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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

On the power sum problem of Lucas polynomials and its divisible property

Wang Xiao
Published Online: 2018-06-22 | DOI: https://doi.org/10.1515/math-2018-0063

Abstract

The main purpose of this paper is to use the mathematical induction and the properties of Lucas polynomials to study the power sum problem of Lucas polynomials. In the end, we obtain an interesting divisible property.

Keywords: Lucas polynomials; Power sum problem; Mathematical induction; Divisible property

MSC 2010: 11B39

1 Introduction

For any integer n ≥ 0, the famous Fibonacci polynomials {Fn(x)} and Lucas polynomials {Ln(x)} are defined by F0(x) = 0, F1(x) = 1, L0(x) = 2, L1(x) = x and Fn+2(x) = xFn+1(x) + Fn(x), Ln+2(x) = xLn+1(x) + Ln(x). The general terms of Fn(x) and Ln(x) are given by

Fn+1(x)=k=0[n2](nkk)xn2k

and

Ln(x)=k=0[n2]nnk(nkk)xn2k,(1)

where (mn)=m!n!(mn)!, and [x] denotes the greatest integer ≤ x.

It is easy to prove the identities

Fn(x)=1x2+4[(x+x2+42)n(xx2+42)n]

and

Ln(x)=(x+x2+42)n+(xx2+42)n.(2)

If we take x = 1, then {Fn(x)} becomes Fibonacci sequences {Fn(1)}, and {Ln(x)} becomes Lucas sequences {Ln(1)}.

Since these sequences and polynomials have very important positions in the theory and application of mathematics, many scholars have studied their various properties, and obtained a series of important results, some of which can be found in references [1, 2, 3, 4, 5], and some other related papers can also be found in [6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]. For example, in a private communication with Curtis Cooper, R. S. Melham suggested that it would be interesting to discover an explicit expansion for

L1L2L2m+1k=1nF2k2m+1

as a polynomial in F2n+1. Wiemann and Cooper [1] reported some conjectures of Melham related to the sum k=1nF2k2m+1. Kiyota Ozeki [2] proved that

k=1nF2k2m+1=15mj=0m(1)jL2m+12j(j2m+1)(F(2m+12j)(2n+1)F2m+12j).

Helmut Prodinger [3] studied the more general summation k=0nF2k+δ2m+1+ϵ, where δ, ϵ ∈ {0, 1}, and obtained many interesting identities.

For the sum of powers of Lucas numbers, Helmut Prodinger [3] also obtained some similar conclusions.

R. S. Melham [4] also proposed the following two conjectures:

Conjecture A

Let m be a positive integer. Then the sum

L1L3L5L2m+1k=1nF2k2m+1

can be expressed as (F2n+1 − 1)2 P2m−1(F2n+1), where P2m−1(x) is a polynomial of degree 2m − 1 with integer coefficients.

Conjecture B

Let m ≥ 0 be an integer. Then the sum

L1L3L5L2m+1k=1nL2k2m+1

can be expressed as (L2n+1 − 1) Q2m(L2n+1), where Q2m(x) is a polynomial of degree 2m with integer coefficients.

Wang Tingting and Zhang Wenpeng [5] studied these problems, and proved several conclusions for k=1hFkn(x)andk=1hLkn(x), where h and n are any positive integers.

As some applications of Theorem 2 in reference [5], they deduced the following:

Corollary A

Let h ≥ 1 and n ≥ 0 be two integers. Then the sum

L1(x)L3(x)L5(x)L2n+1(x)m=1hL2m2n+1(x)

can be expressed as (L2h+1(x) − x) Q2n(x, L2h+1(x)), where Q2n(x, y) is a polynomial in two variables x and y with integer coefficients and degree 2n of y.

Corollary B

Let h ≥ 1 and n ≥ 0 be two integers. Then the sum

L1(x)L3(x)L5(x)L2n+1(x)m=1hF2m2n+1(x)

can be expressed as (F2h+1(x) − 1) H2n(x, F2h+1(x)), where H2n(x, y) is a polynomial in two variables x and y with integer coefficients and degree 2n of y.

Therefore, Wang Tingting and Zhang Wenpeng [5] solved Conjecture B completely. They also obtained some substantial progress for Conjecture A.

It is easy to see that in the above Conjecture A and Conjecture B, the subscripts of Fn and Ln in sums k=1nF2k2m+1andk=1nL2k2m+1 are even numbers. Inspired by the above researches, we naturally ask that for any positive integers n and h with h ≥ 2, whether there exists a polynomial U(x) with degree > 1 such that the congruence

L1(x)L3(x)L2n+1(x)m=0h1L2m+12n+1(x)0modU(x).(3)

If so, what would U(x) look like? How is it different from the form in Conjecture A and Conjecture B ? Obviously, it is hard to guess the exact form of U(x) from (d) of Theorem 1 in reference [5]. Maybe that is why there is no such formal conjecture in [4].

In this paper, we shall use the mathematical induction and the properties of Lucas polynomials to study this problem, and give an exact polynomial U(x) in (3). The result is detailed in the following theorems.

Theorem A

For any positive integers n and h with h ≥ 2, we have

L1(x)L3(x)L2n+1(x)m=0h1L2m+12n+1(x)0modx(L2h(x)2).

Theorem B

For any positive integers n and h with h ≥ 2, we have

L1(x)L3(x)L2n+1(x)m=0h1F2m+12n+1(x)0modxF2h(x).

Taking x = 1, from Theorem 1 and Theorem 2 we may immediately deduce the following two corollaries:

Corollary A

For any positive integers n and h with h ≥ 2, we have

L1L3L2n+1m=0h1L2m+12n+10mod(L2h2).

Corollary B

For any positive integers n and h with h ≥ 2, we have

L1L3L2n+1m=0h1F2m+12n+10modF2h.

2 Two Lemmas

Lemma A

For any non-negative integer n and k, we have the identity

Ln(L2k+1(x))=Ln(2k+1)(x).

Proof

Let α=x+x2+42,β=xx2+42, and replace x by L2k+1(x) in (2). Since we have α2k+1β2k+1 = −1,

L2k+1(x)+L2k+12(x)+4=α2k+1+β2k+1+(α2k+1+β2k+1)2+4=α2k+1+β2k+1+α2(2k+1)+β2(2k+1)2+4=α2k+1+β2k+1+(α2k+1β2k+1)2=2α2k+1

and

L2k+1(x)L2k+12(x)+4=α2k+1+β2k+1(α2k+1+β2k+1)2+4=α2k+1+β2k+1(α2k+1β2k+1)2=2β2k+1,

from (2) we have the identity

Ln(L2k+1(x))=(L2k+1+L2k+12+42)n+(L2k+1L2k+12+42)n=αn(2k+1)+βn(2k+1)=Ln(2k+1)(x).

This completes the proof of Lemma 1. □

Lemma B

For any positive integers n and h, we have the congruence

L1(x)(L2h(2n+1)(x)2)(2n+1)L2n+1(x)(L2h(x)2)0modx(L2h(x)2).

Proof

From (1) we know that xL2n+1(x). Because that L1(x) = x, so in order to prove Lemma 2, we only need to prove the congruence

L2h(2n+1)(x)20mod(L2h(x)2).(4)

Next we prove (4) by complete induction. It is clear that (4) is true for n = 0. If n = 1, then apply the identity

L2h3(x)=(α2h+β2h)3=L6h(x)+3L2h(x),

we have

L6h(x)2=L2h3(x)3L2h(x)2=(L2h(x)2)(L2h(x)+1)20mod(L2h(x)2).

That is to say, the congruence (4) is true for n = 1.

Suppose that the congruence (4) is true for all integers 0 ≤ ns. That is,

L2h(2n+1)(x)20mod(L2h(x)2)(5)

holds for all integers 0 ≤ ns.

Then for positive integer n = s + 1, we have the identities

L4h(x)=L2h2(x)22mod(L2h(x)2)

and

L4h(x)L2h(2s+1)(x)=L2h(2s+3)(x)+L2h(2s1)(x),

from (5) we can deduce the congruence equations as follows

L2h(2s+3)(x)2=L4h(x)L2h(2s+1)(x)L2h(2s1)(x)22L2h(2s+1)(x)L2h(2s1)(x)22(L2h(2s+1)(x)2)(L2h(2s1)(x)2)0mod(L2h(x)2).

That is to say, the congruence (4) is true for n = s + 1.

Now Lemma 2 follows from (4) and completes the induction. □

3 Proofs of the theorems

In this section, we shall use mathematical induction to complete the proofs of our theorems. Here we only prove Theorem 1. Similarly, we can also deduce Theorem 2 and thus we omit its proving process here. After replacing x by L2m+1(x) in (1), we obtain the following expression with Lemma 1

L(2n+1)(2m+1)(x)=L2n+1(L2m+1(x))=k=0n2n+12n+1k(2n+1kk)L2m+12n+12k(x)

or

L(2n+1)(2m+1)(x)(2n+1)L2m+1(x)=k=0n12n+12n+1k(2n+1kk)L2m+12n+12k(x).(6)

For any positive integer h ≥ 2, we first introduce that the identities

m=0h1L(2n+1)(2m+1)(x)=m=0h1(α(2n+1)(2m+1)+β(2n+1)(2m+1))=α2n+1α2h(2n+1)1α2(2n+1)1+β2n+1β2h(2n+1)1β2(2n+1)1=α2h(2n+1)1α2n+1+β2n+1+β2h(2n+1)1β2n+1+α2n+1=L2h(2n+1)(x)2L2n+1(x),(7)

m=0h1L2m+1(x)=m=0h1(α2m+1+β2m+1)=L2h(x)2L1(x).(8)

Then, combining (6), (7) and (8) we have

m=0h1[L(2n+1)(2m+1)(x)(2n+1)L2m+1(x)]=L2h(2n+1)(x)2L2n+1(x)(2n+1)L2h(x)2L1(x)=k=0n12n+12n+1k(2n+1kk)m=0h1L2m+12n+12k(x).(9)

Now we apply (9) and mathematical induction to prove the congruence

L1(x)L3(x)L2n+1(x)m=0h1L2m+12n+1(x)0modx(L2h(x)2).(10)

If n = 1, then from (9) we have

L1(x)L3(x)[L6h(x)2L3(x)3L2h(x)2L1(x)]=L1(x)L3(x)m=0h1L2m+13(x).(11)

From Lemma 2 we know that

L1(x)L3(x)[L6h(x)2L3(x)3L2h(x)2L1(x)]0modx(L2h(x)2).(12)

Combining (11) and (12) we know that the congruence (10) is true for n = 1.

Suppose that (10) is true for all 1 ≤ ns. That is,

L1(x)L3(x)L2n+1(x)m=0h1L2m+12n+1(x)0modx(L2h(x)2)(13)

holds for all 1 ≤ ns.

Then for n = s + 1, from (9) we have

L2h(2s+3)(x)2L2s+3(x)(2s+3)L2h(x)2L1(x)=k=0s2s+32s+3k(2s+3kk)m=0h1L2m+12s+32k(x)=m=0h1L2m+12s+3(x)+k=1s2s+32s+3k(2s+3kk)m=0h1L2m+12s+32k(x).(14)

Applying Lemma 2 we have the congruence

L1(x)L3(x)L2s+3(x)[L2h(2s+3)(x)2L2s+3(x)(2s+3)L2h(x)2L1(x)]0modx(L2h(x)2).(15)

If 1 ≤ ks, then 3 ≤ 2s + 3 − 2k ≤ 2s + 1. From the inductive assumption (13) we have

L1(x)L3(x)L2s+1(x)L2s+3(x)k=1s2s+32s+3k(2s+3kk)m=0h1L2m+12s+32k(x)0modx(L2h(x)2).(16)

Combining (14), (15) and (16) we may immediately deduce the congruence

L1(x)L3(x)L2s+1(x)L2s+3(x)m=0h1L2m+12s+3(x)0modx(L2h(x)2).

This completes the proof of our theorem by mathematical induction.

Acknowledgement

This work is supported by the N. S. F. (Grant No. 11771351) of P. R. China. The author would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper.

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About the article

Received: 2017-11-15

Accepted: 2018-05-15

Published Online: 2018-06-22


Competing interests: The author declare that they have no competing interests.


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 698–703, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0063.

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© 2018 Xiao, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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