International Journal of Combinatorics

Volume 2012 (2012), Article ID 894380, 6 pages

http://dx.doi.org/10.1155/2012/894380

## Combinatorial Proofs of Some Identities for Nonregular Continued Fractions

^{1}Department of Mathematics, Faculty of Science, King Mongkut's University of Technology Thonburi, Bangkok 10140, Thailand^{2}Centre of Excellence in Mathematics, Si Ayutthaya RD, Bangkok 10400, Thailand

Received 15 July 2012; Accepted 2 September 2012

Academic Editor: Toufik Mansour

Copyright © 2012 Oranit Panprasitwech. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

A combinatorial interpretation of nonregular continued fractions is studied. Using a modification of a tiling technique due to Benjamin and Quinn, combinatorial proofs of some identities for nonregular continued fractions are obtained.

#### 1. Introduction

In the recently popular book *Proofs that Really Count* [1], many identities involving linear recurrences were proved by using beautiful tiling interpretations. Many researches provided tiling proofs for a variety of identities. See, for example, [2–4]. By making use of the combinatorial interpretation of continued fractions presented by Benjamin and Quinn in [1], Benjamin and Zeilberger [5] introduced a combinatorial proof of the statement related to prime numbers.

In this research, a combinatorial interpretation of nonregular continued fractions is investigated. The major part of this work is devoted to establishing combinatorial proofs of some identities for nonregular continued fractions.

A continued fraction of the form where for , , , and are positive integers, is called a nonregular continued fraction. It is more convenient to use the notation for the above continued fraction. If for every , we denote and is said to be regular.

Corresponding to each continued fraction , two sequences and are defined inductively by is called the th convergent. An importance property of these numerators and denominators of continued fractions is

#### 2. Combinatorial Interpretation of Nonregular Continued Fractions

As mentioned by Benjamin and Quinn in [1], a simple combinatorial interpretation of nonregular continued fractions can be realized.

Let be the number of ways to tile an with dominoes and single tiles. All cells of the are labeled with from left to right, respectively. Figure 1 illustrates a 3-board, a single tile and a domino. Tiling must satisfy the following three conditions.(1)All cells of the must be covered.(2)For , the cell can be covered by a stack of as many as single tiles.(3)For , two consecutive cells and can be covered by a stack of as many as dominoes.

It is obvious that and by focusing on the last cell covering, it follows that and Since the sequence satisfy the same initial conditions and recurrence relation as the sequence defined by (1.2),

Next, we define .

Similar to the case of , we obtain

Equations (1.3), (2.2), and (2.3) yield

For example, described by the Figures 2 and 3, we get Thus, .

#### 3. Combinatorial Proofs of Some Identities

In this section, the combinatorial interpretation presented in the previous is adopted to reach combinatorial proofs of some identities for nonregular continued fractions.

The reversal identity for the generalization of regular continued fractions recently investigated by Anselm and Weintraub in [6] can easily be verified as Theorem 3.1.

Theorem 3.1. *Let be an arbitrary positive integer and the convergent of . Then for all , one has
*

*Proof. *This reversal identity follows immediately from (2.2)–(2.4) and the fact that the ways to tile an board with the height conditions equals the ways to tile with the height conditions , which leads
and the fact that the ways to tile an board with the height conditions equals the ways to tile with the height conditions , which implies

Theorems 3.2 and 3.3 are proved by modifying the proofs of Identity 110 and Identity 111 in [1] for regular continued fractions to nonregular continued fractions.

Theorem 3.2. *The difference between consecutive convergents of is
**
Equivalently, after multiply both sides by , we have
*

*Proof. *Denote the set of tiling of two boards, where on the top board has cells with height conditions , and the bottom board has cells with height conditions and the set of tiling of two boards, where on the top board has cells with height conditions , and the bottom board has cells with height conditions .

Any element in or is said to has a fault at cell , if have tiles that end at .

It can be seen from the definitions that(1),
(2),
(3)when contains a single tile, must have a fault.

Next, we consider the numbers of fault-free elements of and .

*Case 1 ( is odd). *There are fault-free elements of as shown in Figure 4 and no fault-free element of , since for all , and both cover an odd number of cells that mean and must contain a single tile.

*Case 2 ( is even). * Similar to the case of is odd, there are no fault-free elements of and fault-free elements of .

Finally, let be an element in with has a fault. If we swap the tails of after the rightmost fault, then we get the element in that has the same rightmost fault as . Hence, by using this swapping, a one-to-one correspondence between the set of the elements that has a fault in and the set of the elements that has a fault in can be constructed.

Therefore, .

Theorem 3.3. *For , let be the convergent of . Then
*

*Proof. *Denote the set of tiling of two boards, where on the top board has cells with height conditions , and the bottom board has cells with height conditions and the set of tiling of two boards, where on the top board has cells with height conditions , and the bottom board has cells with height conditions .

Hence and .

Similar to the proof of Theorem 3.2, we can construct a one-to-one correspondence between the set of the elements that has a fault in and the set of the elements that has a fault in by swapping the tails after the rightmost fault.

Thus, it suffices to consider the numbers of fault-free elements of and .

*Case 1 ( is odd). *There are no fault-free elements of and fault-free elements of , described by the examples illustrated in Figures 5 and 6.

*Case 2 ( is even). *There are fault-free elements of and no fault-free element of , see Figures 7 and 8.

Therefore, .

#### Acknowledgment

This research is supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.

#### References

- A. T. Benjamin and J. J. Quinn,
*Proofs that Really Count: The Art of Combinatorial Proof*, Mathematical Association of America, Washington, DC, USA, 2003. - A. T. Benjamin, H. Derks, and J. J. Quinn, “The combinatorialization of linear recurrences,”
*Electronic Journal of Combinatorics*, vol. 18, no. 2, pp. 12–18, 2011. View at Google Scholar · View at Zentralblatt MATH - A. T. Benjamin, A. K. Eustis, and S. S. Plott, “The 99th Fibonacci identity,”
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