FIBONACCI NUMBERS AND SEMISIMPLE CONTINUED FRACTION. and Ln

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1 Commun Korean Math Soc 29 (204), No 3, pp FIBONACCI NUMBERS AND SEMISIMPLE CONTINUED FRACTION Eunmi Choi Abstract The ratios of any two Fibonacci numbers are expressed by means of semisimple continued fraction Introduction The Fibonacci sequence { } n is a series of numbers that begins with F F 2 and each next is the sum of the previous two terms The Lucas sequence {L n } n is a modified Fibonacci sequence starting from L and L 2 3 A number of properties of these sequences were studied by many F researchers Among them, the ratios n and Ln L n of two successive terms of each sequence were investigated by means of simple continued fraction ([2], [3], [4] and [6]) This work is devoted to studying the ratio Fn F k for any n and k For the purpose, a semisimple continued fraction will be defined and compared to a simple continued fractions We shall show that Fn F k is expressed by a semisimple continued fraction more efficiently than by a simple continued fraction And it will be seen that semisimple continued fraction may yield any large Fibonacci numbers, like F 05 a 22 digit number 2 Semisimple continued fractions We begin with a lemma that provides a motivation of this work Lemma 2 ([5]) Let k, t and r be positive integers If t 2 and r k, then we have the following relations () L kt+r L k L k(t )+r + ( ) k L k(t 2)+r, so L n (n kt + r) is expressed by only three Lucas numbers L k, L r and L k+r (2) F kt+r L k F k(t )+r + ( ) k F k(t 2)+r, so (n kt + r) is expressed by L k, F r and F k+r Received February, 204; Revised June 9, Mathematics Subject Classification Primary B37, B39 Key words and phrases Fibonacci sequence, Lucas sequence, continued fraction 387 c 204 Korean Mathematical Society

2 388 E CHOI Note that r may be considered as the remainder of n when divided by k, but we assume r k because the sequences start from F L If k 5, then L 7 L 5 L 2 +L 7 and L 2 L 5 L 7 +L 2, so and again L 7 L 2 L 5L 2 +L 7 L 2 L 5 + L 2 L 7 L 5 + L 5L 7+L 2 L 5 + L 7 L 22 L 5L 7 +L 2 L 5 + L 7 L L 7 L L 2 L 5 + L 5+ L 7 L 2 L 5 + L 7 L 2 These fractions yield a motivation to define a sort of continued fraction composed of only Lucas numbers Let n 2 For real numbers a 0,b 0 0 and a i > 0,b i > (i,,n), the fractions a 0 + a + an 2 + and b 0 b bn 2 b n b n are called (minus) semisimple continued fractions denote by a 0 ;a,, the former and [[b 0 ;b,,b n ]] the latter respectively We also define a 0 ;a [[a 0 ;a ]] a 0 a When every a i are Lucas [resp Fibonacci] numbers, the semisimple continued fraction is called Lucas [resp Fibonacci] continued fraction For instance, L22 L 7 equals the Lucas continued fraction L 5 ;L 5,L 5,L 7,L 2 This providesagood reason to define the semisimple continued fraction We may compare these fractions to the (minus) simple continued fractions a 0 + a + + and b 0, b b n b n where a 0,b 0 0 and a i > 0, b i > (i,,n), denoted by a 0 ;a,, and [b 0 ;b,,b n ], respectively (refer to []) Theorem 22 Let n 2 For the (minus) semisimple continued fractions, () a 0 ;, a 0 + 0;a,, a 0 + a ;,

3 (2) Proof Since FIBONACCI NUMBERS AND SEMISIMPLE CONTINUED FRACTION 389 [[a 0 ;, ]] a 0 +[[0;a,, ]] a 0 [[a ;, ]] [[0;a 0,, ]] 0;a 0,, a 0 ;, 3, 2 +, 2 a 0 ;, 2, 2 + a 0 ;, 2,,, a 0 ;, 2,, a 0 + a 0 + a 0 + an an an an x with x an 2an 2 +, () follows immediately Now for (2), write q +r with 0 r < and q Z Then a;, a+ a+ q+ r a;q,,r a;,, Theorem 23 We further have the following identities () a 0 ;,, a 0 ;,, [[a 0 ;,,]] [a 0 ;, ] for n (2) a 0 ;, a 0 ;, 2, an a 0 ;, 3, 2 + an for n 3 (3) a 0 ;a,, a 0 ;a,,a k, a k+ ;, for 0 k n 2

4 390 E CHOI Proof The proof is not hard Example ;2,3,4,5,6 ;2,3,26,5 ;2,83,26 ;92, by Theorems 22 and 23 On the other hand, ;2,3,4,5,6 is also equal to ;2,3,4, ;2,3, 83 ;2, 96 ; Similar to the usual notation a t,b a;,a,b, we denote the t times }{{} t repeated semisimple continued fractions a;,a,b and [[a;,a,b]] by a t,b and [[a t,b]], respectively 3 Convergents of a semisimple continued fraction The section is devoted to investigating successive convergents of a semisimple continued fraction, and to studying relationships with those of a simple continued fraction Given a continued fraction a 0 ;a,,, umerator u k and a denominator v k of the kth convergent C k are given by the recursive formulas u k u k a k +u k 2 and v k v k a k +v k 2 (k ), where u, u 0 a 0 and v 0, v 0 Then (refer to []) C k a 0 ;a,,a k u k v k and u k v k u k v k ( ) k+ The next theorem shows a recursion formula involving the semisimple continued fraction For every a k > 0, let S n a 0 ;a,, Theorem 3 Let n 2 Then for all 2 k n, the following are equivalent for the kth convergent S k of S n () S k a 0 ;a,,a k p k q k (2) p k [ u k 2 ][ a k +u] k 3 a[ k and q k ][ v k 2 a k +v k 3 a k ] a0 a ak 2 ak a (3) k [ 0 ][ 0 0 a ] k 2 [ a k 2 a k ] uk 2 u k 3 ak a k pk p k v k 2 v k 3 a k a k 2 a k 2 a k q k q k Proof If k 2, then S 2 a 0 ;a,a 2 a 0a +a 2 p 2 a q 2 impliesp 2 a 0 a +a 2 u 0 a +u a 2 andq 2 a v 0 a +v a 2 Furthermore [ ][ ] [ ][ ] a0 a a u0 u a a 0 a 2 a 0 a 0 a v 0 v a 2 a 0 a 0 a [ ] a0 a +a 2 a 0 a a [ p2 p q 2 q ]

5 FIBONACCI NUMBERS AND SEMISIMPLE CONTINUED FRACTION 39 On the other hand, Theorem 23(2) together with the consideration about the continued fraction give rise to S 2 a 0 ;a,a 2 a 0 ; a u 0 a 2 a a 2 +u a v 0 a 2 +v (a 0a +a 2 )/a 2 a /a 2 a 0a +a 2 a p 2 q 2 We now suppose that (), (2) and (3) hold for S k for k < n Then S k a 0 ;a,,a k,a k a 0 ;a,,a k 2, a k a k u k u a k k 2 a k +u k 3 v k a v k k 2 a k +v k 3 u k 2a k +u k 3 a k v k 2 a k +v k 3 a k p k q k Moreover it follows inductively that [ ] [ ][ ][ ] a0 ak 3 ak 2 ak a k a k a k 2 a k 2 a k [ ][ ] uk 3 a k 2 +u k 4 u k 3 ak a k v k 3 a k 2 +v k 4 v k 3 a k a k 2 a k 2 a k [ ][ ] [ ] uk 2 u k 3 ak a k pk p k v k 2 v k 3 a k a k 2 a k 2 a k q k q k if and only if p k u k 2 a k +u k 3 a k and q k v k 2 a k +v k 3 k n Example Consider p5 q 5 ;2,3,4,5,6 Since ;2,3 0 7 u2 ;2,3, u3 v 3, [ p5 q 5 ] [ ][5 6 ] [ ] so ;2,3,4,5, v 2 and From now on, we denote the semisimple continued fraction a 0 ;a,, by S n and the continued fraction a 0 ;a,, by C n where 0 < a i Z Then for every k < n, S k p k q k and C k u k v k are the kth convergents of S n and C n respectively, where u k,v k,p k and q k are in Theorem 3 Lemma 32 C k C k ( )k+ v k v k and C k C k 2 ( )k+ v k 2 v k a k for k Proof It is clear to see that C k C k u k v k u k v k u kv k u k v k v k v k Furthermore, since v k v k a k +v k 2 we have ( )k+ v k v k C k C k 2 ( )k+ v k v k + ( )k v k v k 2 ( )k v k a k v k 2 v k v k ( )k a k v k 2 v k

6 392 E CHOI It shows C 0 < C 2 < C 4 < < C 5 < C 3 < C and {C k } convergesto C n for large enough n ([]) Similarly, we shall investigate the differences S k S k and S k S k 2 Theorem 33 Let δ k a k a k a k +a k+ Then for k, () S k S k ( )k q k q k a k δ k (2) S k S k 2 ( )k q k 2 q k (a k δ k a k 2 δ k 2 δ k 2 δ k ) Proof By Theorem 3, we have [ ][ ] [ ] uk 2 u k 3 ak a k pk p k v k 2 v k 3 a k a k 2 a k 2 a k q k q k Then the determinants of both sides yield p k q k p k q k ( ) k a k (a k 2 a k a k 2 +a k ) Now by setting δ k a k 2 a k a k 2 +a k, we have Again S k S k p k q k p k q k p kq k p k q k q k q k S k S k 2 ( )k a k δ k q k q k + ( )k a k 2 δ k 2 q k 2 q k ( )k a k δ k q k q k ( )k q k 2 q k q k (a k δ k q k 2 a k 2 δ k 2 q k ) Fromthe identitiesv k v k a k +v k 2 andq k v k 2 a k +v k 3 a k intheorem 3, we have q k (v k 3 a k 2 +v k 4 )a k +v k 3 a k Thus v k 3 (a k 2 a k +a k )+v k 4 a k v k 3 (a k 2 a k +a k )+(q k v k 3 a k 2 ) v k 3 (a k 2 a k a k 2 +a k )+q k v k 3 δ k +q k so a k δ k q k 2 a k 2 δ k 2 q k q k (a k δ k a k 2 δ k 2 ) δ k 2 δ k (v k 3 a k 2 +v k 4 a k ) q k (a k δ k a k 2 δ k 2 ) δ k 2 δ k q k q k (a k δ k a k 2 δ k 2 δ k 2 δ k ), S k S k 2 ( )k q k 2 q k q k q k (a k δ k a k 2 δ k 2 δ k 2 δ k ) ( )k q k 2 q k (a k δ k a k 2 δ k 2 δ k 2 δ k )

7 FIBONACCI NUMBERS AND SEMISIMPLE CONTINUED FRACTION 393 Theorem 34 S 3 < S 5 < S 7 < < S 6 < S 4 < S 2, so {S k } converges to S n for large enough n Proof Obviously q k > 0 and δ k a k (a k )+a k+ > 0 for every k Thus S k S k ( )k q k q k a k δ k shows S k < S k if k is even, otherwise S k < S k Furthermore we have a k δ k a k 2 δ k 2 δ k 2 δ k a k 2 a k a k 3 a 2 k 2 a k +a k 3 a k 2 a k a k 3 a k 2 a k +a k 3 a k a k 2 a k (+a k 3 (a k 2 )) a k 3 a k (a k 2 ) (a k 3 (a k 2 )(a k 2 a k +a k )+a k 2 a k ) < 0 Thus from S k S k 2 ( )k+ q k q k 2 (a k 3 (a k 2 )(a k 2 a k +a k )+a k 2 a k ), we have S k < S k 2 for even k, otherwise S k 2 < S k Therefore so S k converges when k gets larger S 3 < S 5 < S 7 < < S 6 < S 4 < S 2, For example, from π a 0 ;a,a 2, 3;7,5,,292,,,,2,,3,,4,, the first few convergents of continued fraction are Moreover C u 0 v 0, C 3; u v C 2 3;7,5 u a 2 +u v a 2 +v u 2, v 2 and similarly C 3 3;7,5, u 3 v 3, C , C , Let us consider S a 0 ;a, 3;7,5,,292,,,,2, Then S 3;7 3 7 p and S 2 3;7,5 36 q 7 p 2 q 2 And Theorem 3 shows that S 3 3;7,5, p 3 u a 2 +u 0 a q 3 v a 2 +v 0 a and S , S , Now the differences of convergents of C n are, for instance C 4 C 3 (3)(3302), C 5 C 4 (3302)(3325) and C C 2 (06)(3302)

8 394 E CHOI Moreover Theorem 33 shows and S 6 S 4 S 5 S 4 292(292+ ) ((292+)(292 )+292) Corollary 35 Let S n a 0 ;a,, pn q n Assume a k for k < n Then S k+ S k ( )k+ a k+ q k q k+ and S k+2 S k ( )k+ a k+ q k q k+2 Proof From Theorem 33, we have S k+ S k ( )k+ q k q k+ (a k (a k (a k )+a k+ )) ( )k+ a k+ q k q k+ Similarly, S k+2 S k ( )k+ q k q k+2 (a k (a k )(a k a k+ +a k+2 )+a k a k+ ) ( )k+ a k+ q k q k+2 From S 3;7,5,,292,,, in the above example, since a 3 and a 4 292, we have S 4 S q 3q 4 and S 5 S q 3q 5 4 Ratios of Fibonacci numbers in semisimple continued fraction It is well known that the ratio Fn+ is equal to n ;,, This section is devoted to study Fn F k for any n and k The fractions, [ ], or [[ ]] will be put together to express the ratios effectively Lemma 4 Let n 3 Then and L 2 2 Proof Obviously ( 2 ) 2 2, while + + ( + 2 ) And (2 2 )+ 3 2 Moreover due to Lemma 2, +k L k + ( ) k k, so we have +2 L 2 2 if k 2 Theorem 42 Let n 3 and { write n 2t+r with r 2 () +2 2; n [[3 t+,]] [3 t+ ] if r 2, [[3 t,2,]] [3 t,2] if r (2) + n { [[2;3 t,]] if r 2, [[2;3 t,2,]] if r

9 FIBONACCI NUMBERS AND SEMISIMPLE CONTINUED FRACTION 395 Proof Due to Lemma 4, it is obvious that n 2;n, and (v ) 2v [[3;,3,F }{{} n 2(v ), 2v ]] v for v t Therefore, for n 2t+r ( r 2), Theorem 23 yields +2 [[3 t,f 2+r,F r ]] { [[3 t,f 4,F 2 ]] [[3 t,3,]] [3 t+ ] if r 2, [[3 t,f 3,F ]] [[3 t,2,]] [3 t,2] if r On the other hand, since 2 2 +, it follows from () that { 2 [[3 t,]] [[2;3t,]] if r 2, 2 [[3 t,2,]] [[2;3t,2,]] if r It shows that if n 2t + r (r,2), then Fn+2 2; n equals either [[3 t+,]]or [[3 t,2,]], and Fn+ n equals either[[2;3 t,]]or[[2;3 t,2,]] This means that while n repeated computations are needed in the simple continued fraction, only about t n 2 repeated computations are required in the semisimple continued fractions For example, if n 6, then [[2;3,3]] F7 F 6 ;,,,, Hence if n is large, a semisimple continued fraction is more convenient than a simple continued fraction Theorem 43 Let n 4 and write n 3t+r with r 3 Then Fn+3 L t 3,F 3+r,F r, which is equal to one of 4 t,3, 4 t,5 or 4 t,4 according to r,2,3 Proof Lemma 2 with k 3 implies +3 L 3 + 3, thus +3 L (t ) 3t

10 396 E CHOI L t 3, 3(t ), 3t 4 t,f 3+r,F r Hence by making use of Theorem 23, it follows that F 4 t,3, 4 t,3 if r, n+3 4 t,5, 4 t,5 if r 2, 4 t,8,2 4 t,4 if r 3 Now we generalize Theorem 43 as follows Theorem 44 Let k 3 Let n k+ and write n kt+r with r k Then +k equals L t k F,F k+r,f r if k is odd, and [[L t k,f k+r,f r ]] if k is even n So F L t k,f k+ or [L t k,f k+] if r, n+k L t k,f k+2 or [L t k,f k+2] if r 2, L t+ k or [L t+ k ] if r k Proof Let k be odd Since +k L k + k in Lemma 2, we have +k L k + L t k,f k+r,f r L k + Lk + F k+r F r If r or 2, then +k equals L t k,f k+, L t k,f k+ or L t k,f k+2, L t k,f k+2 If r k, then since F 2k F k L k, it follows from Theorem 23 that +k L t k F,F 2k,F k L t k,f kl k,f k L t k,l k L t+ k n On the other hand, if k is even, then +k L k k thus +k L k L F k [[L n t k,f k+r,f r ]], L F k n k k 2k and the rest follows similarly In particular, in case of n kt+r with r 3 k, the next theorem follows Theorem 45 With the same notations as in Theorem 44, let r 3 and write k 3u+v with v 3 Set P 0 and P u 4P u +P u 2 If k is odd, then F L t k, 3P u ;2 with P 4;3, if v, n+k L t F k, 5P u ;2 with P 4;5, if v 2, n L t k, 4P u ;2 with P 4;4, if v 3

11 FIBONACCI NUMBERS AND SEMISIMPLE CONTINUED FRACTION 397 If k is even, then +k is either [L t k, 3P u;2 ],[L t k, 5P u;2 ] or [L t k, 4P u;2 ] Proof Assume k is odd Theorem 44 together with Theorem 23 yields +k L t k,f k+r,f r L t k, F k+3;f 3 Since 3 k, if we write k 3u+v with v 3, then we have F k+3 F 3(u+)+v L u F k F 3,F v+3,f v 4 u,f v+3,f v 4 u,λ, 3u+v where λ 3,5 or 4 according to v,2 or 3 by Theorem 43 Let P 0 and P 4;λ Then we see that 4 2,λ 4+ 4;λ 4P + P P 2 P, where P 2 4P +P 0, 4 3,λ ,λ 4P 2 +P P 2 P 3 P 2, where P 3 4P 2 +P, 4 4,λ ,λ 4P 3 +P 2 P 3 P 4 P 3, where P 4 4P 3 +P 2 Thus we have, for any 0 i u, F 3(i+)+v F 3i+v 4 i,λ P i P i, where P i 4P i +P i 2 with P 0 and P 4;λ It then follows that F k+3 ;F 3 F 3(u+)+v F 3 F 3(u+)+v F 3u+v F 3u+v F 3(u )+v F 3 2+v F 3+v F 3+v F 3 4 u,λ 4 u,λ 4 u 2,λ 4;λ F v+3 ;F 3 P u P u P2 P P u P u 2 P F v+3;f 3 P u F v+3 ;F 3 If v, then λ 3 and F k+3 F k 4 u,3 so F k+3 ;F 3 P u F 4 ;F 3 P u 3;2 3P u ;2, with P 4;3 and P i 4P i +P i 2 Thus +k L t k, 3P u;2 If v 2, then λ 5 and F k+3 F k 4 u,5 so +k L t k, 5P u;2 with P 4;5 Similarly if v 3, then λ 4 so +k L t k, 4P u;2 with P 4;4 On the other hand, assume that k is even Then Theorem 44 implies +k [[L t F k,f k+r,f r ]], n

12 398 E CHOI which is equal to [L t k,[[f k+3;f 3 ]]] [L t k, F k+3;f 3 ] Hence for k 3u + v with v,2,3, the above calculations yield 3P u ;2 with P 4;3, if v, F k+3,f 3 P u F v+3,f 3 5P u ;2 with P 4;5, if v 2, 4P u ;2 with P 4;4, if v 3 Example In order to find F 05 and F 88, let n 88 and k 7 Then F 05 F L 5 F 88 F 7,F 20,F 3 L 5 7, F 20 ;F Now for F 20 ;F 3, since 7 3u+v with u 5, v 2, F 20 F 7 F F L 5 3,F 5,F 2 4 5,5 Thus by letting P 0, P 4;5 2 5 and P i 4P i +P i 2 (i 5), we get P , P , P and P Hence Therefore F 20 ;F 3 P 5 F 5 ;F ; F 05 F 88 L 5 7, which is 3,928,43,764,606,87,65,730,00,087,778,366,0,93 Here the numerator is F 05 and the denominator is F 88, which are 22 digit and 9 digit numbers respectively Corollary 46 Let n 6 and write n 5t+r with r 5 Then we have L t 5,8, t,8 if r, F L t 5,3, t,3 if r 2, n+5 L t F 5,F 5+r,F r L t 5,2,2 t,0,2 if r 3, n L t 5,34,3 t+,3 if r 4, L t 5,55,5 t+ if r 5 Proof This is mainly due to Theorem 44 References [] D Burton, Elementary Number Theory, 3rd ed WCB, Oxford, England, 994 [2] M Drmota, Fibonacci numbers and continued fraction expansions, in G E Vergum et al (eds) Applications of Fibonacci numbers, vol 5, 2nd ed Kluwer Academic Publishers, Netherlands, 993 [3] S Kalia, Fibonacci numbers and continued fractions, MIT PRIMES, 20 (retrived from webmitedu/primes/materials/20/20-conf-bookletpdf),

13 FIBONACCI NUMBERS AND SEMISIMPLE CONTINUED FRACTION 399 [4] S Katok, Continued fractions, hyperbolic geometry, quadratic froms, in MASS Selecta, Teaching and learning advanced undergraduate mathematics (S Katok, A Sossinsky, S Tabachnikov eds) American Math Soc 2003 [5] F Koken and D Bozkurt, On Lucas numbers by the matrix method, Hacet J Math Stat 39 (200), no 4, [6] T E Phipps, Fibonacci and continued fractions, Aperion 5 (2008), no 4, Department of Mathematics Hannam University Daejeon , Korea address: emc@hnukr