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$: Combinatorial Proofs of Some Identities for Nonregular … · 2014-04-24 · A combinatorial interpretation of nonregular continued fractions is studied. Using a modiﬁcation of$
Hindawi Publishing CorporationInternational Journal of CombinatoricsVolume 2012, Article ID 894380, 6 pagesdoi:10.1155/2012/894380

Research ArticleCombinatorial Proofs of Some Identities forNonregular Continued Fractions

Oranit Panprasitwech1, 2

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

Correspondence should be addressed to Oranit Panprasitwech, [email protected]

Received 15 July 2012; Accepted 2 September 2012

Academic Editor: Toufik Mansour

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

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

1. Introduction

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

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

A continued fraction of the form

b0 +a1

b1 +a2

b2 + . . .+

an

bn + . . .

,

,

(1.1)

2 International Journal of Combinatorics

where for i > 0, ai, b0, and bi are positive integers, is called a nonregular continued fraction. Itis more convenient to use the notation [b0;a1, b1;a2, b2; . . . ;an, bn; . . .] for the above continuedfraction. If for every i, ai = 1 we denote [b0; b1, b2, . . .] := [b0;a1, b1;a2, b2; . . .] and [b0; b1,b2, . . .] is said to be regular.

Corresponding to each continued fraction [b0;a1, b1;a2, b2; . . .], two sequences {pn}and {qn} are defined inductively by

p0 = b0, p1 = b1b0 + a1, pn = bnpn−1 + anpn−2 (n ≥ 2),

q0 = 1, q1 = b1, qn = bnqn−1 + anqn−2 (n ≥ 2),(1.2)

pn/qn is called the nth convergent. An importance property of these numerators and denom-inators of continued fractions is

pnqn

= [b0;a1, b1; . . . ;an, bn] (n ≥ 0). (1.3)

2. Combinatorial Interpretation of Nonregular Continued Fractions

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

Let P(b0;a1, b1; . . . ;an, bn) be the number of ways to tile an (n+1)-boardwith dominoesand single tiles. All n + 1 cells of the (n + 1)-board are labeled with 0, 1, 2, . . . , n from left toright, respectively. Figure 1 illustrates a 3-board, a single tile and a domino. Tilingmust satisfythe following three conditions.

(1) All n + 1 cells of the (n + 1)-board must be covered.

(2) For 0 ≤ i ≤ n, the ith cell can be covered by a stack of as many as bi single tiles.

(3) For 1 ≤ i ≤ n, two consecutive cells i − 1 and i can be covered by a stack of as manyas ai dominoes.

It is obvious that P(b0) = b0 and by focusing on the last cell covering, it follows thatP(b0;a1, b1) = b1b0 + a1 and

P(b0;a1, b1; . . . ;an, bn) = bnP(b0;a1, b1; . . . ;an−1, bn−1)

+ anP(b0;a1, b1; . . . ;an−2, bn−2) (n ≥ 2).(2.1)

Since the sequence {P(b0;a1, b1; . . . ;an, bn)} satisfy the same initial conditions and recurrencerelation as the sequence {pn} defined by (1.2),

P(b0;a1, b1; . . . ;an, bn) = pn, ∀n ≥ 0. (2.2)

Next, we define Q(b0;a1, b1; . . . ;an, bn) := P(b1;a2, b2; . . . ;an, bn).Similar to the case of {P(b0;a1, b1; . . . ;an, bn)}, we obtain

Q(b0;a1, b1; . . . ;an, bn) = qn, ∀n ≥ 0. (2.3)

International Journal of Combinatorics 3

1

Single tile Domino3-board

20

Figure 1: An empty 3-board, a single tile, and a domino.

3 ways

5 ways

1 way 1 way

4 ways5 ways

2 ways

1 20 1 20 1 20

Figure 2: The ways to tile a 3-board with the height conditions 1; 2, 3; 4, 5.

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

P(b0;a1, b1; . . . ;an, bn)Q(b0;a1, b1; . . . ;an, bn)

= [b0;a1, b1; . . . ;an, bn] (n ≥ 0). (2.4)

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

P(1; 2, 3; 4, 5) = (1 × 3 × 5) + (1 × 4) + (2 × 5) = 29,

Q(1; 2, 3; 4, 5) = P(3; 4, 5) = (3 × 5) + 4 = 19.(2.5)

Thus, [1; 2, 3; 4, 5] = 29/19.

3. Combinatorial Proofs of Some Identities

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

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

Theorem 3.1. Let N be an arbitrary positive integer and pn/qn the nth convergent of [b0;N, b1;. . . ;N, bn]. Then for all n ≥ 1, one has

[bn;N, bn−1; . . . ;N, b0] =pnpn−1

. (3.1)

Proof. This reversal identity follows immediately from (2.2)–(2.4) and the fact that the waysto tile an (n + 1)-board with the height conditions bn;N, bn−1; . . . ;N, b0 equals the ways to tilewith the height conditions b0;N, b1; . . . ;N, bn, which leads

P(bn;N, bn−1; . . . ;N, b0) = P(b0;N, b1; . . . ;N, bn) = pn, (3.2)


4 ways5 ways

3 ways

0 1 0 1

Figure 3: The ways to tile a 2-board with the height conditions 3; 4, 5.

a1

a2

a3

a4

a5

a6

a7

a8

a9

0 1 2 3 4 5 6 7 8 9

Figure 4: An illustration of the fault-free elements of Pn × Qn−1 for the case n is odd.

and the fact that the ways to tile an n-board with the height conditions bn−1;N, bn−2; . . . ;N, b0equals the ways to tile with the height conditions b0;N, b1; . . . ;N, bn−1, which implies

Q(bn;N, bn−1; . . . ;N, b0) = P(bn−1;N, bn−2; . . . ;N, b0) = P(b0;N, b1; . . . ;N, bn−1) = pn−1.(3.3)

Theorems 3.2 and 3.3 are proved by modifying the proofs of Identity 110 and Identity111 in [1] for regular continued fractions to nonregular continued fractions.

Theorem 3.2. The difference between consecutive convergents of [b0;a1, b1; . . . ;an, bn] is

pnqn

− pn−1qn−1

=(−1)n−1∏n

i = 1ai

qnqn−1(n ≥ 1). (3.4)

Equivalently, after multiply both sides by qnqn−1, we have

pnqn−1 − pn−1qn = (−1)n−1n∏

i = 1

ai (n ≥ 1). (3.5)

Proof. Denote Pn × Qn−1 := the set of tiling of two boards, where on the top board has cells0, 1, 2, . . . , n with height conditions b0;a1, b1; . . . ;an, bn, and the bottom board has cells1, 2, . . . , n − 1 with height conditions b1; . . . ;an−1, bn−1 and Pn −1 × Qn := the set of tilingof two boards, where on the top board has cells 0, 1, 2, . . . , n − 1 with height conditionsb0;a1, b1; . . . ;an−1, bn−1, and the bottom board has cells 1, 2, . . . , nwith height conditions b1; . . . ;an, bn.

Any element (A,B) in Pn×Qn−1 or Pn−1×Qn is said to has a fault at cell i ≥ 1, ifA and Bhave tiles that end at i.

International Journal of Combinatorics 5

0 1 2 3 4 5 6 7 8 9

Figure 5: There are no fault-free elements of Pn × Qn−2 for the case n is odd.

a1

a2

a3

a4

a5

a6

a7

a8

0 1 2 3 4 5 6 7 8 9

b9

Figure 6: An illustration of the fault-free elements of Pn−2 × Qn for the case n is odd.

It can be seen from the definitions that

(1) |Pn × Qn−1| = pnqn−1,

(2) |Pn−1 × Qn | = pn−1qn,

(3) when A or B contains a single tile, (A,B) must have a fault.Next, we consider the numbers of fault-free elements of Pn × Qn−1 and Pn−1 × Qn.

Case 1 (n is odd). There are∏n

i = 1ai fault-free elements of Pn ×Qn−1 as shown in Figure 4 andno fault-free element of Pn−1 × Qn, since for all (A,B) ∈ Pn−1 ×Qn, A and B both cover an oddnumber of cells that mean A and B must contain a single tile.

Case 2 (n is even). Similar to the case of n is odd, there are no fault-free elements of Pn ×Qn−1and

∏ni = 1ai fault-free elements of Pn−1 × Qn.

Finally, let (S, T) be an element in Pn ×Qn−1 with (S, T) has a fault. If we swap the tailsof S and T after the rightmost fault, then we get the element (S′, T ′) in Pn−1 × Qn that has thesame rightmost fault as (S, T). Hence, by using this swapping, a one-to-one correspondencebetween the set of the elements that has a fault in Pn × Qn−1 and the set of the elements thathas a fault in Pn−1 × Qn can be constructed.

Therefore, pnqn−1 − pn−1qn = |Pn × Qn−1| − |Pn−1 × Qn | = (−1)n−1∏ni = 1ai.

Theorem 3.3. For n ≥ 2, let pn/qn be the nth convergent of [b0;a1, b1; . . . ;an, bn]. Then

pnqn−2 − pn−2qn = (−1)nbnn−1∏

i = 1

ai. (3.6)

Proof. Denote Pn × Qn−2 := the set of tiling of two boards, where on the top board hascells 0, 1, 2, . . . , n with height conditions b0;a1, b1; . . . ;an, bn, and the bottom board has cells1, 2, . . . , n − 2 with height conditions b1; . . . ;an−2, bn−2 and Pn−2 × Qn := the set of tiling oftwo boards, where on the top board has cells 0, 1, 2, . . . , n − 2 with height conditionsb0;a1, b1; . . . ;an−2, bn−2, and the bottom board has cells 1, 2, . . . , n with height conditionsb1; . . . ;an, bn.


a1

a2

a3

a4

a5

a6

a7

a8

a9

0 1 2 3 4 5 6 7 8 109

b10

Figure 7: An illustration of the fault-free elements of Pn × Qn−2 for the case n is even.

0 1 2 3 4 5 6 7 8 109

Figure 8: There are no fault-free elements of Pn−2 × Qn for the case n is even.

Hence |Pn × Qn−2| = pnqn−2 and |Pn−2 × Qn | = pn−2qn.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 Pn × Qn−2 and the set of the elements thathas a fault in Pn−2 × Qn by swapping the tails after the rightmost fault.

Thus, it suffices to consider the numbers of fault-free elements of Pn × Qn−2 and Pn−2 ×Qn.

Case 1 (n is odd). There are no fault-free elements of Pn × Qn−2 and bn∏n−1

i = 1ai fault-freeelements of Pn−2 × Qn, described by the examples illustrated in Figures 5 and 6.

Case 2 (n is even). There are bn∏n−1

i = 1ai fault-free elements of Pn × Qn −2 and no fault-freeelement of Pn −2 × Qn, see Figures 7 and 8.

Therefore, pnqn−2 − pn−2qn = |Pn × Qn−2| − |Pn−2 × Qn | = (−1)nbn∏n−1

i=1 ai.

Acknowledgment

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

References

[1] A. T. Benjamin and J. J. Quinn, Proofs that Really Count: The Art of Combinatorial Proof, MathematicalAssociation of America, Washington, DC, USA, 2003.

[2] A. T. Benjamin, H. Derks, and J. J. Quinn, “The combinatorialization of linear recurrences,” ElectronicJournal of Combinatorics, vol. 18, no. 2, pp. 12–18, 2011.

[3] A. T. Benjamin, A. K. Eustis, and S. S. Plott, “The 99th Fibonacci identity,” Electronic Journal ofCombinatorics, vol. 15, no. 1, Article ID 34, p. 13, 2008.

[4] A. Eustis and M. Shattuck, “Combinatorial proofs of some identities for the Fibonacci and Lucasnumbers,” Integers, vol. 11, Article ID A23, p. 15, 2011.

[5] A. T. Benjamin and D. Zeilberger, “Pythagorean primes and palindromic continued fractions,” Integers,vol. 5, no. 1, Article ID A30, p. 5, 2005.

[6] M. Anselm and S. H. Weintraub, “A generalization of continued fractions,” Journal of Number Theory,vol. 131, no. 12, pp. 2442–2460, 2011.

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