# Elementary Proofs of Infinitely Many Congruences for $k$-Elongated Partition
Diamonds

Robson da Silva  Universidade Federal de São Paulo, São José dos Campos, SP
12247–014, Brazil<EMAIL_ADDRESS>,  Michael D. Hirschhorn  School of
Mathematics, UNSW, Sydney, Australia 2052<EMAIL_ADDRESS>and  James
A. Sellers  Department of Mathematics and Statistics, University of Minnesota
Duluth, Duluth, MN 55812, USA<EMAIL_ADDRESS>
###### Abstract.

In 2007, Andrews and Paule published the eleventh paper in their series on
MacMahon’s partition analysis, with a particular focus on broken $k$-diamond
partitions. On the way to broken $k$-diamond partitions, Andrews and Paule
introduced the idea of $k$-elongated partition diamonds. Recently, Andrews and
Paule revisited the topic of $k$–elongated partition diamonds. Using partition
analysis and the Omega operator, they proved that the generating function for
the partition numbers $d_{k}(n)$ produced by summing the links of
$k$–elongated plane partition diamonds of length $n$ is given by
$\frac{(q^{2};q^{2})_{\infty}^{k}}{(q;q)_{\infty}^{3k+1}}$ for each $k\geq 1.$
A significant portion of their recent paper involves proving several
congruence properties satisfied by $d_{1},d_{2}$ and $d_{3}$, using modular
forms as their primary proof tool. In this work, our goal is to extend some of
the results proven by Andrews and Paule in their recent paper by proving
infinitely many congruence properties satisfied by the functions $d_{k}$ for
an infinite set of values of $k.$ The proof techniques employed are all
elementary, relying on generating function manipulations and classical
$q$-series results.

###### Key words and phrases:

congruences, partitions, $k$–elongated diamonds, generating functions

###### 2010 Mathematics Subject Classification:

11P83, 05A17

##  1\. Introduction

In 2007, Andrews and Paule [1] published the eleventh paper in their series on
MacMahon’s partition analysis, with a particular focus on the combinatorial
objects that they called broken $k$-diamond partitions. In that paper, Andrews
and Paule introduced the idea of $k$-elongated partition diamonds. Recently,
Andrews and Paule [2] revisited the topic of $k$–elongated partition diamonds.
Using partition analysis and the Omega operator, they proved an elegant
representation for the generating function for the partition numbers
$d_{k}(n)$ produced by summing the links of $k$–elongated plane partition
diamonds of length $n.$ They then proceeded to prove several congruence
properties satisfied by $d_{1},d_{2}$ and $d_{3}$ using modular forms as their
primary proof tool.

More recently, Smoot [8] extended the congruence work of Andrews and Paule,
refining a conjectured infinite family of congruences that appears in [2] and
proving the following via modular forms:

######  Theorem 1.1.

For all $\alpha\geq 0$ and for all $n\geq 0$ such that $8n\equiv
1\pmod{3^{\alpha}},$

$d_{2}(n)\equiv 0\pmod{3^{2\lfloor\alpha/2\rfloor+1}}.$

In this work, our goal is to extend some of the results proven by Andrews and
Paule in [2] by proving infinitely many congruence properties satisfied by the
functions $d_{k}$ for an infinite set of values of $k.$ The proof techniques
employed below are all elementary, relying on generating function
manipulations and classical $q$-series results.

As noted in Andrews and Paule [2], we know that the generating functions in
question are of the form

(1) $\sum_{n=0}^{\infty}d_{k}(n)q^{n}=\frac{f_{2}^{k}}{f_{1}^{3k+1}}$

where $f_{r}=(1-q^{r})(1-q^{2r})(1-q^{3r})(1-q^{4r})\dots$ is the usual
$q$–Pochhammer symbol.

In order to provide the elementary proofs that we desire, we will require a
few well–known $q$–series results. These results include the following:

######  Lemma 1.2.

$\frac{f_{2}^{2}}{f_{1}}=\sum_{m\geq 0}q^{m(m+1)/2}$

###### Proof.

This result appears as (1.5.3) in Hirschhorn [6]. ∎

######  Lemma 1.3.

$f_{1}^{3}=\sum_{m\geq 0}(-1)^{m}(2m+1)q^{m(m+1)/2}.$

###### Proof.

This result appears as (1.7.1) in Hirschhorn [6]. ∎

######  Lemma 1.4.

$f_{1}=\sum_{m=-\infty}^{\infty}(-1)^{m}q^{m(3m-1)/2}.$

###### Proof.

This result appears as (1.6.1) in Hirschhorn [6]. ∎

######  Lemma 1.5.

$\frac{f_{1}^{5}}{f_{2}^{2}}=\sum_{m=-\infty}^{\infty}(6m+1)q^{m(3m+1)/2}.$

###### Proof.

See Berndt [4], Corollary 1.3.21 as well as Hirschhorn [6], (10.7.3). ∎

######  Lemma 1.6.

(2) $\displaystyle\frac{f_{1}^{2}}{f_{2}}$
$\displaystyle=\sum_{j=-\infty}^{\infty}(-1)^{j}q^{j^{2}},$ (3)
$\displaystyle=\frac{f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{16}^{2}}{f_{8}},$
(4)
$\displaystyle=\frac{f_{9}^{2}}{f_{18}}-2q\frac{f_{3}f_{18}^{2}}{f_{6}f_{9}}.$

###### Proof.

Identity (2) is equation (22.4) in Berndt [3]; see also Hirschhorn [6],
(1.5.8). The 2-dissection (3) follows immediately from (1.9.4) in Hirschhorn
[6]. For a proof of (4) see [6, (14.3.4)]. ∎

######  Lemma 1.7.

$f_{1}f_{2}=\frac{f_{6}f_{9}^{4}}{f_{3}f_{18}^{2}}-qf_{9}f_{18}-2q^{2}\frac{f_{3}f_{18}^{4}}{f_{6}f_{9}^{2}}.$

###### Proof.

See Hirschhorn and Sellers [7]. ∎

######  Lemma 1.8.

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

###### Proof.

This identity is (18) in Brietzke, da Silva, and Sellers [5]. ∎

######  Lemma 1.9.

$\frac{f_{1}^{2}f_{4}^{2}}{f_{2}}=\sum_{m=-\infty}^{\infty}(3m+1)q^{3m^{2}+2m}.$

###### Proof.

This result appears as (10.7.6) in Hirschhorn [6]. ∎

######  Lemma 1.10.

$\frac{f_{2}^{5}}{f_{1}^{2}}=\sum_{m=-\infty}^{\infty}(-1)^{m}(3m+1)q^{3m^{2}+2m}.$

###### Proof.

This result appears as (10.7.7) in Hirschhorn [6]. ∎

##  2\. Elementary Proofs of Several Congruences from Andrews and Paule

In their work [2], Andrews and Paule used significant tools based on the work
of Smoot, which are derived from modular forms, in order to prove a number of
congruences for the functions $d_{2}$ and $d_{3}.$ In this section, we wish to
prove the majority of those results from Andrews and Paule using very
elementary tools.

######  Theorem 2.1 (Corollary 5, [2]).

For all $n\geq 0,$ $d_{2}(3n+2)\equiv 0\pmod{3}.$

###### Proof.

We have the following:

$\displaystyle\sum_{n=0}^{\infty}d_{2}(n)q^{n}$ $\displaystyle=$
$\displaystyle\frac{f_{2}^{2}}{f_{1}^{7}}$ $\displaystyle=$
$\displaystyle\frac{f_{2}^{2}}{f_{1}}\frac{1}{f_{1}^{6}}$
$\displaystyle\equiv$ $\displaystyle\frac{1}{f_{3}^{2}}\left(\sum_{m\geq
0}q^{m(m+1)/2}\right)\pmod{3}$

using Lemma 1.2. Now we simply need to determine whether $3n+2=m(m+1)/2$ for
some $m$ and $n.$ Completing the square means this is equivalent to
determining whether

$8(3n+2)+1=(2m+1)^{2}$

or

$24n+17=(2m+1)^{2}$

or

$2\equiv(2m+1)^{2}\pmod{3}.$

This congruence never holds because 2 is a quadratic nonresidue modulo 3. Our
result follows. ∎

######  Theorem 2.2 (Corollary 10, [2]).

For all $n\geq 0,$ $d_{3}(2n+1)\equiv 0\pmod{2}.$

###### Proof.

Note that

$\displaystyle\sum_{n=0}^{\infty}d_{3}(n)q^{n}$ $\displaystyle=$
$\displaystyle\frac{f_{2}^{3}}{f_{1}^{10}}$ $\displaystyle\equiv$
$\displaystyle\frac{f_{2}^{3}}{f_{2}^{5}}\pmod{2}$ $\displaystyle\equiv$
$\displaystyle\frac{1}{f_{2}^{2}}\pmod{2}.$

Since $\frac{1}{f_{2}^{2}}$ is an even function of $q,$ the result follows. ∎

######  Theorem 2.3 (Corollary 12, [2]).

For all $n\geq 0,$ $d_{3}(4n+2)\equiv 0\pmod{2}.$

###### Proof.

Thanks to the proof of Theorem 2.2, we know

$\displaystyle\sum_{n=0}^{\infty}d_{3}(n)q^{n}$ $\displaystyle\equiv$
$\displaystyle\frac{1}{f_{2}^{2}}\pmod{2}$ $\displaystyle\equiv$
$\displaystyle\frac{1}{{f_{4}}}\pmod{2}$

Since $\frac{1}{{f_{4}}}$ is a function of $q^{4},$ the result follows. ∎

######  Theorem 2.4 (Corollary 13, [2]).

For all $n\geq 0,$ $d_{3}(4n+3)\equiv 0\pmod{4}.$

###### Proof.

Using the generating function for $d_{3}$, we have

$\displaystyle\sum_{n=0}^{\infty}d_{3}(n)q^{n}$ $\displaystyle=$
$\displaystyle\frac{f_{2}^{3}}{f_{1}^{10}}=\frac{f_{2}^{4}}{f_{1}^{12}}\frac{f_{1}^{2}}{f_{2}}$
$\displaystyle\equiv$
$\displaystyle\frac{f_{2}^{4}}{f_{2}^{6}}\frac{f_{1}^{2}}{f_{2}}\pmod{4}$
$\displaystyle=$
$\displaystyle\frac{1}{f_{2}^{2}}\left(\frac{f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{16}^{2}}{f_{8}}\right)$

using (3) in Lemma 1.6 above. Extracting the odd parts, dividing by $q$ and
replacing $q^{2}$ by $q$, we are left with

$\sum_{n=0}^{\infty}d_{3}(2n+1)q^{n}\equiv
2\frac{f_{8}^{2}}{f_{1}^{2}f_{4}}\equiv 2f_{2}^{5}\pmod{4}.$

Since $f_{2}$ is a function of $q^{2}$ the result follows. ∎

######  Theorem 2.5 (Corollary 14, [2]).

For all $n\geq 0,$ $d_{3}(5n+1)\equiv 0\pmod{5}.$

###### Proof.

Using the generating function for $d_{3}$ which we have seen above, we have

$\displaystyle\sum_{n=0}^{\infty}d_{3}(n)q^{n}$ $\displaystyle=$
$\displaystyle\frac{f_{2}^{3}}{f_{1}^{10}}$ $\displaystyle\equiv$
$\displaystyle\frac{f_{2}^{3}}{f_{5}^{2}}\pmod{5}$ $\displaystyle=$
$\displaystyle\frac{1}{f_{5}^{2}}\left(\sum_{m=0}^{\infty}(-1)^{m}(2m+1)q^{m(m+1)}\right)$

using Lemma 1.3 above. We now need to ask whether $5n+1$ can be represented as
$m(m+1),$ and this is equivalent to asking whether $4(5n+1)+1$ or $20n+5$ can
be represented as $(2m+1)^{2}.$ If this is the case, then we know

$(2m+1)^{2}=20n+5\equiv 0\pmod{5}$

which implies that $2m+1\equiv 0\pmod{5}.$ Thanks to the presence of the
coefficient of $2m+1$ in front of the term $q^{m(m+1)}$ in the series above,
and the fact that this $2m+1$ must be divisible by 5, we know that our result
follows. ∎

######  Theorem 2.6 (Corollary 15, [2]).

For all $n\geq 0,$

$\displaystyle d_{3}(5n+3)$ $\displaystyle\equiv$ $\displaystyle 0\pmod{5},$
$\displaystyle d_{3}(5n+4)$ $\displaystyle\equiv$ $\displaystyle 0\pmod{5}.$

###### Proof.

In the proof of Theorem 2.5, we noted that

$\displaystyle\sum_{n=0}^{\infty}d_{3}(n)q^{n}$ $\displaystyle\equiv$
$\displaystyle\frac{1}{{f_{5}}^{2}}\left(\sum_{m=0}^{\infty}(-1)^{m}(2m+1)q^{m(m+1)}\right)\pmod{5}.$

We now need to ask whether $5n+3$ can be represented as $m(m+1),$ and this is
equivalent to asking whether $4(5n+3)+1$ or $20n+13$ can be represented as
$(2m+1)^{2}.$ This would mean that $(2m+1)^{2}\equiv 3\pmod{5}.$ However,
since 3 is a quadratic nonresidue modulo 5, we know that this cannot be the
case. Similarly, note that

$4(5n+4)+1=20n+17\equiv 2\pmod{5}$

and 2 is the other quadratic nonresidue modulo 5. The two congruences which
appear in the statement of the theorem follow. ∎

##  3\. New individual Congruences

In this section, we state and prove a set of new individual congruences. We
begin with an extremely unexpected congruence modulo 11 satisfied by the
function $d_{2}$ (one of the functions which received a great deal of
attention in Andrews and Paule [2] and was the primary focus in Smoot’s recent
work [8]).

######  Theorem 3.1.

For all $n\geq 0,$ $d_{2}(11n+7)\equiv 0\pmod{11}.$

###### Proof.

We begin by noting that

$\displaystyle\sum_{n=0}^{\infty}d_{2}(n)q^{n}$ $\displaystyle=$
$\displaystyle\frac{f_{2}^{2}}{f_{1}^{7}}$ $\displaystyle=$
$\displaystyle\frac{f_{1}^{4}f_{2}^{2}}{f_{1}^{11}}$ $\displaystyle\equiv$
$\displaystyle\frac{f_{1}^{4}f_{2}^{2}}{f_{11}}\pmod{11}.$

Replacing $q$ by $q^{2}$ yields

$\displaystyle\sum_{n=0}^{\infty}d_{2}(n)q^{2n}$ $\displaystyle\equiv$
$\displaystyle\frac{f_{2}^{4}f_{4}^{2}}{f_{22}}\pmod{11}.$

Hence, in order to prove our result, we simply need to focus on
$f_{2}^{4}f_{4}^{2}$ mod 11. Thanks to Lemma 1.9 and Lemma 1.10, we know

$\displaystyle f_{2}^{4}f_{4}^{2}$ $\displaystyle=$
$\displaystyle\frac{f_{2}^{5}}{f_{1}^{2}}\frac{f_{1}^{2}f_{4}^{2}}{f_{2}}$
$\displaystyle=$
$\displaystyle\sum_{j=-\infty}^{\infty}(-1)^{j}(3j+1)q^{3j^{2}+2j}\sum_{k=-\infty}^{\infty}(3k+1)q^{3k^{2}+2k}.$

Thus, at this point, we need to ask whether $2(11n+7)$ can be represented as

$2(11n+7)=3j^{2}+2j+3k^{2}+2k.$

This is equivalent to

$24(11n+7)+8=(6j+2)^{2}+(6k+2)^{2}$

which implies

$(6j+2)^{2}+(6k+2)^{2}\equiv 0\pmod{11}$

and this implies

$(3j+1)^{2}+(3k+1)^{2}\equiv 0\pmod{11}.$

Thus, if there is a representation of $2(11n+7)$ as $3j^{2}+2j+3k^{2}+2k,$
then we know that $11$ divides $3j+1$ and $11$ divides $3k+1.$ (Assume this is
not true. Then neither $3j+1$ nor $3k+1$ is divisible by 11. However, we know
that the quadratic residues modulo 11 are 1, 3, 4, 5, and 9. So,
$(3j+1)^{2}+(3k+1)^{2}\not\equiv 0\pmod{11},$ which is a contradiction.) Thus,
we now know that the coefficient of $q^{3j^{2}+2j+3k^{2}+2k}$ in the series
representation for $f_{2}^{4}f_{4}^{2},$ which is $(-1)^{j}(3j+1)(3k+1),$ must
be congruent to 0 modulo 11. ∎

We next prove a theorem connecting $d_{\ell}(n)$ and $p(n)$, for each prime
$\ell$, where $p(n)$ denotes the number of unrestricted partitions of $n.$
This theorem provides congruences for $d_{5},d_{7},$ and $d_{11}$ which are
closely related to Ramanujan’s congruences for $p(n).$

######  Theorem 3.2.

Let $\ell$ be a prime and let $r$, $1\leq r\leq\ell-1,$ be an integer such
that $p(\ell n+r)\equiv 0\pmod{\ell}$, for all $n\geq 0$. Then, for all $n\geq
0$,

$d_{\ell}(\ell n+r)\equiv 0\pmod{\ell}.$

###### Proof.

The generating function for $d_{\ell}(n)$ satisfies

$\displaystyle\sum_{n=0}^{\infty}d_{\ell}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{\ell}}{f_{1}^{3\ell+1}}\equiv\frac{f_{2\ell}}{f_{\ell}^{3}}\frac{1}{f_{1}}$
$\displaystyle\equiv\frac{f_{2\ell}}{f_{\ell}^{3}}\sum_{n=0}^{\infty}p(n)q^{n}\pmod{\ell}.$

Since $f_{2\ell}/f_{\ell}^{3}$ is a function of $q^{\ell}$ and $p(\ell
n+r)\equiv 0\pmod{\ell}$, the result follows. ∎

######  Corollary 3.3.

For all $n\geq 0,$

$\displaystyle d_{5}(5n+4)$ $\displaystyle\equiv$ $\displaystyle 0\pmod{5},$
$\displaystyle d_{7}(7n+5)$ $\displaystyle\equiv$ $\displaystyle 0\pmod{7},$
$\displaystyle d_{11}(11n+6)$ $\displaystyle\equiv$ $\displaystyle
0\pmod{11}.$

Next, we focus our attention specifically on the function $d_{7}(n).$

######  Theorem 3.4.

For all $n\geq 0,$

(5) $\displaystyle d_{7}(4n+2)$ $\displaystyle\equiv$ $\displaystyle
0\pmod{4},$ (6) $\displaystyle d_{7}(8n+5)$ $\displaystyle\equiv$
$\displaystyle 0\pmod{4},$ (7) $\displaystyle d_{7}(16n+9)$
$\displaystyle\equiv$ $\displaystyle 0\pmod{4},$ (8) $\displaystyle
d_{7}(4n+3)$ $\displaystyle\equiv$ $\displaystyle 0\pmod{8},$ (9)
$\displaystyle d_{7}(8n+4)$ $\displaystyle\equiv$ $\displaystyle 0\pmod{8}.$

###### Proof.

Using the generating function for $d_{7}(n)$ and (3), it follows that

(10) $\displaystyle\sum_{n=0}^{\infty}d_{7}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{7}}{f_{1}^{22}}=\frac{f_{2}^{8}}{f_{1}^{24}}\frac{f_{1}^{2}}{f_{2}}$
(11)
$\displaystyle\equiv\frac{1}{f_{4}^{2}}\left(\frac{f_{8}^{5}}{f_{4}^{2}f_{16}^{2}}-2q\frac{f_{16}^{2}}{f_{8}}\right)\pmod{4}.$

Extracting the even parts from (11) and replacing $q^{2}$ by $q$, we are left
with

$\sum_{n=0}^{\infty}d_{7}(2n)q^{n}\equiv\frac{f_{4}^{5}}{f_{2}^{4}f_{8}^{2}}\equiv\frac{1}{f_{4}}\pmod{4},$

from which (5) follows since $1/f_{4}$ is a function of $q^{4}$.

Extracting the odd parts of (11), dividing by $q$ and replacing $q^{2}$ by
$q$, we obtain

(12) $\sum_{n=0}^{\infty}d_{7}(2n+1)q^{n}\equiv
2\frac{f_{8}^{2}}{f_{2}^{2}f_{4}}\equiv 2f_{8}\pmod{4}.$

This means

(13) $\sum_{n=0}^{\infty}d_{7}(4n+1)q^{n}\equiv 2{f_{4}}\pmod{4},$

from which (6) follows since $f_{4}$ is a function of $q^{4}.$

In a similar way, (13) implies

$\sum_{n=0}^{\infty}d_{7}(8n+1)q^{n}\equiv 2{f_{2}}\pmod{4}.$

Since $f_{2}$ is a function of $q^{2}$, (7) follows.

In order to prove (8) and (9), we take (10) modulo $8$:

(14)
$\sum_{n=0}^{\infty}d_{7}(n)q^{n}\equiv\frac{1}{f_{2}^{4}}\frac{f_{1}^{2}}{f_{2}}\pmod{8}.$

Thanks to (3) in Lemma 1.6, we see that the odd part of (14) is

$\displaystyle\sum_{n=0}^{\infty}d_{7}(2n+1)q^{n}$ $\displaystyle\equiv$
$\displaystyle-2\frac{f_{8}^{2}}{f_{1}^{4}f_{4}}\pmod{8}$ $\displaystyle=$
$\displaystyle-2\frac{f_{8}^{2}}{f_{4}}\left(\frac{f_{4}^{14}}{f_{2}^{14}f_{8}^{4}}+4q\frac{f_{4}^{2}f_{8}^{4}}{f_{2}^{10}}\right)$

using Lemma 1.8. It follows that

$\sum_{n=0}^{\infty}d_{7}(4n+3)q^{n}\equiv 0\pmod{8},$

which proves (8).

Using (3), we can extract the even part of (14):

$\sum_{n=0}^{\infty}d_{7}(2n)q^{n}\equiv\frac{f_{4}^{5}}{f_{1}^{4}f_{2}^{2}f_{8}^{2}}\pmod{8}.$

Now, using Lemma 1.8 we extract the even part of the last expression to obtain

$\sum_{n=0}^{\infty}d_{7}(4n)q^{n}\equiv\frac{f_{2}^{19}}{f_{1}^{16}f_{4}^{6}}\equiv\frac{f_{2}^{11}}{f_{4}^{6}}\pmod{8},$

from which (9) follows. ∎

######  Theorem 3.5.

Let $p\geq 5$ be a prime and let $r$, $1\leq r\leq p-1$, be such that $3r+1$
is a quadratic nonresidue modulo $p$. Then, for all $n\geq 0$,

$d_{7}(2pn+2r+1)\equiv 0\pmod{4}.$

###### Proof.

Thanks to Lemma 1.4, we can rewrite (12) as

(15) $\sum_{n=0}^{\infty}d_{7}(2n+1)q^{n}\equiv
2\sum_{m=-\infty}^{\infty}q^{4m(3m-1)}\pmod{4}.$

If $pn+r=4m(3m-1)$, completing square we obtain $3pn+3r+1=(6m-1)^{2}$, from
which it follows that $3r+1\equiv(6m-1)^{2}\pmod{p}$. Since $3r+1$ is a
quadratic nonresidue modulo $p$, there is no $m$ such that $pn+r=4m(3m-1)$.
Thus, the coefficient of $q^{pn+r}$ on the right-hand side of (15) is $0$
modulo $4$, which implies $d_{7}(2(pn+r)+1)\equiv 0\pmod{4}.$ ∎

We next consider a pair of congruences satisfied by the function $d_{8}(n).$

######  Theorem 3.6.

For all $n\geq 0$,

(16) $\displaystyle d_{8}(3n+2)$ $\displaystyle\equiv$ $\displaystyle
0\pmod{9},$ (17) $\displaystyle d_{8}(9n+3)$ $\displaystyle\equiv$
$\displaystyle 0\pmod{9}.$

###### Proof.

Using the generating function for $d_{8}(n)$ and (4) in Lemma 1.6, we obtain

$\displaystyle\sum_{n=0}^{\infty}d_{8}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{8}}{f_{1}^{25}}=\frac{f_{2}^{9}}{f_{1}^{27}}\frac{f_{1}^{2}}{f_{2}}$
$\displaystyle\equiv\frac{f_{6}^{3}}{f_{3}^{9}}\left(\frac{f_{9}^{2}}{f_{18}}-2q\frac{f_{3}f_{18}^{2}}{f_{6}f_{9}}\right)\pmod{9}$
(18)
$\displaystyle=\frac{f_{6}^{3}f_{9}^{2}}{f_{3}^{9}f_{18}}-2q\frac{f_{6}^{2}f_{18}^{2}}{f_{3}^{8}f_{9}},$

from which we immediately deduce (16).

We now extract the terms of the form $q^{3n}$ from both sides of (18) to
obtain

$\sum_{n=0}^{\infty}d_{8}(3n)q^{3n}\equiv\frac{f_{6}^{3}f_{9}^{2}}{f_{3}^{9}f_{18}}\pmod{9}.$

Replacing $q^{3}$ by $q$, we are left with

$\displaystyle\sum_{n=0}^{\infty}d_{8}(3n)q^{n}$
$\displaystyle\equiv\frac{f_{2}^{3}f_{3}^{2}}{f_{1}^{9}f_{6}}\equiv\frac{f_{2}^{3}}{f_{3}f_{6}}\pmod{9}$
$\displaystyle=\frac{1}{f_{3}f_{6}}\sum_{m\geq 0}(-1)^{m}(2m+1)q^{m(m+1)},$

using Lemma 1.3. Since $1/f_{3}f_{6}$ is a function of $q^{3}$ and
$m(m+1)\not\equiv 1\pmod{3}$, we see that the coefficient of $q^{3n+1}$ on the
right-hand side of the congruence above is 0 modulo $9$. Thus, (17) follows. ∎

##  4\. New Infinite Families of Congruences

In this section, we provide multiple extensions of the results of Andrews and
Paule above which, in turn, provide new infinite families of congruences
satisfied by the functions $d_{k}(n).$

The next result is rather surprising, primarily because the moduli in question
range across all primes $p\geq 5.$ The proof of this theorem follows easily
via elementary generating function manipulations along with a $q$–series
identity (Lemma 1.5 above) of Ramanujan.

######  Theorem 4.1.

Let $p\geq 5$ be a prime and let $r$, $1\leq r\leq p-1$, be such that $24r+1$
is a quadratic nonresidue modulo $p$. Then, for all $n\geq 0$,

$d_{p-2}(pn+r)\equiv 0\pmod{p}.$

###### Proof.

The generating function for $d_{p-2}(n)$ is

$\displaystyle\sum_{n=0}^{\infty}d_{p-2}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{p-2}}{f_{1}^{3p-5}}\equiv\frac{f_{1}^{5}}{f_{2}^{2}}\frac{f_{2p}}{f_{p}^{3}}\pmod{p}.$

Thanks to Lemma 1.5, we have

(19)
$\sum_{n=0}^{\infty}d_{p-2}(n)q^{n}\equiv\frac{f_{2p}}{f_{p}^{3}}\left(\sum_{m=-\infty}^{\infty}(6m+1)q^{m(3m+1)/2}\right)\pmod{p}.$

We want to know whether $m(3m+1)/2=pn+r$, for some $m$ and $n$. This is
equivalent to asking whether $24pn+24r+1=(6m+1)^{2}$, which implies
$24r+1\equiv(6m+1)^{2}\pmod{p}$. However $24r+1$ is a quadratic nonresidue
modulo $p$. Therefore $d_{p-2}(pn+r)\equiv 0\pmod{p}.$ ∎

The theorem above is a fairly standard and classic result, providing exactly
$(p-1)/2$ congruences for each prime $p.$ Interestingly enough, numerical
evidence indicates that there are actually $(p-1)/2+1$ or $(p+1)/2$ such
congruences for each prime $p.$ We explain this additional “special”
congruence for each prime $p\geq 5$ here.

######  Theorem 4.2.

Let $p\geq 5$ be a prime and let $t$, $1\leq t\leq p-1$, be the unique value
such that $24t+1\equiv 0\pmod{p}$. Then, for all $n\geq 0$,

$d_{p-2}(pn+t)\equiv 0\pmod{p}.$

###### Proof.

Thanks to Lemma 1.5 and (19) above, we need to ask whether $pn+t=m(3m+1)/2$
for some integers $m$ and $n$. Completing the square and looking the result
modulo $p$ yields $(6m+1)^{2}\equiv 24t+1\equiv 0\pmod{p}$. So $p$ divides
$(6m+1)^{2}$, which implies that $p$ divides $6m+1$. Since the coefficient of
$q^{m(3m+1)/2}$ in the series that appears in Lemma 1.5 is exactly $6m+1$, it
follows that the coefficient in question is congruent to $0$ modulo $p$. ∎

######  Theorem 4.3.

Let $p\geq 5$ be a prime and let $r$, $1\leq r\leq p-1$, be a quadratic
nonresidue modulo $p.$ Then, for all $n\geq 0$,

$d_{p-1}(pn+r)\equiv 0\pmod{p}.$

###### Proof.

Using (1), we know that, for prime $p\geq 5,$

$\displaystyle\sum_{n=0}^{\infty}d_{p-1}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{p-1}}{f_{1}^{3p-2}}$
$\displaystyle=\frac{f_{2}^{p}}{f_{1}^{3p}}\frac{f_{1}^{2}}{f_{2}}$
$\displaystyle\equiv\frac{f_{2p}}{f_{p}^{3}}\frac{f_{1}^{2}}{f_{2}}\pmod{p}$
$\displaystyle=\frac{f_{2p}}{f_{p}^{3}}\left(\sum_{j=-\infty}^{\infty}q^{j^{2}}\right)$

using Lemma 1.6 above. The result immediately follows by recognizing that
$\frac{f_{2p}}{f_{p}^{3}}$ is a function of $q^{p}$ and that $r$ has been
defined to be a quadratic nonresidue modulo $p.$ ∎

We next provide an overarching result which allows us to naturally generalize
Theorems 2.1–2.6 above as well as Theorems 4.1, 4.2, and 4.3.

######  Theorem 4.4.

Let $p$ be a prime, $k\geq 1,$ $j\geq 0,$ and $r$ be an integer such that
$1\leq r\leq p-1.$ If, for all $n\geq 0,$

$d_{k}(pn+r)\equiv 0\pmod{p},$

then for all $n\geq 0,$

$d_{pj+k}(pn+r)\equiv 0\pmod{p}.$

###### Proof.

Replacing $k$ by $pj+k$ in (1) we obtain

$\displaystyle\sum_{n=0}^{\infty}d_{pj+k}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{pj+k}}{f_{1}^{3pj+3k+1}}=\frac{f_{2}^{k}}{f_{1}^{3k+1}}\frac{f_{2}^{pj}}{f_{1}^{3pj}}$
$\displaystyle\equiv\frac{f_{2p}^{j}}{f_{p}^{3j}}\sum_{m=0}^{\infty}d_{k}(m)q^{m}\pmod{p}.$

All the exponents of $q$ coming from ${f_{2p}^{j}}/{f_{p}^{3j}}$ are of the
form of $pN$. Since $d_{k}(pm+r)\equiv 0\pmod{p}$, it follows that
$d_{pj+k}(pn+r)\equiv 0\pmod{p}$. ∎

Theorem 4.4 provides a tool for writing down infinitely many new congruences
with ease. We exhibit such a list of new congruences below, using a shorthand
notation to consolidate the statement of the results. In what follows, the
notation

$An+B_{1},B_{2},\dots,B_{t}$

means we are considering the set of arithmetic progressions

$An+B_{1},An+B_{2},\dots,An+B_{t}.$

######  Corollary 4.5.

For all $j\geq 0$ and $n\geq 0$,

(20) $\displaystyle{d_{2j+1}(2n+1)}$ $\displaystyle\equiv 0\pmod{2},$ (21)
$\displaystyle d_{3j+2}(3n+2)$ $\displaystyle\equiv 0\pmod{3},$ (22)
$\displaystyle d_{5j+3}(5n+1,3,4)$ $\displaystyle\equiv 0\pmod{5},$ (23)
$\displaystyle d_{5j+4}(5n+2,3)$ $\displaystyle\equiv 0\pmod{5},$ (24)
$\displaystyle d_{5j+5}(5n+4)$ $\displaystyle\equiv 0\pmod{5},$ (25)
$\displaystyle d_{7j+5}(7n+2,3,4,6)$ $\displaystyle\equiv 0\pmod{7},$ (26)
$\displaystyle d_{7j+6}(7n+3,5,6)$ $\displaystyle\equiv 0\pmod{7},$ (27)
$\displaystyle d_{7j+7}(7n+5)$ $\displaystyle\equiv 0\pmod{7},$ (28)
$\displaystyle d_{11j+2}(11n+7)$ $\displaystyle\equiv 0\pmod{11},$ (29)
$\displaystyle d_{11j+9}(11n+3,5,6,8,9,10)$ $\displaystyle\equiv 0\pmod{11},$
(30) $\displaystyle d_{11j+10}(11n+2,6,7,8,10)$ $\displaystyle\equiv
0\pmod{11},$ (31) $\displaystyle d_{11j+11}(11n+6)$ $\displaystyle\equiv
0\pmod{11},$ (32) $\displaystyle d_{13j+11}(13n+3,4,6,7,8,10,11)$
$\displaystyle\equiv 0\pmod{13},$ (33) $\displaystyle
d_{13j+12}(13n+2,5,6,7,8,11)$ $\displaystyle\equiv 0\pmod{13}.$

######  Remark 4.6.

The corollary above does not provide an exhaustive list of congruences
satisfied by these functions. Our goal in writing these here is to provide the
reader with a representative set of the kinds of congruences that arise within
this family of partition functions.

###### Proof.

(of Corollary 4.5) Thanks to Theorem 4.4, we only have to check the basis
case, $j=0$, for all of the above results. To prove (20), we note that taking
$k=1$ in (1) we are left with

$\sum_{n=0}^{\infty}d_{1}(m)q^{m}=\frac{f_{2}}{f_{1}^{4}}\equiv\frac{1}{f_{2}}\pmod{2},$

from which we see that $d_{1}(2n+1)\equiv 0\pmod{2}$ and then (20) follows.

The basis case for (21) is proven in Theorem 2.1.

The basis cases for the congruences listed in (22) are proven in Theorems 2.5
and 2.6.

The basis cases for (23) and (26) follow from the $p=5$ and $p=7$ cases of
Theorem 4.3 above, respectively.

The basis cases for (24), (27), and (31) follow from Corollary 3.3 above.

The basis cases for (25)–(33) follow immediately from Theorems 4.1, 4.2, and
4.3.

The basis case for (28) follows from Theorem 3.1 above.

∎

We close this section with two infinite families of congruences modulo 9 which
are closely related to congruences that appear in Andrews and Paule [2].

######  Theorem 4.7.

For all $j\geq 0$ and $n\geq 0$,

$\displaystyle d_{9j+2}(9n+5)$ $\displaystyle\equiv$ $\displaystyle
0\pmod{9},$ $\displaystyle d_{9j+2}(9n+8)$ $\displaystyle\equiv$
$\displaystyle 0\pmod{9}.$

###### Proof.

Using the generating function for $d_{2}$, we have

$\displaystyle\sum_{n=0}^{\infty}d_{2}(n)q^{n}$ $\displaystyle=$
$\displaystyle\frac{f_{2}^{2}}{f_{1}^{7}}=\frac{1}{f_{1}^{9}}f_{1}^{2}f_{2}^{2}$
$\displaystyle\equiv$
$\displaystyle\frac{1}{f_{3}^{3}}f_{1}^{2}f_{2}^{2}\pmod{9}$
$\displaystyle\equiv$
$\displaystyle\frac{1}{f_{3}^{3}}\left(\frac{f_{6}f_{9}^{4}}{f_{3}f_{18}^{2}}-qf_{9}f_{18}-2q^{2}\frac{f_{3}f_{18}^{4}}{f_{6}f_{9}^{2}}\right)^{2}\pmod{9}$

using Lemma 1.7 above. Extracting the terms involving $q^{3n+2}$, dividing by
$q^{2}$ and replacing $q^{3}$ by $q$, we obtain

$\sum_{n=0}^{\infty}d_{2}(3n+2)q^{n}\equiv-3f_{3}^{2}f_{6}^{2}\pmod{9}.$

Since $f_{3}$ and $f_{6}$ are functions of $q^{3}$ it follows that

$\displaystyle d_{2}(9n+5)$ $\displaystyle\equiv$ $\displaystyle 0\pmod{9},$
$\displaystyle d_{2}(9n+8)$ $\displaystyle\equiv$ $\displaystyle 0\pmod{9},$

which provides the basis case for an inductive proof. Thus, let us assume that
$d_{9j+2}(9n+5)\equiv 0\pmod{9}$ and $d_{9j+2}(9n+8)\equiv 0\pmod{9}$ for some
$j$. In order to complete the proof, we note that

$\displaystyle\sum_{n=0}^{\infty}d_{9(j+1)+2}(n)q^{n}$
$\displaystyle=\frac{f_{2}^{9(j+1)+2}}{f_{1}^{27(j+1)+7}}=\frac{f_{2}^{9}}{f_{1}^{27}}\frac{f_{2}^{9j+2}}{f_{1}^{27j+7}}$
$\displaystyle=\frac{f_{2}^{9}}{f_{1}^{27}}\sum_{m=0}^{\infty}d_{9j+2}(m)q^{m}$
(34)
$\displaystyle\equiv\frac{f_{6}^{3}}{f_{9}^{3}}\sum_{m=0}^{\infty}d_{9j+2}(m)q^{m}\pmod{9}.$

Using Lemma 1.3, we have

$f_{6}^{3}=\sum_{k\geq 0}(-1)^{k}(2k+1)q^{3k(k+1)}.$

For all $k\geq 0$, we know that $3k(k+1)\equiv 0,6\pmod{9}$, with
$3k(k+1)\equiv 6\pmod{9}$ if and only if $k\equiv 1,4,7\pmod{9}$, which is
equivalent to $2k+1\equiv 3,9,15\pmod{9}$. Thus, modulo $9$, $f_{6}^{3}$
becomes

$f_{6}^{3}\equiv\sum_{k\geq 0}a_{k}q^{9k}+3\sum_{k\geq
0}b_{k}q^{9k+6}\pmod{9},$

for certain integer coefficients $a_{k}$ and $b_{k}$. Hence, using (34) we
obtain

$\displaystyle\sum_{n=0}^{\infty}d_{9(j+1)+2}(n)q^{n}$
$\displaystyle\equiv\frac{1}{f_{9}^{3}}\left(\sum_{k\geq
0}a_{k}q^{9k}\sum_{m=0}^{\infty}d_{9j+2}(m)q^{m}\right.$ $\displaystyle\ \ \ \
\left.+3\sum_{k\geq
0}b_{k}q^{9k+6}\sum_{m=0}^{\infty}d_{9j+2}(m)q^{m}\right)\pmod{9}.$

By the hypothesis, we see that the coefficients of $q^{9n+5}$ and $q^{9n+8}$
in the first term of the congruence above are congruent to $0$ modulo $9$. For
the second term in the congruence above, namely

(35) $\sum_{m,k\geq 0}3b_{k}d_{9j+2}(m)q^{9k+m+6},$

we note that $9k+m+6\equiv 5\pmod{9}$ if and only if $m\equiv 8\pmod{9}$. In
this case, we know that $m=9N+8$ for some $N$. By (21) in Corollary 4.5, we
have $d_{9j+2}(9n+8)\equiv 0\pmod{3}$, for all $n$, which yields
$3b_{k}d_{9j+2}(9N+8)\equiv 0\pmod{9}$. Thus the coefficients of $q^{9n+5}$ in
(35) are congruent to $0$ modulo $9$. Analogously, we see that $9k+m+6\equiv
8\pmod{9}$ if and only if $m\equiv 2\pmod{9}$. So, $m=9N+2$ for some $N$.
However, we also know from (21) in Corollary 4.5 that $d_{9j+2}(9n+2)\equiv
0\pmod{3}$, for all $n$. Thus, $3b_{k}d_{9j+2}(9N+2)\equiv 0\pmod{9}$, from
which we conclude that the coefficients of $q^{9n+8}$ in (35) are congruent to
$0$ modulo $9$. ∎

##  5\. Concluding Remarks

We readily admit that the above results are not an exhaustive list of all of
the congruences satisfied by the functions $d_{k}(n).$ Indeed, there are many
other congruences to consider. For example, computational evidence hints at
the possibility of infinite families of congruences modulo arbitrarily high
powers of 3 for the function $d_{11}(n)$ as well as the possibility of
infinite families of congruences modulo powers of 2 for $d_{7}(n)$ and
$d_{15}(n)$ (in the spirit of the work completed by Smoot for $d_{2}$ modulo
powers of 3). Indeed, for $k=2^{j}-1$ for some $j,$ it is clear that the
generating function for $d_{k}(n)$ will have a structure that allows for a
number of congruences to hold for small moduli. One must wonder whether an
infinite family of congruences, modulo powers of 2, like the family in Theorem
1.1 holds for these special values of $k.$ We leave such an investigation to
the interested reader.

## Acknowledgments

The first author was supported by São Paulo Research Foundation (FAPESP)
(grant no. 2019/14796-8).

## References

  * [1] G. E. Andrews and P. Paule, MacMahon’s partition analysis XI: Broken diamonds and modular forms, Acta Arithmetica 126, no. 3 (2007), 281–294
  * [2] G. E. Andrews and P. Paule, MacMahon’s partition analysis XIII: Schmidt type partitions and modular forms, J. Number Theory, https://doi.org/10.1016/j.jnt.2021.09.008.
  * [3] B. C. Berndt, Ramanujan’s notebooks, Part III, Springer, 1991.
  * [4] B. C. Berndt, Number Theory in the Spirit of Ramanujan, Amer. Math. Soc., 2006.
  * [5] E. H. M. Brietzke, R. da Silva, J. A. Sellers, Congruences related to an eighth order mock theta function of Gordon and McIntosh. J. Math. Anal. Appl. 479 (2019), 62–89.
  * [6] M. D. Hirschhorn, The power of $q$, a personal journey, Developments in Mathematics, v. 49, Springer, 2017.
  * [7] M. D. Hirschhorn and J. A. Sellers, A Congruence Modulo 3 for Partitions into Distinct Non-Multiples of Four, J. Integer Seq. 17 (2014), Article 14.9.6.
  * [8] N. A. Smoot, A Congruence Family for 2–Elongated Plane Partitions: An Application of the Localization Method, submitted