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Some Periodicities in the Continued Fraction Expansions of Fibonacci and Lucas Dirichlet Series

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Some Periodicities in the Continued Fraction Expansions of Fibonacci and Lucas Dirichlet Series SOME PERIODICITIES IN THE CONTINUED FRACTION EXPANSIONS OF FIBONACCI AND LUCAS DIRICHLET SERIES CHRIS K. CALDWELL AND TAKAO KOMATSU P1 ¡s Abstract. In this paper we consider the Fibonacci Zeta functions ³F (s) = n=1 Fn and P1 ¡s the Lucas Zeta functions ³L(s) = n=0 Ln . The sequences fAº gº¸0 and fBº gº¸0, which Pn ¡s are derived from º=1 Fº = An=Bn, satisfy certain recurrence formulas. We examine some properties of the periodicities of An and Bn. For example, let m and k be positive m integers. If n ¸ mk, then Bn ´¥ 0¦ (mod Fk¥) (with¦ a similar result holding for An). The n P1 n power of 2 which divides Bn is 6 + i=0 3¢2i . 1. Introduction Consider the so-called Fibonacci and Lucas Zeta functions: X1 1 X1 1 ³ (s) = ; ³ (s) = : F F s L Ls n=1 n n=0 n In [13] the analytic continuation of these series is discussed. In [4] it is shown that the numbers ³F (2); ³F (4); ³F (6) ( respectively, ³L(2); ³L(4); ³L(6)) are algebraically independent, and that each of ³F (2s) ( respectively, ³L(2s)) (s = 4; 5; 6;::: ) may be written as a rational (respectively, algebraic) function of these three numbers over Q, e.g. ³ 15 1 6 5 4 3 ³F (8) ¡ ³F (4) = 256u ¡ 3456u + 2880u + 1792u v 14 378(4u + 5)2 ´ ¡ 11100u3 + 20160u2v ¡ 10125u2 + 7560uv + 3136v2 ¡ 1050v ; where u = ³F (2) and v = ³F (6). Similar results are obtained in [4] for the alternating sums à ! X1 (¡1)n+1 X1 (¡1)n+1 ³¤ (2s) := respectively, ³¤ (2s) := (s = 1; 2; 3;::: ): F F 2s L L2s n=1 n n=1 n From the main theorem in [5] it follows that for any positive distinct integers s1; s2; s3 the numbers ³F (2s1), ³F (2s2), and ³F (2s3) are algebraically independent if and only if at least one of s1; s2; s3 is even. Other types of algebraic independence, including the functions The second author was supported in part by the Grant-in-Aid for Scienti¯c Research (C) (No. 18540006), the Japan Society for the Promotion of Science. FEBRUARY 2010 47 THE FIBONACCI QUARTERLY X1 1 X1 1 X1 1 X1 1 ; ; ; F s F s Ls Ls n=1 2n¡1 n=1 2n n=1 2n¡1 n=1 2n are discussed in [9]. On the other hand, in [8] Fibonacci zeta functions and Lucas zeta functions including ¤ ¤ ³F (1); ³F (2); ³F (3); ³F (1); ³L(1); ³L(2); ³L(1) are expanded as non-regular continued fractions whose components are Fibonacci or Lucas numbers. For example, [8, Theorem 1] says Lemma 1.1. We have 1 ³ (1) = F F 2 F ¡ 1 2 F 2 F ¡ 2 3 F 2 F ¡ 3 4 . .. F 2 ¡ n¡1 Fn+1 ¡ ::: and Xn 1 A = n ; F B º=1 º n where fAºgº¸0 and fBºgº¸0 are determined by the recurrence formulas: 2 Aº = Fº+1Aº¡1 ¡ Fº¡1Aº¡2 (º ¸ 2);A0 = 0;A1 = 1; 2 Bº = Fº+1Bº¡1 ¡ Fº¡1Bº¡2 (º ¸ 2);B0 = 1;B1 = 1: Similar continued fraction expansions with corresponding recurrence relations hold for ¤ ¤ ³F (2), ³F (3), ³F (1), ³L(1), ³L(2), ³L(1) and related Fibonacci and Lucas Dirichlet series [8, Table 1]. In [8, Theorem 5.1] the periodicity of the sequences fAngn¸0 and fBngn¸0 modulo t for any integer t ¸ 2 is considered using a result recently obtained in [7]. Lemma 1.2. Let t ¸ 2 be any integer, and let fYngn¸0 be a sequence of integers satisfying the recurrence relation Yº = T (º)Yº¡1 + U(º)Yº¡2 (º ¸ 2) with sequences fT (º)gº¸2 and fU(º)gº¸2 of integers, which are periodic modulo t. Then the sequence fYngn¸0 is ultimately periodic modulo t. If U(º) = 1 for all º ¸ 2, then the sequence fYngn¸0 is periodic modulo t. By applying this lemma to the recurrence formulas for An and Bn in [8, Table 1], the following result is obtained in [8, Theorem 5.2]. Lemma 1.3. For any integer t ¸ 2, the sequences (An)n¸0 and (Bn)n¸0 are ultimately periodic modulo t. However, the exact period has not been known. In this paper we discuss the details about periodicity. For example, let m and k be positive integers. If n ¸ mk, then Bn ´ 0 m ¥(mod¦ Fk ) (with¥ ¦ a similar result holding for An). The power of 2 which divides Bn is n P1 n 6 + i=0 3¢2i . 48 VOLUME 48, NUMBER 1 FRACTION EXPANSIONS OF FIBONACCI AND LUCAS DIRICHLET SERIES 2. Fibonacci-type Zeta Functions Consider Fibonacci-type numbers fGngn¸1 de¯ned by Gn = Gn¡1 + Gn¡2 (n ¸ 2) with positive integral initial values G1 and G2. Let X1 1 X1 (¡1)n¡1 ³ (s) = ; ³¤ (s) = : G Gs G Gs n=1 n n=1 n ¤ Continued fraction expansions of ³G(s) and ³G(s) are obtained in [8, Lemma 2, Lemma 3]. Namely, 1 ³G(s) = 2s s G1 G1 ¡ 2s s s G2 G1 + G2 ¡ 2s s s G3 G2 + G3 ¡ Gs + Gs ¡ . G2s 3 4 .. n¡1 ¡ s s Gn¡1 + Gn ¡ ¢ ¢ ¢ and ¤ 1 ³G(s) = 2s s G1 G1 + 2s s s G2 ¡G1 + G2 + 2s s s G3 ¡G2 + G3 + ¡Gs + Gs + . G2s 3 4 .. n¡1 + s s : ¡Gn¡1 + Gn + ¢ ¢ ¢ Now An (respectively Bn) are de¯ned as the numerator (respectively denominator) con- vergent of the continued fraction expansion given for ³G(s): A 1 n = B G2s n Gs ¡ 1 1 G2s Gs + Gs ¡ 2 1 2 G2s Gs + Gs ¡ 3 2 3 Gs + Gs ¡ G2s 3 4 ... n¡1 ¡ s s : Gn¡1 + Gn Hence, fAºgº¸0 and fBºgº¸0 satisfy the following recurrence formulas. s s 2s Aº = (Gº¡1 + Gº)Aº¡1 ¡ Gº¡1Aº¡2 (º ¸ 2);A0 = 0;A1 = 1; s s 2s s Bº = (Gº¡1 + Gº)Bº¡1 ¡ Gº¡1Bº¡2 (º ¸ 2);B0 = 1;B1 = G1: In fact, Aº and Bº can be expressed explicitly as follows. Lemma 2.1. For n = 1; 2;::: Xn 1 A = (G G :::G )s ;B = (G G :::G )s: n 1 2 n Gs n 1 2 n º=1 º FEBRUARY 2010 49 THE FIBONACCI QUARTERLY s Proof. By induction we have Bn = (G1G2 :::Gn) . Thus, Xn 1 Xn 1 A = B = (G G :::G )s : n n Gs 1 2 n Gs º=1 º º=1 º ¤ ¤ ¤ Similarly, if An (respectively Bn) are de¯ned as the numerator (respectively denominator) ¤ ¤ ¤ convergent of the continued fraction expansion given for ³G(s), then fAºgº¸0 and fBº gº¸0 satisfy the following recurrence formulas. ¤ s s ¤ 2s ¤ ¤ ¤ Aº = (¡Gº¡1 + Gº)Aº¡1 + Gº¡1Aº¡2 (º ¸ 2);A0 = 0;A1 = 1; ¤ s s ¤ 2s ¤ ¤ ¤ s Bº = (¡Gº¡1 + Gº)Bº¡1 + Gº¡1Bº¡2 (º ¸ 2);B0 = 1;B1 = G1: Similar to Lemma 2.1, we have the following. Lemma 2.2. For n = 1; 2;::: Xn (¡1)º¡1 A¤ = (G G :::G )s ;B¤ = (G G :::G )s: n 1 2 n Gs n 1 2 n º=1 º Some reciprocal sums of consecutive Fibonacci or Lucas numbers have been studied (e.g. s s [1, 14]). For example, the reciprocal sum of GnGn+1 has the following continued fraction expansion. Corollary 2.3. X1 1 1 = Gs Gs G2sGs n=1 n n+1 GsGs ¡ 1 2 1 2 GsGs Gs + Gs ¡ 2 3 1 3 GsGs Gs + Gs ¡ 3 4 2 4 Gs + Gs ¡ . Gs Gs 3 5 .. n¡1 n ¡ s s ; Gn¡1 + Gn+1 ¡ ::: 50 VOLUME 48, NUMBER 1 FRACTION EXPANSIONS OF FIBONACCI AND LUCAS DIRICHLET SERIES Proof. By [8, Lemma 2.1], we have X1 1 Gs Gs n=1 n n+1 ¡s ¡s G1 G2 = ¡s ¡s G2 G3 1 ¡ ¡s ¡s ¡s ¡s ¡s ¡s ¡s ¡s G1 G2 G3 G4 G1 G2 + G2 G3 ¡ ¡s ¡s ¡s ¡s ¡s ¡s ¡s ¡s G2 G3 G4 G5 G2 G3 + G3 G4 ¡ ¡s ¡s ¡s ¡s G3 G4 + G4 G5 ¡ ::: 1 = s ¡s s s G1G3 G1G2 ¡ ¡s ¡s ¡s ¡s ¡s ¡s ¡s ¡s G1 G2 G3 G4 G1 G2 + G2 G3 ¡ ¡s ¡s ¡s ¡s ¡s ¡s ¡s ¡s G2 G3 G4 G5 G2 G3 + G3 G4 ¡ ¡s ¡s ¡s ¡s G3 G4 + G4 G5 ¡ ::: 1 = 2s s s s G1 G2 G1G2 ¡ ¡s s s G4 G3 + G1 ¡ ¡s ¡s ¡s ¡s ¡s ¡s ¡s ¡s G2 G3 G4 G5 G2 G3 + G3 G4 ¡ ¡s ¡s ¡s ¡s G3 G4 + G4 G5 ¡ ::: 1 = 2s s s s G1 G2 G1G2 ¡ s s s s G2G3 G1 + G3 ¡ ¡s s s G5 G4 + G2 ¡ ¡s ¡s ¡s ¡s : G3 G4 + G4 G5 ¡ ::: ¤ 3. Fibonacci Zeta Functions Let G1 = G2 = 1. Then Gn = Fn are reduced to Fibonacci numbers. Consider the continued fraction expansion 1 ³F (s) = 2s s F1 F1 ¡ 2s s s F2 F1 + F2 ¡ 2s s s F3 F2 + F3 ¡ .. 2s . Fn¡1 ¡ s s Fn¡1 + Fn ¡ ::: and Xn 1 A = n ; F s B º=1 º n FEBRUARY 2010 51 THE FIBONACCI QUARTERLY where fAºgº¸0 and fBºgº¸0 are determined by the recurrence formulas: s s 2s Aº = (Fº¡1 + Fº )Aº¡1 ¡ Fº¡1Aº¡2 (º ¸ 2);A0 = 0;A1 = 1; s s 2s s Bº = (Fº¡1 + Fº )Bº¡1 ¡ Fº¡1Bº¡2 (º ¸ 2);B0 = 1;B1 = F1 : Theorem 3.1.
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  • Dirichlet's Theorem on Arithmetic Progressions
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