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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 14, Issue 1

## Volume 1 (2003)

Inci Gültekin
/ Ömür Deveci
Published Online: 2016-12-30 | DOI: https://doi.org/10.1515/math-2016-0100

## Abstract

In this paper, we define the arrowhead-Fibonacci numbers by using the arrowhead matrix of the characteristic polynomial of the k-step Fibonacci sequence and then we give some of their properties. Also, we study the arrowhead-Fibonacci sequence modulo m and we obtain the cyclic groups from the generating matrix of the arrowhead-Fibonacci numbers when read modulo m. Then we derive the relationships between the orders of the cyclic groups obtained and the periods of the arrowhead-Fibonacci sequence modulo m.

MSC 2010: 11K31; 11B50; 11C20; 15A15

## 1 Introduction

It is well-known that a square matrix is called an arrowhead matrix if it contain zeros all in entries except for the first row, first column, and main diagonal. In other words, an arrowhead matrix M = [mi,j](n)×(n) is defined as follows: $m1,1m1,2m1,3m1,4⋯m1,nm2,1m2,200⋯0m3,10m3,30⋯0⋮00⋱⋮mn−1,1⋮⋮0mn−1,n−10mn,10⋯00mn,n.$

The k-step Fibonacci sequence $\left\{{F}_{n}^{k}\right\}$ is defined recursively by the equation $Fn+kk=Fn+k−1k+Fn+k−2k+⋯+Fnk$

for n ≥ 0, where ${F}_{0}^{k}={F}_{1}^{k}=\cdots ={F}_{k-2}^{k}=0\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}{F}_{k-1}^{k}=1.$

For detailed information about the k-step Fibonacci sequence, see [1, 2]. It is clear that the characteristic polynomial of the k-step Fibonacci sequence is as follows: $PkF(x)=xk−xk−1−⋯−x−1.$

Suppose that the (n+k)th term of a sequence is defined recursively by a linear combination of the preceding k terms: $an+k=c0an+c1an+1+⋯+ck−1an+k−1$

where c0c1,…,ck−1 are real constants. In [1], Kalman derived a number of closed-form formulas for the generalized sequence by the companion matrix method as follows:

Let the matrix A be defined by $A = [ai,j]k×k = 010⋯00001⋯00000⋱00⋮⋮⋮⋮⋮000⋯01c0c1c2ck−2ck−1,$

then $AnaOa1⋮ak−1=anan+1⋮an+k−1$

for n ≥ 0.

Many of the obtained numbers by using homogeneous linear recurrence relations and their miscellaneous properties have been studied by many authors; see, for example, [311]. Arrowhead-Fibonacci numbers for the 2-step Pell and Pell-Lucas sequences were illustrated in [12]. In Section 2, we define the arrowhead-Fibonacci numbers by using the arrowhead matrix N, which is defined by the aid of the characteristic polynomial of the k-step Fibonacci sequence. Then we derive their miscellaneous properties such as the generating matrix, the combinatorial representation, the Binet formula, the permanental representations, the exponential representation and the sums.

The study of recurrence sequences in groups began with the earlier work of Wall [13], where the ordinary Fibonacci sequence in cyclic groups were investigated. The concept extended to some special linear recurrence sequences by some authors; see, for example, [3, 1416]. In [3, 15, 17], the authors obtained the cyclic groups via some special matrices. In Section 3, we study the arrowhead-Fibonacci sequence modulo m. Also in this section, we obtain the cyclic groups from the multiplicative orders of the generating matrix of the arrowhead-Fibonacci sequence such that the elements of the generating matrix when read modulo m. Then we obtain the rules for the orders of the obtained cyclic groups and we give the relationships between the orders of those cyclic groups and the periods of the arrowhead-Fibonacci sequence modulo m.

We next define the arrowhead matrix N = [ni,j](k+1)×(k+1) by using the characteristic polynomial of the k-step Fibonacci sequence ${P}_{k}^{F}\left(x\right)$ as follows: $1−1−1⋯−1−1−1−10⋯00−10−10⋯0⋮⋮⋱⋱⋱⋮−10⋯0−10−10⋯00−1.$

Now we consider a new (k+1)-step sequence which is defined by using the matrix N and is called the arrowhead-Fibonacci sequence. The sequence is defined by integer constants ak+1(1) = … =ak+1(k) =0 and ak+1(k+1)=1 and the recurrence relation $ak+1(n+k+1)=ak+1(n+k)−ak+1(n+k−1)−⋯−ak+1(n)$(1)

for n ≥ 1, where k ≥ 2.

Let k = 3, then we have the sequence ${a4(n)}={0,0,0,1,1,0,−2,−4,−3,3,12,16,4,−27,−59,...}.$

By (1), we can write a generating matrix for the arrowhead-Fibonacci numbers as follows: $Gk+1 = 1−1−1⋯−1−1100⋯00010⋯00⋮⋱⋮00⋯10000⋯010k+1 × k+1$

The matrix Gk+1 is said to be a arrowhead-Fibonacci matrix. It is clear that $Gk + 1n ak+1 (k + 1) ak+1 (k) ⋮ ak+1 (1) = ak+1 (n + k + 1) ak+1 (n + k) ⋮ ak+1 (n + 1)$(2)

for n ≥ 0. Again by an inductive argument, we may write $(Gk+1)n=ak+1n+k+1−ak+1n+kak+1n+k−ak+1n+k−1ak+1n+k−1G′−ak+1n+k−2⋮⋮ak+1n+1−ak+1n,$

where nk, ak+1(u) is denoted by ${a}_{k\text{\hspace{0.17em}}+\text{\hspace{0.17em}}1}^{u}$ and ${G}_{k+1}^{\prime }$ is a (k+1)×(k−1) matrix as follows: $(Gk+1′ )=− ak+1n+k+ak+1n+k− 1+⋯ +ak+1n+1⋯ − ak+1n+k+ak+1n+k− 1+ak+1n+k− 2− ak+1n+k+ak+1n+k− 1− ak+1n+k− 1+ak+1n+k− 2+⋯ +ak+1n⋯ − ak+1n+k− 1+ak+1n+k− 2+ak+1n+k− 3− ak+1n+k− 1+ak+1n+k− 2− ak+1n+k− 2+ak+1n+k− 3+⋯ +ak+1n− 1⋯ − ak+1n+k− 2+ak+1n+k− 3+ak+1n+k− 4− ak+1n+k− 2+ak+1n+k− 3⋮ ⋮ ⋮ − ak+1n+ak+1n− 1+⋯ +ak+1n− k+1⋯ − ak+1n+ak+1n− 1+ak+1n− 2− ak+1n+ak+1n− 1.$

It is important to note that det Gk+1=(−1)k+1 and the Simpson identity for a recursive sequence can be obtained from the determinant of its generating matrix. From this point of view, we can easily derive the Simpson formulas of the arrowhead-Fibonacci sequences for every k ≥ 2.

Since det G3 = −1 and $(G3)n=a3n+3−a3n+2+a3n+1−a3n+2a3n+2−a3n+1+a3n−a3n+1a3n+1−a3n+1+a3n−1−a3n$

for n ≥ 2, the Simpson formula of sequence {a3(n)} is $a3n+3(a3n)2−2a3n+2a3n+1a3n+(a3n+2)2a3n−1+(a3n+1)3−a3n+3a3n+1a3n−1=1.$

Let C(c1,c2,…;cν) be a ν × ν companion matrix as follows: $C(c1,c2,⋯,cv)=c1c2⋯cυ100⋮⋱⋮0⋯10.$

(Chen and Louck [20]). The (i,j) entry ${c}_{{i}_{},j}^{\left(u\right)}\left({c}_{1},{c}_{2},\cdots ,{c}_{v}\right)$ in the matrix Cu (c1,c2,…,cv) is given by the following formula: $Ci,j(u)(c1,c2,⋯,cv)=∑(t1,t2,⋯,tv)tj+tj+1+⋯+tvt1+t2+⋯+tv×t1+⋯+tvt1,⋯,tvc1t1⋯cvtv$

where the summation is over nonnegative integers satisfying t1 + 2t2 + … + vtv = ui + j, $\left(\begin{array}{l}{t}_{1}+\cdots +{t}_{v}\\ \phantom{\rule{1em}{0ex}}{t}_{1},\cdots ,{t}_{v}\end{array}\right)=\frac{\left({t}_{1}+\cdots +{t}_{v}\right)!}{{t}_{1}!\cdots {t}_{v}!}$ is a multinomial coefficient, and the coefficients in (3) are defined to be 1 if u = ij.

Then we can give a combinatorial representation for the arrowhead-Fibonacci numbers by the following Corollary.

Let ak+1(n) be the nth the arrowhead-Fibonacci numberfor k ≥ 2. Then $am,n(n)=∑(t1,t2⋯,tk+1)tk+1t1+t2+⋯+tk+1×t1+⋯+tk+1t1,⋯,tk+1(−1)k+1$

where the summation is over nonnegative integers satisying t1+2t2+…+(k+1)tk+1 = n.

In Theorem 2.3, if we choose v = k+1, u = n, i = j = k+1, c1 = 1 and c2 = … ck+1 = −1, then the proof is immediately seen from (Gk+1)n.

Now we consider the Binet formulas for the arrowhead-Fibonacci numbers by using the determinantal representation.

The characteristic equation of the arrowhead-Fibonacci sequence xk+1xk + xk − 1 + xk−2 + …+ 1 = 0 does not have multiple roots.

Let f(x) = xk+1xk + xk−1 + xk−2 + … + 1, then $f\left(x\right)={x}^{k+1}-{x}^{k}+\frac{{x}^{k}-1}{x-1}=\frac{{x}^{k}\left(\left(x-1{\right)}^{2}+1\right)-1}{x-1}.$ It is clear that f(0) = 1 and f(1) = k for all k ≥ 2. Let u be a multiple root of f(x), then u ∉ {0,1}. If possible, u is a multiple root of f(x) in which case f(u) = 0 and f′(u) = 0. Now f′(u) = 0 and u ≠ 0 give ${u}_{1}=1+i\frac{k+1}{k+2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{u}_{1}=1-i\frac{k+1}{k+2}$ while f(u) = 0 shows uk((u−1)2 + 1) = 1 so that $\left({u}_{1}{\right)}^{k}=\frac{\left(k+2{\right)}^{2}}{2k+3}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left({u}_{2}{\right)}^{k}=\frac{\left(k+2{\right)}^{2}}{\left(k+1{\right)}^{2}+\left(k+2{\right)}^{2}}$ which are contradictions since k ≥ 2.

If x1, x2, …, xk+1 are roots of the equation xk+1xk + xk−1 + xk−2 + … + 1= 0, then by Lemma 2.5, it is known that x1, x2, …, xk+1 are distinct. Let Vk+1 be (k+1) × (k+1) Vandermonde matrix as follows: $Vk+1=(x1)k(x2)k⋯(xk+1)k(x1)k−1(x2)k−1⋯(xk+1)k−1⋮⋮⋮x1x2xk+111⋯1.$

Let $Uik+1=(x1)n+k+1−i(x2)n+k+1−i⋮(xk+1)n+k+1−i$

and supoose that ${V}_{{i}_{},j}^{k+1}$ is a(k+1) × (k+1) matrix obtained from Vk+1 by replacing the j th column of Vk+1 by ${U}_{i}^{k+1}.$

For nk ≥ 2, $gi,j(n)=detVi,jk+1detVk+1,$

where $\left({G}_{k+1}{\right)}^{n}=\left[{g}_{{i}_{},j}^{\left(n\right)}\right].$

Since the eigenvalues of the matrix Gk+1, x1, x2, …, xk+1 are distinct, the matrix Gk+1 is diagonalizable. Let Dk+1 = (x1, x2 …, xk+1), we easily see that Gk+1 Vk+1 = Vk+1 Dk+1. Since det Vk+1 ≠ 0, the matrix Vk+1 is invertible. Then it is clear that (Vk+1)−1 Gk+1 Vk+1 = Dk+1. Thus, the matrix Gk+1 is similar to Dk+1. So we get (Gk+1)n Vk+1 = Vk+1 (Dk+1)n for nk ≥ 2. Then we can write the following linear system of equations: $gi,1(n)(x1)k+gi,2(n)(x1)k−1+⋯+gi,k+1(n)=(x1)n+k+1−igi,1(n)(x2)k+gi,2(n)(x2)k−1+⋯+gi,k+1(n)=(x2)n+k+1−i⋮gi,1(n)(xk+1)k+gi,2(n)(xk+1)k−1+⋯+gi,k+1(n)=(xk+1)n+k+1−i$

for nk ≥ 2. So, for each i, j = 1,2,…, k + 1, we obtain ${g}_{{i}_{},j}^{\left(n\right)}$ as follows $gi,j(n)=detVi,jk+1detVk+1.$

Then we can give the Binet formulas for the arrowhead-Fibonacci numbers by the following corollary.

Let ak+1(n) be the nth the arrowhead-Fibonacci numberfor k ≥ 2. Then $ak+1(n)=−detVk+1,k+1k+1detVk+1.$

Now we consider the permanent representations of the arrowhead-Fibonacci numbers.

A u × v real matrix M = [mi,j] is called a contractible matrix in the kth column (resp. row) if the kth column (resp. row) contains exactly two non-zero entries.

Suppose that x1 x2, …, xu are row vectors of the matrix M. If M is contractible in the kth column such that mi,k ≠ 0,mj,k ≠ 0 and ij, then the (u − 1)×(v − 1) matrix Mij:k obtained from M by replacing the ith row with mi,k xj + mj,k xi and deleting the jth row. The kth column is called the contraction in the kth column relative to the ith row and the jth row.

In [21], Brualdi and Gibson obtained that per (M) = per(N) if M is a real matrix of order α >1 and N is a contraction of M.

Let Mk+1(n) = $\left[{m}_{{i}_{},j}^{\left(n\right)}\right]$ be the n × n super-diagonal matrix, defined by $mi,j(n)=ifi=j=αfor1<_α<_n1andi=α+1andj=αfor1<_α<_n−1,ifi=α−uandj=αforu+1<_α<_n−1suchthat1<_u<_k,0otherwise.$

that is, $(k+1)th↓Mk+1(n)=1−1−1⋯−10⋯00011−1−1⋯−10⋯00011−1−1⋯−10⋯0⋮⋱⋱⋱⋱⋱⋮0⋯011−1−1⋯−1000⋯011−1−1⋯−100⋯0011−1⋯−1⋮⋮⋱⋱⋱⋱⋮00⋯0011−100⋯0011$

Then we have the following theorem.

For and n ≥ 1 and k ≥ 2, $perMk+1(n)=ak+1(n+k+1),$

where perMk+1(1) = 1.

First we start with considering the case n < 4. The matrices Mk+1(2) and Mk+1(3) are reduced to the following forms: $perMk+1(2)=1−111$

and $perMk+1(3)=1−1−111−1011.$

It is easy to see that perMk+1(2) = 0 and perMk+1(3) = −2. From definition of arrowhead-Fibonacci sequence it is clear that ak+1 (k + 2) = 1, ak+1 (k + 3) = 1 and ak+1 (k + 4) = −2. So we have the conclusion for n < 4. Let the equation hold for n ≥ 4, then we show that the equation holds for n + 1. If we expand the perMk+1(n) by the Laplace expansion of permanent with respect to the first row, then we obtain $perMk+1(n+1)=perMk+1(n)−perMk+1(n−1)−⋯−perMk+1(n−k).$

Since perMk+1(n) = ak+1(n+k+1), perMk+1(n−1) = ak+1(n+k), …, perMk+1(nk) = ak+1(n+1) , we easily obtain that perMk+1(n+1) = ak+1(n+k+2).So the proof is complete.

Let n > k + 1 such that k ≥ 2 and let ${R}_{k+1}\left(n\right)=\left[{r}_{{i}_{},j}^{\left(n\right)}\right]$ be the n × n matrix, defined by $ri,j(n)=ifi=j=αfor1<_α<_n,i=α+1andj=αfor1<_α<_n−k−11andi=n−k+αandn−k+α<_j<_nfor1<_α<_k−1,ifi=αandj=α+ufor1<_α<_n−k−1suchthat1<_u<_k,0otherwise.$

that is, $(k+1)th↓Rk+1(n)=1−1⋯−100⋯00011−1⋯−100⋯00011−1⋯−100⋯0⋮⋱⋱⋱⋱⋱⋱⋮0⋯011−1⋯−1−1000⋯011−1⋯−1−1000⋯0011⋯1⋮⋱⋱⋱⋱⋱⋱⋮0000⋯0001100000⋯0001←(n−k)th$

Assume that the n × n matrix ${T}_{k+1}\left(n\right)=\left[{t}_{{i}_{},j}^{\left(n\right)}\right]$ is defined by $(k+1)th↓Tk+1(n)=1⋯10⋯010Rk+1(n−1)⋮0,$

where n > k + 2 such that k ≥ 2.

Then we can give more general results by using other permanent representations than the above.

Let ak+1(n) be the nth the arrowhead-Fibonacci number for k ≥ 2. Then

1. For n > k + 1, $perRk+1(n)=ak+1(n+1).$

2. For n > k + 2, $perTk+1(n)=∑i=1nak+1n(i).$

1. Let the equation hold for n > k + 1, then we show that the equation holds for n + 1. If we expand the perRk+1(n) by the Laplace expansion of permanent with respect to the first row, then we obtain $perRk+1(n+1)=perRk+1(n)−perRk+1(n−1)−⋯−perRk+1(n−k).$

Also, since $perRk+1(n)=ak+1(n+1),perRk+1(n−1)=ak+1(n),…,perRk+1(n−k)=ak+1(n−k+1),$

it is clear that $perRk+1(n+1)=ak+1(n+2).$

2. It is clear that expanding the perTk+1(n) by the Laplace expansion of permanent with respect to the first row, gives us $perTk+1(n)=perTk+1(n−1)+perRk+1(n).$

Then, by the result of Theorem 2.10. (i) and an induction on n, the conclusion is easily seen.

It is easy to show that the generating function of the arrowhead-Fibonacci sequence {ak+1(n)}is as follows: $g(x)=xk1−x+x2+⋯+xk+1,$

where k ≥ 2.

Then we can give an exponential representation for the arrowhead-Fibonacci numbers by the aid of the generating function with the following theorem.

The arrowhead-Fibonacci numbers have the following exponential representation: $g(x)=xkexp∑i=1∞xii1−x−⋯xki,$

where k ≥ 2.

Since $ln⁡g(x)=ln⁡xk−ln⁡(1−x+x2+⋯+xk+1)$

and $−ln(1−x+x2+⋯+xk+1)=−ln⁡(1−(x−x2−⋯−xk+1))=∑i=1∞xii(1−x−⋯xk)i,$

it is clear that $ln⁡g(x)−ln⁡xk=ln⁡g(x)xk=∑i=1∞xii(1−x−⋯xk)i.$

Thus we have the conclusion.

Now we consider the sums of arrowhead-Fibonacci numbers.

Let $Sn=∑i=1nak+1(i)$

for n ≥ 1 and k ≥ 2, and suppose that Ak+2 is the (k + 2)×(k + 2) matrix such that $Ak+2=10⋯010Gk+1⋮0.$

Then it can be shown by induction that $(Ak+2)n=10⋯0Sn+kSn+k−1(Gk+1)n⋮Sn.$

## 3 The arrowhead-Fibonacci sequence modulo n

If we reduce the arrowhead-Fibonacci sequence {ak+1(n)} by a m, then we get the repeating sequence, denoted by ${ak+1m(n)}={ak+1m(1),ak+1m(2),⋯,ak+1m(i),…}$

where we denote ak+1(i)(modm) by ${a}_{k+1}^{m}\left(i\right).$ It has the same recurrence relation as in (1).

The sequenc $\left\{{a}_{k+1}^{m}\left(n\right)\right\}$ is simply periodic for k ≥2. That is, the sequence is periodic and repeats by returning to its starting values.

Since $\left\{{a}_{4}^{2}\left(n\right)\right\}=\left\{{0}_{},{0}_{},{0}_{},{1}_{},1,0,0,0,1,1,...\right\},{P}_{4}\left(2\right)=5.$

Given an integer matrix X=[xi,j], X(mod m) means that all entries of X are modulo m, that is, X(mod m)= (xi,j(mod m)) . Let us consider the set $〈X{〉}_{m}=\left\{{X}^{i}\left(\mathrm{m}\mathrm{o}\mathrm{d}m\right)|i\underset{_}{>}0\right\}.$ If gcd (m,det X)=1, then 〈Xm is a cyclic group. Let $|〈X{〉}_{m}|$ denote the cardinality of the set 〈Xm. Since det Gk+1=(−1)k+1, 〈Gk+1m is a cyclic group for every positive integer m. By (2), it is easy to show that ${P}_{k+1}\left(m\right)=|〈{G}_{k+1}{〉}_{m}|$

Now we give some useful properties for the period Pk+1(m) by the following theorem.

1. Let p be a prime and suppose that u is the smallest positive integer with Pk+1(pu+1) ≠ Pk+1(pu). Then Pk+1(pv) = pv−u· Pk+1(p) for every v>u and k≥2.

2. If m has the prime factorization $m=∏i=1u(pi)αi,(u≥1),thenPk+1(m)$

equals the least common multiple of the Pk+1((pi)αi)’ s for k≥2.

3. If k is A even integer such that k≥2, then Pk+1(m) is even for every positive integer m.

1. If I is the (k+1)×(k+1) identity matrix and t is a positive integer such that $(Gk+1)Pk+1(pt+1)≡I(modpt+1),thenGk+1)Pk+1(pt+1)≡I(modpt).$

Then, it is clear that Pk+1(pt) divides Pk+1(pt+1). On the other hand, if we denote $\left({G}_{k+1}{\right)}^{{P}_{k+1}\left({p}^{t}\right)}=I+\left({a}_{i;j}^{\left(t\right)}\cdot {p}^{t}\right),$ then by the binomial expansion, we may write $(Gk+1)Pk+1(pt)⋅p=(I+(ai,j(t)⋅pt))p=∑i=0ppi(ai,j(t)⋅pt)i≡I(modpt+1).$

This yields that Pk+1(ptp is divisible by Pk+1(pt). Then, Pk+1(pt+1) = Pk+1(pt) or Pk+1(pt+1) = Pk+1(ptp, and the latter holds if and only if there is a ${a}_{{i}_{},j}^{\left(t\right)}$ which is not divisible by p. Due to fact that we assume u is the smallest positive integer such that Pk+1(pu+1)≠ Pk+1(pu), there is an ${a}_{{i}_{},j}^{\left(u\right)}$ which is not divisible by p. Since there is a ${a}_{{i}_{},j}^{\left(u\right)}$ such that p does not divide ${a}_{{i}_{},j}^{\left(u\right)}$ it is easy to see that there is an ${a}_{{i}_{},j}^{\left(\mathrm{u+1}\right)}$ which is not divisible by p. This shows that Pk+1(pu+2)≠ Pk+1(pu+1). Then we get that ${P}_{k+1}\left({p}^{u+2}\right)=p\cdot {P}_{k+1}\left({p}^{u+1}\right)=p\cdot \left(p\cdot {P}_{k+1}\left({p}^{u}\right)\right)={p}^{2}\cdot {P}_{k+1}\left({p}^{u}\right).$ So by induction on u we obtain Pk+1(pv) = pv−u ·Pk+1(p) for every v>u. In particular, if ${P}_{k+1}\left({p}^{2}\right)\ne {P}_{k+1}\left(p\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{then}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{P}_{k+1}\left({p}^{v}\right)={p}^{v-1}\cdot {P}_{k+1}\left(p\right).$

2. It is clear that the sequence $\left\{{a}_{k+1}^{\left({p}_{i}{\right)}^{{\alpha }_{i}}}\left(n\right)\right\}$ repeats only after blocks of length $\lambda \cdot {P}_{k+1}\left(\left({p}_{i}{\right)}^{{\alpha }_{i}}\right)$ where λ is a natural number. Since Pk+1(m) is period of the sequence $\left\{{a}_{k+1}^{m}\left(n\right)\right\},$ the sequence $\left\{{a}_{k+1}^{\left({p}_{i}{\right)}^{{\alpha }_{\mathrm{i}}}}\left(n\right)\right\}$ repeats after Pk+1(m) terms for all values i. Thus, we easily see that Pk+1(m) is of the form $\lambda \cdot {P}_{k+1}\left(\left({p}_{i}{\right)}^{{\alpha }_{i}}\right)$ for all values of i, and since any such number gives a period of Pk+1(m). Therefore, we conclude that $Pk+1(m)=lcm[Pk+1((p1)α1),…,Pk+1((pu)αu)].$

3. Since det Gk+1 = −1 when k is a even integer and ${P}_{k+1}\left(m\right)=|{〈{G}_{k+1}〉}_{m}|,$ we have the conclusion.

## Acknowledgement

This Project was supported by the Commission for the Scientific Research Projects of Kafkas University. The Project number. 2015-FM-45.

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Accepted: 2016-11-16

Published Online: 2016-12-30

Published in Print: 2016-01-01

Citation Information: Open Mathematics, Volume 14, Issue 1, Pages 1104–1113, ISSN (Online) 2391-5455,

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