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Ratios of Polygonal Numbers As Continued Fractions A International Journal of Engineering Science, Advanced Computing and Bio-Technology Vol. 8, No. 3, July – September 2017, pp. 143 - 155 Ratios of Polygonal Numbers as Continued Fractions A. Gnanam1, *S. Krithika2 1Department of Mathematics, Government Arts College, Trichy-22. 2Department of Mathematics, Seethalakshmi Ramaswami College, Trichy-2. Email: [email protected], [email protected] Abstract: Representations of rational numbers as continued fraction always exist. In number theory study of polygonal numbers has various approaches. Here in this paper we identify the patterns of continued fractions of ratios of polygonal numbers of consecutive order. Keywords: Continued fractions, Simple continued fraction, Euclidean algorithm, Square numbers, Hexagonal numbers, Centered Hexagonal numbers, Star numbers. Notations: 1. a0 ,a1 ,a2 ,a3 ,an : Continued fraction expansion. n 2. : Integer part of the rational number n 2 . 2 nn1 th 3. Tn : n Triangular number 2 2 th 4. Sn n : n Square number th 5. H n 3n(n 1) 1 : n Centered Hexagonal number * th 6. Sn 6n(n 1)1 : n Star number 1. Introduction The Indian mathematician Aryabhata used a continued fraction to solve a linear indeterminate equation. For more than a thousand years, any work that used continued fractions was restricted to specific examples. Throughout Greek and Arab mathematical writing, we can find examples and traces of continued fractions. Euler showed that every rational can be expressed as a terminating simple continued fraction. He also provided an expression for e in continued fraction form. He used this expression to show that e and e 2 are irrational [3]. Polygonal numbers have graphical representation. Golden ratio which is an irrational also has a graphical representation. This idea motivated me to create a set of rational numbers using polygonal numbers and represent them in terms of continued Received: 14 March, 2017; Revised: 28 June, 2017; Accepted: 03 August, 2017 *Corresponding Author 144 International Journal of Engineering Science, Advanced Computing and Bio-Technology fractions. First we give different representations of a rational number as a continued fraction [2, 3, 4, 5]. An expression of the form p b 0 a 0 q b 1 a 1 b 2 a 2 b 3 a 3 Where ai ,bi are real or complex numbers is called a continued fraction. An expression of the form p 1 a 0 1 q a 1 1 a 2 1 a 3 Where bi 1 i, and a0 ,a1 ,a2 , are each positive integers also represents a simple continued fraction. The continued fraction is commonly expressed as p 1 1 1 a0 or simply as a0 ,a1 ,a2 ,a3 , . q a1 a2 a3 The elements a0 ,a1 ,a2 ,a3 , are called the partial quotients. If there are finite number of partial quotients, we call it finite simple continued fraction, otherwise it is infinite. We have to use either Euclidean algorithm[1] or continued fraction algorithm to find such partial quotients. The finite simple continued fraction is denoted by a ,a ,a ,a , a and it has an alternate form a ,a ,a ,a , a 1,1. 0 1 2 3 n 0 1 2 3 n 1.1 The Continued Fraction Algorithm Suppose we wish to find continued fraction expansion of x R. 1 Let x x and set a x . Define x and set a x . 0 0 0 , 1 1 1 x 0 x 0 1 1 and x 2 a2 x 2 ,....., x k ak x k ,.... x1 x1 x k1 x k1 This process is continued infinitely or to some finite stage till an xi N exists such that ai x i . Ratios of Polygonal Numbers as Continued Fractions 145 2. Triangular Numbers Definition 2.1[5, 6]: Triangular Numbers nn 1 The numbers 1, 3, 6, 10, 15, 21, 28, 36, 45,…, T are called triangular n 2 ,… numbers, since the nth number counts the number of dots in an equilateral triangular array with n dots to the side. T 1 T 3 T 6 T 10 T 15 T 21 1 2 3 4 5 6 Theorem: 2.1: n 0;1, ,2 when n is odd. Tn 2 Forn 3, T n n1 0; 1, when n is even. 2 Proof: Using algorithm 1.1 the proof follows n(n 1) (n 1)(n 2) Since T and T n n1 , 2 2 T n n Tn1 n 2 T n 1 1 1 When n is even, n 0 0 0 T n 2 n 2 2 1 . n1 1 1 n n n 2 T n 1 1 1 When n is odd , n 0 0 0 T n 2 n 2 2 1 n1 1 1 n 1 n n . 2 2 T n Hence For each n 3 the continued fraction expansion of n is0; 1, when n is , Tn1 2 n even and is 0;1, ,2 when n is odd. 2 146 International Journal of Engineering Science, Advanced Computing and Bio-Technology 3. Square Numbers Definition 3.1[5, 6]: Square Numbers A square number or perfect square is an integer that is the square of an integer or it is the product of some integer with itself. In other wards the numbers 1,4,9,16,25,36,49,64,81,…, S n2 are called square n ,… numbers, since the nth number counts the number of dots in a square array with n dots to each side. S 1 S 4 S 9 S 16 S 25 1 2 3 4 5 Theorem: 3 .1: n n 0;1, ,3,1, whennisodd. Sn 2 2 For n 3, S n n n1 0; 1, 1,1,3, whenniseven. 2 2 Proof: Using algorithm 1.1 the proof follows Case (i): nis odd S 9 Taking n 3 , 3 S4 16 9 Let x0 ,soa0 0. 16 1 16 7 Then x1 1 a1 1. x 0 x 0 9 9 1 9 2 x 2 1 a2 1. x1 x 1 7 7 1 7 1 x 3 3 a3 3. x 2 x 2 2 2 1 2 x 4 2 a3 2. x 3 x 3 1 9 9 Therefore 0;1,1,3,2or 0;1,1,3,1,1. 16 16 Ratios of Polygonal Numbers as Continued Fractions 147 S Hence 3 0;1,1,3,1,1.The result is true when n 3 .The result is true for S4 n 2k 1, where3 k n. S Then, 2k1 0;1,k 1,3,1,k 1 S2k We prove the result for n 2k 1. 2 S2k1 (2k 1) Here we have to find the continued fraction of 2 . S2k2 (2k 2) (2k 1)2 4k 2 4k1 Let x 0 2 2 , so a0 0. (2k 2) 4k 8k4 1 4k 2 8k 4 4k 3 Then x1 2 1 2 a1 1. x 0 x 0 4k 4k 1 4k 4k 1 1 4k 2 4k 1 k 1 x 2 k a2 k. x1 x1 4k 3 4k 3 1 4k 3 k x 3 3 a3 3. x 2 x 2 k 1 k 1 1 k 1 1 x 4 1 a4 1. x 3 x 3 k k 1 x 5 k a5 k. x 4 x 4 S2k1 Since x 5 k, 0;1,k,3,1,k. S2k2 Hence by induction the result is true for all values of n where nis odd. Case(ii): nis even S 16 Taking n 4 , 4 S5 25 16 Let x0 ,soa0 0. 25 1 25 9 Then x 1 1 a1 1. x 0 x 0 16 16 1 16 7 x 2 1 a2 1. x1 x1 9 9 148 International Journal of Engineering Science, Advanced Computing and Bio-Technology 1 9 2 x 3 1 a3 1. x 2 x 2 7 7 1 7 1 x 4 3 a4 3. x 3 x 3 2 2 1 x 5 2 a5 2. x 4 x 4 S4 Since x 5 2, 0;1,1,1,3,2. S5 The result is true when n 4. The result is true for n 2k 2, where3 k n. S Then, 2k2 0;1,k 2,1,3,k 1. S2k1 We prove the result forn 2k. 2 S2k (2k) Here we have to find the continued fraction of 2 . S2k1 (2k 1) (2k)2 4k 2 Let x0 2 2 , so a0 0. (2k 1) 4k 4k1 1 4k 2 4k 1 4k 1 Then x1 2 1 2 a1 1. x 0 x 0 4k 4k 1 4k 2 3k 1 x 2 k 1 a2 k 1. x 1 x1 4k 1 4k 1 1 4k 1 k x 3 1 a3 1. x 2 x 2 3k 1 3k 1 1 3k 1 1 x 4 3 a4 3. x 3 x 3 k k 1 x 5 k a5 k. x 4 x 4 S2k Since x 5 k, 0;1,k 1,1,3,k. S2k1 Hence by induction the result is true for all values of n where n is even. Hence from case(i) and (ii) we have For each n 3 , the continued fraction expansion of Ratios of Polygonal Numbers as Continued Fractions 149 S n n n n n is0;1, ,3,1, when n is odd and is 0; 1, 1,1,3, when n is Sn1 2 2 2 2 even. 4. Centered Hexagonal Number Definition 4.1[7]: Centered Hexagonal Number A centered hexagonal number, or hex number, is a centered figurate number that represents a hexagon with a dot in the center and all other dots surrounding the center dot in a hexagonal lattice.
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