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\begin{center}
\vskip 1cm{\LARGE\bf Generalizing Recent Results of \\
Kathiravan, Majumdar,  Sangale, and Srinivas\\
\vskip .08in
on $(\ell, k)$-regular Partitions
}
\vskip 1cm
\large
Manjil P.\ Saikia\\
Mathematical and Physical Sciences Division\\
School of Arts and Sciences\\
Ahmedabad University\\
Navrangpura, Ahmedabad, Pin 380009\\
India\\
\href{manjil.saikia@ahduni.edu.in}{\tt manjil.saikia@ahduni.edu.in} \\
\ \\
James A.\ Sellers\\
Department of Mathematics and Statistics\\
University of Minnesota Duluth\\
Duluth, MN 55812\\
USA\\
\href{jsellers@d.umn.edu}{\tt jsellers@d.umn.edu} \\
\end{center}

\vskip .2 in
\begin{abstract}
    In two recent papers, Kathiravan, Srinivas, and Sangale and Kathiravan, Sangale, and Majumdar proved several congruences satisfied by $(\ell,k)$-regular partitions; that is, partitions whose parts are not divisible by integers $\ell$ 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.
\end{abstract}


\section{Introduction}\label{sec:intro}

An integer partition of $n$ is a non-increasing sequence $\lambda=(\lambda_1, \lambda_2, \ldots, \lambda_k)$ such that~$\lambda_i\geq \lambda_{i+1}$ for all $1\leq i \leq k-1$ and $\sum\limits_{i=1}^k\lambda_i=n$. The $\lambda_i$'s are called parts of the partition $\lambda$. For instance, there are $7$ partitions of $5$, and they are
\[
(5), (4,1), (3,2), (3,1,1), (2,2,1), (2,1,1,1), \text{ and } (1,1,1,1,1).
\]
We let $p(n)$ denote the number of partitions of $n$, and we have the following generating function, due to Euler:
\[
\sum_{n\geq 0}p(n)q^n=\frac{1}{\prod_{i\geq 1}(1-q^i)}=\frac{1}{(q;q)_\infty}=\frac{1}{f_1},
\]
where we have used the shorthand notation
\[
(a;q)_\infty:=\prod_{i\geq 0}(1-aq^i) \quad \text{and} \quad f_k:=(q^k;q^k)_\infty.
\]

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 \cite{gea1}. One area of study in partitions is that of generalized partition functions. For instance, if we do not allow parts divisible by $\ell$, then the resulting partitions are called $\ell$-regular partitions (see \cite[\seqnum{A000009}, \seqnum{A000726}, \seqnum{A001935}, \seqnum{A035959}, \seqnum{A219601}]{OEIS}). We let $b_\ell(n)$ denote the number of $\ell$-regular partitions of $n$, and it is known that
\[
\sum_{n\geq 0}b_\ell(n)q^n=\frac{f_\ell}{f_1}.
\]
This class of partitions is very well-studied; see for instance the work of Andrews, Hirschhorn, and Sellers \cite{AndHirSel}, Cui and Gu \cite{CuiGu}, Wang \cite{WangL}, Du and Tang \cite{DuTang}, and the references therein.


Generalizing this idea further, Kathiravan, Srinivas, and Sangale \cite{KathiravanSrinivasSangale} introduced the concept of $(\ell,k)$-regular partitions, which are partitions such that no part is divisible by $\ell$ or $k$, where $(\ell, k)=1$. They let $b_{\ell, k}(n)$ denote the number of such partitions of $n$ and gave the  generating function
\begin{equation}\label{eq:gf-blm}
    \sum_{n\geq 0}b_{\ell, k}(n)q^n=\frac{f_\ell f_k}{f_1f_{\ell k}}.
\end{equation}
Kathiravan, Srinivas, and Sangale \cite{KathiravanSrinivasSangale} 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 \cite{KathiravanSangaleMajumdar} 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 \cite{NaikaHemanthkumarSumanth} 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 $(\ell, k)$-regular partitions.  For example, 
Kathiravan, Sangale, and Majumdar \cite[Theorem 3.2]{KathiravanSangaleMajumdar} proved the following congruences: for all $n\geq 0$, we have
    \[
    b_{4,11}(22n+i) \equiv 0 \pmod 2, \quad i\in \{ 2, 8, 16, 18, 20\}.
    \]
The result below strengthens their result. 

\begin{theorem}\label{thm:modp}
    Let $p\geq 5$ be a prime, and $r$ is an integer such that $8r+1$ is a quadratic nonresidue modulo $p$.  Then we have, for all $n\geq 0$,
    \[
    b_{4,p}(pn+r)\equiv 0 \pmod 2.
    \]
\end{theorem}

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

\begin{theorem}\label{thm:modp1}
        Let $p\geq 5$ be a prime, and $r$ is an integer such that $24r+1$ is a quadratic nonresidue modulo $p$.  Then we have, for all $n\geq 0$,
    \[
    b_{2,p}(pn+r)\equiv 0 \pmod 2.
    \]
\end{theorem}


Theorems \ref{thm:modp} and \ref{thm:modp1} are proved in Section \ref{sec:proof1} using elementary techniques.  In Section \ref{sec:b49_mod_12}, we provide a proof of the following infinite family of congruences (related to \cite[\seqnum{A187020}]{OEIS}) which significantly generalizes a result in \cite{KathiravanSrinivasSangale}.

\begin{theorem}
\label{thm_b49_mod_12}
    For all $n\geq 0$ and $\alpha \geq 3$, we have
    \begin{equation}\label{eq:al}
        b_{4,9}(2^\alpha n+6\cdot 2^{\alpha -3}+1)\equiv 0\pmod {12}.
    \end{equation}
\end{theorem}
\begin{remark}
 Kathiravan, Srinivas, and Sangale \cite[Theorem 1.3]{KathiravanSrinivasSangale} proved the $\alpha=3$ case of \eqref{eq:al}.
\end{remark}

All of the proofs which appear below are elementary in nature, relying on well--known $q$--series identities and generating function manipulations.  
\section{Proof of Theorems \ref{thm:modp} and \ref{thm:modp1}}
\label{sec:proof1}

In order to prove Theorems \ref{thm:modp} and \ref{thm:modp1}, we simply need two well--known identities that date back to Euler and Jacobi.  

\begin{lemma}[Euler]
\label{Euler_PNT}
    We have 
    $$f_1 = \sum_{k=-\infty}^\infty (-1)^kq^{(3k^2-k)/2}.
    $$
\end{lemma}
\begin{proof}
See \cite[Section 1.6]{Power}.
\end{proof}

\begin{lemma}[Jacobi]
\label{Jacobi_f1cubed}
We have 
$$
f_1^3 = \sum_{k=0}^\infty (-1)^k (2k+1)q^{k(k+1)/2}.
$$
\end{lemma}
\begin{proof}
See \cite[Section 1.7]{Power}.
\end{proof}


With the above in hand, we can now prove Theorems \ref{thm:modp} and \ref{thm:modp1} relatively quickly.  


\begin{proof}[Proof of Theorem \ref{thm:modp}]
From \eqref{eq:gf-blm} and Lemma \ref{Jacobi_f1cubed}, we have
\begin{align*}
\sum_{n\geq 0}b_{4, p}(n)q^n
&= 
\frac{f_4 f_{p}}{f_1f_{4p}} \\
&\equiv
\frac{f_1^3}{f_{p}^3}\pmod{2} \\
&\equiv 
\frac{1}{f_{p}^3}\sum_{k=0}^\infty q^{k(k+1)/2} \pmod{2}.    
\end{align*}
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 $8r+1 \equiv (2k+1)^2 \pmod{p}$.  However, this cannot be the case since we defined $r$ such that $8r+1$ is a quadratic nonresidue modulo $p$.  This completes the proof.  
\end{proof}

We next prove Theorem \ref{thm:modp1} in very similar fashion. 

    \begin{proof}[Proof of Theorem \ref{thm:modp1}]
    From \eqref{eq:gf-blm} and Lemma \ref{Euler_PNT}, we know
    \begin{align*}
\sum_{n\geq 0}b_{2, p}(n)q^n
&= 
\frac{f_2 f_{p}}{f_1f_{2p}} \\
&\equiv
\frac{f_1}{f_{p}}\pmod{2} \\
&\equiv 
\frac{1}{f_{p}}\sum_{k=-\infty}^\infty q^{(3k^2-k)/2} \pmod{2}.    
\end{align*}
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 $24r+1 \equiv (6k+1)^2 \pmod{p}$.  However, this cannot be the case since we defined $r$ such that $24r+1$ is a quadratic nonresidue modulo $p$.
\end{proof}    

\section{An elementary proof of Theorem \ref{thm_b49_mod_12} }
\label{sec:b49_mod_12}  
Our goal in this section is to prove Theorem \ref{thm_b49_mod_12} 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. 

\begin{lemma}
\label{lemma:f9_over_f1}
We have 
$$
\frac{f_9}{f_1} = \frac{f_{12}^3f_{18}}{f_2^2f_6f_{36}} + q\frac{f_4^2f_6f_{36}}{f_2^3f_{12}}.
$$
\end{lemma}
\begin{proof}
See \cite[Lemma 3.5]{XiaYao}.
\end{proof}

\begin{lemma}
\label{lemma:f3_over_f1cubed}
We have 
$$
\frac{f_3}{f_1^3} = \frac{f_4^6f_6^3}{f_2^9f_{12}^2} + 3q\frac{f_4^2f_6f_{12}^2}{f_2^7}.
$$
\end{lemma} 
\begin{proof}
  See \cite[Eq.\ (49)]{NDB}. (Also see \cite[\seqnum{A273845}]{OEIS}.)
\end{proof}

\begin{lemma}
\label{f3squared_over_f1squared}
We have 
$$
\frac{f_3^2}{f_1^2} = \frac{f_4^4f_6f_{12}^2}{f_2^5f_8f_{24}} +2q\frac{f_4f_6^2f_8f_{24}}{f_2^4f_{12}}.
$$
\end{lemma}
\begin{proof}
See \cite[Eq.\ (3.29)]{XiaYao2}. (Also see \cite[\seqnum{A328547}]{OEIS}.)
\end{proof}

\begin{lemma}
\label{1_over_f1to4}
We have 
\begin{align*}
\frac{1}{f_1^4} 
&= 
\frac{f_4^{14}}{f_2^{14}f_8^4} + 4q\frac{f_4^2f_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_4^4f_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_4^6f_8^{12}}{f_2^{30}}.
\end{align*} 
\end{lemma}
\begin{proof}
See \cite[Lemma 2.2]{SellersJMAA}.  
\end{proof}
\begin{lemma}
\label{1_over_f1f3}
We have     
$$
\frac{1}{f_1f_3} = \frac{f_8^2f_{12}^5}{f_2^2f_4f_6^4f_{24}^2} + q\frac{f_4^5f_{24}^2}{f_2^4f_6^2f_8^2f_{12}}.
$$
\end{lemma}
\begin{proof}
  See \cite[Lemma 2.2]{NDB}.  (Also see \cite[\seqnum{A318026}]{OEIS}.)
\end{proof}

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 \ref{thm_b49_mod_12} are odd, so we do not need the generating function for $b_{4,9}(2n)$, although we could easily obtain it as well.)
\begin{theorem}
\label{thm:2n1}
We have 
$$
\sum_{n\geq 0} b_{4,9}(2n+1)q^n = \frac{f_2^3f_3}{f_1^3f_6}.
$$
\end{theorem}
\begin{proof}
Using Lemma \ref{lemma:f9_over_f1}, we can rewrite the generating function for $b_{4,9}(n)$ in the following way: 
\begin{align*}
\sum_{n\geq 0} b_{4,9}(n)q^n 
&= 
\frac{f_4f_9}{f_1f_{36}} \\
&= \frac{f_4}{f_{36}} \left( \frac{f_{12}^3f_{18}}{f_2^2f_6f_{36}} + q\frac{f_4^2f_6f_{36}}{f_2^3f_{12}}\right) 
\end{align*}
which allows us to see immediately that 
$$
\sum_{n\geq 0} b_{4,9}(2n+1)q^{2n+1} 
= q\frac{f_4}{f_{36}}\cdot \frac{f_4^2f_6f_{36}}{f_2^3f_{12}}
$$
or 
$$
\sum_{n\geq 0} b_{4,9}(2n+1)q^{n} 
= \frac{f_2}{f_{18}}\cdot \frac{f_2^2f_3f_{18}}{f_1^3f_{6}} = \frac{f_2^3f_3}{f_1^3f_{6}}. 
$$
\end{proof}
From here, we can take care of the modulo 3 portion of Theorem \ref{thm_b49_mod_12} almost immediately.  Namely, note that 
\begin{align*} 
\sum_{n\geq 0} b_{4,9}(2n+1)q^{n} 
&= 
\frac{f_2^3f_3}{f_1^3f_{6}} \\
&\equiv 
\frac{f_6f_3}{f_3f_{6}} \pmod{3} \\
&= 1.
\end{align*}
This means that, for all $n\geq 1$, 
\begin{equation}\label{eqmod3}
b_{4,9}(2n+1) \equiv 0 \pmod{3}.\end{equation}  (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 \ref{thm_b49_mod_12} is now proven.  

Thus, we need to focus our attention now on the mod 4 aspects of the congruences in Theorem \ref{thm_b49_mod_12}. In order to complete the proof of Theorem \ref{thm_b49_mod_12}, 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 \ref{thm_b49_mod_12}.  

\begin{theorem}
\label{thm:internal_cong_mod4}
For all $n\geq 0$, we have $b_{4,9}(16n+1) \equiv b_{4,9}(4n+1) \pmod{4}$.
\end{theorem}
\begin{proof}
Using Theorem \ref{thm:2n1} and Lemma \ref{lemma:f3_over_f1cubed}, we know that 
\begin{align*}
\sum_{n\geq 0} b_{4,9}(2n+1)q^n 
&= 
\frac{f_2^3f_3}{f_1^3f_6} \\
&= 
\frac{f_2^3}{f_6}\left( \frac{f_4^6f_6^3}{f_2^9f_{12}^2} + 3q\frac{f_4^2f_6f_{12}^2}{f_2^7} \right)
\end{align*}
which means 
$$
\sum_{n\geq 0} b_{4,9}(2(2n)+1)q^{2n} 
=
\frac{f_2^3}{f_6}\cdot \frac{f_4^6f_6^3}{f_2^9f_{12}^2} 
$$
or 
\begin{equation}
\label{genfn4n1} 
\sum_{n\geq 0} b_{4,9}(4n+1)q^{n} 
=
\frac{f_1^3}{f_3}\cdot \frac{f_2^6f_3^3}{f_1^9f_{6}^2} 
= 
\frac{f_2^6f_3^2}{f_1^6f_6^2}.   
\end{equation}
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: 
\begin{align*}
\sum_{n\geq 0} b_{4,9}(4n+1)q^{n} 
&= 
\frac{f_2^6f_3^2}{f_1^6f_6^2} \\
&= 
\frac{f_2^6}{f_6^2}\cdot \frac{f_3^2}{f_1^2}\cdot\frac{1}{f_1^4}.   
\end{align*}
Using Lemma \ref{f3squared_over_f1squared} and Lemma \ref{1_over_f1to4},  we see that 
$$
\sum_{n\geq 0} b_{4,9}(8n+1)q^{2n} 
\equiv 
\frac{f_2^6}{f_6^2}\cdot \frac{f_4^{18}f_6f_{12}^2}{f_2^{19}f_8^5f_{24}} \pmod{4}
$$
or 
\begin{align}\label{eq8n+1}
\sum_{n\geq 0} b_{4,9}(8n+1)q^{n} 
&\equiv 
\frac{f_1^6}{f_3^2}\cdot \frac{f_2^{18}f_3f_{6}^2}{f_1^{19}f_4^5f_{12}} \pmod{4} \\\nonumber
&= 
\frac{f_2^{18}f_{6}^2}{f_1^{13}f_3f_4^5f_{12}}.
\end{align}
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 
\begin{align*}
\sum_{n\geq 0} b_{4,9}(8n+1)q^{n} 
&\equiv 
\frac{f_2^{18}f_{6}^2}{f_4^5f_{12}}\cdot \frac{1}{f_1f_3}\cdot \frac{1}{f_1^{12}} \pmod{4}.
\end{align*}
Lemma \ref{1_over_f1to4} and Lemma \ref{1_over_f1f3} can now be utilized to show that 
\begin{align*}
\sum_{n\geq 0} b_{4,9}(8(2n)+1)q^{2n} 
&\equiv 
\frac{f_2^{18}f_{6}^2}{f_4^5f_{12}}\cdot \frac{f_8^2f_{12}^5}{f_2^2f_4f_6^4f_{24}^2}\cdot \frac{f_4^{42}}{f_2^{42}f_8^{12}} \pmod{4} \\
&= 
\frac{f_4^{36}f_{12}^4}{f_2^{26}f_6^2f_8^{10}f_{24}^2}
\end{align*}
which means 
\begin{equation}
\label{genfn16n1} 
\sum_{n\geq 0} b_{4,9}(16n+1)q^{n} 
\equiv 
\frac{f_2^{36}f_{6}^4}{f_1^{26}f_3^2f_4^{10}f_{12}^2} \pmod{4}.
\end{equation}
The goal now is to show that the expressions in (\ref{genfn4n1}) and (\ref{genfn16n1}) are congruent to one another modulo 4.  The only tool necessary to prove this is the fact that $f_1^4 \equiv f_2^2 \pmod{4}$ which easily follows from the Binomial Theorem.  We have 
\begin{align*}
\sum_{n\geq 0} b_{4,9}(16n+1)q^{n} 
& \equiv 
\frac{f_2^{36}f_{6}^4}{f_1^{26}f_3^2f_4^{10}f_{12}^2} \pmod{4} \\
& \equiv 
\frac{f_2^{36}f_{12}^2}{f_1^{26}f_3^2f_4^{10}f_{12}^2} \pmod{4} \\
& \equiv 
\frac{f_2^{36}}{f_1^{6}f_2^{10}f_3^2f_2^{20}} \pmod{4} \\
& \equiv 
\frac{f_2^{6}}{f_1^{6}f_3^2} \pmod{4} \\
&= 
\frac{f_2^{6}f_3^2}{f_1^{6}f_3^4} \\
&\equiv 
\frac{f_2^{6}f_3^2}{f_1^{6}f_6^2} \pmod{4} \\
&\equiv 
\sum_{n\geq 0} b_{4,9}(4n+1)q^{n} \pmod{4}. 
\end{align*}
\end{proof}
Theorem \ref{thm:internal_cong_mod4} will serve as the ``engine'' for a proof of Theorem \ref{thm_b49_mod_12} by mathematical induction.  In order to complete that proof, we need two base cases.  We now prove those two individual congruences modulo 4.  

\begin{theorem}
\label{thm:16n13}
For all $n\geq 0$, we have $b_{4,9}(16n+13) \equiv 0 \pmod{4}.$ 
\end{theorem}
\begin{proof}
    We note from \eqref{genfn4n1} that
    \[
    \sum_{n\geq 0}b_{4,9}(4n+1)q^n=\frac{f_2^6}{f_6^2}\left(\frac{f_3}{f_1^3}\right)^2.
    \]
    Applying Lemma \ref{lemma:f3_over_f1cubed}, we obtain
\begin{align*}
     \sum_{n\geq 0}b_{4,9}(4n+1)q^n&=\frac{f_2^6}{f_6^2}\left(\frac{f_4^6f_6^3}{f_2^9f_{12}^2}+3q\frac{f_4^2f_6f_{12}^2}{f_2^7}\right)^2\\\nonumber
     &=\frac{f_2^6}{f_6^2}\left(\frac{f_4^{12}f_6^6}{f_2^{18}f_{12}^4}+9q^2\frac{f_4^4f_6^2f_{12}^4}{f_2^{14}}+6q\frac{f_4^8f_6^4}{f_2^{16}}\right).
\end{align*}
Extracting the terms involving the odd powers of $q$, dividing by $q$, and replacing $q^2$ by $q$ throughout, we obtain
\begin{equation}\label{eq8n+5}
    \sum_{n\geq 0}b_{4,9}(8n+5)q^n=6\frac{f_2^8f_3^2}{f_1^{10}}=6f_2^8\cdot \left(\frac{f_3^2}{f_1^2}\right)\cdot \left(\frac{1}{f_1^{8}}\right).
\end{equation}
Applying Lemma \ref{f3squared_over_f1squared} and the second equation of Lemma \ref{1_over_f1to4} to \eqref{eq8n+5}, we see that
\[
\sum_{n\geq 0}b_{4,9}(8n+5)q^n\equiv 2f_2^8\left(\frac{f_4^4f_6f_{12}^2}{f_2^5f_8f_{24}} +2q\frac{f_4f_6^2f_8f_{24}}{f_2^4f_{12}}\right)\frac{f_4^{28}}{f_2^{28}f_8^8} \pmod 4.
\]
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
\[
\sum_{n\geq 0}b_{4,9}(16n+13)q^n\equiv 0 \pmod 4.
\]
This implies the theorem.
\end{proof}
\begin{theorem}
\label{thm:32n25}
For all $n\geq 0$, we have $b_{4,9}(32n+25) \equiv 0 \pmod{4}.$ 
\end{theorem}
\begin{proof}
    From \eqref{eq8n+1}, we have
    \[
    \sum_{n\geq 0}b_{4,9}(8n+1)q^n\equiv \frac{f_2^{18}f_6^2}{f_4^5f_{12}}\cdot \frac{1}{f_1f_3}\cdot \frac{1}{f_1^{12}} \pmod 4.
    \]
    Using Lemma \ref{1_over_f1f3} and the third equation of Lemma \ref{1_over_f1to4} in the above, we obtain
    \[
  \sum_{n\geq 0}b_{4,9}(8n+1)q^n\equiv \frac{f_2^{18}f_6^2}{f_4^5f_{12}}  \left(\frac{f_8^2f_{12}^5}{f_2^2f_4f_6^4f_{24}^2} + q\frac{f_4^5f_{24}^2}{f_2^4f_6^2f_8^2f_{12}}\right)\frac{f_4^{42}}{f_2^{42}f_8^{12}}\pmod 4.
    \]
    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
    \[
    \sum_{n\geq 0}b_{4,9}(16n+9)q^n \equiv \frac{f_2^{42}}{f_1^{28}f_6^2f_4^{14}} =\frac{f_2^{42}}{f_6^2f_4^{14}}\cdot \left(\frac{1}{f_1^{12}}\right)^2\cdot \frac{1}{f_1^4} \pmod 4.
    \]
    It is easy to see that, applying Lemma \ref{1_over_f1to4} in the above equation, we obtain
    \[
     \sum_{n\geq 0}b_{4,9}(16n+9)q^n \equiv \frac{f_2^{42}}{f_6^2f_4^{14}}\left(\frac{f_4^{42}}{f_2^{42}f_8^{12}} \right)^2\frac{f_4^{14}}{f_2^{14}f_8^4}  \pmod 4.
    \]
Since the right-hand-side of the above equation involves only even powers of $q$, we readily obtain
    \[
    \sum_{n\geq 0}b_{4,9}(32n+25)q^n\equiv 0\pmod 4.
    \]
    Our result now follows.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm_b49_mod_12}]
As a reminder, we are trying to prove that, for all $n\geq 0$ and $\alpha \geq 3$, we have
    \begin{equation*}
        b_{4,9}(2^\alpha n+6\cdot 2^{\alpha -3}+1)\equiv 0\pmod {12}.
    \end{equation*}    
As was noted above, the $\alpha=3$ case of the above was proven by Kathiravan, Srinivas, and Sangale \cite[Theorem 1.3]{KathiravanSrinivasSangale}.  Our goal is to now prove the remaining cases via induction. Also, thanks to \eqref{eqmod3}, it is enough to prove the congruence modulo $4$.

First, note that the $\alpha=4$ case corresponds to the arithmetic progression $2^4 n+6\cdot 2^{1}+1 = 16n+13$, and Theorem \ref{thm:16n13} provides the result in this case.  Next, the $\alpha=5$ case corresponds to the arithmetic progression $2^5 n+6\cdot 2^{2}+1 = 32n+25$, and Theorem \ref{thm:32n25} 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 $\alpha$ and once for odd $\alpha$.  
First, let $\alpha = 2\beta$ for some integer $\beta$.  We assume that, for some $\beta\geq 2$ and all $n\geq 0$,
$$
b_{4,9}(2^{2\beta}n + 6\cdot 2^{2\beta - 3} +1) \equiv 0 \pmod{4}.
$$
We then want to prove that, for all $n\geq 0$,
$$
b_{4,9}(2^{2\beta+2}n + 6\cdot 2^{2\beta - 1} +1) \equiv 0 \pmod{4}.
$$
Note that 
\begin{align*}
b_{4,9}(2^{2\beta+2}n + 6\cdot 2^{2\beta - 1} +1) 
&= 
b_{4,9}(16(2^{2\beta-2}n + 6\cdot 2^{2\beta - 5}) +1) \\
&\equiv 
b_{4,9}(4(2^{2\beta-2}n + 6\cdot 2^{2\beta - 5}) +1) \pmod{4} \textrm{\ \ from Theorem \ref{thm:internal_cong_mod4}} \\
&= 
b_{4,9}(2^{2\beta}n + 6\cdot 2^{2\beta - 3}) +1) \\
&\equiv 
0 \pmod{4} 
\end{align*}
by the induction hypothesis.  This completes the proof for even $\alpha$.

Now, let $\alpha=2\beta+1$ for some integer $\beta$. We assume that, for some $\beta\geq 1$ and all $n\geq 0$,
\[
b_{4,9}(2^{2\beta+1}n+6\cdot 2^{2\beta-2}+1)\equiv 0 \pmod{4}.
\]
We then want to prove that, for all $n\geq 0$,
\[
b_{4,9}(2^{2\beta+3}+6\cdot 2^{2\beta}+1)\equiv 0 \pmod{4}.
\]
This is readily seen, as before
\begin{align*}
    b_{4,9}(2^{2\beta+3}+6\cdot 2^{2\beta}+1) &= b_{4,9}(16(2^{2\beta -1}n+6\cdot 2^{2\beta -4})+1) \\
&\equiv 
b_{4,9}(4(2^{2\beta-1}n + 6\cdot 2^{2\beta - 4}) +1) \pmod{4} \textrm{\ \ from Theorem \ref{thm:internal_cong_mod4}} \\
&= 
b_{4,9}(2^{2\beta+1}n + 6\cdot 2^{2\beta - 2}) +1) \\
&\equiv 
0 \pmod{4}.
\end{align*}
\end{proof}

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

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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11P83; Secondary 11P81, 05A17.

\noindent \emph{Keywords: } partition, congruence, generating function, dissection.

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

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  August 1 2025;
revised versions received  August 2 2025; October 17 2025.
Published in {\it Journal of Integer Sequences}, October 19 2025.

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\noindent
Return to \href{https://cs.uwaterloo.ca/journals/JIS/}{Journal of Integer Sequences home page}.
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