Binomial Transform and Dold Sequences

Transcription

1 Journa of Integer Sequences, Vo. 18 (2015), Artice Binomia Transform and Dod Sequences Kaudiusz Wójcik Department of Mathematics and Computer Science Jagieonian University Lojasiewicza Kraków Poand Kaudiusz.Wojcik@uj.edu.p Abstract We prove that the binomia transform T(a) of a Dod sequence a is a Dod sequence itsef. We aso show that if a and T(a) are bounded Dod sequences then they are both periodic with period 6. 1 Introduction Let a = (a n ) n 0 be a sequence of compex numbers. The binomia transform of the sequence a is the sequence T(a) = (T(a) n ) n 0 defined in [16] by T(a) n = n ( 1) i a i, n 0. i i=0 The binomia transform is a inear invoution and the origina sequence a can be recovered from the reation n a n = ( 1) i T(a) i, n 0. i i=0 We say that a sequence of integers a = (a n ) n 0 is a Dod sequence [11] if µ(k)a n/k 0 (mod n), n 1, (1) k n 1

2 where µ : N Z is the Möbius function given by 1, if n = 1; µ(n) = ( 1) k, if n = p 1 p k, p i, distinct primes; 0, otherwise. It is easy to verify that a sequence of integers a is a Dod sequence a if and ony if for every prime p and natura numbers m,α such that gcd(p,m) = 1 we have a mp α a mp α 1 (mod p α ). We wi say that a sequence of integers a is a weak Dod sequence if for every prime number p a p a 1 (mod p). Obviousy a Dod sequence is a weak Dod sequence. Babienko and Bogatyi proved that if a Dod sequence is bounded, then it is periodic [11]. Among the Dod sequences, the Lefschetz sequences pay an important roe from the point of view of appications in dynamica systems. A sequence of integers a is a Lefschetz sequence if there exist a compact ENR (i.e., Eucidean neighborhood retract) X and a continuous map f : X X such that a n is the Lefschetz number L(f n ) of the n-th iteration of f. Let us reca that a Lefschetz number of f is defined by L(f) = k 0( 1) k trh k (f), where H k (f) : H k (X) H k (X) is the endomorphism induced by f on the k-th singuar homoogy of X (with rationa coefficients). One can prove [11] that a Dod sequence a = (a n ) n 1 is a Lefschetz sequence if and ony if there are integer matrices A M k (Z), B M m (Z) such that a n = tra n trb n, n 1. In the seque we wi treat a Lefschetz sequence a = (a n ) n 1 as a sequence a = (a n ) n 0 with a 0 = tra 0 trb 0. If a is a Lefschetz sequence with B = 0 then we ca it a sequence of traces of integer matrix, i.e., a n = tra n. The aim of this note is to study the binomia transforms of Dod sequences. We show that a is a Dod sequence if and ony if its binomia transform T(a) is a Dod sequence (Theorem 7). The proof of Theorem 7 is trivia if a is a Lefschetz sequence. Indeed, if a is a sequence of traces then we have a cosed formua for its binomia transform T(a), namey T(a) n = n ( 1) i tra i = tr(i A) n, i i=1 2

3 so T(a) is a sequence of traces. In particuar, binomia transform of the sequence of traces is a Dod sequence. Obviousy the same argument works in the case of a Lefschetz sequence a with matrices A and B. Then T(a) is a Lefschetz sequence itsef with matrices I A and I B. Unfortunatey, there are Dod sequences that are not Lefschetz sequences. Let us observe that if a is a Lefschetz sequence then a n kρ n, where ρ is the spectra radius of H(f) and k is a dimension of H(X). In particuar, the sequence a n = k n kk cannot be obtained as a Lefschetz sequence athough it is Dod sequence [11]. We aso prove that if a is a Dod sequence such that a and T(a) are bounded then both a and T(a) are 6-periodic. In particuar, if a is -periodic (non-constant) Dod sequence with 6 then T(a) is unbounded (Theorem 9). This resut has interesting consequences from the point of view of the geometric method for detecting chaotic dynamics generated by non-autonomous periodic in time ordinary differentia equations based on the notion of isoating segments [19, 20, 15, 17]. The above method gives a sufficient conditions for the existence of the compact invariant set I for the Poincaré map P such that P restricted to I is semi-conjugated to the shift map σ : Σ 2 Σ 2 where Σ 2 = {0,1} Z i.e., there is a continuous surjective map g : I Σ 2 such that g P I = σ g. The important dynamica question is if for n-periodic sequence c Σ 2 there exists n-periodic point x of the Poincaré map P such that x g 1 (c), i.e., trajectory of x is coded by the sequence c. In the context of resuts in [20] it appears that there is a Dod sequence a with the property that T(a) k 0 impies that g 1 (c) contains n-periodic point of P for every n-periodic sequence c Σ 2 with k-symbos 1. Our Theorem 9 guarantees the existence of infinitey many periodic points of the Poincaré map. It seems that the resuts we obtained can be aso interesting from the point of view of eementary number theory and combinatorics, because many cassica integer sequences are Dod sequences. For exampe, the k-lucas sequences [6, 18], the generaized k-fibonacci sequences [5], sequences generated by Lucas functions [4, 6], and sequences generated by Tchebycheff poynomias of the first kind [6, 9]. In particuar, the cassica Lucas sequence L given by L n = ( 1+ ) n 5 +( 2 1 ) n 5, 2 is a Dod sequence (even sequence of traces with the matrix A = congruences 1. if p is odd prime then L 2p 3 (mod p), 2. for k 1 we have L 2 k 3 (mod 4). 3. if r 2 then L 2 r 7 (mod 8). considered by the authors in [1, 12, 13] easiy foow by Eq. (1). [ ] ) and the foowing 3

4 We wi finish this paper with the resuts concerning the sequences of the form a n (,r) = ( 1) s, b n (,r) = a n (,r), n 0. s s r (mod ) where 2 and r {0,..., 1} are fixed. The congruences reated to these sequences were studiedin[21,22]. AstheappicationofTheorem7wegetthatfor 2andr {0,..., 1} the sequence b(,r) is a Dod sequence exacty if one of the foowing conditions hods 1. 2 and r = 0, 2. = 2r and r 1. We aso show that a n (,0) = 0 if and ony if is odd and n 2N+1. It is easy to check that if a is -periodic then T(a) n = 1 r=0 a ra n (,r). We prove that if a is -periodic (weak) Dod sequence and is prime then T(a) n = a n (,0)(a 0 a 1 ), n 0. 2 Main resuts. Lemma 1. Let a be a sequence of integers. Then a is a weak Dod sequence if and ony if T(a) is a weak Dod sequence. Proof. Since T is an invoution so it is sufficient to show that T(a) is a weak Dod sequence provided a is a weak Dod sequence. We have and a 1 a 2 (mod 2) so Let p be odd prime. Then T(a) p = T(a) 1 = a 0 a 1, T(a) 2 = a 0 2a 1 +a 2, T(a) 2 a 0 +a 2 a 0 +a 1 a 0 a 2 T(a) 1 (mod 2). p ( ) p ( 1) i a i a 0 a p a 0 a 1 T(a) 1 (mod p), i i=0 because p ( p i) for i = 1,...,p 1. Lemma 2. Let a be a -periodic weak Dod sequence ( 2). If gcd(n,) = 1 then a n = a 1. 4

5 Proof. It foows by Dirichet s theorem that there exist infinitey many primes in the sequence k + n. Let p be a prime of the form k + n such that p > a n a 1. Since a is a -periodic weak Dod sequence, we have Hence p a n a 1 and consequenty a n = a 1. a n = a p a 1 (mod p). Coroary 3. Let a be a -periodic weak Dod sequence with prime. Then where a n (,0) = s 0 (mod ) ( 1)s( n s). T(a) n = a n (,0)(a 0 a 1 ), n 0, Proof. It foows by Lemma 2 that the -periodic sequence a has the form a 1 =... = a 1, a 0 = a, hence T(a) n = ( s) n ( 1) s a 0 + ( s) n ( 1) s a 1 s 0 (mod ) s 0 (mod ) = a n (,0)(a 0 a 1 ) because n s=0 ( 1)s( n s) = 0. We reca the characterization of Dod sequences given in [6]. Let us note that Dod sequence in [6] is caed a generaized Fermat sequence. Let c = (c n ) n 1 be a sequence of integers. We say that sequence a is a Newton sequence generated by c if a n = c 1 a n 1 +c 2 a n 2 + +c n 1 a 1 +nc n, n 1. Remark 4. Assume that c is a finite sequence, i.e., there exists m 1 such that c n = 0 for n > m. One can check [6] that then a n = trm n m, n 1 wherem m isthecompanionmatrixofthepoynomiax m c 1 x m 1 c 2 x m 2 c m 1 x c m, i.e., c m c m 1 M m = c m 2 M m (Z) c 1 5

6 Proposition 5. Let a = (a n ) n 1 be a sequence of integers. Then a is a Dod sequence if and ony if there exists an integer sequence c such that a is a Newton sequence generated by c. Proof. It foows by [6, Thm. 5, Thm. 6]. Lemma 6. Let a = (a n ) n 1 be a sequence of integers. Then a is a Dod sequence if and ony if for every m 1 there exists a matrix A m M m (Z) such that a n = tra n m, 1 n m. Proof. We first prove the part if, so we assume that a is a Dod sequence. Let c be a sequence of integers such that a is a Newton sequence generated by c. Let m 1 be fixed. Let us consider a finite sequence c(m) such that { c n, if 1 n m; c(m) n = 0, otherwise. Let b be a Newton sequence generated by c(m). Then b n = trm n m, n 1, and a n = b n for 1 n m by definition of the Newton sequence, so the resut foows. We now prove the ony if part. We have to show that a is a Dod sequence. Let m 1 be fixed. We show that µ(k)a m/k 0 (mod m). By assumption there exists a matrix A m M m (Z) such that k m a n = trm n m, 1 n m. Let b be a sequence defined by b n = tra n m, n 1. Then b is a Dod sequence and a n = b n for 1 n m. In particuar, µ(k)b m/k 0 (mod m), so the proof is compete. k m µ(k)a m/k = k m Theorem 7. Assume that a = (a n ) n 0 is a sequence of integers. Then a is a Dod sequence if and ony if T(a) is a Dod sequence. Proof. Since T is an invoution, it is sufficient to show that T(a) is a Dod sequence provided that a is a Dod sequence. Let m 2 be fixed. We have to show that µ(k)t(a) m/k 0 (mod m). k m 6

7 By Lemma 6 there exists a matrix A m M m (Z) such that Let ã be a sequence defined by Then and T(ã) is a Dod sequence given by In particuar, i=0 a n = tra n m, 1 n m. ã n = tra n m. a 1 = ã 1, a m = ã m, ã 0 = m, T(ã) n = tr(i A m ) n, n 0. µ(k)t(ã) m/k 0 (mod m). k m Let 1 n m. Then n T(ã) m = ( 1) i ã i = m a 0 + i Consequenty, we get that k m µ(k)t(a) m/k = k m n ( 1) i a i = m a 0 +T(a) m. i i=0 µ(k)t(ã) m/k +(m a 0 ) µ(k) k m = k m µ(k)t(ã) m/k 0 (mod m), because k m µ(k) = 0 for m 2. The proof is compete since m 2 was arbitrary. Let a be a Lefschetz sequence with matrices A and B. We ca a compex number λ C an essentia eigenvaue of the pair (A,B) if the agebraic mutipicity of λ as the eigenvaue of A is different from the agebraic mutipicity of λ as the eigenvaue of B. By σ ess (A,B) we denote the set of a essentia eigenvaues of (A,B). Let ρ ess := ρ ess (A,B) be the essentia spectra radius of (A, B), i.e., ρ ess = max{ λ : λ σ ess (A,B)}. Remark 8. Assume that a Dod sequence a = (a n ) n 0 is bounded. It foows by [11, Theorem ] that a is a periodic Lefschetz sequence with -periodic matrices A and B. We treat a as the sequence (a n ) n 0 with a 0 = tra 0 trb 0. Then a 0 = a and binomia transform of a is given by T(a) n = tr(i A) n tr(i B) n, n 0. 7

8 Theorem 9. If a is a Dod sequence and a, T(a) are bounded then (a n ) n 1, (T(a)) n 1 are 6-periodic. Proof. Since a is bounded Dod sequence, it foows by [11, Theorem ] that a is - periodic Lefschetz sequence with -periodic matrices A, B (i.e., A = I k, A = I m ). In particuar, σ(a),σ(b) {1,ω,...,ω 1 }, where ω = e 2πi so is a primitive root of unity. We treat a as a sequence a n = tra n trb n, n 0, T(a) n = tr(i A) n tr(i B) n, n 0. Observe that λ σ ess (A,B) if and ony if 1 λ σ ess (I A,I B) and T(a) is periodic, because it is a bounded Dod sequence. Let 1 λ σ ess (I A,I B) S 1. Then λ = 1 λ = 1, so one can easiy check that λ = e ±πi 3. Observe that then 1 λ = λ. It foows that σ ess (A,B) (S 1 \{1}) = σ ess (I A,I B) S 1 {e ±πi 3 }. Suppose that ρ := ρ ess (I A,I B) > 1. Then there exists exacty one pair λ,λ {1,ω,...,ω 1 } such that 1 λ = 1 λ = ρ. Then T(a) n im = g 0, n ρ n so T(a) is unbounded, a contradiction. So we get that ρ 1. We show that We have that T(a) n = so the sequence 1 λ σ ess(i A,I B) {0,e ±πi 3 } σ ess (I A,I B) {0,e ±πi 3 }. (1 λ) n + 1 λ σ ess(i A,I B) {z:0< z <1} 1 λ σ ess(i A,I B) {z:0< z <1} (1 λ) n is the periodic integer sequence that converge to 0, so we get that σ ess (I A,I B) {z : 0 < z < 1} =. (1 λ) n, Coroary 10. If a is -periodic (non-constant) Dod sequence with 6 then T(a) is unbounded. 8

9 3 Appications to binomia sums Assume that 2 and r Z. We define the -periodic sequence a[,r] = (a n [,r]) n 0 by { 1, if n r (mod ); a n [,r] = 0, otherwise. Then a(,r) = (a n (,r)) n 0 = T(a[,r]) is its binomia transform where a n (,r) = ( 1) s, n 0. s We define s r (mod ) b[,r] := a[,r], b(,r) := a(,r). Remark 11. It foows by Coroary 3 we have T(a) = a(,0)(a 0 a 1 ), n 0, if a is -periodic Dod sequence with prime. Moreover, by Eq. (1) so T(a) = kb(,0) for some k Z. a 0 = a a 1 (mod ), Remark 12. The sequence b(, r) is particuary interesting from the point of view of appications in thedetection of chaotic dynamics [19, 20]. It turnsout that it is cosey reated to the sequence of indices of fixed points aowing to understand the nature of symboic dynamics for the Poincaré mapping associated with the periodic in time ordinary differentia equations [15, 17]. More specificay, et P be the Poincaré map for the panar periodic equation ż = z k (1+ z 2 e iκt ), z C. For sufficienty sma κ > 0 there exist a compact invariant set I for P and a continuous surjective map g : I Σ 2 such that g (P I ) = σ g, where σ : Σ 2 Σ 2 is the shift map [20]. It foows by [20] that if b n (k + 1,0) 0 and c Σ 2 is m-periodic sequence with n-symbos 1 then the Poincaré map P has m-periodic point x I such that g(x) = c. In this section we appy Theorem 7 to prove that for 2 and r {0,..., 1} the sequence b(,r) is a Dod sequence if and ony if 2 and r = 0 or = 2r and r 1. We aso prove Gaisher-type congruences for the sequence a(, r). Moreover, we show that a n (,0) = 0 if and ony if is odd and n is an odd mutipicity of. 9

10 Remark 13. Let us observe that 1 b n (,r) = ω tr (1 ω t ) n, n 0, where ω = e 2πi so we get that is a primitive -root of unity. Indeed, it is we-known that { 1 ω ts, if s; = 0, otherwise. 1 1 n ω tr (1 ω t ) n = ω tr ( 1) n k ω t(n k) n k k=0 n ( ) ( ) n 1 = ( 1) n k ω t(n k r) n k k=0 n ( ) ( ) n 1 = ( 1) s ω t(s r) = b n (,r). s s=0 Theorem 14. The sequence b(,0) = (b n (,0)) n 0 is a sequence of traces. In particuar, b(,0) is a Dod sequence. Proof. Let 2 and ω = e 2πi. We consider the matrix A M (Z) given by A = M (Z) One can check that A has eigenvaues 1,ω,...,ω 1, hence 1 tr(i A ) n = (1 ω t ) n = b n (,0). Theorem 15. Let 2 and r {1,..., 1}. The sequence b(,r) is a Dod sequence if and ony if = 2r. Lemma 16. If = 2r then b(2r,r) is a Dod sequence (even a Lefschetz sequence). 10

11 Proof. For = 2r and ω = e 2πi we have 1 1 b n (2r,r) = ω tr (1 ω t ) n = ( 1) t (1 ω t ) n = t even(1 ω t ) n t odd (1 ω t ) n. Let us observe that 1,ω 2,...,ω 2 are roots of the poynomia λ r 1 = 0 and the eigenvaues of the matrix A = M r (Z) and ω,ω 3,...,ω 1 are roots of the poynomia λ r +1 = 0 and the eigenvaues of the matrix B = M r (Z) hence b n (2r,r) = tr(i A) n tr(i B) n, so b(2r, r) is a Dod sequence (even Lefschetz sequence). Remark 17. If n > 2 then there exists a prime number p < n such that gcd(n,p) = 1. Indeed, if n is odd then one can take p = 2. If n = 2k is even then by Bertrand s postuate there is a prime p such that k < p < 2k. Lemma 18. If > 2 and r {1,..., 1} then the sequence b(,r) is a weak Dod sequence if and ony if gcd(r,) > 1. In particuar, b(,1) is not a Dod sequence for > 2. Proof. Assume that > 2. Since b(,r) = T(b[,r]), so by Theorem 7 it is sufficient to show that b[,r] is a weak Dod sequence if and ony if gcd(r,) > 1. Assume that gcd(r,) > 1. In particuar, r > 1 so b 1 [,r] = 0 by definition of the sequence b[,r]. Since gcd(r,) > 1, there is no prime p of the form k +r. Hence b p [,r] = 0 for every prime p, and b[,r] is a weak Dod sequence. Assume that b[,r] is a weak Dod sequence and suppose that gcd(r,) = 1. If r > 1 then b 1 [,r] = 0 and b r [,r] = by definition. On the other hand b 1 [,r] = b r [,r] by Lemma 2, a contradiction. Let r = 1. Then b 1 [,1] = by definition. By Remark 17 there is a prime number p such that p < and gcd(,p) = 1. Then b p [,1] = b 1 [,1] = by Lemma 2 and b p [,1] = 0 by definition, a contradiction. 11

12 Lemma 19. Let > 2 and r {1,..., 1}. If b(,r) is a Dod sequence then r > 1 and r. Proof. Assume that b(,r) = a(,r) is a Dod sequence. We may assume that gcd(r,) > 1. In particuar, 2 r < 1. Then for 1 n < r we have b n (,r) = 0 and b r (,r) = ( 1) r. By Eq. (1) we get that ( 1) r = b r (,r) = ( r µ b k (,r) 0 (mod r), k) k r so the resut foows. Lemma 20. Assume > 2 and r {1,..., 1}. If b(,r) is a Dod sequence then = 2r. Proof. It foows by Theorem 7 that b[,r] is a Dod sequence. By Lemma 19 we may assume that r, so = mr for some m > 1. By definition we have that b r [,r] = and b n [,r] = 0 for n {1,...,r(m + 1) 1} \ {r}. Let p < m be a prime number. It foows by Eq. (1) with n = pr that µ(k)b n/k [,r] = b r [,r] = 0 (mod n). k n Hence pr = mr, and p m. By Remark 17 we get that m = 2. Proof of Theorem 15: It foows by Lemmas 16 and 20. Lemma 21. The sequence b n (,r) has properties 1. b n+1 (,r) = b n (,r) b n (,r 1) for n 0, 2. b n (,r) = b n (,r +) for n 0. Proof. A direct cacuation shows that b n+1 (,r) = ω tr (1 ω t ) n+1 = ω tr (1 ω t ) n ω t(r 1) (1 ω t ) n = b n (,r) b n (,r 1). Since ω = 1 so 1 b n (,r +)) = ω t(r+) (1 ω t ) n = b n (,r). 12

13 Theorem 22. Let 2. Assume that q > 1 and k 1 are such that 1 (mod q), if r 0 (mod ); a k+1 (,r) 1 (mod q), if r 1 (mod ); 0 (mod q), if r s {2,..., 1}. Then for every r Z we have Proof. We have a n+k (,r) a n (,r) (mod q), n > 0. 1, r 0 (mod ); a 1 (,r) = 1, r 1 (mod ); 0, otherwise, so the resut hods for n = 1. Assume that it hods for some n 1. Then by inductive step and Lemma 21 we have a n+1+k (,r) = a n+k (,r) a n+k (,r 1) a n (,r) a n (,r 1) = a n+1 (,r) (mod q). Exampe 23 (Gaisher). Assume that p is prime and b 1 is such that p b 1 (mod ). Then for k = p b 1 and q = p we get a n+p b 1(,r) a n (,r) (mod p). In particuar, for = p 1 and b = 1 we get Gaisher s congruence a n+p 1 (p 1,r) a n (p 1,r) (mod p), n > 0. Exampe 24. Assume that q > 1 is an integer reativey prime to Z +. Let q = t s=1 pαs s, where p 1,...,p t are distinct primes and α s Z +. Put ν (q) = LCM ( p α (p β 1 1 1),...,p αt 1 t ) (p βt t 1), where β s is the smaest positive integer with p βs s 1 (mod m). Then we have [22] 1 (mod q), if r 0 (mod ); a 1+ν (q)(,r) 1 (mod q), if r 1 (mod ); 0 (mod q), if r s {2,..., 1}. Lemma 25. For n 0 and k Z we have 13

14 1. b 2n (,n+k) = b 2n (,n k), 2. b 2n+1 (,n+k +1) = b 2n+1 (,n k). Proof. The resut is obvious for n = 0, because b 0 (,m) = b (0, m) and {, r = k; b 0 (,r) = 0, otherwise. For n 1 we get 1 1 b 2n (,n+k) = ω t(n+k) (1 ω t ) 2n = ω t(n+k) (1 2ω t +ω 2t ) n 1 1 = ω t(n+k) (ω t ω t 2ω t +ω 2t ) n = ω tk (ω t +ω t 2) n. Since b 2n (,n+k) Z and ω t +ω t 2 R, so 1 b 2n (,n+k) = b (2n,n+k) = ω tk (ω t +ω t 2) n = b 2n (,n k). By Lemma 21 and just proved formua we get b 2n+1 (,n+k +1) = b 2n (,n+k +1) b 2n (,n+k) so the resut foows. = b 2n (,n k 1) b 2n (,n k) = (b 2n (,n k) b 2n (,n k 1)) = b 2n+1 (,n k), Coroary 26. For n 0 we have b n (,n) = ( 1) n b n (,0). Lemma 27. If is even then ( 1) r b n (,r) > 0 for n 0. Proof. If is even then ( 1) r b n (,r) = ( 1) r = s r (mod ) s r (mod ) s ( 1) s = ( 1) r s > 0. s r (mod ) ( 1) r ( n s ) Lemma 28. If is odd and n 2r 2N+1 then b n (,r) = 0. 14

15 Proof. Assume that n 2r k = r n 1 we get that 2 By Lemma 21 we have = 2k + 1. It foows that n 1 is even, so by Lemma 25 with b n 1 (,r) = b n 1 (,n 1 r). b n 1 (,r 1) = b n 1 (,n 1 r+(2r n)) = b n 1 (,n 1 r (2k +1)) = b n 1 (,n 1 r) hence by Lemma 25 we get b n (,r) = b n 1 (,r) b n 1 (,r 1) = b n 1 (,n 1 r) b n 1 (,n 1 r) = 0. We wi finish this paper with the resut obtained in [17]. Theorem 29. Assume that is odd. Then for n 1 = 0, if n 2r 2N+1; ( 1) r b n (,r) > 0, if n 2r (4k 1,4k +1); < 0, if n 2r (4k +1,4k +3). Proof. We wi use the induction with respect to n. Let n = 1. Observe that since is odd hence n 2r cannot be an odd number. By definition ( ) ( ) 1 1 ( 1) r b 1 (,r) = ( 1) s+r = ( 1) s r, s s where s {0,..., 1} is such that s r (mod ). If is odd then ( 1) s r = ( 1) r s, so ( 1) r b ( 1,r) > 0 r s [ r 2N 2N ] r Since 1 is even, so [2m,2m+1) 1 2r ( 4m 1 1, 4m+1 1 ]. hence 1 2r 1 2r / ( 4m 1 1, 4m 1+ 1 ), ( 4m 1 1, 4m+1 1 ] 1 2r ( 4m 1, 4m+1). 15

16 Simiary, ( 1) r b 1 (,r) < 0 1 2r ( 4m+1 1, 4m+3 1 ] 1 2r ( 4m+1, 4m+3), so the resut is true for n = 1. Assume that the concusion is true for n 1. We consider three cases. If n 2r = 2k +1 then it foows by Lemma 28 that b n (,r) = 0. (4k 1,4k +1). Then the numbers Let n 2r (n 1) 2(r 1) = n 2r + 1, (n 1) 2r = n 2r ie in the interva [4k 1,4k+1] and at east one of them is in the interior of [4k 1,4k+1]. It foows by the inductive step that ( 1) r 1 b n 1 (,r 1) 0, ( 1) r b n 1 (,r) 0 and at most one of them is an equaity. Hence ( 1) r b n (,r) = ( 1) r b n 1 (,r)+( 1) r 1 b n 1 (,r 1) > 0. 1 In a simiar way, for n 2r (4k +1,4k +3), we get that ( 1) r b n (,r) = ( 1) r b n 1 (,r)+( 1) r 1 b n 1 (,r 1) < 0. Coroary 30. Let 2. Then b n (,0) = 0 if and ony if is odd and n 2N+1. References [1] A. Barbuescu and D. Savin, Some congruences of Fibonacci and Lucas numbers and properties of Fibonacci functions, in N. E. Mastorakis and Z. Bojkovic, eds., Recent Researches in Appied Mathematics and Informatics, WSEAS Press, 2011, pp [2] L. Caritz, A specia congruence, Proc. Amer. Math. Soc. 4 (1953), [3] I. Chaendar and K. Keay, T. Ransford, Binomia sums, moments and invariant subspaces, Israe J. Math. 115 (2000), [4] L. E. Dickson, History of the Theory of Numbers, Vo. I, AMS Chesea Pub.,

17 [5] B.-S. Du, A simpe method which generates infinitey many congruence identities, Fibonacci Quart. 27 (1989), [6] B.-S. Du, S.-S. Huang, and M.-C. Li, Generaized Fermat, Doube Fermat and Newton Sequences J. Number Theory 98 (2003), [7] J. Franks, Homoogy and Dynamica Systems, CBMS Regiona Conference Series in Mathematics, No. 49, American Mathematica Society, [8] I. M. Gesse and T. Lengye, On the order of Stiring numbers and aternating binomia coefficient sums, Fibonacci Quart. 39 (2001), [9] F. S. Giespie, A generaization of Fermat s itte theorem, Fibonacci Quart. 27 (1989), [10] A. Granvie, Arithmetic properties of binomia coefficients. I. Binomia coefficients moduo prime powers, in Organic Mathematics, CBMS Conf. Proc., 20, American Math. Society, 1997, pp [11] J. Jezierski and W. Marzantowicz, Homotopy Methods in Topoogica Fixed and Periodic Point Theory, Springer, [12] R. Keskin and B. Demirturk, Fibonacci and Lucas congruences and its appications, Acta Math. Sin (Eng. Ser.) 27 (2011), [13] R. Keskin and Z. Yosma, On Fibonacci and Lucas numbers of the form cx 2, J. Integer Seq. 14 (2011), Artice [14] J. Mashreghi and T. Ransford, Binomia sums and functions of exponentia type, Bu. Lond. Math. Soc. 37 (2005), [15] W. Marzantowicz and K. Wójcik, Periodic segment impies infinitey many periodic soutions, Proc. Amer. Math. Soc. 135 (2007), [16] H. Prodinger, Some information about binomia transform, Fibonacci Quart. 32 (1994), [17] L. Pieni ażek and K. Wójcik, Compicated dynamics in non-autonomous ODE s, Univ. Iag. Acta Math. 41 (2003), [18] A. Rotkiewicz, Probems on Fibonacci numbers and their generaizations, in G. E. Bergum, A. N. Phiippou, and A. F. Horadam, eds., Fibonacci Numbers and Their Appications, D. Reide Pubishing Company, 1986, pp [19] R. Srzednicki and K. Wójcik, A geometric method for detecting chaotic dynamics, J. Differentia Equations 135 (1997),

18 [20] R. Srzednicki, K. Wójcik, and P. Zgiczyński, Fixed point resuts based on Ważewski method, in R. Brown, M. Furi, L. Górniewicz, and B. Jiang, eds., Handbook of Topoogica Fixed Point Theory, Kuwer, 2004, pp [21] Zhi-Wei Sun, On the sum k r (mod m)( n k) and reated congruences, Israe J. Math. 128 (2002), [22] Zhi-Wei Sun and R. Tauraso, Congruences for sums of binomia coefficients, J. Number Theory 126 (2007), [23] C. S. Weisman, Some congruences for binomia coefficients, Michigan Math. J. 24 (1977), [24] H. S. Wif, Generatingfunctionoogy, 2nd edition, Academic Press, Mathematics Subject Cassification: Primary 05A10; Secondary 11A07. Keywords: binomia transform, trace, Lefschetz number, integer matrix, congruence. Received September ; revised versions received September ; December Pubished in Journa of Integer Sequences, December Return to Journa of Integer Sequences home page. 18