Table Of ContentFactors of alternating sums of products of
binomial and q-binomial coefficients
Victor J. W. Guo, Fr´ed´eric Jouhet and Jiang Zeng
Abstract. In this paper we study the factors of some alternating sums of prod-
7 uctsofbinomialandq-binomialcoefficients. Weprovethatforallpositiveintegers
0 n ,...,n , n = n , and 0 j m 1,
1 m m+1 1
0 ≤ ≤ −
n 2 n1+nm −1 n1 ( 1)kqjk2+(k2) m ni+ni+1 N[q],
n − n +k ∈
Ja (cid:20) 1 (cid:21) k=X−n1 Yi=1(cid:20) i (cid:21)
7 which generalizes a result of Calkin [Acta Arith. 86 (1998), 17–26]. Moreover,
we show that for all positive integers n, r and j,
]
T
N 2n −1 2j n ( 1)n−kqA 1−q2k+1 2n k+j r N[q],
. n j − 1 qn+k+1 n k k j ∈
h (cid:20) (cid:21) (cid:20) (cid:21)k=j − (cid:20) − (cid:21)(cid:20) − (cid:21)
X
t
a
m where A = (r 1) n + r j+1 + k rjk, which solves a problem raised by
− 2 2 2 −
Zudilin [Electron. J. Combin. 11 (2004), #R22].
[ (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
4 AMS Subject Classifications (2000): 05A10, 05A30, 11B65.
v
5
3 1 Introduction
6
1
1 In 1998, Calkin [4] proved that for all positive integers m and n,
5
0 −1 n m
2n 2n
/ ( 1)k (1.1)
h
n − n+k
t (cid:18) (cid:19) k=−n (cid:18) (cid:19)
a X
m
is an integer by arithmetical techniques. For m = 1,2 and 3, by the binomial
:
v theorem, Kummer’s formula and Dixon’s formula, it is easy to see that (1.1) is
i equal to 0, 1 and 3n , respectively. Recently in the study of finite forms of the
X n
Rogers-Ramanujan identities [9] we stumbled across (1.1) for m = 4 and m = 5,
r (cid:0) (cid:1)
a which gives
n 2 n 2
2n+k 2n 3n k 2n+k 2n
and − ,
k n+k n k k n+k
k=0(cid:18) (cid:19)(cid:18) (cid:19) k=0(cid:18) − (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)
X X
respectively. Indeed, de Bruijn [3] has shown that for m 4 there is no closed
≥
formfor(1.1)byasymptotictechniques. Ourfirstobjectiveistogiveaq-analogue
of Calkin’s result, which also implies that (1.1) is positive for m 2.
≥
In 2004, Zudilin [14] proved that for all positive integers n, j and r,
−1 n r
2n 2j 2k+1 2n k+j
( 1)n−k Z, (1.2)
n j − n+k+1 n k k j ∈
(cid:18) (cid:19) (cid:18) (cid:19)k=j (cid:18) − (cid:19)(cid:18) − (cid:19)
X
which was originally observed by Strehl[12]in1994. Infact, Zudilin’smotivation
was to solve the following problem, which was raised by Schmidt [11] in 1992 and
was apparently not related to Calkin’s result.
1
Problem1.1(Schmidt[11]). Foranyintegerr 2, defineasequenceofnumbers
≥
(r)
{ck }k∈N, independent of the parameter n, by
n r r n
n n+k n n+k
(r)
= c ,
k k k k k
k=0(cid:18) (cid:19) (cid:18) (cid:19) k=0(cid:18) (cid:19)(cid:18) (cid:19)
X X
(r)
Is it true that all the numbers c are integers?
k
At the end of his paper, Zudilin[14] raised the problem of findingand solving
aq-analogueofProblem1.1. Oursecondobjectiveistoprovidesuchaq-analogue.
For any integer n, define the q-shifted factorial (a) by (a) = 1 and
n 0
(1 a)(1 aq) (1 aqn−1), n = 1,2,...,
(a) = − − ··· −
n ((1 aq−1)(1 aq−2) (1 aqn)−1, n = 1, 2,....
− − ··· − − −
We will also use the compact notations for m 1:
≥
(a ,...,a ) := (a ) (a ) , (a ,...,a ) := lim (a ,...,a ) .
1 m n 1 n m n 1 m ∞ 1 m n
··· n→∞
The q-binomial coefficients are defined as
n n (q)
n
:= = .
k k (q) (q)
(cid:20) (cid:21) (cid:20) (cid:21)q k n−k
Since 1 = 0 if n < 0, we have n = 0 if k > n or k < 0.
(q)n k
The following is our first generalization of Calkin’s result.
(cid:2) (cid:3)
Theorem 1.2. For m 3 and all positive integers n ,...,n , there holds
1 m
≥
n1 ( 1)kq(m−1)k2+(k2) m ni+ni+1
− n +k
k=X−n1 Yi=1(cid:20) i (cid:21)
m−2
n +n λ n +n
= 1 m qλ2i i−1 i+1 i+2 , (1.3)
n λ n λ
(cid:20) 1 (cid:21) λ i=1 (cid:20) i (cid:21)(cid:20) i+1− i (cid:21)
X Y
where n = λ = n and the sum is over all sequences λ = (λ ,...,λ ) of
m+1 0 1 1 m−2
nonnegative integers such that λ λ λ .
0 1 m−2
≥ ≥ ··· ≥
Calkin[4]hasgivenapartialq-analogueof (1.1)byconsideringthealternating
sum n ( 1)kqjk n m. In this respect, besides (1.3), we shall also prove the
k=0 − k
following divisibility result.
P (cid:2) (cid:3)
Theorem 1.3. For all positive integers n ,...,n , n = n , the alternating
1 m m+1 1
sum
S(n1,...,nm;j,q) := n1+nm −1 n1 ( 1)kqjk2+(k2) m ni+ni+1
n − n +k
(cid:20) 1 (cid:21) k=X−n1 Yi=1(cid:20) i (cid:21)
is a polynomial in q with nonnegative integral coefficients for 0 j m 1.
≤ ≤ −
WeshallgivetwoproofsofTheorem1.2: Thefirstoneisbasedonarecurrence
relationformulaforS(n ,...,n ;j,q),whichalsoleadstoaproofofTheorem1.3.
1 m
The second one follows directly from Andrews’ basic hypergeometric identity
between a single sum and a multiple sum [1, Theorem 4].
2
Theorem 1.4 (Andrews [1]). For every integer m 0, the following identity
≥
holds:
(a,q√a, q√a,b ,c ,...,b ,c ,q−N) amqm+N k
1 1 m m k
−
(q,√a, √a,aq/b ,aq/c ,...,aq/b ,aq/c ,aqN+1) b c b c
k≥0 − 1 1 m m k (cid:18) 1 1··· m m(cid:19)
X
= (aq,aq/bmcm)N (aq/b1c1)l1···(aq/bm−1cm−1)lm−1
(aq/b ,aq/c ) (q) (q)
m m N l1,...,Xlm−1≥0 l1··· lm−1
(b ,c ) ...(b ,c )
2 2 l1 m m l1+···+lm−1
× (aq/b ,aq/c ) ...(aq/b ,aq/c )
1 1 l1 m−1 m−1 l1+···+lm−1
(q−N)l1+···+lm−1 (aq)lm−2+···+(m−2)l1ql1+···+lm−1 . (1.4)
× (bmcmq−N/a)l1+···+lm−1 (b2c2)l1···(bm−1cm−1)l1+···+lm−2
It is interesting to note that Theorem 1.4 is also a key ingredient in Zudilin’s
approach to Problem 1.1. Therefore, using Theorem 1.4 in its full generality we
are able to formulate and prove a q-analogue of Problem 1.1.
(r)
Theorem 1.5. For any integer r 1, define rational fractions c (q) of the
≥ k
variable q, independent of n, by writing
n r r n
qr(n−2k)+(1−r)(n2) n n+k = q(n−2k)+(1−r)(k2) n n+k c(r)(q). (1.5)
k k k k k
k=0 (cid:20) (cid:21) (cid:20) (cid:21) k=0 (cid:20) (cid:21)(cid:20) (cid:21)
X X
Then c(r)(q) N[q].
n
∈
Since the r = 1 case is trivial, we may supposethat r 2 in what follows. As
≥
n n+k = 2k n+k , invoking the q-Legendre transform, which is a special case
k k k n−k
of Carlitz’s q-Gould-Hsu inverse formula [5] (see also [10]):
(cid:2) (cid:3)(cid:2) (cid:3) (cid:2) (cid:3)(cid:2) (cid:3)
an = n q(n−2k) nn+kk bk ⇐⇒ bn = n (−1)n−k11−qqn2+kk++11 n2nk ak,
k=0 (cid:20) − (cid:21) k=0 − (cid:20) − (cid:21)
X X
we derive immediately from (1.5) that
n r
q(1−r)(n2) 2n c(r)(q) = 2j t(r)(q),
n n j n,j
(cid:20) (cid:21) j=0(cid:20) (cid:21)
X
where
t(r)(q) = qr(j+21) n ( 1)n−k 1−q2k+1 2n k+j rq(k2)−rjk.
n,j − 1 qn+k+1 n k k j
k=j − (cid:20) − (cid:21)(cid:20) − (cid:21)
X
Therefore, Theorem 1.5 is a consequence of the following theorem, which is our
q-analogue of Zudilin’s result (1.2).
Theorem 1.6. For any integer r 2, we have
≥
−1
q(r−1)(n2) 2j 2n t(r)(q) N[q].
j n n,j ∈
(cid:20) (cid:21)(cid:20) (cid:21)
As will be shown, Theorem 1.6 follows directly from Andrews’ identity (1.4).
This paper is organized as follows. We will prove Theorems 1.2 and 1.3 in the
next section. The proof of Theorem 1.6 is given in Section 3. Some interesting
divisibility results are given in Section 4. In the last section we will present four
related conjectures.
3
2 Proof of Theorems 1.2 and 1.3
We will need two known identities in q-series. One is the q-Pfaff-Saalschu¨tz
identity [6, Appendix (II.12)] (see also [7, 13]):
n +n n +n n +n n1−k qk2+2kr(q)
1 2 2 3 3 1 = n1+n2+n3−k−r ,
n +k n +k n +k (q) (q) (q) (q) (q)
(cid:20) 1 (cid:21)(cid:20) 2 (cid:21)(cid:20) 3 (cid:21) r=0 r r+2k n1−k−r n2−k−r n3−k−r
X
(2.1)
where 1 = 0 if n < 0, and the other is the q-Dixon identity:
(q)n
n1 n +n n +n n +n (q)
( 1)kq(3k2−k)/2 1 2 2 3 3 1 = n1+n2+n3 . (2.2)
− n +k n +k n +k (q) (q) (q)
k=X−n1 (cid:20) 1 (cid:21)(cid:20) 2 (cid:21)(cid:20) 3 (cid:21) n1 n2 n3
A short proof of (2.2) is given in [8].
We first establish the following recurrence formula.
Lemma 2.1. Let m 3. Then for all positive integers n ,...,n and any
1 m
≥
integer j, the following recurrence holds:
n1 n n +n
S(n ,...,n ;j,q) = ql2 1 2 3 S(l,n ,...,n ;j 1,q). (2.3)
1 m 3 m
l n l −
l=0 (cid:20) (cid:21)(cid:20) 2− (cid:21)
X
Proof. For any integer k and positive integers a ,...,a , let
1 l
l
a +a
i i+1
C(a ,...,a ;k) = ,
1 l
a +k
i=1(cid:20) i (cid:21)
Y
where a = a . Then
l+1 1
S(n1,...,nm;j,q) = (q)n1(q)nm n1 ( 1)kqjk2+(k2)C(n1,...,nm;k). (2.4)
(q) −
n1+nm k=X−n1
We observe that for m 3, we have
≥
(q) (q) n +n n +n
C(n ,...,n ;k) = n2+n3 nm+n1 1 2 1 2 C(n ,...,n ;k),
1 m 3 m
(q) (q) n +k n +k
n1+n2 nm+n3(cid:20) 1 (cid:21)(cid:20) 2 (cid:21)
and, by letting n in (2.1),
3
→ ∞
n +n n +n n1−k qr2+2kr(q)
1 2 1 2 = n1+n2 .
n +k n +k (q) (q) (q) (q)
(cid:20) 1 (cid:21)(cid:20) 2 (cid:21) r=0 r r+2k n1−k−r n2−k−r
X
Plugging these into (2.4) we can write its right-hand side as
R := n1 n1−k( 1)kC(n ,...,n ;k) q(r+k)2+(j−1)k2+(k2)(q)n2+n3(q)n1(q)nm .
3 m
− (q) (q) (q) (q) (q)
k=X−n1 Xr=0 r r+2k n1−k−r n2−k−r nm+n3
Setting l = r +k, then n l n , but if l < 0, at least one of the indices
1 1
− ≤ ≤
l+k and l k is negative for any integer k, which implies that 1 = 0
− (q)l−k(q)l+k
by convention. Therefore, exchanging the order of summation, we have
R = n1 ql2(q)n2+n3(q)n1(q)nm l ( 1)kC(n ,...,n ;k)q(j−1)k2+(k2).
3 m
(q) (q) (q) − (q) (q)
l=0 n1−l n2−l nm+n3 k=−l l−k l+k
X X
4
Now, in the last sum making the substitution
(q) (q) (q)
C(n ,...,n ;k) = l−k l+k nm+n3C(l,n ,...,n ;k),
3 m 3 m
(q) (q)
n3+l nm+l
we obtain the right-hand side of (2.3).
First proof of Theorem 1.2. Letting n in (2.2) yields that
3
→ ∞
S(n ,n ;1,q) = 1. (2.5)
1 2
Theorem 1.2 then follows by iterating (m 2) times formula (2.3).
−
Second proof of Theorem 1.2. Since
M =( 1)kq(M−N)k−(k2) M (q−M+N)k,
N +k − N (qN+1)
(cid:20) (cid:21) (cid:20) (cid:21) k
by collecting the terms of index k and k, the left-hand side of (1.3) can be
−
written as
L := m ni+ni+1 + n1 (1+qk)( 1)kq(m−1)k2+(k2) m ni+ni+1
n − n +k
i=1(cid:20) i (cid:21) k=1 i=1(cid:20) i (cid:21)
Y X Y
= m ni+ni+1 1+ n1 (1+qk)( 1)(m−1)kq(m−1)(k+21) m qnik(q−ni+1)k .
i=1(cid:20) ni (cid:21)( k=1 − i=1 (qni+1)k )
Y X Y
Letting c = c = = c = c and a 1 in Andrews’ formula (1.4) we
1 2 m
··· → ∞ →
get
1+ (1+qk) (b1,...,bm,q−N)k ( 1)mkqm(k2) qm+N k
(q/b ,...,q/b ,qN+1) − b b ...b
k≥1 1 m k (cid:18) 1 2 m(cid:19)
X
(q) (q)
N N
=
(q/b ) (q) (q) (q)
m N l1,...,Xlm−1≥0 l1··· lm−1 N−l1−···−lm−1
m−1 (bi+1)l1+···+li −1 l1+···+liq(l1+·2··+li)+(m−i)li. (2.6)
× (q/b ) b
i=1 i l1+···+li (cid:18) i+1(cid:19)
Y
Now, shifting m to m 1 in (2.6), setting
−
N = n , b = q−ni for i= 1,...,m 1,
m i
−
and λ = l + +l for i= 1,...,m 2, one sees that L equals
i 1 i
··· −
m ni+ni+1 (q)2nm m−2(q−ni+1)λi(−1)λiq(λi2+1)+ni+1λi
Yi=1(cid:20) ni (cid:21)(q1+nm−1)nm 0≤λ1≤X···≤λm−2 Yi=1 (q1+ni)λi(q)λi−λi−1
m−2
n +n λ n +n
= 1 m qλ2i i+1 i i+1 ,
n λ n +λ
(cid:20) 1 (cid:21)0≤λ1≤X···≤λm−2 Yi=1 (cid:20) i (cid:21)(cid:20) i i (cid:21)
where λ = 0 and λ = n . The latter identity is clearly equivalent to
0 m−1 m
Theorem 1.2.
5
In order to prove Theorem 1.3, we shall need the following relation:
S(n ,...,n ;0,q) = S(n ,...,n ;m 1,q−1)qn1n2+n2n3+···+nm−1nm. (2.7)
1 m 1 m
−
As n = n qk(k−n), Eq. (2.7) can be verified by substituting q by q−1 and
k q−1 k
then replacing k by k in the definition of S(n ,...,n ;m 1,q).
(cid:2) (cid:3) (cid:2) (cid:3) − 1 m −
Proof of Theorem 1.3. We proceed by induction on m 1. By the q-binomial
≥
theorem [6, (II.3)], we have
S(n1;0,q) = n1 ( 1)kq(k2) 2n = 0.
− n+k
k=X−n1 (cid:20) (cid:21)
In view of (2.5), it follows from (2.7) that
S(n ,n ;0,q) = S(n ,n ;1,q−1)qn1n2 = qn1n2.
1 2 1 2
So the theorem is valid for m 2.
≤
Now suppose that the expression S(n ,...,n ;j,q) is a polynomial in q
1 m−1
with nonnegative integral coefficients for some m 3 and 0 j m 2.
≥ ≤ ≤ −
Then by the recurrence formula (2.3), so is S(n ,...,n ;j,q) for 1 j m
1 m
≤ ≤ −
1. It remains to show that S(n ,...,n ;0,q) has the required property. By
1 m
Theorem 1.2 we know that S(n ,...,n ;m 1,q) is a polynomial in q. Since the
1 m
−
q-binomial coefficient n is a polynomial in q of degree k(n k) (see [2, p. 33]),
k −
it is easy to see from the definition of S(n ,...,n ;m 1,q) that the degree of
1 m
(cid:2) (cid:3) −
the polynomial S(n ,...,n ;m 1,q) is less than or equal to n n +n n + +
1 m 1 2 2 3
− ···
n n . It follows from (2.7) that S(n ,...,n ;0,q) is also a polynomial in q
m−1 m 1 m
with nonnegative integral coefficients. This completes the inductive step of the
proof.
Remark. Thoughitisnotnecessarytocheckthem = 3casetovalidourinduction
argument, wethink it is convenient to includeheretheformulas for m = 3. First,
the q-Dixon identity (2.2) implies that
n +n +n
1 2 3
S(n ,n ,n ;1,q) = .
1 2 3
n
(cid:20) 2 (cid:21)
From (2.3) and (2.5) we derive
n1 n n +n
S(n ,n ,n ;2,q) = ql2 1 2 3 .
1 2 3
l n l
l=0 (cid:20) (cid:21)(cid:20) 2− (cid:21)
X
Finally, applying (2.7) we get
S(n ,n ,n ;0,q) = S(n ,n ,n ;2,q−1)qn1n2+n2n3
1 2 3 1 2 3
n1 n n +n
= q(n1−l)(n2−l)+n3l 1 2 3 .
l n l
l=0 (cid:20) (cid:21)(cid:20) 2− (cid:21)
X
3 Proof of Theorem 1.6
We will distinguish the cases where r 2 is even or odd, and treat separately
≥
the values r = 2 and r = 3.
6
For r = 2, apply (1.4) specialized with m = 1, a = q−(2n+1), N = n j,
• −
b = q−n and c = q−(n−j). The left-hand side of (1.4) is then equal to
1 1
−2
n+j q−2(n−2j)+(n2)t(2)(q).
2j n,j
(cid:20) (cid:21)
Equating this with the right-hand side gives
t(2)(q) = (q)2n(q)2j q2(n−2j)−(n2),
n,j (q) (q) (q) (q)2
n 2j 2j−n n−j
which shows that 2j 2n −1q(n2)t(2)(q) N[q].
j n n,j ∈
(cid:2) (cid:3)(cid:2) (cid:3)
For r = 3, apply (1.4) specialized with m = 1, a = q−(2n+1), N = n j
• −
and b = c = q−(n−j). This yields in that case
1 1
t(3)(q)= (q)2n q3(n−2j)−2(n2),
n,j (q) (q)3
3j−n n−j
which shows that 2j 2n −1q2(n2)t(3)(q) N[q].
j n n,j ∈
(cid:2) (cid:3)(cid:2) (cid:3)
For r = 2s 4, apply (1.4) with m = s 2, a = q−(2n+1), N = n j,
• ≥ ≥ −
b = q−n and c = b = c = q−(n−j), i 2,...,s to get
1 1 i i
∀ ∈ { }
q(2s−1)(n2)−2s(n−2j)t(2s)(q) =
n,j
(q)2n(q)j j n−l1 n−l1+j q(l21)+2j(s−1)l1+(j+1−n)l1
(q) (q) (q) l j n l j
n 2j n−j lX1≥0(cid:20) 1(cid:21)(cid:20) (cid:21)(cid:20) − 1− (cid:21)
2
2j n−l1−l2+j q(l22)+2j(s−2)l2+(j+1−n)l2
× l n l l j ×···
lX2≥0(cid:20) 2(cid:21)(cid:20) − 1− 2− (cid:21)
2
2j n−l1−···−ls−1+j q(ls−21)+2jls−1+(j+1−n)ls−1
×lsX−1≥0(cid:20)ls−1(cid:21)(cid:20)n−l1−···−ls−1−j(cid:21)
2j (l1+···+ls−1)
q 2 .
× n l l j
(cid:20) − 1−···− s−1− (cid:21)
As the condition l + +l n j holds in the last summation, we
1 s−1
can see that for s 2,··2·j 2n −1≤q(2s−−1)(n2)t(2s)(q) N[q].
≥ j n n,j ∈
For r = 2s+1 5, app(cid:2)ly(cid:3)((cid:2)1.4(cid:3)) with m = s 2, a = q−(2n+1), N = n j,
• ≥ ≥ −
7
and b = c = q−(n−j), i 1,...,s to get
i i
∀ ∈ { }
q2s(n2)−(2s+1)(n−2j)t(2s+1)(q) =
n,j
2
(q)2n 2j n−l1+j q(l21)+2j(s−1)l1+(j+1−n)l1
(q) (q)2 l n l j
2j n−j lX1≥0(cid:20) 1(cid:21)(cid:20) − 1− (cid:21)
2
2j n−l1−l2+j q(l22)+2j(s−2)l2+(j+1−n)l2
× l n l l j ×···
lX2≥0(cid:20) 2(cid:21)(cid:20) − 1− 2− (cid:21)
2
2j n−l1−···−ls−1+j q(ls−21)+2jls−1+(j+1−n)ls−1
×lsX−1≥0(cid:20)ls−1(cid:21)(cid:20)n−l1−···−ls−1−j(cid:21)
2j (l1+···+ls−1)
q 2 .
× n l l j
(cid:20) − 1−···− s−1− (cid:21)
As the condition l + +l n j holds in the last summation, we
1 s−1
can see that for s 2,··2·j 2n −1≤q2s(n2−)t(2s+1)(q) N[q].
≥ j n n,j ∈
Remark. In the special case r(cid:2) =(cid:3)(cid:2)2,(cid:3)our proof gives the following expression for
(2)
the coefficients c (q):
n
n 2
c(2)(q) = 2j n q2(n−2j). (3.1)
n n j
j=0(cid:20) (cid:21)(cid:20) (cid:21)
X
(2)
These coefficients are q-analogues of the famous c (1) involved in Apery’s proof
n
of the irrationality of ζ(3):
n 2 n 3
2j n n
c(2)(1) = = . (3.2)
n n j j
j=0(cid:18) (cid:19)(cid:18) (cid:19) j=0(cid:18) (cid:19)
X X
As explained in [12], when q = 1, one can derive the last expression from (3.1)
in an elementary way (by two iteration of the Chu-Vandermonde formula). But
our q-analogue (3.1) does not lead to a natural q-analogue of (3.2).
4 Consequences of Theorems 1.2 and 1.3
Letting q = 1 in Theorem 1.2 we obtain a direct generalization of Calkin’s re-
sult (1.1).
Theorem 4.1. For m 3 and all positive integers n ,...,n , there holds
1 m
≥
n1 m n +n n +n m−2 λ n +n
( 1)k i i+1 = 1 m i−1 i+1 i+2 ,
− n +k n λ n λ
k=X−n1 Yi=1(cid:18) i (cid:19) (cid:18) 1 (cid:19)Xλ Yi=1 (cid:18) i (cid:19)(cid:18) i+1− i (cid:19)
(4.1)
where n = λ = n and the sum is over all sequences λ = (λ ,...,λ ) of
m+1 0 1 1 m−2
nonnegative integers such that λ λ λ .
0 1 m−2
≥ ≥ ··· ≥
Remark. For m = 1 and 2, it is easy to see that the left-hand side of (4.1) is
equal to 0 and n1+n2 , respectively. Calkin’s result follows from (4.1) by setting
n1
n = n for all i = 1,...,m.
i
(cid:0) (cid:1)
8
Letting n = = n = n in Theorem 1.3, we obtain a complete q-analogue
1 m
···
of Calkin’s result.
Corollary 4.2. For all positive m, n and 0 j m 1,
≤ ≤ −
−1 n m
2n ( 1)kqjk2+(k2) 2n
n − n+k
(cid:20) (cid:21) k=−n (cid:20) (cid:21)
X
is a polynomial in q with nonnegative integral coefficients.
Letting n = m and n = n for 1 i r in Theorem 1.3, we obtain
2i−1 2i
≤ ≤
Corollary 4.3. For all positive m, n, r and 0 j 2r 1,
≤ ≤ −
−1 m r r
m+n ( 1)kqjk2+(k2) m+n m+n
m − m+k n+k
(cid:20) (cid:21) k=−m (cid:20) (cid:21) (cid:20) (cid:21)
X
is a polynomial in q with nonnegative integral coefficients. In particular,
m r r
m+n m+n
( 1)k
− m+k n+k
k=−m (cid:18) (cid:19) (cid:18) (cid:19)
X
is divisible by m+n .
m
Letting n (cid:0) =(cid:1)l, n = m and n = n for 1 i r in Theorem 1.3, we
3i−2 3i−1 3i
≤ ≤
obtain
Corollary 4.4. For all positive l, m, n, r and 0 j 3r 1,
≤ ≤ −
−1 l r r r
l+n ( 1)kqjk2+(k2) l+m m+n n+l
n − l+k m+k n+k
(cid:20) (cid:21) k=−l (cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21)
X
is a polynomial in q with nonnegative integral coefficients. In particular,
l r r r
l+m m+n n+l
( 1)k
− l+k m+k n+k
k=−l (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
is divisible by l+m , m+n and n+l .
l m n
Letting m(cid:0)= 2r(cid:1)+(cid:0)s, n(cid:1)= n (cid:0)= (cid:1) = n = n+1 and let all the other n
1 3 2r−1 i
···
be n in Theorem 4.1, we get
Corollary 4.5. For all positive r, s and n,
n r r s
2n+1 2n+1 2n
( 1)k
− n+k+1 n+k n+k
k=−n (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
is divisible by both 2n and 2n+1 , and is therefore divisible by (2n+1) 2n .
n n n
However, the fo(cid:0)llow(cid:1)ing re(cid:0)sult i(cid:1)s not a special case of Theorem 4.1. (cid:0) (cid:1)
9
Corollary 4.6. For all nonnegative r and s and positive t and n,
n r s t
2n+1 2n+1 2n
( 1)k .
− n+k+1 n+k n+k
k=−n (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
is divisible by 2n .
n
Proof. We pro(cid:0)cee(cid:1)d by induction on r s . The r = s case is clear from Corol-
| − |
lary 4.5. Suppose the statement is true for r s m 1. By Theorem 4.1, one
| − |≤ −
sees that
n m s s t
2n+2 2n+1 2n+1 2n
( 1)k
− n+k+1 n+k+1 n+k n+k
k=−n (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
n m−1 s+1 s+1 t−1
2n+2 2n+2 2n+1 2n+1 2n
= ( 1)k ,
2n+1 − n+k+1 n+k+1 n+k n+k
k=−n (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
(4.2)
where m,t 1, is divisible by
≥
2n+2 2n+1 2n
= 2 .
2n+1 n n
(cid:18) (cid:19) (cid:18) (cid:19)
By the binomial theorem, we have
m m
2n+2 2n+1 2n+1
= +
n+k+1 n+k+1 n+k
(cid:18) (cid:19) (cid:18)(cid:18) (cid:19) (cid:18) (cid:19)(cid:19)
m i m−i
m 2n+1 2n+1
= . (4.3)
i n+k+1 n+k
i=0(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
X
Substituting (4.3) into the left-hand of (4.2) and using the induction hypothesis
and symmetry, we find that
n m+s s t
2n+1 2n+1 2n
( 1)k
− n+k+1 n+k n+k
k=−n (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
n s m+s t
2n+1 2n+1 2n
+ ( 1)k
− n+k+1 n+k n+k
k=−n (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
is divisible by 2 2n . However, replacing k by k, one sees that
n −
n(cid:0) (cid:1) m+s s t
2n+1 2n+1 2n
( 1)k
− n+k+1 n+k n+k
k=−n (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
n s m+s t
2n+1 2n+1 2n
= ( 1)k .
− n+k+1 n+k n+k
k=−n (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
X
This proves that the statement is true for r s = m.
| − |
It is clear that Theorems 1.3 and 4.1 can be restated in the following forms.
10
