Generalizing Recent Results of Kathiravan, Majumdar, Sangale, and Srinivas

on (ℓ, k)-regular Partitions

Manjil P. Saikia Mathematical and Physical Sciences Division School of Arts and Sciences Ahmedabad University Navrangpura, Ahmedabad, Pin 380009 India

              manjil.saikia@ahduni.edu.in

James A. Sellers Department of Mathematics and Statistics University of Minnesota Duluth Duluth, MN 55812 USA

                  jsellers@d.umn.edu

Abstract

In two recent papers, Kathiravan, Srinivas, and Sangale and Kathiravan, Sangale, and Majumdar proved several congruences satisfied by (ℓ, k)-regular partitions; that is, partitions whose parts are not divisible by integers ℓ or k. In this short note, we generalize some of their results, and prove two infinite family of congruences modulo 2 and one infinite family of congruences modulo 12. Our proofs involve only elementary techniques.

1 Introduction

An integer partition of n is a non-increasing sequence λ = (λ_{1}, λ_{2}, …, λ_{k}) such that λ_{i} ≥ λ_{i+1}

for all 1 ≤ i ≤ k −1 and ∑_{i=1}^{k} λ_{i} = n. The λ_{i}’s are called parts of the partition λ. For instance,

there are 7 partitions of 5, and they are

(5), (4, 1), (3, 2), (3, 1, 1), (2, 2, 1), (2, 1, 1, 1), and (1, 1, 1, 1, 1).

We let p(n) denote the number of partitions of n, and we have the following generating function, due to Euler:

∑_{n≥0} p(n)q^{n} = \frac{1}{∏_{i≥1}(1 − q^{i})} = \frac{1}{(q; q)_{∞}} = \frac{1}{f_{1}} ,

where we have used the shorthand notation

(a; q)_{∞} := ∏_{i≥0}(1 − aq^{i}) and f_{k} := (q^{k}; q^{k})_{∞}.

Partitions have been very well-studied since the time of Euler, and prominent mathematicians such as Euler, Jacobi, and Ramanujan are associated with the study of partitions. For a general overview of the area, we refer to the book of Andrews [2]. One area of study in partitions is that of generalized partition functions. For instance, if we do not allow parts divisible by ℓ, then the resulting partitions are called ℓ-regular partitions (see [11, \underline{A000009}, \underline{A000726}, \underline{A001935}, \underline{A035959}, \underline{A219601}]). We let b_{ℓ}(n) denote the number of ℓ-regular partitions of n, and it is known that

∑_{n≥0} b_{ℓ}(n)q^{n} = \frac{f_{ℓ}}{f_{1}} .

This class of partitions is very well-studied; see for instance the work of Andrews, Hirschhorn, and Sellers [1], Cui and Gu [5], Wang [12], Du and Tang [6], and the references therein.

Generalizing this idea further, Kathiravan, Srinivas, and Sangale [9] introduced the concept of (ℓ, k)-regular partitions, which are partitions such that no part is divisible by ℓ or k, where (ℓ, k) = 1. They let b_{ℓ,k}(n) denote the number of such partitions of n and gave the generating function

∑_{n≥0} b_{ℓ,k}(n)q^{n} = \frac{f_{ℓ}f_{k}}{f_{1}f_{ℓk}} .  (1)

Kathiravan, Srinivas, and Sangale [9] proved congruences modulo 2 for (3, 8) and (4, 7)- regular partitions and modulo 8, 9 and 12 for (4, 9)-regular partitions. Kathiravan, Sangale, and Majumdar [8] proved congruences modulo 2 for (2, 7), (4, 11), and (5, 8)-regular partitions and modulo 4 for (4, 5)-regular partitions. Naika, Hemanthkumar, and Sumanth Bharadwaj [10] proved congruences modulo 2 for (3, 5)-regular partitions.

Our goal in this note is to generalize the recent congruence results of Kathiravan, Majumdar, Sangale, and Srinivas on (ℓ, k)-regular partitions. For example, Kathiravan, Sangale, and Majumdar [8, Theorem 3.2] proved the following congruences: for all n ≥ 0, we have

b_{4,11}(22n + i) ≡ 0 (mod 2), i ∈ {2, 8, 16, 18, 20}.

The result below strengthens their result.

Theorem 1. Let p ≥ 5 be a prime, and r is an integer such that 8r + 1 is a quadratic nonresidue modulo p. Then we have, for all n ≥ 0,

b_{4,p}(pn + r) ≡ 0 (mod 2).

A similar result also holds for the functions b_{2,p}(n).

Theorem 2. Let p ≥ 5 be a prime, and r is an integer such that 24r + 1 is a quadratic nonresidue modulo p. Then we have, for all n ≥ 0,

b_{2,p}(pn + r) ≡ 0 (mod 2).

Theorems 1 and 2 are proved in Section 2 using elementary techniques. In Section 3, we provide a proof of the following infinite family of congruences (related to [11, \underline{A187020}]) which significantly generalizes a result in [9].

Theorem 3. For all n ≥ 0 and α ≥ 3, we have

b_{4,9}(2^{α}n + 6 · 2^{α−3} + 1) ≡ 0 (mod 12).  (2)

Remark 4. Kathiravan, Srinivas, and Sangale [9, Theorem 1.3] proved the α = 3 case of (2).

All of the proofs which appear below are elementary in nature, relying on well–known q–series identities and generating function manipulations.

2 Proof of Theorems 1 and 2

In order to prove Theorems 1 and 2, we simply need two well–known identities that date back to Euler and Jacobi.

Lemma 5 (Euler). We have

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

Proof. See [7, Section 1.6].

Lemma 6 (Jacobi). We have

f_{1}^{3} = ∑_{k=0}^{∞}(−1)^{k}(2k + 1)q^{k(k+1)/2}.

Proof. See [7, Section 1.7].

With the above in hand, we can now prove Theorems 1 and 2 relatively quickly.

Proof of Theorem 1. From (1) and Lemma 6, we have

∑_{n≥0} b_{4,p}(n)q^{n} = \frac{f_{4}f_{p}}{f_{1}f_{4p}}

≡ \frac{f_{1}^{3}}{f_{p}^{3}} (mod 2)

≡ \frac{1}{f_{p}^{3}} ∑_{k=0}^{∞} q^{k(k+1)/2} (mod 2).

Since f_{p}^{3} is a function of q^{p}, and since we are interested in arguments of the form pn + r where r is not divisible by p, we can focus our attention on the f_{1}^{3}. In particular, we need to ask whether

pn + r = \frac{k(k + 1)}{2}

for some nonnegative integer k. This is equivalent to asking whether 8(pn+r)+1 = (2k+1)^{2} by completing the square. This would mean that 8r + 1 ≡ (2k + 1)^{2} (mod p). However, this cannot be the case since we defined r such that 8r + 1 is a quadratic nonresidue modulo p. This completes the proof.

We next prove Theorem 2 in very similar fashion.

Proof of Theorem 2. From (1) and Lemma 5, we know

∑_{n≥0} b_{2,p}(n)q^{n} = \frac{f_{2}f_{p}}{f_{1}f_{2p}}

≡ \frac{f_{1}}{f_{p}} (mod 2)

≡ \frac{1}{f_{p}} ∑_{k=−∞}^{∞} q^{(3k^{2}−k)/2} (mod 2).

Since f_{p} is a function of q^{p}, and since we are interested in arguments of the form pn + r where r is not divisible by p, we can focus our attention on the f_{1}. In particular, we need to ask whether

pn + r = \frac{(3k^{2} − k)}{2}

for some nonnegative integer k. This is equivalent to asking whether 24(pn+r)+1 = (6k+1)^{2} by completing the square. This would mean that 24r + 1 ≡ (6k + 1)^{2} (mod p). However, this cannot be the case since we defined r such that 24r +1 is a quadratic nonresidue modulo p.

3 An elementary proof of Theorem 3

Our goal in this section is to prove Theorem 3 as noted above. In order to complete the proof, we need a few more q–series dissection results that appear in the literature which we list here.

Lemma 7. We have

\frac{f_{9}}{f_{1}} = \frac{f_{12}^{3}f_{18}}{f_{2}^{2}f_{6}f_{36}} + q \frac{f_{4}^{2}f_{6}f_{36}}{f_{2}^{3}f_{12}} .

Proof. See [13, Lemma 3.5].

Lemma 8. We have

\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}} .

Proof. See [4, Eq. (49)]. (Also see [11, \underline{A273845}].)

Lemma 9. We have

\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}} .

Proof. See [14, Eq. (3.29)]. (Also see [11, \underline{A328547}].)

Lemma 10. We have

\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}} ,
\frac{1}{f_{1}^{8}} = \frac{f_{4}^{28}}{f_{2}^{28}f_{8}^{8}} + 8q \frac{f_{4}^{16}}{f_{2}^{24}} + 16q^{2} \frac{f_{4}^{4}f_{8}^{8}}{f_{2}^{20}} ,
\frac{1}{f_{1}^{12}} = \frac{f_{4}^{42}}{f_{2}^{42}f_{8}^{12}} + 12q \frac{f_{4}^{30}}{f_{2}^{38}f_{8}^{4}} + 48q^{2} \frac{f_{4}^{18}f_{8}^{4}}{f_{2}^{34}} + 64q^{3} \frac{f_{4}^{6}f_{8}^{12}}{f_{2}^{30}} .

Proof. See [3, Lemma 2.2].

Lemma 11. We have

\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}} .

Proof. See [4, Lemma 2.2]. (Also see [11, \underline{A318026}].)

Our proof relies on various generating function dissections and manipulations (utilizing the lemmas above), along with a very straightforward application of mathematical induction.

We begin by completing a 2–dissection to obtain the generating function for b_{4,9}(2n + 1). (Note that all of the arguments in the statement of Theorem 3 are odd, so we do not need the generating function for b_{4,9}(2n), although we could easily obtain it as well.)

Theorem 12. We have

∑_{n≥0} b_{4,9}(2n + 1)q^{n} = \frac{f_{2}^{3}f_{3}}{f_{1}^{3}f_{6}} .

Proof. Using Lemma 7, we can rewrite the generating function for b_{4,9}(n) in the following way:

∑_{n≥0} b_{4,9}(n)q^{n} = \frac{f_{4}f_{9}}{f_{1}f_{36}}

= \frac{f_{4}}{f_{36}} (\frac{f_{12}^{3}f_{18}}{f_{2}^{2}f_{6}f_{36}} + q \frac{f_{4}^{2}f_{6}f_{36}}{f_{2}^{3}f_{12}} )

which allows us to see immediately that

∑_{n≥0} b_{4,9}(2n + 1)q^{2n+1} = q \frac{f_{4}}{f_{36}} · \frac{f_{4}^{2}f_{6}f_{36}}{f_{2}^{3}f_{12}}

or

∑_{n≥0} b_{4,9}(2n + 1)q^{n} = \frac{f_{2}}{f_{18}} · \frac{f_{2}^{2}f_{3}f_{18}}{f_{1}^{3}f_{6}} = \frac{f_{2}^{3}f_{3}}{f_{1}^{3}f_{6}} .

From here, we can take care of the modulo 3 portion of Theorem 3 almost immediately. Namely, note that

∑_{n≥0} b_{4,9}(2n + 1)q^{n} = \frac{f_{2}^{3}f_{3}}{f_{1}^{3}f_{6}}

≡ \frac{f_{6}f_{3}}{f_{3}f_{6}} (mod 3)

= 1.

This means that, for all n ≥ 1,

b_{4,9}(2n + 1) ≡ 0 (mod 3).  (3)

(Note that the 1 that appears at the end of the string of congruences above arises because b_{4,9}(1) = 1.) Hence, the mod 3 portion of all of the congruences associated with Theorem 3 is now proven.

Thus, we need to focus our attention now on the mod 4 aspects of the congruences in Theorem 3. In order to complete the proof of Theorem 3, we need a few specific results mod 4. We now prove those results, which serve as the building blocks of our elementary proof of Theorem 3.

Theorem 13. For all n ≥ 0, we have b_{4,9}(16n + 1) ≡ b_{4,9}(4n + 1) (mod 4).

Proof. Using Theorem 12 and Lemma 8, we know that

∑_{n≥0} b_{4,9}(2n + 1)q^{n} = \frac{f_{2}^{3}f_{3}}{f_{1}^{3}f_{6}}

= \frac{f_{2}^{3}}{f_{6}} (\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}} )

which means

∑_{n≥0} b_{4,9}(2(2n) + 1)q^{2n} = \frac{f_{2}^{3}}{f_{6}} · \frac{f_{4}^{6}f_{6}^{3}}{f_{2}^{9}f_{12}^{2}}

or

∑_{n≥0} b_{4,9}(4n + 1)q^{n} = \frac{f_{1}^{3}}{f_{3}} · \frac{f_{2}^{6}f_{3}^{3}}{f_{1}^{9}f_{6}^{2}} = \frac{f_{2}^{6}f_{3}^{2}}{f_{1}^{6}f_{6}^{2}} .  (4)

We now 2–dissect this expression again in order to find a result for the generating function for b_{4,9}(8n + 1) in the following way:

∑_{n≥0} b_{4,9}(4n + 1)q^{n} = \frac{f_{2}^{6}f_{3}^{2}}{f_{1}^{6}f_{6}^{2}}

= \frac{f_{2}^{6}}{f_{6}^{2}} · \frac{f_{3}^{2}}{f_{1}^{2}} · \frac{1}{f_{1}^{4}} .

Using Lemma 9 and Lemma 10, we see that

∑_{n≥0} b_{4,9}(8n + 1)q^{2n} ≡ \frac{f_{2}^{6}}{f_{6}^{2}} · \frac{f_{4}^{18}f_{6}f_{12}^{2}}{f_{2}^{19}f_{8}^{5}f_{24}} (mod 4)

or

∑_{n≥0} b_{4,9}(8n + 1)q^{n} ≡ \frac{f_{1}^{6}}{f_{3}^{2}} · \frac{f_{2}^{18}f_{3}f_{6}^{2}}{f_{1}^{19}f_{4}^{5}f_{12}} (mod 4)  (5)

= \frac{f_{2}^{18}f_{6}^{2}}{f_{1}^{13}f_{3}f_{4}^{5}f_{12}} .

We now perform one additional 2–dissection in order to obtain an expression, modulo 4, for the generating function for b_{4,9}(16n + 1). In order to do so, we rewrite the above as

∑_{n≥0} b_{4,9}(8n + 1)q^{n} ≡ \frac{f_{2}^{18}f_{6}^{2}}{f_{4}^{5}f_{12}} · \frac{1}{f_{1}f_{3}} · \frac{1}{f_{1}^{12}} (mod 4).

Lemma 10 and Lemma 11 can now be utilized to show that

∑_{n≥0} b_{4,9}(8(2n) + 1)q^{2n} ≡ \frac{f_{2}^{18}f_{6}^{2}}{f_{4}^{5}f_{12}} · \frac{f_{8}^{2}f_{12}^{5}}{f_{2}^{2}f_{4}f_{6}^{4}f_{24}^{2}} · \frac{f_{4}^{42}}{f_{2}^{42}f_{8}^{12}} (mod 4)

= \frac{f_{4}^{36}f_{12}^{4}}{f_{2}^{26}f_{6}^{2}f_{8}^{10}f_{24}^{2}}

which means

∑_{n≥0} b_{4,9}(16n + 1)q^{n} ≡ \frac{f_{2}^{36}f_{6}^{4}}{f_{1}^{26}f_{3}^{2}f_{4}^{10}f_{12}^{2}} (mod 4).  (6)

The goal now is to show that the expressions in (4) and (6) are congruent to one another modulo 4. The only tool necessary to prove this is the fact that f_{1}^{4} ≡ f_{2}^{2} (mod 4) which easily follows from the Binomial Theorem. We have

∑_{n≥0} b_{4,9}(16n + 1)q^{n} ≡ \frac{f_{2}^{36}f_{6}^{4}}{f_{1}^{26}f_{3}^{2}f_{4}^{10}f_{12}^{2}} (mod 4)

≡ \frac{f_{2}^{36}f_{12}^{2}}{f_{1}^{26}f_{3}^{2}f_{4}^{10}f_{12}^{2}} (mod 4)

≡ \frac{f_{2}^{36}}{f_{1}^{6}f_{2}^{10}f_{3}^{2}f_{2}^{20}} (mod 4)

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

= \frac{f_{2}^{6}f_{3}^{2}}{f_{1}^{6}f_{3}^{4}}

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

≡ ∑_{n≥0} b_{4,9}(4n + 1)q^{n} (mod 4).

Theorem 13 will serve as the “engine” for a proof of Theorem 3 by mathematical induction. In order to complete that proof, we need two base cases. We now prove those two individual congruences modulo 4.

Theorem 14. For all n ≥ 0, we have b_{4,9}(16n + 13) ≡ 0 (mod 4).

Proof. We note from (4) that

∑_{n≥0} b_{4,9}(4n + 1)q^{n} = \frac{f_{2}^{6}}{f_{6}^{2}} (\frac{f_{3}}{f_{1}^{3}} )^{2} .

Applying Lemma 8, we obtain

∑_{n≥0} b_{4,9}(4n + 1)q^{n} = \frac{f_{2}^{6}}{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}} )^{2}

= \frac{f_{2}^{6}}{f_{6}^{2}} (\frac{f_{4}^{12}f_{6}^{6}}{f_{2}^{18}f_{12}^{4}} + 9q^{2} \frac{f_{4}^{4}f_{6}^{2}f_{12}^{4}}{f_{2}^{14}} + 6q \frac{f_{4}^{8}f_{6}^{4}}{f_{2}^{16}} ) .

Extracting the terms involving the odd powers of q, dividing by q, and replacing q^{2} by q throughout, we obtain

∑_{n≥0} b_{4,9}(8n + 5)q^{n} = 6\frac{f_{2}^{8}f_{3}^{2}}{f_{1}^{10}} = 6f_{2}^{8} · (\frac{f_{3}^{2}}{f_{1}^{2}} ) · (\frac{1}{f_{1}^{8}} ) .  (7)

Applying Lemma 9 and the second equation of Lemma 10 to (7), we see that

∑_{n≥0} b_{4,9}(8n + 5)q^{n} ≡ 2f_{2}^{8} (\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}} ) \frac{f_{4}^{28}}{f_{2}^{28}f_{8}^{8}} (mod 4).

Now, from the above equation, extracting the terms involving odd powers of q, dividing both sides by q and then replacing q^{2} by q, we obtain

∑_{n≥0} b_{4,9}(16n + 13)q^{n} ≡ 0 (mod 4).

This implies the theorem.

Theorem 15. For all n ≥ 0, we have b_{4,9}(32n + 25) ≡ 0 (mod 4).

Proof. From (5), we have

∑_{n≥0} b_{4,9}(8n + 1)q^{n} ≡ \frac{f_{2}^{18}f_{6}^{2}}{f_{4}^{5}f_{12}} · \frac{1}{f_{1}f_{3}} · \frac{1}{f_{1}^{12}} (mod 4).

Using Lemma 11 and the third equation of Lemma 10 in the above, we obtain

∑_{n≥0} b_{4,9}(8n + 1)q^{n} ≡ \frac{f_{2}^{18}f_{6}^{2}}{f_{4}^{5}f_{12}} (\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}} ) \frac{f_{4}^{42}}{f_{2}^{42}f_{8}^{12}} (mod 4).

Extracting the terms involving the odd powers of q, dividing throughout by q and then replacing q^{2} by q in the above equation, we arrive at

∑_{n≥0} b_{4,9}(16n + 9)q^{n} ≡ \frac{f_{2}^{42}}{f_{1}^{28}f_{6}^{2}f_{4}^{14}} = \frac{f_{2}^{42}}{f_{6}^{2}f_{4}^{14}} · (\frac{1}{f_{1}^{12}} )^{2} · \frac{1}{f_{1}^{4}} (mod 4).

It is easy to see that, applying Lemma 10 in the above equation, we obtain

∑_{n≥0} b_{4,9}(16n + 9)q^{n} ≡ \frac{f_{2}^{42}}{f_{6}^{2}f_{4}^{14}} (\frac{f_{4}^{42}}{f_{2}^{42}f_{8}^{12}} )^{2} \frac{f_{4}^{14}}{f_{2}^{14}f_{8}^{4}} (mod 4).

Since the right-hand-side of the above equation involves only even powers of q, we readily obtain

∑_{n≥0} b_{4,9}(32n + 25)q^{n} ≡ 0 (mod 4).

Our result now follows.

Proof of Theorem 3. As a reminder, we are trying to prove that, for all n ≥ 0 and α ≥ 3, we have

b_{4,9}(2^{α}n + 6 · 2^{α−3} + 1) ≡ 0 (mod 12).

As was noted above, the α = 3 case of the above was proven by Kathiravan, Srinivas, and Sangale [9, Theorem 1.3]. Our goal is to now prove the remaining cases via induction. Also, thanks to (3), it is enough to prove the congruence modulo 4.

First, note that the α = 4 case corresponds to the arithmetic progression 2^{4}n+6·2^{1} +1 = 16n + 13, and Theorem 14 provides the result in this case. Next, the α = 5 case corresponds to the arithmetic progression 2^{5}n + 6 · 2^{2} + 1 = 32n + 25, and Theorem 15 handles this case. We use these two specific cases as base cases for arguments by induction.

We now prove our result in two stages, once for even α and once for odd α. First, let α = 2β for some integer β. We assume that, for some β ≥ 2 and all n ≥ 0,

b_{4,9}(2^{2β}n + 6 · 2^{2β−3} + 1) ≡ 0 (mod 4).

We then want to prove that, for all n ≥ 0,

b_{4,9}(2^{2β+2}n + 6 · 2^{2β−1} + 1) ≡ 0 (mod 4).

Note that

b_{4,9}(2^{2β+2}n + 6 · 2^{2β−1} + 1) = b_{4,9}(16(2^{2β−2}n + 6 · 2^{2β−5}) + 1)
≡ b_{4,9}(4(2^{2β−2}n + 6 · 2^{2β−5}) + 1) (mod 4) from Theorem 13
= b_{4,9}(2^{2β}n + 6 · 2^{2β−3}) + 1)
≡ 0 (mod 4)

by the induction hypothesis. This completes the proof for even α.

Now, let α = 2β + 1 for some integer β. We assume that, for some β ≥ 1 and all n ≥ 0,

b_{4,9}(2^{2β+1}n + 6 · 2^{2β−2} + 1) ≡ 0 (mod 4).

We then want to prove that, for all n ≥ 0,

b_{4,9}(2^{2β+3} + 6 · 2^{2β} + 1) ≡ 0 (mod 4).

This is readily seen, as before

b_{4,9}(2^{2β+3} + 6 · 2^{2β} + 1) = b_{4,9}(16(2^{2β−1}n + 6 · 2^{2β−4}) + 1)

≡ b_{4,9}(4(2^{2β−1}n + 6 · 2^{2β−4}) + 1) (mod 4) from Theorem 13

= b_{4,9}(2^{2β+1}n + 6 · 2^{2β−2}) + 1)
≡ 0 (mod 4).

4 Acknowledgments

The first author is partially supported by a Start-Up Grant from Ahmedabad University, India (Reference No. URBSASI24A5).

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2020 Mathematics Subject Classification: Primary 11P83; Secondary 11P81, 05A17. Keywords: partition, congruence, generating function, dissection.

(Concerned with sequences \underline{A000009}, \underline{A000726}, \underline{A001935}, \underline{A035959}, \underline{A187020}, \underline{A219601}, \underline{A273845}, \underline{A318026}, and \underline{A328547}.)

Received August 1 2025; revised versions received August 2 2025; October 17 2025. Published in Journal of Integer Sequences, October 19 2025.

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