Factors of sums involving q-binomial coefficients
and powers of q-integers
Victor J. W. Guo1 and Su-Dan Wang2
arXiv:1701.07016v1 [math.CO] 24 Jan 2017
1
School of Mathematical Sciences, Huaiyin Normal University, Huai’an, Jiangsu 223300,
People’s Republic of China
[email protected]
2
Department of Mathematics, East China Normal University, Shanghai 200062,
People’s Republic of China
[email protected]
Abstract. We show that, for all positive integers n1 , . . . , nm , nm+1 = n1 , and any nonnegative integers j and r with j 6 m, the expression
−1 X
m n1
Y
ni + ni+1
1 n1 + nm
2r jk 2 −(r+1)k
[2k][k] q
ni + k
n1
[n1 ]
i=1
k=1
is
in q with integer cofficients, where [n] = 1 + q + · · · + q n−1 and
na Laurent
Qk polynomial
n−i+1
= i=1 (1 − q
)/(1 − q i ). This gives a q-analogue of a divisibility result on the
k
Catalan triangle obtained by the first author and Zeng, and also confirms a conjecture of
the first author and Zeng. We further propose several related conjectures.
Keywords: q-binomial coefficients; q-Catalan triangle; q-Pfaff-Saalschütz identity; q-Narayana
numbers; q-super Catalan numbers
AMS Subject Classifications (2000): 05A30, 05A10, 11B65.
1
Introduction
A few years ago, motivated by the divisibility results in [3, 5, 6, 12, 18, 20, 22, 31], the first
author and Zeng [16] proved that
−1 X
n 2n
2 2n
k 2r+1 ∈ Z for r > 1,
2
n−k
n n
(1.1)
−1 X
m n1
Y
2 n1 + nm
ni + ni+1
2r+1
∈ Z for r > 0,
k
n
+
k
n1
n1
i
i=1
k=1
(1.2)
k=1
and more general
where nm+1 = n1 .
1
The first objective of this paper is to give a q-analogue of (1.1) and (1.2). Recall that
n
and the q-shifted factorials (see [9]) are defined
the q-integers are defined by [n] = 1−q
1−q
by (a; q)0 = 1 and (a; q)n = (1 − a)(1 − aq) · · · (1 − aq n−1 ) for n = 1, 2, . . . . The q-binomial
coefficients are defined as

(q; q)n


, if 0 6 k 6 n,
n
= (q; q)k (q; q)n−k

k
0,
otherwise.
Our main results can be stated as follows.
Theorem 1.1. For j = 0, 1 and all positive integers n and r, the expression
−1 X
n
2n
1 2n
2r (r+1)(n−k)+jk 2
[2k][k] q
n+k
[n]2 n
k=1
(1.3)
is a polynomial in q with integer coefficients.
Theorem 1.2. Let n1 , . . . , nm , nm+1 = n1 be positive integers. Then for any non-negative
integers j and r with j 6 m, the expression
−1 X
m n1
Y
ni + ni+1
1 n1 + nm
2r jk 2 −(r+1)k
(1.4)
[2k][k] q
ni + k
n1
[n1 ]
i=1
k=1
is a Laurent polynomial in q with integer coefficients.
One may wonder why to consider [2k][k]2r /[2] as a q-analogue of k 2r+1 rather than
[k]2r+1 . The idea comes from Schlosser’s work [21] on q-analogues of sums of powers of
the first n consecutive natural numbers and the first author and Zeng’s work [14] which
positively answers Schlosser’s question.
The first author and Zeng [16] also introduced the q-Catalan triangle with entries
given by
[k] 2n
Bn,k (q) :=
, 1 6 k 6 n.
[n] n − k
The second objective is to give the following congruence related to the q-Catalan triangle.
Theorem 1.3. Let n be a positive integer. Let r > 0 and s > 1 such that r 6≡ s (mod 2).
Then, for 0 6 j 6 s, the expression
−1 X
n
2n
2
s
(1 + q k )[k]r Bn,k
(q)q jk −(r+s+1)k/2
n
k=1
is a Laurent polynomial in q with integer coefficients.
It is easy to see that Theorem 1.3 is a generalization of [16, Theorem 1.4]. Letting
q = 1 in Theorem 1.3, we confirm a conjecture in [16].
The paper is organized as follows. We shall prove Theorems 1.1–1.3 in Sections 2–4,
respectively. The q-Pfaff-Saalschütz identity will play an important role in our proof. We
give some consequences of Theorem 1.1 in Section 5. Finally, some open problems will be
proposed in Section 6.
2
2
Proof of Theorem 1.1
We first consider the j = 0 case. Let
Sr (n; q) =
n
X
2r (r+1)(n−k)
[2k][k] q
k=1
2n
.
n+k
It is easy to see that
S0 (n; q) =
n
X
[2k]q
n−k
k=1
X
X
n
n
2n
2n
2n
[n − k]
−
[n − k + 1]
=
n−k
n
−
k
+
1
n−k
k=1
k=1
2n
.
= [n]
n
For r > 1, noticing the relation
2n − 2
2n
2n
n−k
2
2
,
− [2n][2n − 1]
q
= [n]
[k]
n+k−1
n+k
n+k
we have
Sr (n; q) = [n]2 Sr−1 (n; q) − q r [2n][2n − 1]Sr−1 (n − 1; q), n = 1, 2, . . . .
(2.1)
It follows that
2n − 2
2n
2 2n
.
= [n]
− q[2n][2n − 1][n − 1]
S1 (n; q) = [n]
n
n−1
n
This proves that S1 (n; q) ≡ 0 (mod [n]2 2n
). Applying the recurrence relation (2.1) and
n
by induction on r, we can prove that, for all positive integers r, there holds
2 2n
).
(2.2)
Sr (n; q) ≡ 0 (mod [n]
n
3
For j = 1, let
Tr (n; q) =
n
X
2r (r+1)(n−k)+k 2
[2k][k] q
k=1
It is well known that
(2.2), we have
n
k
q −1
=
2n
.
n+k
n k(k−n)
2
q
. Hence, Tr (n; q) = q n +2rn+2n−2r−1 Sr (n; q −1 ). By
k
2n
).
Tr (n; q) ≡ 0 (mod [n]
n
2
Since both Sr (n; q) and Tr (n; q) are polynomials in q, we obtain the desired conclusion.
3
3
Proof of Theorem 1.2
We will need the q-Pfaff-Saalschütz identity (see [9, (II.12)] or [15, 30]):
n1 + n2
n1 + k
where
1
(q;q)n
n2 + n3
n2 + k
nX
2
1 −k
n3 + n1
q s +2ks (q; q)n1 +n2 +n3 −k−s
,
=
(q;
q)
(q;
q)
(q;
q)
(q;
q)
(q;
q)
n3 + k
s
s+2k
n
−k−s
n
−k−s
n
−k−s
1
2
3
s=0
(3.1)
= 0 if n < 0. Denote (1.4) by S(n1 , . . . , nm ; r, j, q). Namely,
S(n1 , . . . , nm ; r, j, q) =
n1
(q; q)n1 −1 (q; q)nm X
2
[2k][k]2r q jk −(r+1)k C(n1 , . . . , nm ; k),
(q; q)n1+nm
k=1
where
C(a1 , . . . , al ; k) =
l Y
ai + ai+1
ai + k
i=1
(3.2)
(al+1 = a1 ).
It is easy to see that, for m > 3,
(q; q)n2 +n3 (q; q)nm +n1 n1 + n2 n1 + n2
C(n3 , . . . , nm ; k).
C(n1 , . . . , nm ; k) =
n2 + k
(q; q)n1 +n2 (q; q)nm +n3 n1 + k
(3.3)
Letting n3 → ∞ in (3.1), we get
n1 + n2
n1 + k
nX
2
1 −k
n1 + n2
q s +2ks (q; q)n1+n2
=
.
n2 + k
(q; q)s (q; q)s+2k (q; q)n1 −k−s (q; q)n2−k−s
s=0
(3.4)
Substituting (3.3) and (3.4) into (3.2), we obtain
S(n1 , . . . , nm ; r, j, q)
2
2
n1 nX
1 −k
(q; q)n2+n3 (q; q)n1 −1 (q; q)nm X
q s +2ks+jk −(r+1)k [2k][k]2r C(n3 , . . . , nm ; k)
=
(q; q)nm+n3
(q; q)s (q; q)s+2k (q; q)n1 −k−s (q; q)n2−k−s
k=1 s=0
2
n1
l
X
(q; q)n2+n3 (q; q)n1 −1 (q; q)nm X
q (j−1)k −(r+1)k [2k][k]2r C(n3 , . . . , nm ; k)
l2
=
,
q
(q; q)nm+n3
(q; q)l−k (q; q)l+k (q; q)n1 −l (q; q)n2−l
l=1
k=1
where l = s + k. Observing that, for m > 3,
(q; q)nm +n3
C(n3 , . . . , nm ; k)
=
C(l, n3 , . . . , nm ; k),
(q; q)l−k (q; q)l+k
(q; q)n3 +l (q; q)nm+l
we get the following recurrence relation
n1
X
n2 + n3
l2 n1 − 1
S(l, n3 , . . . , nm ; r, j − 1, q).
S(n1 , . . . , nm ; r, j, q) =
q
n
l
−
1
2−l
l=1
4
(3.5)
On the other hand, for m = 2, applying (3.4) we may deduce that
n1
X
n2
l2 n1 − 1
S(l; r, j − 1, q).
q
S(n1 , n2 ; r, j, q) =
l
l−1
(3.6)
l=1
We now proceed by induction on m. For m = 1, the conclusion follows readily from
the proof of Theorem 1.1. Suppose that the expression S(n1 , . . . , nm−1 ; r, j, q) is a Laurent
polynomial in q with integer coefficients for some m > 2 and 0 6 j 6 m − 1. Then by the
recurrence (3.5) or (3.6), so is S(n1 , . . . , nm ; r, j, q) for 1 6 j 6 m. It is easy to see that
S(n1 , . . . , nm ; r, 0, q) = S(n1 , . . . , nm ; r, m, q −1 )q n1 n2 +n2 n3 +···+nm−1 nm −n1 −2r
for m > 2.
Therefore, the expression S(n1 , . . . , nm ; r, 0, q) is also a Laurent polynomial in q with
integer coefficients. This completes the inductive step of the proof.
4
Proof of Theorem 1.3
Let Φn (x) be the n-th cyclotomic polynomial. We need the following result (see [19,
Equation (10)] or [7, 17]).
Proposition 4.1. The q-binomial coefficient m
can be factorized into
k
Y
m
Φd (q),
=
k
d
where the product is over all positive integers d 6 m such that ⌊k/d⌋ + ⌊(m − k)/d⌋ <
⌊m/d⌋.
By Proposition 4.1, for any positive integer n, we have
2n
, [n] = 1.
gcd
n
(4.1)
Let r + s ≡ 1 (mod 2), m = s > 1, and 0 6 j 6 s. Setting n1 = · · · = ns = n in
Theorem 1.2, we see that
s
−1 X
n
1 2n
2n
r+s−1 jk 2 −(r+s+1)k/2
[2k][k]
q
n+k
[n] n
k=1
is a Laurent polynomial in q with integer coefficients.
2nNote
that Bn,k (q) is a polynomial
in q with integer coefficients (see [13, 16]). Hence, [k] n+k = [n]Bn,k (q) is clearly divisible
by [n]. It follows that
n
gcd 2n
, [n]s X
2
s
n
2n
(1 + q k )[k]r Bn,k
(q)q jk −(r+s+1)k/2
n
k=1
is a Laurent polynomial in q with integer coefficients. The proof then follows from (4.1).
5
5
Some consequences of Theorem 1.2
In this section, we will give some consequences of Theorem 1.2 and confirm some conjectures in [16, Section 7]. Letting n2i−1 = m and n2i = n for i = 1, . . . , r in Theorem 1.2
and observing the symmetry of m and n, we obtain
Corollary 5.1. Let m, n, r and s be positive integers, and let a and j be non-negative
integers with j 6 r + s. Then the expression
r
r −1 X
m
gcd([m], [n]) m + n
m+n
2a jk 2 −(a+1)k m + n
[2k][k] q
n+k
m+k
m
[m][n]
k=1
is a Laurent polynomial in q with integer coefficients.
Letting n3i−2 = l, n3i−1 = m and n3i = n for i = 1, . . . , r in Theorem 1.2, we get
Corollary 5.2. Let l, m, n and r be positive integers, and let a and j be non-negative
integers with j 6 3r. Then the expression
r
r r −1 X
m
n+l
m+n
gcd([m], [n]) m + n
2a jk 2 −(a+1)k l + m
[2k][k] q
m+k n+k
l+k
m
[m][n]
k=1
is a Laurent polynomial in q with integer coefficients.
Taking m = 2r + s and letting ni = n + 1 if i = 1, 3, . . . , 2r − 1 and ni = n otherwise
in Theorem 1.2, we get
Corollary 5.3. Let n, r and s be positive integers, and let a and j be non-negative integers
with j 6 2r + s. Then the expression
s
r r −1 X
m
2n
2n + 1
2n + 1
1
2n + 1
2a jk 2 −(a+1)k
[2k][k] q
n+k
n+k
n+k+1
n
[n][n + 1]
k=1
is a Laurent polynomial in q with integer coefficients.
Let [n]! = [n][n − 1] · · · [1]. It is clear that Theorem 1.2 can be restated as follows.
Theorem 5.4. Let n1 , . . . , nm be positive integers. Let j and r be non-negative integers
with j 6 m. Then the expression
n1
m m
Y
Y
2ni
[ni + ni+1 ]! X
2r jk 2 −(r+1)k
,
[2k][k] q
[n1 − 1]!
n
+
k
[2n
]!
i
i
i=1
i=1
k=1
where nm+1 = 0, is a Laurent polynomial in q with integer coefficients.
Letting n1 = · · · = nr = m and nr+1 = · · · = nr+s = n in Theorem 5.4 and noticing
the symmetry of m and n, we have
6
Corollary 5.5. Let m, n, r and s be positive integers, and let a and j be non-negative
integers with j 6 r + s. Then the expression
s
r m
2n
2m
gcd([m], [n])[m + n]![m − 1]![n − 1]! X
2a jk 2 −(a+1)k
[2k][k] q
m+k n+k
[2m]![2n]!
k=1
is a Laurent polynomial in q with integer coefficients.
In particular, the expression
s
r −1 X
m
1 4n
2n
4n
2a jk 2 −(a+1)k
[2k][k] q
2n + k n + k
[2n] n
k=1
is a Laurent polynomial in q with integer coefficients.
The q-Narayana numbers Nq (n, k) and the q-Catalan numbers Cn (q) may be defined
as follows:
1 n
1
n
2n
Nq (n, k) =
, Cn (q) =
.
[n] k k − 1
[n + 1] n
It is well known that both q-Narayana numbers and q-Catalan numbers are polynomials
in q with non-negative integer coefficients (see [2, 8]). It should be mentioned that the
definition of Nq (n, k) here differs a factor q k(k−1) from that in [2]. Motivated mainly by
Z.-W. Sun’s work on congruences for combinatorial numbers [24–28], the first author and
Jiang [11] proved that, for 0 6 j 6 2r − 1,
n
X
k
2
(−1)k q jk +(2) Nq (2n + 1, n + k + 1)r ≡ 0 (mod Cn (q)).
(5.1)
k=−n
Exactly similarly to the proof of (5.1) in [11], we can deduce the following result from
Theorem 5.4 by considering four special cases (n1 , . . . , n2r ) = (n, . . . , n, n + 1, . . . , n +
1), (n+1, . . . , n+1, n, . . . , n), (n, n+1, n, n+1, . . . , n, n+1), (n+1, n, n+1, n, . . . , n+1, n)
and noticing that [n] and [n + 1] are relatively prime.
Corollary 5.6. Let n and r be positive integers, and let a and j be non-negative integers
with j 6 2r. Then the expression
−1 X
n
1 2n
2
[2k][k]2a q jk −(a+1)k Nq (2n + 1, n + k + 1)r
(5.2)
[n] n
k=1
is a Laurent polynomial in q with integer coefficients.
6
Some open problems
In this section we propose several related conjectures for further study. Note that some
similar conjectures were raised in [11, Section 3].
7
[2m]![2n]!
are polynomials in q
It is easy to see that the q-super Catalan numbers [m+n]![m]![n]!
with integer coefficients. Warnaar and Zudilin [29, Proposition 2] proved that the q-super
Catalan numbers are in fact polynomials in q with non-negative integer coefficients. The
following conjecture related to the q-super Catalan numbers is a refinement of Corollary 5.1 and is also a q-analogue of [16, Conjecture 7.5].
Conjecture 6.1. Let m, n, s and t be positive integers. Let r be a non-negative integer
with r + s + t ≡ 1 (mod 2) and let j be a integer. Then the expression
m
[m + n]![m]![n]! X
2
s
t
(1 + q k )[k]r q jk −(r+s+t+1)k/2 Bm,k
(q)Bn,k
(q)
[2m]![2n]!
k=1
is a Laurent polynomial in q, and is a Laurent polynomial in q with non-negative integer
coefficients for 0 6 j 6 s + t.
It seems that Corollary 5.6 can be further generalized as follows.
Conjecture 6.2. Corollary 5.6 is still true for any integer j, and (5.2) is a Laurent
polynomial in q with non-negative integer coefficients for 0 6 j 6 2r.
The following is a generalization of Theorem 1.2.
Conjecture 6.3. Let n1 , . . . , nm , nm+1 = n1 be positive integers. Then for any integer j
and non-negative integer r, the expression
−1 X
n1
m Y
ni + ni+1
1 n1 + nm
2r jk 2 −(r+1)k
[2k][k] q
ni + k
n1
[n1 ]
i=1
k=1
is a Laurent polynomial in q, and is a Laurent polynomial in q with non-negative integer
coefficients if 0 6 j 6 m.
We end the paper with the following conjecture. Note that when all the ni ’s are equal
to n, it reduces to Corollary 5.6.
Conjecture 6.4. Let n1 , . . . , nm , nm+1 = n1 be positive integers. Then for any integer j
and non-negative integer r, the expression
n1
m m
X
Y
Y
1
ni + ni+1 + 1 ni + ni+1 + 1
1
2r jk 2 −(r+1)k
[2k][k] q
m
ni + k + 1
ni + k
[ni + ni+1 + 1]
[n1 ] n1 +n
n1
i=1
i=1
k=1
is a Laurent polynomial in q, and is a Laurent polynomial in q with non-negative integer
coefficients if 0 6 j 6 2m.
Acknowledgments. The first author was partially supported by the National Natural
Science Foundation of China (grant 11371144), the Natural Science Foundation of Jiangsu
Province (grant BK20161304), and the Qing Lan Project of Education Committee of
Jiangsu Province.
8
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