# On New Polynomial Sequences Constructed to Each Vertex in an n-Gon

Authors: Abdul Hamid Ganie; Engin ŏzkan; Mine Uysal; Afroza Akhter
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

In this work, we bring to light the properties of newly formed polynomial sequences at each vertex of Pell polynomial sequences placed clockwise at each vertex in the n-gon. We compute the relation among the polynomials with such vertices. Moreover, in an n-gon, we generate a recurrence relation for a sequence giving themth term formed at the kth. Similar to the situations we are talking about, we applied to Pell-Lucas polynomials, Jacobsthal polynomials, and Jacobsthal-Lucas polynomials and have obtained new relations recurrence.


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## Body

## 1. Introduction

It is an interesting tool for many researchers to know about number sequences, especially thrust is given to Fibonacci numbers. In this direction, many researchers work on it and study its wings, and the Lucas numbers became a vital role for them to study as can be seen in [1–8]. While framing these sequences, other sequences came into existence such as Pell, Pell-Lucas, Jacobsthal and Jacobsthal-Lucas by replacing the initial conditions [9–15].As in [16–23], in engineering, science, and technology, the structure of such sequences with generalizations plays a vital significance art and architecture. In studying such discipline, Pell, Pell-Lucas, Jacobsthal, and Jacobsthal-Lucas polynomials have shown their significance as in [24–27].Recently, the study of triangles with coordinates of vertices have been studied in [14, 19].We have made a new study for polynomials of Pell, Pell-Lucas, Jacobsthal, and Jacobsthal by reference to [19] articles in this study.The motivation of this work is to study the clockwise nature of the Pell, Pell-Lucas, Jacobsthal, and Jacobsthal-Lucas polynomials to be located on the vertices of the polygons.We have computed some interesting relations in n-gon concerning the vertexBk corresponding to mth term for 0≤k<n. Also, a relation corresponding to a vertex among the numbers will be determined.


## 2. Materials and Methods

Definition 1.
The polynomials of the given by linear recurrence relation(1)Pn+2x=2xPn+1x+Pnx,with P0x=0 and P1x=1 called as Pell polynomials.
The negative Pell polynomials are defined by the recurrence relation,(2)P−nx=−1n+1Pnx.
The first few terms of this sequence are1,2x,4x2+1,8x3+4x,16x4+12x2+1,32x5+32x3+6x,….
For Fibonacci sequence, the characteristic equation is(3)r2−2xr−1=0.
The roots of this equation areαx and βv.(4)αx=x+x2+1andβx=x−x2+1.
The Binet formula of this sequence is(5)Pnx=αnx−βnxαx−βx.Definition 2.
The linear reoccurrence relation for Pell-Lucas polynomials is(6)Qn+2x=2xQn+1x+Qnx,with Q0x=2 and Q1x=2x.
The first few terms of this sequence are2,2x,4x2+2,8x3+6x,16x4+16x2+2,32x5+40x3+10x,….
The negative Pell-Lucas polynomials are defined by the recurrence relation,(7)Q−nx=−1nQnx.
The Binet formula of this sequence is(8)Qnx=αnx+βnx.Definition 3.
The recurrence relation for Jacobsthal polynomials is(9)Jnx=Jn−1x+2xJn−2x,with J0x=0 and J1x=1.
Note that(10)J−nx=−−1n2x−nJnx.
The first few terms of this sequence are0,1,1,1+2x,1+4x,1+6x+4x2,1+8x+12x2,….
The characteristic equation for the Jacobsthal polynomials sequence is(11)k2−k−2x=0.
The roots of this equation areγx and δx,(12)γx=1+1+8x2andδx=1−1+8x2.
The Binet formula of this sequence is(13)Jnx=γnx−δnxγx−δx.Definition 4.
The recurrence relation for Jacobsthal-Lucas polynomials is(14)jnx=Jn−1x+2xJn−2x,with j0x=2 and j1x=1.
Note that(15)j−nx=−1n2x−njnx.
The first few terms of this sequence are2,1,1+4x,1+8x+8x2,1+10x+20x2,….
The Binet formula of this sequence is(16)jnx=γnx+δnx.


## 3. Results and Discussion

### 3.1. New Recurrence Relations for the Pell and Pell-Lucas Numbers

From here onwards, we shall consider a continuously clockwise sense approach to the polygon of the Pell polynomials and Pell-Lucas polynomials. We will continue to n-gon starting this placement from a point. For such a polygon, let us call each vertex point asB0,B1,B2,… for k=0,1,2,…. Here, we aim at finding a relationship that locates any term of new Bk sequences to every vertex of the polygon. We establish such a relationship in Theorem 3.1.1 to Pell polynomials and in Theorem 3.1.3 to Pell-Lucas polynomials.Let the vertex number be represented byk and m represents the order of any term of the new sequence occurring at any vertex in any n-gon, with 0≤k<n and m≥1.Takingn=1, the Pell polynomials in succession over a dot are given, as depicted in Figure 1.Figure 1
Pell polynomials placed over a dot.Forn=1 yields k=0 and giving a recurrence relation,(17)Pm−1n+kx=Pm−1x=2xPm−2x+Pm−3x,with P0x=0 and P1x=1 of the form of Pell polynomials.Forn=2, choose the Pell polynomials in succession at the endpoints of a line, as depicted in Figure 2.Figure 2
Pell polynomials kept at the endpoints of a line.So, it determines a recurrence relation of the form(18)Pm−1n+kx=Pm−12+kx=4x2+2Pm−22+kx−Pm−32+kx.Forn=3, the vertices of the triangle will be labeled by Pell polynomials clockwise, as given in Figure 3.Figure 3
Pell polynomials kept in clockwise to vertices of a triangle.So, it determines a recurrence relation of the form(19)Pm−1n+kx=Pm−13+kx=8x3+6xPm−23+kx+Pm−33+kx.Now, forn=4, the Pell numbers are taken clockwise on the vertices of a tetragon, as given in Figure 4.Figure 4
On tetragon representing Pell polynomials which are kept clockwise at its vertices.So, it determines a recurrence relation of the form(20)Pm−1n+kx=Pm−14+kx=16x4+16x2+2Pm−24+kx−Pm−34+kx.Similarly, for an n-gon, we take clockwise the Pell numbers on its vertices, as given in Figure5.Figure 5
Onn-gon representation of Pell polynomials kept clockwise at its vertices.For example, in a new sequence to determine the second term of it occurring at the vertexB1 in a 3-gon.Thus, the value forn=3, m=3, and k=1 corresponds to the 3rd term of B1=1,8x3+4x,64x6+80x4+24x2+1,… sequence in Figure 3.This term must be64x6+80x4+24x2+1.Now, to determine this value from the recurrence relation withn=3 from(21)Pm−1n+kx=Pm−13+kx=8x3+6xPm−23+kx+Pm−33+kx,so that(22)P3−13+1=8x3+6xP3−23+1x+P3−33+1x,P7x=8x3+6xP4x+P1x,64x6+80x4+24x2+1=8x3+6x8x3+4x+1=64x6+32x4+48x4+24x2+1.In Figure5, we define a new sequence created at the vertex Bk.Bk=x1,x2,…=Pk,Pn+k,P2n+k,P3n+k,…. As Pk=P0n+k is the first term to this sequence, the second term is Pn+kx=P1n+kx and the third term is P2n+kx, and if it continues like this, it is observed that the mth term will be Pm−1n+kx of the new sequence formed in an n-gon.In addition, the coefficient ofPm−2n+kx is Pell-Lucas polynomials Qnx and −1n is the coefficient of Pm−3n+kx.So, we have the following result.Theorem 1.
For0≤k<n, the relation(23)Pm−1n+kx=QnxPm−2n+kx−−1nPm−3n+kxis satisfied, giving mth term of the sequence with respect to vertex Bk.Proof.
Using Binet formulas of Pell and Pell-Lucas polynomials, we see(24)QnxPm−2n+kx−−1nPm−3n+kx=αnx+βnxαm−2n+kx−βm−2n+kxαx−βx−−1nαm−3n+kx−βm−3n+kxαx−βx=αm−1n+kx−βm−1n+kxαx−βx−αnxβm−2n+kxαx−βx+βnxαm−2n+kxαx−βx−−1n.αm−3n+kxαx−βx+−1n.βm−3n+kxαx−βx=αm−1n+kx−βm−1n+kxαx−βx+αm−3n+kxβnxαnx−−1nαx−βx−βm−3n+kxαnxβnx−−1nαx−βx.
Because ofαxβx=−1, we have(25)=αm−1n+kx−βm−1n+kxαx−βx+αm−3n+kx−1n−−1nαx−βx−βm−3n+kx−1n−−1nα−β=αm−1n+k−βm−1n+kα−β=Pm−1n+kx.
Let us indicate the coefficients corresponding to this sequence in Table1.
Table1 clearly indicates that the first components are Pell-Lucas polynomials, 2x,4x2+1,8x3+6x,16x4+16x2+2,32x5+40x3+10x,64x6+96x4+36x2+2,128x7+224x5+112x3+14x,….
Table1 clearly indicates that the second are {1, −1, 1, −1, 1, −1, 1, −1, …}. {1, −1, 1, −1, 1, −1, 1, −1, …} is numbered as sequences A033999 in the On-Line Encyclopedia of Integer Sequences [25].Table 1
The coefficients.
DotSegmentTriangleTetragonPentagonHexagonHeptagon(2x, 1)(4x2+1,− 1)(8x3+6x, 1)(16x4+16x2+2, −1)(32x5+40x3+10x, 1)(64x6+96x4+36x2+2, −1)128x7+224x5+112x3+14xTheorem 2.
The relation for0≤k<n between the numbers with respect to the vertex Bk is(26)Pmn+kx=Pn+kx∑t=02m−3−−1m/4−1n+1tm−v−1vQnm−2t−1x+Pkx∑t=02m−5−−1m−1/4−1n+1t+1m−v−2vQnm−2t−2x.Proof.
To establish this result, we induct onm.
Thus, whenm=1, we see(27)Pn+kx=Pn+kx∑v=00−1n+1v−vvQn−2vx+Pkx∑v=0−1−1n+1v+1−v−1vQn−2v−2x=Pn+kx−1000Qn0+Pnx−1n+1−10Qn−2x=Pn+kx.
Form=s, we assume the claim is true, that is,(28)Psn+kx=Pn+kx∑v=02s−3−−1s4−1n+1vs−v−1vQns−2v−1x+Pkx∑v=02s−5−−1s−14−1n+1v+1s−v−2vQns−2v−2x.
Form=s+1, let us show it is also true.
For this, if we use (23), then we see(29)Ps+1n+kx=Psn+kxQnx−−1nPs−1n+kx.
By using (28), we have(30)Ps+1n+k=Pn+kx∑v=02m−3−−1s/4−1n+1vs−v−1vQns−2v−1xQnx+Pkx∑v=02m−5−−1s−1/4−1n+1v+1s−v−2vQns−2v−2xQnx−−1nPn+kx∑v=02m−5−−1s−1/4−1n+1vs−v−2vQns−2v−2x−−1nPkx∑v=02m−7−−1s−2/4−1n+1v+1s−v−3vQns−2v−3x.
Whens=2r, we have(31)Ps+1n+k=Pn+kx∑v=0r−1−1n+1v2r−v−1vQn2r−2vx+Pn+kx∑v=0r−1−1n+1v+12r−v−2vQn2r−2v−2x+Pkx∑v=0r−1−1n+1v+12r−v−2vQn2r−2v−1x+Pkx∑v=0r−2−1n+1v+22r−v−3vQn2r−2v−3x=Pn+kx2r−10Qn2rx+−1n+12r−21Qn2r−2x+2r−32Qn2r−4x+…+−1n+1r−1rr−1Qn2x−1n+12r−20Qn2r−2x+−1n+122r−31Qn2r−4x+…+−1n+1rr−1r−1+Pk−1n+12r−20Qn2r−2x+−12n+12r−31Qn2r−3x+…+−1n+1rr−1r−1Qnx+−12n+12r−30Qn2r−3x+−13n+12r−41Qn2r−5x+…+−1n+1rr−1r−2Qnx=Pn+kxQn2rx+−1n+12r−10Qn2r−2x+−12n+12r−22Qn2r−4x+…+−1n+1r−1r+1r−2Qn2x+−1n+1r+Pk−1n+1Qn2r−1x+−12n+12r−20Qn2r−3x+…+−1n+1rrr−1Qnx=Pn+kx∑v=0r−1n+1v2r−vvQn2r−2vx+Pkx∑v=0r−1−1n+1v+12r−v−1vQn2r−2v−1x=Pn+k∑v=02s−1−−1s+14−1n+1vs−vvQns−2vx+Pk∑v=02s−3−−1s4−1n+1v+1s−v−1vQns−2v−1x.
The equality has been shown fors=2r.
Now, whens=2r−1, we get(32)=Pn+kx∑v=0r−1−1n+1v2r−v−2vQn2r−2v−1x+Pn+kx∑t=0r−2−1n+1v+12r−v−3vQn2r−2v−3x+Pk∑v=0r−2−1n+1v+12r−v−3vQn2r−2v−2x+Pk∑v=0r−2−1n+1v+22r−v−4vQn2r−2v−4x=Pn+kx2r−20Qn2r−1x+−1n+12r−31Qn2r−3x+−12n+12r−42Qn2r−5x+…+−1n+1r−1r−1r−1Qnx+−1n+12r−30Qn2r−3x+−12n+12r−41Qn2r−5x+…+−1n+1r−1r−1r−2Qnx+Pk−1n+12r−30Qn2r−2x+−12n+12r−41Qn2r−4x+−13n+12r−52Qn2r−6x+…+−1n+1r−1r−1r−1Qn2x+−12n+12r−40Qn2r−4x−13n+12r−51Qn2r−6x+…+−1n+1rr−1r−1=Pn+kxQn2r−1x+−1n+12r−21Qn2r−3x+−12n+12r−32Qn2r−5x+…+−1n+1r−1rr−2Qnx+Pk−1n+1Qn2r−2+−12n+12r−31Qn2r−4x+−13n+12r−42Qn2r−6x+…+−1n+1r=Pn+kx∑v=0r−1−1n+1t2r−v−1vQn2r−2v−1x+Pkx∑v=0r−1−1n+1v+12r−v−2vQn2r−2v−2x=Pn+kx∑v=02s−1−−1s+14−1n+1vs−vvQns−2vx+Pkx∑v=023−3−−1s4−1n+1v+1s−v−1vQns−2v−1.
Thus, the proof is obtained.
Let us give example as an application. Forn=3,k=2, and m=2, we have(33)P8x=P5x∑v=00−14v1−vvQ31−2vx+P2x∑v=00−14v+1−vvQ3−2vx,P8x=P5xQ3x+P2x1,128x7+192x5+80x3+8x=16x4+12x2+18x3+6x+2x=128x7+96x5+96x5+72x3+8x3+6x+2x=128x7+192x5+80x3+8x.
Now, let us place the Pell-Lucas polynomials like the Pell polynomials continuously clockwise at the vertices of a regular polygon.
For this new sequence, themth term with respect to the vertex Bk of an n-gon is Qm−1n+kx.
So, we have the following theorem about this situation.Theorem 3.
If on the vertices of an n-gon, the Pell-Lucas polynomials are placed clockwise, then for0≤k<n, the mth term of sequence with respect to the vertex Bk is(34)Qm−1n+kx=Qm−2n+kxQnx−−1nQm−3n+kx.Proof.
To prove the result, by employing Binet formula for Pell-Lucas polynomials, we see(35)Qm−2n+kxQnx−−1nQm−3n+kx=αm−2n+kx+βm−2n+kxαnx+βnx−−1nαm−3n+kx+βm−3n+kx=αm−1n+kx+βm−1n+kx+αm−2n+kxβnx+αnxβm−2n+kx−−1nαm−3n+kx−−1nβm−3n+kx=αm−1n+k+βm−1n+k−αm−3n+kx−1n−αnxβnx−βm−3n+kx−1n−αnxβnx.
Because ofαβ=−1, we have(36)=αm−1n+kx+βm−1n+kx,=Qm−1n+kx.Theorem 4.
For an n-gon, if Pell-Lucas polynomials are kept on the vertices clockwise sense. Then, for0≤k<n, there exists a relation of the form(37)Qmn+kx=Qn+kx∑v=02m−3−−1m/4−1n+1vm−v−1vQnm−2v−1x+Qkx∑v=02m−5−−1m−1/4−1n+1v+1m−v−2vQnm−2v−2x,between the numbers with respect to the vertex Bk.Proof.
The result can be proved in a similar fashion as Theorem 3.1.2.
For example, whenn=4,k=1 and m=2, we have(38)Q9x=Q5x∑v=00−15v1−vvQ41−2vx+Q1x∑v=00−15v+1−vvQ4−2vxQ9x=Q5xQ4x−Q1x512x9+1152x7+864x5+240x3+18x=32x5+40x3+10x16x4+16x2+2−2x=512x9+512x7+64x5+640x7+640x5+80x3+160x5+160x3+20x−2x=512x9+1152x7+864x5+240x3+18x.


### 3.2. Recurrences Notion Corresponding to Jacobsthal and Jacobsthal-Lucas Polynomials

In this direction, we involve the clockwise sense of the Jacobsthal polynomials on the point, at extreme points, at vertices of the triangle, and lastly at the vertices of an n-gon.Since for this new sequence, the mth term computed at vertexBk is Jm−1n+kx for an n-gon. Also, the coefficient of Jm−2n+kx is Jacobsthal-Lucas polynomials jnx and the coefficient of Jm−3n+kx is −2xn.In this case, we have the following result.Theorem 5.
On an n-gon, suppose Jacobsthal numbers are kept on its vertices clockwise. Then, for0≤k<n, we have(39)Jm−1n+kx=jnxJm−2n+kx−−2xnJm−3n+kx,and it determines the mth term of the sequence with respect to vertex Bk.Proof.
By using the Binet formulas of Jacobsthal and Jacobsthal-Lucas polynomials, we have the following:(40)jnxJm−2n+kx−−2xnJm−3n+kx=γnx+δnxγm−2n+kx−δm−2n+kxγx−δx−−2xnγm−3n+kx−δm−3n+kxγx−δx=γm−1n+kx−δm−1n+kxγx−zx+δnxγm−2n+kx−γnxδm−2n+kxγx−δx−−2xnγm−3n+kxγx−δx+−2xnδm−3n+kxγx−δx=γm−1n+kx−δm−1n+kxγx−δx+αm−3n+kxδnxγnx−−2xnγx−δx−γm−3n+kxγnxδnx−−2xnγx−δx.
Sinceγxδx=−2x, γnxδnx−−2xn=0, thus(41)jnxJm−2n+kx−−2xnJm−3n+kx=Jm−1n+kx.
In Table2, we write the coefficients concerning the above sequences.
So, clearly in Table2, the Jacobsthal-Lucas polynomials 1,1+4x,1+6x,1+8x+8x2,1+10x+20x2,1+12x+36x2+16x3,1+14x+56x2+56x3 are first components.
The second components in Table2 are {2x, −2x2,2x3,−2x4, 25,−2x6,2x7, …}. {1, −1, 1, −1, 1, −1, 1, −1, …} is enumerate as sequences A084633 in the On-Line Encyclopedia of Integer Sequences [25].Table 2
The coefficients.
DotSegmentTriangleTetragonPentagonHexagonHeptagon1,2x1+4x,−4x21+6x,8x31+8x+8x2,−16x41+10x+20x2,32x51+12x+36x2+16x3,−64x61+14x+56x2+56x3,128x7Theorem 6.
On an n-gon, let the Pell-Lucas polynomials be kept on its vertices clockwise. Then, for0≤k<n, there exists a relation corresponding to numbers with respect to the vertex Bk as(42)Jmn+kx=Jn+kx∑t=02m−3−−1m4−1n+1t2xntm−t−1tjnm−2t−1x+Jkx∑t=02m−5−−1m−14−1n+1t+12xnt+1m−t−2tjnm−2t−2x.Proof.
As in Theorem2, the proof can be performed.
Now, for ann-gon, taking the Jacobsthal-Lucas polynomials on the vertices clockwise and in this n-gon, the mth term formed to this new sequence at the vertex Bk is jm−1n+kx.
We have the following result in this direction.Theorem 7.
On an n-gon, if the Jacobsthal-Lucas polynomials are kept on their vertices clockwise. Then, for0≤k<n, the mth term with respect to the vertex Bk is(43)jm−1n+kx=jnxjm−2n+kx−−2xnjm−3n+kx.Proof.
To establish this result, Binet formula for Jacobsthal-Lucas numbers is employed and gives(44)jnxjm−2n+kx−−2xnjm−3n+kx=γnx+δnxγm−2n+kx+δm−2n+kx−−2xnγm−3n+kx+γm−3n+kx=γm−1n+kx+δm−1n+kx+γnxδm−2n+kx+δnxγm−2n+kx−−2xnγm−3n+kx−−2xnδm−3n+kx=γm−1n+kx+δm−1n+kx+γm−3n+kxδnxαnx−−2xn+δm−3n+kxγnxδnx−−2xn.
Sinceγxδx=−2x, γnxδnx−−2xn=0, thus(45)jnxjm−2n+kx−−2xnjm−3n+kx=jm−1n+kx.Theorem 8.
On an n-gon, let the Jacobsthal-Lucas numbers be kept on its vertices clockwise. Then, for0≤k<n, the numbers with respect to the vertex Ak are(46)jmn+kx=jn+kx∑v=02m−3−−1m4−1n+1v2xnvm−v−1vjnm−2v−1x+jkx∑v=02m−5−−1m−14−1n+1v+12xnv+1m−v−2vjnm−2v−2c.
Proof. The result can be proved in a similar fashion as Theorem2.
For example, whenn=2,k=1 and m=2, we get(47)j5x=j3x∑v=00−13v2x2v1−vvj21−2vx+j1x∑v=00−13v+12x2v+1−vvj2−2vxj5x=j3xj2x−j1x4x220x2+10x+1=1+6x1+4x−4x2=1+4x+6x+24x2−4x2=20x2+10x+1.


## 3.1. New Recurrence Relations for the Pell and Pell-Lucas Numbers

From here onwards, we shall consider a continuously clockwise sense approach to the polygon of the Pell polynomials and Pell-Lucas polynomials. We will continue to n-gon starting this placement from a point. For such a polygon, let us call each vertex point asB0,B1,B2,… for k=0,1,2,…. Here, we aim at finding a relationship that locates any term of new Bk sequences to every vertex of the polygon. We establish such a relationship in Theorem 3.1.1 to Pell polynomials and in Theorem 3.1.3 to Pell-Lucas polynomials.Let the vertex number be represented byk and m represents the order of any term of the new sequence occurring at any vertex in any n-gon, with 0≤k<n and m≥1.Takingn=1, the Pell polynomials in succession over a dot are given, as depicted in Figure 1.Figure 1
Pell polynomials placed over a dot.Forn=1 yields k=0 and giving a recurrence relation,(17)Pm−1n+kx=Pm−1x=2xPm−2x+Pm−3x,with P0x=0 and P1x=1 of the form of Pell polynomials.Forn=2, choose the Pell polynomials in succession at the endpoints of a line, as depicted in Figure 2.Figure 2
Pell polynomials kept at the endpoints of a line.So, it determines a recurrence relation of the form(18)Pm−1n+kx=Pm−12+kx=4x2+2Pm−22+kx−Pm−32+kx.Forn=3, the vertices of the triangle will be labeled by Pell polynomials clockwise, as given in Figure 3.Figure 3
Pell polynomials kept in clockwise to vertices of a triangle.So, it determines a recurrence relation of the form(19)Pm−1n+kx=Pm−13+kx=8x3+6xPm−23+kx+Pm−33+kx.Now, forn=4, the Pell numbers are taken clockwise on the vertices of a tetragon, as given in Figure 4.Figure 4
On tetragon representing Pell polynomials which are kept clockwise at its vertices.So, it determines a recurrence relation of the form(20)Pm−1n+kx=Pm−14+kx=16x4+16x2+2Pm−24+kx−Pm−34+kx.Similarly, for an n-gon, we take clockwise the Pell numbers on its vertices, as given in Figure5.Figure 5
Onn-gon representation of Pell polynomials kept clockwise at its vertices.For example, in a new sequence to determine the second term of it occurring at the vertexB1 in a 3-gon.Thus, the value forn=3, m=3, and k=1 corresponds to the 3rd term of B1=1,8x3+4x,64x6+80x4+24x2+1,… sequence in Figure 3.This term must be64x6+80x4+24x2+1.Now, to determine this value from the recurrence relation withn=3 from(21)Pm−1n+kx=Pm−13+kx=8x3+6xPm−23+kx+Pm−33+kx,so that(22)P3−13+1=8x3+6xP3−23+1x+P3−33+1x,P7x=8x3+6xP4x+P1x,64x6+80x4+24x2+1=8x3+6x8x3+4x+1=64x6+32x4+48x4+24x2+1.In Figure5, we define a new sequence created at the vertex Bk.Bk=x1,x2,…=Pk,Pn+k,P2n+k,P3n+k,…. As Pk=P0n+k is the first term to this sequence, the second term is Pn+kx=P1n+kx and the third term is P2n+kx, and if it continues like this, it is observed that the mth term will be Pm−1n+kx of the new sequence formed in an n-gon.In addition, the coefficient ofPm−2n+kx is Pell-Lucas polynomials Qnx and −1n is the coefficient of Pm−3n+kx.So, we have the following result.Theorem 1.
For0≤k<n, the relation(23)Pm−1n+kx=QnxPm−2n+kx−−1nPm−3n+kxis satisfied, giving mth term of the sequence with respect to vertex Bk.Proof.
Using Binet formulas of Pell and Pell-Lucas polynomials, we see(24)QnxPm−2n+kx−−1nPm−3n+kx=αnx+βnxαm−2n+kx−βm−2n+kxαx−βx−−1nαm−3n+kx−βm−3n+kxαx−βx=αm−1n+kx−βm−1n+kxαx−βx−αnxβm−2n+kxαx−βx+βnxαm−2n+kxαx−βx−−1n.αm−3n+kxαx−βx+−1n.βm−3n+kxαx−βx=αm−1n+kx−βm−1n+kxαx−βx+αm−3n+kxβnxαnx−−1nαx−βx−βm−3n+kxαnxβnx−−1nαx−βx.
Because ofαxβx=−1, we have(25)=αm−1n+kx−βm−1n+kxαx−βx+αm−3n+kx−1n−−1nαx−βx−βm−3n+kx−1n−−1nα−β=αm−1n+k−βm−1n+kα−β=Pm−1n+kx.
Let us indicate the coefficients corresponding to this sequence in Table1.
Table1 clearly indicates that the first components are Pell-Lucas polynomials, 2x,4x2+1,8x3+6x,16x4+16x2+2,32x5+40x3+10x,64x6+96x4+36x2+2,128x7+224x5+112x3+14x,….
Table1 clearly indicates that the second are {1, −1, 1, −1, 1, −1, 1, −1, …}. {1, −1, 1, −1, 1, −1, 1, −1, …} is numbered as sequences A033999 in the On-Line Encyclopedia of Integer Sequences [25].Table 1
The coefficients.
DotSegmentTriangleTetragonPentagonHexagonHeptagon(2x, 1)(4x2+1,− 1)(8x3+6x, 1)(16x4+16x2+2, −1)(32x5+40x3+10x, 1)(64x6+96x4+36x2+2, −1)128x7+224x5+112x3+14xTheorem 2.
The relation for0≤k<n between the numbers with respect to the vertex Bk is(26)Pmn+kx=Pn+kx∑t=02m−3−−1m/4−1n+1tm−v−1vQnm−2t−1x+Pkx∑t=02m−5−−1m−1/4−1n+1t+1m−v−2vQnm−2t−2x.Proof.
To establish this result, we induct onm.
Thus, whenm=1, we see(27)Pn+kx=Pn+kx∑v=00−1n+1v−vvQn−2vx+Pkx∑v=0−1−1n+1v+1−v−1vQn−2v−2x=Pn+kx−1000Qn0+Pnx−1n+1−10Qn−2x=Pn+kx.
Form=s, we assume the claim is true, that is,(28)Psn+kx=Pn+kx∑v=02s−3−−1s4−1n+1vs−v−1vQns−2v−1x+Pkx∑v=02s−5−−1s−14−1n+1v+1s−v−2vQns−2v−2x.
Form=s+1, let us show it is also true.
For this, if we use (23), then we see(29)Ps+1n+kx=Psn+kxQnx−−1nPs−1n+kx.
By using (28), we have(30)Ps+1n+k=Pn+kx∑v=02m−3−−1s/4−1n+1vs−v−1vQns−2v−1xQnx+Pkx∑v=02m−5−−1s−1/4−1n+1v+1s−v−2vQns−2v−2xQnx−−1nPn+kx∑v=02m−5−−1s−1/4−1n+1vs−v−2vQns−2v−2x−−1nPkx∑v=02m−7−−1s−2/4−1n+1v+1s−v−3vQns−2v−3x.
Whens=2r, we have(31)Ps+1n+k=Pn+kx∑v=0r−1−1n+1v2r−v−1vQn2r−2vx+Pn+kx∑v=0r−1−1n+1v+12r−v−2vQn2r−2v−2x+Pkx∑v=0r−1−1n+1v+12r−v−2vQn2r−2v−1x+Pkx∑v=0r−2−1n+1v+22r−v−3vQn2r−2v−3x=Pn+kx2r−10Qn2rx+−1n+12r−21Qn2r−2x+2r−32Qn2r−4x+…+−1n+1r−1rr−1Qn2x−1n+12r−20Qn2r−2x+−1n+122r−31Qn2r−4x+…+−1n+1rr−1r−1+Pk−1n+12r−20Qn2r−2x+−12n+12r−31Qn2r−3x+…+−1n+1rr−1r−1Qnx+−12n+12r−30Qn2r−3x+−13n+12r−41Qn2r−5x+…+−1n+1rr−1r−2Qnx=Pn+kxQn2rx+−1n+12r−10Qn2r−2x+−12n+12r−22Qn2r−4x+…+−1n+1r−1r+1r−2Qn2x+−1n+1r+Pk−1n+1Qn2r−1x+−12n+12r−20Qn2r−3x+…+−1n+1rrr−1Qnx=Pn+kx∑v=0r−1n+1v2r−vvQn2r−2vx+Pkx∑v=0r−1−1n+1v+12r−v−1vQn2r−2v−1x=Pn+k∑v=02s−1−−1s+14−1n+1vs−vvQns−2vx+Pk∑v=02s−3−−1s4−1n+1v+1s−v−1vQns−2v−1x.
The equality has been shown fors=2r.
Now, whens=2r−1, we get(32)=Pn+kx∑v=0r−1−1n+1v2r−v−2vQn2r−2v−1x+Pn+kx∑t=0r−2−1n+1v+12r−v−3vQn2r−2v−3x+Pk∑v=0r−2−1n+1v+12r−v−3vQn2r−2v−2x+Pk∑v=0r−2−1n+1v+22r−v−4vQn2r−2v−4x=Pn+kx2r−20Qn2r−1x+−1n+12r−31Qn2r−3x+−12n+12r−42Qn2r−5x+…+−1n+1r−1r−1r−1Qnx+−1n+12r−30Qn2r−3x+−12n+12r−41Qn2r−5x+…+−1n+1r−1r−1r−2Qnx+Pk−1n+12r−30Qn2r−2x+−12n+12r−41Qn2r−4x+−13n+12r−52Qn2r−6x+…+−1n+1r−1r−1r−1Qn2x+−12n+12r−40Qn2r−4x−13n+12r−51Qn2r−6x+…+−1n+1rr−1r−1=Pn+kxQn2r−1x+−1n+12r−21Qn2r−3x+−12n+12r−32Qn2r−5x+…+−1n+1r−1rr−2Qnx+Pk−1n+1Qn2r−2+−12n+12r−31Qn2r−4x+−13n+12r−42Qn2r−6x+…+−1n+1r=Pn+kx∑v=0r−1−1n+1t2r−v−1vQn2r−2v−1x+Pkx∑v=0r−1−1n+1v+12r−v−2vQn2r−2v−2x=Pn+kx∑v=02s−1−−1s+14−1n+1vs−vvQns−2vx+Pkx∑v=023−3−−1s4−1n+1v+1s−v−1vQns−2v−1.
Thus, the proof is obtained.
Let us give example as an application. Forn=3,k=2, and m=2, we have(33)P8x=P5x∑v=00−14v1−vvQ31−2vx+P2x∑v=00−14v+1−vvQ3−2vx,P8x=P5xQ3x+P2x1,128x7+192x5+80x3+8x=16x4+12x2+18x3+6x+2x=128x7+96x5+96x5+72x3+8x3+6x+2x=128x7+192x5+80x3+8x.
Now, let us place the Pell-Lucas polynomials like the Pell polynomials continuously clockwise at the vertices of a regular polygon.
For this new sequence, themth term with respect to the vertex Bk of an n-gon is Qm−1n+kx.
So, we have the following theorem about this situation.Theorem 3.
If on the vertices of an n-gon, the Pell-Lucas polynomials are placed clockwise, then for0≤k<n, the mth term of sequence with respect to the vertex Bk is(34)Qm−1n+kx=Qm−2n+kxQnx−−1nQm−3n+kx.Proof.
To prove the result, by employing Binet formula for Pell-Lucas polynomials, we see(35)Qm−2n+kxQnx−−1nQm−3n+kx=αm−2n+kx+βm−2n+kxαnx+βnx−−1nαm−3n+kx+βm−3n+kx=αm−1n+kx+βm−1n+kx+αm−2n+kxβnx+αnxβm−2n+kx−−1nαm−3n+kx−−1nβm−3n+kx=αm−1n+k+βm−1n+k−αm−3n+kx−1n−αnxβnx−βm−3n+kx−1n−αnxβnx.
Because ofαβ=−1, we have(36)=αm−1n+kx+βm−1n+kx,=Qm−1n+kx.Theorem 4.
For an n-gon, if Pell-Lucas polynomials are kept on the vertices clockwise sense. Then, for0≤k<n, there exists a relation of the form(37)Qmn+kx=Qn+kx∑v=02m−3−−1m/4−1n+1vm−v−1vQnm−2v−1x+Qkx∑v=02m−5−−1m−1/4−1n+1v+1m−v−2vQnm−2v−2x,between the numbers with respect to the vertex Bk.Proof.
The result can be proved in a similar fashion as Theorem 3.1.2.
For example, whenn=4,k=1 and m=2, we have(38)Q9x=Q5x∑v=00−15v1−vvQ41−2vx+Q1x∑v=00−15v+1−vvQ4−2vxQ9x=Q5xQ4x−Q1x512x9+1152x7+864x5+240x3+18x=32x5+40x3+10x16x4+16x2+2−2x=512x9+512x7+64x5+640x7+640x5+80x3+160x5+160x3+20x−2x=512x9+1152x7+864x5+240x3+18x.


## 3.2. Recurrences Notion Corresponding to Jacobsthal and Jacobsthal-Lucas Polynomials

In this direction, we involve the clockwise sense of the Jacobsthal polynomials on the point, at extreme points, at vertices of the triangle, and lastly at the vertices of an n-gon.Since for this new sequence, the mth term computed at vertexBk is Jm−1n+kx for an n-gon. Also, the coefficient of Jm−2n+kx is Jacobsthal-Lucas polynomials jnx and the coefficient of Jm−3n+kx is −2xn.In this case, we have the following result.Theorem 5.
On an n-gon, suppose Jacobsthal numbers are kept on its vertices clockwise. Then, for0≤k<n, we have(39)Jm−1n+kx=jnxJm−2n+kx−−2xnJm−3n+kx,and it determines the mth term of the sequence with respect to vertex Bk.Proof.
By using the Binet formulas of Jacobsthal and Jacobsthal-Lucas polynomials, we have the following:(40)jnxJm−2n+kx−−2xnJm−3n+kx=γnx+δnxγm−2n+kx−δm−2n+kxγx−δx−−2xnγm−3n+kx−δm−3n+kxγx−δx=γm−1n+kx−δm−1n+kxγx−zx+δnxγm−2n+kx−γnxδm−2n+kxγx−δx−−2xnγm−3n+kxγx−δx+−2xnδm−3n+kxγx−δx=γm−1n+kx−δm−1n+kxγx−δx+αm−3n+kxδnxγnx−−2xnγx−δx−γm−3n+kxγnxδnx−−2xnγx−δx.
Sinceγxδx=−2x, γnxδnx−−2xn=0, thus(41)jnxJm−2n+kx−−2xnJm−3n+kx=Jm−1n+kx.
In Table2, we write the coefficients concerning the above sequences.
So, clearly in Table2, the Jacobsthal-Lucas polynomials 1,1+4x,1+6x,1+8x+8x2,1+10x+20x2,1+12x+36x2+16x3,1+14x+56x2+56x3 are first components.
The second components in Table2 are {2x, −2x2,2x3,−2x4, 25,−2x6,2x7, …}. {1, −1, 1, −1, 1, −1, 1, −1, …} is enumerate as sequences A084633 in the On-Line Encyclopedia of Integer Sequences [25].Table 2
The coefficients.
DotSegmentTriangleTetragonPentagonHexagonHeptagon1,2x1+4x,−4x21+6x,8x31+8x+8x2,−16x41+10x+20x2,32x51+12x+36x2+16x3,−64x61+14x+56x2+56x3,128x7Theorem 6.
On an n-gon, let the Pell-Lucas polynomials be kept on its vertices clockwise. Then, for0≤k<n, there exists a relation corresponding to numbers with respect to the vertex Bk as(42)Jmn+kx=Jn+kx∑t=02m−3−−1m4−1n+1t2xntm−t−1tjnm−2t−1x+Jkx∑t=02m−5−−1m−14−1n+1t+12xnt+1m−t−2tjnm−2t−2x.Proof.
As in Theorem2, the proof can be performed.
Now, for ann-gon, taking the Jacobsthal-Lucas polynomials on the vertices clockwise and in this n-gon, the mth term formed to this new sequence at the vertex Bk is jm−1n+kx.
We have the following result in this direction.Theorem 7.
On an n-gon, if the Jacobsthal-Lucas polynomials are kept on their vertices clockwise. Then, for0≤k<n, the mth term with respect to the vertex Bk is(43)jm−1n+kx=jnxjm−2n+kx−−2xnjm−3n+kx.Proof.
To establish this result, Binet formula for Jacobsthal-Lucas numbers is employed and gives(44)jnxjm−2n+kx−−2xnjm−3n+kx=γnx+δnxγm−2n+kx+δm−2n+kx−−2xnγm−3n+kx+γm−3n+kx=γm−1n+kx+δm−1n+kx+γnxδm−2n+kx+δnxγm−2n+kx−−2xnγm−3n+kx−−2xnδm−3n+kx=γm−1n+kx+δm−1n+kx+γm−3n+kxδnxαnx−−2xn+δm−3n+kxγnxδnx−−2xn.
Sinceγxδx=−2x, γnxδnx−−2xn=0, thus(45)jnxjm−2n+kx−−2xnjm−3n+kx=jm−1n+kx.Theorem 8.
On an n-gon, let the Jacobsthal-Lucas numbers be kept on its vertices clockwise. Then, for0≤k<n, the numbers with respect to the vertex Ak are(46)jmn+kx=jn+kx∑v=02m−3−−1m4−1n+1v2xnvm−v−1vjnm−2v−1x+jkx∑v=02m−5−−1m−14−1n+1v+12xnv+1m−v−2vjnm−2v−2c.
Proof. The result can be proved in a similar fashion as Theorem2.
For example, whenn=2,k=1 and m=2, we get(47)j5x=j3x∑v=00−13v2x2v1−vvj21−2vx+j1x∑v=00−13v+12x2v+1−vvj2−2vxj5x=j3xj2x−j1x4x220x2+10x+1=1+6x1+4x−4x2=1+4x+6x+24x2−4x2=20x2+10x+1.


## 4. Conclusions

In this paper, we have described the clockwise sense of Pell, Pell-Lucas, Jacobsthal, and Jacobsthal-Lucas polynomials on the vertices of polygons. We have computed the recurrence relations forPmn+kx numbers by sorting the Pell polynomials around the polygons. Also, we have determined new correlations between the terms with respect to each vertex and generated new findings by employing these results. The work can be further extended to k−Jacobsthal and Jacobsthal-Lucas numbers and k−Fibonacci and Lucas numbers. [20-27]



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