On q-congruences involving harmonic numbers
We give some congruences involving $q$-harmonic numbers and alternating $q$-harmonic numbers of order $m$. Some of them are $q$-analogues of several known congruences.
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507632552378368 |
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| author | Ge, B. Ге, Б. |
| author_facet | Ge, B. Ге, Б. |
| author_sort | Ge, B. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
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| datestamp_date | 2019-12-05T09:26:20Z |
| description | We give some congruences involving $q$-harmonic numbers and alternating $q$-harmonic numbers of order $m$. Some of them
are $q$-analogues of several known congruences. |
| first_indexed | 2026-03-24T02:12:24Z |
| format | Article |
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UDC 512.5
B. He (College Sci., Northwest A&F Univ., Yangling, Shaanxi, China)
ON \bfitq -CONGRUENCES INVOLVING HARMONIC NUMBERS*
ПРО \bfitq -КОНГРУЕНЦIЇ, ЩО ВКЛЮЧАЮТЬ ГАРМОНIЧНI ЧИСЛА
We give some congruences involving q-harmonic numbers and alternating q-harmonic numbers of order m. Some of them
are q-analogues of several known congruences.
Наведено деякi конгруенцiї, що включають q-гармонiчнi числа та знакозмiннi q-гармонiчнi числа m-го порядку.
Деякi з цих конгруенцiй є q-аналогами кiлькох вiдомих конгруенцiй.
1. Introduction. For arbitrary positive integer n, the q-integer can be defined by
[n]q =
1 - qn
1 - q
.
It is easy to see that \mathrm{l}\mathrm{i}\mathrm{m}q\rightarrow 1[n]q = n. Supposing that a \equiv b (\mathrm{m}\mathrm{o}\mathrm{d} p), we have
[a]q =
1 - qa
1 - q
=
1 - qb + qb(1 - qa - b)
1 - q
\equiv 1 - qb
1 - q
= [b]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]q).
Here and in what follows, each congruence is considered over the polynomial ring \BbbZ [q] in the variable
q with integral coefficients.
For m = 1, 2, 3, . . . and n = 0, 1, 2, . . . , we define
H
(m)
0 = 0, H(m)
n =
n\sum
j=1
1
jm
for n \geq 1
and call it a harmonic number of order m. Those Hn = H
(1)
n are usually called the classical harmonic
numbers. Similarly, the alternating harmonic numbers of order m are given by
I
(m)
0 = 0, I(m)
n =
n\sum
j=1
( - 1)j
jm
for n \geq 1.
In this paper, we define
Hn(q) =
n\sum
j=1
1
[j]q
, \widetilde Hn(q) =
n\sum
j=1
qj
[j]q
,
H(2)
n (q) =
n\sum
j=1
1
[j]2q
, \widetilde H(2)
n (q) =
n\sum
j=1
qj
[j]2q
,
H(3)
n (q) =
n\sum
j=1
1
[j]3q
, \widetilde H(3)
n (q) =
n\sum
j=1
qj
[j]3q
* This work was supported by the Natural Science Basic Research Plan in Shaanxi Province of China (No. 2017JQ1001),
the Initial Foundation for Scientific Research of Northwest A&F University (No. 2452015321) and the Fundamental
Research Funds for the Central Universities (No. 2452017170).
c\bigcirc B. HE, 2017
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 9 1257
1258 B. HE
and
In(q) =
n\sum
j=1
( - 1)j
[j]q
,
I(2)n (q) =
n\sum
j=1
( - 1)j
[j]2q
,
where
H0(q) = \widetilde H0(q) = H
(2)
0 (q) = \widetilde H(2)
0 (q) = H
(3)
0 (q) = \widetilde H(3)
0 (q) = I0(q) = I
(2)
0 (q) = 0.
They are q-analogues of harmonic numbers of order m. So we call them q-harmonic numbers and
alternating q-harmonic numbers of order m.
In view of the q-analogue of Glaishers congruence, Andrews [1] (Theorem 4) showed that
Hp - 1(q) \equiv
p - 1
2
(1 - q) (\mathrm{m}\mathrm{o}\mathrm{d} [p]q)
and
\widetilde Hp - 1(q) \equiv
p - 1
2
(q - 1) (\mathrm{m}\mathrm{o}\mathrm{d} [p]q).
L. L. Shi and H. Pan obtained (see [6], Theorem 1)
Hp - 1(q) \equiv
p - 1
2
(1 - q) +
p2 - 1
24
(1 - q)2[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q) (1.1)
for each p \geq 5, which is equivalent to
\widetilde Hp - 1(q) \equiv
1 - p
2
(1 - q) +
p2 - 1
24
(1 - q)2[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q).
Pan established (see [5], Theorem 1.1) that for each odd prime p, there holds
2
p - 1
2\sum
j=1
1
[2j]q
+ 2Qp(2, q) - Qp(2, q)
2[p]q \equiv
\equiv
\biggl(
Qp(2, q)(1 - q) +
p2 - 1
8
(1 - q)2
\biggr)
[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q), (1.2)
where Qp(2, q) =
( - q; q)p - 1 - 1
[p]q
and (x; q)n =
\prod n - 1
k=0
(1 - xqk). For some material on congruences
of q-harmonic numbers, see, for example, [3]. Some other q-congruences were obtained by different
authors, see, for example [2, 4, 9].
Our aim of this paper is to give some congruences involving q-harmonic numbers and alternating
q-harmonic numbers of order m which are q-analogues of several known identities.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 9
ON q-CONGRUENCES INVOLVING HARMONIC NUMBERS 1259
Theorem 1.1. Let p \geq 5 be a prime. Then
p - 1\sum
k=1
qkH
(2)
k (q) \equiv - p - 1
2
(1 - q) - (p - 1)(p - 3)
8
(1 - q)2[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q), (1.3)
p - 1\sum
k=1
qk \widetilde H(2)
k (q) \equiv p - 1
2
(1 - q) - p2 - 1
8
(1 - q)2[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q), (1.4)
p - 1\sum
k=1
qkH
(3)
k (q) \equiv (p - 1)(p - 5)
12
(1 - q)2 (\mathrm{m}\mathrm{o}\mathrm{d} [p]q), (1.5)
p - 1\sum
k=1
qk \widetilde H(3)
k (q) \equiv p2 - 1
12
(1 - q)2 (\mathrm{m}\mathrm{o}\mathrm{d} [p]q). (1.6)
When q \rightarrow 1, the first two q-congruences in Theorem 1.1 reduce to the following result [7]
(Lemma 2.1):
p - 1\sum
k=1
H
(2)
k \equiv 0 (\mathrm{m}\mathrm{o}\mathrm{d} p2)
while the last two q-congruences reduce to
p - 1\sum
k=1
H
(3)
k \equiv 0 (\mathrm{m}\mathrm{o}\mathrm{d} p).
Theorem 1.2. Let p \geq 5 be a prime. Then
Ip - 1(q) \equiv - 2Qp(2, q) -
p - 1
2
(1 - q)+
+
\biggl(
Qp(2, q)
2 +Qp(2, q)(1 - q) +
p2 - 1
12
(1 - q)2
\biggr)
[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q) (1.7)
and
I
(2)
p - 1(q) \equiv
1 - p
2
(1 - q)2 - 2Qp(2, q)(1 - q) (\mathrm{m}\mathrm{o}\mathrm{d} [p]q). (1.8)
It is clear that (1.7) and (1.8) are respectively q-analogues of
Ip - 1 \equiv - 2qp(2) + qp(2)
2p (\mathrm{m}\mathrm{o}\mathrm{d} p2)
and
I
(2)
p - 1 \equiv 0 (\mathrm{m}\mathrm{o}\mathrm{d} p)
(see [8], Lemma 2.1), where qp(2) =
2p - 1 - 1
p
.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 9
1260 B. HE
Theorem 1.3. Let p \geq 5 be a prime. Then
p - 1\sum
k=1
qkIk(q) \equiv
\biggl(
- 2Qp(2, q) -
p - 1
2
(1 - q)
\biggr)
[p]q +
+
\biggl(
Qp(2, q)
2 +Qp(2, q)(1 - q) +
p2 - 1
12
(1 - q)2
\biggr)
[p]2q (\mathrm{m}\mathrm{o}\mathrm{d} [p]3q),
p - 1\sum
k=1
qkI
(2)
k (q) \equiv 2Qp(2, q) +
p - 1
2
(1 - q) -
-
\biggl(
Qp(2, q)
2 + 3Qp(2, q)(1 - q) +
(7 + p)(p - 1)
12
(1 - q)2
\biggr)
[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q).
When q \rightarrow 1, the two q-congruences in Theorem 1.3 reduce respectively to the following two
congruences:
p - 1\sum
k=1
Ik \equiv - 2qp(2)p+ qp(2)
2p2 (\mathrm{m}\mathrm{o}\mathrm{d} p3),
p - 1\sum
k=1
I
(2)
k \equiv 2qp(2) - qp(2)
2p (\mathrm{m}\mathrm{o}\mathrm{d} p2).
Theorem 1.4. Let p \geq 5 be a prime. Then
p - 1\sum
k=1
( - 1)k
[k]q
Ik(q) \equiv 2Q2
p(2, q) + (p - 1)(1 - q) +
p2 - 1
12
(1 - q)2 (\mathrm{m}\mathrm{o}\mathrm{d} [p]q).
The congruence in Theorem 1.4 is a q-analogue of\sum
1\leq i\leq j\leq p - 1
( - 1)i+j
ij
\equiv 2q2p(2) (\mathrm{m}\mathrm{o}\mathrm{d} p).
Our method of proving Theorems 1.1 – 1.4 is to write the finite sums involving (alternating) q-
harmonic numbers into a linear combination of at most two (alternating) q-harmonic sums. We will
provide one lemma in the next section. Section 3 is devoted to our proof of Theorems 1.1 – 1.4.
2. Auxiliary result. To prove Theorems 1.1 – 1.4, we need the following auxiliary result.
Lemma 2.1. For any prime p \geq 5, there hold
p - 1\sum
j=1
1
[j]2q
\equiv - (p - 1)(p - 5)
12
(1 - q)2 (\mathrm{m}\mathrm{o}\mathrm{d} [p]q), (2.1)
p - 1\sum
j=1
qj
[j]2q
\equiv - p2 - 1
12
(1 - q)2 (\mathrm{m}\mathrm{o}\mathrm{d} [p]q), (2.2)
p - 1
2\sum
j=1
1
[2j]2q
\equiv - p2 - 1
24
(1 - q)2 - Qp(2, q)(1 - q) (\mathrm{m}\mathrm{o}\mathrm{d} [p]q). (2.3)
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 9
ON q-CONGRUENCES INVOLVING HARMONIC NUMBERS 1261
Proof. See [6] (Lemma 2) for the proof of (2.1) and (2.2).
We now prove (2.3). It is obvious that
1
[p - 2j]q
\equiv - q2j
[2j]q
(\mathrm{m}\mathrm{o}\mathrm{d} [p]q). (2.4)
Then by (2.2) and (2.4), we get
- p2 - 1
12
(1 - q)2 \equiv
p - 1\sum
j=1
qj
[j]2q
=
p - 1
2\sum
j=1
q2j
[2j]2q
+
p - 1
2\sum
j=1
qp - 2j
[p - 2j]2q
\equiv
\equiv 2
p - 1
2\sum
j=1
q2j
[2j]2q
(\mathrm{m}\mathrm{o}\mathrm{d} [p]q),
namely,
p - 1
2\sum
j=1
q2j
[2j]2q
\equiv - p2 - 1
24
(1 - q)2 (\mathrm{m}\mathrm{o}\mathrm{d} [p]q). (2.5)
By (1.2), we have
p - 1
2\sum
j=1
1
[2j]q
\equiv - Qp(2, q) (\mathrm{m}\mathrm{o}\mathrm{d} [p]q). (2.6)
Hence, with the help of
q2j
[2j]2q
=
1
[2j]2q
- (1 - q)
1
[2j]q
,
(2.5) and (2.6), we obtain
p - 1
2\sum
j=1
1
[2j]2q
=
p - 1
2\sum
j=1
q2j
[2j]2q
+ (1 - q)
p - 1
2\sum
j=1
1
[2j]q
\equiv
\equiv - p2 - 1
24
(1 - q)2 - Qp(2, q)(1 - q) (\mathrm{m}\mathrm{o}\mathrm{d} [p]q).
Lemma 2.1 is proved.
3. Proofs of Theorems 1.1 – 1.4. Proof of Theorem 1.1. We first prove (1.3) and (1.4). Observe
that
p - 1\sum
k=1
qkH
(2)
k (q) =
p - 1\sum
j=1
1
[j]2q
p - 1\sum
k=j
qk =
p - 1\sum
j=1
1
[j]2q
qj - qp
1 - q
=
=
p - 1\sum
j=1
1
[j]2q
\biggl(
1 - qp
1 - q
- 1 - qj
1 - q
\biggr)
=
= [p]qH
(2)
p - 1(q) - Hp - 1(q).
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 9
1262 B. HE
In view of (1.1) and (2.1), we obtain
p - 1\sum
k=1
qkH
(2)
k (q) \equiv - p - 1
2
(1 - q) - (p - 1)(p - 3)
8
(1 - q)2[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q).
This proves (1.3).
By [3] (Theorem 1.2),
p - 1\sum
k=1
qkHk(q) \equiv 1 - p+
p - 1
2
(1 - q)[p]q +
p2 - 1
24
(1 - q)2[p]2q (\mathrm{m}\mathrm{o}\mathrm{d} [p]3q)
which implies that
p - 1\sum
k=1
qkHk(q) \equiv 1 - p+
p - 1
2
(1 - q)[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q). (3.1)
Then (1.4) follows from (1.3), (3.1) and the fact H(2)
k (q) - \widetilde H(2)
k (q) = (1 - q)Hk(q).
We now show (1.5) and (1.6). Similarly, we can arrive at
p - 1\sum
k=1
qkH
(3)
k (q) = [p]qH
(3)
p - 1(q) - H
(2)
p - 1(q) \equiv
\equiv - H
(2)
p - 1(q) (\mathrm{m}\mathrm{o}\mathrm{d} [p]q).
We use the above and (2.1) to get
p - 1\sum
k=1
qkH
(3)
k (q) \equiv (p - 1)(p - 5)
12
(1 - q)2 (\mathrm{m}\mathrm{o}\mathrm{d} [p]q),
which proves (1.5). Then (1.6) follows from (1.5) and the fact H(3)
k (q) - \widetilde H(3)
k (q) = (1 - q)H
(2)
k (q).
Theorem 1.1 is proved.
Proof of Theorem 1.2. We first prove (1.7). Notice that
Ip - 1(q) =
p - 1\sum
k=1
( - 1)k + 1
[k]q
-
p - 1\sum
k=1
1
[k]q
=
= 2
p - 1
2\sum
k=1
1
[2k]q
-
p - 1\sum
k=1
1
[k]q
.
By (1.1) and (1.2), we obtain
Ip - 1(q) \equiv - 2Qp(2, q) -
p - 1
2
(1 - q)+
+
\biggl(
Qp(2, q)
2 +Qp(2, q)(1 - q) +
p2 - 1
12
(1 - q)2
\biggr)
[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q).
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 9
ON q-CONGRUENCES INVOLVING HARMONIC NUMBERS 1263
We now show (1.8). Note that
I
(2)
p - 1(q) =
p - 1\sum
k=1
( - 1)k + 1
[k]2q
-
p - 1\sum
k=1
1
[k]2q
=
= 2
p - 1
2\sum
k=1
1
[2k]2q
-
p - 1\sum
k=1
1
[k]2q
.
With the help of (2.1) and (2.3), we get
I
(2)
p - 1(q) \equiv
1 - p
2
(1 - q)2 - 2Qp(2, q)(1 - q) (\mathrm{m}\mathrm{o}\mathrm{d} [p]q).
Theorem 1.2 is proved.
Proof of Theorem 1.3. Observe that
p - 1\sum
k=1
qkIk(q) =
p - 1\sum
j=1
( - 1)j
[j]q
p - 1\sum
k=j
qk =
p - 1\sum
j=1
( - 1)j
[j]q
qj - qp
1 - q
=
=
p - 1\sum
j=1
( - 1)j
[j]q
\biggl(
1 - qp
1 - q
- 1 - qj
1 - q
\biggr)
= [p]qIp - 1(q)
and
p - 1\sum
k=1
qkI
(2)
k (q) =
p - 1\sum
j=1
( - 1)j
[j]2q
p - 1\sum
k=j
qk =
p - 1\sum
j=1
( - 1)j
[j]2q
qj - qp
1 - q
=
=
p - 1\sum
j=1
( - 1)j
[j]2q
\biggl(
1 - qp
1 - q
- 1 - qj
1 - q
\biggr)
=
= [p]qI
(2)
p - 1(q) - Ip - 1(q).
By (1.7) and (1.8), we arrive at
p - 1\sum
k=1
qkIk(q) \equiv
\biggl(
- 2Qp(2, q) -
p - 1
2
(1 - q)
\biggr)
[p]q+
+
\biggl(
Qp(2, q)
2 +Qp(2, q)(1 - q) +
p2 - 1
12
(1 - q)2
\biggr)
[p]2q (\mathrm{m}\mathrm{o}\mathrm{d} [p]3q)
and
p - 1\sum
k=1
qkI
(2)
k (q) \equiv 2Qp(2, q) +
p - 1
2
(1 - q) -
-
\biggl(
Qp(2, q)
2 + 3Qp(2, q)(1 - q) +
(7 + p)(p - 1)
12
(1 - q)2
\biggr)
[p]q (\mathrm{m}\mathrm{o}\mathrm{d} [p]2q),
which completes the proof of Theorem 1.3.
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 9
1264 B. HE
Proof of Theorem 1.4. Note that
p - 1\sum
k=1
( - 1)k
[k]q
Ik(q) =
p - 1\sum
j=1
( - 1)j
[j]q
p - 1\sum
k=j
( - 1)k
[k]q
=
=
p - 1\sum
j=1
( - 1)j
[j]q
\Biggl(
p - 1\sum
k=1
( - 1)k
[k]q
-
j\sum
k=1
( - 1)k
[k]q
+
( - 1)j
[j]q
\Biggr)
.
Hence,
p - 1\sum
k=1
( - 1)k
[k]q
Ik(q) =
1
2
\left( I2p - 1(q) +
p - 1\sum
j=1
1
[j]2q
\right) . (3.2)
By (1.7), we have
Ip - 1(q) \equiv - 2Qp(2, q) -
p - 1
2
(1 - q) (\mathrm{m}\mathrm{o}\mathrm{d} [p]q),
which implies that
I2p - 1(q) \equiv 4Q2
p(2, q) + 2(p - 1)(1 - q) +
(p - 1)2
4
(1 - q)2 (\mathrm{m}\mathrm{o}\mathrm{d} [p]q). (3.3)
With the help of (2.1), (3.2) and (3.3), we obtain
p - 1\sum
k=1
( - 1)k
[k]q
Ik(q) \equiv 2Q2
p(2, q) + (p - 1)(1 - q) +
p2 - 1
12
(1 - q)2 (\mathrm{m}\mathrm{o}\mathrm{d} [p]q),
which completes the proof of Theorem 1.4.
References
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145. – P. 301 – 316.
3. He B. Some congruences involving q-harmonic numbers // Ars Combin. (to appear).
4. He B., Wang K. Some congruences on q-Catalan numbers // Ramanujan J. – 2016. – 40. – P. 93 – 101.
5. Pan H. A q-analogue of Lehmers congruence // Acta Arith. – 2007. – 128. – P. 303 – 318.
6. Shi L. L., Pan H. A q-analogue of Wolstenholmes harmonic series congruence // Amer. Math. Monthly. – 2007. –
114. – P. 529 – 531.
7. Sun Z.-W. Supercongruences motivated by e // J. Number Theory. – 2015. – 147. – P. 326 – 341.
8. Sun Z.-W., Zhao L.-L. Arithmetic theory of harmonic numbers (II) // Colloq. Math. – 2013. – 130. – P. 67 – 78.
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Received 28.12.15,
after revision — 25.03.16
ISSN 1027-3190. Укр. мат. журн., 2017, т. 69, № 9
|
| id | umjimathkievua-article-1776 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:12:24Z |
| publishDate | 2017 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/2d/917031455594d749bb0b3a788b2d122d.pdf |
| spelling | umjimathkievua-article-17762019-12-05T09:26:20Z On q-congruences involving harmonic numbers Про \bfitq -конгруенцiї, що включають гармонiчнi числа Ge, B. Ге, Б. We give some congruences involving $q$-harmonic numbers and alternating $q$-harmonic numbers of order $m$. Some of them are $q$-analogues of several known congruences. Наведено деякi конгруенцiї, що включають $q$-гармонiчнi числа та знакозмiннi $q$-гармонiчнi числа $m$-го порядку. Деякi з цих конгруенцiй є $q$-аналогами кiлькох вiдомих конгруенцiй. Institute of Mathematics, NAS of Ukraine 2017-09-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1776 Ukrains’kyi Matematychnyi Zhurnal; Vol. 69 No. 9 (2017); 1257-1265 Український математичний журнал; Том 69 № 9 (2017); 1257-1265 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1776/758 Copyright (c) 2017 Ge B. |
| spellingShingle | Ge, B. Ге, Б. On q-congruences involving harmonic numbers |
| title | On q-congruences involving harmonic numbers |
| title_alt | Про \bfitq -конгруенцiї, що включають гармонiчнi числа |
| title_full | On q-congruences involving harmonic numbers |
| title_fullStr | On q-congruences involving harmonic numbers |
| title_full_unstemmed | On q-congruences involving harmonic numbers |
| title_short | On q-congruences involving harmonic numbers |
| title_sort | on q-congruences involving harmonic numbers |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1776 |
| work_keys_str_mv | AT geb onqcongruencesinvolvingharmonicnumbers AT geb onqcongruencesinvolvingharmonicnumbers AT geb probfitqkongruenciíŝovklûčaûtʹgarmoničničisla AT geb probfitqkongruenciíŝovklûčaûtʹgarmoničničisla |