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The power sum of balancing polynomials and their divisible properties

  • In recent years, many scholars have studied the division properties of polynomials and sequence power sums. In this paper, we use Girard-Waring formula and combinatorial method to study the power sum problem of balancing polynomials and Lucas-balancing polynomials, and then study the division of balancing polynomials and Lucas-balancing polynomials by mathematical induction and the properties of polynomials.

    Citation: Hong Kang. The power sum of balancing polynomials and their divisible properties[J]. AIMS Mathematics, 2024, 9(2): 2684-2694. doi: 10.3934/math.2024133

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  • In recent years, many scholars have studied the division properties of polynomials and sequence power sums. In this paper, we use Girard-Waring formula and combinatorial method to study the power sum problem of balancing polynomials and Lucas-balancing polynomials, and then study the division of balancing polynomials and Lucas-balancing polynomials by mathematical induction and the properties of polynomials.



    Behera and Panda [1] introduced the concept of balancing numbers Bn, a positive integer n is a balancing number if

    1+2++(n1)=(n+1)+(n+2)++(n+r), n, rN.

    That r is the balancer corresponding to the balancing number n. The balancing numbers Bn satisfy the relation Bn+1=6BnBn1 ,n1 with B0=0, B1=1. The sequence Cn=8B2n+1 is called a Lucas-balancing number. The Lucas-balancing number satisfies same relation Cn+1=6CnCn1 ,n1 with C0=1, C1=3. Some conclusions about these two sequences can be found in the references [2,3]. The balancing polynomial and the Lucas-balancing polynomial are natural extensions of balancing numbers and Lucas-balancing numbers.

    For any integer n0, the balancing polynomials Bn(x) and Lucas-balancing polynomials Cn(x) are defined as follows (see Frontczak and Goy [4]):

    Bn(x)=n12k=0(1)k(n1kk)(6x)n12k,
    Cn(x)=n2n2k=0(1)knk(nkk)(6x)n2k,

    where x denotes the greatest integer x.

    Bn(x) and Cn(x) are the second-order linear recurrence polynomials, they satisfy the recurrence formulae (see Frontczak and Goy [4]):

    Bn+1(x)=6xBn(x)Bn1(x) for all n1, with B0(x)=0, B1(x)=1,

    Cn+1(x)=6xCn(x)Cn1(x) for all n1, with C0(x)=1, C1(x)=3x.

    The closed forms which are also called Binets formulas for balancing polynomials and Lucas-balancing polynomials are given by

    Bn(x)=αn(x)βn(x)29x21, Cn(x)=αn(x)+βn(x)2,

    where α(x)=3x+9x21, β(x)=3x9x21. The relations Bn(x)=(1)n+1Bn(x) and Cn(x)=(1)nCn(x) follow from α(x)=β(x) and α(x)=β(x). If we take x=1, then {Bn(x)} becomes balancing sequences {Bn}, and {Cn(x)} becomes Lucas-balancing sequences {Cn}. Such balancing numbers and balancing polynomials have been widely studied in recent years. Frontczak [5] proves the sum of powers of balancing polynomials and Lucas balancing polynomials:

    B2m+1n(x)=22m(9x21)mmk=0(2m+1mk)(1)mkB(2k+1)n(x),C2m+1n(x)=22mmk=0(2m+1mk)C(2k+1)n(x).

    Kim and Kim [6] used nine orthogonal polynomials to represent the sum of the finite product of balancing polynomials to obtained the following result:

    i1+i2++ir+1=nBi1+1(x)Bi2+1(x)Bir+1+1(x)=(2)nr!nk=0(2)kΓ(k+α+β+1)Γ(2k+α+β+1)[nk2]l=0(14)l(n+rl)!l!(nk2l)!×2F1(k+2ln,k+β+1;2k+α+β+2;2)P(α, β)k(3x).

    Ray [7] studied the divisible property of balancing numbers and Lucas-balancing number obtained the congruence:

    B2mn+k(1)nBk(modCm),C2mn+k(1)nCk(modCm).

    For any integer n0, the famous Fibonacci polynomials Fn(x) and Lucas polynomials Ln(x) are defined as follows (see Wang and Zhang [8]) :

    Fn+2(x)=xFn+1(x)+Fn(x), with F0(x)=0, F1(x)=1,

    Ln+2(x)=xLn+1(x)+Ln(x), with L0(x)=2, L1(x)=x.

    These polynomials and sequences have some similarities in structure and properties. Kim [9,10,11,12,13] obtained many meaningful results by studying connections between polynomials. Mathematics has a wide range of applications in other disciplines, see [14,15,16]. We can obtain some divisible properties of polynomials and sequences in references [17,18,19]. For example, Wang and Zhang [8] proved the congruence of the sum of powers of Fibonacci numbers. That is

    L1L3L5L2m+1nk=1L2m+12k0 mod (L2n+11).

    In this paper, we use the properties of balancing polynomials and Lucas balancing polynomials to study the divisible properties of hm=0B2n+12sml(x) and hm=0C2n+12sml(x) to get more general results. That is, we shall prove the following two theorems.

    Theorem 1. Let n and h be non-negative integer with h1, s and l be positive integers. Then we have the congruence

    22n+1(9x21)n+1B2s1l(x)B2s13l(x)B2s1(2n+1)l(x)hm=0B2n+12sml(x)0 mod(C2s1l(2h+1)(x)C2s1l(x)).

    Theorem 2. Let n and h be non-negative integers with h1, s and l be positive integers. Then we have the congruence

    22n+1B2s1l(x)B2s13l(x)B2s1(2n+1)l(x)hm=0C2n+12sml(x)0 mod(B2s1l(2h+1)(x)+B2s1l(x)).

    From the two theorems, we can obtain the following corollaries.

    Corollary 1. For any non-negative integers n and h with h1, we have

    22n+1(9x21)n+1B1(x)B3(x)B(2n+1)(x)hm=0B2n+12m(x)0 mod(C2h+1(x)3x).

    Corollary 2. For any non-negative integers n and h with h1, we have

    22n+1B2(x)B6(x)B2(2n+1)(x)hm=0C2n+14m(x)0 mod(B2(2h+1)(x)+6x).

    Corollary 3. For any non-negative integers n and h with h1, and s and l be positive integers, we have

    25n+4B2s1lB2s13lB2s1(2n+1)lhm=0B2n+12sml0 mod(C2s1l(2h+1)C2s1l).

    Corollary 4. For any non-negative integers n and h with h1, and s and l be positive integers, we have

    22n+1B2s1lB2s13lB2s1(2n+1)lhm=0C2n+12sml0 mod(B2s1l(2h+1)+B2s1l).

    For Chebyshev polynomials of the first kind Tn+1(x)=2xTn(x)Tn1(x) with T0(x)=1, T1(x)=x and Chebyshev polynomials of the second kind Un+1(x)=2xUn(x)Un1(x) with U0(x)=1, U1(x)=2x. The balancing polynomials possess a simple connection to Chebyshev polynomials of the first and second kind Tn(x) and Un(x), specifically Bn(x)=Un1(3x) and Cn(x)=Tn(3x).

    Taking x=13x in Theorem 1, we can get the following,

    Corollary 5. For any non-negative integers n and h with h1, and s and l be positive integers, we have

    22n+1(x21)n+1U2s1l1(x)U2s13l1(x)U2s1(2n+1)l1(x)hm=0U2n+12sml1(x)0 mod(T2s1l(2h+1)(x)T2s1l(x)).

    Taking s=1 and x=13x in Theorem 2, we can get the following,

    Corollary 6. For any non-negative integers n and h with hl, and l be positive integers, we have

    22n+1Ul1(x)U3l1(x)U(2n+1)l1(x)hm=0T2n+12ml(x)0 mod(Ul(2h+1)1(x)+Ul1(x)).

    In the following, we use the properties of balancing polynomials and Lucas-balancing polynomials to prove our next several lemmas, which will help us better complete the proofs of the theorems.

    Lemma 1. Let s and h be positive integers. Then, for any integers n and l, we have the identity

    C2s1l(2n+1)(2h+1)(x)C2s1l(2n+1)(x)0 mod(C2s1l(2h+1)(x)C2s1l(x)).

    Proof. We prove this polynomial congruence by complete induction for n0. It is clear that Lemma 1 is true for n=0. If n=1, then note C2s13l(2h+1)(x)=4C32s1l(2h+1)(x)3C2s1l(2h+1)(x), we have

    C2s13l(2h+1)(x)C2s13l(x)=4C32s1l(2h+1)(x)3C2s1l(2h+1)(x)4C32s1l(x)+3C2s1l(x)=(C2s1l(2h+1)(x)C2s1l(x))(4C22s1l(2h+1)(x)=+4C2s1l(2h+1)(x)C2s1l(x)+4C22s1l(x)3)0 mod(C2s1l(2h+1)(x)C2s1l(x)). 

    That is to say, Lemma 1 is true for n=1.

    Suppose that Lemma 1 is true for all positive integers 0nj. That is,

    C2s1l(2n+1)(2h+1)(x)C2s1l(2n+1)(x)0 mod(C2s1l(2h+1)(x)C2s1l(x)), (2.1)

    for all 0nj.

    Then, for n=j+12, we have

    C2s12l(2h+1)(x)C2s1l(2n+1)(2h+1)(x)=14(α2s12l(2h+1)(x)+β2s12l(2h+1)(x))(α2s1l(2n+1)(2h+1)(x)+β2s1l(2n+1)(2h+1)(x))=14(α2s1l(2h+1)(2n+3)(x)+β2s1l(2h+1)(2n1)(x)+α2s1l(2h+1)(2n1)(x)+β2s1l(2h+1)(2n+3)(x))=12(C2s1l(2h+1)(2n+3)(x)+C2s1l(2h+1)(2n1)(x))

    and

    C2s12l(2h+1)(x)=12(α2s12l(2h+1)(x)+β2s12l(2h+1)(x))=12(α2s1l(2h+1)(x)+β2s1l(2h+1)(x))212C22s1l(x)1 mod(C2s1l(2h+1)(x)C2s1l(x)). 

    Applying inductive hypothesis (2.1), we have

    C2s1l(2n+1)(2h+1)(x)C2s1l(2n+1)(x)=C2s1l(2j+3)(2h+1)(x)C2s1l(2j+3)(x)=2C2s12l(2h+1)(x)C2s1l(2j+1)(2h+1)(x)C2s1l(2j1)(2h+1)(x)2C2s12l(x)C2s1l(2j+1)(x)+C2s1l(2j1)(x)=2(2C22s1l(2h+1)(x)1)C2s1l(2j+1)(2h+1)(x)C2s1l(2j1)(2h+1)(x)2(2C22s1l(x)1)C2s1l(2j+1)(x)+C2s1l(2j1)(x)2(2C22s1l(x)1)(C2s1l(2j+1)(2h+1)(x)C2s1l(2j+1)(x))(C2s1l(2j1)(2h+1)(x)C2s1l(2j1)(x))0 mod(C2s1l(2h+1)(x)C2s1l(x)). 

    That is to say, the Lemma 1 is true for n=j+1.

    Now Lemma 1 follows from complete induction.

    Lemma 2. Let s and h be positive integers. Then, for any integers n and l, we have the identity

    B2s1l(2n+1)(2h+1)(x)+B2s1l(2n+1)(x)0 mod(B2s1l(2h+1)(x)+B2s1l(x)).

    Proof. We can also prove Lemma 2 by complete induction. If n=0, then it is clear that Lemma 2 is true. If n=1, then note

    B2s13l(2h+1)(x)=4(9x21)B32s1l(2h+1)(x)+3B2s1l(2h+1)(x),

    we have

    B2s13l(2h+1)(x)+B2s13l(x) = 4(9x21)B32s1l(2h+1)(x)+3B2s1l(2h+1)(x)+4(9x21)B32s1l(x)+3B2s1l(x) = 4(9x21)(B2s1l(2h+1)(x)+B2s1l(x))(B22s1l(2h+1)(x)+B22s1l(x)=B2s1l(2h+1)(x)B2s1l(x))+3(B2s1l(2h+1)(x)+B2s1l(x)) 0 mod(B2s1l(2h+1)(x)+B2s1l(x)). 

    So Lemma 2 is true for n=1. Suppose that Lemma 2 is true for positive integers 0nj. That is,

    B2s1l(2n+1)(2h+1)(x)+B2s1l(2n+1)(x)0 mod(B2s1l(2h+1)(x)+B2s1l(x)) (2.2)

    for all 0nj.

    Then, for n=k+1, note the identities

    2C2s12l(2h+1)(x)B2s1l(2n+1)(2h+1)(x)=129x21(α2s12l(2h+1)(x)+β2s12l(2h+1)(x))(α2s1l(2n+1)(2h+1)(x)β2s1l(2n+1)(2h+1)(x))=129x21(α2s1l(2h+1)(2n+3)(x)β2s1l(2h+1)(2n1)(x)+α2s1l(2h+1)(2n1)(x)β2s1l(2h+1)(2n+3)(x))=B2s1l(2n+3)(2h+1)(x)+B2s1l(2n1)(2h+1)(x)

    and

    C2s12l(2h+1)(x)=12(α2s12l(2h+1)(x)+β2s12l(2h+1)(x))=12(α2s1l(2h+1)(x)β2s1l(2h+1)(x))2+1=2(9x21)B22s1l(2h+1)(x)+12(9x21)B22s1l(x)+1 mod(B2s1l(2h+1)(x)+B2s1l(x)),

    applying inductive hypothesis (2.2), we have

    B2s1l(2n+1)(2h+1)(x)+B2s1l(2n+1)(x)=B2s1l(2j+3)(2h+1)(x)+B2s1l(2j+3)(x)=2C2s12l(2h+1)(x)B2s1l(2j+1)(2h+1)(x)B2s1l(2j1)(2h+1)(x)+2C2s12l(x)B2s1l(2j+1)(x)B2s1l(2j-1)(x)=[4(9x21)B22s1l(2h+1)(x)+2]B2s1l(2j+1)(2h+1)(x)B2s1l(2j1)(2h+1)(x) +[4(9x21)B22s1l(x)+2]B2s1l(2j+1)(x)B2s1l(2j1)(x)2[2(9x21)B22s1l(x)+1](B2s1l(2j+1)(2h+1)(x)+B2s1l(2j+1)(x)) (B2s1l(2j1)(2h+1)(x)+B2s1l(2j1)(x))0mod(B2s1l(2h+1)(x)+B2s1l(x)). 

    This completes the proof of Lemma 2.

    Lemma 3. For all non-negative integers r and real numbers w,v, we have the identity

    [r2]k=0(1)krrk(rkk)(w+v)r2k(wv)k=wr+vr,

    in which [x] denotes the greatest integer x.

    Proof. The formula due to E. Waring [20] and can be found in H. W. Gould [21].

    We shall prove our theorems by mathematical induction. Taking w=α2sml(x), v=β2sml(x) and r=2n+1 in Lemma 3, we notice that wv=1, from the expression of Bn(x) we have

    B2sml(2n+1)(x)=nk=0(1)k22n2k2n+12n+1k(2n+1kk)B2n+12k2sml(x)(9x21)2n2k(1)k=nk=022n2k(9x21)2n2k2n+12n+1k(2n+1kk)B2n+12k2sml(x). (3.1)

    For any integer h1, from (3.1) we have

    hm=0(B2sml(2n+1)(x)(2n+1)B2sml(x))=n1k=02n+12n+1k22n2k(9x21)nk(2n+1kk)hm=0B2n+12k2sml(x). (3.2)

    Note the identities

    hm=0B2sml(2n+1)(x)=hm=0129x21(α2sml(2n+1)(x)β2sml(2n+1)(x)) = 129x21(1α2sl(2n+1)(h+1)(x)1α2sl(2n+1)(x)1β2sl(2n+1)(h+1)(x)1β2sl(2n+1)(x)) = 129x21(α2s1l(2n+1)(2h+1)(x)β2s1l(2n+1)(x)α2s1l(2n+1)(x)β2s1l(2n+1)(x)=α2s1l(2n+1)(x)β2s1l(2n+1)(2h+1)(x)α2s1l(2n+1)(x)β2s1l(2n+1)(x)) = C2s1l(2h+1)(2n+1)(x)C2s1l(2n+1)(x)2(9x21)B2s1l(2n+1)(x). (3.3)

    And

    hm=0C2sml(2n+1)(x)=12B2s1l(2n+1)(2h+1)(x)+B2s1l(2n+1)(x)B2s1l(2n+1)(x). (3.4)

    Combining (3.2) and (3.3) we have

    hm=0(B2sml(2n+1)(x)(2n+1)B2sml(x)) = C2s1l(2h+1)(2n+1)(x)C2s1l(2n+1)(x)2(9x21)B2s1l(2n+1)(x)(2n+1)C2s1l(2h+1)(x)C2s1l(x)2(9x21)B2s1l(x) = n1k=02n+12n+1k22n2k(9x21)nk(2n+1kk)hm=0B2n+12k2sml(x). (3.5)

    Now we apply (3.5) and mathematical induction to prove Theorem 1. If n=1, then from (3.5) we have

    2(9x21)B2s1l(x)B2s13l(x)(C2s13l(2h+1)(x)C2s13l(x)2(9x21)B2s13l(x)=3C2s1l(2h+1)(x)C2s1l(x)2(9x21)B2s1l(x))=8(9x21)2B2s1l(x)B2s13l(x)hm=0B32sml(x). (3.6)

    From Lemma 1 we know that

    2(9x21)B2s1l(x)B2s13l(x)(C2s13l(2h+1)(x)C2s13l(x)2(9x21)B2s13l(x)=3C2s1l(2h+1)(x)C2s1l(x)2(9x21)B2s1l(x))0 mod(C2s1l(2h+1)(x)C2s1l(x)). (3.7)

    Combining (3.6) and (3.7) we know that Theorem 1 is true for n=1.

    Suppose that Theorem 1 is true for all integers 1nj. Then, for n=j+1, from (3.5) we have

    C2s1l(2h+1)(2j+3)(x)C2s1l(2j+3)(x)2(9x21)B2s1l(2j+3)(x)(2j+3)C2s1l(2h+1)(x)C2s1l(x)2(9x21)B2s1l(x) = (2j+3)jk=012j+3k(2j+3kk)22j+22k(9x21)j+1khm=0B2j+32k2sml(x) = (2j+3)jk=112j+3k(2j+3kk)22j+22k(9x21)j+1khm=0B2j+32k2sml(x) +22j+2(9x21)j+1hm=0B2j+32sml(x). (3.8)

    From Lemma 1 we have

    2(9x21)B2s1l(x)B2s13l(x)B2s1(2n+1)l(x)C2s1l(2h+1)(2n+1)(x)C2s1l(2n+1)(x)2(9x21)B2s1l(2n+1)(x) 0 mod(C2s1l(2h+1)(x)C2s1l(x)). (3.9)

    Applying inductive hypothesis (3.8), we have

    B2s1l(x)B2s13l(x)B2s1(2n+1)l(x)jk=12j+32j+3k(2j+3kk) (3.10)
     × 22j+32k(9x21)j+2khm=0B2j+32k2sml(x)0 mod(C2s1l(2h+1)(x)C2s1l(x)). (3.11)

    Combining (3.7)–(3.10) and Lemma 1, we have the congruence

    22n+1(9x21)n+1B2s1l(x)B2s13l(x)B2s1(2n+1)l(x)hm=0B2n+12sml(x)0 mod(C2s1l(2h+1)(x)C2s1l(x)). 

    This completes the proof of Theorem 1 by mathematical induction.

    Now we prove Theorem 2, we have

    hm=0(C2sml(2n+1)(x)(1)n(2n+1)C2sml(x)) = B2s1l(2h+1)(2n+1)(x)+B2s1l(2n+1)(x)2B2s1l(2n+1)(x)(1)n(2n+1)B2s1l(2h+1)(x)+B2s1l(x)2B2s1l(x) = n1k=0(1)k2n+12n+1k22n2k(2n+1kk)hm=0C2n+12k2sml(x). (3.12)

    Applying (3.11), Lemma 2 and the method of proving Theorem 1, we can deduce the congruence

    22n+1B2s1l(x)B2s13l(x)B2s1(2n+1)l(x)hm=0C2n+12sml(x)0 mod(B2s1l(2h+1)(x)+B2s1l(x)).

    In this paper, we study the divisible property of the general power sum of balancing polynomials and Lucas-balancing polynomials. By taking specific values for s and l in the Theorems 1 and 2, similar results can be obtained as studied in the literature. In this paper, we take x=1 and obtain the divisible property of the sequence hm=0B2n+12sml and hm=0C2n+12sml. We apply a simple relation between the balancing polynomials and the Chebyshev polynomials to further obtain the divisibility properties of hm=0U2n+12sml1(x) and hm=0T2n+12ml(x) in the Corollaries 5 and 6. This paper can help us to investigate the properties of polynomials and explore further relations between polynomials.

    The author declares he/she has not used Artificial Intelligence (AI) tools in the creation of this article.

    The author would like to express his/her sincere thanks to anonymous reviewers for their helpful comments and suggestions.

    This work is supported by the N. S. F. (12126357) of P. R. China.

    The author declares no conflict of interest.



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