FIBONACCI NUMBERS AND SEMISIMPLE CONTINUED FRACTION. and Ln

Commun Korean Math Soc 29 (204), No 3, pp 387 399 http://dxdoiorg/0434/ckms204293387 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, 204 200 Mathematics Subject Classification Primary B37, B39 Key words and phrases Fibonacci sequence, Lucas sequence, continued fraction 387 c 204 Korean Mathematical Society

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 5 + 7 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 ;,

(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 3 + 2 + an an 3 + 2 + an 3 + 2 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

390 E CHOI Proof The proof is not hard Example ;2,3,4,5,6 ;2,3,26,5 ;2,83,26 ;92,83 275 92 by Theorems 22 and 23 On the other hand, ;2,3,4,5,6 is also equal to ;2,3,4, 0 78 83 92 275 3 ;2,3, 83 ;2, 96 ; 275 92 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 ]

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 0 0 0 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 ;2,3,4 43 30 u3 v 3, [ p5 q 5 ] [ 40 0 30 7 ][5 6 ] [275 92 ] so ;2,3,4,5,6 275 92 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

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 )

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 0 3 3 u 0 v 0, C 3;7 22 7 u v C 2 3;7,5 u a 2 +u 0 22 5+3 v a 2 +v 0 7 5+ 333 06 u 2, v 2 and similarly C 3 3;7,5, 355 3 u 3 v 3, C 4 03993 3302, C 5 04348 3325, 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 3 22 5+3 q 3 v a 2 +v 0 a 3 7 5+ 333 06 and S 4 333 +22 292 06 +7 292 6757 250, S 5 355 292+333 3 292+06 03993 3302, 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 292 4 C 2 (06)(3302)

394 E CHOI Moreover Theorem 33 shows and S 6 S 4 S 5 S 4 292(292+ ) 250 3302 2922 769300 85555 ((292+)(292 )+292) 250 3325 742250 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 3 292 q 3q 4 and S 5 S 3 292 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 + 2 2 2 + 2 and +2 3 2 L 2 2 Proof Obviously + + + ( 2 ) 2 2, while + + ( + 2 ) + 2 + 2 And +2 + + (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

FIBONACCI NUMBERS AND SEMISIMPLE CONTINUED FRACTION 395 Proof Due to Lemma 4, it is obvious that +2 2 + 2+ 2+ n 2;n, and +2 3 2 3 2 3 3 3 3 2(v ) 2v 3 2 4 2 3 [[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 3 2 4 On the other hand, since 2 2 +, it follows from () that + 2 2 2 2 { 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 3 + 3 4+ 4+ 4+ 3 6 4+ 4+ 3(t ) 3t

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

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 4+ 4 2,λ 4P 2 +P P 2 P 3 P 2, where P 3 4P 2 +P, 4 4,λ 4+ 4 3,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

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 7 6+3 L 5 F 88 F 7,F 20,F 3 L 5 7, F 20 ;F 3 7 5+3 Now for F 20 ;F 3, since 7 3u+v with u 5, v 2, F 20 F 7 F 3 6+2 F 3 5+2 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 2 89 5, P 3 377 5, P 4 597 5 and P 5 6765 5 Hence Therefore F 20 ;F 3 P 5 F 5 ;F 3 6765;2 6765 2 F 05 F 88 L 5 7, 6765 2 357+ 357 + 357+ 357+ 357+ 6765 2 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),

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, 47 475 [6] T E Phipps, Fibonacci and continued fractions, Aperion 5 (2008), no 4, 534 550 Department of Mathematics Hannam University Daejeon 306-79, Korea E-mail address: emc@hnukr