Congruences modulo 8 for \((2,\, k)\) regular overpartitions for odd \(k > 1\)
Arabian Journal of Mathematics
Congruences modulo 8 for (2, k )regular overpartitions for odd k > 1
Chandrashekar Adiga 0 1 2
M. S. Mahadeva Naika 0 1 2
D. Ranganatha 0 1 2
C. Shivashankar 0 1 2
Mathematics Subject Classification 0 1 2
0 D. Ranganatha (
1 M. S. Mahadeva Naika
2 C. Adiga Department of Studies in Mathematics, University of Mysore , Manasagangotri, Mysore, Karnataka 570006 , India
In this paper, we study various arithmetic properties of the function p2, k (n), which denotes the number of (2, k)regular overpartitions of n with odd k > 1. We prove several infinite families of congruences modulo 8 for p2, k (n). For example, we find that for all nonnegative integers β, n and k ≡ 1 (mod 8), 5, 4 + 1, 2 + 2 + 1, 2 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1.

05A15 · 05A17 · 11P83
1 Introduction
Let N denote the set of natural numbers and N0 = N ∪ {0}. A partition of an integer n ∈ N is a nonincreasing
sequence of positive integers that sum to n. For a positive integer > 1, a partition is called regular if none
of the parts is divisible by . For example, the 3regular partitions of 5 are
An overpartition of n ∈ N0 is a partition of n in which the first occurrence of a number may be overlined.
For example, the overpartitions of 3 are
3, 3, 2 + 1, 2 + 1, 2 + 1, 2 + 1, 1 + 1 + 1, 1 + 1 + 1.
The generating function for p(n), the number of overpartitions of n with p(0) = 1 is given by [
5
]
∞
n=0
p(n)qn =
n=1
An extensive study of overpartitions can be found in the work of Corteel and Lovejoy [
5
].
Let po(n) denote the number of overpartitions of n into odd parts. The generating function for po(n) is
given by
Many mathematicians have extensively studied the arithmetic properties of po(n) and they have also established
several Ramanujantype congruences satisfied by po(n) (for example, one can see [
4,7
]).
Let A (n) denote the number of regular overpartitions of n. The generating function for A (n) is given
by
where
and for m ∈ N
∞
n=0
A (n)qn =
f2 f 2
,
This function was introduced and investigated by Lovejoy [
9
]. Later, Shen [
13
] discovered several
Ramanujanlike congruences for A3(n) and A4(n). Since then, a number of congruence properties for various regular
overpartition functions have been proved. (For example, one can see [
2,3,10,12
].) Very recently, the arithmetic
properties of regular overpartition pairs have been studied in [11].
Definition 1.1 A partition of n is said to be a (2, k)regular overpartition of n, if it is both 2 and kregular
overpartition of n.
Motivated by the above works, in this paper we prove infinite families of congruences modulo 8 for p2, k (n)
which enumerate the number of (2, k)regular overpartitions of n, for infinitely many values of odd k > 1.
For example, we prove the following theorems:
Theorem 1.2 If n, α ∈ N0 and k ≡ 1 (mod 4), then
p2, k (22+αn) ≡ p2, k (22n) (mod 8),
p2, k (21+α(4n + 2)) ≡ 7 p2, k (4n + 2) (mod 8).
Theorem 1.3 Let p ≡ 5, 7 (mod 8) and k ≡ 5 (mod 8) such that k
p
= 1. Then for all n, β ∈ N0, we have
It is easy to see that the generating function for p2, k (n) is
= (q; q2)∞(−qk ; q2k )∞
(1.1)
(1.2)
(1.3)
(1.4)
For example, setting k = 3 in (1.4), we obtain
∞
n=0
p2, 3(n)qn
The (2, 3)regular overpartitions of the integer 6 are
We note that work of Lin [
8
] on overpartition pairs into odd parts gives numerous congruences for p2,3(n)
modulo 3.
2 Set of preliminary results
In order to prove the main congruences of this paper, we collect some dissection formulas in this section.
For ab < 1, Ramanujan’s general theta function f (a, b) is defined by [
1
]
The following lemma is a consequence of Entry 25 of (v) and (vi) in [1, pp. 35–36].
Lemma 2.1 The following 2dissection formulas are true:
f 5
Ranganatha [
12
] established the pdissection formula for f12 which can be stated as follows:
2
Lemma 2.2 [12, Theorem 3.2] If p ≥ 5 is a prime and
± p − 1
6
:=
p−61 ,
if p ≡ 1 (mod 6),
− p−1 , if p ≡ −1 (mod 6),
6
From [
6
], we recall the following pdissection formula for ff212 :
Lemma 2.3 [6, Theorem 2.1] For any odd prime p, we have
Furthermore, m22+m ≡ p28−1 (mod p) for 0 ≤ m ≤ p−23 .
f (a, b) =
∞
n=−∞
an(n+1)/2bn(n−1)/2.
For notational convenience, in this section, we assume that all congruences are modulo 8, k ∈ N is odd and p
is a prime, unless stated otherwise. We first establish the following generating function for p2, k (2n) modulo
8.
Lemma 3.1 For any odd k > 1 ∈ N and n ∈ N0, we have
Proof Substituting (2.1) and (2.2) in (1.4), we find that
(3.1)
(3.2)
(3.3)
(3.4)
(3.5)
(3.6)
(3.7)
(3.8)
(3.9)
∞
n=0
p2, k (n)q
n
=
=
− 4q k+1 f4k f4 f126 f126k
+ 4q k f43 f83k + 4q f49 + 6q k −21 f8k f4k f8
Extracting the terms involving even powers of q in the above identity where k > 1 ∈ N is odd, we obtain
In view of (2.2), we have for any s ∈ N
Using (3.4), we can rewrite (3.3) as follows:
1
f 2 f 2 =
1 s
It is easy to check that, for any m,
Employing (3.6)–(3.8) into (3.5), we obtain (3.1).
Lemma 3.2 For any odd k > 1 ∈ N and n ∈ N0, we have
p2, k (2n)q
n
≡
f 5 f 5 f 3 f 3
4 4k 8 8k
Proof From (3.2) with odd k > 1 , we have
By (3.4), we find that
∞
n=0
p2, k(2n + 1)qn = 2
In view of (3.6), (3.7), (3.10) and (3.11), we obtain the required congruence.
Theorem 3.3 If n,α ∈ N0 and k ≡ 1 (mod 4), then
Proof Extracting the odd and even powers of q in (3.1), where k ≡ 1 (mod 4), we obtain
p2, k(22+αn) ≡ p2, k(22n) (mod 8),
p2, k(21+α(4n + 2)) ≡ 7p2, k(4n + 2) (mod 8).
∞
n=0
∞
n=0
Applying (2.1), we deduce that
Substituting (3.16) into (3.14) and then using (3.6)–(3.8) in the resulting congruence, we can rewrite (3.14) as
Congruences (3.12) and (3.13) follow from the above two congruences and by induction on α ≥ 0.
Lemma 3.4 For any k ≡ 1 (mod 8) and n ∈ N0, we have
∞
n=0
p2, k(8n + 6)qn ≡ 4q 3k4−3 f156k f25k
f322k f42k + 4q k−41 f43 f23k + 4q k−21 f43 f23k + 4 f156 f25 (mod 8). (3.17)
f322 f42
(3.10)
(3.11)
(3.12)
(3.13)
(3.14)
(3.15)
(3.16)
Proof Substituting (3.4) into (3.15), we see that
p2, k (4n + 2)qn ≡ 2q k −21 f4 f43k f85 f85k
and for 1 ≤ j ≤ p − 1,
Proof If k ≡ 1, then terms appearing on the right side of (3.17) are powers of q2 and thus equating the odd
Congruence (3.22) follows from the above congruence and (3.13).
Next, we turn to prove (3.23). From (3.17), we have
,
Invoking (3.20) and (3.21) into (3.19), we obtain the required congruence.
prime. The Legendre symbol
is defined by
In order to state the congruence properties for p2, k (n), we need the following definition: Let p ≥ 3 be a
a
p
if a is a quadratic residue modulo p and p a,
:=
−1 if a is a quadratic nonresidue modulo p and p a,
Theorem 3.5 Let p ≡ 5, 7 (mod 8) and k ≡ 1 (mod 8) such that
k
p
= 1. Then for all n and β ∈ N0, we
Extracting the odd powers of q in (3.18) for k ≡ 1 (mod 8), we find that
where k ≡ 1 and
Now, consider the following congruence relations:
∞
n=0
a(n)q
n
p−1
where − 2
≤ m1, r1 ≤ p−21 , 0 ≤ m2, r2 ≤ p−23 . Because
−2
p
= −1, the first congruence holds if and
only if m1 = r1 = ± p6−1 and the second congruence holds if and only if m2 = r2 = p−21 , since −p2k
= −1.
p2−1
Substituting Lemmas 2.2 and 2.3 into (3.25), extracting the terms involving q pn+3 8 in the resulting identity,
canceling q3 p28−1 on both sides and then replacing q p by q, we deduce
∞
n=0
a pn + 3 ·
The terms appearing on the right side are powers of q p and thus for 1 ≤ j ≤ p − 1, we have
and
From (3.26) and by induction on β ∈ N0, we find that for all n ∈ N0
2
a p n + 3 ·
Finally, replacing n by p2β+2n + p2β+1 j + 3 · p2β+2−1 in (3.24) and then using (3.29), we obtain (3.23). This
8
Theorem 3.6 Let p ≡ 5, 7 (mod 8) and k ≡ 5 (mod 8) such that k
p
= 1. Then for all n, β ∈ N0, we have
Proof Extracting the terms involving odd powers of q in (3.18) where k ≡ 5, we see that
Now, consider the following congruence relations:
8 · (3m12 + m1) + (3r12 + r1) ≡ 3 ·
Substituting Lemmas 2.2 and 2.3 into (3.31), extracting the terms involving qpn+3· p24−1, we obtain
∞
n=0
b pn + 3 ·
p2 − 1 qn = p2qp3k8−3 f156kp f25kp 3
f322kp f42kp + qpk−41 f43p f2kp + qpk−85 fp3 fk3p
4
where 1 ≤ j ≤ p − 1. From (3.33) and by induction on β, we find that for n ∈ N0,
By (3.35) and (3.34), we find that
which implies that
and
we have
and
≡ 0 (mod 2).
Theorem 3.7 Let p ≥ 3andk ≡ 3 (mod 8)suchthat −k
p = −1.Thenforα,β,n ∈ N0 and1 ≤ j ≤ p−1,
p2, k(22+αn) ≡ p2, k(22n) (mod 8)
p2, k 22+α 2p2β+2n + 2p2β+1 j + p2β+2 ≡ 3p2,k 4p2β+2n + 4p2β+1 j + 2p2β+2 (mod 8).
(3.32)
(3.33)
(3.34)
(3.35)
(3.36)
(3.37)
(3.38)
Proof Extracting the even and odd powers of q in (3.1), where k ≡ 3 (mod 4), we find that
p2, k (8n) ≡ p2, k (4n),
∞
n=0
p2, k (8n + 4)q ≡ 3
n
p2, k (4n + 2)q
n
Extracting the even and odd powers of q in (3.41) where k ≡ 3 and then using (3.39) and (3.40), we obtain
which implies that
∞
n=0
∞
n=0
(3.39)
(3.40)
(3.41)
(3.42)
(3.43)
(3.44)
(3.45)
(3.46)
Congruence (3.37) follows from (3.42) and by induction on α ∈ N0.
∞
n=0
c(n)q
n
Now, consider the congruence relation
where 0 ≤ m, r ≤ p−23 and p ≥ 3. The above congruence relation holds if and only if m = r = p−21 , since
p2−1
= −1. Applying Lemma 2.3 into (3.44) and then extracting the terms of the form q pn+ 2 , we obtain
∞
n=0
c pn +
where 1 ≤ j ≤ p − 1. From the above two identities and by induction, we find that
From (3.43), (3.44) and (3.46), we have
p2, k 8 p2β+2n + 8 p2β+1 j + 4 p2β+2
≡
3 p2, k 4 p2β+2n + 4 p2β+1 j + 2 p2β+2 .
Proof In view of (3.4), (3.6) and (3.7), we deduce that
Combining (3.40) and (3.49) and then extracting the odd and even powers of q where k
obtain
≡ 3 (mod 4), we
n
n
p2, k (8n + 2)q
,
Substituting (3.52) in (3.50) and then extracting the terms involving odd powers of q where k ≡ 3, we obtain
Lemma 3.9 For all n ∈ N0 and k ≡ 7 (mod 8), we have
Lemma 3.8 If k ≡ 3 (mod 8), then for all n ∈ N0, we have
and
and
∞
(mod 4).
(3.49)
(3.47)
(3.48)
(3.50)
(3.51)
(3.52)
(3.53)
(3.54)
(3.55)
p2, k (16n + 6)q
n
k+1
≡ 4 f8 f1 + 4q 8 fk3 f43
we obtain (3.54). Congruence (3.55) follows from (3.52) and (3.50).
Proof Substituting (3.53) in (3.51) and then extracting the terms involving the even powers of q where k ≡ 7,
1 ≤ j ≤ p − 1, we have
and
Proof From (3.47), we have
where k ≡ 3 and
where 1 ≤ j ≤ p − 1. By (3.60) and (3.61), we have
The congruence (3.56) follows from (3.58) and (3.62). The proof of the congruence (3.57) is analogous to the
proof of (3.56), except that in place of (3.47), (3.48) is used.
Theorem 3.11 Let p ≡ 5,7 (mod 8) and k ≡ 7 (mod 8) such that kp = 1. Then for all n,β ∈ N0 and
For p ≡ 1,7 and 0 ≤ m,r ≤ p−23, the following congruences
hold true if and only if m = r = p−21, since −2k
p = −1. Substituting Lemma 2.3 into (3.59) and then
extracting the terms involving qpn+5· p28−1, we obtain
∞
n=0
p2 − 1 qn = qpk−83 f23p fk3p + qpk−21 fp3 f43kp,
d pn + 5 · 8
and so we have
∞
n=0
p2, k(16n + 10) ≡ 4d(n),
Proof Rewrite (3.54) in the form
where k ≡ 7 and
Now, consider the following congruence relations:
2
≡ 3 ·
p−23 . The first congruence relation holds true if and only if
= 1, the second congruence relation holds true if and only if
m2 = r2 = p−21 . Substituting Lemmas 2.2, 2.3 into (3.66) and then extracting the terms involving q pn+3· p28−1 ,
we obtain
which implies that
∞
n=0
f pn + 3 ·
p2 − 1 qn = p2 f8 p f p + q p k +81 fk3p f43p,
8
where 1 ≤ j ≤ p − 1. From the above two identities, we deduce that
Finally, replacing n by p2β+2n + p2β+1 j + 3 · p2β+82−1 in (3.65) and then using the above identity, we arrive
at (3.63). The proof of the congruence (3.64) is similar to the proof of (3.63), except that in place of (3.54),
(3.55) is used.
Theorem 3.12 Let p ≡ 5, 7 (mod 8) and k ≡ 1 (mod 8) such that
have
k
p
= 1. Then for all n, β ∈ N0, we
p2, k (8n + 7) ≡ 0 (mod 8)
p2, k 8 p2β+2n + p2β+1 + 3 p2β+2
Proof Extracting the terms involving the odd powers of q in (3.9) where k ≡ 1 (mod 4), we obtain
p2, k (4n + 3)qn ≡ 4q k −21 f23 f43k + 4 f29 + 4q 3k4−3 f29k + 4q k −41 f43 f23k .
(3.65)
(3.66)
(3.67)
(3.68)
(3.69)
From (3.25) and (3.71), we have
which implies that
∞
n=0
p2, k (8n + 3)q
n
≡ 4
a(n)qn,
∞
n=0
p2, k (8n + 3) ≡ 4a(n).
Now, the congruence (3.68) follows from the above congruence and (3.29).
If we extract the even and odd powers of q in (3.69) where k ≡ 5, we obtain the following lemma:
Lemma 3.13 If k ≡ 5 (mod 8) and n ∈ N0, then
and
p2, k (8n + 3)q
n
For k ≡ 1, equating the coefficients of odd powers of q, we arrive at (3.67). Again, from (3.69), we have
Employing above congruence in (3.70), we obtain
p2, k (8n + 3)q ≡
n
Lemma 3.14 If k ≡ 1 (mod 8) and for n ∈ N0, then
Proof Let k ≡ 1 (mod 4). By (3.9), we see that
p2, k (8n + 5)q ≡
n
4q k −21 f13 f43k + 4q(k−1)/8 fk3 f43
In view of (3.52), (3.53) and (3.75), we have
∞
n=0
p2, k (4n + 1)q
n
≡ 2
Extracting the terms involving q2n+1 in (3.76) where k ≡ 1, we obtain the required congruence.
and
and
≡
∞
n=0
p2, k (16n + 6)qn,
p2, k (8n + 3) ≡ p2, k (16n + 6).
p2, k (16n + 10) ≡ p2, k (8n + 5).
(3.77)
(3.78)
(3.79)
(3.80)
(3.81)
(3.82)
(3.83)
(3.84)
(3.85)
n=0
which yields
Changing n to 2n + 1 in (3.79), we obtain
Finally, from (3.63), (3.84) and (3.64), (3.85), we obtain (3.82) and (3.83), respectively.
It follows from (3.81) with k ≡ 3 and (3.48),
In view of (3.57) and the above congruence, we obtain (3.78).
Theorem 3.16 Let p ≡ 5, 7 (mod 8) and k ≡ 7 (mod 8) such that k
p
n, β ∈ N0, we have
p2, k (16n + 14) ≡ p2, k (8n + 7).
= 1. Then for 1 ≤ j ≤ p − 1 and
p2, k 8 p2β+2n + 8 p2β+1 j + 3 p2β+2
≡ 0 (mod 8)
p2, k 8 p2β+2n + 8 p2β+1 j + 5 p2β+2
Proof Let k ≡ 7. From (3.6), (3.54) and (3.81), we see that
∞
n=0
We can easily prove the following theorem with the help of Lemmas 3.13, 2.2 and 2.3:
≡ 0
In view of Lemmas 2.3 and 3.14, we can prove the following congruence:
= 1. Then for n, β ∈ N0 and
1 ≤ j ≤ p − 1, we have
≡ 0
and
Acknowledgements The authors would like to thank the anonymous referees for their valuable suggestions.
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