# On Third-Order Linear Recurrent Functions

Authors: Kodjo Essonana Magnani
Journal: Discrete Dynamics in Nature and Society
 (2019)

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## Abstract

A functionψ:R→R is said to be a Tribonacci function with period p if ψ(x+3p)=ψ(x+2p)+ψ(x+p)+ψ(x), for all x∈R. In this paper, we present some properties on the Tribonacci functions with period p. We show that if ψ is a Tribonacci function with period p, then limx→∞ψ(x+p)/ψ(x)=β, where β is the root of the equation x3-x2-x-1=0 such that 1<β<2.


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## Body

## 1. Introduction

The most popular numbers studied in many different forms for centuries are Fibonacci numbers. Fibonacci sequence is famous for their amazing properties (see [1–5]). Many research articles talk about these numbers. The third-order linear recurrence of these sequences are what we callTribonacci sequences {Tn} by the recurrence relation Tn+2=Tn+1+Tn+Tn-1 with T0=a, T1=b, T2=c, a,b,c∈N and n as an integer. Han, Kim, and Neggers studied Fibonacci numbers [6–8] and introduced the concept of Fibonacci functions with Fibonacci numbers in [8] which were later extended by B. Sroysang [9] to Fibonacci functions with period p. In the same order as in [8], Parizi and Gordji [10] studied Tribonacci functions. They gave some properties of Tribonacci functions: a function ψ:R→R is said to be a Tribonacci function if ψ(x+3)=ψ(x+2)+ψ(x+1)+ψ(x), for all x∈R. They also showed that if ψ is a Tribonacci function, then limx→∞ψ(x+1)/ψ(x)=β=1/319+3331/3+19-3331/3+1 which is a root of x3-x2-x-1=0.In this paper, for any positive integerp, a function ψ:R→R is said to be a Tribonacci function with period p if ψ(x+3p)=ψ(x+2p)+ψ(x+p)+ψ(x), for all x∈R. In Sections 2 and 3, we present some properties of these functions. In Section 4, we develop the notions of these functions using the concept of even and odd functions discussed in [8, 10]. We also show that if ψ is a Tribonacci function with period p, then limx→∞ψ(x+p)/ψ(x)=β.


## 2. Tribonacci Functions with Periodp

In this section, we present some properties of Tribonacci functions with periodp.Definition 1.
Letp be a positive integer. A function ψ:R→R is said to be a Tribonacci function with period p if ψ(x+3p)=ψ(x+2p)+ψ(x+p)+ψ(x), for all x∈R.Example 2.
Letβ be a positive root of the equation x3-x2-x-1=0. Then, β3=β2+β+1. Define a map φ:R→R by φ(x)=βx/p, where p∈N. Then, φ is a Tribonacci function with period p.Proposition 3.
Letψ:R→R be a Tribonacci function with period p∈N and define ht(x)=ψ(x+t), for all x∈R, where t∈R. Then, ht is also a Tribonacci function with period p.Proof.
Letx∈R. Then,(1)htx+3p=ψx+3p+t=ψx+t+3p=ψx+t+2p+ψx+t+p+ψx+t=ψx+2p+t+ψx+p+t+ψx+t=htx+2p+htx+p+htx.Theorem 4.
Letψ:R→R be a Tribonacci function with period p and let {Tn}, {Tn′}, and {Tn′′} be the sequences of Tribonacci numbers with T1=1,T2=2,T3=3,Tn+3=Tn+2+Tn+1+Tn and T1′=0,T2′=T3′=1,Tn+3′=Tn+2′+Tn+1′+Tn′ and T1′′=T2′′=0T3′′=1,Tn+3′′=Tn+2′′+Tn+1′′+Tn′′. Then, ψ(x+np)=Tn′ψ(x+2p)+Tn-2ψ(x+p)+Tn′′ψ(x) for any x∈R and n≥3 an integer.Proof.
ψ ( x + 3 p ) = ψ ( x + 2 p ) + ψ ( x + p ) + ψ ( x ). The assertion holds for n=3.
So, fixn∈N and assume that the assertion is valid for every s∈{3,…,n+2}. Then,(2)ψx+n+3p=ψx+np+2p+ψx+np+p+ψx+np.=ψx+n+2p+ψx+n+1p+ψx+np.=Tn+2′ψx+2p+Tnψx+p+Tn+2′′ψx+Tn+1′ψx+2p+Tn-1ψx+p+Tn+1′′ψx++Tn′ψx+2p+Tn-2ψx+p+Tn′′ψx.ψx+n+3p=Tn+2′+Tn+1′+Tn′ψx+2p+Tn+Tn-1+Tn-2ψx+p+Tn+2′′+Tn+1′′+Tn′′ψx=Tn+3′ψx+2p+Tn+1ψx+p+Tn+3′′ψx.Corollary 5.
Let{Tn}, {Tn′}, and {Tn′′} be the sequences of Tribonacci numbers with T1=1,T2=2,T3=3,Tn+3=Tn+2+Tn+1+Tn and T1′=0, T2′=T3′=1, Tn+3′=Tn+2′+Tn+1′+Tn′ and T1′′=T2′′=0T3′′=1, Tn+3′′=Tn+2′′+Tn+1′′+Tn′′. Let β be the root of the equation x3-x2-x-1=0 such that 1<β<2. Then, βn=Tn′β2+Tn-2β+Tn′′.Proof.
Letψ(x)=βx/p. We have seen in Example 2 that ψ(x)=βx/p is a Tribonacci function with period p. Applying Theorem 4, we have βx/p+n=Tn′βx/p+2+Tn-2βx/p+1+Tn′′βx/p, for all x∈R. We obtain βn=Tn′β2+Tn-2β+Tn′′.Remark 6.
Consider the Tribonacci sequenses{Tn}, {Tn′}, and {Tn′′} as in Theorem 4. We have the relations Tn=Tn′+Tn+1′ and Tn′′=Tn-1′.


## 3. Odd Tribonacci Functions with Periodp

Here we discuss odd Tribonacci functions with periodp defined as follows.Definition 7.
Letp be a positive integer. A function ψ:R→R is said to be an odd Tribonacci function with period p if ψ(x+3p)=-ψ(x+2p)-ψ(x+p)+ψ(x), for all x∈R.Example 8.
Letψ(x)=αx/p be an odd Tribonacci function with period p∈N, α>0. It is clear that αx/p+3=-αx/p+2-αx/p+1+αx/p, for all x∈R. We have α3=-α2-α+1. Then, α=β1 such that β1 is root of the equation x3+x2+x-1=0, 0<β1<1. Thus, ψ(x)=β1x/p is an odd Tribonacci function with period p on R.Proposition 9.
Letψ:R→R be an odd Tribonacci function with period p∈N and define ht(x)=ψ(x+t), for all x∈R, where t∈R. Then, ht is also an odd Tribonacci function with period p.Proof.
Letx∈R. Then,(3)htx+3p=ψx+3p+t=ψx+t+3p=-ψx+t+2p-ψx+t+p+ψx+t=-ψx+2p+t-ψx+p+t+ψx+t=-htx+2p-htx+p+htx.Theorem 10.
Letψ:R→R be an odd Tribonacci function with period p and let {T-n}, {T-n′}, and {T-n′′} be the sequences of Tribonacci numbers with T0=0,T1=1,T2=2,T-n-1=-T-n-T-n+1+T-n+2 and T0′=T1′=0,T2′=1,T-n-1′=-T-n′-T-n+1′+T-n+2′ and T0′′=1,T1′′=T2′′=0,T-n-1′′=-T-n′′-T-n+1′′+T-n+2′′. Then, ψ(x+np)=T-n+1′ψ(x+2p)+T-nψ(x+p)+T-n+3′′ψ(x) for any x∈R.Proof.
ψ ( x + 3 p ) = - ψ ( x + 2 p ) - ψ ( x + p ) + ψ ( x ). The assertion holds for n=3.
So, fixn∈N and assume that the assertion is valid for every s∈{3,…,n+2}. Then,(4)ψx+n+3p=ψx+np+2p+ψx+np+p+ψx+np.=ψx+n+2p+ψx+n+1p+ψx+np.=T-n-1′ψx+2p+T-n-2ψx+p+T-n+1′′ψx+T-n′ψx+2p+T-n-1ψx+p+T-n+2′′ψx++T-n+1′ψx+2p+T-nψx+p+T-n+3′′ψx.ψx+n+3p=-T-n-1′-T-n′+T-n+1′ψx+2p+-T-n-2-T-n-1+T-nψx+p++-T-n+1′′-T-n+2′′+T-n+3′′ψx=T-n+3+1′ψx+2p+T-n+3ψx+p+T-n+3+3′′ψx.


## 4. Even and Odd Functions with Periodp

In this section, we will talk about the notion of Tribonacci functions using even and odd functions. Here we get results obtained in [10] with third-order linear recurrence. We give the limit of the quotient of a Tribonacci function with period p, extending the results of [11] in third-order linear recurrence.Definition 11.
Letp be a positive integer and λ:R→R a function such that the preimage of 0 by λ has the empty interior. The function λ is said to be an even (resp., odd) function with period p if λ(x+p)=λ(x) (resp., λ(x+p)=-λ(x)), for all x∈R.Example 12.
Ifλ(x)=x-⌊x⌋, then λ(x)h(x)≡0 implies that h(x)≡0 if x∉Z. Due to the fact that R∖Z is dense in R and h is continuous, it follows that h=0. Let p∈N and x∈R. Then, λ(x+p) = x+p-⌊x+p⌋ = x+p-⌊x⌋-p = x-⌊x⌋=λ(x). Hence, λ is an even function with period p.Theorem 13.
Letp∈N and λ:R→R be an even function with period p and let v:R→R be a continuous function. Then, v is a (resp., an odd) Tribonacci function with period p if and only if λv is a (resp., an odd) Tribonacci function with period p.Proof.
Letv be a Tribonacci function with period p. For any x∈R, we have(5)λvx+3p=λx+3pvx+3p=λx+2pvx+3p=λx+2pvx+2p+vx+p+vx=λx+2pvx+2p+λx+pvx+p+λxvx=λvx+2p+λvx+p+λvx.Hence, λv is a Tribonacci function with period p.
Now assume thatλv is a Tribonacci function with period p. For x∈R, we have(6)λx+pvx+3p=λx+2pvx+3p=λx+3pvx+3p=λvx+3p=λvx+2p+λvx+p+λvx=λx+2pvx+2p+λx+pvx+p+λxvx=λx+pvx+2p+λx+pvx+p+λx+pvx=λx+pvx+2p+vx+p+vx.For all x∈R, we get v(x+3p)=v(x+2p)+v(x+p)+v(x). Hence, v is a Tribonacci function with period p.We give the proof for the case wherev is a Tribonacci function with period p. The case where v is an odd Tribonacci function with period p is similar and left to the reader.Now we give the limit of the quotient of a Tribonacci function with periodp.Theorem 14.
Letp∈N and ψ:R→R be a Tribonacci function with period p. Then, the limit of the quotient ψ(x+p)/ψ(x) is β, the root of the equation x3-x2-x-1=0 such that 1<β<2.Proof.
Givenx∈R, there exists y∈R and n∈N such that x=y+np. Let us set u=ψ(y),v=ψ(y+p) and w=ψ(y+2p). By Theorem 4, we have ψ(y+np)=Tn′ψ(y+2p)+Tn-2ψ(y+p)+Tn′′ψ(y) and hence ψ(y+np)=Tn′w+Tn-2v+Tn′′u. Hence,(7)ψx+pψx=ψy+n+1pψy+np=Tn+1′ψy+2p+Tn-1ψy+p+Tn+1′′ψyTn′ψy+2p+Tn-2ψy+p+Tn′′ψy=Tn+1′w+Tn-1v+Tn+1′′uTn′w+Tn-2v+Tn′′u.By Remark 6, we have(8)ψx+pψx=Tn+1′w+Tn-1′+Tn′v+Tn′uTn′w+Tn-2′+Tn-1′v+Tn-1′u=Tn′Tn-1′Tn+1′/Tn′w+Tn-1′/Tn′+1v+uTn′/Tn-1′w+Tn-2′/Tn-1′+1v+u.
Now, the well-known result thatlimn→+∞Tn+1′/Tn′=β yields(9)limx→+∞⁡ψx+pψx=ββw+1/β+1v+uβw+1/β+1v+u.Hence, we obtain(10)limx→+∞⁡ψx+pψx=β.



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