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Articles

Some identities for generalized Fibonacci and Lucas numbers

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Abstract

In this paper we study one parameter generalization of the Fibonacci numbers, Lucas numbers which generalizes the Jacobsthal numbers, Jacobsthal–Lucas numbers simultaneously. We present some their properties and interpretations also in graphs. Presented results generalize well-known results for Fibonacci numbers, Lucas numbers, Jacobsthal numbers and Jacobsthal–Lucas numbers.

1 Introduction and preliminary results

The Fibonacci numbers Fn are defined by the recurrence relation Fn=Fn1+Fn2, for n2 with F0=0, F1=1.

The nth Lucas number Ln is defined recursively by Ln=Ln1+Ln2 for n2 with initial terms L0=2, L1=1.

Apart from the Fibonacci numbers and the Lucas numbers the well-known are the Jacobsthal numbers and the Jacobsthal–Lucas numbers.

For an integer n0 the nth Jacobsthal number Jn is defined recursively by Jn=Jn1+2Jn2, for n2 with J0=0,J1=1. The nth Jacobsthal–Lucas number jn is defined by jn=jn1+2jn2, for n2 with j0=2,j1=1.

Let us consider one parameter generalization of the Fibonacci numbers.

Let n0, t1 be integers. The nth generalized Fibonacci number Jt,n is defined recursively as follows (1) Jt,n=Jt,n1+tJt,n2,(1) for n2 with initial conditions Jt,0=0 and Jt,1=1.

It is interesting to note that Jt,n generalizes the Fibonacci numbers and the Jacobsthal numbers. If t=1 then J1,n=Fn and for t=2 holds J2,n=Jn. For these reasons numbers Jt,n also are named as generalized Jacobsthal numbers.

In the same way we can define the generalized Lucas numbers, which are a generalization of Lucas numbers and Jacobsthal–Lucas numbers.

Let n0, t1 be integers. The nth generalized Lucas number jt,n is defined recursively as follows (2) jt,n=jt,n1+tjt,n2,(2) for n2 with initial conditions jt,0=2 and jt,1=1.

If t=1 then j1,n=Ln and for t=2 holds j2,n=jn.

Since the characteristic equation of relations (1) and (2) is r2rt=0, so roots of it are α=1+1+4t2, β=11+4t2. Consequently for n0 the direct formulas (named also as Binet formulas) for Jt,n and jt,n have the forms (3) Jt,n=11+4t1+1+4t2n11+4t2n,(3) (4) jt,n=1+1+4t2n+11+4t2n.(4)

The Fibonacci numbers have many interpretations also in graph theory, see Citation[1–6]. It was shown that the generalized Fibonacci numbers Jt,n are related to the Merrifield–Simmons index of a special graph product, see for details Citation[7]. In this paper we show another graph interpretation of Jt,n.

Only undirected, connected simple graphs are considered. A subset SV(G) is independent if for each x,yS, there is no edge between them. Moreover the empty set and every subset containing exactly one vertex are independent. Let n1 be an integer. Let us consider n copies of edgeless graph Nt of order t, t1, denoted by Nti, with V(Nti)={x1i,,xti} for i=1,,n. Then Gn is a graph such that V(Gn)=i=1nV(Nti) and E(Gn)=i=1n{xjixki+1;1in,1j,kt}. Let σ(Gn) be the number of all independent sets S of Gn such that |SV(Nti)|1 for i=1,,n.

Theorem 1

Let n1 , t1 be integers. Then (5) σ(Gn)=Jt,n+2.(5)

Proof

By Induction on n Let n,t be as in the statement of the theorem. If n=1,2 then the result is obvious. Suppose that n3 and let σ(Gk)=Jt,k+2 for k<n. We shall show that σ(Gn)=Jt,n+2. Let SV(Gn) be an arbitrary independent set such that |SV(Nti)|1 for i=1,,n. We consider the following cases.

1.

SV(Ntn)=. Then S is an arbitrary independent set of a graph Gn1 with the condition |SV(Ntj)|1 for j=1,,n1. By the induction hypothesis there are σ(Gn1)=Jt,n+1 subsets S in this case.

2.

SV(Ntn). Clearly |SV(Ntn)|=1 and by the definition of the graph Gn holds SV(Ntn1)=. Since the unique vertex belonging to SV(Ntn) can be chosen in t ways and by induction hypothesis there are tσ(Gn2)=tJt,n subsets S in this case.

Finally σ(Gn)=σ(Gn1)+tσ(Gn2)=Jt,n+1+tJt,n=Jt,n+2 which ends the proof. □

2 Identities for generalized Fibonacci and Lucas numbers

In this section we give some properties of generalized Fibonacci numbers and generalized Lucas numbers.

Theorem 2

Let n0 , t1 be integers. Then (6) Jt,n+2+tJt,n=jt,n+1.(6)

Proof

By Induction on n For n=0 we have Jt,2+tJt,0=1+t0=1=jt,1.

Let k0 be given and suppose that (6) is true for all n=0,1,2,,k. We shall show that (6) holds for n=k+1. Using induction’s assumption for n=k and n=k1 and (1), (2) we have Jt,(k+1)+2+tJt,k+1==Jt,k+3+tJt,k+1==(Jt,k+2+tJt,k+1)+tJt,k+tJt,k1==(Jt,k+2+tJt,k)+tJt,k+1+tJt,k1==jt,k+1+tjt,k=jt,k+2=jt,(k+1)+1.

Thus, (6) holds for n=k+1, and the proof of the induction step is complete. □

Theorem 3

Let n0 , t1 be integers. Then (7) Jt,n+jt,n=2Jt,n+1.(7)

Proof

By Induction on n For n=0 we have Jt,0+jt,0=0+2=21=2Jt,1.

Let k0 be given and suppose that (7) is true for all n=0,1,2,,k. We shall show that (7) holds for n=k+1. Using induction’s assumption for n=k and n=k1 and (1), (2) we have Jt,k+1+jt,k+1==(Jt,k+tJt,k1)+(jt,k+tjt,k1)==(Jt,k+jt,k)+t(Jt,k1+jt,k1)==2Jt,k+1+t2Jt,k==2Jt,k+1+tJt,k=2Jt,k+2=2Jt,(k+1)+1.

Thus, (7) holds for n=k+1, and the proof of the induction step is complete. □

Theorem 4

Let n0 , t1 be integers. Then (8) i=0nJt,i=Jt,n+2Jt,1t.(8)

Proof

Using (1) we have Jt,n=1t(Jt,n+2Jt,n+1).

For integers 0,1,,n we obtain Jt,0=1t(Jt,2Jt,1)Jt,1=1t(Jt,3Jt,2)Jt,2=1t(Jt,4Jt,3)Jt,n1=1t(Jt,n+1Jt,n)Jt,n=1t(Jt,n+2Jt,n+1).

Adding these equalities we obtain (8). □

In the same way one can easily prove the next theorem.

Theorem 5

Let n0 , t1 be integers. Then (9) i=0njt,i=jt,n+2jt,1t.(9)

From the above theorems we obtain the well-known identities for Fibonacci numbers, Lucas numbers, Jacobsthal numbers and Jacobsthal–Lucas numbers.

Corollary 6

Let n0 be an integer. Then i=0nFi=Fn+2F1=Fn+21,i=0nLi=Ln+2L1=Ln+21,i=0nJi=Jn+2J12=Jn+212,i=0nji=jn+2j12=jn+212.

Some identities for numbers Jt,n and jt,n can be found using their matrix generators.

For integers n1 and t1 let J(t,n)=Jt,nJt,n1Jt,n+1Jt,nbe a matrix with entries being generalized Fibonacci numbers.

Theorem 7

Let n1 , t1 be integers. Then Jt,nJt,n1Jt,n+1Jt,n=Jt,1Jt,0Jt,2Jt,111t0n1.

Proof

By Induction on n If n=1 then the result is obvious. Assume that Jt,nJt,n1Jt,n+1Jt,n=Jt,1Jt,0Jt,2Jt,111t0n1.We shall show that Jt,n+1Jt,nJt,n+2Jt,n+1=Jt,1Jt,0Jt,2Jt,111t0n.By simple calculation using induction’s hypothesis we have Jt,1Jt,0Jt,2Jt,111t0n111t0=Jt,nJt,n1Jt,n+1Jt,n11t0= =Jt,n+tJt,n1Jt,nJt,n+1+tJt,nJt,n+1=Jt,n+1Jt,nJt,n+2Jt,n+1,which ends the proof. □

This generator immediately gives the Cassini formula for the generalized Fibonacci numbers.

Corollary 8

Let n1 , t1 be integers. Then Jt,n2Jt,n1Jt,n+1=Jt,12Jt,0Jt,2(t)n1=(t)n1.

If t=1 and t=2 then we obtain the well-known Cassini formulas for the Fibonacci numbers and the Jacobsthal numbers, respectively.

Corollary 9

Let n1 be an integer. Then Fn2Fn1Fn+1=(1)n1Jn2Jn1Jn+1=(2)n1.

Analogously we can define the matrix generator and the Cassini formula for the generalized Lucas numbers.

For integers n1 and t1 let j(t,n)=jt,njt,n1jt,n+1jt,nbe a matrix with entries being generalized Lucas numbers.

Theorem 10

Let n1 , t1 be integers. Then jt,njt,n1jt,n+1jt,n=jt,1jt,0jt,2jt,111t0n1.

Corollary 11

Let n1 , t1 be integers. Then jt,n2jt,n1jt,n+1=(14t)(t)n1.

If t=1 and t=2 then we have the well-known Cassini formulas for the Lucas numbers and the Jacobsthal–Lucas numbers, respectively.

Corollary 12

Let n1 be an integer. Then Ln2Ln1Ln+1=5(1)n1,jn2jn1jn+1=9(2)n1.

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