JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 10.1155/2014/428020 428020 Research Article On k-Distance Pell Numbers in 3-Edge-Coloured Graphs Piejko Krzysztof Włoch Iwona Cen Song 1 Faculty of Mathematics and Applied Physics Rzeszow University of Technology Aleja Powstańców Warszawy 12, 35-959 Rzeszów Poland prz.edu.pl 2014 582014 2014 27 05 2014 17 07 2014 19 07 2014 5 8 2014 2014 Copyright © 2014 Krzysztof Piejko and Iwona Włoch. 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.

We define in this paper new distance generalizations of the Pell numbers and the companion Pell numbers. We give a graph interpretation of these numbers with respect to a special 3-edge colouring of the graph.

1. Introduction

The Fibonacci sequence is defined by the following recurrence relation Fn=Fn-1+Fn-2 for n2 with F0=F1=1. Among sequences of the Fibonacci type there is the Pell sequence defined by Pn=2Pn-1+Pn-2 for n2 with the initial conditions P0=0 and P1=1. The companion Pell sequence is closely related to the Pell sequence and is given by formula Q0=Q1=2 and Qn=2Qn-1+Qn-2 for n2. The Pell sequences play an important role in the number theory and they have many interesting interpretations. We recall some of them.

The number of lattice paths from the point O(0,0) to the line x=n consisting of a=[1,1], b=[1,-1], and c=[2,0] steps is equal to Pn+1; see .

The number of compositions (i.e., ordered partitions) of a number n into two sorts of of 1’s and one sort of 2’s is equal to Pn+1, see .

The number of n-step non-self-intersecting paths starting at the point O(0,0) with steps of types p=[1,0], q=[-1,0], or r=[0,1] is equal to (1/2)Qn+1, see .

Interesting generalizations of the numbers of the Fibonacci type (also generalizations of the Pell sequences) are studied, for instance, by Kilic in . Among others Kilic in  introduced the generalization of the Pell numbers. He defined the generalized Pell (p,i)-numbers for any given p, where p1 and for n>p+1 and 0ip in the following way Pp(i)(n)=2Pp(i)(n-1)+Pp(i)(n-p-1) with initial conditions and Pp(i)(1)==Pp(i)(i)=0,  Pp(i)(i+1)==Pp(i)(p+1)=1. If i=0 then Pp(i)(1)==Pp(i)(p+1)=1.

In this paper we describe new kinds of generalized Pell sequence and the companion Pell sequence. Our generalization is closely related to the recurrence given in  by Kilic. By other initial conditions we obtain other generalized Pell sequences. We give their graph interpretations which are closely related to a concept of edge colouring in graphs. Graph interpretations of the Fibonacci numbers and the like are study intensively; see, for example, .

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Let k2, n0 be integers. The k-distance Pell sequence Pk(n) is defined by the kth order linear recurrence relation: (1)Pk(n)=2Pk(n-1)+Pk(n-k)fornk, with the following initial conditions: (2)Pk(n)={0ifn=0,2n-1ifn=1,2,3,,k-1. If k=2, then this definition reduces to the classical Pell numbers; that is, P2(n)=Pn.

Table 1 includes a few first words of the Pk(n) for special values of k.

The k-distance Pell sequence Pk(n).

 n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 P n 0 1 2 5 12 29 70 169 408 985 2378 5741 13860 33461 P 3 ( n ) 0 1 2 4 9 20 44 97 214 472 1041 2296 5064 11169 P 4 ( n ) 0 1 2 4 8 17 36 76 160 337 710 1496 3152 6641 P 5 ( n ) 0 1 2 4 8 16 33 68 140 288 592 1217 2502 5144 P 6 ( n ) 0 1 2 4 8 16 32 65 132 268 514 1104 2240 4545

Firstly we give an interpretation of the k-distance Pell sequence, which generalizes result given in (i).

Theorem 1.

Let k2 and n1 be integers. Then the number of lattice paths from the point O(0,0) to the line x=n consisting of a=[1,1], b=[1,-1], and c=[k,0] steps is equal to Pk(n+1).

Proof.

Let s(k,n) be the number of paths from the point O(0,0) to the line x=n. It is easy to notice that s(k,n)=Pk(n+1) for n=1,2,,k.

Let nk+1 and let sa(k,n), sb(k,n), and sc(k,n) denote the number of such paths from the point O(0,0) to the line x=n, for which the last step is of the form a=[1,1], b=[1,-1], and c=[k,0], respectively.

It can be easily seen that sa(k,n)=s(k,n-1), sb(k,n)=s(k,n-1), and sc(k,n)=s(k,n-k). Since s(k,n)=sa(k,n)+sb(k,n)+sc(k,n) then we have s(k,n)=2s(k,n-1)+s(k,n-k) and consequently s(k,n)=Pk(n+1), for all n1.

In the same way we can prove the generalization of the result (ii).

Theorem 2.

Let k2 and n1 be integers. Then the number of all compositions of the number n into two sorts of 1’s and one sort of k’s is equal to Pk(n+1).

By analogy to the Pell sequence we introduce a generalization of the companion Pell sequence which generalizes the classical companion Pell sequence in the distance sense.

Let k2, n0 be integers. The k-distance companion Pell sequence Qk(n) is defined by the kth order linear recurrence relation: (3)Qk(n)=2Qk(n-1)+Qk(n-k)fornk, with the initial conditions (4)Qk(n)={kifn=0,2nifn=1,2,3,,k-1. If k=2 then Q2(n) gives the classical companion Pell numbers Qn; that is, Q2(n)=Qn.

Table 2 includes a few first words of the Qk(n) for special values of k.

The k-distance companion Pell sequence Qk(n).

 n 0 1 2 3 4 5 6 7 8 9 10 11 Q n 2 2 6 14 34 82 198 478 1154 2786 6726 16238 Q 3 ( n ) 3 2 4 11 24 52 115 254 560 1235 2724 6008 Q 4 ( n ) 4 2 4 8 20 42 88 184 388 818 1724 3632 Q 5 ( n ) 3 2 4 8 16 37 76 156 320 656 1349 2774 Q 6 ( n ) 6 2 4 8 16 32 70 142 288 584 1184 2400

The following theorem gives the basic relation between Pk(n) and Qk(n).

Theorem 3.

Let k2 and nk-1 be integers. Then (5)Qk(n)=2Pk(n)+kPk(n-k+1).

Proof (by induction on <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M275"><mml:mrow><mml:mi>n</mml:mi></mml:mrow></mml:math></inline-formula>).

For n=k-1 the result follows immediately by the definitions of Pk(n) and Qk(n). Let nk. Assume that formula (5) is true for t=k, k+1,,n. We will prove that Qk(n+1)=2Pk(n+1)+kPk(n-k+2).

By the induction hypothesis and the definitions of Pk(n) and Qk(n), we have that (6)2Pk(n+1)+kPk(n-k+2)=2[2Pk(n)+Pk(n-k+1)]+k[2Pk(n-k+1)+Pk(n-2k+2)]=2[2Pk(n)+kPk(n-k+1)]+2Pk(n-k+1)+kPk(n-2k+2)=2Qk(n)+Qk(n-k+1)=Qk(n+1), which ends the proof.

For k=2 Theorem 3 gives the well-known relation between the classical Pell numbers and the companion Pell numbers; namely, Qn=2(Pn+Pn-1).

Theorem 4.

Let k2, m1, and n0 be integers. Then (7)Pk(n)+2i=1mPk(n+ki-1)=Pk(n+km),(8)Qk(n)+2i=1mQk(n+ki-1)=Qk(n+km).

Proof of (<xref ref-type="disp-formula" rid="EEq2">7</xref>) (by induction on <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M292"><mml:mrow><mml:mi>m</mml:mi></mml:mrow></mml:math></inline-formula>).

If m=1 then the equation is obvious. Let m2. Assume that formula (7) is true for an arbitrary m. We will prove that Pk(n)+2i=1m+1Pk(n+ik-1)=Pk(n+(m+1)k).

By the induction hypothesis and the definition of Pk(n) we have (9)Pk(n)+2i=1m+1Pk(n+ik-1)=Pk(n)+2i=1mPk(n+ik-1)+2Pk(n+(m+1)k-1)=Pk(n+mk)+2Pk(n+(m+1)k-1)=Pk(n+(m+1)k), which ends the proof of (7).

In the same way we can prove the equality (8), so we omit the proof.

If n=0 or n=1, respectively, then from Theorem 4 the following follows.

Corollary 5.

Let k2 and m1 be integers. Then (10)2i=1mPk(ki-1)=Pk(mk),2i=1mPk(ki)=Pk(mk+1)-1,2i=1mQk(ki-1)=Qk(mk)-k,2i=1mQk(ki)=Qk(mk+1)-2.

For k=2 we obtain the well-known identities for the classical Pell numbers and the companion Pell numbers; namely, (11)2i=1mP2i-1=P2m,2i=1mP2i=P2m+1-1,2i=1mQ2i-1=Q2m-2,2i=1mQ2i=Q2m+1-2.

Theorem 6.

Let k2 and m=0,1,2,,k. Then (12)Pk(k+m)=2k+m-1+m2m-1,(13)Qk(k+m)=2k+m+1+k2m+m2m.

Proof of (<xref ref-type="disp-formula" rid="EEq4">12</xref>) (by induction on <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M309"><mml:mrow><mml:mi>m</mml:mi></mml:mrow></mml:math></inline-formula>).

If m=0 then the result is obvious. Assume that Pk(k+m)=2k+m-1+m2m-1 for an arbitrary m, such that 1mk-1. We will prove that Pk(k+m+1)=2k+m+(m+1)2m.

By the induction hypothesis and the definition of Pk(n), we have (14)Pk(k+m+1)=2Pk(k+m)+Pk(m+1)=2[2k+m-1+m2m-1]+2m=2k+m+(m+1)2m, which ends the proof.

In the same way we can prove the equality (13).

If m=0, m=1, or m=k-1, respectively, then from Theorem 6 we obtain the following corollary.

Corollary 7.

Let k2. Then (15)Pk(k)=2k-1,Pk(k+1)=2k+1,Pk(2k-1)=2k-2(2k+k-1),Qk(k)=2k+1+k,Qk(k+1)=2k+2+2k+2,Qk(2k-1)=22k+2k-1(2k-1).

Theorem 8.

Let k2 and n2k-1 be integers. Then (16)Pk(n)=(2k+1)Pk(n-k)+i=1k-12iPk(n-k-i),(17)Qk(n)=(2k+1)Qk(n-k)+i=1k-12iQk(n-k-i).

Proof of (<xref ref-type="disp-formula" rid="EEq6">16</xref>) (by induction on <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M325"><mml:mrow><mml:mi>n</mml:mi></mml:mrow></mml:math></inline-formula>).

For n=2k-1 we have the equation (18)Pk(2k-1)=(2k+1)Pk(k-1)+i=1k-12iPk(k-1-i). By the initial conditions of the sequence Pk(n) and Theorem 6 we can see that this equation is an identity (because it is equivalent to third identity from Corollary 7).

Let n2k. Assume that formula (16) is true for an arbitrary t where 2ktn. We will prove that Pk(n+1)=(2k+1)Pk(n+1-k)+i=1k-12iPk(n+1-k-i). By the induction hypothesis and the definition of Pk(n), we have (19)(2k+1)Pk(n+1-k)+i=1k-12iPk(n+1-k-i)=(2k+1)[2Pk(n-k)+Pk(n+1-2k)]+i=1k-12i[2Pk(n-k-i)+Pk(n+1-2k-i)]=2[(2k+1)Pk(n-k)+i=1k-12iPk(n-k-i)]+(2k+1)Pk(n-2k+1)+i=1k-12iPk(n+1-2k-i)=2Pk(n)+Pk(n+1-k)=Pk(n+1), which ends the proof of (16). In the same way we can prove (17) which completes the proof of the theorem.

If k=2 then from Theorem 8 we obtain the following formula for the classical Pell numbers and companion Pell numbers: (20)Pn=5Pn-2+2Pn-3,Qn=5Qn-2+2Qn-3,n3.

3. <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M337"><mml:mo stretchy="false">(</mml:mo><mml:mi>A</mml:mi><mml:mo>,</mml:mo><mml:mi>B</mml:mi><mml:mo>,</mml:mo><mml:mi>k</mml:mi><mml:mi>C</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>-Coloured Graphs

For concepts not defined here see . The numbers of the Fibonacci type have many applications in distinct areas of mathematics. There is a large interest of modern science in the applications of the numbers of the Fibonacci type. These numbers are studied intensively in a wide sense also in graphs and combinatorials problem. In graphs Prodinger and Tichy initiated studying the Fibonacci numbers and the like. In  they showed the relations between the number of independent sets in Pn and Cn with the Fibonacci numbers and the Lucas numbers, where Pn and Cn denote an n-vertex path and an n-vertex cycle, respectively. This short paper gave an impetus for counting problems related to the numbers of the Fibonacci type. Many of these problems and results are closely related with the Merrifield-Simmons index σ(G) and the Hosoya index Z(G) in graphs; see [6, 12]. The Pell numbers also have a graph interpretation. It is well-known that Z(PnK1)=Pn+1, where GH denotes the corona of two graphs.

In this section we give a graph interpretation of the k-distance Pell numbers with respect to special edge colouring of a graph.

Let G be a 3-edge coloured graph with the set of colours {A,B,C}. Let M{A,B,C}. We say that a path is M-monochromatic if all its edges are coloured alike by colour M. By l(M) we denote the length of the M-monochromatic path. For xyE(G) notation M(xy) means that the edge xy has the colour M.

Let k1 be an integer. In the graph G we define a (A,B,kC)-edge colouring, such that l(A)0, l(B)0, and l(C)=qk, where q0 is an integer.

Theorem 9.

Let k2 and n2 be integers. The number of all (A,B,kC)-edge colouring of the graph Pn is equal to Pk(n).

Proof.

Let V(Pn)={x1,x2,,xn} be the set of vertices of a graph Pn with the numbering in the natural fashion. Then E(Pn)={x1x2,x2x3,,xn-1xn}. Let σ(k,n) be the number of all (A,B,kC)-edge colouring of the graph Pn. By inspection we obtain that σ(k,n)=Pk(n) for n=2,,k+1.

Let nk+2 and let σA(k,n), σB(k,n), and σC(k,n) denote the number of (A,B,kC)-edge colouring of the graph Pn, with A(xn-1xn), B(xn-1xn), and C(xn-1xn), respectively. It is obvious that σA(k,n)=σB(k,n).

Clearly σA(k,n) and σB(k,n) are equal to the number of all (A,B,kC)-edge colouring of the graph Pn-1 and σC(k,n) is equal to the number of all (A,B,kC)-edge colouring of the graph Pn-k. In the other words σA(k,n)=σ(k,n-1), σB(k,n)=σ(k,n-1), and σC(k,n)=σ(k,n-k). Since (21)σ(k,n)=σA(k,n)+σB(k,n)+σC(k,n), then we have σ(k,n)=2σ(k,n-1)+σ(k,n-k). By the initial conditions we have that σ(k,n)=Pk(n), for all n2, which ends the proof.

Corollary 10.

Let n2 be integer. The number of all (A,B,2C)-edge colouring of the graph Pn is equal to Pn.

Using the concept of (A,B,kC)-edge colouring of the graph Pn we can obtain the direct formula for the numbers Pk(n) and Qk(n).

Let k1, n2, and 0t[(n-1)/k] be integers and let pk(n,t) be the number of (A,B,kC)-edge colouring of graph Pn, such that C-monochromatic path appears in this colouring exactly t times. In the other words tk edges of Pn have colour C.

Theorem 11.

Let n2 and 0tn-1 be integers. Then p1(n,t)=(n-1t)2n-1-t.

Proof.

Since Pn has n-1 edges, then for t=0,1,2, we obtain that (22)p1(n,0)=2n-1,p1(n,1)=(n-11)2n-2,p1(n,2)=(n-12)2n-3,, and so p1(n,t)=(n-1t)2n-1-t.

Theorem 12.

Let k1, n2, and 0t[(n-1)/k] be integers. Then for all s=0,1,2,,k-1 holds pk(n,t)=pk-s(n-ts,t).

Proof.

Consider the graph Pn with the (A,B,kC)-edge colouring. By contracting the C-monochromatic paths of length k to the C-monochromatic paths of length k-s we can see that the number of (A,B,kC)-edge colouring of graph Pn, such that C-monochromatic path appears in this colouring exactly t times is equal to the number of (A,B,(k-s)C)-edge colouring of graph Pn-ts, such that C-monochromatic path appears in this colouring exactly t times and the Theorem follows.

If s=k-1 or s=k-2 then Theorem 12 gives the following result.

Corollary 13.

Let k1, n2, and 0t[(n-1)/k] be integers. Then (23)pk(n,t)=p1(n-t(k-1),t),(24)pk(n,t)=p2(n-t(k-2),t).

By (23) and Theorem 11 we obtain the direct formula for pk(n,t).

Corollary 14.

Let k1, n2, 0t[(n-1)/k] be integers. Then (25)pk(n,t)=(n-1-(k-1)tt)2n-1-kt.

Moreover Theorem 9 immediately gives that (26)t0pk(n,t)=σ(k,n)=Pk(n).

Using (26) and Corollary 14 we can give the direct formula for the number Pk(n).

Theorem 15.

Let k2 and n1 be integers. Then (27)Pk(n)=i=0[(n-1)/k](n-1-(k-1)ii)2n-1-ki.

If k=2 then we obtain the direct formula for the classical Pell numbers of the form (28)Pn=i=0[(n-1)/2](n-1-ii)2n-1-2i=i=0[(n-1)/2](n2i+1)2i.

From Theorems 3 and 15 follows the direct formula for the sequence Qk(n).

Theorem 16.

Let k2 and n1 be integers. Then (29)Qk(n)=2i=0[(n-1)/k](n-1-(k-1)ii)2n-1-ki+ki=0[(n-k)/k](n-k-(k-1)ii)2n-k-ki.

If k=2 then using the above theorem and after some calculations we obtain that (30)Qn=2(i=0[(n-1)/2](n-1-ii)2n-1-2i0000000+i=0[(n-2)/2](n-2-ii)2n-2-2i).

Now we give the recurrence relation for the number pk(n,k).

Theorem 17.

Let k1, nk+2, and 0t[(n-1)/k] be integers. Then (31)pk(n,t)=2pk(n-1,t)+pk(n-k,t-1).

Proof.

Let V(Pn)={x1,x2,,xn} be the set of vertices of a graph Pn with the numbering in the natural fashion. Then E(Pn)={x1x2,x2x3,,xn-1xn}.

Let pkA(n,t), pkB(n,t), and pkC(n,t) denote the number of (A,B,kC)-edge colouring of the graph Pn, such that C-monochromatic path appears in this colouring exactly t times with A(xn-1xn), B(xn-1xn), and C(xn-1xn), respectively.

It can be easily seen that both pkA(n,t) and pkB(n,t) are equal to the number of all (A,B,kC)-edge colouring of the graph Pn-1, such that C-monochromatic path appears in this colouring exactly t times and pkC(n,t) is equal to the number of all (A,B,kC)-edge colouring of the graph Pn-k, such that C-monochromatic path appears in this colouring exactly t-1 times. In the other words pkA(n,t)=pk(n-1,t), pkB(n,t)=pk(n-1,t), and pkC(k,n)=pk(n-k,t-1). Since (32)pk(n,t)=pkA(n,t)+pkB(n,t)+pkC(n,t), so the result immediately follows.

By (24) and by Theorem 17 we obtain the following.

Corollary 18.

Let k1, nk+2, and 0t[(n-1)/k] be integers. Then (33)pk(n,t)=2p2(n-t(k-2)-1,t)+p2(n-t(k-2)-k,t-1).

Using (26) and Corollary 18 we have the following.

Corollary 19.

Let k2, nk+2, and 0t[(n-1)/k] be integers. Then (34)Pk(n)=t=0[(n-1)/k](2p2(n-t(k-2)-1,t)0000000000+p2(n-t(k-2)-k,t-1)).

Now we consider the (A,B,kC)-edge colouring of the cycle Cn with the numbering of edges in the natural fashion. For explanation for cycle C4 with E(C4)={e1,e2,e3,e4} there are two distinct (A,B,2C)-edge colouring using only colour C. The first (A,B,2C)-edge colouring gives a partition of the set E(C4)={e1,e2}{e3,e4} and the second (A,B,2C)-edge colouring gives a partition E(C4)={e2,e3}{e4,e1}.

Let ρ(k,n) be the number of all (A,B,2C)-edge colouring of the cycle Cn.

Theorem 20.

Let k2 and n3 be integers. The number of all (A,B,kC)-edge colouring of the cycle Cn is equal to Qk(n).

Proof.

Let V(Cn)={x1,x2,,xn} be the set of vertices of a graph Cn with the numbering in the natural fashion. Then E(Cn)={x1x2,x2x3,,xn-1xn,xnx1}.

It is easy to check that ρ(k,n)=Qk(n) for n=3,,k.

Let nk+1 and let ρA(k,n), ρB(k,n), and ρC(k,n) denote the number of (A,B,kC)-edge colouring of the cycle Cn with A(xnx1), B(xnx1), and C(xnx1), respectively.

It can be easily seen that ρA(k,n) and ρB(k,n) are equal to the number of all (A,B,kC)-edge colouring of the graph Pn and ρC(k,n) is equal to the number of all (A,B,kC)-edge colouring of the graph Pn-k+1 multiplied by k. By Theorem 9 we have ρA(k,n)=Pk(n), ρB(k,n)=Pk(n), and ρC(k,n)=kPk(n-k+1). Since (35)ρ(k,n)=ρA(k,n)+ρB(k,n)+ρC(k,n), then we have ρ(k,n)=2Pk(n)+kPk(n-k+1) and so by Theorem 3 we have ρ(k,n)=Qk(n) for all n3.

Corollary 21.

Let n3 be an integer. Then the number of all (A,B,2C)-edge colouring of the cycle Cn is equal to Qn, n3.

4. Concluding Remarks

The interpretation of the generalized Pell numbers with respect to (A,B,kC)-colouring of a 3-edge coloured graph gives a motivation for studying this type of colouring in graphs. For an arbitrary k2 this problem seems to be difficult and more interesting results can be obtained for special value of k (e.g., if we study (A,B,2C)-colouring in graphs). In the class of trees some interesting results can be obtained.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The authors would like to thank the referee for helpful valuable suggestions which resulted in improvements to this paper.

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