Melham’s sums for some Lucas polynomial sequences

: A Lucas polynomial sequence is a pair of generalized polynomial sequences that satisfy the Lucas recurrence relation. Special cases include Fibonacci polynomials, Lucas polynomials, and Balancing polynomials. We define the ( a, b ) -type Lucas polynomial sequences and prove that their Melham’s sums have some interesting divisibility properties. Results in this paper generalize the original Melham’s conjectures.


Introduction
Melham [8] made two conjectures on sums involving odd powers of certain Fibonacci numbers and Lucas numbers with even index.Conjecture 1.Let m ≥ 1 be a positive integer.Then the sum can be expressed as (F 2n+1 − 1) 2 M 2m−1 (F 2n+1 ), where M 2m−1 (x) is a polynomial of degree 2m − 1 with integer coefficients.
Conjecture 2. Let m ≥ 0 be a positive integer.Then the sum 2k can be expressed as (L 2n+1 − 1)N 2m (L 2n+1 ), where N 2m (x) is a polynomial of degree 2m with integer coefficients.
Ozeki [9] was the first one to give an expression for the sum n k=1 F 2m+1 2k as a polynomial in power of F 2n+1 .Subsequently, Prodinger [11] did the same thing independently and obtained more summation formulae of similar type.Indeed, they gave the following so-called Ozeki-Prodinger identity: In 2004, Wiemann and Cooper [16] proved that the summation is a multiple of 5 m .Hence, the second term of Ozeki-Prodinger identity is an integer.Sun et al. [12] gave proof for Melham's two conjectures.In fact, in a spirit of Wang and Zhang's previous work [13], they proved the polynomial version of Melham's conjectures as follows.
Theorem 1.1 ( [12]).Let F n (x), L n (x) be the n-th Fibonacci polynomial and the n-th Lucas polynomial, respectively.For any positive integers m and n, the Melham's sum can be expressed as (F 2n+1 (x) − 1) 2 P 2m−1 (x, F 2n+1 (x)), where P 2m−1 (x, y) is a polynomial in two variables x and y with integer coefficients and of degree 2m − 1 in y.
Theorem 1.2 ( [12,13]).For any positive integers m and n, the Melham's sum can be expressed as (L 2n+1 (x) − 1)Q 2m (x, L 2n+1 (x)), where Q 2m (x, y) is a polynomial in two variables x and y with integer coefficients and of degree 2m in y.
In 2019, Chen and Wang [3] used a different approach to conclude that The above congruence with x = 1 also gives a partial answer of the Conjecture 1.
For any two non-zero polynomials p(x) and q(x) with integer coefficients, we introduce the Lucas polynomial sequence W n (x) [4,5] as the sequence of polynomials satisfying the Lucas recurrence relation with the initials W 0 (x) = 0, and W 1 (x) = 1.Its companion sequence w n (x) was defined in [4,5] by satisfying the same recurrence relation with slightly different initials w 0 (x) = 2, and w 1 (x) = p(x).
A (a, b)-type Lucas polynomial sequence is a pair of generalized polynomials V (a,b) n (x) := V n (x), and v (a,b) (x) := v n (x) which is defined as p(x) = ax and q(x) = b for integers a, b with a 2 + b 2 ̸ = 0.That is, they satisfy recurrence relations with initials V 0 (x) = 0, V 1 (x) = 1, v 0 (x) = 2, and v 1 (x) = ax, respectively.For simplicity, we denote the (a, 1)-type Lucas polynomial sequence by W n (x) and w n (x), and the (a, −1)-type Lucas polynomial sequence by W n (x) and w n (x).It is worth noting that many important polynomial sequences satisfy the recurrence relation (1), such as Fibonacci polynomial, Lucas polynomial, Pell polynomial, and so on.See Table 1 on the next page.
Our main results are listed as follows.
Theorem 1.3.For any positive integers m and n, the Melham's sum for the (a, b)-type Lucas polynomial sequence Theorem 1.4.For any positive integers m and n, the Melham's sum , where H 2m−1 (x, y) is a polynomial in two variables x and y with integer coefficients and of degree 2m − 1 in y.
Table 1.Special cases of Lucas polynomials Theorem 1.5.For any positive integers m and n, the Melham's sum , where M 2m−1 (x, y) is a polynomial in two variables x and y with integer coefficients and of degree 2m − 1 in y.
An outline of this paper is as follows.In Section 2, we derive basic properties of (a, b)-type Lucas polynomials and derive the expansions of sum involving odd powers of (a, b)-type Lucas polynomials with even index.This paper also gives some analog results of Wiemann and Cooper in [16].We prove Theorem 1.3 in Section 3, and then discuss the case b = 1 and prove Theorem 1.4 in Section 4. In light of expressions W mn (x) (respectively, w mn (x)) as a polynomial in W n (x) (respectively, polynomial in w n (x)) for odd m, we derive the Ozeki-Prodinger-like identities for some Lucas polynomial sequences.The discussion of the case b = −1 and the proof of Theorem 1.5 will be given in Section 5. Some remarks are made in the concluding section.

Preliminaries
Let a and b be integers.According to the recurrence relation (1), we define the negative index of (a, b)-type Lucas polynomial sequences as below: for any positive integer n.
. Also it is easy to derive, for n ≥ 1, the Binet formula: Proposition 2.1.For integers m and n, we have (e) For a ̸ = 0, , where i 2 = −1.

y) ay
The proof is similar when a < 0, so we omit here.
It is also easy to obtain the following relation between V and v-polynomials, for any integer m.For instance, one can prove this by induction on m.
Proposition 2.2.There are explicit formulae for V n (x) and v n (x): and Proof.By part (d) of Proposition 2.1, it is sufficient to show the explicit formula for v n (x).We prove by induction on n.For n = 1, the formula holds obviously.Suppose that it is true for n < k with k ≥ 2. If k is even, say k = 2t, by (1) we have Here we have used the combinatorial identity: This proves the explicit formula for v k (x) when k is even.Similarly, the formula holds when k = 2t − 1 is odd.So our proof is done by induction.
For integers m ≥ 0 and n ≥ 1, by the Binet formula for V n (x), we have This implies Similarly, we have The next two propositions deal with a divisibility of V −v type convolution which is motivated by a result in [16].

Proposition 2.3 ([16]
).For any non-negative integer m, we let f m (y) be a polynomial in y defined by Then f m (y) has a polynomial factor (y − 1) 2m+1 .
Lemma 2.1 (See also Lemma 3 in [16]).Let m and k be integers with k < m.Then where h(j) is any polynomial of odd degree 2k+1 and h(i Proof.Let ∆ denote the forward-difference operator.That is, ∆h(x) = h(x + 1) − h(x).We have The second equality holds by the hypothesis of the polynomial h(j).However, ∆ 2m+1 h(0) = 0 since h(j) is of degree 2k + 1 < 2m + 1.Thus, we conclude our assertion.
Proposition 2.4.For a non-negative integer m, we let g m (y) be a polynomial in y defined by Then g m (y) has a polynomial factor (y − 1) m (y + 1) 2m+1 .
Proof.It is clear that g m (y) has a polynomial factor (y−1) m .(See the paragraph after Proposition 2.3.)Let h m (y) := m j=0 (−1) m−j 2m+1 m−j (y 2j+1 + 1)/(y 2j+1 − 1).We rewrite h m (y) as We prove that g m (y) has a polynomial factor (y + 1) 2m+1 by claiming that Clearly h m (−1) = 0 and this proves (7) when p = 0. Now we apply a result given by Leslie [7]: For any n times continuously differentiable function g(x), we have Then we have So we have for p ≥ 1 Following [16], we let a p,r := p k=r (−1) k+r 2 p−k p+1 k+1 k r , and have (see Lemma 5 in [16]) for two positive integers p and t with p ≥ 2t.
We need the expansion where s(p, ℓ) is the signed Stirling number of the first kind.Now we should rewrite the above evaluation as This expression vanishes by dividing the summation into three cases by ℓ = 0, ℓ ≥ 1 odd, and ℓ ≥ 1 even.The result is zero when ℓ = 0 since s(p, 0) = 0.When ℓ is a odd number, by Lemma 2.1 since ℓ ≤ 2m − 1 (ℓ ≤ p ≤ 2m and ℓ is odd), we see that the inner sum When ℓ is an even number, say ℓ = 2t for some integer t ≥ 1, we just apply (8).Therefore, we have showed that d p dy p h m (y) y=−1 = 0 for p = 0, 1, . . ., 2m and complete the proof.
We remark that all coefficients of both f m (y) and g m (y) are even for m ≥ 1.It is because of 3 Proof of Theorem 1.3 For |b| > 1, suppose that . We consider the whole identity under modulo |b| and have It can not happen based on the discussion in the previous paragraph.
Remark.If b = −1, for our convenience, we consider the Balancing polynomial B n (x) and Lucas-Balancing polynomial C n (x) [10].Assume that n = m = 1, we get Altogether, in view of discussion in the next section, Theorem 1.3 may be rephrased as the Melham's sum 4 The case b = 1 Throughout this section, we assume that b = 1.Recall that, the (a, 1)-type Lucas polynomial sequences are denoted by {W n (x)} n∈Z and {w n (x)} n∈Z .We begin with a simple lemma.
Lemma 4.1.For any positive integers n and m, we have if m is even, and Using the Binet formula and noting that α(x)β(x) = −1, we have The first identity follows by Proposition 2.1 and the others can be proved similarly.
From (5), it follows by Lemma 4.1 that where Now we need another important result, which is in the spirit of Jennings' theorem [6].
Lemma 4.2.For any non-negative integers n and q, we have and Proof.Let p = 2q + 1 be an odd integer and and note that yz = (−1) n .Write We need the following two identities: and By identity (12) we have for n is odd that When n is even, identity (13) gives So the first expression of (10) follows.Differentiating both sides of the first expression of (11) with respect to x, and by the part (d) of Proposition 2.1, we obtain the second expression of (10).Now for (11), consider w pn (x) When n is odd, the identity (13) gives When n is even, by (12) we have Therefore we obtain the first identity of (11).
In addition, we apply the following two identities and to the expression of w (2q+1)n (x)/w n (x) to yield Remark.If we apply identities ( 14) and (15) to the expression of W (2q+1)n (x) /W n (x) in the proof of Lemma 4.2, we also get the second expression of (10).By the way, all identities (12), ( 13), ( 14) and ( 15) can be proved easily by induction on q.
Taking n = 1 into Lemma 4.2, it immediately infer to the following.
We remark here that, from Lemma 4.2, expansions of W 2j+1 (x) and w 2j+1 (x) in (ax) can be derived, namely respectively.Also these two identities can be derived from Proposition 2.2.Another way is to use part (e) of Proposition 2.1.By (16) we have Thus, by the part (e) of Proposition 2.1, it gives the expansion of w 2j+1 (x) in (ax).
Corollary 4.2.Let the (a, 1)-type W sequence be {W n } n≥0 := {W n (1)} n≥0 .For any two odd primes p, q and a positive integer n, we have where * p is the Legendre symbol and Proof.In light of identity (10), the coefficient of the right hand side divides by p = 2q + 1 for ℓ = 0 to q − 1.Hence, by the Euler's criterion, we have Letting n = 1 and n = q into the above congruence, we have respectively.Thus, W pq ≡ W p W q (modp).By symmetry, W pq ≡ W p W q (modq).
Let {w n (1)} n≥0 be the (a, 1)-type w sequence.Similarly, by identity (11), we conclude that for any odd prime p and any positive integer n.Back to the proof of Lemma 4.2, we note that and with y = α n (x), z = β n (x) and yz = (−1) n .Now we need two slightly different formulae given by and Both identities ( 18) and ( 19) follow easily by induction on q.
Repeating the same argument, we have Combining (20) with identity (10), it leads to the divisible relation for any positive integers m, n with m | n.
Theorem 4.1.We have Ozeki-Prodinger-like identities for the (a, 1)-type Lucas polynomial sequences: Proof.From identity (9) and by Lemma 4.2, we have The first term of the right hand side can be rewritten as If we start with identity (9) and use identity (10) to expand W (2n+1)(2j+1) (x) as a polynomial in power of w n (x), the second expression of n k=1 W 2m+1 2k (x) follows.Similar proof for (ii), we omit here.
Remark.There are certainly formulae of power sums involving (a, 1)-type Lucas polynomial sequences (e.g. for , and they can be derived easily through the same way. Let H 2m+1 (x, y) be a polynomial in two variables x and y with the degree 2m + 1 of y defined by with .
Thus, we rewrite the desired identity as where h(j; ℓ, m) = b(j)p(j; ℓ, m).Now h(j; ℓ, m) meets all conditions listed in Lemma 2.1 and the desired identity follows by Lemma 2.1.
Remark.One may compare the above result with Lemma 2.6 in [12].Unfortunately, we feel that the statement of Lemma 2.6 in [12] is wrong.The correct version requests an extra condition on the polynomial p(j) as mentioned in Lemma 4.3.
We are now at the stage to give a proof of Theorem 1.4.
The case ℓ = 0 is trivial.For ℓ = 1, we have and T 1,m (2i/a) = 0 since w 2m+1 (2i/a) = (−1) m 2i by ( 17), where i 2 = −1.This implies that w 2m−1 (x)w 2m+1 (x)T 1,m (x) has a polynomial factor a 2 x 2 + 4. In the following we claim that T ℓ,m (2i/a) = 0 for 2 ≤ ℓ ≤ m and By the definition of T ℓ,m (x) we have Hence, by Lemma 4.3 with h(j) = 2m − 2j + 1, we have that T ℓ,m (2i/a) = 0 for 2 ≤ ℓ ≤ m.Now apply a result of Leslie [7] (see page 8 in this paper), we obtain For two positive integers n, p with n ≥ p, by (4) we note that To see why the last step holds, let In light of the identity we have Thus, we obtain Our assertion follows if we can prove that the inner sum vanishes.That is, where It is routine that one expresses H(j) as a product of integers and check that H(i) = −H(2m−i+ 1) for i = 0, 1, . . ., m.Also we note that H(j) is a polynomial in j of degree 2p+1 < 2(p+1)+1, an odd number.Thus, (24) holds by Lemma 4.3.
Another consideration leads to the extension of Melham's Conjecture 2 or Theorem 1.2.
Theorem 4.2.For any non-negative integers n and m, the Melham's sum can be expressed as (ax) m (w 2n+1 (x) − ax) S 2m (x, w 2n+1 (x)), where S 2m (x, y) is a polynomial in two variables x and y with integer coefficients and of degree 2m in y.
Proof.The fact that (ax) m is a polynomial factor of the Melham's sum follows similarly by viewing the proof of our Theorem 1.4.We note that this polynomial factor (ax) m actually dues to the part of the product In view of (ii) in Theorem 4.1, we let Then by ( 4), From this and the binomial theorem, we see S 2m+1 (x, ax) = 0 and this implies the sum has a polynomial factor (w 2n+1 (x) − ax).
To see that S 2m (x, y) is a polynomial with integer coefficients, it is suffices to show is an integer.
Corollary 4.3.Let Q n (x) be n-th Pell-Lucas polynomial and the Melham's sum for Q n (x) be define by Then for any positive integers n and m, we have the quotient Q(n, m; x)/(Q 2n+1 (x) − 2x) is an integer polynomial in x.
One can substitute the Melham's sum in Theorem 4.2 with The only slightly difference is that the value k begins with zero under the summation sign.It is easy to derive that n k=0 Thus, we conclude a kind of variation of Theorem 4.2.
Theorem 4.3.For two non-negative integer n, m, the sum can be expressed as (ax) m (w 2n+1 (x) + ax)S 2m (x, w 2n+1 (x)), where S 2m (x, y) is a polynomial in two variables x and y with integer coefficients and of degree 2m in y. 5 The case b = −1 In this section, we discuss the case when b = −1.Recall that, in our notation, V (a,−1) n (x) := W n (x).That is, the polynomial sequence {W n (x)} n≥0 satisfies the recurrence relation with initial values W 0 (x) = 0 and W 1 (x) = 1.Let v (a,−1) n (x) := w n (x) by analogy.The following identities are easy to prove by the Binet formula for W (x) and w(x).Lemma 5.1.For any positive integers n and m, we have and From (6) and Lemma 5.1, we get Lemma 5.2.For two non-negative integers n and q, we have and Proof.Let y = α n (x), z = β n (x) and note that yz = 1.We compute (−1) q+ℓ q + ℓ q − ℓ (y + z) 2ℓ by identity (14).
Our second assertion follows and the proof completes.
Theorem 5.1.We have Ozeki-Prodinger-like identities for the (a, 1)-type Lucas polynomial sequences: Proof.We should only prove (ii), There is an interesting implication when y is substituted for α(x)/β(x) in the Proposition 2.4.Then we obtain where It is time to prove our main theorem in this section.
Proof of Theorem 1.To see that M 2m−1 (x, y) ∈ Z[x, y], we just notice that (25) and the fact that (28).From this, we conclude that the Melham's sum can be divisible by (W 2n+1 (x)−1).To see that N 2m (x, y) is an integer polynomial, we just notice (25).
We raise an open question in studying Melham's sum for some Lucas polynomial sequences.The polynomials H 2m−1 (x, y), S 2m (x, y), M 2m−1 (x, y) and N 2m (x, y) can be viewed as polynomials in y of a suitable degree with integer coefficients.Are they irreducible polynomials in y for all m ≥ 1? We hope to attract the attention of interested readers.