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 nn1 3. T  : nth Triangular number n 2 2 th 4. Sn  n : n Square number 5. H  3n(n 1) 1 : nth Centered Hexagonal number n * 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  a  x ,....., x  a  x ,.... 2 2 2 k k k x1  x1 x k1 x k1 This process is continued infinitely or to some finite stage till an x N exists such that i

ai x i .

Ratios of Polygonal Numbers as Continued Fractions 145

2. Triangular Numbers Definition 2.1[5, 6]: Triangular Numbers nn 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 n1 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  n1  , 2 2

Tn n  Tn1 n  2 T n 1 1 1 When n is even, n  0  0   0  T n  2 n  2 2 1 . n1 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 n1 1 1 n 1 n n   . 2 2

Tn n Hence For each n  3 , the continued fraction expansion of is0; 1,  when n is Tn1 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 n1 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;1,1,3,2or 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, 2k1 0;1,k 1,3,1,k 1 S2k We prove the result for n  2k  1. S (2k 1)2 2k1 Here we have to find the continued fraction of  2 . S2k2 (2k  2) (2k 1)2 4k 2 4k1 Let x 0  2  2 , so a0 0. (2k  2) 4k 8k4 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 S Since x  k,  2k1 0;1,k,3,1,k. 5 S2k2 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, 2k2  0;1,k  2,1,3,k 1. S2k1 We prove the result forn  2k. 2 S2k (2k) Here we have to find the continued fraction of  2 . S2k1 (2k 1) (2k)2 4k 2 Let x0  2  2 , so a0 0. (2k 1) 4k 4k1 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. S2k1 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 is0;1,  ,3,1,  when n is odd and is 0; 1, 1,1,3,  when n is     Sn1 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. The nth centered hexagonal number is given by the formula

H n  3n(n 1) 1 Another way of expressing the hex number as 1 H  6[ n(n 1)]1 n 2 Shows that the centered hexagonal number for n is 1 more than 6 times the (n − 1)th triangular number. The first few centered hexagonal numbers are 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919

H 1 H 7 H 19 H 37 1  2  3  4 

Theorem: 4.1:  n 0; 1,  , 6n whennisodd. H n  2 For n  3,  H n 3n n1 0; 1, 1, 1,1, 1,2 whenniseven.  2 2 Proof: Using algorithm 1.1 the proof follows Case (i): nis odd H 19 Takingn  3 , 3  H 4 37 19 Let x0  ,soa0 0. 37 150 International Journal of Engineering Science, Advanced Computing and Bio-Technology

1 37 18 Then x 1   1 a1 1. x 0 x 0 19 19 1 19 1 x 2   1 a2 1. x1 x1 18 18 1 x 3  18 a3 18. x 2 x 2 19 Therefore 0;1,1, 18 37 H Hence 3 0;1,1,18. H 4 The result is true when n  3 . The result is true for n 2k 1, where 3  k  n. H Then, 2k1 0;1, k 1, 6(2k 1). H 2k We prove the result for n  2k  1. 12k 2 6k 1 H 2k1   Here we have to find the continued fraction of  2 . H 2k2 12k 18k  7 (2k 1)2 12k 2  6k1 Let x 0  2  2 , so a0 0. (2k  2) 12k 18k7 1 12k 2 18k  7 12k  6 Then x1   2 1 2 a1 1. x 0 x 0 12k  6k 1 12k  6k 1 1 12k 2  6k  1 1 x 2    k  a2 k. x 1 x1 12k  6 12k  6 1 x 3  12k  6a3 12k  6. x 2 x 2

H Since x  12k  6,  2k1 0;1,k,6(2k  1). 3 H 2k2 Hence by induction the result is true for all values of n where nis odd. H 37 Case(ii): nis even. Taking n  4 then 4  H 5 61 37 Let x0  ,soa0 0. 61 Ratios of Polygonal Numbers as Continued Fractions 151

1 61 24 Then x 1   1 a1 1. x 0 x 0 37 37 1 37 13 x 2   1 a2 1. x1 x 1 24 24 1 24 11 x 3   1 a3 1. x 2 x 2 13 13 1 13 2 x 4    1  a4 1. x 3 x 3 11 11 1 11 1 x 5   5  a5 5. x 4 x 4 2 2 1 x 6   2a6 2. x 5  x 5

H 4 Since x 2  2,  0;1,1,1,1,5,2. H 5 The result is true when n  4. . The result is true for n 2k  2, where3  k  n. H Then, 2k2 0;1,k  2,1, 1,3k  4,2. H 2k1 We prove the result for n  2k. 2 H 2k 12k  6k 1 Here we have to find the continued fraction of  2 . H 2k1 12k  6k 1 (2k)2 12k 2  6k 1 Let x 0  2  2 , so a0 0. (2k 1) 12k  6k 1 1 12k 2  6k  1 12k Then x 1   2 1 2 a1 1. x 0 x 0 12k  6k  1 12k  6k  1 1 12k 2  6k  1 6k 1 x 2    k 1  a2 k 1. x1 x1 12k 12k 1 12k 6k 1 x 3   1  a3 1. x 2 x 2 6k  1 6k  1 1 6k  1 2 x 4    1  a4 1. x 3 x 3 6k 1 6k 1 1 6k 1 1 x 5    3k 1 a5 3k 1. x 4 x 4 2 2 152 International Journal of Engineering Science, Advanced Computing and Bio-Technology

1 x 6   2 a6 2. x 5 x 5

H 2k Since x 6  2,  0;1,k 1,1,1,3k 1,2. H 2k1 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 H n n 3n n is0;1,  ,6n when n is odd and is 0; 1, 1,1,1, 1, 2 when n is   H n1 2 2 2 even.

5. Star Numbers Definition 5.1[7]: Star Numbers A star number is a centered figurate number that represents a centered hexagram (six- pointed star), such as the one that Chinese checkers is played on.

The nth star number is given by the formula * Sn  6n(n 1)1 The first few star numbers are 1, 13, 37, 73, 121, 181, 253, 337, 433, 541 The digital root of a star number is always 1 or 4, and progresses in the sequence 1, 4, 1.

S* 1 S * 13 S * 37 1  2  3 

Theorem: 5.1:  n * 0; 1, , 12n whennisodd. Sn  2 For n  3, *  S n n1 0; 1, 1, 1,1, 3n 1,2 whenniseven.  2 Proof : Similar to the proof of theorem 2

Ratios of Polygonal Numbers as Continued Fractions 153

6. The following table summarizes the continued fraction expansion of consecutive fraction of some polygonal numbers. Types of Numbers Consecutive Continued Fractions Nature of Fraction of n Numbers Triangular number T n Odd n 0;1, ,2   Tn1 2 Triangular number T n Even n 0;1,  Tn1 2 Square number S n n Odd n 0;1,  ,3,1,       Sn1 2 2 Square number S n n Even n 0; 1, 1,1,3,  Sn1 2 2 Centered Hexagonal H n Odd n 0;1,  ,6n Number   H n1 2

Centered Hexagonal H n n 3n Even Number 0; 1, 1,1,1, 1, 2 H n1 2 2 Star Number S* n Odd n 0;1, ,12n * 2  Sn1   Star Number S * n Even n 0; 1, 1,1,1,3n 1, 2 * 2 Sn1

7. Illustration The following table gives the patterns of continued fraction of consecutive fractions of some polygonal numbers. Consecutive fraction of Continued fraction expansion numbers T 0;1, 3,2 7 T8 T 0;1, 47 94 T93 S 0;1,4,1,3,5 10 S11 154 International Journal of Engineering Science, Advanced Computing and Bio-Technology

S 0;1,13,3,1,13 27 S28 H 0;1, 2, 30 5 H 6 H 0;1,4,1,1,14,2 10 H 11 * S9 0;1, 4,108 * S10 * S18 0;1, 8,1,1,53,2 * S19

8. Conclusion In this paper we have identified various patterns of continued fractions of ratios of polygonal numbers of consecutive order and rank. This work may be extended to higher order figurate numbers like pyramidal numbers.

References: [1] George E. Andrews, “Number Theory”, W.B. Saunders Company.

[2] Ivan Niven , Herbert S. Zuckerman, Hugh L. Montgomery, “An introduction to theory of numbers”, Fifth edition, Wiley Student Edition.

[3] Jonathan Browein, Alfvander Poorten, Jeffrey Shallit, Wadim Zudilin, “Neverending Fractions. An Introduction to continued fractions”, Cambridge University Press.

[4] Joseph H. Silverman, “A friendly introduction to Number Theory”, fourth edition.

[5] Neville Robbins, “Beginning Number Theory”, Second edition, Narosa Publishing House.

[6] www.maths.surrey.ac.uk/hosted-sites/R.Knott/Figurate/figurate.html

[7] https://en.wikipedia.org/wiki/Polygonal_number

Authors’ Profile: Gnanam A. was born in Tiruchirappalli, Tamil Nadu, India in 1967. She received M.Sc. and M.Phil. degree in Annamalai University, Chithambaram. She got Ph.D degee in Mathematics from Bharathidasan University, Tiruchirappalli. She started her career as a lecturer in 1993. Now she is working as an Assistant Professor in Mathematics in Government Arts College, Thuvakudi, Tiruchirappalli. Her area of specialization is Number Theory. Dr. Gnanam published seventeen papers in National and International journals. She produced one ph.D, scholar and five more students are doing research under her guidance.

Ratios of Polygonal Numbers as Continued Fractions 155

Krithika S. was born in Tiruchirappalli, Tamil Nadu, India in 1978. She received M.Sc. and M. Phil. Degree in Seethalakshmi Ramaswami Collelge, Thiruchirappalli. She started her career as a lecturer in 2002. Now she is working as an Assistant Professor in Mathematics in Seethalakshmi Ramaswami College, Tiruchirappalli. She published one paper in International journal.