On the divisibility of sums involving powers of multi-variable Schmidt polynomials Qi-Fei Chen1 and Victor J. W. Guo2*
arXiv:1412.5734v1 [math.NT] 18 Dec 2014
Department of Mathematics, Shanghai Key Laboratory of PMMP, East China Normal University, 500 Dongchuan Rd., Shanghai 200241, People’s Republic of China 1
[email protected], 2
[email protected], http://math.ecnu.edu.cn/~jwguo
Abstract. The multi-variable Schmidt polynomials are defined by Sn(r) (x0 , . . . , xn )
r n X 2k n+k xk . := k 2k k=0
We prove that, for any positive integers m, n, r, and ε = ±1, all the coefficients in the polynomial n−1 X (r) εk (2k + 1)Sk (x0 , . . . , xk )m k=0
are multiples of n. This generalizes a recent result of Pan on the divisibility of sums of Ap´ery polynomials. Keywords: Schmidt numbers, generalized Schmidt polynomials, Pfaff-Saalsch¨ utz identity AMS Subject Classifications: 11A07, 11B65, 05A10, 05A19
1
Introduction
Schmidt [7] introduced the numbers r n r X n+k n k=0
k
k
,
which, for r = 2, are called the Ap´ery numbers since Ap´ery [2] published his ingenious proof of the irrationality of ζ(3). Motivated by the work of Z.-W. Sun [9] on the Ap´ery polynomials, Guo and Zeng [5] introduced the Schmidt polynomials Sn(r) (x)
=
r n r X n+k n k=0
*
k
Corresponding author.
1
k
xk ,
and obtained the following congruence: n−1 X
(r)
εk (2k + 1)Sk (x) ≡ 0 (mod n),
k=0
where ε = ±1. Here and in what follows, we say that a polynomial P (perhaps in multivariables) is congruent to 0 modulo n, if all the coefficients in P are multiples of n. Recently, Pan [6] proved that, for any positive integers r, m and n, there holds n−1 X
(r)
εk (2k + 1)Sk (x)m ≡ 0 (mod n),
(1.1)
k=0
where ε = ±1, and thus proved the related conjectures by Sun [9,10] and Guo and Zeng [5]. Note that, it is not easy to prove (1.1) for m > 2 directly, because there are no simple formula to calculate the coefficients in the left-hand side of (1.1). In fact, Pan [6] proved (1.1) by establishing a q-analogue. Let r n X 2k n+k (r) xk . Sn (x0 , . . . , xn ) = k 2k k=0
We shall give a generalization of (1.1) as follows.
Theorem 1.1. For any positive integers r, m, n and nonnegative integer a, we have n−1 X
(r)
(1.2)
(r)
(1.3)
(2k + 1)Sk (x0 , . . . , xk )m ≡ 0 (mod n),
k=0
n−1 X
(−1)k (2k + 1)Sk (x0 , . . . , xk )m ≡ 0 (mod n).
k=0
It is clear that, if we take xk = obtain the congruence (1.1).
2
2k r−1 k x k
in the congruences (1.2) and (1.3), then we
Proof of Theorem 1.1 for m = 1
We first establish the following lemma. (r)
Lemma 2.1. For ℓ, m ∈ N and r > 1, there exist integers bm,k (m 6 k 6 rm) divisible k by m such that r X rm 2k ℓ+k 2m ℓ+m (r) . = bm,k k 2k m 2m k=m 2
(2.1)
Proof. We proceed by mathematical induction on r. It is clear that the identity (2.1) (1) holds for r = 1 with bm,m = 1 = m . Suppose that the statement is true for r. A special m case of the Pfaff-Saalsch¨ utz identity (see [8, p. 44, Exercise 2.d] and [1]) reads
ℓ+m m+k
X k ℓ+m+i m ℓ−m ℓ+k , = ℓ m+i−k i k i=0
which can be rewritten as X k (m + k)!i! m 2k ℓ+k ℓ−m ℓ+m+i = . k 2k i i k − i (m + i)!k! i=0 Multiplying both sides of (2.1) by
ℓ+m 2m
(2.2)
and applying (2.2), we obtain
r+1 2m ℓ+m m 2m rm X (r) ℓ + k 2k ℓ + m bm,k = 2m k 2k k=m k rm X X (m + k)!i! m ℓ−m ℓ+m+i ℓ+m (r) = bm,k 2m i i (m + i)!k! k − i i=0 k=m −1 rm X k X m + k ℓ + m + i 2(m + i) m m+i k (r) . = bm,k m + i 2(m + i) 2m k − i m m k=m i=0
(2.3)
Letting m + i = j and exchanging the summation order in the right-hand side of (2.3), we see that the identity (2.1) holds for r + 1 with (r+1) bm,j
−1 X rm m+k m k j (r) , b = 2m j−k m k=m m,k m
m 6 j 6 (r + 1)m.
(2.4)
(r) (r+1) k By the induction hypothesis, each bm,k is divisible by m , and so bm,j is an integer divisible by mj . This proves that the identity (2.1) is also true for r + 1. Proof of Theorem 1.1 for m = 1. Applying (2.1) and the identity n−1 X
n+k n 2k ℓ+k , =n (2ℓ + 1) k k+1 k 2k ℓ=k
3
(2.5)
we have n−1 X
(2k +
(r) 1)Sk (x0 , . . . , xk )
k=0
r k X 2i k+i xi (2k + 1) = i 2i i=0 k=0 n−1 n−1 ri X X X (r) 2j k+j = xi bi,j (2k + 1) j 2j i=0 k=i j=i ri n−1 X X n+j n (r) . bi,j xi =n j j+1 j=i i=0 n−1 X
Similarly, applying (2.1) and the identity n−1 X ℓ=k
n−1 n+k 2k ℓ+k n−1 , = (−1) n (−1) (2ℓ + 1) k k k 2k ℓ
(2.6)
we get n−1 X
k
(−1) (2k +
(r) 1)Sk (x0 , . . . , xk )
= (−1)
n−1
n
n−1 X
xi
i=0
k=0
ri X j=i
(r) bi,j
n−1 n+j . j j
This completes the proof.
3
Proof of Theorem 1.1 for m > 1
It is easy to see that r !m k X 2i k + i (r) xi εk (2k + 1) εk (2k + 1)Sk (x0 , . . . , xk )m = i 2i i=0 k=0 k=0 r n−1 m X X Y 2ij k + ij k xij , (3.1) = ε (2k + 1) i 2i j j k=0 06i ,...,i 6k j=1 n−1 X
n−1 X
1
m
Exchanging the summation order, we may rewrite the right-hand side of (3.1) as X
n−1 X
06i1 ,...,im 6n−1 k=max{i1 ,...,im }
By Lemma 2.1, we see that
k+ij r 2ij , 2ij ij
r m Y 2ij k + ij xij . ε (2k + 1) i 2i j j j=1 k
(3.2)
as a polynomial in k, is a linear combination of
2rij k + rij k + ij + 1 2ij + 2 2ij k + ij ,..., , rij 2rij ij + 1 2ij + 2 ij 2ij 4
with integer coefficients. Note that (2.2) with m = j and k = i also gives X i 2j + 2k j+k ℓ+j+k j i+j 2j ℓ + i 2i ℓ + j . = j+k 2j + 2k k i−k i j 2j i 2i k=0
(3.3)
Therefore, Q by applying 2.1 and then repeatedly using (3.3), we find that the Lemma m k+ij r 2ij expression j=1 2ij , as a polynomial in k, can be written as a linear combination ij of k + r(i1 + · · · + im ) 2r(i1 + · · · + im ) k + i + 1 2i + 2 k + i 2i , ,..., , r(i1 + · · · + im ) 2r(i1 + · · · + im ) i+1 2i + 2 i 2i with integer coefficients, where i = max{i1 , . . . , im }. Finally, using (2.5) and (2.6), we deduce that the coefficients in (3.2) are all multiples of n. Namely, Theorem 1.1 holds for m > 1. Remark. The identity (3.3) was also utilized by the second author [3] to confirm some conjectures of Z.-W. Sun [9, 10] on sums of powers of Delannoy polynomials.
4
Concluding remarks
Theorem 1.1 can be further generalized as follows. Let (x)0 = 1 and (x)n = x(x+1) · · · (x+ n − 1) for all n > 1. Then, for any nonnegative integer a, the proof of [4, Lemma 2.1] gives X a k+j+i a a k+j (2j + 1)2i , (4.1) = ci (j, a) k (k + 1) j+i 2j i=0 where c0 (j, a), . . . , ca (j, a) are integers independent of k. Using the identity (4.1) and the same idea in the previous section, we can show that (see the proof of [4, Theorem 1.2]) n−1 X
(r)
εk (2k + 1)k a (k + 1)a Sk (x0 , . . . , xk )m ≡ 0
(mod n),
k=0
where ε = ±1. Noticing that (2k + 1)
2a
a X a i i 4 k (k + 1)i , = (4k + 4k + 1) = i i=0 2
a
we immediately get n−1 X
(r)
εk (2k + 1)2a+1 Sk (x0 , . . . , xk ) ≡ 0
k=0
5
(mod n),
(4.2)
where ε = ±1. Acknowledgments. This work was partially supported by the Fundamental Research Funds for the Central Universities and the National Natural Science Foundation of China (grant 11371144).
References [1] G.E. Andrews, Identities in combinatorics, I: on sorting two ordered sets, Discrete Math. 11 (1975), 97–106. [2] R. Ap´ery, Irrationalit´e de ζ(2) et ζ(3), Ast´erisque 61 (1979), 11–13. [3] V.J.W. Guo, Some congruences involving powers of Delannoy polynomials, preprint, 2014. [4] V.J.W. Guo and J. Zeng, New congruences for sums involving Apery numbers or central Delannoy numbers, preprint, Int. J. Number Theory 8 (2012), 2003–2016. [5] V.J.W. Guo and J. Zeng, Proof of some conjectures of Z.-W. Sun on congruences for Ap´ery polynomials, J. Number Theory, 132 (2012), 1731–1740. [6] Hao Pan, On divisibility of sums of Ap´ery polynomials, J. Number Theory 143 (2014), 214–223. [7] A.L. Schmidt, Generalized q-Legendre polynomials, J. Comput. Appl. Math. 49 (1993), 243–249. [8] R.P. Stanley, Enumerative Combinatorics, Vol. I, Cambridge University Press, Cambridge, 1997. [9] Z.-W. Sun, On sums of Ap´ery polynomials and related congruences, J. Number Theory, 132 (2012), 2673–2699. [10] Z.-W. Sun, Congruences involving generalized trinomial coefficients, Sci. China 57 (2014), 1375-1400.
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