Some symmetric $q$-congruences

0.12 MB·

by  He-Xia Ni

SOME SYMMETRIC q-CONGRUENCES
HE-XIA NI AND HAO PAN
6
1
0 Abstract. We prove some symmetric q-congruences.
2

n
a
J
1. Introduction
5

] Suppose that f0,f1,f2,... are a sequence of integers. Let
T
N k k
fˆ(q) = (−1)j f .
. k j
h j
t Xj=0 (cid:18) (cid:19)
a
m In [2, Theorem 2.4], Sun proved the following symmetric congruence:
[
p−1 α −1−α p−1 α −1−α
1 f ≡ (−1)hαip fˆ (mod p2), (1.1)
k k
v k k k k
k=0(cid:18) (cid:19)(cid:18) (cid:19) k=0(cid:18) (cid:19)(cid:18) (cid:19)
4 X X
3 where p is an odd prime, α ∈ Q is p-integral and hαi is the least non-negative
p
9
residue of α modulo p. With the help of (1.1), Sun obtained many interesting
3
0 congruences modulo p.
.
1 In this paper, we shall give a q-analogue of (1.1). Define
0
6 (1−x)(1−xq)···(1−xqn−1), if n ≥ 1,
1 (x;q) =
n
: (1, if n = 0.
v
i
X Define the q-binomial coefficient
r
a α (qα−k+1;q)k
= .
k (q;q)
(cid:20) (cid:21)q k
In particular, we set α = 0 if k < 0. Suppose that f (q),f (q),... are a sequence
k q 1 2
of polynomials in q with integral coefficients. Let
(cid:2) (cid:3)
k
k
fˆk(q) = (−1)jq(j+21) fj(q)
j
j=0 (cid:20) (cid:21)q
X
for 0 ≤ k ≤ n.
2010 Mathematics Subject Classification. Primary 11B65; Secondary 05A10, 05A30, 11A07.
Key words and phrases. congruence; q-binomial coefficient.
1
2 HE-XIANIAND HAOPAN
Theorem 1.1. Letn ≥ 2 andd ≥ 1 with(n,d) = 1. Supposethat f (q),...,f (q) ∈
1 n
Z[q] and r is an integer. Then if n is odd,
qd(a+21)+(ad+r)(n2−1−2a) n−1 (qr;q(dq)kd(;qqdd−)2r;qd)k ·qdkfk(qd)
k=0 k
X
n−1 (qr;qd) (qd−r;qd)
≡(−1)a k k ·qdkfˆ(qd) (mod Φ (q)2), (1.2)
(qd;qd)2 k n
k=0 k
X
where a = h−r/di and Φ is the n-th cyclotomic polynomial. And when n is even,
n n
qd(a+21)+(ad+r)(n2−1−2a) n−1 (qr;q(dq)kd(;qqdd−)2r;qd)k ·qdkfk(qd)
k=0 k
X
n−1 (qr;qd) (qd−r;qd)
≡(−1)a+adn+r (qkd;qd)2 k ·qdkfˆk(qd) (mod Φn(q)2). (1.3)
k=0 k
X
Now for odd n ≥ 3, replacing q by q−1 in (1.4) and noting that
Φ (q−1) = q−φ(n)Φ (q)
n n
where φ is the Euler totient function, we get
q−d(a+21)−(ad+r)(n2−1−2a) n−1 (q−r;q(−qd−)dk;(qq−r−d)d2;q−d)k ·q−dkfk(q−d)
k=0 k
X
n−1 (q−r;q−d) (qr−d;q−d)
≡(−1)a k k ·q−dkfˆ(q−d) (mod Φ (q)2).
(q−d;q−d)2 k n
k=0 k
X
Notice that
k k
k k
fˆk(q−1) = (−1)jq−(j+21) fj(q−1) = (−1)jq(2j)−kj fj(q−1)
j j
j=0 (cid:20) (cid:21)q−1 j=0 (cid:20) (cid:21)q
X X
and
(q−r;q−d) (qr−d;q−d) (qr;qd) (qd−r;qd)
k k = qdk · k k.
(q−d;q−d)2 (qd;qd)2
k k
We have
n−1 (qr;qd) (qd−r;qd)
k k ·g (qd)
(qd;qd)2 k
k=0 k
X
≡(−1)aqd(a+21)+(ad+r)(n2−1−2a) n−1 (qr;q(dq)kd(;qqdd−)2r;qd)k ·g˜k(qd) (mod Φn(q)2),
k=0 k
X
SOME SYMMETRIC q-CONGRUENCES 3
where
k
k
g˜k(q) = (−1)jq(2j)−kj gj(q).
j
j=0 (cid:20) (cid:21)q
X
The similar discussion is also valid when n is even. Thus
Theorem 1.2. Letn ≥ 2 andd ≥ 1 with(n,d) = 1. Supposethat f (q),...,f (q) ∈
1 n
Z[q] and r is an integer. Then if n is odd,
n−1 (qr;qd) (qd−r;qd)
k k ·f (qd)
(qd;qd)2 k
k=0 k
X
≡(−1)aqd(a+21)+(ad+r)(n2−1−2a) n−1 (qr;q(dq)kd(;qqdd−)2r;qd)k ·f˜k(qd) (mod Φn(q)2), (1.4)
k=0 k
X
where a = h−r/di and Φ is the n-th cyclotomic polynomial. And when n is even,
n n
n−1 (qr;qd) (qd−r;qd)
k k ·f (qd)
(qd;qd)2 k
k=0 k
X
≡(−1)a+adn+rqd(a+21)+(ad+r)(n2−1−2a) n−1 (qr;q(dq)kd(;qqdd−)2r;qd)k ·f˜k(qd) (mod Φn(q)2).
k=0 k
X
(1.5)
2. Proof of Theorem 1.1
Below we slightly extend the notion of congruence. Let
s
X(q) = c qβj : s ≥ 1, c ,...,c ∈ C, β ,...,β ∈ R .
j 1 s 1 s
nXj=1 o
For f(q),g(q),h(q) ∈ X(q), we say
f(q) ≡ g(q) (mod h(q)),
provided
f(q)−g(q)
∈ X(q).
h(q)
Theorem 2.1. Suppose that n ≥ 2, α ∈ Q and the denominator of α is prime to
n. Then when n is odd,
n−1
α −1−α
(−1)aq(a+21)+sna−s(n2) qk2+k fk(q)
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1
α −1−α
≡ qk2+k fˆ(q) (mod Φ (q)2),
k n
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
4 HE-XIANIAND HAOPAN
where a = hαi and s = (α−a)/n. And if n is even, then
n
n−1
α −1−α
(−1)a+sq(a+21)+sna−s(n2) qk2+k fk(q)
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1
α −1−α
≡ qk2+k fˆ(q) (mod Φ (q)2).
k n
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
Let
k
k
fˆk(q) = (−1)iq(i+21) fi(q).
i
i=0 (cid:20) (cid:21)q
X
Suppose that p is an odd prime and α are positive integer,0 ≤< α > ≤ p−1. Let
p
a = hαi . Then
n
Proof. Using the q-Chu-Vandemonde identity, we have
k
a+sn sn a
= q(sn−j)(k−j)
k j k −j
(cid:20) (cid:21)q j=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
k
a sn a
≡ qsnk + q−j(k−j) (mod Φ (q)2).
n
k j k −j
(cid:20) (cid:21)q j=1 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
Notice that
sn [sn] sn−1
q
= ≡ 0 (mod Φ (q))
n
j [j] j −1
(cid:20) (cid:21)q q (cid:20) (cid:21)q
and
sn−1 −1
≡ = (−1)j−1q−(2j) (mod Φn(q)).
j −1 j −1
(cid:20) (cid:21)q (cid:20) (cid:21)q
So
k
a+sn a [sn] a
≡ qsnk + q ·(−1)j−1q−(2j)−j(k−j) (mod Φn(q)2).
k k [j] k −j
(cid:20) (cid:21)q (cid:20) (cid:21)q j=1 q (cid:20) (cid:21)q
X
Similarly, we also have
−1−a−sn
k
(cid:20) (cid:21)q
k
−1−a [−sn] −1−a
≡q−snk + q ·(−1)j−1q−(2j)−j(k−j) (mod Φn(q)2).
k [j] k −j
(cid:20) (cid:21)q j=1 q (cid:20) (cid:21)q
X
SOME SYMMETRIC q-CONGRUENCES 5
Thus
n−1
α −1−α
qk2+k ((−1)aq(a+21)fk(q)−fˆk(q))
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1
a −1−a
≡ qk2+k ((−1)aq(a+21)fk(q)−fˆk(q)) (2.1)
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1 k
a [n] −1−a
−s qk2+k fk(q) (−1)a+j−1q(a+21)−(2j)−j(k−j)· q (2.2)
k [j] k −j
k=1 (cid:20) (cid:21)q j=1 q(cid:20) (cid:21)q
X X
n−1 k
a [n] −1−a
+s qk2+k fˆk(q) (−1)j−1q−(2j)−j(k−j)· q (2.3)
k [j] k −j
k=1 (cid:20) (cid:21)q j=1 q(cid:20) (cid:21)q
X X
n−1 k
−1−a [n] a
+s qk2+k fk(q) (−1)a+j−1q(a+21)−(2j)−j(k−j)· q (2.4)
k [j] k −j
k=1 (cid:20) (cid:21)q j=1 q(cid:20) (cid:21)q
X X
n−1 k
−1−a [n] a
−s qk2+k fˆk(q) (−1)j−1q−(2j)−j(k−j) · q (mod Φn(q)2).
k [j] k −j
k=1 (cid:20) (cid:21)q j=1 q(cid:20) (cid:21)q
X X
(2.5)
First, we consider (2.1). Clearly
n−1
a −1−a
qk2+k fˆ(q)
k
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
a k
a −1−a k
= qk2+k (−1)jq(j+21)fj(q)
k k j
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q j=0 (cid:20) (cid:21)q
X X
a a
a −1−a −1−a−j
= (−1)jq(j+21)fj(q) qk2+k
k j k −j
j=0 k=j (cid:20) (cid:21)q(cid:20) (cid:21)q(cid:20) (cid:21)q
X X
a a
−1−a a −1−a−j
= (−1)jq(j+21)fj(q) qa2+a q(a−k)(−1−a−k)
j a−k k −j
j=0 (cid:20) (cid:21)q k=j (cid:20) (cid:21)q(cid:20) (cid:21)q
X X
a
−1−a −1−j
= (−1)jq(j+21)fj(q) qa2+a
j a−j
j=0 (cid:20) (cid:21)q (cid:20) (cid:21)q
X
a
−1−a a
= (−1)aqj2+j+(a+21)fj(q) ,
j j
j=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
where we use the q-Chu-Vandemonde identity in the fourth equality. So (2.1)
always vanishes.
6 HE-XIANIAND HAOPAN
Note that
k k
[n] −1−a n−1 [n] −1−a
(−1)j−1q−(2j)−j(k−j)· q ≡ ·q−j(k−j)· q
[j] k −j j −1 [j] k −j
j=1 q(cid:20) (cid:21)q j=1 (cid:20) (cid:21)q q(cid:20) (cid:21)q
X X
k
n −1−a n−1−a −1−a
≡ q(n−j)(k−j) = −qnk (mod Φ (q)2).
n
j k −j k k
j=1 (cid:20) (cid:21)q(cid:20) (cid:21)q (cid:20) (cid:21)q (cid:20) (cid:21)q
X
Hence
n−1
a n−1−a −1−a
(2.2) ≡−s(−1)aq(a+21) qk2+k fk(q)· −qnk
k k k
k=1 (cid:20) (cid:21)q (cid:18)(cid:20) (cid:21)q (cid:20) (cid:21)q(cid:19)
X
(2.6)
n−1
a n−1−a
≡−s(−1)aq(a+21) qk2+k fk(q) (2.7)
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1
a −1−a
+s(−1)aq(a+21) qk2+k+nk fk(q) (mod Φn(q)2). (2.8)
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
Similarly, we have
k
[n] a−n
(−1)j−1q−(2j)−j(k−j) q
[j] k −j
j=1 q(cid:20) (cid:21)q
X
k
n a−n a a−n
≡ q(n−j)(k−j) = −qnk (mod Φ (q)2).
n
j k −j k k
j=1 (cid:20) (cid:21)q(cid:20) (cid:21)q (cid:20) (cid:21)q (cid:20) (cid:21)q
X
So
n−1−a k
n−1−a [n] a−n
(2.4) ≡s qk2+k+(a+21) fk(q) (−1)a+j−1q−(2j)−j(k−j)· q
k [j] k −j
k=1 (cid:20) (cid:21)q j=1 q(cid:20) (cid:21)q
X X
n−1−a
n−1−a a
≡s(−1)aq(a+21) qk2+k fk(q) (2.9)
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1−a
n−1−a a−n
−s(−1)aq(a+21) qk2+k+nk fk(q) (mod Φp(q)2).
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
(2.10)
SOME SYMMETRIC q-CONGRUENCES 7
On the other hand, clearly
n−1 k
a [n] −1−a
qk2+k fˆk(q) (−1)j−1q−(2j)−j(k−j)· q
k [j] k −j
k=1 (cid:20) (cid:21)q j=1 q(cid:20) (cid:21)q
X X
n−1 k k
a k [n] −1−a
= qk2+k (−1)iq(i+21) fi(q) (−1)j−1q−(2j)−j(k−j)· q
k i [j] k −j
k=1 (cid:20) (cid:21)q i=0 (cid:20) (cid:21)q j=1 q(cid:20) (cid:21)q
X X X
= a (−1)iq(i+21) a fi(q) a (−1)j−1q−(2j)[n]q a qk2+k−j(k−j) a−i −1−a
i [j] a−k k −j
i=0 (cid:20) (cid:21)q j=1 q k=j (cid:20) (cid:21)q(cid:20) (cid:21)q
X X X
= a (−1)iq(i+21) a fi(q) a (−1)j−1q−(2j)[n]q ·qa2+a+j2−aj −1−i ,
i [j] a−j
i=0 (cid:20) (cid:21)q j=1 q (cid:20) (cid:21)q
X X
where in the last step we use the q-Chu-Vandemonde identity. Thus
(2.3) =s a (−1)iq(i+21) a fi(q) a (−1)j−1qa2+a+j2−aj−(2j)[n]q −1−i
i [j] a−j
i=0 (cid:20) (cid:21)q j=1 q (cid:20) (cid:21)q
X X
a a
a n −1−i
≡s (−1)iq(i+21) fi(q) qa2+a+j2−aj
i j a−j
i=0 (cid:20) (cid:21)q j=1 (cid:20) (cid:21)q(cid:20) (cid:21)q
X X
a a
a n −1−i
≡sqa2+a (−1)iq(i+21) fi(q) q(n−j)(a−j)
i j a−j
i=0 (cid:20) (cid:21)q j=1 (cid:20) (cid:21)q(cid:20) (cid:21)q
X X
a
a n−1−i
=sqa2+a (−1)iq(i+21) fi(q)· (2.11)
i a
i=0 (cid:20) (cid:21)q (cid:20) (cid:21)q
X
a
a −1−i
−sqa2+a (−1)iq(i+21) fi(q)·qna (mod Φn(q)2). (2.12)
i a
i=0 (cid:20) (cid:21)q (cid:20) (cid:21)q
X
Similarly, noting that
n−1−a
n−1−a−i a
qk2+k−j(k−j)
k −i k −j
k=j (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1−a
n−1−a−i a
≡ qa2+a+j2+j+aj+(n−1−a−k)(a−k+j)
n−1−a−k k −j
k=j (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1−i
=qa2+a+j2+j+aj (mod Φ (q)),
n
n−1−a−j
(cid:20) (cid:21)q
8 HE-XIANIAND HAOPAN
we can get
(2.5) ≡−sn−1−a n−1−a fˆ(q) k (−1)j−1q−(2j)[n]q ·qk2+k−j(k−j) a
k
k [j] k −j
k=1 (cid:20) (cid:21)q j=1 q (cid:20) (cid:21)q
X X
n−1−a n−1−a
n−1−a n n−1−i
≡−s (−1)iq(i+21) fi(q) ·qa2+a+j2+j+aj
i j n−1−a−j
i=0 (cid:20) (cid:21)q j=1 (cid:20) (cid:21)q (cid:20) (cid:21)q
X X
n−1−a n−1−a
n−1−a n −1−i
≡−s (−1)iq(i+21) fi(q) qa2+a+(n−j)(n−1−a−j)
i j n−1−a−j
i=0 (cid:20) (cid:21)q j=1 (cid:20) (cid:21)q(cid:20) (cid:21)q
X X
n−1−a
n−1−a n−1−i
=−s (−1)iqa2+a+(i+21) fi(q)· (2.13)
i n−1−a
i=0 (cid:20) (cid:21)q (cid:20) (cid:21)q
X
n−1−a
n−1−a −1−i
+s (−1)iqa2+a+(i+21) fi(q)·qn(n−1−a) (mod Φn(q)2).
i n−1−a
i=0 (cid:20) (cid:21)q (cid:20) (cid:21)q
X
(2.14)
Clearly
(2.7)+(2.9) = 0. (2.15)
And we also get
(2.11)+(2.13) = 0, (2.16)
since
a n−1−i n−1−i n−1−2i
=
i a i a−i
(cid:20) (cid:21)q(cid:20) (cid:21)q (cid:20) (cid:21)q(cid:20) (cid:21)q
n−1−i n−1−2i n−1−a n−1−i
= = .
i n−1−a−i i n−1−a
(cid:20) (cid:21)q(cid:20) (cid:21)q (cid:20) (cid:21)q(cid:20) (cid:21)q
Noting that
−1−k a+k a+k −1−a
(−1)aqak+(a+21) = = = (−1)kqak+(k+21) ,
a a k k
(cid:20) (cid:21)q (cid:20) (cid:21)q (cid:20) (cid:21)q (cid:20) (cid:21)q
we have
n−1
a −1−a
(2.8)+(2.12) = s(−1)aq(a+21) qk2+k fk(q)·(qnk −qna). (2.17)
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
Similarly, from
(−1)n−1−aq(n−2a) −1−k = (−1)kq(k+21) a−n ,
n−1−a k
(cid:20) (cid:21)q (cid:20) (cid:21)q
SOME SYMMETRIC q-CONGRUENCES 9
it follows that
(2.10)+(2.14)
n−1−a
n−1−a a−n
=s(−1)aq(a+21) qk2+k fk(q)·((−1)n−1q(n2) −qnk).
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
(2.18)
Noting that
1−qn
n
1+q2 = n ≡ 0 (mod Φn(q))
1−q
2
for those even n, we always have
(−1)n−1q(n2) ≡ 1 (mod Φn(q)).
Thus by (2.15), (2.16), (2.17) and (2.18), we get
n−1
α −1−α
qk2+k ((−1)aq(a+21)fk(q)−fˆk(q))
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1
−1−a a
≡(−1)asq(a+21) qk2+k((−1)n−1q(n2) −qna) fk(q)
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1
−1−α α
≡(−1)asq(a+21) qk2+k(1−(−1)n−1qna−(n2)) fk(q) (mod Φn(q)2).
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
When n is odd, we have
s(1−qna−s(n2)) ≡ 1−qsna−s(n2) (mod Φn(q)2).
So
n−1
α −1−α
(−1)aq(a+21)+sna−s(n2) qk2+k fk(q)
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1
α −1−α
≡ qk2+k fˆ(q) (mod Φ (q)2).
k n
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
Suppose that n is even. Then
s(1−(−1)n−1qna−s(n2)) ≡ 1−(−1)sqsna−s(n2) (mod Φn(q)2).
10 HE-XIANIAND HAOPAN
Then we have
n−1
α −1−α
(−1)a+sq(a+21)+sna−s(n2) qk2+k fk(q)
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
n−1
α −1−α
≡ qk2+k fˆ(q) (mod Φ (q)2).
k n
k k
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q
X
(cid:3)
Remark. Let
f (q) = xk.
k
Then
k
k
fˆk(q) = (−1)jq(j+21) xj = (xq;q)k.
j
j=0 (cid:20) (cid:21)q
X
Then Theorem 1.1 implies a conjecture of Guo and Zeng [1, Conjecture 7.1].
Remark. Let
qk(x;q)
k
f (q) = .
k
(q;q)
k
Then it is not difficult to prove that
k k qj(x;q) (xq−1;q−1)
f˜k(q) = (−1)jq(2j)−kj · j = k.
j (q;q) (q−1;q−1)
j=0 (cid:20) (cid:21)q j k
X
Define
n−1
α −1−α (x;q)
P (α,x;q) = qk2+k · k.
n
k k (q;q)
k=0 (cid:20) (cid:21)q(cid:20) (cid:21)q k
X
Then in view of Theorem 1.2, we can get
Pn(−r/d,x;qd) ≡ (−1)aqd(a+21)+(ad+r)(n2−1−2a)Pn(−r/d,xq−d;q−d) (mod Φn(q)2)
for odd n ≥ 3, where a = h−r/di . This is a q-analogue of [2, Theorem 2.2].
n
References
[1] VictorJ.W.Guo andJ.Zeng,Some q-analogues of supercongruences of Rodriguez-Villegas,
J. Number Theory, 145(2014), 301–316.
[2] Z.-H. Sun, Generalized Legendre polynomials and related supercongruences, J. Number The-
ory, 143(2014), 293–319.