Arithmetic Properties of Partition k-tuples with Odd Parts Distinct

M. S. Mahadeva Naika and D. S. Gireesh^{1} Department of Mathematics Bangalore University Central College Campus Bangalore-560 001 Karnataka India

                msmnaika@rediffmail.com
                 gireeshdap@gmail.com

Abstract

Let pod_{−k}(n) denote the number of partition k-tuples of n wherein odd parts are distinct (and even parts are unrestricted). We establish some interesting infinite families of congruences and internal congruences modulo 4, 16, and 5 for pod_{−2}(n), pod_{−4}(n), and pod_{−6}(n), respectively. We also find Ramanujan-type congruences modulo 5 for pod_{−3}(n) and densities of pod_{−2}(n), pod_{−3}(n), pod_{−4}(n), and pod_{−6}(n) modulo 4, 5, 16, and 5, respectively.

1 Introduction

For |q| < 1, Ramanujan’s theta functions φ(q) and ψ(q) are defined by

φ (q) := 1 + 2 ∑_{n=1}^{∞} q^{n^{2}} = ∑_{n=−∞}^{∞} q^{n^{2}} = (−q; q^{2})_{∞}^{2}(q^{2}; q^{2})_{∞}  (1)

1This research was partially supported by DST Govt. of India (SR/S4/MS:739/11).

and

ψ (q) := ∑_{n=0}^{∞} q^{(n^{2}+n)/2} = ∑_{n=−∞}^{∞} q^{2n^{2}+n} = \frac{(q^{2}; q^{2})_{∞}}{(q; q^{2})_{∞}} ,  (2)

where (a; q)_{∞} = (1 − a)(1 − aq)(1 − aq^{2}) · · ·.

Let pod(n) denote the number of partitions of n wherein odd parts are distinct (and even parts are unrestricted). The generating function of pod(n) is

∑_{n=0}^{∞} pod(n)q^{n} = \frac{(−q; q^{2})_{∞}}{(q^{2}; q^{2})_{∞}} = \frac{1}{ψ(−q)}.

In 2010, Hirschhorn and Sellers [5] proved that, for all α ≥ 0 and n ≥ 0,

pod (3^{2α+3}n + \frac{23 × 3^{2α+2} + 1}{8}) ≡ 0 (mod 3).

They also found some internal congruences such as

pod(81n + 17) ≡ 5pod(9n + 2) (mod 27).

Recently, Wang [10] established new congruences for pod(n). For example, for each α ≥ 1 and n ≥ 0,

pod (5^{2α+2}n + \frac{11 × 5^{2α+1} + 1}{8}) ≡ 0 (mod 5).

Let pod_{−k}(n) denote the number of partition k-tuples of n wherein odd parts are distinct (and even parts are unrestricted). The generating function of pod_{−k}(n) is

∑_{n=0}^{∞} pod_{−k}(n)q^{n} = \frac{(−q; q^{2})_{∞}^{k}}{(q^{2}; q^{2})_{∞}^{k}} = \frac{1}{ψ(−q)^{k}} .  (3)

Chen and Lin [3] established congruences modulo 3 and 5 for pod_{−2}(n). For example, for α ≥ 1 and n ≥ 0,

pod_{−2} (5^{α+1}n + \frac{11 × 5^{α} + 1}{4}) ≡ 0 (mod 5).

Wang [8, 9] has established congruences modulo 7, 9, and 11 satisfied by pod_{−3}(n) and congruences modulo 5, 9, and 81 satisfied by pod_{−4}(n) by employing theta function identities. For example, for α ≥ 1 and n ≥ 0,

pod_{−3} (3^{2α+2}n + \frac{23 × 3^{2α+1} + 3}{8}) ≡ 0 (mod 9)

and

pod_{−4} (3^{α+1}n + \frac{5 × 3^{α} + 1}{2}) ≡ 0 (mod 9).

He also found some internal congruences such as

pod_{−4}(27n + 5) ≡ −pod_{−4}(9n + 2) (mod 9).

In this paper, we establish congruences modulo powers of 2 and modulo 5 for pod_{−k}(n) for k ∈ {2, 3, 4, 6}. In this vein, in Section 3, we find infinite family of congruences and internal congruences modulo 4 satisfied by pod_{−2}(n) and we also find density of pod_{−2}(n) modulo 4. In Section 4, we prove Ramanujan-type congruences modulo 5 for pod_{−3}(n) and that pod_{−3}(n) is divisible by 5 at least 1/30 of the time. In Section 5, we establish infinite family of congruences and internal congruences modulo 16 satisfied by pod_{−4}(n) following density of pod_{−4}(n) modulo 16. In Section 6, we determine infinite family of congruences and internal congruences modulo 5 satisfied by pod_{−6}(n) and we also determine density of pod_{−6}(n) modulo 5.

2 Preliminaries

The following results are useful in proving our main results.

Lemma 1. [2, pp. 40–49] We have

φ(q) = φ(q^{4}) + 2qψ(q^{8}),  (4)
φ(q)^{2} = φ(q^{2})^{2} + 4qψ(q^{4})^{2},  (5)
ψ(q) = f(q^{3}, q^{6}) + qψ(q^{9})  (6)
= f(q^{10}, q^{15}) + qf(q^{5}, q^{20}) + q^{3}ψ(q^{25}),  (7)
ψ(q)^{2} = φ(q)ψ(q^{2}).  (8)

Lemma 2. [1, Eq. 1.6.7, p. 26] We have

f(q, q^{4})f(q^{2}, q^{3}) = ψ(q)^{2} − qψ(q^{5})^{2}.  (9)

Lemma 3. Let ∑_{n=0}^{∞} h(n)q^{n} = qψ(q)^{4}. Then

∑_{n=0}^{∞} h(5n + 3)q^{n} ≡ ψ(q)^{4} (mod 5).  (10)

Proof. From (7), it follows that

∑_{n=0}^{∞} h(n)q^{n} = qψ(q)^{4}
= q (f(q^{10}, q^{15}) + qf(q^{5}, q^{20}) + q^{3}ψ(q^{25}))^{4}
= 12q^{5}f(q^{10}, q^{15})^{2}f(q^{5}, q^{20})ψ(q^{25}) + 4q^{10}f(q^{10}, q^{15})ψ(q^{25})^{3}
+ 12q^{8}f(q^{10}, q^{15})f(q^{5}, q^{20})ψ(q^{25})^{2} + 6q^{7}f(q^{10}, q^{15})^{2}ψ(q^{25})^{2}
+ 4q^{2}f(q^{10}, q^{15})^{3}f(q^{5}, q^{20}) + 4q^{4}f(q^{10}, q^{15})^{3}ψ(q^{25})
+ 6q^{3}f(q^{10}, q^{15})^{2}f(q^{5}, q^{20})^{2} + q^{5}f(q^{5}, q^{20})^{4} + q^{13}ψ(q^{25})^{4}
+ 4q^{4}f(q^{10}, q^{15})f(q^{5}, q^{20})^{3} + 6q^{9}f(q^{5}, q^{20})^{2}ψ(q^{25})^{2}
+ 4q^{7}f(q^{5}, q^{20})^{3}ψ(q^{25}) + 12q^{6}f(q^{10}, q^{15})f(q^{5}, q^{20})^{2}ψ(q^{25})
+ 4q^{11}f(q^{5}, q^{20})ψ(q^{25})^{3} + qf(q^{10}, q^{15})^{4},

which yields

∑_{n=0}^{∞} h(5n + 3)q^{n} ≡ 2qf(q^{2}, q^{3})f(q, q^{4})ψ(q^{5})^{2} + f(q^{2}, q^{3})^{2}f(q, q^{4})^{2}
+ q^{2}ψ(q^{5})^{4} (mod 5).

Using (9) in the above equation, we arrive at (10). Lemma 4. Let ∑_{n=0}^{∞} g(n)q^{n} = ψ(q)^{2}. Then

∑_{n=0}^{∞} g(5n + 1)q^{n} = 2ψ(q)^{2} − qψ(q^{5})^{2}.  (11)

Proof. It follows from (7) that

∑_{n=0}^{∞} g(n)q^{n} = ψ(q)^{2}
= (f(q^{10}, q^{15}) + qf(q^{5}, q^{20}) + q^{3}ψ(q^{25}))^{2}
= q^{6}ψ(q^{25})^{2} + 2q^{4}ψ(q^{25})f(q^{5}, q^{20}) + 2q^{3}ψ(q^{25})f(q^{10}, q^{15})
+ q^{2}f(q^{5}, q^{20})^{2} + 2qf(q^{10}, q^{15})f(q^{5}, q^{20}) + f(q^{10}, q^{15})^{2},

which yields

∑_{n=0}^{∞} g(5n + 1)q^{n} = 2f(q^{2}, q^{3})f(q, q^{4}) + qψ(q^{5})^{2}.

Using (9) in the above equation, we arrive at (11).

Lemma 5. [4, Theorem 2.1] For any odd prime, p,

ψ(q) = ∑_{m=0}^{\frac{p−3}{2}} q^{\frac{m^{2}+m}{2}} f (q^{\frac{p^{2}+(2m+1)p}{2}} , q^{\frac{p^{2}−(2m+1)p}{2}} ) + q^{\frac{p^{2}−1}{8}} ψ(q^{p^{2}}).  (12)

Furthermore, \frac{m^{2}+m}{2} ≢ \frac{p^{2}−1}{8} (mod p), for 0 ≤ m ≤ \frac{p−3}{2}.

3 Arithmetic properties of pod_{−2}(n)

In this section, we prove the infinite family of congruences and internal congruences modulo 4 for pod_{−2}(n).

3.1 Infinite family of congruences modulo 4

Theorem 6. Let p be any odd prime such that (\frac{−2}{p}) = −1 and α ≥ 0. Then

∑_{n=0}^{∞} pod_{−2} (2p^{2α}n + \frac{3p^{2α} + 1}{4}) q^{n} ≡ 2ψ(q)ψ(q^{2}) (mod 4)  (13)

and, for all n ≥ 0 and 1 ≤ ξ ≤ p − 1,

pod_{−2} (2p^{2α+1}(pn + ξ) + \frac{3p^{2α+2} + 1}{4}) ≡ 0 (mod 4).  (14)

Proof. We have

∑_{n=0}^{∞} pod_{−2}(n)q^{n} = \frac{1}{ψ(−q)^{2}} .  (15)

Invoking (8) and (15),

∑_{n=0}^{∞} pod_{−2}(n)q^{n} = \frac{1}{ψ(q^{2})φ(−q)}

= \frac{(1 − (1 − φ(−q)))^{−1}}{ψ(q^{2})}

= \frac{1 + (1 − φ(−q)) + (1 − φ(−q))^{2} + ⋯}{ψ(q^{2})}

≡ \frac{2 − φ(−q)}{ψ(q^{2})} (mod 4) from (1).

Using (4) in the above equation, we find that

∑_{n=0}^{∞} pod_{−2}(n)q^{n} ≡ \frac{2 − φ(q^{4}) + 2qψ(q^{8})}{ψ(q^{2})} (mod 4),

which yields

∑_{n=0}^{∞} pod_{−2}(2n + 1)q^{n} ≡ 2\frac{ψ(q^{4})}{ψ(q)} (mod 4).  (16)

From the binomial theorem, we can see that for any prime p and for each positive integer ℓ,

(q; q)^{p^{ℓ}} ≡ (q^{p}; q^{p})^{p^{ℓ−1}} (mod p^{ℓ}).  (17)

In view of (17), (16) can be expressed as

∑_{n=0}^{∞} pod_{−2}(2n + 1)q^{n} ≡ 2ψ(q)ψ(q^{2}) (mod 4),  (18)

which is the α = 0 case of (13). If we assume that (13) holds for some α ≥ 0, then, substituting (12) in (13),

∑_{n=0}^{∞} pod_{−2} (2p^{2α}n + \frac{3p^{2α} + 1}{4}) q^{n}

≡ 2( ∑_{m=0}^{\frac{p−3}{2}} q^{\frac{m^{2}+m}{2}} f (q^{\frac{p^{2}+(2m+1)p}{2}} , q^{\frac{p^{2}−(2m+1)p}{2}} ) + q^{\frac{p^{2}−1}{8}} ψ(q^{p^{2}}))

×( ∑_{m=0}^{\frac{p−3}{2}} q^{m^{2}+m}f (q^{p^{2}+(2m+1)p}, q^{p^{2}−(2m+1)p}) + q^{\frac{p^{2}−1}{4}} ψ(q^{2p^{2}})) (mod 4).  (19)

For any odd prime, p, and 0 ≤ m_{1}, m_{2} ≤ (p − 3)/2, consider the congruence

\frac{m_{1}^{2} + m_{1}}{2} + 2 × \frac{m_{2}^{2} + m_{2}}{2} ≡ \frac{3p^{2} − 3}{8} (mod p),

which implies that

(2m_{1} + 1)^{2} + 2(2m_{2} + 1)^{2} ≡ 0 (mod p).  (20)

Since (\frac{−2}{p}) = −1, the only solution of the congruence (20) is m_{1} = m_{2} = \frac{p − 1}{2}. Therefore,

equating the coefficients of q^{pn+ \frac{3p^{2}−3}{8}} from both sides of (19), dividing throughout by q^{\frac{3p^{2}−3}{8}} and then replacing q^{p} by q, we obtain

∑_{n=0}^{∞} pod_{−2} (2p^{2α+1}n + \frac{3p^{2α+2} + 1}{4}) q^{n} ≡ 2ψ(q^{p})ψ(q^{2p}) (mod 4).  (21)

Equating the coefficients of q^{pn} on both sides of (21) and then replacing q^{p} by q, we obtain

∑_{n=0}^{∞} pod_{−2} (2p^{2α+2}n + \frac{3p^{2α+2} + 1}{4}) q^{n} ≡ 2ψ(q)ψ(q^{2}) (mod 4),

which is the α + 1 case of (13).

Equating the coefficients of q^{pn+ξ} for 1 ≤ ξ ≤ p − 1 from (21), we arrive at (14).

Corollary 7. Let p be any odd prime such that (\frac{−2}{p}) = −1. Then pod_{−2}(n) is divisible by 4 for at least \frac{1}{2(p+1)} of all nonnegative integers n.

Proof. The arithmetic sequences \{2p^{2α+1}(pn + ξ) + \frac{3p^{2α+2}+1}{4} : α ≥ 0\} for 1 ≤ ξ ≤ p − 1, on which pod_{−2}(·) is 0 modulo 4, do not intersect. These sequences account for

(p − 1) (\frac{1}{2p^{2}} + \frac{1}{2p^{4}} + \frac{1}{2p^{6}} + ⋯) = \frac{1}{2(p + 1)}

of all nonnegative integers.

3.2 Some internal congruences

Theorem 8. For each n ≥ 0,

pod_{−2}(54n + 25) ≡ pod_{−2}(6n + 3) (mod 4),  (22)
pod_{−2}(54n + 43) ≡ pod_{−2}(6n + 5) (mod 4),  (23)
pod_{−2}(162n + 7) ≡ 2pod_{−2}(18n + 1) (mod 4),  (24)
pod_{−2}(162n + 115) ≡ 2pod_{−2}(18n + 13) (mod 4).  (25)

Proof. If ∑_{n=0}^{∞} a(n)q^{n} = ψ(q)ψ(q^{2}), then the authors [6] found that

∑_{n=0}^{∞} a(3n)q^{n} = ψ(q)φ(q) − qψ(q^{3})ψ(q^{6}).  (26)

Using (26), we can express (18) as

∑_{n=0}^{∞} pod_{−2}(6n + 1)q^{n} ≡ 2ψ(q)φ(q) − 2qψ(q^{3})ψ(q^{6}) (mod 4).  (27)

Invoking (1) and (27),

∑_{n=0}^{∞} pod_{−2}(6n + 1)q^{n} ≡ 2ψ(q) + 2qψ(q^{3})ψ(q^{6}) (mod 4).  (28)

Substituting (6) into (28) and extracting the terms involving q^{3n+1},

∑_{n=0}^{∞} pod_{−2}(18n + 7)q^{n} ≡ 2ψ(q^{3}) + 2ψ(q)ψ(q^{2}) (mod 4),

which is equivalent to

∑_{n=0}^{∞} pod_{−2}(18n + 7)q^{n} ≡ 2ψ(q^{3}) + ∑_{n=0}^{∞} pod_{−2}(2n + 1)q^{n} (mod 4).  (29)

Equating the coefficients of q^{3n+1} and q^{3n+2} from (29), we arrive at (22) and (23), respectively. Equating the coefficients of q^{3n} and then replacing q^{3} by q from (29),

∑_{n=0}^{∞} pod_{−2}(54n + 7)q^{n} ≡ 2ψ(q) + ∑_{n=0}^{∞} pod_{−2}(6n + 1)q^{n} (mod 4).  (30)

Invoking (28) and (30),

∑_{n=0}^{∞} pod_{−2}(54n + 7)q^{n} ≡ 2 ∑_{n=0}^{∞} pod_{−2}(6n + 1)q^{n} − 2qψ(q^{3})ψ(q^{6}) (mod 4),  (31)

Equating the coefficients of q^{3n} and q^{3n+2} from (31), we arrive at (24) and (25), respectively.

4 Ramanujan-type congruences for pod_{−3}(n)

In this section, we prove the Ramanujan-type congruences modulo 5 for pod_{−3}(n). Theorem 9. For each α ≥ 1,

pod_{−3} (5^{2α+1}n + \frac{μ × 5^{2α} + 3}{8}) ≡ 0 (mod 5),  (32)

where μ = 13, 21, 29, and 37. Proof. We have

∑_{n=0}^{∞} pod_{−3}(n)(−1)^{n}q^{n} = \frac{1}{ψ(q)^{3}} .

It follows from (17) that

∑_{n=0}^{∞} pod_{−3}(n)(−1)^{n}q^{n} ≡ \frac{ψ(q)^{2}}{ψ(q^{5})} (mod 5)

≡ \frac{1}{ψ(q^{5})} ∑_{n=0}^{∞} g(n)q^{n} (mod 5).

Extracting the terms involving q^{5n+1}, dividing throughout by q and then replacing q^{5} by q,

∑_{n=0}^{∞} pod_{−3}(5n + 1)(−1)^{n+1}q^{n} ≡ \frac{1}{ψ(q)} ∑_{n=0}^{∞} g(5n + 1)q^{n} (mod 5)

≡ \frac{1}{ψ(q)} (2ψ(q)^{2} − qψ(q^{5})^{2}) (mod 5) from (11)

≡ 2ψ(q) − qψ(q^{5})ψ(q)^{4} (mod 5) using (17).  (33)

Substituting (7) into (33) and from the Lemma (3) , we find that

∑_{n=0}^{∞} pod_{−3}(5n + 1)(−1)^{n+1}q^{n} ≡ 2f(q^{10}, q^{15}) + 2qf(q^{5}, q^{20}) + 2q^{3}ψ(q^{25})

− ψ(q^{5}) ∑_{n=0}^{∞} h(n)q^{n} (mod 5),

which implies that

∑_{n=0}^{∞} pod_{−3}(25n + 16)(−1)^{n}q^{n} ≡ 2ψ(q^{5}) − ψ(q) ∑_{n=0}^{∞} h(5n + 3)q^{n} (mod 5)
≡ 2ψ(q^{5}) − ψ(q)^{5} (mod 5) using (10).  (34)

Using (17), (34) can be expressed as

∑_{n=0}^{∞} pod_{−3}(25n + 16)(−1)^{n}q^{n} ≡ ψ(q^{5}) (mod 5).  (35)

Extracting the terms involving q^{5n} from (35),

∑_{n=0}^{∞} pod_{−3}(125n + 16)(−1)^{n}q^{n} ≡ ψ(q) (mod 5)
≡ f(q^{10}, q^{15}) + qf(q^{5}, q^{20}) + q^{3}ψ(q^{25}) (mod 5),

which yields

∑_{n=0}^{∞} pod_{−3}(625n + 391)(−1)^{n+1}q^{n} ≡ ψ(q^{5}) (mod 5).  (36)

From (35), (36), and by induction, we find that for each α ≥ 1,

∑_{n=0}^{∞} pod_{−3} (5^{2α}n + \frac{5^{2α+1} + 3}{8}) (−1)^{n+1}q^{n} ≡ (−1)^{α}ψ(q^{5}) (mod 5).  (37)

Equating the coefficients of q^{5n+ξ} for 1 ≤ ξ ≤ 4 from (37), we arrive at (32).

Corollary 10. The function pod_{−3}(n) is divisible by 5 for at least \frac{1}{30} of all nonnegative integers n.

Proof. The arithmetic sequences \{5^{2α+1}n + \frac{μ×5^{2α}+3}{8} : α ≥ 1\} for μ = 13, 21, 29, and 37, on which pod_{−3}(·) is 0 modulo 5, do not intersect. These sequences account for

4 (\frac{1}{5^{3}} + \frac{1}{5^{5}} + \frac{1}{5^{7}} + ⋯) = \frac{1}{30}

of all nonnegative integers.

5 Arithmetic properties of pod_{−4}(n)

In this section, we prove the infinite family of congruences and internal congruences modulo 16 for pod_{−4}(n).

5.1 Infinite family of congruences modulo 16

Theorem 11. Let p be any prime such that p ≡ 3 (mod 4) and α ≥ 0. Then

∑_{n=0}^{∞} pod_{−4} (2p^{2α}n + \frac{p^{2α} + 1}{2}) q^{n} ≡ 4ψ(q)^{2} (mod 16)  (38)

and, for all nonnegative integers n and 1 ≤ ξ ≤ p − 1,

pod_{−4} (2p^{2α+1}(pn + ξ) + \frac{p^{2α+2} + 1}{2}) ≡ 0 (mod 16).  (39)

Proof. We have

∑_{n=0}^{∞} pod_{−4}(n)q^{n} = \frac{1}{ψ(−q)^{4}} .  (40)

Invoking (8) and (40),

∑_{n=0}^{∞} pod_{−4}(n)q^{n} = \frac{1}{ψ(q^{2})^{2}φ(−q)^{2}}

= \frac{(1 − (1 − φ(−q)^{2}))^{−1}}{ψ(q^{2})^{2}}

= \frac{1 + (1 − φ(−q)^{2}) + (1 − φ(−q)^{2})^{2} + ⋯}{ψ(q^{2})^{2}}

≡ \frac{2 − φ(−q)^{2}}{ψ(q^{2})^{2}} (mod 16) using (1)

≡ \frac{2 − φ(q^{2})^{2} + 4qψ(q^{4})^{2}}{ψ(q^{2})^{2}} (mod 16) from (5),

which implies that

∑_{n=0}^{∞} pod_{−4}(2n + 1)q^{n} ≡ 4\frac{ψ(q^{2})^{2}}{ψ(q)^{2}} (mod 16),  (41)

In view of (17), (41) can be expressed as

∑_{n=0}^{∞} pod_{−4}(2n + 1)q^{n} ≡ 4ψ(q)^{2} (mod 16),  (42)

which is the α = 0 case of (38). If we assume that (38) holds for some α ≥ 0, then, substituting (12) into (38),

∑_{n=0}^{∞} pod_{−4} (2p^{2α}n + \frac{p^{2α} + 1}{2}) q^{n}

≡ 4( ∑_{m=0}^{\frac{p−3}{2}} q^{\frac{m^{2}+m}{2}} f (q^{\frac{p^{2}+(2m+1)p}{2}} , q^{\frac{p^{2}−(2m+1)p}{2}} ) + q^{\frac{p^{2}−1}{8}} ψ(q^{p^{2}}))^{2}.  (43)

For any odd prime, p, and 0 ≤ m_{1}, m_{2} ≤ (p − 3)/2, consider the congruence

\frac{m_{1}^{2} + m_{1}}{2} + \frac{m_{2}^{2} + m_{2}}{2} ≡ \frac{2p^{2} − 2}{8} (mod p),

which implies that

(2m_{1} + 1)^{2} + (2m_{2} + 1)^{2} ≡ 0 (mod p).  (44)

Since (\frac{−1}{p}) = −1 for p ≡ 3 (mod 4), the only solution of the congruence (44) is m_{1} = m_{2} =

\frac{p − 1}{2}. Therefore, equating the coefficients of q^{pn+ \frac{2p^{2}−2}{8}} from both sides of (43), dividing

throughout by q^{\frac{2p^{2}−2}{8}} and then replacing q^{p} by q, we obtain

∑_{n=0}^{∞} pod_{−4} (2p^{2α} (pn + \frac{2p^{2} − 2}{8}) + \frac{p^{2α} + 1}{2}) q^{n} ≡ 4ψ(q^{p})^{2} (mod 16).  (45)

Equating the coefficients of q^{pn} on both sides of (45) and then replacing q^{p} by q, we obtain

∑_{n=0}^{∞} pod_{−4} (2p^{2α+2}n + \frac{p^{2α+2} + 1}{2}) q^{n} ≡ 4ψ(q)^{2} (mod 16),

which is the α + 1 case of (38).

Equating the coefficients of q^{pn+ξ} for 1 ≤ ξ ≤ p − 1 from (45), we arrive at (39).

Corollary 12. Let p be a prime such that p ≡ 3 (mod 4). Then pod_{−4}(n) is divisible by 16 for at least \frac{1}{2(p+1)} of all nonnegative integers n.

Proof. The arithmetic sequences \{2p^{2α+1}(pn + ξ) + \frac{p^{2α+2}+1}{2} : α ≥ 0\} for 1 ≤ ξ ≤ p − 1, on which pod_{−4}(·) is 0 modulo 16, do not intersect. These sequences account for

(p − 1) (\frac{1}{2p^{2}} + \frac{1}{2p^{4}} + \frac{1}{2p^{6}} + ⋯) = \frac{1}{2(p + 1)}

of all nonnegative integers.

5.2 Some internal congruences

Theorem 13. For each n ≥ 0,

pod_{−4}(50n + 3) ≡ 2pod_{−4}(10n + 1) (mod 16),
pod_{−4}(50n + 23) ≡ 2pod_{−4}(10n + 5) (mod 16),
pod_{−4}(50n + 33) ≡ 2pod_{−4}(10n + 7) (mod 16),
pod_{−4}(50n + 43) ≡ 2pod_{−4}(10n + 9) (mod 16).

Proof. If ∑_{n=0}^{∞} g(n)q^{n} = ψ(q)^{2}, then (42) can be expressed as

∑_{n=0}^{∞} pod_{−4}(2n + 1)q^{n} ≡ 4 ∑_{n=0}^{∞} g(n)q^{n} (mod 16),

which yields

∑_{n=0}^{∞} pod_{−4}(10n + 3)q^{n} ≡ 4 ∑_{n=0}^{∞} g(5n + 1)q^{n} (mod 16).  (46)

Invoking (11) and (46),

∑_{n=0}^{∞} pod_{−4}(10n + 3)q^{n} ≡ 8ψ(q)^{2} − 4qψ(q^{5})^{2} (mod 16).  (47)

Substituting (42) into (47),

∑_{n=0}^{∞} pod_{−4}(10n + 3)q^{n} ≡ 2 ∑_{n=0}^{∞} pod_{−4}(2n + 1)q^{n} − 4qψ(q^{5})^{2} (mod 16),

equating the coefficients of q^{5n+i} for i = 0, 2, 3, and 4 from the above equation, we obtain the desired results.

6 Arithmetic properties of pod_{−6}(n)

In this section, we prove the infinite family of congruences and internal congruences modulo 5 for pod_{−6}(n).

6.1 Infinite family of congruences modulo 5

Theorem 14. Let p be any prime such that p ≡ 3 (mod 4) and α ≥ 0. Then

∑_{n=0}^{∞} pod_{−6} (5p^{2α}n + \frac{5p^{2α} + 3}{4}) q^{n} ≡ ψ(q)^{2} (mod 5)

and, for each n ≥ 0 and 1 ≤ ξ ≤ p − 1,

pod_{−6} (5p^{2α+1}(pn + ξ) + \frac{5p^{2α+2} + 3}{4}) ≡ 0 (mod 5).

Proof. We have

∑_{n=0}^{∞} pod_{−6}(n)q^{n} = \frac{1}{ψ(−q)^{6}} .  (48)

In view of (17), (48) can be expressed as

∑_{n=0}^{∞} pod_{−6}(n)(−1)^{n}q^{n} ≡ \frac{ψ(q)^{4}}{ψ(q^{5})^{2}} (mod 5).  (49)

Substituting (7) into (49),

∑_{n=0}^{∞} pod_{−6}(n)(−1)^{n}q^{n} ≡ \frac{(f(q^{10}, q^{15}) + qf(q^{5}, q^{20}) + q^{3}ψ(q^{25}))^{4}}{ψ(q^{5})^{2}} (mod 5),

which yields

∑_{n=0}^{∞} pod_{−6}(5n + 2)(−1)^{n}q^{n} ≡ \frac{f(q^{2}, q^{3})^{2}f(q, q^{4})^{2}}{ψ(q)^{2}} + \frac{2qf(q^{2}, q^{3})f(q, q^{4})ψ(q^{5})^{2}}{ψ(q)^{2}}
+ \frac{q^{2}ψ(q^{5})^{4}}{ψ(q)^{2}} (mod 5).  (50)

Invoking (9) and (50),

∑_{n=0}^{∞} pod_{−6}(5n + 2)(−1)^{n}q^{n} ≡ ψ(q)^{2} + q^{2} \frac{ψ(q^{5})^{4}}{ψ(q)^{2}} − 2qψ(q^{5})^{2} + 2qψ(q^{5})^{2}
− 2q^{2} \frac{ψ(q^{5})^{4}}{ψ(q)^{2}} + q^{2} \frac{ψ(q^{5})^{4}}{ψ(q)^{2}} (mod 5),  (51)

∑_{n=0}^{∞} pod_{−6}(5n + 2)(−1)^{n}q^{n} ≡ ψ(q)^{2} (mod 5).  (52)

The remainder of the proof is similar to that of Theorem 11, but rather than (42), we use (52).

Corollary 15. Let p be a prime such that p ≡ 3 (mod 4). Then pod_{−6}(n) is divisible by 5 for at least \frac{1}{5(p+1)} of all nonnegative integers n.

Proof. The arithmetic sequences \{5p^{2α+1}(pn + ξ) + \frac{5p^{2α+2}+3}{4} : α ≥ 0\} for 1 ≤ ξ ≤ p − 1, on which pod_{−6}(·) is 0 modulo 5, do not intersect. These sequences account for

(p − 1) (\frac{1}{5p^{2}} + \frac{1}{5p^{4}} + \frac{1}{5p^{6}} + ⋯) = \frac{1}{5(p + 1)}

of all nonnegative integers.

6.2 Some internal congruences

Theorem 16. For each n ≥ 0,

pod_{−6}(125n + 7) ≡ 3pod_{−6}(25n + 2) (mod 5),
pod_{−6}(125n + 57) ≡ 3pod_{−6}(25n + 12) (mod 5),
pod_{−6}(125n + 82) ≡ 3pod_{−6}(25n + 17) (mod 5),
pod_{−6}(125n + 107) ≡ 3pod_{−6}(25n + 22) (mod 5).

Proof. The proof is similar to that of Theorem 13, but rather than (42), we use (52).

7 Acknowledgments

The authors would like to thank anonymous referee of his/her helpful suggestions and comments.

References

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[6] M. S. Mahadeva Naika and D. S. Gireesh, Arithmetic properties arising from Ramanujan’s theta functions, Ramanujan J., (2016), to appear.

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2010 Mathematics Subject Classification: Primary 05A17; Secondary 11P83. Keywords: congruence, partition k-tuple, partition with odd parts distinct.

(Concerned with sequences \underline{A006950}, \underline{A273225}, \underline{A273226}, \underline{A273228}, and \underline{A273236}.)

Received November 4 2015; revised version received April 18 2016; May 28 2016. Published in Journal of Integer Sequences, June 4 2016.

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