# The Polytopic-k-Step Fibonacci Sequences in Finite Groups

Authors: Ömür Deveci
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

We study the polytopic-k-step Fibonacci sequences, the polytopic-k-step Fibonacci sequences modulo m, and the polytopic-k-step Fibonacci sequences in finite groups. Also, we examine the periods of the polytopic-k-step Fibonacci sequences in semidihedral group SD2m.


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## Body

## 1. Introduction

The well- knownk-step Fibonacci sequence {Fnk}(k≥2) is defined as(1.1)F0(k)=0,…,Fk-2(k)=0,Fk-1(k)=1,Fn+k(k)=Fn+k-1(k)+Fn+k-2(k)+⋯+Fn(k)forn≥0.
Let {aj}j=0k-1  (k≥2,ak-1≠0) be a sequence of real numbers. A k-generalized Fibonacci sequence {Vn}n=0+∞ is defined by the following linear recurrence relation of order k:(1.2)Vn+1=a0Vn+a1Vn-1+⋯+ak-1Vn-k-1,forn≥k-1,
where V0,…,Vk-1 are specified by the initial conditions.Thek-step Fibonacci sequence, the k-generalized Fibonacci sequence, and their properties have been studied by several authors; see, for example, [1–5].Thek-step Fibonacci sequence is a special case of a sequence which is defined as a linear combination by Kalman as follows(1.3)an+k=c0an+c1an+1+⋯+ck-1an+k-1,
where c0,c1,…,ck-1 are real constants. In [6], Kalman derived a number of closed-form formulas for the generalized sequence by companion matrix method as follows:(1.4)Ak=[aij]k×k=[c0c1c2⋯ck-2ck-1100⋯00010⋯00001⋯00⋮⋮⋮⋮⋮000⋯10].
Then, by an inductive argument he obtained (1.5)Akn[a0a1⋮ak-1]=[anan+1⋮an+k-1].
A sequence of group elements is periodic if, after a certain point, it consists only of repetitions of a fixed subsequence. The number of elements in the repeating subsequence is called the period of the sequence. For example, the sequence a,b,c,d,e,b,c,d,e,b,c,d,e,… is periodic after the initial element a and has period 4. A sequence of group elements is simply periodic with period k if the first k elements in the sequence form a repeating subsequence. For example, the sequence a,b,c,d,e,f,a,b,c,d,e,f,a,b,c,d,e,f,… is simply periodic with period 6.Definition 1.1.
For a finitely generated groupG=〈A〉, where A={a1,a2,…,an}, the sequence xi=ai+1, 0≤i≤n-1, xi+n=∏j=1nxi+j-1, i≥0, is called the Fibonacci orbit of G with respect to the generating set A, denoted by FA(G). If  FA(G) is periodic, then the length of the period of the sequence is called the Fibonacci length of G with respect to generating set A, written as LENA(G) [7].Definition 1.2.
For every integerk, where 2≤k≤LENA(G), the sequence {yi}1∞ of the elements of G defined by
(1.6)yi=xi,i=1,…,k,yi=(yi-k)α1(yi-k+1)α2⋯(yi-1)αk,i≥k+1
is called a k-step generalized Fibonacci sequence of G, for some positive integers α1,α2,…,αk [8].Definition 1.3.
Ak-nacci sequence in a finite group is a sequence of group elements x0,x1,x2,x3,…,xn,… for which, given an initial (seed) set x0,…,xj-1, each element is defined by
(1.7)xn={x0x1⋯xn-1forj≤n<k,xn-kxn-k+1⋯xn-1forn≥k.
We also require that the initial elements of the sequence, x0,…,xj-1, generate the group, thus forcing the k-nacci sequence to reflect the structure of the group. The k-nacci sequence of a group G seeded by x0,…,xj-1 is denoted by Fk(G;x0,…,xj-1) and its period is denoted by Pk(G;x0,…,xj-1) [9].The Fibonacci sequence, thek-nacci sequence, and the generalized order-k Pell sequence in finite groups have been studied by some authors, and different periods of these sequences in different finite groups have been obtained; see, for example, [7, 9–16]. Formulas which classified according to certain rules for this periods are critical to be used in cryptography, see, for example, [17–19]. Because the exponents of each term in the generalized Fibonacci sequence are determined randomly, classification according to certain rule of periods is resulting from application of this sequence in groups is possible, only if the exponent of each term are determined integers obtained according to a certain rule. Therefore, In this paper, by expanding the k-step Fibonacci sequence which is special type of the generalized Fibonacci sequences with polytopic numbers which are a well-known family of integers, we conveyed the sequence named the polytopic-k-step Fibonacci sequence that exponent of  n+tnd term is determined that (α+k-t-1k-t) formula to finite groups and named the polytopic-k-step Fibonacci sequence in finite groups as polytopic-k-nacci sequence. Because of varying both α and according to the number of step and the exponent of each term of this is determined according to a certain rule, the polytopic-k-step Fibonacci sequence is more useful and more general than the k-nacci sequences and the generalized order-k Pell sequence which varying only by the number of step. So that considered by different α value, different step values and different initial (seed) sets, different lineer recurrence sequences which are a special type of generalized Fibonacci sequences occur, and thus by conveying the polytopic-k-step Fibonacci sequence to finite groups, more useful and more general formulas than formulas used to obtain periods of the k-nacci and the generalized order-k Pell sequence in finite groups are obtained to be used in cryptography.In this paper, the usual notationp is used for a prime number.


## 2. The Polytopic-k-Step Fibonacci Sequences

The well-knownk-topic numbers are defined as(2.1)Pk(n)=n(n+1)(n+2)⋯(n+r-1)k!=(n+k-1k).
When k=2, the k-topic numbers, Pk(n), are reduced to the triangular numbers. In [20], Gandhi and Reddy obtained triangular numbers in the generalized Pell sequence {Pn(α)} and generalized associated Pell sequence {Qn(α)} which are defined for a fixed α>0, respectively, as (2.2)P0(α)=0,P1(α)=1,Pn+2(α)=(α+1)Pn+1(α)+α(α+1)2Pn(α)forn≥0,Q0(α)=Q1(α)=1,Qn+2(α)=(α+1)Qn+1(α)+α(α+1)2Qn(α)forn≥0.
Now we define for a fixed integer α>0, a new sequence called the polytopic-k-step Fibonacci sequence {Fn(k,α)}, by(2.3)F0(k,α)=0,…,Fk-2(k,α)=0,Fk-1(k,α)=1,Fn+k(k,α)=αFn+k-1(k,α)+(α+12)Fn+k-2(k,α)+⋯+(α+k-2k-1)Fn+1(k,α)+(α+k-1k)Fn(k,α)forn≥0.
Obviously, if we take α=1 in (2.3), then this sequence reduces to the well-known k-step Fibonacci sequence. When α≥2 and k=2 in (2.3), we call {Fn(2,α)} the polytopic Fibonacci sequence.By (2.3), we can write (2.4)[Fn+k(k,α)Fn+k-1(k,α)Fn+k-2(k,α)⋮Fn+1(k,α)]=[α(α+12)⋯(α+k-2k-1)(α+k-1k)10⋯0001⋯00⋮⋮⋮⋮00⋯10][Fn+k-1(k,α)Fn+k-2(k,α)Fn+k-3(k,α)⋮Fn(k,α)]
for the polytopic-k-step Fibonacci sequence. Let(2.5)M=[mij]k×k=[α(α+12)⋯(α+k-2k-1)(α+k-1k)10⋯0001⋯00⋮⋮⋮⋮00⋯10].
The matrix M is called the polytopic-k-step Fibonacci matrix.We obtain that the polytopic Fibonacci sequences{Fn(2,α)} are generated by a matrix Qα for a fixed integer α≥2:(2.6)Qα=[αα(α+1)210],(Qα)n=[Fn+1(2,α)α(α+1)2Fn(2,α)Fn(2,α)α(α+1)2Fn-1(2,α)]
which can be proved by mathematical induction.


## 3. The Polytopic-k-Step Fibonacci Sequences Modulo m

In this section we examine the polytopic-k-step Fibonacci sequences modulo m for α≥2 and k≥2.Reducing the polytopic-k-step Fibonacci sequence by a modulus m, we can get a repeating sequence denoted by(3.1){F(k,α)(m)}={F0(k,α)(m),F1(k,α)(m),F2(k,α)(m),…,Fi(k,α)(m),…},
where Fi(k,α)(m)=Fi(k,α)(modm). It has the same recurrence relation as in (2.3).Theorem 3.1.
{F(k,α)(m)} is a periodic sequence for k≥2 and α≥2.Proof.
LetUk={(x1,x2,…,xk)∣0≤xi≤m-1}. Then we have that |Uk|=mk is finite, that is, for any a≥0, there exist b≥a such that Fa+1(k,α)(m)≡Fb+1(k,α)(m),…,Fa+k(k,α)(m)≡Fb+k(k,α)(m). From the definition of the polytopic-k-step Fibonacci sequence {Fn(k,α)} we have Fn+k(k,α)=αFn+k-1(k,α)+(α+12)Fn+k-2(k,α)+⋯+(α+k-2k-1)Fn+1(k,α)+(α+k-1k)Fn(k,α), that is, (α+k-1k)Fn(k,α)=Fn+k(k,α)-αFn+k-1(k,α)-(α+12)Fn+k-2(k,α)-⋯-(α+k-2k-1)Fn+1(k,α).Then we can easily get that Fa(k,α)(m)≡Fb(k,α)(m), Fa-1(k,α)(m)≡Fb-1(k,α)(m),…,F2(k,α)(m)≡Fb-a+2(k,α)(m)  and  F1(k,α)(m)≡Fb-a+1(k,α)(m), which implies that {Fn(k,α)} is a periodic sequence.
Lethk(α)(m) denote the smallest period of {F(k,α)(m)}, called the period of the polytopic-k-step Fibonacci sequence modulo m. When k=2, h2(α)(m) is the period of the polytopic Fibonacci sequence modulo m.Example 3.2.
We have{F(3,4)(3)}={0,0,1,1,2,2,0,0,1,…} and then repeat. So we get h3(4)(3)=6.By elementary number theory it is easy to prove that ifm=∏i=1tpiei, (t≥1), where pi’s are distinct primes, then hk(α)(m)=Icm[hk(α)(piei)].For a given matrixA=[aij] with aij’s being integers, A(modm) means that every entry of A is reduced modulo m, that is, A(modm)=(aij(modm)). Let 〈M〉pa={Mi(modpa)∣i≥0} be a cyclic group, and let |〈M〉pa| denote the order of 〈M〉pa with p∤(α+k-1k) (where by p∤(α+k-1k) we mean that (α+k-1k) is not divided by p) and T the transpose of a matrix. It is clear that
(3.2)(Mi[1,0,0,…,0]T)T(mod⁡m)=[Fi+k-1(k,α)(m),Fi+k-2(k,α)(m),…,Fi(k,α)(m)].
We then obtain that hk(α)(m) is least positive integer h(α) such that
(3.3)(Mh(α)[1,0,0,…,0]T)T(mod⁡m)=[1,0,0,…,0].Theorem 3.3.
Letα≥2. If p∤(α+k-1k), then hk(α)(pa)=|〈M〉pa|.Proof.
It is clear that|〈M〉pa| is divisible by hk(α)(pa). Then we need only to prove that hk(α)(pa) is divisible by |〈M〉pa|. Let hk(α)(pa)=n. Then we have
(3.4)Mn=[m11m12⋯m1km21m22⋯m2k⋮⋮⋮mk1mk2⋯mkk].
The elements of the matrix Mn are in the following forms:
(3.5)m11=Fn+k-1(k,α),m21=Fn+k-2(k,α),…,mk1=Fn(k,α),mii=β1Fn+k-2(k,α)+β2Fn+k-3(k,α)+⋯+βk-1Fn(k,α)+1,for2≤i≤k,β1,β2,…,βk-1≥0,mij=η1Fn+k-2(k,α)+η2Fn+k-3(k,α)+⋯+ηk-1Fn(k,α)fori≠j,1≤i≤k,2≤j≤k,η1,η2,…,ηk-1≥0.
We thus obtain that
(3.6)mii≡1(mod⁡pa),for1≤i≤k,mij≡0(mod⁡pa),for1≤i,j≤ksuchthati≠j.
So we get that Mn≡I(modpa), which yields that n is divisible by |〈M〉pa|. We are done.Theorem 3.4.
Letα≥2, and let t be the largest positive integer such that hk(α)(p)=hk(α)(pt). Then hk(α)(pa)=pa-thk(α)(p) for every a≥t. In particular, if hk(α)(p)≠hk(α)(p2), then hk(α)(pa)=pa-1hk(α)(p) holds for every a>1.Proof.
Letq be a positive integer. Since Mhk(α)(pq+1)≡I(modpq+1), that is, Mhk(α)(pq+1)≡I(modpq), we get that hk(α)(pq+1) is divided by hk(α)(pq). On the other hand, writing Mhk(α)(pq)=I+(aij(q)pq), we have
(3.7)Mhk(α)(pq)p=(I+(aij(q)pq))p=∑i=0p(pi)(aij(q)pq)i≡I(mod⁡pq+1),
which yields that hk(α)(pq)p is divided by hk(α)(pq+1). Therefore, hk(α)(pq+1)=hk(α)(pq) or hk(α)(pq+1)=hk(α)(pq)p, and the latter holds if, and only if, there is an aij(q) which is not divisible by p. Since hk(α)(pt)≠hk(α)(pt+1), there is an aij(t+1) which is not divisible by p, thus, hk(α)(pt+1)≠hk(α)(pt+2). The proof is finished by induction on t.Conjecture 3.5.
Letα≥2. If p≥k, then there exists a σ with 0≤σ≤k such that (pk+1-pσ) is divided by hk(α)(p).Table1 list some primes for which the conjecture is true when k=5 and α=5.Table 1
The length ofh5(5)(p).
ph5(5)(p)Result55h5(5)(p)∣p6-p51180525h5(5)(p)∣p6-p1315372h5(5)(p)∣p6-123145992h5(5)(p)∣p6-12924388h5(5)(p)∣p6-p331461760h5(5)(p)∣p6-p2431749132h5(5)(p)∣p6-1471661152h5(5)(p)∣p6-1532808h5(5)(p)∣p6-p459205378h5(5)(p)∣p6-1674030224h5(5)(p)∣p6-p2731419912h5(5)(p)∣p6-p29744264640h5(5)(p)∣p6-p210113136325h5(5)(p)∣p6-122352856154h5(5)(p)∣p6-139778804h5(5)(p)∣p6-p441912914277518098h5(5)(p)∣p6-p52347685222h5(5)(p)∣p6-p360716969333200h5(5)(p)∣p6-p271989206789920h5(5)(p)∣p6-1821454331269680h5(5)(p)∣p6-p2853529414856880h5(5)(p)∣p6-p21009518758082640h5(5)(p)∣p6-115232319528h5(5)(p)∣p6-p416132601768h5(5)(p)∣p6-p420114044120h5(5)(p)∣p6-p4301127298090330h5(5)(p)∣p6-p34021262790931413426025h5(5)(p)∣p6-p505943159140126h5(5)(p)∣p6-p36037132826492154616h5(5)(p)∣p6-p2


## 4. The Polytopic-k-Nacci Sequences in Finite Groups

Definition 4.1.
For a finitely generated groupG=〈A〉, where A={a1,a2,…,an}, we define the polytopic Fibonacci orbit FAα(G) with respect to the generating set A to be the sequence {xi} of the elements of G such that
(4.1)xi=ai+1,for0≤i≤n-1,xi+n=(xi)(α+n-1n)(xi+1)(α+n-2n-1)⋯(xi+n-2)(α+12)(xi+n-1)α,fori≥0,Example 4.2.
LetG=〈A〉, where A={a1,a2,a3}. FAα(G) is
(4.2)x0=a1,x1=a2,x2=a3,xi+3=(xi)α(α+1)(α+2)/6(xi+1)α(α+1)/2(xi+2)α,fori≥0.Definition 4.3.
A polytopic-k-nacci sequence in a finite group is a sequence of group elements x0,x1,…xn,… for which, given an initial (seed) set x0,…,xj-1, each element is defined by
(4.3)xn={x0(α+n-1n)x1(α+n-2n-1)⋯(xn-1)αforj≤n<k,xn-k(α+k-1k)xn-k+1(α+k-2k-1)⋯(xn-1)αforn≥k.
It is required that the initial elements of the sequence, x0,…,xj-1, generate the group, thus, forcing the polytopic-k-nacci sequence to reflect the structure of the group. We denote the polytopic-k-nacci sequence of a group G generated by x0,…,xj-1 by Fkα(G;x0,…,xj-1).Example 4.4.
LetG=〈A〉, where A={a1,a2,a3}. F4α(G;a1,a2,a3) is
(4.4)x0=a1,x1=a2,x2=a3,x3=(x0)α(α+1)(α+2)/6(x1)α(α+1)/2(x2)α,xi+4=(xi)α(α+1)(α+2)(α+3)/24(xi+1)α(α+1)(α+2)/6(xi+2)α(α+1)/2(xi+3)αfori≥0.
It is important to note that the polytopic Fibonacci orbit of a k-generated group is a polytopic-k-nacci sequence.
The classic polytopic Fibonacci sequence in the integers modulom can be written as F2α(ℤm;0,1). We call a polytopic-2-nacci sequence of a group of elements a polytopic Fibonacci sequence of a finite group.Theorem 4.5.
A polytopic-k-nacci sequence in a finite group is periodic.Proof.
The proof is similar to the proof of  Theorem  1 in [6] and is omitted.
We denote the period of a polytopic-k-nacci sequence Fkα(G;x0,…,xj-1) by Pkα(G;x0,…,xj-1). When α=1, Fkα(G;x0,…,xj-1) and Pkα(G;x0,…,xj-1) are reduced to Fk(G;x0,…,xj-1) and Pk(G;x0,…,xj-1), respectively.
From the definition, it is clear that the period of a polytopic-k-nacci sequence in a finite group depends on the chosen generating set and the order in which the assignments of x0,x1,…xn-1 are made.Definition 4.6.
LetG be a finite group. If there exists a polytopic-k-nacci sequence of the group G such that every element of the group G appears in the sequence, then the group G is called polytopic-k-nacci sequenceable.It is important to note that the direct product of polytopic-k-nacci sequenceable groups is not necessarily polytopic-k-nacci sequenceable. Consider that the group C2×C4 is defined by the presentation
(4.5)〈x,y∣x2=y4=e,xy=yx〉.
The polytopic Fibonacci sequences of the group C2×C4 for α=2 are
(4.6)F22(C2×C4;x,y)=x,y,xy2,y3,x,y,…,F22(C2×C4;y,x)=y,x,y3,xy2,y,x,….
Since the elements e,xy,andxy3 do not in either sequences, the group C2×C4 is not polytopic-2-nacci sequenceable.The group〈x〉 has a polytopic Fibonacci sequence
(4.7)F22(〈x〉;e,x)=e,x,e,x,…
and hence is polytopic-2-nacci sequenceable. The group 〈y〉 has a polytopic Fibonacci sequence
(4.8)F22(〈y〉;e,y)=e,y,y2,y3,e,y,…
and hence is polytopic-2-nacci sequenceable.We will now address the periods of the polytopic-k-nacci sequences in specific classes of groups. A group SD2m is semidihedral group of order 2m if
(4.9)SD2m=〈a,b∣a2m-1=b2=e,b-1ab=a-1+2m-2〉
for every m≥4. Note that the orders a and b are 2m-1 and 2, respectively.Theorem 4.7.
The periods of the polytopic-k-nacci sequences in the group SD2m for initial (seed) set, a,b, and α=2 are as follows:(i)
Pk2(SD2m;a,b)=hk(2)(2m-2),for2≤k≤4.(ii)
Pk2(SD2m;a,b)=hk(2)(2m-1),fork≥5.Proof.
(i) Ifk=2, we have the polytopic-2-nacci sequence for α=2:(4.10)x0=a,x1=b,x2=a3,x3=a(2m-2-1)⋅2⋅3b,x4=a32,x5=a(2m-2-1)⋅2⋅3+(2m-2-1)⋅2⋅32b,…,x2m-2=a32m-3,x2m-2+1=a(2m-2-1)⋅2⋅3+(2m-2-1)⋅2⋅32+⋯+(2m-2-1)⋅2⋅32m-3b,….
By mathematical induction, it is easy to prove that
(4.11)32m-3≡1(mod⁡2m-1),(2m-2-1)⋅2⋅3+(2m-2-1)⋅2⋅32+⋯+(2m-2-1)⋅2⋅32m-3≡0(mod⁡2m-1).
So we get x2m-2=a32m-3=a, x2m-2+1=a(2m-2-1)·2·3+(2m-2-1)·2·32+⋯+(2m-2-1)·2·32m-3b=b.It is easy to see that h2(2)(2m-2)=2m-3·h2(2)(2)=2m-3·2=2m-2. Since the elements succeeding xh2(2)(2m-2), xh2(2)(2m-2)+1, depend on a and b for their values, the cycle begins again with the h2(2)(2m-2)nd, that is, xh2(2)(2m-2)=x0  and xh2(2)(2m-2)+1=x1. Thus, the period of F22(SD2m;a,b) is h2(2)(2m-2).
Ifk=3, we have the polytopic-3-nacci sequence for α=2:
(4.12)x0=a,x1=b,x2=a3,x3=a4+(2m-2-1)⋅2⋅3b,x4=a32,x5=a3⋅4+4+(2m-2-1)⋅2⋅3+(2m-2-1)⋅2⋅32b,x6=a33,…,x2m-2=a32m-3,x2m-2+1=a32m-3-1⋅4+32m-3-2⋅4+⋯+4+(2m-2-1)⋅2⋅3+(2m-2-1)⋅2⋅32+⋯+(2m-2-1)⋅2⋅32m-3b,x2m-2+2=a32m-3+1⋯.
By mathematical induction, it is easy to prove that 32m-3-1·4+32m-3-2·4+⋯+4≡0(mod2m-1), 32m-3+1≡3(mod2m-1).So we get x2m-2=a32m-3=a, x2m-2+1=a32m-3-1·4+32m-3-2·4+⋯+4+(2m-2-1)·2·3+(2m-2-1)·2·32+⋯+(2m-2-1)·2·32m-3b=b, x2m-2+2=a32m-3+1=a3. It is easy to see that h3(2)(2m-2)=2m-3·h3(2)(2)=2m-3·2=2m-2. Since the elements succeeding xh3(2)(2m-2),xh3(2)(2m-2)+1,xh3(2)(2m-2)+2 depend on a, b, and a3 for their values, the cycle begins again with h3(2)(2m-2)nd, that is xh3(2)(2m-2)=x0,xh3(2)(2m-2)+1=x1, and xh3(2)(2m-2)+1=x2. Thus, the period of F32(SD2m;a,b) is h3(2)(2m-2). The proof for k=4 is similar and is omitted.
(ii) Ifk≥5, we have the polytopic-k-nacci sequence for α=2:(4.13)x0=a,x1=b,x2=a3,x3=a2m-1-2b,x4=a14,x5=au1,x6=au2,…,xk=auk-4,x4hk(2)(2)=a25,x4hk(2)(2)+1=a8b,x4hk(2)(2)+2=a27,x4hk(2)(2)+3=a2m-1-2b,x4hk(2)(2)+4=a30,x4hk(2)(2)+5=au1+8⋅λ1,x4hk(2)(2)+6=au2+8⋅λ2,…,x4hk(2)(2)+k=auk-4+8⋅λk-4,…,xi⋅4hk(2)(2)=a1+24⋅i,xi⋅4hk(2)(2)+1=a8⋅ib,xi.4hk(2)(2)+2=a3+24⋅i,xi⋅4hk(2)(2)+3=a2m-1-2b,xi⋅4hk(2)(2)+4=a14+16⋅i,xi⋅4hk(2)(2)+5=au1+8⋅i⋅λ1,xi⋅4hk(2)(2)+6=au2+8⋅i⋅λ2,…,xi⋅4hk(2)(2)+k=auk-4+8⋅i⋅λk-4,…,
where λ1,…,λk-4 are natural numbers and u1,…,uk-4 are even natural numbers. So we need the smallest i∈ℕ such that 8·i=2m-1. If we choose i=2m-4, we obtain xhk(2)(2m-1)=a=x0, xhk(2)(2m-1)+1=b=x1, xhk(2)(2m-1)+2=a3=x2, xhk(2)(2m-1)+3=a2m-1-2b=x3, xhk(2)(2m-1)+4=a14=x4, xhk(2)(2m-1)+5=au1=x5, xhk(2)(2m-1)+6=au2=x6, …, xhk(2)(2m-1)+k=auk-4=xk since 2m-2·hk(2)(2)=hk(2)(2m-1). So we get Pk2(SD2m;a,b)=hk(2)(2m-1) for k≥5.Theorem 4.8.
The periods of the the polytopic-k-nacci sequences in the group SD2m for initial (seed) sets b,a, and α=2 are as follows:(i)
Pk2(SD2m;b,a)=hk(2)(2m-2)for2≤k≤3,(ii)
Pk2(SD2m;b,a)=hk(2)(2m-1)fork≥4.Proof.
The proof is similar to the proof of Theorem4.5 and is omitted.



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*Source: 431840-2011-10-10.xml*