Selçuk J. Appl. Math. Selçuk Journal of Vol. 12. No. 2. Pp. 71-80, 2011 Applied Mathematics

Selçuk J. Appl. Math. Selçuk Journal of Vol. 12. No. 2. pp. 71-80, 2011 Applied Mathematics

On the Periods of Some Figurate Numbers

Omur Deveci1, Erdal Karaduman2

1Department of Mathematics, Faculty of Science and Letters, Kafkas University, 36100 Kars, Turkiye e-mail: [email protected] Department of Mathematics, Faculty of Science, Atatürk University, 25240 Erzurum, Turkiye e-mail: [email protected]

Received Date:December 7, 2010 Accepted Date: October 19, 2011

Abstract. The number which can be represented by a regular geometrical arrangement of equally spaced point is defined as figurate number. Each of polygonal, centered polygonal and pyramidal numbers is a class of the series of figurate numbers. In this paper, we obtain the periods of polygonal, centered polygonal and pyramidal numbers by reducing each element of these numbers modulo m.

Key words: Period, polygonal number, centered polygonal number, pyramidal number. 2000 Mathematics Subject Classiﬁcation: 11B75, 11B50.

1. Introduction

The ﬁgurate numbers have a very important role to solve some problems in number theory and to determine speciality of some numbers, see for example, [8,9]. The polygonal numbers, the centered polygonal numbers, the pyramidal numbers and their properties have been studied by some authors, see for example, [1,3,6,15]. The study of Fibonacci numbers by reducing modulo m began with the earlier work of Wall [13] where the periods of Fibonacci numbers according to modulo m were obtained. The theory is expanded to 3-step Fibonacci sequence by Özkan, Aydin and Dikici [11]. Lü and Wang [10] contributed to study of the Wall number for the k step Fibonacci sequence. Deveci and Karaduman [5] extended the concept to− Pell numbers. Now we extend the concept to the polygonal numbers, the centered polygonal numbers and the pyramidal numbers which are classes of the series of ﬁgurate numbers.

71 A sequence 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, b, c, b, c, is periodic after the initial element a and has period 2. A sequence is simply··· periodic with period n if the ﬁrst n elements in the sequence form a repeating subsequence. For example, the sequence a, b, c, a, b, c, a, b, c, , is simply periodic with period 3. ··· We have the following formulas for the polygonal numbers, the centered polygonal numbers and the pyramidal numbers: Let k be the number of sides in a polygon. The nth k gonal number is obtained by the formula − k k P (k, n)=( 1)n2 2 n. 2 − − 2 − µ ¶ A polygonal number is a number represented as dots or pebbles arranged in the shape of a regular polygon. We obtain nth triangular number, nth square number, nth pentagonal number, for k =3, 4, 5, . For more information on k gonal numbers, see [12]. ··· ··· The −nth centered k gonal number is obtained by the formula − kn C (k, n)= (n 1) + 1. 2 − A centered polygonal number formed by a central dot, surrounded by polygonal layers with a constant number of sides. Each side of a polygonal layer contains one dot more than a side in the previous layer, so starting from the second polygonal layer each layer of a centered k gonal number contains k more points than the previous layer. We obtain nth centered− triangular number, nth centered square number, nth centered pentagonal number, for k =3, 4, 5, .For more information on centered k gonal numbers, see··· [2]. ··· The nthk gonal pyramidal number− is obtained by formula − n2 k 1 k 5 P (k) = + n3 n − . n 2 6 − 3 − 6 µ ¶ µ ¶ A pyramidal number represents a pyramid with a base and given number of sides. We obtain nth triangular pyramidal number, nth square pyramidal number, nth pentagonal pyramidal number, for k =3, 4, 5, . For more information on k gonal pyramidal numbers,··· see [14]. ··· In this paper,− the usual notation p is used for a prime number.

2. Polygonal Numbers

In this section, we obtain the lengths of the periods of k gonal numbers modulo − m. The notation LP (m) denote the length of the smallest period which the period is obtained each element of the polygonal numbers by reducing modulo m for k 3. ≥

72 Theorem 2.1. Let k 0(mod4). The lengths of the periods of the polygonal numbers are as follows:≡ u 2foru =1, i. For u N, LP (2 )= u 1 ∈ 2 − for u>2. ½ θ θ ii. If p =2and θ N,thenLP p = p . 6 t ∈ ei iii. If m = i=1 pi (t 1) where pi’s are distinct primes, then LP (m)= ei ≥ ¡ ¢ lcm [LP (p )]. i Q Proof. We prove this by direct calculation. Since k 0(mod 4) and k 3, ≡ ≥ k =4, ( N). i. Since ∈ P (4, n)(mod2) (2 1) n2 (2 2) n (mod 2) 0(mod 2)for n is even and ≡ − − − ≡ P (4, n)(mod2) £(2 1) n2 (2 2) n¤ (mod 2) 1(mod 2)for n is odd, ≡ − − − ≡ LP (2) = 2.Ifu>1,then £ ¤ u 1 u u 1 2 u 1 u P 4, 2 − (mod 2 ) (2 1) 2 − +(2 2) 2 − (mod 2 ) ≡ − − ≡ u u 1 u h u 2 u ui ¡2 2 −¢ 1 +2 1 2 − ¡(mod¢ 2 ) 0(mod 2 ) , ≡ − − ≡ u 1 u P £4, 2¡ − +1 ¢(mod¡ 2 ) ¢¤ ≡ u 1 2 u 1 u ¡(2 1) 2 ¢− +1 +(2 2) 2 − +1 (mod 2 ) ≡ − − ≡ h u u 1 u 2 u i u 2 +2¡ − 2 −¢ +1 (mod¡ 2 ) 1(mod¢ 2 ) , , ≡ − ≡ ··· u 1 u P £4,¡2 − + n (mod¢ 2 ) ¤ ≡ u 1 2 u 1 u ¡(2 1) 2 −¢ + n +(2 2) 2 − + n (mod 2 ) ≡ − − ≡ h u u 1 u 2 i u 2 2 − ¡ +2n ¢2 − n ¡+1 +1 ¢(mod 2 )+ ≡ − − − + £(2 ¡ 1) n2 +(2 2) n (mod d 2u)¢ ¤ − − ≡ £ (2 1) n2 +(2 2) n¤ (mod 2u) P (4, n)(mod2u) . ≡ − − ≡ u u 1 So, we£ get LP (2 )=2 − . ¤ ii. The proof is similar to the proof of i. and is omitted. ei iii. Let lcm [LP (pi )] = β. P (4, β)(modm) (2 1) β2 +(2 2) β (mod m) 0(mod m) , ≡ − − ≡ P (4, β +1)(mod m£) (2 1) (β +1)2 +(2¤ 2) (β +1) (mod m) ≡ − − ≡ β2 (2 1) + 1 (mod hm) 1(mod m) , , i ≡ − ≡ ··· P (4£ , β)(modm)¤ (2 1) β2 +(2 2) β (mod m) 0(mod m) , ≡ − − ≡ P (4, β +1)(mod m)£ (2 1) (β +1)2 +(2¤ 2) (β +1) (mod m) ≡ − − ≡ β2 (2 1) + 1 (mod hm) 1(mod m) , , i ≡ − ≡ ··· £ ¤ 73 P (4, β + n)(modm) (2 1) (β + n)2 +(2 2) (β + n) (mod m) ≡ − − ≡ [β (2β +4n β 2n h 2 +2)+1](mod m)+ i ≡ − − − + (2 1) n2 +(2 2) n (mod m) − − ≡ £ (2 1) n2 +(2 2) n¤ (mod m) P (4, n)(modm) . ≡ − − ≡ ei So, we£ get LP (m)=lcm[LP¤ (pi )].

Theorem 2.2: Let k 2(mod4).ThenLP (m)=m for m 2. ≡ ≥ Proof: We prove this by direct calculation. Since k 2(mod4)and k 3, ≡ ≥ k =4 +2, ( N). ∈ P (4 +2,m+ n)(modm) (2)(m + n)2 +(2 1) (m + n) (mod m) ≡ − ≡ h i m (2m +2n 2 1) (mod m)+ (2)(n)2 +(2 1) (n) (mod m) ≡ − − − ≡ P (4 +2,n)(modm) . h i ≡

So, we get LP (m)=m.

Theorem 2.3. Let k 1(mod4)or k 3(mod4). The lengths of the periods of the polygonal≡ numbers are as follows:≡ u u+1 i. For u N, LP (2 )=2 . ∈ θ θ ii. If p =2and θ N,thenLP p = p . 6 t ∈ ei iii. If m = i=1 pi (t 1) where pi’s are distinct primes, then LP (m)= ei ≥ ¡ ¢ lcm [LP (p )]. i Q Proof. The proof is similar to the proof of Theorem 2.1. and is omitted.

3. Centered Polygonal Numbers

In this section, we obtain the lengths of the periods of centered k gonal numbers − modulo m.ThenotationLCP (m) denote the length of the smallest period which the period is obtained each element of the centered polygonal numbers by reducing modulo m for k 3. ≥ Theorem 3.1. Let k 0(mod4)that is k =2α.pe1 .pe2 . .pet , α 2 such ≡ 1 2 ··· t ≥ that p1,p2, ,pt are distinct primes and e1,e2, ,et are 0 or positive integers. The lengths··· of the periods of the centered polygonal··· numbers are as follows: u 1foru α, i. For u N, LCP (2 )= u 1 ≤ ∈ 2 − for u>α. ½ λ 1forλ ev, ii. If v N such that 1 v t and λ N,thenLCP pv = λ ev ≤ ∈ ≤ ≤ ∈ pv− for λ>ev. θ θ ½ iii. If p =2,p1,p2, ,pt and θ N,thenLCP p ¡= p¢ . 6 ··· ∈ ¡ ¢ 74 Proof. We prove this by direct calculation. Since k 0(mod4)and k 3, α e1 e2 et ≡ ≥ k =2 .p .p . .p , α 2 such that p1,p2, ,pt are distinct primes and 1 2 ··· t ≥ ··· e1,e2, ,et are 0 or positive integers. i. If u···α,then ≤ k C (k,n)(mod2u) n (n 1) + 1 (mod 2u) 1(mod 2u) . ≡ 2 − ≡ ∙ ¸ u So, we get LCP (2 )=1.Ifu>α,then

u 1 u k u 1 u 1 u u C k, 2 − (mod 2 ) 2 − 2 − 1 +1 (mod 2 ) 1(mod 2 ) , ≡ 2 − ≡ u 1 u k u 1 u 1 u u C ¡k, 2 − ¢+1 (mod 2 )£ ¡2 − +1¢2 − ¤+1 (mod 2 ) 1(mod 2 ) , , ≡ 2 ≡ ··· u 1 u k u 1 u 1 u C ¡k, 2 − + n¢ (mod 2 ) £ ¡2 − + n¢ 2 − + n¤ 1 +1 (mod 2 ) ≡ 2 − ≡ k u u 1 u 1 u k u ¡ 2 n +2¢− 2 − 1 £ (mod¡ 2 )+¢¡ n (n 1) +¢ 1 (mod¤ 2 ) ≡ 2 − 2 − ≡ £ k n¡ (n 1) + 1¡ (mod 2u¢¢¤) C (k, n)(mod2£ u) . ¤ ≡ 2 − ≡ u u 1 So,£ we get LCP (2 ¤)=2 − . ii. If λ ev and 1 v t,then ≤ ≤ ≤ k C (k, n) mod pλ n (n 1) + 1 mod pλ 1 mod pλ . v ≡ 2 − v ≡ v ∙ ¸ ¡ λ ¢ ¡ ¢ ¡ ¢ So, we get LCP p =1.Ifλ>ev and 1 v t,then v ≤ ≤ ¡ ¢ λ ev λ k λ ev λ ev λ λ C k, p − mod p p − p − 1 +1 mod p 1 mod p , v v ≡ 2 v v − v ≡ v λ ev λ C ¡k, p − ¢¡+1 mod¢ p £ ¡ ¢ ¤¡ ¢ ¡ ¢ v v ≡ k λ ev λ ev λ λ ¡ p − +1¢¡p − +1¢ mod p 1 mod p , , ≡ 2 v v v ≡ v ··· λ ev λ £ ¡ C k,¢ p − + n¤¡mod p ¢ ¡ ¢ v v ≡ k λ ev λ ev λ ¡ p − + n¢¡p − +¢n 1 +1 mod p ≡ 2 v v − v ≡ k λ ev λ ev λ ev λ £ ¡2np − +¢¡p − p − ¢1 ¤¡mod p ¢+ ≡ 2 v v v − v + £k n¡(n 1) + 1 mod¡ pλ ¢¢¤ ¡ ¢ 2 − v ≡ £ k n (n 1) + 1¤¡mod pλ¢ C (k, n) mod pλ . ≡ 2 − v ≡ v λ λ ev So, we get LCP £pv = pv− . ¤¡ ¢ ¡ ¢ iii. If p =2,p1,p2, ,pt,then 6 ¡ ···¢ C k, pθ mod pθ k pθ pθ 1 +1 mod pθ 1 mod pθ , ≡ 2 − ≡ C ¡k, pθ¢¡+1 mod¢ pθ £ ¡k pθ +1¢ pθ¤¡+1 mod¢ pθ ¡ 1 mod¢ pθ , , ≡ 2 ≡ ··· C ¡k, pθ + n¢¡mod pθ¢ £ k ¡pθ + n¢ pθ +¤¡n 1 +1¢ mod¡ pθ ¢ ≡ 2 − ≡ ¡ k 2npθ +¢¡pθ pθ 1¢ £ mod¡ pθ ¢¡+ k n (n 1)¢ + 1¤¡mod pθ¢ ≡ 2 − 2 − ≡ £ k n¡ (n 1) +¡ 1 mod¢¢¤pλ ¡pθ C¢(k, n£ ) mod pθ . ¤¡ ¢ ≡ 2 − v ≡ £ ¤¡ ¢ ¡ ¢ 75 θ θ So, we get LCP p = p .

Theorem 3.2.¡Let¢ k 2(mod4)that is k =2.pe1 .pe2 . .pet such that ≡ 1 2 ··· t p1,p2, ,pt are distinct primes and e1,e2, ,et are positive integers. The lengths··· of the periods of the centered polygonal··· numbers are as follows: u 1foru =1, i. For u N, LCP (2 )= u ∈ 2 for u>1. ½ λ 1forλ ev, ii. If v N such that 1 v t and λ N,thenLCP pv = λ ev ≤ ∈ ≤ ≤ ∈ pv− for λ>ev. θ θ ½ iii. If p =2,p1,p2, ,pt and θ N,thenLCP p ¡= p¢ . 6 ··· ∈ Proof. We prove this by direct calculation.¡ Since¢ k 2(mod4)and e1 e2 et ≡ k 3, k =2.p .p . .p such that p1,p2, ,pt are distinct primes and ≥ 1 2 ··· t ··· e1,e2, ,et are positive integers. i. C (k,··· n)(mod2) k n (n 1) + 1 (mod 2) 1(mod 2). ≡ 2 − ≡ So, we get LCP (2) = 1.Ifu>1,then £ ¤ C (k,2u)(mod2u) k 2u (2u 1) + 1 (mod 2u) 1(mod 2u) , ≡ 2 − ≡ C (k,2u +1)(mod2u)£ k (2u +1)2u¤+1 (mod 2u) 1(mod 2u) , , ≡ 2 ≡ ··· C (k,2u + n)(mod2u) £ k (2u + n)(2u + ¤n 1) + 1 (mod 2u) ≡ 2 − ≡ k 2u+1n +2u (2u 1) £ (mod 2u)+ k n (n 1) +¤ 1 (mod 2u) ≡ 2 − 2 − ≡ £ k n¡ (n 1) + 1 (mod¢¤ 2u) C (k, n)(mod2£ u) . ¤ ≡ 2 − ≡ u u So, we£ get LCP (2 )=2¤ . The proofs of ii. and iii. are similar to the proofs of Theorem 3.1. ii. and Theorem 3.1.iii., respectively and are omitted.

Theorem 3.3. Let k 3(mod4)that is k = pe1 .pe2 . .pet such that ≡ 1 2 ··· t p1,p2, ,pt are distinct primes and e1,e2, ,et are positive integers. The lengths··· of the periods of the centered polygonal··· numbers are as follows: u u+1 i. For u N, LCP (2 )=2 . ∈ λ 1forλ ev, ii. If v N such that 1 v t and λ N,thenLCP pv = λ ev ≤ ∈ ≤ ≤ ∈ pv− for λ>ev. θ θ ½ iii. If p =2,p1,p2, ,pt and θ N,thenLCP p ¡= p¢ . 6 ··· ∈ Proof. We prove this by direct calculation.¡ Since¢ k 3(mod4), k = e1 e2 et ≡ p1 .p2 . .pt such that p1,p2, ,pt are distinct primes and e1,e2, ,et are positive··· integers. ··· ··· C k, 2u+1 (mod 2u) k 2u+1 2u+1 1 +1 (mod 2u) 1(mod 2u) , ≡ 2 − ≡ u+1 u k u+1 u+1 u u C ¡k, 2 ¢+1 (mod 2 )£ 2 ¡2 +1¢2 ¤+1 (mod 2 ) 1(mod 2 ) , , C k, 2u+1 + n (mod 2u) ≡ k 2u+1 + n 2u+1 + n 1 +1 ≡(mod 2u) ··· ¡ ¢ ≡ £ 2 ¡ ¢ ¤ − ≡ ¡ k 2u+2n +2¢ u+1 2u+1 1£ ¡ (mod 2u¢¡)+ k n (n 1)¢ + 1 ¤ (mod 2u) ≡ 2 − 2 − ≡ £ k n¡ (n 1) + 1 (mod¡ 2u) ¢¢¤C (k, n)(mod2£ u) . ¤ ≡ 2 − ≡ £ ¤ 76 u u+1 So, we get LCP (2 )=2 . The proofs of ii. and iii. are similar to the proofs of Theorem 3.1. ii. and Theorem 3.1.iii., respectively and are omitted. If k 1(mod4), then the rules are the same the rules of Theorem 3.3. The proof≡ is similar to the proof of Theorem 3.3 and is omitted.

t ei Theorem 3.4. If m = i=1 pi (t 1) where pi’s are distinct primes, then ei ≥ LCP (m)=lcm[LCP (p )]. i Q ei Proof. We prove this by direct calculation. Let lcm [LCP (pi )] = β. C (k, β)(modm) k β (β 1) + 1 (mod m) 1(mod m) , ≡ 2 − ≡ C (k, β +1)(mod m)£ k (β +1)β¤+1 (mod m) 1(mod m) , , ≡ 2 ≡ ··· C (k, β + n)(modm) £ k (β + n)(β +¤n 1) + 1 (mod m) ≡ 2 − ≡ k β (β +2n 1) (mod£ m)+ k n (n 1) + 1 (mod¤ m) ≡ 2 − 2 − ≡ £ k n (n 1) + 1 ¤(mod m) C£(k,n)(mod2u¤) . ≡ 2 − ≡ ei So, we£ get LCP (m)=lcm[¤ LCP (pi )]. 4. Pyramidal Numbers

In this section, we obtain the lengths of the periods of k gonal pyramidal − numbers modulo m. The notation LPY (m) denote the length of the smallest period which the period is obtained each element of the pyramidal numbers by reducing modulo m for k 3. ≥ Theorem 4.1. Let k 0(mod3)or k 1(mod3). The lengths of the periods of the pyramidal≡ numbers are as follows:≡ u u+1 i. If p =2, 3,thenLPY (p )=p for u N. θ ∈θ ii. If p =2, 3 and θ N,thenLPY p = p . 6 t e∈i iii. If m = i=1 pi (t 1) where pi’s are distinct primes, then LPY (m)= ei ≥ ¡ ¢ lcm [LPY (p )]. i Q Proof. We prove this by direct calculation. Let k 0(mod3)that is k =3, ≡ ( N). ∈ i. If p =2,thenwehaveforu N ∈

(3) u P2u+1 1 (mod 2 ) − ≡ u+1 2 (2 1) u+1 3 3 1 u+1 3 5 u − + 2 1 2 1 − (mod 2 ) ≡ 2 − 6 − 3 − − 6 ≡ ∙ ¸ u 2 5 22u+3¡ (mod¢ 2u¡) 0(mod¢ ¡ 2u) , ¢¡ ¢ ≡ 3 − ≡ u+1 2 (3£) ¡ u ¢¤ (2 ) u+1 3 3 1 u+1 3 5 u P u+1 (mod 2 ) + 2 2 − (mod 2 ) 2 ≡ 2 6 − 3 − 6 ≡ ∙ ¸ u 2 5 22u+3 (mod 2u) ¡ 0(mod¢ ¡ 2u) , ¢ ¡ ¢¡ ¢ ≡ 3 − ≡ £ ¡ ¢¤ 77 P (3) (mod 2u) 2u+1+1 ≡ u+1 2 (2 +1) u+1 3 3 1 u+1 3 5 u + 2 +1 2 +1 − (mod 2 ) ≡ 2 6 − 3 − 6 ≡ ∙ ¸ u 2 5 22u+3¡ +1 (mod2¢ ¡ u) ¢1(mod¡ 2u) , ¢¡ , ¢ ≡ 3 − ≡ ··· P (3£) ¡ (mod 2u¢¤) 2u+1+n ≡ u+1 2 (2 +n) u+1 3 3 1 u+1 3 5 u + 2 + n 2 + n − (mod 2 ) ≡ 2 6 − 3 − 6 ≡ ∙ ¸ 2u 2u+3¡ ¢u ¡ n2 ¢ ¡3 3 1 ¢¡ 3¢ 5 u 5 2 (mod 2 ) + + n n − (mod 2 ) ≡ 3 − 2 6 − 3 − 6 ≡ £ (3¡) u¢ ¤ h ¡ ¢ ¡ ¢i Pn (mod 2 ) . ≡ u u+1 So, we get LPY (2 )=2 . If p =3, then the proof is similar to the proof of the case p =2and is omitted. ii. If p =2, 3 and θ N,thenwehave 6 ∈

(3) u Ppu 1 (mod p ) − ≡ u 2 (p 1) u 3 3 1 u 3 5 u − +(p 1) (p 1) − (mod p ) ≡ 2 − 6 − 3 − − 6 ≡ h u i p (5 2.pu)(pu +1)¡ (mod¢ pu) 0(mod¡ pu¢) , ≡ 6 − ≡ u 2 (3h) u (p ) i u 3 3 1 u 3 5 u P u (mod p ) +(p ) (p ) − (mod p ) p ≡ 2 6 − 3 − 6 ≡ u h i p (5 2.pu)(pu +1) (mod ¡pu) 0(mod¢ pu¡) , ¢ ≡ 6 − ≡ (3h) u i P u (mod p ) p +1 ≡ u 2 (p +1) u 3 3 1 u 3 5 u +(p +1) (p +1) − (mod p ) ≡ 2 6 − 3 − 6 ≡ h u i p (5 2.pu)(pu +1)+1¡ ¢(mod pu) ¡1(mod¢ pu) , , ≡ 6 − ≡ ··· (3h) u i P u (mod p ) p +n ≡ u 2 (p +n) u 3 3 1 u 3 5 u +(p + n) (p + n) − (mod p ) ≡ 2 6 − 3 − 6 ≡ h u i p (5 2.pu)(pu +1)¡ (mod¢ pu)+ ¡ ¢ ≡ 6 − hn2 3 3 1 i3 5 u (3) u + + n n − (mod 2 ) Pn (mod 2 ) . 2 6 − 3 − 6 ≡ h i ¡ θ ¢ θ¡ ¢ So, we get LPY p = p . ¡ ¢

78 ei iii. Let lcm [LPY (pi )] = β.

(3) (β 1)2 3 3 1 3 5 Pβ 1 (mod m) −2 +(β 1) 6 3 (β 1) 6− (mod m) − ≡ − − − − ≡ h i β (5 2.β)(β +1) (mod m) 0(mod¡ m¢ ) , ¡ ¢ ≡ 6 − ≡ (3h) (β)i2 3 3 1 3 5 P (mod m) +(β) (β) − (mod m) β ≡ 2 6 − 3 − 6 ≡ h i β (5 2.β)(β +1) (mod m¡ ) 0(mod¢ m¡) , ¢ ≡ 6 − ≡ h i (3) (β+1)2 3 3 1 3 5 P (mod m) +(β +1) (β +1) − (mod m) β+1 ≡ 2 6 − 3 − 6 ≡ h i β (5 2.β)(β +1)+1 (mod m)¡ 1(mod¢ m) , ,¡ ¢ ≡ 6 − ≡ ··· (3h) (β+n)2 i 3 3 1 3 5 P (mod m) +(β + n) (β + n) − (mod m) β+n ≡ 2 6 − 3 − 6 ≡ β h n2 ¡ 3 3¢ 1 ¡3 5 ¢i (5 2.β)(β +1) (mod m)+ + n n − (mod m) 6 − 2 6 − 3 − 6 ≡ h (3) i h ¡ ¢ ¡ ¢i Pn (mod m) . ≡ ei So, we get LPY (m)=lcm[LPY (pi )].

Theorem 4.2. Let k 2(mod3). The lengths of the periods of the pyramidal numbers are as follows:≡ u u+1 i. For u N, LPY (2 )=2 . ∈ θ θ ii. If p =2and θ N,thenLPY p = p . 6 t ∈ei iii. If m = i=1 pi (t 1) where pi’s are distinct primes, then LPY (m)= ei ≥ ¡ ¢ lcm [LPY (p )]. i Q Proof. The proof is similar to the proof of Theorem 4.1. and is omitted.

5. Open Question

Wall in [13] proved that the lengths of the periods of the recurring sequences obtained by reducing a Fibonacci sequences by a modulo m are equal to the lengths of the of ordinary 2 step Fibonacci recurrences in cyclic groups. The theory is expanded to 3-step− Fibonacci sequence by Özkan, Aydin and Dikici [11]. Lü and Wang contributed to the study of the Wall number for the k- step Fibonacci sequence [10]. Some works on the concept have been made, for example, [4,7]. Deveci and Karaduman [5] extended the concept to Pell sequences in ﬁnite groups. Are there groups such that the lengths of the periods of recurrence sequences obtained by reducing according to a modulo m anyone of polygonal numbers, centered polygonal numbers and pyrimidal numbers are equal to the lengths of the periods of 2 step or k step Fibonacci recurrences in these groups? − −

79 6. Acknowledgment

The authors thank the referee for her/his valuable suggestions which improved the presentation of the paper.

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