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 n≥2 with F0=F1=1. Among sequences of the Fibonacci type there is the Pell sequence defined by Pn=2Pn-1+Pn-2 for n≥2 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 n≥2. 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 [1].
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 [1].
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 [1].
Interesting generalizations of the numbers of the Fibonacci type (also generalizations of the Pell sequences) are studied, for instance, by Kilic in [2–4]. Among others Kilic in [4] introduced the generalization of the Pell numbers. He defined the generalized Pell (p,i)-numbers for any given p, where p≥1 and for n>p+1 and 0≤i≤p 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 [4] 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, [5–9].
Let k≥2, n≥0 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)forn≥k,
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
Pn
0
1
2
5
12
29
70
169
408
985
2378
5741
13860
33461
P3(n)
0
1
2
4
9
20
44
97
214
472
1041
2296
5064
11169
P4(n)
0
1
2
4
8
17
36
76
160
337
710
1496
3152
6641
P5(n)
0
1
2
4
8
16
33
68
140
288
592
1217
2502
5144
P6(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 k≥2 and n≥1 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 n≥k+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 n≥1.
In the same way we can prove the generalization of the result (ii).
Theorem 2.
Let k≥2 and n≥1 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 k≥2, n≥0 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)forn≥k,
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
Qn
2
2
6
14
34
82
198
478
1154
2786
6726
16238
Q3(n)
3
2
4
11
24
52
115
254
560
1235
2724
6008
Q4(n)
4
2
4
8
20
42
88
184
388
818
1724
3632
Q5(n)
3
2
4
8
16
37
76
156
320
656
1349
2774
Q6(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 k≥2 and n≥k-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 n≥k. 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 k≥2, m≥1, and n≥0 be integers. Then
(7)Pk(n)+2∑i=1mPk(n+ki-1)=Pk(n+km),(8)Qk(n)+2∑i=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 m≥2. Assume that formula (7) is true for an arbitrary m. We will prove that Pk(n)+2∑i=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)+2∑i=1m+1Pk(n+ik-1)=Pk(n)+2∑i=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 k≥2 and m≥1 be integers. Then
(10)2∑i=1mPk(ki-1)=Pk(mk),2∑i=1mPk(ki)=Pk(mk+1)-1,2∑i=1mQk(ki-1)=Qk(mk)-k,2∑i=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)2∑i=1mP2i-1=P2m,2∑i=1mP2i=P2m+1-1,2∑i=1mQ2i-1=Q2m-2,2∑i=1mQ2i=Q2m+1-2.
Theorem 6.
Let k≥2 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 1≤m≤k-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 k≥2. 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 k≥2 and n≥2k-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 n≥2k. Assume that formula (16) is true for an arbitrary t where 2k≤t≤n. 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,n≥3.
For concepts not defined here see [10]. 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 [11] 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(Pn∘K1)=Pn+1, where G∘H 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 xy∈E(G) notation M(xy) means that the edge xy has the colour M.
Let k≥1 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 q≥0 is an integer.
Theorem 9.
Let k≥2 and n≥2 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 n≥k+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 n≥2, which ends the proof.
Corollary 10.
Let n≥2 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 k≥1, n≥2, and 0≤t≤[(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 n≥2 and 0≤t≤n-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 k≥1, n≥2, and 0≤t≤[(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 k≥1, n≥2, and 0≤t≤[(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 k≥1, n≥2, 0≤t≤[(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)∑t≥0pk(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 k≥2 and n≥1 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 k≥2 and n≥1 be integers. Then
(29)Qk(n)=2∑i=0[(n-1)/k](n-1-(k-1)ii)2n-1-ki+k∑i=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 k≥1, n≥k+2, and 0≤t≤[(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 k≥1, n≥k+2, and 0≤t≤[(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 k≥2, n≥k+2, and 0≤t≤[(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 k≥2 and n≥3 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 n≥k+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 n≥3.
Corollary 21.
Let n≥3 be an integer. Then the number of all (A,B,2C)-edge colouring of the cycle Cn is equal to Qn, n≥3.
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 k≥2 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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