Congruences Modulo Small Powers of 2 and 3 for Partitions into Odd Designated Summands

B. Hemanthkumar Department of Mathematics M. S. Ramaiah University of Applied Sciences Bengaluru-560 058 India

              hemanthkumarb.30@gmail.com

H. S. Sumanth Bharadwaj and M. S. Mahadeva Naika^{1} Department of Mathematics Central College Campus Bangalore University Bengaluru-560 001 India

              sumanthbharadwaj@gmail.com
                msmnaika@rediffmail.com

Abstract

Andrews, Lewis and Lovejoy introduced a new class of partitions, partitions with designated summands. Let PD(n) denote the number of partitions of n with designated summands and PDO(n) denote the number of partitions of n with designated summands in which all parts are odd. Andrews et al. established many congruences modulo 3 for PDO(n) by using the theory of modular forms. Baruah and Ojah obtained numerous congruences modulo 3, 4, 8 and 16 for PDO(n) by using theta function identities. In this paper, we prove several infinite families of congruences modulo 9, 16 and 32 for PDO(n).

1Corresponding author.

1 Introduction

A partition of a positive integer n is a non increasing sequence of positive integers λ_{1}, λ_{2}, …, λ_{m} such that n = λ_{1} + λ_{2} + ⋯ + λ_{m}, where the λ_{i}’s (i = 1, 2, …, m) are called parts of the partition. For example, the partitions of 4 are

4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.

Let p(n) denote the number of partitions of n. Thus p(4) = 5.

Andrews, Lewis and Lovejoy [1] studied partitions with designated summands, which are constructed by taking ordinary partitions and tagging exactly one of each part size. Thus the partitions of 4 with designated summands are given by

4′, 3′ + 1′, 2′ + 2, 2 + 2′, 2′ + 1′ + 1, 2′ + 1 + 1′,
1′ + 1 + 1 + 1, 1 + 1′ + 1 + 1, 1 + 1 + 1′ + 1, 1 + 1 + 1 + 1′.

Let PD(n) denote the number of partitions of n with designated summands and PDO(n) denote the number of partitions of n with designated summands in which all parts are odd. Thus PD(4) = 10 and PDO(4) = 5.

Recently, Chen et al. [5] obtained the generating functions for PD(3n), PD(3n + 1) and PD(3n + 2) and gave a combinatorial interpretation of the congruence PD(3n + 2) ≡ 0 (mod 3). Xia [11] proved infinite families of congruences modulo 9 and 27 for PD(n). For example, for all n ≥ 0 and k ≥ 1

PD(2^{18k−1}(12n + 1)) ≡ 0 (mod 27).

Throughout this paper, we use the notation

f_{k} := (q^{k}; q^{k})_{∞} (k = 1, 2, 3, …), where (a; q)_{∞} := ∏_{m=0}^{∞}(1 − aq^{m}).

The generating function for PDO(n) satisfies

∑_{n=0}^{∞} PDO(n)q^{n} = \frac{f_{4}f_{6}^{2}}{f_{1}f_{3}f_{12}} .  (1)

Using the theory of q-series and modular forms Andrews et al. [1] derived

∑_{n=0}^{∞} PDO(3n)q^{n} = \frac{f_{2}^{2}f_{6}^{4}}{f_{1}^{4}f_{12}^{2}} ,  (2)

∑_{n=0}^{∞} PDO(3n + 1)q^{n} = \frac{f_{2}^{4}f_{3}^{3}f_{12}}{f_{1}^{5}f_{4}f_{6}^{2}} ,  (3)

and

∑_{n=0}^{∞} PDO(3n + 2)q^{n} = 2\frac{f_{2}^{3}f_{6}f_{12}}{f_{1}^{4}f_{4}} .  (4)

They also established, for all n ≥ 0

PDO(9n + 6) ≡ 0 (mod 3)

and

PDO(12n + 6) ≡ 0 (mod 3).

Baruah and Ojah [3] proved several congruences modulo 3, 4, 8 and 16 for PDO(n). For instance,

PDO(8n + 7) ≡ 0 (mod 8)

and

PDO(12n + 9) ≡ 0 (mod 16).

The aim of this paper is to prove several new infinite families of congruences modulo 9, 16 and 32 for PDO(n). In particular, we prove the following

Theorem 1. For all nonnegative integers α, β and n, we have

PDO(2^{α+2}3^{β}(72n + 66)) ≡ 0 (mod 144)  (5)

and

PDO(2^{α+2}3^{β}(144n + 138)) ≡ 0 (mod 288).  (6)

In Section 2, we list some preliminary results. We prove several infinite families of congruences modulo 9 for PDO(n) in Section 3, and Theorem 1 and many infinite families of congruences modulo 16 and 32 for PDO(n) in Section 4.

2 Definitions and preliminaries

We will make use of the following definitions, notation and results.

Let f(a, b) be Ramanujan’s general theta function [2, p. 34] given by

f(a, b) := ∑_{n=−∞}^{∞} a^{\frac{n(n+1)}{2}} b^{\frac{n(n−1)}{2}} .

Jacobi’s triple product identity can be stated in Ramanujan’s notation as follows:

f(a, b) = (−a; ab)_{∞}(−b; ab)_{∞}(ab; ab)_{∞}.

In particular,

φ(q) := f(q, q) = ∑_{k=−∞}^{∞} q^{k^{2}} = \frac{f_{2}^{5}}{f_{1}^{2}f_{4}^{2}} ,  φ(−q) := f(−q, −q) = \frac{f_{1}^{2}}{f_{2}} ,  (7)

ψ(q) := f(q, q^{3}) = ∑_{k=0}^{∞} q^{k(k+1)/2} = \frac{f_{2}^{2}}{f_{1}}  (8)

and

f(−q) := f(−q, −q^{2}) = ∑_{k=−∞}^{∞}(−1)^{k}q^{k(3k−1)/2} = f_{1}.  (9)

For any positive integer k, let k(k + 1)/2 be the k^{th} triangular number and k(3k ± 1)/2 be a generalized pentagonal number.

Lemma 2. The following 2-dissections hold:

f_{1}^{2} = \frac{f_{2}f_{8}^{5}}{f_{4}^{2}f_{16}^{2}} − 2q \frac{f_{2}f_{16}^{2}}{f_{8}} ,  (10)
\frac{1}{f_{1}^{2}} = \frac{f_{8}^{5}}{f_{2}^{5}f_{16}^{2}} + 2q \frac{f_{4}^{2}f_{16}^{2}}{f_{2}^{5}f_{8}} ,  (11)
f_{1}^{4} = \frac{f_{4}^{10}}{f_{2}^{2}f_{8}^{4}} − 4q \frac{f_{2}^{2}f_{8}^{4}}{f_{4}^{2}}  (12)

and

\frac{1}{f_{1}^{4}} = \frac{f_{4}^{14}}{f_{2}^{14}f_{8}^{4}} + 4q \frac{f_{4}^{2}f_{8}^{4}}{f_{2}^{10}} .  (13)

Proof. Lemma 2 is an immediate consequence of dissection formulas of Ramanujan, collected in Berndt’s book [2, Entry 25, p. 40].

Lemma 3. The following 2-dissections hold:

\frac{f_{1}^{3}}{f_{3}} = \frac{f_{4}^{3}}{f_{12}} − 3q \frac{f_{2}^{2}f_{12}^{3}}{f_{4}f_{6}^{2}} ,  (14)
\frac{f_{3}^{3}}{f_{1}} = \frac{f_{4}^{3}f_{6}^{2}}{f_{2}^{2}f_{12}} + q \frac{f_{12}^{3}}{f_{4}} ,  (15)
\frac{f_{3}}{f_{1}^{3}} = \frac{f_{4}^{6}f_{6}^{3}}{f_{2}^{9}f_{12}^{2}} + 3q \frac{f_{4}^{2}f_{6}f_{12}^{2}}{f_{2}^{7}}  (16)

and

\frac{f_{1}}{f_{3}^{3}} = \frac{f_{2}f_{4}^{2}f_{12}^{2}}{f_{6}^{7}} − q \frac{f_{2}^{3}f_{12}^{6}}{f_{4}^{2}f_{6}^{9}} .  (17)

Proof. Hirschhorn et al. [6] established (14) and (15). Replacing q by −q in (14) and (15), and using the relation

(−q; −q)_{∞} = \frac{f_{2}^{3}}{f_{1}f_{4}} ,

we obtain (16) and (17).

Lemma 4. The following 2-dissections hold:

f_{1}f_{3} = \frac{f_{2}f_{8}^{2}f_{12}^{4}}{f_{4}^{2}f_{6}f_{24}^{2}} − q \frac{f_{4}^{4}f_{6}f_{24}^{2}}{f_{2}f_{8}^{2}f_{12}^{2}} ,  (18)
\frac{1}{f_{1}f_{3}} = \frac{f_{8}^{2}f_{12}^{5}}{f_{2}^{2}f_{4}f_{6}^{4}f_{24}^{2}} + q \frac{f_{4}^{5}f_{24}^{2}}{f_{2}^{4}f_{6}^{2}f_{8}^{2}f_{12}}  (19)

and

\frac{f_{3}^{2}}{f_{1}^{2}} = \frac{f_{4}^{4}f_{6}f_{12}^{2}}{f_{2}^{5}f_{8}f_{24}} + 2q \frac{f_{4}f_{6}^{2}f_{8}f_{24}}{f_{2}^{4}f_{12}} .  (20)

Proof. Baruah and Ojah [4] derived the above identities.

Lemma 5. The following 3-dissections hold:

φ(−q) = φ(−q^{9}) − 2qf(−q^{3}, −q^{15})  (21)

and

ψ(q) = f(q^{3}, q^{6}) + qψ(q^{9}).  (22)

Proof. See Berndt’s book [2, p. 49] for a proof of (21) and (22).

Lemma 6. The following 3-dissection holds:

f_{1}f_{2} = \frac{f_{6}f_{9}^{4}}{f_{3}f_{18}^{2}} − qf_{9}f_{18} − 2q^{2} \frac{f_{3}f_{18}^{4}}{f_{6}f_{9}^{2}} .  (23)

Proof. Hirschhorn and Sellers [8] have proved the above identity.

Let t be a positive integer. A partition of n is called a t-core partition of n if none of the hook numbers of its associated Ferrers-Young diagram are multiples of t. Let a_{t}(n) denote the number of t-core partitions of n. Then the generating function of a_{t}(n) satisfies

∑_{n=0}^{∞} a_{t}(n)q^{n} = \frac{f_{t}^{t}}{f_{1}} .  (24)

Many mathematicians have studied arithmetic properties of a_{3}(n). See for example, Keith [9], and Lin and Wang [10]. Hirschhorn and Sellers [7] obtained an explicit formula for a_{3}(n) by using elementary methods and proved Lemma 7. Let 3n + 1 = ∏_{i=1}^{k} p_{i}^{α_{i}} ∏_{j=1}^{m} q_{j}^{β_{j}}, where p_{i} ≡ 1 (mod 3) and q_{j} ≡ 2 (mod 3) with α_{i}, β_{j} ≥ 0 be the prime factorization of 3n + 1. Then

a_{3}(n) = \begin{cases}∏_{i=1}^{k}(α_{i} + 1), & if all β_{j}are even; \\ 0, & otherwise. \end{cases}

3 Congruences modulo 9

In this section, we prove the following infinite families of congruences modulo 9 for PDO(n). Theorem 8. For all nonnegative integers α, β and n, we have

PDO(4^{α}(24n + 16)) ≡ PDO(24n + 16) (mod 9),  (25)
PDO(2^{α}3^{β}(24n + 24)) ≡ (−1)^{α} PDO(24n + 24) (mod 9),  (26)
PDO(4^{α}(48n + 40)) ≡ 0 (mod 9)  (27)

and

PDO(2^{α}3^{β}(144n + 120)) ≡ 0 (mod 9).  (28)

Theorem 9. For any nonnegative integer n, let 3n+1 = ∏_{i=1}^{k} p_{i}^{α_{i}} ∏_{j=1}^{m} q_{j}^{β_{j}}, where p_{i} ≡ 1 (mod 3) and q_{j} ≡ 2 (mod 3) are primes with α_{i}, β_{j} ≥ 0. Then,

PDO(48n + 16) ≡ \begin{cases}6 ∏_{i=1}^{k}(α_{i} + 1) (mod 9), & if all β_{j} are even; \\ 0 (mod 9), & otherwise. \end{cases}  (29)

and

PDO(72n + 24) ≡ \begin{cases}3 ∏_{i=1}^{k}(α_{i} + 1) (mod 9), & if all β_{j} are even; \\ 0 (mod 9), & otherwise. \end{cases}  (30)

Corollary 10. Let p ≡ 2 (mod 3) be a prime. Then for all nonnegative integers α and n with p ∤ n, we have

PDO(48p^{2α+1}n + 16p^{2α+2}) ≡ 0 (mod 9)  (31)

and

PDO(72p^{2α+1}n + 24p^{2α+2}) ≡ 0 (mod 9).  (32)

Theorem 11. If n cannot be represented as the sum of a triangular number and three times a triangular number, then

PDO(48n + 24) ≡ 0 (mod 9).

Corollary 12. For any positive integer k, let p_{j} ≥ 5, 1 ≤ j ≤ k be primes. If (−3/p_{j}) = −1 for every j, then for all nonnegative integers n with p_{k} ∤ n we have

PDO(48p_{1}^{2}p_{2}^{2} ⋯ p_{k−1}^{2}p_{k}n + 24p_{1}^{2}p_{2}^{2} ⋯ p_{k}^{2}) ≡ 0 (mod 9).  (33)

By the binomial theorem, it is easy to see that for any positive integer m,

f_{m}^{3} ≡ f_{3m} (mod 3)  (34)

and

f_{m}^{9} ≡ f_{3m}^{3} (mod 9).  (35)

Proof of Theorem 8. From (35), it follows that

\frac{f_{3}^{3}}{f_{1}^{5}} ≡ f_{1}^{4} (mod 9).  (36)

In view of (36), we rewrite (3) as

∑_{n=0}^{∞} PDO(3n + 1)q^{n} ≡ \frac{f_{1}^{4}f_{2}^{4}f_{12}}{f_{4}f_{6}^{2}} (mod 9).  (37)

Substituting (12) in (37) and extracting the terms containing odd powers of q, we get

∑_{n=0}^{∞} PDO(6n + 4)q^{n} ≡ −4\frac{f_{1}^{6}f_{4}^{4}f_{6}}{f_{2}^{3}f_{3}^{2}} (mod 9).  (38)

Employing (14) in (38) and extracting the terms containing even powers of q, we derive

∑_{n=0}^{∞} PDO(12n + 4)q^{n} ≡ −4\frac{f_{2}^{10}f_{3}}{f_{1}^{3}f_{6}^{2}} (mod 9).  (39)

Substituting (16) in (39) and extracting the terms containing odd powers of q, we get

∑_{n=0}^{∞} PDO(24n + 16)q^{n} ≡ −12\frac{f_{1}^{3}f_{2}^{2}f_{6}^{2}}{f_{3}} (mod 9).

From (34),

\frac{f_{1}^{3}f_{2}^{2}f_{6}^{2}}{f_{3}} ≡ \frac{f_{6}^{3}}{f_{2}} (mod 3).

In view of the above two identities,

∑_{n=0}^{∞} PDO(24n + 16)q^{n} ≡ 6\frac{f_{6}^{3}}{f_{2}} (mod 9),  (40)

which implies that

PDO(48n + 40) ≡ 0 (mod 9)  (41)

for all n ≥ 0 and

∑_{n=0}^{∞} PDO(48n + 16)q^{n} ≡ 6\frac{f_{3}^{3}}{f_{1}} (mod 9).  (42)

Invoking (15) in (42) and extracting the terms containing odd powers of q,

∑_{n=0}^{∞} PDO(96n + 64)q^{n} ≡ 6\frac{f_{6}^{3}}{f_{2}} (mod 9).  (43)

By (40) and (43),

PDO(96n + 64) ≡ PDO(24n + 16) (mod 9).  (44)

Congruence (25) follows from (44) and mathematical induction. Congruence (27) follows from (41) and (25).

Employing (13) in (2),

∑_{n=0}^{∞} PDO(3n)q^{n} = \frac{f_{4}^{14}f_{6}^{4}}{f_{2}^{12}f_{8}^{4}f_{12}^{2}} + 4q \frac{f_{4}^{2}f_{6}^{4}f_{8}^{4}}{f_{2}^{8}f_{12}^{2}} ,

which yields

∑_{n=0}^{∞} PDO(6n)q^{n} = \frac{f_{2}^{14}f_{3}^{4}}{f_{1}^{12}f_{4}^{4}f_{6}^{2}}  (45)

and

∑_{n=0}^{∞} PDO(6n + 3)q^{n} = 4\frac{f_{2}^{2}f_{3}^{4}f_{4}^{4}}{f_{1}^{8}f_{6}^{2}} .  (46)

Applying (16) and (45),

∑_{n=0}^{∞} PDO(6n)q^{n} = \frac{f_{2}^{14}}{f_{4}^{4}f_{6}^{2}} (\frac{f_{4}^{6}f_{6}^{3}}{f_{2}^{9}f_{12}^{2}} + 3q \frac{f_{4}^{2}f_{6}f_{12}^{2}}{f_{2}^{7}} )^{4} ,

which implies that

∑_{n=0}^{∞} PDO(12n)q^{n} ≡ \frac{f_{2}^{20}f_{3}^{10}}{f_{1}^{22}f_{6}^{8}} (mod 9).

From (35),

\frac{f_{2}^{20}f_{3}^{10}}{f_{1}^{22}f_{6}^{8}} ≡ \frac{f_{2}^{2}f_{3}^{4}}{f_{1}^{4}f_{6}^{2}} (mod 9).

In view of the above two identities,

∑_{n=0}^{∞} PDO(12n)q^{n} ≡ \frac{f_{2}^{2}f_{3}^{4}}{f_{1}^{4}f_{6}^{2}} (mod 9).  (47)

Substituting (15) and (16) in (47), and extracting the terms containing even powers of q, we have

∑_{n=0}^{∞} PDO(24n)q^{n} ≡ \frac{f_{2}^{9}f_{3}^{3}}{f_{1}^{9}f_{6}^{3}} + 3q \frac{f_{2}f_{6}^{5}}{f_{1}^{5}f_{3}} (mod 9).  (48)

Using (34) and (35) in (48), we get

∑_{n=0}^{∞} PDO(24n + 24)q^{n} ≡ 3\frac{f_{1}f_{2}f_{6}^{5}}{f_{3}^{3}} (mod 9).  (49)

Substituting (17) in (49) and using (34), we have

∑_{n=0}^{∞} PDO(24n + 24)q^{n} ≡ 3\frac{f_{2}^{2}f_{4}^{2}f_{12}^{2}}{f_{6}^{2}} − 3q \frac{f_{2}^{4}f_{12}^{6}}{f_{4}^{2}f_{6}^{4}}

≡ 3\frac{f_{4}^{2}f_{12}^{2}}{f_{2}f_{6}} − 3q \frac{f_{2}f_{4}f_{12}^{5}}{f_{6}^{3}} (mod 9),

which yields

∑_{n=0}^{∞} PDO(48n + 24)q^{n} ≡ 3ψ(q)ψ(q^{3}) (mod 9)  (50)

and

∑_{n=0}^{∞} PDO(48n + 48)q^{n} ≡ −3\frac{f_{1}f_{2}f_{6}^{5}}{f_{3}^{3}} (mod 9).  (51)

By (49) and (51),

PDO(2(24n + 24)) ≡ − PDO(24n + 24) (mod 9).  (52)

Employing (23) in (49) and using (34),

∑_{n=0}^{∞} PDO(24n + 24)q^{n} ≡ 3\frac{f_{6}^{6}f_{9}^{4}}{f_{3}^{4}f_{18}^{2}} − 3q \frac{f_{6}^{5}f_{9}f_{18}}{f_{3}^{3}} − 6q^{2} \frac{f_{6}^{4}f_{18}^{4}}{f_{3}^{2}f_{9}^{2}}

≡ 3\frac{f_{9}^{3}}{f_{3}} + 6q \frac{f_{18}^{3}}{f_{6}} + 3q^{2} \frac{f_{3}f_{6}f_{18}^{5}}{f_{9}^{3}} (mod 9)

which implies that

∑_{n=0}^{∞} PDO(72n + 24)q^{n} ≡ 3\frac{f_{3}^{3}}{f_{1}} (mod 9),  (53)

∑_{n=0}^{∞} PDO(72n + 48)q^{n} ≡ 6\frac{f_{6}^{3}}{f_{2}} (mod 9)  (54)

and

∑_{n=0}^{∞} PDO(72n + 72)q^{n} ≡ 3\frac{f_{1}f_{2}f_{6}^{5}}{f_{3}^{3}} (mod 9).  (55)

From (54),

PDO(144n + 120) ≡ 0 (mod 9).  (56)

By (49) and (55),

PDO(3(24n + 24)) ≡ PDO(24n + 24) (mod 9).  (57)

Congruence (26) follows from (52), (57), and mathematical induction. Congruence (28) follows from (56) and (26). Proof of Theorem 9. From (24), (42) and (53), it is clear that for all n ≥ 0

PDO(48n + 16) ≡ 6a_{3}(n) (mod 9),  (58)

and

PDO(72n + 24) ≡ 3a_{3}(n) (mod 9).  (59)

Congruence (29) follows from Lemma 7 and (58). Congruence (30) follows from Lemma 7 and (59).

For any prime p and any positive integer N, let υ_{p}(N) denote the exponent of the highest power of p dividing N. Proof of Corollary 10. Suppose α ≥ 0, p ≡ 2 (mod 3) and p ∤ n, then it is clear that

υ_{p} (3 (p^{2α+1}n + \frac{p^{2α+2} − 1}{3}) + 1) = υ_{p} (3p^{2α+1}n + p^{2α+2}) = 2α + 1.  (60)

Congruences (31) and (32) follow from (29), (30) and (60). Proof of Theorem 11. From (8) and (50), we have

∑_{n=0}^{∞} PDO(48n + 24)q^{n} ≡ 3 ∑_{k=0}^{∞} ∑_{m=0}^{∞} q^{k(k+1)/2+3m(m+1)/2} (mod 9).  (61)

Theorem 11 follows from (61). Proof of Corollary 12. By (61),

∑_{n=0}^{∞} PDO(48n + 24)q^{48n+24} ≡ 3 ∑_{k=0}^{∞} ∑_{m=0}^{∞} q^{6(2k+1)^{2}+2(6m+3)^{2}} (mod 9),

which implies that if 48n+24 is not of the form 6(2k+1)^{2}+2(6m+3)^{2}, then PDO(48n+24) ≡ 0 (mod 9). Let k ≥ 1 be an integer and let p_{i} ≥ 5, 1 ≤ i ≤ k be primes with (\frac{−3}{p_{i}}) = −1. If N is of the form 2x^{2} + 6y^{2}, then v_{p_{i}}(N) is even since (\frac{−3}{p_{i}}) = −1. Let

N = 48 (p_{1}^{2}p_{2}^{2} ⋯ p_{k−1}^{2}p_{k}n + \frac{p_{1}^{2}p_{2}^{2} ⋯ p_{k−1}^{2}p_{k}^{2} − 1}{2}) + 24
= 48p_{1}^{2}p_{2}^{2} ⋯ p_{k−1}^{2}p_{k}n + 24p_{1}^{2}p_{2}^{2} ⋯ p_{k−1}^{2}p_{k}^{2}.

If p_{k} ∤ n, then v_{p_{k}}(N) is an odd number and hence N is not of the form 2x^{2} + 6y^{2}. Therefore (33) holds.

4 Congruences modulo 2^{4} and 2^{5}

In this section, we establish the following infinite families of congruences modulo 16 and 32 for PDO(n).

Theorem 13. For all nonnegative integers α and n, we have

PDO(4^{α}(12n + 8)) ≡ PDO(12n + 8) (mod 2^{4}),  (62)
PDO(4^{α}(24n + 23)) ≡ 0 (mod 2^{4}),  (63)
PDO(4^{α}(48n + 14)) ≡ 0 (mod 2^{4})  (64)

and

PDO(24n + 17) ≡ 0 (mod 2^{4}).  (65)

Theorem 14. For all nonnegative integers α and n, we have

PDO(2^{α}(12n)) ≡ PDO(12n) (mod 2^{4}),  (66)
PDO(2^{α}(72n + 42)) ≡ 0 (mod 2^{4})  (67)

and

PDO(2^{α}(72n + 66)) ≡ 0 (mod 2^{4}).  (68)

Theorem 15. For all nonnegative integers α and n, we have

PDO(4^{α}(24n) ≡ PDO(24n) (mod 2^{5}),  (69)
PDO(9^{α}(6n + 3)) ≡ PDO(6n + 3) (mod 2^{5}),  (70)
PDO(24n + 9) ≡ 0 (mod 2^{5}),  (71)
PDO(9^{α}(216n + 117)) ≡ 0 (mod 2^{5}),  (72)
PDO(2^{α}(72n + 69)) ≡ 0 (mod 2^{5}),  (73)
PDO(3^{α}(72n + 69)) ≡ 0 (mod 2^{5})  (74)

and

PDO(2^{α}(144n + 42)) ≡ 0 (mod 2^{5}).  (75)

Theorem 16. If n cannot be represented as the sum of two triangular numbers, then for all nonnegative integers α and r ∈ {1, 6} we have

PDO(2^{α}r(12n + 3)) ≡ 0 (mod 2^{4}).

Corollary 17. If p is a prime, p ≡ 3 (mod 4)), 1 ≤ j ≤ p − 1 and r ∈ {1, 6}, then for all nonnegative integers α, β and n, we have

PDO(2^{α}p^{2β+1}r(12pn + 12j + 3p)) ≡ 0 (mod 2^{4}).  (76)

For example, taking p = 3, we deduce that for all α, β, n ≥ 0,

PDO(2^{α}3^{β}(216n + 126)) ≡ 0 (mod 2^{4})  (77)

and

PDO(2^{α}3^{β}(216n + 198)) ≡ 0 (mod 2^{4}).  (78)

Combining (77) and (67),

PDO(2^{α}3^{β}(72n + 42)) ≡ 0 (mod 2^{4}).  (79)

Combining (78) and (68),

PDO(2^{α}3^{β}(72n + 66)) ≡ 0 (mod 2^{4}).  (80)

Theorem 18. If n cannot be represented as the sum of a triangular number and four times a triangular number, then for all nonnegative integers α and r ∈ {1, 3} we have

PDO(2^{α}r(48n + 30)) ≡ 0 (mod 2^{5}).

Corollary 19. If p is any prime with p ≡ 3 (mod 4)), 1 ≤ j ≤ p − 1 and r ∈ {1, 3}, then for all nonnegative integers α, β and n, we have

PDO(2^{α}p^{2β+1}r(48pn + 48j + 30p)) ≡ 0 (mod 2^{5}).  (81)

For example, taking p = 3 we find that for all α, n ≥ 0 and β ≥ 1,

PDO(2^{α}3^{β}(144n + 42)) ≡ 0 (mod 2^{5})  (82)

and

PDO(2^{α}3^{β}(144n + 138)) ≡ 0 (mod 2^{5}).  (83)

Combining (82) and (75), for all α, β, n ≥ 0,

PDO(2^{α}3^{β}(144n + 42)) ≡ 0 (mod 2^{5})  (84)

and combining (83), (73) and (74), for all α, β, n ≥ 0,

PDO(2^{α}3^{β}(72n + 69)) ≡ 0 (mod 2^{5}).  (85)

Theorem 20. If n cannot be represented as the sum of twice a pentagonal number and three times a triangular number, then for any nonnegative integer α we have

PDO(4^{α}(24n + 11)) ≡ 0 (mod 2^{4}).

Corollary 21. For any positive integer k, let p_{j} ≥ 5, 1 ≤ j ≤ k be primes. If (−2/p_{j}) = −1 for every j, then for all nonnegative integers α and n with p_{k} ∤ n we have

PDO(6 · 4^{α+1}p_{1}^{2}p_{2}^{2} ⋯ p_{k−1}^{2}p_{k}n + 11 · 4^{α}p_{1}^{2}p_{2}^{2} ⋯ p_{k}^{2}) ≡ 0 (mod 2^{4}).

Theorem 22. If n cannot be represented as the sum of a pentagonal number and six times a triangular number, then for any nonnegative integer α we have

PDO(4^{α}(48n + 38)) ≡ 0 (mod 2^{4}).

Corollary 23. For any positive integer k, let p_{j} ≥ 5, 1 ≤ j ≤ k be primes. If (−2/p_{j}) = −1 for every j, then for all nonnegative integers n with p_{k} ∤ n we have

PDO(3 · 4^{α+2}p_{1}^{2}p_{2}^{2} ⋯ p_{k−1}^{2}p_{k}n + 38 · 4^{α}p_{1}^{2}p_{2}^{2} ⋯ p_{k}^{2}) ≡ 0 (mod 2^{4}).

Theorem 24. If n cannot be represented as the sum of a pentagonal number and four times a pentagonal number, then we have

PDO(24n + 5) ≡ 0 (mod 2^{5}).

Corollary 25. For any positive integer k, let p_{j} ≥ 5, 1 ≤ j ≤ k be primes. If (−1/p_{j}) = −1 for every j, then for all nonnegative integers n with p_{k} ∤ n we have

PDO(24p_{1}^{2}p_{2}^{2} ⋯ p_{k−1}^{2}p_{k}n + 5p_{1}^{2}p_{2}^{2} ⋯ p_{k}^{2}) ≡ 0 (mod 2^{5}).

Theorem 26. If n cannot be represented as the sum of a pentagonal number and sixteen times a pentagonal number, then we have

PDO(24n + 17) ≡ 0 (mod 2^{5}).

Corollary 27. For any positive integer k, let p_{j} ≥ 5, 1 ≤ j ≤ k be primes. If (−1/p_{j}) = −1 for every j, then for all nonnegative integers n with p_{k} ∤ n we have

PDO(24p_{1}^{2}p_{2}^{2} ⋯ p_{k−1}^{2}p_{k}n + 17p_{1}^{2}p_{2}^{2} ⋯ p_{k}^{2}) ≡ 0 (mod 2^{5}).

By the binomial theorem, it is easy to see that for all positive integers k and m,

f_{m}^{2^{k}} ≡ f_{2m}^{2^{k−1}} (mod 2^{k}).  (86)

Proof of Theorem 13. Using (13), we can rewrite (4) as

∑_{n=0}^{∞} PDO(3n + 2)q^{n} = 2\frac{f_{4}^{13}f_{6}f_{12}}{f_{2}^{11}f_{8}^{4}} + 8q \frac{f_{4}f_{6}f_{8}^{4}f_{12}}{f_{2}^{7}} ,

which yields

∑_{n=0}^{∞} PDO(6n + 2)q^{n} = 2\frac{f_{2}^{13}f_{3}f_{6}}{f_{1}^{11}f_{4}^{4}}

and

∑_{n=0}^{∞} PDO(6n + 5)q^{n} = 8\frac{f_{2}f_{3}f_{4}^{4}f_{6}}{f_{1}^{7}} .

From (86) with k = 3 and k = 2, we have

\frac{f_{2}^{13}f_{3}f_{6}}{f_{1}^{11}f_{4}^{4}} ≡ \frac{f_{2}f_{3}f_{6}}{f_{1}^{3}} (mod 2^{3})

and

\frac{f_{2}f_{3}f_{4}^{4}f_{6}}{f_{1}^{7}} ≡ \frac{f_{3}f_{4}^{4}f_{6}}{f_{1}^{3}f_{2}} (mod 2^{2}).

Thus,

∑_{n=0}^{∞} PDO(6n + 2)q^{n} ≡ 2\frac{f_{2}f_{3}f_{6}}{f_{1}^{3}} (mod 2^{4})

and

∑_{n=0}^{∞} PDO(6n + 5)q^{n} ≡ 8\frac{f_{3}f_{4}^{4}f_{6}}{f_{1}^{3}f_{2}} (mod 2^{5}).

Substituting (16) in the above two congruences and extracting the terms containing even and odd powers of q, we get

∑_{n=0}^{∞} PDO(12n + 2)q^{n} ≡ 2\frac{f_{2}^{6}f_{3}^{4}}{f_{1}^{8}f_{6}^{2}} ≡ 2\frac{f_{2}^{2}f_{3}^{4}}{f_{6}^{2}} (mod 2^{4}),  (87)

∑_{n=0}^{∞} PDO(12n + 8)q^{n} ≡ 6\frac{f_{2}^{2}f_{3}^{2}f_{6}^{2}}{f_{1}^{6}} (mod 2^{4}),  (88)

∑_{n=0}^{∞} PDO(12n + 5)q^{n} ≡ 8\frac{f_{2}^{10}f_{3}^{4}}{f_{1}^{10}f_{6}^{2}} ≡ 8\frac{f_{2}^{6}}{f_{1}^{2}} (mod 2^{5})  (89)

and

∑_{n=0}^{∞} PDO(12n + 11)q^{n} ≡ 24\frac{f_{2}^{6}f_{3}^{2}f_{6}^{2}}{f_{1}^{8}} ≡ 24f_{4}f_{6}^{3} (mod 2^{4}).  (90)

It follows from (90) that

∑_{n=0}^{∞} PDO(24n + 11)q^{n} ≡ 8f_{2}f_{3}^{3} ≡ 8f_{2} \frac{f_{6}^{2}}{f_{3}} (mod 2^{4})  (91)

and

PDO(24n + 23) ≡ 0 (mod 2^{4}).  (92)

Employing (12) in (87) and extracting the terms containing odd powers of q,

∑_{n=0}^{∞} PDO(24n + 14)q^{n} ≡ −8q \frac{f_{1}^{2}f_{12}^{4}}{f_{6}^{2}} ≡ 8qf_{2}f_{12}^{3} (mod 2^{4}),

which implies that

∑_{n=0}^{∞} PDO(48n + 38)q^{n} ≡ 8f_{1}f_{6}^{3} ≡ 8f_{1} \frac{f_{12}^{2}}{f_{6}} (mod 2^{4})  (93)

and

PDO(48n + 14) ≡ 0 (mod 2^{4}).  (94)

Substituting (13) and (20) in (88), and extracting the terms containing even and odd powers of q, we have

∑_{n=0}^{∞} PDO(24n + 8)q^{n} ≡ 6\frac{f_{2}^{18}f_{3}^{3}f_{6}^{2}}{f_{1}^{17}f_{4}^{5}f_{12}} ≡ 6\frac{f_{2}^{2}f_{3}^{3}f_{6}^{2}}{f_{1}f_{4}f_{12}} (mod 2^{4})  (95)

and

∑_{n=0}^{∞} PDO(24n + 20)q^{n} ≡ 12\frac{f_{2}^{15}f_{3}^{4}f_{12}}{f_{1}^{16}f_{4}^{3}f_{6}} + 24\frac{f_{2}^{6}f_{3}^{3}f_{4}^{3}f_{6}^{2}}{f_{1}^{13}f_{12}}

≡ 12\frac{f_{2}^{7}f_{6}f_{12}}{f_{4}^{3}} + 24\frac{f_{3}^{3}f_{4}^{3}}{f_{1}} (mod 2^{4}).  (96)

Substituting (15) in (95) and (96), and extracting even and odd powers of q, we have

∑_{n=0}^{∞} PDO(48n + 8)q^{n} ≡ 6\frac{f_{2}^{2}f_{3}^{4}}{f_{6}^{2}} (mod 2^{4}),  (97)

∑_{n=0}^{∞} PDO(48n + 32)q^{n} ≡ 6\frac{f_{1}^{2}f_{3}^{2}f_{6}^{2}}{f_{2}^{2}} (mod 2^{4})  (98)

and

∑_{n=0}^{∞} PDO(48n + 44)q^{n} ≡ 8f_{2}^{2}f_{6}^{3} ≡ 8\frac{f_{4}f_{12}^{2}}{f_{6}} (mod 2^{4}).  (99)

Again employing (12) in (97) and extracting the terms containing odd powers of q,

∑_{n=0}^{∞} PDO(96n + 56)q^{n} ≡ 8q \frac{f_{1}^{2}f_{12}^{4}}{f_{6}^{2}} ≡ 8qf_{2}f_{12}^{3} (mod 2^{4}).  (100)

By (99) and (100),

PDO(96n + 92) ≡ 0 (mod 2^{4})  (101)

and

PDO(192n + 56) ≡ 0 (mod 2^{4}).  (102)

From (86), (88) and (98),

PDO(48n + 32) ≡ PDO(12n + 8) (mod 2^{4}).  (103)

Congruence (62) follows from (103) and mathematical induction. Congruence (63) follows from (92), (101), and (62). Similarly, congruence (64) follows from (94), (102), and (62).

By invoking (11) in (89) and extracting the terms containing even and odd powers of q,

∑_{n=0}^{∞} PDO(24n + 5)q^{n} ≡ 8\frac{f_{1}f_{4}^{5}}{f_{8}^{2}} ≡ 8f_{1}f_{4} (mod 2^{5})  (104)

and

∑_{n=0}^{∞} PDO(24n + 17)q^{n} ≡ 16\frac{f_{1}f_{2}^{2}f_{8}^{2}}{f_{4}} ≡ 16f_{1}f_{16} (mod 2^{5}).  (105)

Congruence (65) follows from (105).

Proofs of Theorem 14 and Theorem 15. From (86),

\frac{f_{2}^{2}f_{3}^{4}f_{4}^{4}}{f_{1}^{8}f_{6}^{2}} ≡ \frac{f_{3}^{4}f_{4}^{4}}{f_{2}^{2}f_{6}^{2}} (mod 2^{3}).  (106)

Using (106), we rewrite (46) as

∑_{n=0}^{∞} PDO(6n + 3)q^{n} ≡ 4\frac{f_{3}^{4}f_{4}^{4}}{f_{2}^{2}f_{6}^{2}} (mod 2^{5}).  (107)

In view of (7) and (8),

∑_{n=0}^{∞} PDO(6n + 3)q^{n} ≡ 4ψ^{2}(q^{2})ϕ^{2}(−q^{3}) (mod 2^{5}).  (108)

Substituting (22) in (108) and extracting the terms involving q^{3n+1}, we get

∑_{n=0}^{∞} PDO(18n + 9)q^{n} ≡ 4qψ^{2}(q^{6})ϕ^{2}(−q) (mod 2^{5}).  (109)

Using (21) in (109) and extracting the terms involving q^{3n+1},

∑_{n=0}^{∞} PDO(54n + 27)q^{n} ≡ 4ψ^{2}(q^{2})ϕ^{2}(−q^{3}) (mod 2^{5}).  (110)

Congruence (70) follows from (108), (110), and mathematical induction. Applying (12) in (107) and using (86),

∑_{n=0}^{∞} PDO(6n + 3)q^{n} ≡ 4\frac{f_{4}^{4}f_{12}^{10}}{f_{2}^{2}f_{6}^{4}f_{24}^{4}} − 16q^{3} \frac{f_{4}^{4}f_{24}^{4}}{f_{2}^{2}f_{12}^{2}}

≡ 4\frac{f_{4}^{4}f_{12}^{2}}{f_{2}^{2}f_{6}^{4}} + 16q^{3} \frac{f_{8}^{2}f_{48}^{2}}{f_{4}f_{24}} (mod 2^{5})

which implies that

∑_{n=0}^{∞} PDO(12n + 3)q^{n} ≡ 4\frac{f_{2}^{4}f_{6}^{2}}{f_{1}^{2}f_{3}^{4}} (mod 2^{5})  (111)

and

∑_{n=0}^{∞} PDO(12n + 9)q^{n} ≡ 16q \frac{f_{4}^{2}f_{24}^{2}}{f_{2}f_{12}} (mod 2^{5}).  (112)

Congruence (71) follows from (112).

From (112) and (8),

∑_{n=0}^{∞} PDO(24n + 21)q^{n} ≡ 16ψ(q)ψ(q^{6}) (mod 2^{5}).  (113)

Invoking (22) in (113) and extracting the terms containing q^{3n+2} and q^{3n+1},

PDO(72n + 69) ≡ 0 (mod 2^{5})  (114)

for all n ≥ 0 and

∑_{n=0}^{∞} PDO(72n + 45)q^{n} ≡ 16ψ(q^{2})ψ(q^{3}) (mod 2^{5}).  (115)

Using (22) in (115) and extracting the terms containing q^{3n+1},

PDO(216n + 117) ≡ 0 (mod 2^{5}).  (116)

Congruence (72) follows from (70) and (116).

Substituting (11) and (13) in (111), and extracting the terms containing odd powers of q, we get

∑_{n=0}^{∞} PDO(24n + 15)q^{n} ≡ 8\frac{f_{2}^{2}f_{6}^{14}f_{8}^{2}}{f_{1}f_{3}^{12}f_{4}f_{12}^{4}} + 16q \frac{f_{4}^{5}f_{6}^{2}f_{12}^{4}}{f_{1}f_{3}^{8}f_{8}^{2}}
≡ 8ψ(q)ψ(q^{4}) + 16qψ(q)ψ(q^{12}) (mod 2^{5}).  (117)

Employing (22) in (117) and extracting the terms involving q^{3n+2}, we get

∑_{n=0}^{∞} PDO(72n + 63)q^{n} ≡ 16ψ(q^{3})ψ(q^{4}) + 8qψ(q^{3})ψ(q^{12}) (mod 2^{5}).  (118)

Again, extracting the terms involving q^{3n+2} in (118), we get

PDO(216n + 207) ≡ 0 (mod 2^{5}).  (119)

Congruence (74) follows from (114), (119), and (70).

Substituting (12) and (13) in (45), and extracting the terms containing even powers of q, we find

∑_{n=0}^{∞} PDO(12n)q^{n} ≡ \frac{f_{2}^{38}f_{6}^{10}}{f_{1}^{28}f_{3}^{4}f_{4}^{12}f_{12}^{4}} ≡ \frac{f_{1}^{4}f_{2}^{6}f_{12}^{4}}{f_{3}^{4}f_{4}^{4}f_{6}^{6}} (mod 2^{4}).  (120)

Using (12) and (13) in (120), and extracting the terms containing even powers of q, we have

∑_{n=0}^{∞} PDO(24n)q^{n} ≡ \frac{f_{1}^{4}f_{2}^{6}f_{6}^{18}}{f_{3}^{20}f_{4}^{4}f_{12}^{4}} ≡ \frac{f_{1}^{4}f_{2}^{6}f_{12}^{4}}{f_{3}^{4}f_{4}^{4}f_{6}^{6}} (mod 2^{4}).  (121)

From (120) and (121),

PDO(2(12n)) ≡ PDO(12n) (mod 2^{4}).  (122)

Congruence (66) follows from (122) and mathematical induction.

Substituting (20) and (13) in (45), and using (86), we get

∑_{n=0}^{∞} PDO(6n)q^{n} = \frac{f_{4}^{32}f_{12}^{4}}{f_{2}^{24}f_{8}^{10}f_{24}^{2}} + 16q^{2} \frac{f_{4}^{8}f_{8}^{6}f_{12}^{4}}{f_{2}^{16}f_{24}^{2}} + 4q^{2} \frac{f_{4}^{26}f_{6}^{2}f_{24}^{2}}{f_{2}^{22}f_{8}^{6}f_{12}^{2}}

+ 64q^{4} \frac{f_{4}^{2}f_{6}^{2}f_{8}^{10}f_{24}^{2}}{f_{2}^{14}f_{12}^{2}} + 32q^{2} \frac{f_{4}^{17}f_{6}f_{12}}{f_{2}^{19}} + 8q \frac{f_{4}^{20}f_{12}^{4}}{f_{2}^{20}f_{8}^{2}f_{24}^{2}}

+ 32q^{3} \frac{f_{4}^{14}f_{6}^{2}f_{8}^{2}f_{24}^{2}}{f_{2}^{18}f_{12}^{2}} + 4q \frac{f_{4}^{29}f_{6}f_{12}}{f_{2}^{23}f_{8}^{8}} + 64q^{3} \frac{f_{4}^{5}f_{6}f_{8}^{8}f_{12}}{f_{2}^{15}}

≡ \frac{f_{2}^{8}f_{4}^{16}f_{12}^{4}}{f_{8}^{10}f_{24}^{2}} + 16q^{2}f_{8}^{2}f_{16}^{2} + 4q^{2} \frac{f_{4}^{2}f_{6}^{2}f_{8}^{2}f_{24}^{2}}{f_{2}^{6}f_{12}^{2}}

+ 8qf_{4}^{2}f_{8}^{2} + 4qf_{2}f_{4}f_{6}f_{12} (mod 2^{5}), which yields

∑_{n=0}^{∞} PDO(12n)q^{n} ≡ \frac{f_{1}^{8}f_{2}^{16}f_{6}^{4}}{f_{4}^{10}f_{12}^{2}} + 16qf_{4}^{2}f_{8}^{2} + 4q \frac{f_{2}^{2}f_{3}^{2}f_{4}^{2}f_{12}^{2}}{f_{1}^{6}f_{6}^{2}} (mod 2^{5})  (123)

and

∑_{n=0}^{∞} PDO(12n + 6)q^{n} ≡ 8f_{2}^{2}f_{4}^{2} + 4f_{1}f_{2}f_{3}f_{6} (mod 2^{5}).  (124)

Substituting (18) in (124) and extracting the terms containing even and odd powers of q, we get

∑_{n=0}^{∞} PDO(24n + 6)q^{n} ≡ 4\frac{f_{1}^{2}f_{4}^{2}f_{6}^{4}}{f_{2}^{2}f_{12}^{2}} + 8f_{1}^{2}f_{2}^{2} (mod 2^{5})  (125)

and

∑_{n=0}^{∞} PDO(24n + 18)q^{n} ≡ −4\frac{f_{2}^{4}f_{3}^{2}f_{12}^{2}}{f_{4}^{2}f_{6}^{2}} (mod 2^{5}).  (126)

By (86) and (8),

\frac{f_{2}^{4}f_{3}^{2}f_{12}^{2}}{f_{4}^{2}f_{6}^{2}} ≡ ψ^{2}(q^{3}) (mod 2^{2})

and

4\frac{f_{1}^{2}f_{4}^{2}f_{6}^{4}}{f_{2}^{2}f_{12}^{2}} + 8f_{1}^{2}f_{2}^{2} ≡ −4ψ^{2}(q) (mod 2^{4}).

In view of above identities,

∑_{n=0}^{∞} PDO(24n + 6)q^{n} ≡ −4ψ^{2}(q) (mod 2^{4})  (127)

and

∑_{n=0}^{∞} PDO(24n + 18)q^{n} ≡ −4ψ^{2}(q^{3}) (mod 2^{4}).  (128)

It follows from (128) that

PDO(72n + 42) ≡ 0 (mod 2^{4})  (129)

and

PDO(72n + 66) ≡ 0 (mod 2^{4})  (130)

for all n ≥ 0 and

∑_{n=0}^{∞} PDO(72n + 18)q^{n} ≡ −4ψ^{2}(q) (mod 2^{4}).  (131)

Employing (10) in (125) and (126) and extracting the terms containing odd powers of q, we get

∑_{n=0}^{∞} PDO(48n + 30)q^{n} ≡ −8\frac{f_{2}^{2}f_{3}^{4}f_{8}^{2}}{f_{1}f_{4}f_{6}^{2}} + 16\frac{f_{1}^{3}f_{8}^{2}}{f_{4}} ≡ 8ψ(q)ψ(q^{4}) (mod 2^{5})  (132)

and

∑_{n=0}^{∞} PDO(48n + 42)q^{n} ≡ 8q \frac{f_{1}^{4}f_{6}^{2}f_{24}^{2}}{f_{2}^{2}f_{3}f_{12}} ≡ 8qψ(q^{3})ψ(q^{12}) (mod 2^{5}).  (133)

In view of (133), we have

PDO(144n + 42) ≡ 0 (mod 2^{5})  (134)

and

PDO(144n + 138) ≡ 0 (mod 2^{5})  (135)

for all n ≥ 0 and

∑_{n=0}^{∞} PDO(144n + 90)q^{n} ≡ 8ψ(q)ψ(q^{4}) (mod 2^{5}).  (136)

Substituting (12), (13), and (20) in (123), and extracting the terms containing even and odd powers of q, we get

∑_{n=0}^{∞} PDO(24n)q^{n} ≡\frac{f_{1}^{12}f_{2}^{10}f_{3}^{4}}{f_{4}^{8}f_{6}^{2}} + 16q \frac{f_{1}^{20}f_{3}^{4}f_{4}^{8}}{f_{2}^{14}f_{6}^{2}} + 16q \frac{f_{2}^{8}f_{4}^{3}f_{6}^{4}}{f_{1}^{13}f_{3}f_{12}} + 8q \frac{f_{2}^{17}f_{6}f_{12}}{f_{1}^{16}f_{4}^{3}}

≡\frac{f_{3}^{4}f_{4}^{8}}{f_{1}^{20}f_{2}^{6}f_{6}^{2}} + 16qf_{4}^{6} + 16q \frac{f_{1}^{3}f_{4}^{3}f_{12}}{f_{3}} + 8qf_{2}f_{4}f_{6}f_{12} (mod 2^{5})  (137)

and

∑_{n=0}^{∞} PDO(24n + 12)q^{n} ≡ −8\frac{f_{1}^{16}f_{3}^{4}}{f_{2}^{2}f_{6}^{2}} + 4\frac{f_{2}^{20}f_{6}^{4}}{f_{1}^{17}f_{3}f_{4}^{5}f_{12}} + 16f_{2}^{2}f_{4}^{2}

≡ 4\frac{f_{2}^{4}f_{6}^{4}}{f_{1}f_{3}f_{4}f_{12}} + 8f_{2}^{2}f_{4}^{2} (mod 2^{5}).  (138)

Employing (19) in (138) and extracting the terms involving even and odd powers of q, we get

∑_{n=0}^{∞} PDO(48n + 12)q^{n} ≡ 4\frac{f_{1}^{2}f_{4}^{2}f_{6}^{4}}{f_{2}^{2}f_{12}^{2}} + 8f_{1}^{2}f_{2}^{2} (mod 2^{5})  (139)

and

∑_{n=0}^{∞} PDO(48n + 36)q^{n} ≡ 4\frac{f_{2}^{4}f_{3}^{2}f_{12}^{2}}{f_{4}^{2}f_{6}^{2}} (mod 2^{5}).  (140)

In view of (125), (126), (139), and (140), we have

PDO(2(24n + 6)) ≡ PDO(24n + 6) (mod 2^{5})  (141)

and

PDO(2(24n + 18)) ≡ − PDO(24n + 18) (mod 2^{5}).  (142)

Congruence (67) follows from (129), (142), and (66). Congruence (68) follows from (130), (142), and (66).

Substituting (13), (14), and (20) in (137), and extracting even and odd powers of q, we get

∑_{n=0}^{∞} PDO(48n)q^{n} ≡ \frac{f_{2}^{72}f_{6}^{4}}{f_{1}^{72}f_{4}^{18}f_{12}^{2}} + 4q \frac{f_{2}^{66}f_{3}^{2}f_{12}^{2}}{f_{1}^{70}f_{4}^{14}f_{6}^{2}} − 48q \frac{f_{1}^{2}f_{2}^{2}f_{6}^{4}}{f_{3}^{2}}

≡ \frac{f_{2}^{8}f_{6}^{4}}{f_{1}^{8}f_{4}^{2}f_{12}^{2}} + 4q \frac{f_{2}^{2}f_{3}^{2}f_{4}^{2}f_{12}^{2}}{f_{1}^{6}f_{6}^{2}} + 16qf_{2}f_{4}f_{6}f_{12} (mod 2^{5})  (143)

and

∑_{n=0}^{∞} PDO(48n + 24)q^{n} ≡ 4\frac{f_{2}^{69}f_{3}f_{6}}{f_{1}^{71}f_{4}^{16}} + 16\frac{f_{2}^{60}f_{6}^{4}}{f_{1}^{68}f_{4}^{10}f_{12}^{2}} + 8f_{1}f_{2}f_{3}f_{6}
≡ 12f_{1}f_{2}f_{3}f_{6} + 16f_{2}^{2}f_{4}^{2} (mod 2^{5}).  (144)

Substituting (18) in (144) and extracting the terms containing even and odd powers of q, we get

∑_{n=0}^{∞} PDO(96n + 24)q^{n} ≡ −4\frac{f_{1}^{2}f_{4}^{2}f_{6}^{4}}{f_{2}^{2}f_{12}^{2}} + 16f_{1}^{2}f_{2}^{2} (mod 2^{5})  (145)

and

∑_{n=0}^{∞} PDO(96n + 72)q^{n} ≡ −12\frac{f_{2}^{4}f_{3}^{2}f_{12}^{2}}{f_{4}^{2}f_{6}^{4}} (mod 2^{5}).  (146)

From (126) and (146), we have

PDO(2^{2}(24n + 18)) ≡ 3 PDO(24n + 18) (mod 2^{5}).  (147)

Employing (10) in (145) and extracting the terms containing odd powers of q, we get

∑_{n=0}^{∞} PDO(192n + 120)q^{n} ≡ 8\frac{f_{2}^{2}f_{3}^{4}f_{8}^{2}}{f_{1}f_{4}f_{6}^{2}} ≡ 8ψ(q)ψ(q^{4}) (mod 2^{5}).  (148)

Substituting (13) and (20) in (143), and extracting the terms containing even and odd power of q, we have

∑_{n=0}^{∞} PDO(96n)q^{n} ≡ \frac{f_{2}^{26}f_{3}^{4}}{f_{1}^{20}f_{4}^{8}f_{6}^{2}} + 16q \frac{f_{2}^{2}f_{3}^{4}f_{4}^{8}}{f_{1}^{12}f_{6}^{2}} + 16q \frac{f_{2}^{8}f_{4}^{3}f_{6}^{4}}{f_{1}^{13}f_{3}f_{12}} + 8q \frac{f_{2}^{17}f_{6}f_{12}}{f_{1}^{16}f_{4}^{3}}

≡ \frac{f_{3}^{4}f_{4}^{8}}{f_{1}^{20}f_{2}^{6}f_{6}^{2}} + 16qf_{4}^{6} + 16q \frac{f_{1}^{3}f_{4}^{3}f_{12}}{f_{3}} + 8qf_{2}f_{4}f_{6}f_{12} (mod 2^{5})  (149)

and

∑_{n=0}^{∞} PDO(96n + 48)q^{n} ≡ 8\frac{f_{3}^{4}f_{2}^{14}}{f_{1}^{16}f_{6}^{2}} + 4\frac{f_{2}^{20}f_{6}^{4}}{f_{1}^{17}f_{3}f_{4}^{5}f_{12}} + 16f_{1}f_{2}f_{3}f_{6}

≡ 8f_{2}^{2}f_{4}^{2} + 4\frac{f_{2}^{4}f_{6}^{4}}{f_{1}f_{3}f_{4}f_{12}} + 16f_{1}f_{2}f_{3}f_{6} (mod 2^{5}).  (150)

By (149) and (137)

PDO(4(24n)) ≡ PDO(24n) (mod 2^{5}).  (151)

Congruence (69) follows from (151) and mathematical induction.

By substituting (19) and (18) in (150), and extracting the terms containing even and odd powers of q,

∑_{n=0}^{∞} PDO(192n + 48)q^{n} ≡ 8f_{1}^{2}f_{2}^{2} + 20\frac{f_{1}^{2}f_{4}^{2}f_{6}^{4}}{f_{2}^{2}f_{12}^{2}} (mod 2^{5})  (152)

and

∑_{n=0}^{∞} PDO(192n + 144)q^{n} ≡ −12\frac{f_{2}^{4}f_{3}^{2}f_{12}^{2}}{f_{4}^{2}f_{6}^{2}} (mod 2^{5}).  (153)

From (125) and (152),

PDO(2^{3}(24n + 6)) ≡ −3 PDO(24n + 6) (mod 2^{5}).  (154)

In view of (126) and (153),

PDO(2^{3}(24n + 18)) ≡ 3 PDO(24n + 18) (mod 2^{5}).  (155)

Congruence (73) follows from (114), (135), (142), (147), (155), and (69). Congruence (75) follows from (134), (142), (147), (155), and (69).

Proof of Theorem 16. Using (86) and (8) in (111),

∑_{n=0}^{∞} PDO(12n + 3)q^{n} ≡ 4ψ^{2}(q) (mod 2^{4}).  (156)

From (156), (127), (141) and (66),

∑_{n=0}^{∞} PDO(2^{α}(24n + 6))q^{n} ≡ −4ψ^{2}(q) (mod 2^{4})  (157)

From (131), (141) and (66),

∑_{n=0}^{∞} PDO(2^{α}(72n + 18))q^{n} ≡ \begin{cases}−4ψ^{2}(q) (mod 2^{4}), & if α = 0; \\ 4ψ^{2}(q) (mod 2^{4}), & if α ≠ 0. \end{cases}  (158)

Combining (156), (157), and (158),

∑_{n=0}^{∞} PDO(2^{α}r(12n + 3))q^{n} ≡ \begin{cases}(−1)^{r+1}4 ∑_{k,m=0}^{∞} q^{k(k+1)/2+m(m+1)/2} (mod 2^{4}), & if α = 0; \\ (−1)^{r}4 ∑_{k,m=0}^{∞} q^{k(k+1)/2+m(m+1)/2} (mod 2^{4}), & if α ≠ 0. \end{cases}

(159)

Theorem 16 follows from (159). Proof of Theorem 18. From (132), (141), (148), (154) and (69), we have

∑_{n=0}^{∞} PDO(2^{α}(48n + 30))q^{n} ≡ 8ψ(q)ψ(q^{4}) (mod 2^{5}).  (160)

From (136), (142), (147), (155) and (69), we have

∑_{n=0}^{∞} PDO(2^{α}(144n + 90))q^{n} ≡ \begin{cases}8ψ(q)ψ(q^{4}) (mod 2^{5}), & if α = 0; \\ −8ψ(q)ψ(q^{4}) (mod 2^{5}), & if α ≠ 0. \end{cases}  (161)

Theorem 18 follows from (160) and (161). Proofs of Theorems 20, 22, 24 and 26. From (99),

∑_{n=0}^{∞} PDO(96n + 44)q^{n} ≡ 8f_{2} \frac{f_{6}^{2}}{f_{3}} (mod 2^{4}).  (162)

Replacing n by 8n + 3 in (62), we see that for all α, n ≥ 0,

PDO(4^{α}(96n + 44)) ≡ PDO(96n + 44) (mod 2^{4}).  (163)

In view of (8), (91), (162) and (163),

∑_{n=0}^{∞} PDO(4^{α}(24n + 11)q^{n} ≡ 8f_{2}ψ(q^{3}) (mod 2^{4}).  (164)

Theorem 20 follows from (164). From (100),

∑_{n=0}^{∞} PDO(192n + 152)q^{n} ≡ 8f_{1} \frac{f_{12}^{2}}{f_{6}} (mod 2^{4}).  (165)

Replacing n by 16n + 12 in (62), we see that for all α, n ≥ 0,

PDO(4^{α}(192n + 152)) ≡ PDO(192n + 152) (mod 2^{4}).  (166)

In view of (8), (93), (165) and (166),

∑_{n=0}^{∞} PDO(4^{α}(48n + 38)q^{n} ≡ 8f_{1}ψ(q^{6}) (mod 2^{4}).  (167)

Theorem 22 follows from (167). Theorem 24 follows from (9) and (104). Theorem 26 follows from (9) and (105).

The proofs of Corollaries 17, 19, 21, 23, 25 and 27 are similar to the proof of Corollary 12; hence we omit the details.

Proof of Theorem 1. Replacing n by 2n + 1 in (28),

PDO(2^{α+2}3^{β}(72n + 66)) ≡ 0 (mod 9).  (168)

Congruence (5) follows readily from (80) and (168). Replacing n by 4n + 3 in (28),

PDO(2^{α+2}3^{β}(144n + 138)) ≡ 0 (mod 9).  (169)

Congruence (6) follows readily from (83) and (169).

5 Acknowledgments

The authors would like to thank the anonymous referee for his/her valuable comments and suggestions. The second author was supported by CSIR Senior Research Fellowship (No. 09/039(0111)/2014-EMR-I).

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2010 Mathematics Subject Classification: Primary 11P83; Secondary 05A17. Keywords: partition with designated summand, congruence, theta function.

(Concerned with sequences \underline{A077285} and \underline{A102186}.)

Received October 25 2016; revised version received January 13 2017; January 23 2017. Published in Journal of Integer Sequences, January 28 2017.

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