On the Arrowhead-Fibonacci Numbers

Open Math. 2016; 14: 1104–1113

Open Mathematics Open Access

Research Article

˙Inci Gültekin* and Ömür Deveci On the arrowhead-Fibonacci numbers

DOI 10.1515/math-2016-0100 Received August 17, 2016; accepted November 16, 2016.

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.

Keywords: The arrowhead-Fibonacci Numbers, Sequence, Matrix, Period

MSC: 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 is defined as D .n/ .n/ follows: 2 m1;1 m1;2 m1;3 m1;4 m1;n 3 6 m2;1 m2;2 0 0 0 7 6 7 6 m 0 m 0 0 7 6 3;1 3;3 7 6 : : 7 6 : :: : 7 . 6 : 0 0 : : 7 6 7 6 : : 7 4 mn 1;1 : : 0 mn 1;n 1 0 5 mn;1 0 0 0 mn;n ˚ k« The k-step Fibonacci sequence Fn is defined recursively by the equation

k k k k Fn k Fn k 1 Fn k 2 Fn C D C C C C C for n 0, where F k F k F k 0 and F k 1. 0 D 1 D D k 2 D k 1 D For detailed information about thek-step Fibonacci sequence, see [1, 2]. It is clear that the characteristic polynomial of the k-step Fibonacci sequence is as follows:

P F .x/ xk xk 1 x 1. k D Suppose that the .n k/th term of a sequence is deﬁned recursively by a linear combination of the preceding k C terms:

an k c0an c1an 1 ck 1an k 1 C D C C C C C

*Corresponding Author: Inci˙ Gültekin: Department of Mathematics, Faculty of Science, Atatürk University, 25240 Erzurum, Turkey, E-mail: [email protected] Ömür Deveci: Department of Mathematics, Faculty of Science and Letters, Kafkas University 36100, Turkey, E-mail: [email protected]

where c0; c1; : : : ; 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 2 0 1 0 0 0 3 6 0 0 1 0 0 7 6 7 6 : 7 6 0 0 0 :: 0 0 7 A a 6 7 , i;j k k 6 : : : : : 7 D D 6 : : : : : 7 6 : : : : : 7 6 7 4 0 0 0 0 1 5 c0 c1 c2 ck 2 ck 1 then 2 3 2 3 a0 an 6 a1 7 6 an 1 7 n 6 7 6 7 A 6 : 7 6 :C 7 6 : 7 D 6 : 7 4 : 5 4 : 5 ak 1 an k 1 C 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, [3–11]. 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, 14–16]. 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.

2 The arrowhead-Fibonacci numbers

We next define the arrowhead matrix N ni;j .k 1/ .k 1/ by using the characteristic polynomial of the k-step F D C C Fibonacci sequence Pk .x/ as follows: 2 1 1 1 1 1 3 6 1 1 0 0 0 7 6 7 6 1 0 1 0 0 7 6 7 6 : : : 7 . 6 : : :: :: :: : 7 6 : : : : : : 7 6 7 4 1 0 : : : 0 1 0 5 1 0 : : : 0 0 1 Now we consider a new .k 1/-step sequence which is defined by using the matrix N and is called the arrowhead- C Fibonacci sequence. The sequence is defined by integer constants ak 1 .1/ ak 1 .k/ 0 and C D D C D ak 1 .k 1/ 1 and the recurrence relation C C D ak 1 .n k 1/ ak 1 .n k/ ak 1 .n k 1/ ak 1 .n/ (1) C C C D C C C C C for n 1, where k 2. 1106 ˙I. Gültekin, Ö. Deveci

Example 2.1. Let k 3, then we have the sequence D a4 .n/ 0; 0; 0; 1; 1; 0; 2; 4; 3; 3; 12; 16; 4; 27; 59; : : : : f g D f g By (1), we can write a generating matrix for the arrowhead-Fibonacci numbers as follows: 2 1 1 1 1 1 3 6 1 0 0 0 0 7 6 7 6 0 1 0 0 0 7 6 7 Gk 1 6 : : : 7 . C D 6 : : : 7 6 : : : 7 6 7 4 0 0 1 0 0 5 0 0 0 1 0 .k 1/ .k 1/ C C The matrix Gk 1 is said to be a arrowhead-Fibonacci matrix. It is clear that C 2 3 2 3 ak 1 .k 1/ ak 1 .n k 1/ C C C C C 6 ak 1 .k/ 7 6 ak 1 .n k/ 7 n 6 7 6 7 .Gk 1/ 6 C: 7 6 C : C 7 (2) C 6 : 7 D 6 : 7 4 : 5 4 : 5 ak 1 .1/ ak 1 .n 1/ C C C for n 0. Again by an inductive argument, we may write 2 n k 1 n k 3 akC1C akC1 Cn k n Ck 1 6 akC1 akC1 7 6 n Ck 1 nCk 2 7 n 6 0 7 .Gk 1/ 6 akC1 G akC1 7 ; C D 6 C C 7 6 : : 7 4 : : 5 n 1 n akC1 ak 1 C C u 0 where n k, ak 1 .u/ is denoted by ak 1 and Gk 1 is a .k 1/ .k 1/ matrix as follows: C C C C 2 n k n k 1 n 1Á n k n k 1 n k 2Á n k n k 1Á 3 a C a C a C a C a C a C a C a C k 1 C k 1 C C k 1 k 1 C k 1 C k 1 k 1 C k 1 n Ck 1 Cn k 2 Cn Á n Ck 1 Cn k 2 Cn k 3Á n Ck 1 Cn k 2Á 6 a C a C a a C a C a C a C a C 7 6 k 1 C k 1 C C k 1 k 1 C k 1 C k 1 k 1 C k 1 7 6 nCk 2 nCk 3 nC1 Á nCk 2 nCk 3 nCk 4Á nCk 2 nCk 3Á 7 G0 6 akC1 akC1 ak 1 akC1 akC1 akC1 akC1 akC1 7 . k 1 6 C C C C C C 7 C D 6 C C: C C C: C C : C 7 6 : : : 7 4 : : : 5 n n 1 n k 1Á n n 1 n 2 Á n n 1 Á ak 1 ak 1 ak 1C ak 1 ak 1 ak 1 ak 1 ak 1 C C C C C C C C C C C C C C k 1 It is important to note that det Gk 1 . 1/ C and the Simpson identity for a recursive sequence can be obtained C D 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.

Example 2.2. Since det G3 1 and D 2 n 3 n 2 n 1Á n 2 3 a C a C a C a C 3 3 C 3 3 n 6 n 2 n 1 nÁ n 1 7 .G3/ 6 a C a C a a C 7 D 6 3 3 C 3 3 7 4 n 1 n 1 n 1Á n 5 a C a C a a 3 3 C 3 3 for n 2, the Simpson formula of sequence a3 .n/ is f g 2 3 n 3 n2 n 2 n 1 n n 2Á n 1 n 1Á n 3 n 1 n 1 a C a 2a C a C a a C a a C a C a C a 1. 3 3 3 3 3 C 3 3 C 3 3 3 3 D

Let C .c1; c2; : : : ; cv/ be a v v companion matrix as follows: 2 3 c1 c2 cv 6 1 0 0 7 6 7 C .c1; c2; : : : ; cv/ 6 : : 7 . D 6 : :: : 7 4 : : : 5 0 1 0 On the arrowhead-Fibonacci numbers 1107

See [18, 19] for more information about the companion matrix.

.u/ u Theorem 2.3 (Chen and Louck [20]). The .i; j / entry ci;j .c1; c2; : : : ; cv/ in the matrix C .c1; c2; : : : ; cv/ is given by the following formula: ! .u/ X tj tj 1 tv t1 tv t1 tv ci;j .c1; c2; : : : ; cv/ C C C C C C c1 cv (3) D t1 t2 tv t1; : : : ; tv .t1;t2;:::;tv / C C C

t1 tv where the summation is over nonnegative integers satisfying t1 2t2 vtv u i j , CC C C C D C t1;:::;tv D .t1 tv /Š is a multinomial coefficient, and the coefficients in (3) are defined to be 1 if u i j . tCC1Š tv Š D Then we can give a combinatorial representation for the arrowhead-Fibonacci numbers by the following Corollary.

Corollary 2.4. Let ak 1 .n/ be the nth the arrowhead-Fibonacci number for k 2. Then C ! X tk 1 t1 tk 1 k 1 am;n .n/ C C C C . 1/ C D t1 t2 tk 1 t1; : : : ; tk 1 .t1;t2:::;tk 1/ C C C C C C where the summation is over nonnegative integers satisfying t1 2t2 .k 1/ tk 1 n: C C C C C D

Proof. In Theorem 2.3 , if we choose v k 1, u n, i j k 1, c1 1 and c2 ck 1 1, then D n C D D D C D D C D the proof is immediately seen from .Gk 1/ . C Now we consider the Binet formulas for the arrowhead-Fibonacci numbers by using the determinantal representation.

Lemma 2.5. The characteristic equation of the arrowhead-Fibonacci sequence xk 1 xk xk 1 xk 2 C C C C C 1 0 does not have multiple roots. D k 2 k 1 k k 1 k 2 k 1 k xk 1 x ..x 1/ 1/ 1 Proof. Let f .x/ x x x x 1, then f .x/ x x C . D C C C C C D C C x 1 D 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 D D … f g possible, u is a multiple root of f .x/ in which case f .u/ 0 and f 0 .u/ 0. Now f 0 .u/ 0 and u 0 give k 1 k 1 D k 2 D Á D k .k ¤2/2 u1 1 i kC2 and u1 1 i kC2 while f .u/ 0 shows u .u 1/ 1 1 so that .u1/ 2kC 3 and D C C D C D C D D C k .k 2/2 .u2/ which are contradictions since k 2. .k 1/2C .k 2/2 D C C C k 1 k k 1 k 2 If x1, x2, :::, xk 1 are roots of the equation x C x x x 1 0, then by Lemma 2.5, it is C k 1 C C C C D known that x1, x2, :::, xk 1 are distinct. Let V C be .k 1/ .k 1/ Vandermonde matrix as follows: C C C 2 k k k 3 .x1/ .x2/ .xk 1/ k 1 k 1 C k 1 6 .x1/ .x2/ .xk 1/ 7 6 C 7 k 1 6 : : : 7 V C 6 : : : 7 . D 6 : : : 7 6 7 4 x1 x2 xk 1 5 C 1 1 1 Let 2 n k 1 i 3 .x1/ C C n k 1 i 6 .x2/ 7 k 1 6 C C 7 U C 6 : 7 i D 6 : 7 4 : 5 n k 1 i .xk 1/ C C C k 1 k 1 k 1 and supoose that Vi;jC is a .k 1/ .k 1/ matrix obtained from V C by replacing the j th column of V C k 1 C C by Ui C . 1108 ˙I. Gültekin, Ö. Deveci

Theorem 2.6. For n k 2, k 1 det V C .n/ i;j , gi;j k 1 D det V C n h .n/i where .Gk 1/ gi;j . C D

Proof. Since the eigenvalues of the matrix Gk 1, x1, x2, :::, xk 1 are distinct, the matrix Gk 1 is diagonalizable. C k 1C k 1 Ck 1 Let Dk 1 .x1; x2; : : : ; xk 1/, we easily see that Gk 1V C V C Dk 1. Since det V C 0, the matrix k 1 C D C k 1 1 C k 1D C ¤ V C is invertible. Then it is clear that V C Gk 1V C Dk 1. Thus, the matrix Gk 1 is similar to n k 1 k 1 nC D C C Dk 1. So we get .Gk 1/ V C V C .Dk 1/ for n k 2. Then we can write the following linear C C D C system of equations:

8 .n/ k .n/ k 1 .n/ n k 1 i ˆ gi;1 .x1/ gi;2 .x1/ gi;k 1 .x1/ C C ˆ .n/ k C .n/ k 1 C C .n/C D n k 1 i <ˆ gi;1 .x2/ gi;2 .x2/ gi;k 1 .x2/ C C C C: C C D ˆ : ˆ :ˆ .n/ k .n/ k 1 .n/ n k 1 i gi;1 .xk 1/ gi;2 .xk 1/ gi;k 1 .xk 1/ C C C C C C C C D C for n k 2. So, for each i; j 1; 2; : : : ; k 1, we obtain g.n/ as follows D C i;j k 1 .n/ det Vi;jC gi;j k 1 . D det V C Then we can give the Binet formulas for the arrowhead-Fibonacci numbers by the following corollary.

Corollary 2.7. Let ak 1 .n/ be the nth the arrowhead-Fibonacci number for k 2. Then C k 1 det V C k 1;k 1 . ak 1 .n/ Ck 1C C D det V C Now we consider the permanent representations of the arrowhead-Fibonacci numbers.

th Deﬁnition 2.8. A u v real matrix M mi;j is called a contractible matrix in the k column (resp. row) if the D kth column (resp. row) contains exactly two non-zero entries.

th Suppose that x1; x2; : : : ; xu are row vectors of the matrix M . If M is contractible in the k column such that th mi;k 0,mj;k 0 and i j , then the .u 1/ .v 1/ matrix Mij k obtained from M by replacing the i row ¤ ¤ ¤ th th W th with mi;kxj mj;kxi and deleting the j row. The k column is called the contraction in the k column relative C to the i th row and the j th row. In [21], Brualdi and Gibson obtained that per .M / per .N / if M is a real matrix of order ˛ > 1 and N is a D contraction of M . h .n/i Let Mk 1 .n/ mi;j be the n n super-diagonal matrix, deﬁned by C D 8 if i j ˛ for 1 ˛ n ˆ ˆ D D Ä Ä ˆ 1 and ˆ ˆ i ˛ 1 and j ˛ for 1 ˛ n 1, .n/ < D C D Ä Ä mi;j if i ˛ u and j ˛ for u 1 ˛ n D ˆ D D C Ä Ä ˆ 1 such that ˆ ˆ 1 u k, ˆ Ä Ä :ˆ 0 otherwise. On the arrowhead-Fibonacci numbers 1109 that is, .k 1/ th C # 2 1 1 1 1 0 0 0 0 3 6 1 1 1 1 1 0 0 0 7 6 7 6 0 1 1 1 1 1 0 0 7 6 7 6 : : : : : : : 7 6 : : : : : : : 7 6 : : : : : : : 7 M .n/ 6 7 . k 1 6 0 0 1 1 1 1 1 0 7 C D 6 7 6 0 0 0 1 1 1 1 1 7 6 7 6 7 6 0 0 0 0 1 1 1 1 7 6 : : : 7 6 : : :: :: :: :: : 7 6 : : : : : : : 7 6 7 4 0 0 0 0 1 1 1 5 0 0 0 0 1 1 Then we have the following theorem.

Theorem 2.9. For n > 1 and k 2,

perMk 1 .n/ ak 1 .n k 1/ ; C D C C C where perMk 1 .1/ 1. C D

Proof. First we start with considering the case n < 4. The matrices Mk 1 .2/ and Mk 1 .3/ are reduced to the C C following forms: " # 1 1 perMk 1 .2/ C D 1 1 and 2 3 1 1 1 perMk 1 .3/ 6 1 1 1 7 . C D 4 5 0 1 1

It is easy to see that perMk 1 .2/ 0 and perMk 1 .3/ 2. From deﬁnition of arrowhead-Fibonacci sequence C D C D 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. C C D C C D C C D 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 C C 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/ . C C D C C C

Since perMk 1 .n/ ak 1 .n k 1/, perMk 1 .n 1/ ak 1 .n k/, :::, perMk 1 .n k/ C D C C C C D C C C D ak 1 .n 1/, we easily obtain that perMk 1 .n 1/ ak 1 .n k 2/. So the proof is complete. C C C C D C C C h .n/i Let n > k 1 such that k 2 and let Rk 1 .n/ ri;j be the n n matrix, deﬁned by C C D 8 if i j ˛ for 1 ˛ n, ˆ D D Ä Ä ˆ i ˛ 1 and j ˛ for 1 ˛ n k 1 ˆ ˆ 1 D C D Ä Ä ˆ and ˆ .n/ < i n k ˛ and n k ˛ j n for 1 ˛ k 1, ri;j D C C Ä Ä Ä Ä D ˆ if i ˛ and j ˛ u for 1 ˛ n k ˆ D D C Ä Ä ˆ 1 such that ˆ ˆ 1 u k, ˆ Ä Ä :ˆ 0 otherwise. that is, 1110 ˙I. Gültekin, Ö. Deveci

.k 1/ th C # 2 1 1 1 0 0 0 0 0 3 6 1 1 1 1 0 0 0 0 7 6 7 6 0 1 1 1 1 0 0 0 7 6 7 6 : : : : : : : : 7 6 : : : : : : : : 7 6 : : : : : : : : 7 6 7 6 0 0 1 1 1 1 1 0 7 Rk 1 .n/ 6 7 C D 6 0 0 0 1 1 1 1 1 7 .n k/ th 6 7 6 7 6 0 0 0 0 0 1 1 1 7 6 : : 7 6 : :: :: :: :: :: :: : 7 6 : : : : : : : : 7 6 7 4 0 0 0 0 0 0 0 1 1 5 0 0 0 0 0 0 0 0 1 h .n/i Assume that the n n matrix Tk 1 .n/ ti;j is deﬁned by C D .k 1/ th C # 2 1 1 0 0 3 6 1 7 6 7 6 0 R .n 1/ 7 Tk 1 .n/ 6 k 1 7 , C D 6 C 7 6 : 7 4 : 5 0 where n > k 2 such that k 2. C Then we can give more general results by using other permanent representations than the above.

Theorem 2.10. Let ak 1 .n/ be the nth the arrowhead-Fibonacci number for k 2. Then C (i) For n > k 1, C perRk 1 .n/ ak 1 .n 1/ . C D C C (ii) For n > k 2, C n X perTk 1 .n/ ak 1 .i/ . C D C i 1 D Proof. (i). Let the equation hold for n > k 1, then we show that the equation holds for n 1. If we expand the C C perRk 1 .n/ by the Laplace expansion of permanent with respect to the ﬁrst row, then we obtain C

perRk 1 .n 1/ perRk 1 .n/ perRk 1 .n 1/ perRk 1 .n k/ . C C D C C C 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/ , C D C C C D C C D C C it is clear that

perRk 1 .n 1/ ak 1 .n 2/ . C C D C C (ii) It is clear that expanding the perTk 1 .n/ by the Laplace expansion of permanent with respect to the ﬁrst row, C gives us

perTk 1 .n/ perTk 1 .n 1/ perRk 1 .n/ . C D C C C 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: f C g xk g .x/ , D 1 x x2 xk 1 C C C C On the arrowhead-Fibonacci numbers 1111 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.

Theorem 2.11. The arrowhead-Fibonacci numbers have the following exponential representation:

i ! X1 x Ái g .x/ xk exp 1 x xk , D i i 1 D where k 2. Proof. Since Á ln g .x/ ln xk ln 1 x x2 xk 1 D C C C C and

i Á ÁÁ X1 x Ái ln 1 x x2 xk 1 ln 1 x x2 xk 1 1 x xk , C C C C D C D i i 1 D it is clear that i g .x/ X1 x Ái ln g .x/ ln xk ln 1 x xk . D xk D i i 1 D Thus we have the conclusion.

Now we consider the sums of arrowhead-Fibonacci numbers. Let n X Sn ak 1 .i/ D C i 1 D for n 1 and k 2, and suppose that Ak 2 is the .k 2/ .k 2/ matrix such that C C C 2 1 0 0 3 6 1 7 6 7 6 0 G 7 Ak 2 6 k 1 7 . C D 6 C 7 6 : 7 4 : 5 0

Then it can be shown by induction that

2 1 0 0 3 6 Sn k 7 6 C 7 n 6 S .G /n 7 .Ak 2/ 6 n k 1 k 1 7 . C D 6 C C 7 6 : 7 4 : 5 Sn

3 The arrowhead-Fibonacci sequence modulo n

If we reduce the arrowhead-Fibonacci sequence ak 1 .n/ by a modulus m, then we get the repeating sequence, f C g denoted by ˚ m « ˚ m m m « ak 1 .n/ ak 1 .1/ ; ak 1 .2/ ; : : : ; ak 1 .i/ ;::: C D C C C m where we denote ak 1 .i/.mod m/ by ak 1 .i/. It has the same recurrence relation as in (1). C C 1112 ˙I. Gültekin, Ö. Deveci

n o Theorem 3.1. The sequence am .n/ is simply periodic for k 2. That is, the sequence is periodic and repeats k 1 by returning to its starting values.C

k 1 Proof. Let X .x1; x2; : : : ; xk 1/ xi ’s be integers such that 0 xi m 1 . Since there are m C distinct D f C j Ä Ä g n m o .k 1/-tuples of elements of Zm, at least one of the .k 1/-tuples appears twice in the sequence a .n/ . Thus, C C k 1 the subsequence following this .k 1/-tuple repeats; hence, the sequence is periodic. Assume that uC > v and C m m m m m m ak 1 .u 1/ ak 1 .v 1/ ; ak 1 .u 2/ ak 1 .v 2/ ; : : : ; ak 1 .u k 1/ ak 1 .v k 1/ , C C Á C C C C Á C C C C C Á C C C then u v .mod k 1/. From the deﬁnition of the arrowhead-Fibonacci sequence, we can easily derive Á C m m m m m ak 1 .n/ ak 1 .n k 1/ ak 1 .n k/ ak 1 .n k 1/ ak 1 .n 1/ C D C C C C C C C C C C So using this equation, we easily obtain that

m m m m m m ak 1 .u/ ak 1 .v/ ; ak 1 .u 1/ ak 1 .v 1/ ; : : : ; ak 1 .u v 1/ ak 1 .1/ , C Á C C Á C C C D C n m o which implies that ak 1 .n/ is a simply periodic sequence. C n m o We next denote the period of the sequence ak 1 .n/ by Pk 1 .m/ : C C ˚ 2 « Example 3.2. Since a .n/ 0; 0; 0; 1; 1; 0; 0; 0; 1; 1; : : : , P4 .2/ 5. 4 D f g D Given an integer matrix X xi;j , X .mod m/ means that all entries of X are modulo m, that is, X .mod m/ D ˚ i « D xi;j .mod m/ . Let us consider the set X m X .mod m/ i 0 . If gcd .m; det X/ 1, then X m is a h i D j k 1 D h i cyclic group. Let X m denote the cardinality of the set X m. Since det Gk 1 . 1/ C , Gk 1 m is a cyclic jh i j h i Cˇ D ˇ h C i group for every positive integer m. By (2), it is easy to show that Pk 1 .m/ ˇ Gk 1 mˇ for k 2. C D h C i Now we give some useful properties for the period Pk 1 .m/ by the following theorem. C Theorem 3.3. u 1 u (i) Let p be a prime and suppose that u is the smallest positive integer with Pk 1 p C Pk 1 .p /. Then v v u C ¤ C Pk 1 .p / p Pk 1 .p/ for every v > u and k 2. C D C u Q ˛i (ii) If m has the prime factorization m .pi / , .u > 1/, then Pk 1 .m/ equals the least common multiple of D i 1 C ˛i D the Pk 1 .pi / ’s for k 2. C (iii) If k is a even integer such that k 2, then Pk 1 .m/ is even for every positive integer m. C t 1 Pk 1.p C / Proof. (i) If I is the .k 1/ .k 1/ identity matrix and t is a positive integer such that .Gk 1/ C C Ct 1 C Á t 1 Pk 1.p C / t t t 1 I modp C ;then .Gk 1/ C I modp . Then, it is clear that Pk 1 p divides Pk 1 p C . On C Á t Á C C Pk 1.p / .t/ t the other hand, if we denote .Gk 1/ C I ai;j p , then by the binomial expansion, we may write C D C p ! t ÁÁp Ái Á Pk 1.p / p .t/ t X p .t/ t t 1 .Gk 1/ C I ai;j p ai;j p I mod p C . C D C D i Á i 0 D t t t 1 t t 1 This yields that Pk 1 p p is divisible by Pk 1 p . Then, Pk 1 p C Pk 1 p or Pk 1 p C t C C .t/ C D C C D Pk 1 p p, and the latter holds if and only if there is a ai;j which is not divisible by p. Due to fact that we assume C u 1 u .u/ u is the smallest positive integer such that Pk 1 p C Pk 1 .p /, there is an ai;j which is not divisible by p. .u/ C .u/ ¤ C .u 1/ Since there is a ai;j such that p does not divide ai;j , it is easy to see that there is an ai;jC which is not divisible u 2 u 1 u 2 u 1 by p. This shows that Pk 1 p C Pk 1 p C . Then we get that Pk 1 p C p Pk 1 p C u 2C u¤ C C v v Du C D p .p Pk 1 .p // p Pk 1 .p /. So by induction on u we obtain Pk 1 .p / p Pk 1 .p/ for every C D C2 v v 1 C D C v > u. In particular, if Pk 1 p Pk 1 .p/, then Pk 1 .p / p Pk 1 .p/. C ¤ C C D C n ˛i o .pi / ˛i (ii) It is clear that the sequence ak 1 .n/ repeats only after blocks of length Pk 1 .pi / where is a C C ˛ n m o n .pi / i o natural number. Since Pk 1 .m/ is period of the sequence ak 1 .n/ ; the sequence ak 1 .n/ repeats after C C C On the arrowhead-Fibonacci numbers 1113

˛i Pk 1 .m/ terms for all values i. Thus, we easily see that Pk 1 .m/ is of the form Pk 1 .pi / for all values C C C of i, and since any such number gives a period of Pk 1 .m/. Therefore, we conclude that C

˛1 ˛u Pk 1 .m/ lcm Pk 1 .p1/ ;:::;Pk 1 .pu/ . C D C C ˇ ˇ (iii) Since det Gk 1 1 when k is a even integer and Pk 1 .m/ ˇ Gk 1 mˇ, we have the conclusion. C D C D h C i

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

References

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