+n,

n= 1,2,3, ... then xj=e.

Also, the following hold:

x2k+2 =x=

2k+1

j=k+2

xj, x2k+3 =y=

2k+2

j=k+3

xj,

x2k+4 =xy =

2k+3

j=k+4

xj.

Observe that the values of consecutive terms x2k+2,

x2k+3, and x2k+4 rely on x, y and the cycle starts

again with the (2k+ 2)th term, which is, x0=

x2k+2, x1=x2k+3, x2=x2k+4, ....

Therefore, P erk(S3;x, y) = 2k+ 2.

From the above sequence, we can see that

BP erk(S3;x, y) = 2k+ 2, since xθ =x,yθ =x2y,

xyθ =y, where θis an outer automorphism.

Also, this theorem is valid in the group D3.

Theorem 3.2. Let A3=hx, y :x3=y2= (xy)3=

eibe the presentation for the group A3. Then the

periods and the basic periods of the basic k-step Fi-

bonacci sequences are given as:

(i) For k= 2,P er2(A3;x, y) = 16 and

BP er2(A3;x, y) = 16.

(ii) For k= 3,P er3(A3;x, y) = 13 and

BP er3(A3;x, y) = 13.

Proof. (i). For k= 2, we have the sequence,

x, y, xy, yxy, x2, xyx2, xyx, x2, xy, y, x, yx, xyx,

xyx2, x2y, yx2, x, y, ...,

which repeats after 16 terms therefore the period is

16. Similarly, for the basic sequence,

x, xyx2, x2yx2, yx, yx2, x2yx, x2y, x2, x2yx2, xyx2,

x, xy, x2y, x2yx, yx2, xyx, x, xyx2, ...,

the basic period is 16, since xθ =x,yθ =xyx2,

where θis the inner automorphism induced by con-

jugation by x. So, P er2(A3;x, y) = 16 and

BP er2(A3;x, y) = 16.

(ii). For k= 3, the first few terms of the sequence

are

x, y, xy, (xy)2, y, y, yx2, yx2, x2yx2, x2y, x2, xyx,

e, yx, y, yxy, xyx, y, y, xyx, xyx, e, x2yx2,

e, x2yx2, xyx, e, e, xyx, xyx, yxy, yx2y, xyx, xyx,

e, yxy, e, yxy, xyx, e, e, xyx, xyx, yxy, ....

Here,

x26 =e, x27 =e, x28 =xyx, x29 =yxy, ...,

x39 =e, x40 =e, x41 =xyx, x42 =yxy, ...,

x52 =e, x53 =e, x54 =xyx, x55 =yxy, ....

Also, for k= 3,

xuhk(3)−(k−4) =e, xuhk(3)−(k−3) =e,

xuhk(3)−(k−2) =xyx, xuhk(3)−(k−1) =xyx,

xuhk(3)−k=yxy,

where u∈Z+and hk(3) refer to the Wall number for

the k-step Fibonacci sequence modulo 3.

So that, P er2(A3;x, y) = 13.

Similarly, BP er2(A3;x, y) = 13 because xθ =x,

yθ =xyx2.

Also, this theorem is valid for h2,3,3ipolyhedral

group.

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Theorem 3.3. Let Q8=hx, y :x4=e, x2=y2,

y−1xy =x−1ibe the presentation for the group Q8.

Then the periods and the basic periods of the basic

k-step Fibonacci sequences are given as:

(i) For k= 2,P er2(Q8;x, y) = 3 and

BP er2(Q8;x, y) = 3.

(ii) For k≥3,P erk(Q8;x, y) = 2k+ 2 and

BP erk(Q8;x, y) = 2k+ 2.

Proof. (i). For k= 2, the sequence is as follows:

x, y, xy, x, y, xy, ...

and it has period P er2(Q8;x, y) = 3.

For the basic period, the sequence is

x, y3, x3y, x, y3, x3y, ...,

which has the period BP er2(Q8;x, y)=3since

xθ =x, yθ =y3where θis an inner automorphism

induced by conjugation by x.

(ii). If k≥3;

For k= 3, the sequence is

x, y, xy, x2, x3, y, xy, e, x, y, xy, x2, x3, ....

For k= 5, the sequence is

x, y, xy, x2, e, e, x3, y2, xy, e, , e, e, x, y, xy, x2, ....

Here, it is seen that the period for each k≥3is 2k+2.

And similarly, BP erk(Q8;x, y) = 2k+ 2.

Theorem 3.4. Let Q8×Z2m=hx, y, z :x4=

e, x2=y2, y−1xyx =e, z2m=e= [x, z] = [y, z]i

be the presentation for the group Q8×Z2m. Then the

periods are given as:

(i) For k= 2,P er2(Q8×Z2m;x, y, z)=

lcm(h2(2m),3).

(ii) For k= 3,P er3(Q8×Z2m;x, y, z) =

2lcm(2m, 2k+ 2).

(iii) For k≥4,P erk(Q8×Z2m;x, y, z) =

lcm(2m,2k+2)(k+1)

Proof. (i). For k= 2, the sequence is as follows,

x, y, z, yz, z2y, z3y2, z5y3, z8y5, z13y8, z21y13,

z34y21, ....

From this sequence, we obtain a sub-sequence as fol-

lows:

y, z, yz, z2y, z3y2, z5y3, ....

For k= 2, the sequence conforms to the following

pattern,

xt+1 =zf2,t+1 yf2,t , xt+2 =zf2,t+2 yf2,t+1 .

We need the smallest t, satisfying xt+1 =y, xt+2 =z

Letting lcm(h2(2m),3) = λ, then we have

2m|f2,t+1,3|f2,t+1 and hence f2,λ ≡1 (mod 2m)

and f2,λ ≡1 (mod 3). Also, f2,λ+2 ≡1 (mod 2m)

and f2,λ+2 ≡1 (mod 3).

If we choose t=λ, then we obtain xλ+1 =yand

xλ+2 =z.

So we get P er2(Q8×Z2m;x, y, z)=lcm(h2(2m),3).

(ii). For k= 3, the first few terms of the sequence

are:

x, y, z, xyz, z2x, z4y, z7y2, z13xy3, z24x, z44y, ....

Here,

x8=z24x, x9=z44y, x10 =z81y4, ...,

x16 =z3136x, x17 =z5768y, x18 =z10609y8, ...,

x24 =z410744x, x25 =z755476y, x26 =z1389537y12, ....

Using the above, we have

x8β=zβ

j4uβx, x8β+1 =zβ

j4uβ+1 y,

x8β+2 =zβ

j4uβ+2+1y4β, ...,

where jis odd and β∈N,β= 2σjand uβ, uβ+1,

uβ+2 ∈N, also gcd(uβ, uβ+1, uβ+2) = 1.

We need the smallest β, satisfying x8β=x, x8β+1 =

yand x8β+2 =z.

If we choose β=lcm(2m,2k+2)

4, then we obtain

x2lcm(2m,2k+2) =x, x2lcm(2m,2k+2)+1 =y,

x2lcm(2m,2k+2)+2 =z.

Thus, we get,

P er3(Q8×Z2m;x, y, z) = 2lcm(2m, 2k+ 2).

(iii). For k≥4, the sequence is as follows,

x, y, z, xyz, z2x2, z4x3, z8y, z15x2, z29yx, z56, z108x,

z208y, z401, z773xy, z1490x2, z2872x3, z5536y,

z10671x2, z20569yx, z39648, ....

Using the above k-step Fibonacci sequence, we have

xβ(2k+2)−k−1=zβ

j4δβ−1x3, xβ(2k+2)−k=zβ

j4δβy,

xβ(2k+2)−k+1 =zβ

j4δβ+1+3x2,

xβ(2k+2)−k+2 =zβ

j4δβ+2+1yx,

xβ(2k+2)−k+3 =zβ

j4δβ+3 ,

xβ(2k+2)−k+4 =zβ

j4δβ+4 x,

xβ(2k+2)−k+5 =zβ

j4δβ+5 y,

xβ(2k+2)−k+6 =zβ

j4δβ+6+1x2, ...,

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where jis a positive odd integer and βis a pos-

itive integer, δβ−1, δβ, δβ+1, ..., δβ+4 ∈Nand

gcd(δβ−1, δβ, δβ+1, ..., δβ+4) = 1.

Here, we need the smallest βsatisfying

xβ(2k+2) =x, xβ(2k+2)+1 =y, xβ(2k+2)+2 =z.

Now, if we choose β=lcm(2m, 2k+ 2)

4, then we

get

xlcm(2m,2k+2)(k+1)

=x, xlcm(2m,2k+2)(k+1)

4+1 =y,

xlcm(2m,2k+2)(k+1)

4+2 =z.

Thus, we obtain

P erk(Q8×Z2m;x, y, z) = lcm(2m, 2k+ 2)(k+ 1)

4 Conclusion

In this study, the k-Fibonacci group was defined and

its algebraic properties were examined. A Fibonacci

sequence was created in groups, with some having

two generators and others having three generators.

They have been shown to be periodic, and their fun-

damental periods have been determined. This work

can be applied to many different groups as well as to

other sequences such as the Lucas and Pell sequences.

Acknowledgment

The authors are very thankful to the anonymous re-

viewer for their care and rapid review. The first

and second authors acknowledge the University Grant

Commission, India for SRF.

References:

[1] D. D. Wall, “Fibonacci series modulo m”, Amer.

Math. Monthly 67 (1960) 525–532.

[2] H. J. Wilcox, “Fibonacci sequences of pe-

riod n in groups”, Fibonacci Quart 24 (1986)

356–361.

[3] C. Campbell, H. Doostie and E. Robertson, “Fi-

bonacci length of generating pairs in groups”,

in Applications of Fibonacci numbers, eds.

G. Bergum, A. Philippou and A. Horadam

(Springer, 1990), volume 3, pp. 27–35.

[4] K. Lu and J. Wang, “k-step Fibonacci sequence

modulo m”, Utilitas Mathematica 71 (2006)

169–177.

[5] C. Campbell and P. Campbell, “The Fibonacci

lengths of binary polyhedral groups and related

groups”, Congr. Numer. 194 (2009) 95–102.

[6] C. Campbell and P. Campbell, “The Fibonacci

length of certain centro-polyhedral groups”, J.

Appl. Math. Comput. 19 (2005) 231–240.

[7] R. Dikici and E. Özkan, “An application of Fi-

bonacci sequences in groups”, Appl Math Com-

put 136 (2003) 323–331.

[8] E. Özkan, “3-step Fibonacci sequences in nilpo-

tent groups”, Applied mathematics and compu-

tation 144 (2003) 517–527.

[9] H. Aydin and R. Dikici, “General Fibonacci se-

quences in finite groups”, Fibonacci Quart 36

(1998) 216–221.

[10] H. Aydin and G. C. Smith, “Finite p-quotients

of some cyclically presented groups”, Journal

of the London Mathematical Society 49 (1994)

83–92.

[11] E. Özkan, H. Aydın and R. Dikici, “3-Step Fi-

bonacci series modulo m”, Applied Mathematics

and Computation 143 (2003) 165–172.

[12] E. Özkan, H. Aydın and R. Dikici, “Applica-

tions of Fibonacci sequences in a finite nilpotent

group”, Applied mathematics and computation

141 (2003) 565–578.

[13] E. Özkan, “Fibonacci sequences in nilpotent

groups with class n”, Chiang Mai J Sci 31 (2004)

205–212.

[14] Ö. Deveci, M. Avcı and E. Karaduman, “The

periods of the k-step Fibonacci and k-step Pell

sequences in D2n×z2iand D2n×ϕZ2i”, An.

S tiint . Univ. Al. I. Cuza Ia si. Mat. (N.S.) Tomul

LXIV.

[15] G.-Y. Lee, S.-G. Lee and H.-G. Shin, “On the

k-generalized Fibonacci matrix qk”, Linear Al-

gebra and its applications 251 (1997) 73–88.

[16] G.-Y. Lee and J.-S. Kim, “The linear algebra of

the k-Fibonacci matrix”, Linear algebra and its

applications 373 (2003) 75–87.

[17] K. Prasad and H. Mahato, “Cryptography us-

ing generalized Fibonacci matrices with affine-

hill cipher”, Journal of Discrete Mathematical

Sciences and Cryptography (2021) 1–12. DOI:

10.1080/09720529.2020.1838744

[18] S. W. Knox, “Fibonacci sequences in finite

groups”, Fibonacci Quart. 30 (1992) 116–120.

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WSEAS TRANSACTIONS on MATHEMATICS

DOI: 10.37394/23206.2022.21.95

Munesh Kumari, Kalika Prasad, Bahar Kuloğlu, Engin Özkan

E-ISSN: 2224-2880

843

Volume 21, 2022