Some $q$-analogs of congruences for central binomial sums

0.13 MB·English

by  Roberto Tauraso

SOME q-ANALOGS OF CONGRUENCES
FOR CENTRAL BINOMIAL SUMS
ROBERTO TAURASO
2
1 Abstract. We establish q-analogs for four congruences involving central binomial coeffi-
0
cients. The q-identitiesnecessary for this purposeare shown viathe q-WZ method.
2

n
a
J 1. Introduction

0
3 Recently, a number of papers have appeared concerning congruences for central binomial
sums (see through the references). Here we would like to draw the attention to one aspect of

]
the matter which has been partly neglected so far: q-analogs. In [5, 11], the authors identified
T
N a first group of such congruences which have a q-counterpart. Among them we mention: for
any prime p > 2
.
h
t
ma p−1 2k ≡ p (mod p2), p−1qk 2k ≡ p q⌊p32⌋ (mod [p]q),
k 3 k 3
[ Xkp−=10(cid:18) (cid:19) (cid:16) (cid:17) Xkp−=10 (cid:20) (cid:21)q (cid:16) (cid:17)
v1 (−1)k 2kk ≡ p5 (mod p), (−1)kq−(k+21) 2kk ≡ p5 q−⌊p54⌋ (mod [p]q),
2 Xk=0 (cid:18) (cid:19) (cid:16) (cid:17) Xk=0 (cid:20) (cid:21)q (cid:16) (cid:17)
15 p−1 21k 2kk ≡ (−1)p−21 (mod p2), p−1 (−qq;kq) 2kk ≡ (−1)p−21q⌊p42⌋ (mod [p]q),
6 k=0 (cid:18) (cid:19) k=1 k(cid:20) (cid:21)q
. X X
1
0 where · denotes the Legendre symbol. It has been conjectured in [5] that the first q-
·
2 congruence holds modulo [p]2, and we claim that the same can be said for the third one.
1 (cid:0) (cid:1) q
However, in this short note, we are not going to refine these q-congruences. Instead, we will
:
v presentafewmoreexamples of thisphenomena. Moreprecisely weshowthatthecongruences
i
X
r p−1 (1/2)k 2k 3 p−1 (−1/2)k 2k −1
a (1) ≡ − ≡ Q (2) (mod p),
p
k k 2 k k
k=1 (cid:18) (cid:19) k=1 (cid:18) (cid:19)
X X
p−1
1 2k
(2) ≡ 0 (mod p2),
k k
k=1 (cid:18) (cid:19)
X
5 p−1 (−1)k 2k p−1 1
(3) ≡ − (mod p3),
2 k2 k k2
k=1 (cid:18) (cid:19) k=1
X X
2000 Mathematics Subject Classification. 11B65, 11A07 (Primary) 05A10, 05A19, 05A30 (Secondary).
Key words and phrases. q-analogs, Gaussian q-binomial coefficients, central binomial coefficients,
congruences.
1
2 ROBERTOTAURASO
where p > 5 is a prime and Q (2) = (2p−1 − 1)/p is the usual Fermat’s quotient, have as
p
q-analogs respectively
p−1 qk 2k 1 p−1 (−1)k(1+qk +q2k) 2k −1
(4) ≡ − ≡ Q (2;q) (mod [p] ),
p q
(−q;q) [k] k 2 (−q;q) [k] k
k=1 k q(cid:20) (cid:21)q k=1 k q (cid:20) (cid:21)q
X X
(5) p−1 (1+qk +q2k)q−(k2) 2k ≡ [p]q(p2−1)(1−q)2 (mod [p]2),
(1+qk)2[k] k 24 q
k=1 q (cid:20) (cid:21)q
X
(6) p−1 (−1)k(1+3qk +q2k)q−(k2) 2k ≡ −p−1 qk − [p]2q(p4−1)(1−q)4 (mod [p]3).
(1+qk)[k]2 k [k]2 240 q
k=1 q (cid:20) (cid:21)q k=1 q
X X
where Q (2;q) = ((−q;q) −1)/[p] . Proofs of (1), (2), (3) can be found in [12, Theorem
p p−1 q
3.1] ((2) appeared first in [10]).
We are optimistically hopeful that there are plenty of interesting q-analogs to discover. For
example, recently in [8], the authors proved that for 0 < q < 1
∞ (1+2qk)qk2 2k −1 ∞ qk
= .
[k]2 k [k]2
k=1 q (cid:20) (cid:21)q k=1 q
X X
By letting q → 1, it gives a well known series identity which then happened to have a
congruence version: in [12] we showed that for any prime p > 3
p−1 −1 p−1
1 2k 1 1
≡ − (mod p3).
k2 k 6 k2
k=1 (cid:18) (cid:19) k=1
X X
Is there a q-analog for the above congruence?
2. Notations a preliminary results
The first two results of this section yield a family of q-analogs of the classical congruence
for the harmonic sums: for any prime p > d+2 where d is a positive integer,
p−1
1 0 (mod p2) if d is odd,
H (d) := ≡
p−1 kd 0 (mod p) if d is even.
k=1 (cid:26)
X
This family depends on two integer parameters a,b and it concerns the sum
p−1 qbk
[ak]d
q
k=1
X
where
1−qn
[n] = = 1+q+···+qn−1.
q
1−q
Several special cases have already been discussed by numerous authors (see [3, 4, 7, 9]). In
particular, K. Dilcher found in [4] a determinant expression in the case when a = 1 and
b ∈ {0,1}. We point out that, in this paper, two rational functions in q are congruent
modulo [p]r for r ≥ 1 if the numerator of their difference is congruent to 0 modulo [p]r in the
q q
polynomial ring Z[q] and the denominator is relatively prime to [p] .
q
SOME q-ANALOGS OF CONGRUENCES FOR CENTRAL BINOMIAL SUMS 3
Theorem 2.1. For any prime p > 2, if a,b,d are integers such that a,d > 0, b ≥ 0 and
gcd(a,p) = 1 then
p−1 qbk (1−q)d d−1 r +sp d d sp
(7) ≡ (−1)dp c 0 − (−1)s (mod [p] )
[ak]d pd s d s 2d q
k=1 q s=0 (cid:18) (cid:19) s=0 (cid:18) (cid:19)(cid:18) (cid:19)!
X X X
where r ≡−b/a (mod p) such that r ∈ {0,1,...,p−1} and
0 0
s
r +kp+d−1 d
c = (−1)s−k 0 .
s
d−1 s−k
k=0 (cid:18) (cid:19)(cid:18) (cid:19)
X
Proof. Let q be a p-root of unity such that q 6= 1. Since
p−1
p if p | (aj +b),
qk(aj+b) = −1+
0 otherwise,
k=1 (cid:26)
X
it follows that
p−1 qbk p−1 j+d−1
= qbk qakjzj
(1−qakz)d d−1
k=1 k=1 j≥0(cid:18) (cid:19)
X X X
p−1
j +d−1
= zj qk(aj+b)
d−1
j≥0(cid:18) (cid:19) k=1
X X
r +sp+d−1 1
= p 0 zr0+sp−
d−1 (1−z)d
s≥0(cid:18) (cid:19)
X
p d−1c zr0+sp 1
= s=0 s − .
(1−zp)d (1−z)d
P
Let z =1+w, then
p−1 qbk p ds=−01cs(1+w)r0+sp− 1−(−1+ww)p d
= lim
(1−qak)d w→0 P (1−(1+w)p)(cid:16)d (cid:17)
k=1
X
p d−1c r0+sp wd+o(wd)−(−1)d d (−1)s d sp wd+o(wd)
= lim s=0 s d s=0 s 2d
w→0 (−pw+o(w))d
P (cid:0) (cid:1) P (cid:0) (cid:1)(cid:0) (cid:1)
d−1 d
1 r +sp d sp
= (−1)dp c 0 − (−1)s .
pd s d s 2d
!
s=0 (cid:18) (cid:19) s=0 (cid:18) (cid:19)(cid:18) (cid:19)
X X
(cid:3)
4 ROBERTOTAURASO
Thefollowing specialcases areworth mentioning. By letting d= 1,2,3 in (7), weobtain these
q-congruences which hold modulo [p] :
q
p−1 qbk p−1
(8) ≡ (1−q) −r ,
0
[ak] 2
k=1 q (cid:18) (cid:19)
X
p−1 qbk (p−1)(p−5) r (p−2−r )
(9) ≡ (1−q)2 − + 0 0 ,
[ak]2 12 2
k=1 q (cid:18) (cid:19)
X
p−1 qbk (p−1)(p−3) r (p2−9p+12−3r (p−3)+2r2)
(10) ≡ (1−q)3 − − 0 0 0 .
[ak]3 8 12
k=1 q (cid:18) (cid:19)
X
Theorem 2.2. Let b,b,a,d be non-negative integers such that ad = b+b > 0.
Then for any prime p > 2 such that gcd(a,p) = 1,
p−1 (−1)d−1qbk +qbk p−1 qbk p−1 qbk
(11) ≡b(1−q)[p] −ad[p] (mod [p]2)
[ak]d q [ak]d q [ak]d+1 q
k=1 q k=1 q k=1 q
X X X
and
(12)
p−1 (−1)k((−1)dqbk +qbk) p−1 (−1)kqbk p−1 (−1)kqbk
≡ b(1−q)[p] −ad[p] (mod [p]2).
[ak]d q [ak]d q [ak]d+1 q
q q q
k=1 k=1 k=1
X X X
Moreover, let b ,b ,b ,b be non-negative integers.
1 1 2 2
If d = b +b > 0 and d = b +b >0 then
1 1 1 2 2 2
(13)
qb1j+b2k +(−1)d1+d2qb1j+b2k p−1 qb1j p−1 qb2k p−1 q(b1+b2)k
≡ · − (mod [p] ).
[j]d1[k]d2 [j]d1 [k]d2 [k]d1+d2 q
1≤j<k≤p−1 q q k=1 q k=1 q k=1 q
X X X X
Proof. As regards (11) and (12), it suffices to note that
(−1)dqb(p−k) (−1)dqb(p−k)+adk qbp+bk
= =
[a(p−k)]d ([ap] −[ak] )d [ak]d(1−[ap] /[ak] )d
q q q q q q
qbk(1−b(1−q)[p] ) [p]
q q
≡ 1+d
[ak]d [ak]
q (cid:18) q(cid:19)
qbk b(1−q)[p] qbk ad[p] qbk
(14) ≡ − q + q (mod [p]2).
[ak]d [ak]d [ak]d+1 q
q q q
Moreover
p−1 qb1j p−1 qb2k p−1 q(b1+b2)k qb1j+b2k qb1j+b2k
· − = +
[j]d1 [k]d2 [k]d1+d2 [j]d1[k]d2 [j]d1[k]d2
k=1 q k=1 q k=1 q 1≤j<k≤p−1 q q 1≤k<j≤p−1 q q
X X X X X
and by (14) we get
qb1j+b2k qb1(p−j)+b2(p−k) (−1)d1+d2qb1j+b2k
= ≡ (mod [p] ).
[j]d1[k]d2 [p−j]d1[p−k]d2 [j]d1[k]d2 q
1≤k<j≤p−1 q q 1≤j<k≤p−1 q q 1≤j<k≤p−1 q q
X X X
Hence the proof of (13) is complete. (cid:3)
SOME q-ANALOGS OF CONGRUENCES FOR CENTRAL BINOMIAL SUMS 5
By letting a = 1, d = 1 and b = 0 in (11), and by using (7), we easily find [9, Theorem 1]:
for any prime p > 3:
p−1 1 1 p−1 1−qk 1 p−1 1+qk (1−q)(p−1) (p2−1)(1−q)2[p]
= ≡ + q (mod [p]2).
[k] 2 [k] 2 [k] 2 24 q
q q q
k=1 k=1 k=1
X X X
In a similar way, for a = 1, d= 3 and b = 1, (11) yields
(15)
p−1 qk +q2k p−1 q2k p−1 q2k [p] (1−q)4(p4−1)
≡ (1−q)[p] −3[p] ≡ − q (mod [p]2).
[k]3 q [k]3 q [k]4 240 q
q q q
k=1 k=1 k=1
X X X
Inorder to show theq-congruences stated in theintroduction, we needsuitable q-identities.
Such identities are not easy to find, but ones they are guessed correctly hopefully they can be
proved via the q-WZ method (see for example [6, 13]).
For n ≥ k ≥ 0, a pair (F(n,k),G(n,k)) is called q-WZ pair if
F(n+1,k)/F(n,k), F(n,k+1)/F(n,k), G(n+1,k)/G(n,k), G(n,k+1)/G(n,k)
are all rational functions of qn and qk, and
F(n+1,k)−F(n,k) = G(n,k+1)−G(n,k).
Let
N−1
S(n) = F(n,k)
k=0
X
then
n−1
S(n+1)−S(n) = F(n+1,n)+ (F(n+1,k)−F(n,k))
k=0
X
n−1
= F(n+1,n)+ (G(n,k+1)−G(n,k))
k=0
X
= F(n+1,n)+G(n,n)−G(n,0)
and, by summing over n from 0 to N −1, we get the identity
N−1 N−1 N−1
(16) F(N,k) = (F(n+1,n)+G(n,n))− G(n,0)
k=0 n=0 n=0
X X X
which can be considered as the finite form of [6, Theorem 7.3].
The q-identities we are interested in involves Gaussian q-binomial coefficients
n (q;q) (q;q)−1(q;q)−1 if 0 ≤ k ≤n,
= n k n−k
(cid:20)k(cid:21)q ( 0 otherwise,
where (a;q) = n−1(1−aqj) (note that n is a polynomial in q). The next lemma allows
n j=0 k q
us to reduce a special class of q-binomial coefficients modulo a power of [p] .
Q (cid:2) (cid:3) q
6 ROBERTOTAURASO
Lemma 2.3. Let p be a prime and let a be a positive integer. For k = 1,...,p−1 we have
k
(17) apk−1 ≡ (−1)kq−(k+21)1−a[p]q [j1]  (mod [p]2q),
(cid:20) (cid:21)q j=1 q
X
 
p−1+k [p] k−1 qj
(18) ≡ q 1+[p] (mod [p]3),
k [k]  q [j]  q
(cid:20) (cid:21)q q j=1 q
X
 
(19) p−1+k p−1 −1 ≡ (−1)kq(k+21)[p]q 1+ [p]q +[p] k−1 1+qj (mod [p]3).
k k [k]  [k] q [j]  q
(cid:20) (cid:21)q(cid:20) (cid:21)q q q j=1 q
X
 
Proof. Since [ap] ≡ a[p] (mod [p]2), it follows that
q q q
k k
apk−1 = (−1)kq−(k+21) 1− [a[jp]]q ≡ (−1)kq−(k+21)1−a[p]q [j1]  (mod [p]2q).
(cid:20) (cid:21)q j=1(cid:18) q (cid:19) j=1 q
Y X
Moreover  
p−1+k [p] k−1 qj[p] [p] k−1 qj
= q 1+ q ≡ q 1+[p] (mod [p]3).
k [k] [j] [k]  q [j]  q
(cid:20) (cid:21)q q j=1(cid:18) q (cid:19) q j=1 q
Y X
Conguences (17) and (18) easily yield (19).   (cid:3)
It should be noted that when p is an odd prime, by using (18) for k = p−1, we recover
the q-congruence [3, (3.2)]:
(20)
p−1
app−−11 ≡ q−(p2)1−a[p]q [j1] ≡ q−(p2) 1− a[p]q(p−21)(1−q) ≡ q(a−1)(p2) (mod [p]2q).
(cid:20) (cid:21)q j=1 q (cid:18) (cid:19)
X
 
3. Proof of (4)
By [2, (5.17)] (see [5, (4.1)] for a generalization), if n is odd then by
n (−1)n−kq(n−2k) n 2k
= 0.
(−q;q) k k
k=0 k (cid:20) (cid:21)q(cid:20) (cid:21)q
X
Hence for n = p we have that
p−1 (−1)k−1q(p−2k)−(p2) p−1 2k = 1 (−q;q)p−1−q−(p2) 2p−1 .
(−q;q) [k] k−1 k (−q;q) [p] p−1
k=1 k q (cid:20) (cid:21)q(cid:20) (cid:21)q p−1 q (cid:20) (cid:21)q!
X
By (17) and (20) we get
p−1 qk 2k p−1 q−pk+k 2k (−q;q) −1
p−1
≡ ≡ ≡Q (2;q) (mod [p] ),
p q
(−q;q) [k] k (−q;q) [k] k (−q;q) [p]
k=1 k q(cid:20) (cid:21)q k=1 k q(cid:20) (cid:21)q p−1 q
X X
and the first congruence is proved.
As regards the second one, we take
(−1)k n+k+1 −1 qn+1F(n,k)
F(n,k) = and G(n,k) = .
(−q;q) [k+1] k+1 1+qn+1
n q(cid:20) (cid:21)q
SOME q-ANALOGS OF CONGRUENCES FOR CENTRAL BINOMIAL SUMS 7
This pair can be found in [1][Subsection 2.1] in connection with the irrationality proof of the
q-series
∞ (−1)k
Ln (2) :=
q qn−1
k=1
X
for |q| 6∈ {0,1}. Hence by (16) we obtain the identity
n (−1)k(1+qk +q2k) 2k −1 1 n (−1)k n+k −1 n qk
(21) = −
(−q;q) [k] k (−q;q) [k] k (−q;q) [k]
k=1 k q (cid:20) (cid:21)q n k=1 q (cid:20) (cid:21)q k=1 k q
X X X
Let n = p−1. Now by (17) and [7, (1.5)]
1 p−1 (−1)k n+k −1 p−1 (−1)k k−1 qj
≡ 1−[p]
q
(−q;q) [k] k [p]  [j] 
p−1 k=1 q (cid:20) (cid:21)q k=1 q j=1 q
X X X
p−1 k−1 qj p−1q2j−1
≡ − (−1)k ≡−
[j] [2j −1]
q q
k=1 j=1 j=1
X X X
p−1 (p−1)/2
(p−1)(1−q) 1 1
≡ − + ≡ −Q (2;q) (mod [p] ).
p q
2 [j] [2j]
q q
k=1 k=1
X X
By the q-binomial theorem,
n k−1
n
(x−qj)= xn
k
k=0(cid:20) (cid:21)qj=0
X Y
and for n = p, x = −1 together with (17), we get
p−1 q−(k2)(−q;q)k−1 p−1 (−1)k−1(−q;q)k−1 p−1
≡ = −Q (2;q).
p
[k] [k] k−1
k=1 q k=1 q (cid:20) (cid:21)q
X X
Hence (see the dual congruence [7, (5.4)])
p−1 qk p−1 qp−k p−1 q−(k2)(−q;q)k−1
≡ ≡ ≡ −Q (2;q) (mod [p] )
p q
(−q;q) [k] (−q;q) [p−k] [k]
k q p−k q q
k=1 k=1 k=1
X X X
where we used
[p−k]q = −q−k[k]q and (−q;q)−p−1k ≡ q−(k2)(−q;q)k−1 (mod [p]q).
Therefore, by identity (21),
p−1 (−1)k(1+qk +q2k) 2k −1
≡ −2Q (2;q) (mod [p] )
p q
(−q;q) [k] k
k=1 k q (cid:20) (cid:21)q
X
and we are done.
8 ROBERTOTAURASO
4. Proof of (5)
Let
q−(k+21) n+k+2 n+1 −1
F(n,k) =
[k+1] n+1 k+1
q(cid:20) (cid:21)q(cid:20) (cid:21)q
and
qn+2(1−qk+1)(1−qn+1−k)F(n,k)
G(n,k) = −
(1+qn+2)(1−qn+2)2
then (16) gives the identity
n (1+qk +q2k)q−(k2) 2k n q−(k2) n+k n −1 n qk
(22) = − .
(1+qk)2[k] k [k] k k [2k]
k=1 q (cid:20) (cid:21)q k=1 q (cid:20) (cid:21)q(cid:20) (cid:21)q k=1 q
X X X
Let n = p−1 then by (19) and (12), we obtain
p−1 q−(k2) p−1+k p−1 −1 p−1 (−1)kqk
≡[p] ≡ 0 (mod [p]2).
[k] k k q [k]2 q
k=1 q (cid:20) (cid:21)q(cid:20) (cid:21)q k=1 q
X X
Moreover, (11) and (9) implies that
p−1 qk p−1 qk [p] (p2−1)(1−q)2
≡ −[p] ≡ − q (mod [p]2),
[2k] q [2k]2 24 q
q q
k=1 k=1
X X
and (5) follows easily from (22).
Note that by letting q → 1 in (22) we obtain the identity
n n −1 n
3 1 2k 1 n+k n 1 1
= − .
4 k k k k k 2 k
k=1 (cid:18) (cid:19) k=1 (cid:18) (cid:19)(cid:18) (cid:19) k=1
X X X
which can be exploited to prove an improvement of [12, Theorem 4.2]:
p−1
1 2k 8
≡− H (1)+2p4B (mod p5)
p−1 p−5
k k 3
k=1 (cid:18) (cid:19)
X
for any prime p > 3.
5. Proof of (6)
By taking
(−1)kq−(k+21)(1+qk+1) n+k+2 n+1 −1
F(n,k) =
[k+1]2 n+1 k+1
q (cid:20) (cid:21)q(cid:20) (cid:21)q
and
qn+2(1−qk+1)2(1−qn+1−k)F(n,k)
G(n,k) = ,
(1+qk+1)(1−qn+2)3
(16) yields the identity
(23)
n (−1)k(1+3qk +q2k)q−(k2) 2k n (−1)k(1+qk)q−(k2) n+k n −1 n qk
= − .
(1+qk)[k]2 k [k]2 k k [k]2
k=1 q (cid:20) (cid:21)q k=1 q (cid:20) (cid:21)q(cid:20) (cid:21)q k=1 q
X X X
SOME q-ANALOGS OF CONGRUENCES FOR CENTRAL BINOMIAL SUMS 9
Let n = p−1. Now by (19)
p−1 (−1)k(1+qk)q−(k2) p−1+k p−1 −1 p−1 qk +q2k p−1 qk +q2k
≡ [p] +[p]2
[k]2 k k q [k]3 q [k]4
k=1 q (cid:20) (cid:21)q(cid:20) (cid:21)q k=1 q k=1 q
X X X
(1+qj)(qk +q2k)
+[p]2 (mod [p]3).
q [j] [k]3 q
q q
1≤j<k≤p−1
X
Then the q-congruence (6) follows from (23) by using (15) and
p−1 qk +q2k (1+qj)(qk +q2k) (1−q)4(p2−1)(p2 −4)
≡ − ≡ (mod [p] ),
[k]4 [j] [k]3 360 q
q q q
k=1 1≤j<k≤p−1
X X
which is a straightforward application of (7) and (13).
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