Congruences modulo powers of 11 for some eta-quotients

TL;DR

This paper establishes infinite families of congruences modulo powers of 11 for a class of partition functions defined via eta-quotients, extending previous results by Atkin and Gordon through modular form techniques.

Contribution

It generalizes Atkin and Gordon's congruences for partition functions by proving new infinite families of congruences modulo powers of 11 for eta-quotient-based partition functions.

Findings

Proves infinite families of congruences modulo powers of 11.

Uses an explicit basis for modular functions of (11).

Extends previous congruence results to broader classes of partition functions.

Abstract

The partition function $ p_{[1^c11^d]}(n)$ can be defined using the generating function, \[\sum_{n=0}^{\infty}p_{[1^c{11}^d]}(n)q^n=\prod_{n=1}^{\infty}\dfrac{1}{(1-q^n)^c(1-q^{11 n})^d}.\] In this paper, we prove infinite families of congruences for the partition function $ p_{[1^c11^d]}(n)$ modulo powers of $11$ for any integers $c$ and $d$, which generalizes Atkin and Gordon's congruences for powers of the partition function. The proofs use an explicit basis for the vector space of modular functions of the congruence subgroup $\Gamma_0(11)$.

Tables

Table 1. Table 1 . Values of α 𝛼 \alpha

	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20	21	22	23	24
0	2	1	2	1	1	1	2	2	1	1	2	2	1	2	1	0	0	1	1	0	0	1	1	0
24	1	1	1	1	2	2	1	1	2	2	1	0	0	0	0	1	1	0	0	1	1	1	0	0
48	1	1	2	2	1	1	1	0	1	0	1	0	0	1	1	0	0	1	0	1	0	1	0	0
72	2	1	1	1	2	1	2	1	2	1	2	2	1	1	1	2	1	2	1	2	1	1	1	0
96	0	0	1	0	1	0	1	0	1	1	0	0	0	1	0	1	0	1	0	1	1	0	0	0

Table 2. Table 2. Values of δ ​ ( μ , ν ) 𝛿 𝜇 𝜈 \delta(\mu,\nu) .

	$\nu$
$\mu$	0	1	2	3	4
0	-1	8	7	6	15
1	0	9	8	2	11
2	1	10	4	13	12
3	2	6	5	4	13
4	3	7	6	5	9

Table 3. Table 3. Values of θ ​ ( λ , μ ) 𝜃 𝜆 𝜇 \theta(\lambda,\mu) .

	$\lambda$
$\mu$	0	1	2	3	4	5	6	7	8	9	10
0	0	1	0	1	0	1	0	1	1	0	0
1	1	1	0	1	0	0	0	1	1	0	0
2	1	1	1	0	0	0	0	1	1	0	0
3	1	0	1	0	0	0	0	1	1	0	0
4	1	0	1	0	1	0	1	1	0	0	0

Equations

$$\sum _{n=0}^{\infty }{p}_{[1^c{11}^d]}(n)q^n=\prod _{n=1}^{\infty }\frac{1}{(1-q^n{)}^c(1-q^{11n}{)}^d}.$$

$$p(5^{j}n+{\delta }_{5,j})$$

$$p(7^{j}n+{\delta }_{7,j})$$

$$p(11^{j}n+{\delta }_{11,j})$$

$$\prod _{n=1}^{\infty }\frac{1}{(1-q^n{)}^c(1-q^{\ell n}{)}^d}=\sum _{n=0}^{\infty }{p}_{[1^c{\ell }^d]}(n)q^n.$$

$$p_{[1^{1}11^{-11}]}\left(11^{k}n+11^k-5\right)$$

$$p_{[1^{1}11^{-1}]}\left(11^{2k-1}n+\frac{7\cdot 11^{2k-1}-5}{12}\right)$$

$$p_{[1^{1}11^1]}\left(11^{k}n+\frac{11^k+1}{2}\right)$$

$$p_{[1^c{11}^d]}({11}^{r}m+n_r)\equiv 0(\mathrm{mod}{11}^{A_r})$$

$$p_{[1^{c}11^d]}(11^{r}m+n_r)\equiv 0(\mathrm{mod}11^{\frac{1}{2}\alpha r+ϵ})$$

$$U_p\left(f(\tau )\right)=\sum _{pn\ge N}a(pn)q^n.$$

$$U_p\left(f(\tau )g(p\tau )\right)=g(\tau )U_p\left(f(\tau )\right).$$

$${\mathrm{ord}}_0{J}_{\nu }(\tau )=\{\begin{array}{ll}-\nu & \text{if }\nu \equiv 0(\mathrm{mod}5),\\ -\nu -1& \text{if }\nu \equiv 1,2\text{ or }3(\mathrm{mod}5),\\ -\nu -2& \text{if }\nu \equiv 4(\mathrm{mod}5).\end{array}$$

$$\varphi (\tau ):=q^5\prod _{n=1}^{\infty }\left(\frac{1-q^{121n}}{1-q^n}\right)=\frac{\eta (121\tau )}{\eta (\tau )}.$$

$$T_{\lambda }:f(\tau )\to U_{11}\left(\varphi (\tau {)}^{\lambda }f(\tau )\right)$$

$$U_{11}\left(\varphi (\tau {)}^{\lambda }{J}_{\mu }(\tau )\right)=\sum _{\nu }{C}_{\mu ,\nu }^{\lambda }{J}_{\nu }(\tau )$$

$$\pi (C_{\mu ,\nu }^{\lambda })\ge \left[\frac{11\nu -\mu -5\lambda +\delta }{10}\right].$$

$$\pi (C_{\mu ,\nu }^{\lambda })\ge \left[\frac{11\nu -\mu -5\lambda -1}{10}\right].$$

$$C_{\mu,\nu}^{\lambda}\equiv 0\pmod{11}\ \mbox{for all $\nu$}.$$

$$\theta (\lambda -11,\mu )=\theta (\lambda +12,\mu -5)=\theta (\lambda ,\mu ).$$

$$A_r(c,d):=\sum _{i=1}^{r-1}\theta ({\lambda }_i,{\mu }_i),$$

$$L_1(\tau ):$$

$$=U_{11}\left(q^{5c}\prod _{n=1}^{\infty }\frac{(1-q^{121n}{)}^c(1-q^{11n}{)}^d}{(1-q^n{)}^c(1-q^{11n}{)}^d}\right),$$

$$=\prod _{n=1}^{\infty }(1-q^{11n}{)}^c(1-q^n{)}^d\sum _{m\ge \lceil \frac{5c}{11}\rceil }^{\infty }{p}_{[1^{c}11^d]}(11m-5c)q^m.$$

$$L_2(\tau ):$$

$$L_2(\tau )$$

$$L_r:=U_{11}\left({\varphi }^{{\lambda }_{r-1}}(\tau )L_{r-1}\right),$$

$$\quad\lambda_{r}=\left\{\begin{array}[]{ll}c&\mbox{if $r$ is even },\\ d&\mbox{if $r$ is odd}.\end{array}\right.$$

$$L_{2r}(\tau )$$

$$L_{2r-1}(\tau )$$

$$n_{2r}=-5\cdot d\cdot 11^{2r-1}+n_{2r-1},$$

$$n_{2r-1}=-5\cdot c\cdot 11^{2r-2}+n_{2r-2}.$$

$$n_{2r-1}$$

$$n_{2r}$$

$$24n_r\equiv (c+11\cdot d)\mathrm{mod}11^r.$$

Full text

[TABLE]

In this paper, we prove infinite families of congruences for the partition function $p_{[1^{c}11^{d}]}(n)$ modulo powers of $11$ for any integers $c$ and $d$, which generalizes Atkin and Gordon’s congruences for powers of the partition function. The proofs use an explicit basis for the vector space of modular functions of the congruence subgroup $\Gamma_{0}(11)$.

1. Introduction

An (integer) partition of $n$ is a non-increasing sequence of positive integers $\lambda_{1}\geq\lambda_{2}\cdots\geq\lambda_{r}\geq 1$ that sum to $n$. Let $p(n)$ be the number of partitions of $n$. By convention, we take $p(0)=1$ and $p(n)=0$ for negative $n$.

This function has been extensively studied in the last century. In the 1920’s Ramanujan discovered amazing congruence properties for $p(n)$.

Theorem**.**

For all positive integers $j$, we have,

[TABLE]

where $24\delta_{\ell,j}\equiv 1\pmod{\ell^{j}}$ for $\ell\in\{5,7,11\}$.

Ramanujan in [9] proved the first two congruences for the case of $j=1$ by using the Jacobi triple product and later in [10] using the theory of modular forms on $SL_{2}(\mathbb{Z})$. For arbitrary $j\geq 1$, Watson in [12] gave a proof using modular equations for prime $5$ and $7$. Ramanujan in [11] stated that he found a proof for the third congruence for $j=1,2$, but did not include the proof. In $1967$, Atkin in [2] gave a proof for the third congruence.

These fascinating congruence properties not only hold for the partition function itself, but also for the restricted partitions. To study a large class of restricted partitions, we study the partition function $p_{[1^{c}\ell^{d}]}(n)$. This partition function also well studied in recent years, for example see Chan and Toh [6], and Liuquan Wang [13].

The partition function $p_{[1^{c}\ell^{d}]}(n)$ is defined using the generating function in the following way.

[TABLE]

To illustrate the importance of studying this partition function, let’s look at the following four examples .

• •

$d=0$ and $c>0$: $c-$color partitions.

This partition function generates partitions of $n$ in to $c$ colors. See Gordon [8] and Atkin [1] for interesting congruence relations for primes less than or equal to 13.

• •

$c=1,d=-1$: $\ell$-regular partitions.

This partition function generates partitions of $n$ with a restriction that no parts divisible by $\ell$. This partitions also well studied in recent years. See Wang [14] and [15] for divisibility properties of $5$- regular and $7$-regular partitions.

• •

$c=1,d=-\ell$: $\ell$-core partitions.

This partition function generates partition of $n$ with a restriction that no hook numbers are divisible by $\ell$. See Wang [13] to see the divisibility properties of this partition function.

• •

$c=1,d=1$ and $\ell=2$: The cubic partition function.

This partition function has a deep connection to the Ramanujan’s cubic continued fraction, in [4] and [5] Chan used this connection to obtain interesting congruences.

In 2016, in [13], Wang proved the following congruences.

Theorem** (Wang, 2016).**

For any integers $n\geq 0$ and $k\geq 1$,

[TABLE]

Furthermore in [13], he stated that it possible to obtain congruences for each value $c,d\in\mathbb{Z}$ separately.The primary goal of this paper is to find a unified way to prove congruences for the partition function $p_{[1^{c}{11}^{d}]}(n)$ for any $c,d\in\mathbf{Z}$.

Theorem 1.1**.**

For any integers $c$, $d$ and for any positive integer $r$,

[TABLE]

where $24n_{r}\equiv(c+{11}d)\pmod{{11}^{r}}$.

Here $A_{r}$ only depends on the integers $c,d$ and it can be calculated explicitly.

Moreover, we can obtain the following corollary, this is similar to the Gordon’s Theorem 1.2 in [8].

Corollary 1.2**.**

For any positive integer $r$,

[TABLE]

where $24n_{r}\equiv(c+11d)\pmod{11^{r}}$, $\epsilon=\epsilon(c,d)=O(\log|c+11d|)$ and when $c+11d\geq 0$ , $\alpha$ depends on the residue of $c+11d\pmod{120}$ which is shown in Table 1.

Here the entry is $\alpha(24i+j)$ where row labelled $24i$ and column labeled $j$. When $c+11d<0$, the last column must be changed to $2,2,2,0,2$.

Remark 1.3*.*

This is the same shape as Gordon’s result for $k-$colored partitions, with k replaced by $c+11d$.

The remainder of the paper is as follows. Section 2 reviews properties of modular forms and operators on their coefficients. In Section 3 the main results are proven. The paper closes with several examples in Section 4.

2. Preliminaries

For a Laurent series $f(\tau)=\sum_{n\geq N}a(n)q^{n}$, we define the $U_{p}$ operator by,

[TABLE]

Let $g(\tau)=\sum_{n\geq N}b(n)q^{n}$ be another Laurent series. The following simple property will play a key role in our proof.

[TABLE]

Proposition 2.1** ([3], lemma 7).**

If $f(\tau)$ is a modular function for $\Gamma_{0}(N)$, if $p^{2}|N$, then U_{p}\Big{(}f(\tau)\Big{)} is a modular function for $\Gamma_{0}(N/p)$.

Let $V$ be the vector space of modular functions on $\Gamma_{0}(11)$, which are holomorphic everywhere except possibly at 0 and $\infty$.

Atkin constructed a basis for $V$, Gordon in [2] slightly modified these basis elements and defined $J_{\nu}(\tau)$. For detailed information about the construction of the basis elements see [2].

Theorem 2.2** (Gordon [8]).**

*For all $\nu\in\mathbf{Z}$, we have:

1. (1)

$J_{\nu}(\tau)=J_{\nu-5}(\tau)J_{5}(\tau)$, 2. (2)

$\{J_{\nu}(\tau)|-\infty<\nu<\infty\}$* is a basis of $V$,* 3. (3)

${\rm ord}_{\infty}J_{\nu}(\tau)=\nu$, 4. (4)

[TABLE] 5. (5)

The Fourier series of $J_{\nu}(\tau)$ has integer coefficients, and is of the form $J_{\nu}(\tau)=q^{\nu}+\dots$.

Now let

[TABLE]

This is a modular function on $\Gamma_{0}(121)$ by proposition 2.1, hence $V$ is mapped to itself by the linear transformation,

[TABLE]

where $\lambda$ is an integer. Let $(C_{\mu,v}^{\lambda})_{\mu,\nu}$ be the matrix of the linear transfomation $T_{\lambda}$ with respect to the basis elements $J_{\nu}$.

[TABLE]

Gordon in [8], proved an inequality (equation (17)) about the $11$-adic orders of the matrix elements (denoted by $\pi(C_{\mu,\nu})$).

[TABLE]

Here $[x]$ means the floor function of the real number $x$ and $\delta=\delta(\mu,\nu)$ depends on the residues of $\mu$ and $\nu\pmod{5}$ according to the Table 2.

We can clearly see from the table that $\delta\geq-1$, so we can rewrite 2.4 as,

[TABLE]

Now by Lemma $2.1(v)$, the Fourier series of $T_{\lambda}(J_{\mu})$ has all coefficients divisible by 11 if and only if,

[TABLE]

Now we define $\theta(\lambda,\mu)=1$ if all the coefficients of $U(\phi^{\lambda}J_{\mu})$ divisible by $11$. Otherwise we put $\theta(\lambda,\mu)=0$.

Now from the recurrences obtained in [8] page 119, we have

[TABLE]

Therefore the values of $\theta(\lambda,\mu)$ is completely determined by its values in the range $0\leq\lambda\leq 10$ and $0\leq\mu\leq 4$. Those values are listed in Table 3.

Now we define,

[TABLE]

for any positive integer $r$ and integers $c,d$. We put $A_{0}:=0$.

We construct a sequence of modular functions that are the generating functions for the $p_{[1^{c}{11}^{d}]}(n)$ restricted to certain arithmetic progressions. This generalizes Gordon’s construction for ”$k-$color” partitions. Here we use (2.2) repeatedly.

$L_{0}:=1$, and

[TABLE]

Similarly we can define,

[TABLE]

Now, to get an equation for higher powers, We define,

[TABLE]

where

[TABLE]

Then by a short calculation using (2.2) gives,

[TABLE]

From (2.5), (2.6) and (2.8) we can see that,

[TABLE]

[TABLE]

Since $n_{0}=0$, using above recurrence relations we have,

• $n_{1}=-5\cdot c$
• $n_{2}=-5\cdot 11\cdot d-5\cdot c$

Using induction,

[TABLE]

From this we have that,

$24\cdot n_{2r-1}\equiv(c+11\cdot d)\mod 11^{2r-1}$ and $24\cdot n_{2r}\equiv(c+11\cdot d)\mod 11^{2r}$ .

Therefore, for each $n_{r}$ are integers such that,

[TABLE]

Now, we need to find $\mu_{r}$ in terms of integers $c,d$. Notice that $\mu_{r}$ is the least integer $m$ such that $11^{r}m+n_{r}\geq 0$, which implies that,

[TABLE]

Following Gordon, we represent these formulas in the following form.

[TABLE]

[TABLE]

3. The proofs

If

[TABLE]

we define,

[TABLE]

Here we follow Gordon’s argument to prove $\pi(L_{r})\geq A_{r}(c,d)$.

Proof of Theorem 1.1.

We see that using Proposition 2.1, $L_{r}\in V$ for all $r$. So we have,

[TABLE]

Now by (2.1), (2.8) and (3.2),

[TABLE]

Now we prove by induction,

[TABLE]

[TABLE]

Since $A_{0}=0$, the result holds for $r=0$. Now assume the result is true for $r-1$. Using (3.4) we have,

[TABLE]

Now from equation (3.3) we have,

[TABLE]

From (2.5) and (3.5) the right hand side of (3.6) is at least equal to,

[TABLE]

This expression cannot decrease if $\mu$ is increased by $2$, so its minimum occurs when $\mu=\mu_{r-1}+1$, therefore at,

[TABLE]

Now from (2.8) we have,

[TABLE]

therefore $\mu_{r}\geq\left(\dfrac{5\lambda_{r-1}+\mu_{r-1}}{11}\right).$

Plugging it in (3.8), the right hand side of (3.5) is at least equal to

[TABLE]

since $A_{r-1}+1\geq A_{r}$.

Now consider $\nu=\mu_{r}$ or $\nu=\mu_{r}+1$.

If $\mu=\mu_{r-1},$

[TABLE]

This also works when $\mu\geq\mu_{r-1}+2$, since by induction hypothesis,

[TABLE]

Now consider $\mu=\mu_{r-1}+1$,

Now we need to show,

[TABLE]

Since $\nu=\mu_{r}$ or $\mu_{r}+1$, it suffices to show that when $\theta(\lambda_{r-1},\mu_{r-1})=1$,

[TABLE]

and

[TABLE]

Now from (2.6), Table 2 and Table 3 we see that the above claims hold.

∎

Proof of Corollary 1.2.

.

Recall (2.7),

[TABLE]

Here we follow Gordon’s argument from Section 4 of [8], again with $k$ replaced by $c+11d$. Note that Gordon’s calculations had terms involving $11k$, which will be written in a more symmetric shape here using the fact that $11(c+11d)\equiv 11c+d\pmod{24}$.

[TABLE]

Here $N_{1}$ is the number of odd integers and $N_{2}$ is the number of even integers in the interval $\left[\log_{11}|c+11d|,r-1\right]$ respectively.

[TABLE]

Now if $r\leq\log_{11}|c+11d|+1$ then $N_{1}=N_{2}=0$,

[TABLE]

If $r>\log_{11}|c+11d|+1$,

[TABLE]

Now consider,

[TABLE]

[TABLE]

So we have $A_{r}=\frac{1}{2}\alpha r+\mathcal{O}\left(\log{}|c+11d|\right)$.

Now we prove the condition for $\alpha$. As in [2], the proof is complete once we show that first $\alpha$ only depends $c+11d$ with the period $120$. Periodicity follows by the fact that $\alpha(c+11d)$ is invariant under the each maps,

[TABLE]

∎

.

4. Examples

Here we give three examples to demonstrate our method. The first two examples are Wang’s results and the final one is a new example.

Example 4.1** ($c=1$ and $d=1$).**

[TABLE]

In this case, $\lambda_{i}=1$, $\mu_{i}=1$ for all $i$ and $\theta(1,1)=1$. So $A_{r}=r$ and $n_{r}=-\frac{11^{r}-1}{2}$. thus we have,

[TABLE]

In [1], Wang obtained $n_{r}=-\frac{11^{r}+1}{2}$, which is slightly different from the $n_{r}$ given here, though one can obtain Wang’s $n_{r}$ by setting $m\mapsto m-1$, thus both congruences are equivalent.

Now, we illustrate corollary 1.2 using this example. Since $c+11d=12$ from table 1, we have $\alpha=2$ so $A_{r}$ is asymptotically equals to $r$.

Example 4.2** ($c=1$ and $d=-1$).**

In this case, $\lambda_{i}$ is 1 if $i$ even or is -1 if $i$ is odd, we also have using (3.9), $\mu_{i}$ is 1 if $i$ is odd and it is 0 if $i$ is even. By (2.10) and table 3, we have $n_{2r}=5\cdot\frac{11^{2r}-1}{12}$ and $A_{2r}=r$.

[TABLE]

In view of corollary 1.2, we have $\alpha=1$ from table 1 since $c+11d=-10$. Notice here we used the fact that $\alpha$ is a periodic function of $c+11d$ with period 120. Now we have $A_{2r}$ asymptotically equals to $r$ with the error ‘$\epsilon’=0$.

Example 4.3** ($c=2$ and $d=7$).**

Then we have $\lambda_{i}$ is 2 if $i$ is even or 7 if $i$ is odd, we also have $n_{2r}=\frac{7\cdot 11^{2r}-79}{24}$ and $A_{2r}=2r-1$. In this case and the most cases $\mu_{r}$ are not immediately periodic. Here $\mu_{0}=0$ and $\mu_{1}=1$ afterwords, $\mu_{i}$ is $4$ if $i$ is odd or 2 if $i$ is even.

[TABLE]

In this case $c+11d=79$, now from the table 1, $\alpha=2$ so $A_{2r}$ asymptotically $2r$ with the error $\epsilon=-1$.

Acknowledgements

The author thanks thesis advisor Karl Mahlburg for suggesting this problem and for his guidance during this project.