$23$-REGULAR PARTITIONS AND MODULAR FORMS WITH COMPLEX MULTIPLICATION

Abstract

A partition of a positive integer n is called $\ell $-regular if none of its parts is divisible by $\ell $. Denote by $b_{\ell }(n)$ the number of $\ell $-regular partitions of n. We give a complete characterisation of the arithmetic of $b_{23}(n)$ modulo $11$ for all n not divisible by $11$ in terms of binary quadratic forms. Our result is obtained by establishing a relation between the generating function for these values of $b_{23}(n)$ and certain modular forms having complex multiplication by ${\mathbb Q}(\sqrt {-69})$.

1 Introduction

A partition of a positive integer n is a nonincreasing sequence of positive integers whose sum is n. For $\ell>1$ , a partition is called $\ell $ -regular if none of its parts is divisible by $\ell $ . Denoting by $b_{\ell }(n)$ the number of $\ell $ -regular partitions of n and adopting the convention $b_{\ell }(0):=1$ , we have the identity

$$ \begin{align*} \sum_{n=0}^{\infty} b_{\ell}(n)q^n = \prod_{n=1}^{\infty} \bigg( \frac{1-q^{\ell n}}{1-q^n} \bigg). \end{align*} $$

A substantial body of results on the arithmetic of $b_{\ell }(n)$ modulo m has been established. Perhaps the most elegant of these concerns the parity of $b_2(n)$ , where Euler’s pentagonal number theorem implies that $b_2(n)$ is odd precisely when $24n+1$ is a perfect square. This result has been extended in a number of directions. For example, Ono and the author [4] gave a description of the behaviour of $b_2(n)$ modulo $8$ in terms of the arithmetic of the ring ${\mathbb Z}[\sqrt {-6}]$ , while Gordon and Ono [3] proved that if p is a prime divisor of $\ell $ such that $p^a \mid \ell $ and $p^a \geq \sqrt {\ell }$ , then for every positive integer j, the congruence $b_{\ell }(n) \equiv 0 \pmod {p^{\,j}}$ holds for almost all $n \geq 0$ , that is, on a set of nonnegative integers of density one.

Addressing the case where $\ell $ and m are coprime, Ahlgren and Lovejoy [1] showed that if $p \geq 5$ is prime, then $b_2(n) \equiv 0 \pmod {p^{\,j}}$ for at least ${(p+1)}/{2p}$ of the values of n. In [5], the author extended this work to odd primes $\ell \leq 23$ , and also proved that $b_{23}(n) \equiv 0 \pmod {11}$ for at least ${10}/{11}$ of the values of n by showing that this congruence holds for almost all n not divisible by $11$ (see, for example, [2, 9, 10] for further results). In this paper, we use the theory of modular forms to refine the latter result, giving a complete characterisation of the arithmetic of $b_{23}(n)$ modulo $11$ for n with $11 \nmid n$ .

Our description can be phrased in terms of binary quadratic forms. To give a flavour here (see Section 3 for complete results), for $a,b,c \in {\mathbb Z}$ , let

$$ \begin{align*} F_{a,b,c}(X,Y) = aX^2+bXY+cY^2. \end{align*} $$

Given an odd prime p with $\big(\frac{-69}{p}\big)=1$ , the classical theory implies that there exist integers x and y with

(1.1) $$ \begin{align} p=F_{a,b,c}(x,y) \end{align} $$

for a unique triple $(a,b,c)$ in the set

$$ \begin{align*} \{(1,0,69),(3,0,23),(2,2,35),(6,6,13),(5,2,14),(7,2,10)\}. \end{align*} $$

Theorem 1.1. Suppose n is a positive integer such that $12n+11=p$ is prime.

1. (i) If $\big(\frac{-69}{p}\big)=-1$ , then $b_{23}(n) \equiv 0 \pmod {11}$ .

2. (ii) If $\big(\frac{-69}{p}\big)=1$ , let x and y be integers satisfying (1.1) and let $\delta _p \in \{-1,1\}$ be defined as in Section 3 (see (3.1)). Then

   $$ \begin{align*} b_{23}(n) \equiv \bigg( \frac{p}{11} \bigg) \cdot \delta_p \cdot 9yp^4 \pmod{11}. \end{align*} $$
   In particular, $b_{23}(n) \equiv 0 \pmod {11}$ if and only if $11 \mid y$ .

Theorem 1.1 addresses the case where $12n+11$ is prime. For general n with $11 \nmid n$ , our description requires consideration of the prime factorisation of $12n+11$ .

The remainder of the paper is organised as follows. In Section 2, we provide the necessary background on modular forms and construct the forms we require. In Section 3, we state and prove our main results and illustrate their use.

2 Background and two CM-forms

Let k and N be positive integers and $\chi $ a Dirichlet character modulo N. We denote by $M_k(\Gamma _0(N),\chi )$ the space of holomorphic modular forms of weight k on $\Gamma _0(N)$ with Nebentypus $\chi $ , and by $S_k(\Gamma _0(N),\chi )$ the subspace of cusp forms (we suppress $\chi $ when it is the trivial character). Denote by $\chi _{23}$ the character defined by $\chi _{23}( \bullet ) = \big(\frac{23}{\bullet}\big)$ and write $q := e^{2 \pi i z}$ for complex z with $\mathrm {Im}(z)>0$ .

We begin by stating a slight refinement of Theorem 2.1 in [5] which implies that the values of $b_{23}(n)$ with $11 \nmid n$ can be realised modulo $11$ as the Fourier coefficients of a modular form of integer weight. (Since this result can be established in the same way as Proposition 2.1 in [6], we do not include the proof.)

Proposition 2.1. There exists a cusp form

$$ \begin{align*} H(z) \in S_{1330}(\Gamma_0(144 \cdot 11^3 \cdot 23),\chi_{23}) \cap {\mathbb Z}[[q]] \end{align*} $$

such that

$$ \begin{align*} H(z) \equiv \sum_{n=0}^{\infty} \bigg( \frac{n}{11} \bigg) b_{23}(n) q^{12n+11} \pmod{11}. \end{align*} $$

Following closely the exposition in [7], we now construct two normalised Hecke eigenforms with complex multiplication that turn out to be related to $H(z)$ modulo $11$ . Let $K = {\mathbb Q}( \sqrt {-69} )$ , a number field of discriminant $d=-276$ , and define the ideal ${\mathcal F}$ of the ring of integers ${\mathcal O}_K = {\mathbb Z}[\sqrt {-69}]$ by

$$ \begin{align*} {\mathcal F} = ( 6, 2 \sqrt{-69} ). \end{align*} $$

We note that $N({\mathcal F})=12$ , and from now on, we write $\theta := \sqrt {-69}$ . One can verify that

$$ \begin{align*} \{ 1,5,2+\theta,4+\theta \} \end{align*} $$

is a set of representatives of $( {\mathcal O}_K / {\mathcal F})^*$ , and since $( {\mathcal O}_K / {\mathcal F})^* \cong {\mathbb Z}_2 \times {\mathbb Z}_2$ , it follows that the function $\rho : ( {\mathcal O}_K / {\mathcal F})^* \to {\mathbb C}^*$ defined by

$$ \begin{align*} \rho(1)=1 , \quad \rho(5)=-1, \quad \rho(2+\theta)=1, \quad \rho(4+\theta)=-1 \end{align*} $$

is a group homomorphism.

Next, we define Hecke characters $c_1$ and $c_2$ on the fractional ideals of K coprime to ${\mathcal F}$ . We begin by defining $c_1$ on principal ideals of ${\mathcal O}_K$ via

(2.1) $$ \begin{align} c_1((a+b \theta)) = \rho(a+b \theta) (a+b \theta)^{49} \end{align} $$

for $a,b \in {\mathbb Z}$ . Since ${\mathcal O}_K$ has class number eight, we must extend $c_1$ . One can check that the class group is isomorphic to ${\mathbb Z}_2 \oplus {\mathbb Z}_4$ and is generated by the nonprincipal ideals $(5,1+\theta )$ and $(23,\theta )$ which satisfy $(5,1+\theta )^4 = (2-3\theta )$ and $(23,\theta )^2=(23)$ . Letting

(2.2) $$ \begin{align} c_1((5,1+\theta)) = \bigg( \frac{1}{2}\bigg(\sqrt{10+3\sqrt{6}} - i \sqrt{10-3\sqrt{6}}\bigg) \bigg)^{49} \end{align} $$

and

(2.3) $$ \begin{align} c_1((23,\theta)) = ( \sqrt{-23})^{49} = 23^{24} \sqrt{-23} \end{align} $$

allows us to complete $c_1$ to a Hecke character of K with exponent $49$ and conductor ${\mathcal F}$ , which in turn yields a Dirichlet character $\omega _1$ defined by

$$ \begin{align*} \omega_1(n) = c_1( n {\mathcal O}_K ) / n^{49} \end{align*} $$

for $n \in {\mathbb Z}$ coprime to $6$ . We define companion characters $c_2$ and $\omega _2$ similarly, with $c_2$ extending to nonprincipal ideals via

$$ \begin{align*} c_2((5,1+\theta)) = i c_1((5,1+\theta)) \end{align*} $$

and

$$ \begin{align*} c_2((23,\theta)) = - c_1((23,\theta)). \end{align*} $$

For $v \in \{1,2\}$ , let

$$ \begin{align*} g_{v}(z) = \sum c_{v}({\mathcal I}) q^{N({\mathcal I})} : = \sum_{n=1}^{\infty} s_{v}(n) q^n, \end{align*} $$

where the first sum is over all ideals ${\mathcal I}$ of ${\mathcal O}_K$ coprime to ${\mathcal F}$ . Then $g_v$ is a normalised Hecke eigenform of weight $50$ , level $|d| \cdot N({\mathcal F}) = 144 \cdot 23$ and character $\epsilon _K \omega _v$ , where $\epsilon _K(p) = \big(\frac{-69}{p}\big)$ for any prime $p \not \in \{2,3,23\}$ . Note that for such a prime, we have $\omega _v(p) = \rho (p) = \big(\frac{p}{3}\big)$ . Then since $\big(\frac{-69}{p}\big) \big(\frac{p}{3}\big) = \big(\frac{23}{p}\big)$ by quadratic reciprocity, it follows that

$$ \begin{align*} g_v(z) \in S_{50}( \Gamma_0(144 \cdot 23),\chi_{23}). \end{align*} $$

The fact that $g_v(z)$ is an eigenform implies that

(2.4) $$ \begin{align} s_v(mn)=s_v(m)s_v(n) \quad\mbox{when } (m,n)=1 \end{align} $$

and

(2.5) $$ \begin{align} s_v(p^{\, j+1}) = s_v(p) s_v(p^{\,j}) - \chi_{23}(p) p^{49} s_v(p^{\,j-1}) \end{align} $$

for all primes $p \geq 5$ and $j \geq 1$ .

3 Proofs of the main results

We begin by finding explicit expressions for the Fourier coefficients $s_v(p)$ of the CM-forms $g_v(z)$ which can be leveraged to evaluate the coefficients of two auxiliary forms (see $h_v(z)$ below) modulo $11$ . An odd prime p satisfying $\big(\frac{-69}{p}\big)=-1$ is inert in K, which implies that $s_v(p)=0$ . Since $23$ ramifies in K, we have $s_v(23)=c_v((23,\theta ))$ . Now suppose p is an odd prime with $\big(\frac{-69}{p}\big)=1$ and that $x,y \in {\mathbb Z}$ are as in (1.1). Define $\delta _p$ by

(3.1) $$ \begin{align} \delta_p = \begin{cases} \rho(x+y \theta) & \mathrm{if}\ (a,b,c)=(1,0,69), \\ \rho((9x+16y)-(x-y) \theta) & \mathrm{if}\ (a,b,c)=(6,6,13), \\ \rho((5x+y)-y \theta) & \mathrm{if}\ (a,b,c)=(5,2,14), \\ \rho((7x+y)+y \theta) & \mathrm{if}\ (a,b,c)=(7,2,10), \\ \rho(23y+x\theta) & \mathrm{if}\ (a,b,c)=(3,0,23), \\ \rho((5x-32y)+(x+3y)\theta) & \mathrm{if}\ (a,b,c)=(2,2,35), \end{cases} \end{align} $$

and for $n \geq 1$ , let

$$ \begin{align*} t_1(n) = \begin{cases} s_1(n) & \mathrm{if}\ n \equiv 1 \pmod{12}, \\ s_1(n) / i\sqrt{10 - 3\sqrt{6}} & \mathrm{if}\ n \equiv 5 \pmod{12}, \\ s_1(n) / \sqrt{14-5\sqrt{6}} & \mathrm{if}\ n \equiv 7 \pmod{12}, \\ s_1(n) / \sqrt{-23} & \mathrm{if}\ n \equiv 11 \pmod{12}. \end{cases} \end{align*} $$

We define $t_2(n)$ in the same way as $t_1(n)$ except that we divide $s_2(n)$ by $i\sqrt {10 + 3\sqrt {6}}$ when $n \equiv 5 \pmod {12}$ , and by $\sqrt {14 + 5\sqrt {6}}$ when $n \equiv 7 \pmod {12}$ . Our next result (in the proof of which we will see that $t_1(n) \in {\mathbb Z}[\sqrt {6}]$ ) gives expressions for $t_1(p)$ modulo $11$ in terms of the six reduced binary quadratic forms of discriminant $-276$ mentioned in the introduction. Recall that $\theta = \sqrt {-69}$ .

Theorem 3.1. Suppose p is an odd prime with $\big(\frac{-69}{p}\big)=1$ . Let $x,y \in {\mathbb Z}$ be defined by (1.1), replacing x and y by $-x$ and $-y$ if necessary so that $-x-2y+ \sqrt {6}y>0$ if $(a,b,c)=(5,2,14)$ and $x-2y - \sqrt {6}y>0$ if $(a,b,c)=(7,2,10)$ . Then, modulo the ideal of ${\mathbb Z}[\sqrt {6}]$ generated by $11$ ,

$$ \begin{align*} \delta_p t_1(p) \equiv \begin{cases} 2xp^4 & \mathrm{if}\ (a,b,c)=(1,0,69), \\ (2x+y)p^4 \cdot \sqrt{6} & \mathrm{if}\ (a,b,c)=(6,6,13), \\ (x+2y+\sqrt{6}y) p^4 & \mathrm{if}\ (a,b,c)=(5,2,14), \\ (x-2y-\sqrt{6}y) p^4 & \mathrm{if}\ (a,b,c)=(7,2,10), \\ 9yp^4 & \mathrm{if}\ (a,b,c)=(3,0,23), \\ yp^4 \cdot \sqrt{6} & \mathrm{if}\ (a,b,c)=(2,2,35). \end{cases} \end{align*} $$

Proof. For the sake of brevity, we provide details for only two of the cases, as the others can be handled similarly.

For $(a,b,c)=(2,2,35)$ , we begin with the ideal calculation

$$ \begin{align*} (5,1+\theta)^2 (23,\theta)(47,5+\theta) = (46-19\theta). \end{align*} $$

Combining this with (2.1), (2.2), (2.3) and the multiplicativity of $c_1$ yields

$$ \begin{align*} c_1((47,5+\theta)) = \bigg( \frac{5+\theta}{\sqrt{2}} \bigg)^{49}. \end{align*} $$

Now suppose that $p>47$ is a prime with $p=2x^2+2xy+35y^2$ for some $x,y \in {\mathbb Z}$ , and denote by ${\mathfrak P}$ and ${\mathfrak P}'$ the prime ideals of ${\mathcal O}_K$ lying above p. It is easy to verify that if $a+b \theta \in (47,5+\theta )$ for some $a,b \in {\mathbb Z}$ , then $a \equiv 5b \pmod {47}$ . Using this and the equalities

$$ \begin{align*} 47p = (5x-32y)^2+69(x+3y)^2 = (5x+37y)^2+69(x-2y)^2, \end{align*} $$

we find that the elements

$$ \begin{align*} \mu = (5x-32y)+(x+3y)\theta \quad \mathrm{and} \quad \nu = (5x+37y)+(x-2y)\theta \end{align*} $$

generate the principal ideals $(47,5+\theta ) {\mathfrak P}$ and $(47,5+\theta ){\mathfrak P}'$ . Since

$$ \begin{align*} \mu \nu = -44x^2-44xy-770y^2+(10x^2+10xy+175y^2)\theta , \end{align*} $$

it follows that

$$ \begin{align*} \rho(\,\mu)\rho(\nu) = \rho(\,\mu \nu) = \rho(4+\theta)=-1, \end{align*} $$

and hence $s_1(p) = c_1({\mathfrak P})+c_1({\mathfrak P}')$ is equal to

$$ \begin{align*} \rho(\,\mu) \bigg[ \bigg( \sqrt{2}x + \frac{\sqrt{2}}{2} y + \frac{y}{2} \sqrt{-138} \bigg)^{49} - \bigg( \sqrt{2}x + \frac{\sqrt{2}}{2} y - \frac{y}{2} \sqrt{-138} \bigg)^{49} \bigg]. \end{align*} $$

A calculation now shows that $s_1(p)$ has the form $\rho (\,\mu ) yC(x,y) \sqrt {-138}$ , where $C(x,y) \in {\mathbb Z}[x,y]$ is a homogeneous polynomial of degree $48$ . Using Fermat’s Little Theorem we find that $yC(x,y)$ is congruent modulo $11$ to $y(2x^2+2xy+35y^2)^4$ , and since $p \equiv 11 \pmod {12}$ , our result follows.

Now let $(a,b,c)=(7,2,10)$ . We begin by using the equality

$$ \begin{align*} (5,1+\theta)(23,\theta)(7,6+\theta) = (23-2\theta) \end{align*} $$

as above to deduce that

$$ \begin{align*} c_1((7,6+\theta)) = \bigg( \frac{1}{2}\bigg(\sqrt{14-5\sqrt{6}} + i \sqrt{14+5\sqrt{6}}\bigg) \bigg)^{49}. \end{align*} $$

Combining this with

$$ \begin{align*} c_1((7,6+\theta))c_1((7,1+\theta)) = c_1((7)) = \rho(7) \cdot 7^{49} = 7^{49} \end{align*} $$

then gives

$$ \begin{align*} c_1((7,1+\theta)) = \bigg( \frac{1}{2}\bigg(\sqrt{14-5\sqrt{6}} - i \sqrt{14+5\sqrt{6}}\bigg) \bigg)^{49}. \end{align*} $$

Next suppose that $p>7$ is a prime with $p=7x^2+2xy+10y^2$ . Then $7p = (7x+y)^2 + 69y^2$ , and by replacing x by $-x$ and y by $-y$ if necessary we may ensure that $x-2y-\sqrt {6}y>0$ . Let z be an integer such that

$$ \begin{align*} z(7x+y) \equiv -69y \pmod{p}. \end{align*} $$

One can verify that $z^2 \equiv -69 \pmod {p}$ (which implies that $(p, \pm z+\theta )$ are the prime ideals of ${\mathcal O}_K$ above p) and

$$ \begin{align*} (7,1+\theta)(p,z+\theta) = ((7x+y)+y\theta). \end{align*} $$

It follows that

$$ \begin{align*} c_1((p,& \,z+\theta)) = \rho((7x+y)+y\theta)) \\ & \cdot\bigg( \frac{1}{2}\bigg(\sqrt{2p+(-5x^2-8xy+6y^2)\sqrt{6}} + s \cdot \sqrt{-2p+(-5x^2-8xy+6y^2)\sqrt{6}}\bigg) \bigg)^{49}, \end{align*} $$

where $s = \mathrm {sign}(x^2-4xy-2y^2)$ .

Since $p \equiv 7 \pmod {12}$ ,

$$ \begin{align*} c_1((p,z+\theta))c_1((p,-z+\theta)) = c_1((p)) = \rho(p) p^{49} = p^{49}. \end{align*} $$

Hence, $c_1((p,-z+\theta )) = \overline {c_1((p,z+\theta ))}$ , and a calculation reveals that

(3.2) $$ \begin{align} s_1(p) = \sqrt{14-5 \sqrt{6}} \cdot \rho((7x+y)+y \theta)(x-2y-\sqrt{6}y) D(x,y) \end{align} $$

with $D(x,y) \in {\mathbb Z}[\sqrt {6}][x,y]$ . Our result now follows as in the previous case.

Remark 3.2. For our claim that $t_1(n) \in {\mathbb Z}[\sqrt {6}]$ , note that if $p=5x^2+2xy+14y^2$ , then $p \equiv 5 \pmod {12}$ , and an argument similar to the one given in our proof of (3.2) for primes represented by $F_{7,2,10}$ shows that

(3.3) $$ \begin{align} s_1(p) = i \sqrt{10 - 3\sqrt{6}} \cdot \rho((5x+y)-y \theta) (x+2y+\sqrt{6}y) E(x,y) \end{align} $$

for some $E(x,y) \in {\mathbb Z}[\sqrt {6}][x,y]$ . Moreover, observe that

$$ \begin{align*} i \sqrt{10 - 3\sqrt{6}} \cdot \sqrt{14-5\sqrt{6}} = (-2+\sqrt{6}) \sqrt{-23}. \end{align*} $$

Remark 3.3. The parallel result to Theorem 3.1 for $t_2(p)$ states that for each odd prime p with $\big(\frac{-69}{p}\big)=1$ ,

$$ \begin{align*} \delta_p t_2(p) \equiv \begin{cases} 2xp^4 & \mathrm{if}\ (a,b,c)=(1,0,69), \\ -(2x+y)p^4 \cdot \sqrt{6} & \mathrm{if}\ (a,b,c)=(6,6,13), \\ (-x-2y+\sqrt{6}y) p^4 & \mathrm{if}\ (a,b,c)=(5,2,14), \\ (-x+2y-\sqrt{6}y) p^4 & \mathrm{if}\ (a,b,c)=(7,2,10), \\ -9yp^4 & \mathrm{if}\ (a,b,c)=(3,0,23), \\ yp^4 \cdot \sqrt{6} & \mathrm{if}\ (a,b,c)=(2,2,35). \end{cases} \end{align*} $$

We now show that the behaviour of $b_{23}(n)$ modulo $11$ for $11 \nmid n$ is controlled by the Fourier coefficients of the CM-forms $g_v(z)$ defined in Section 2. We do this by establishing a relation between the coefficients of the form $H(z)$ from Proposition 2.1 and those of two auxiliary forms built from the $g_v(z)$ which we now define.

For $n \geq 1$ , write

$$ \begin{align*} t_1(n) = \alpha_n + \beta_n \sqrt{6} \end{align*} $$

with $\alpha _n,\beta _n \in {\mathbb Z}$ . Using Theorem 3.1 and Remark 3.3 along with (2.4) and (2.5), it is straightforward to show that $t_2(n) = - \alpha _n + \beta _n \sqrt {6}$ when $n \equiv 11 \pmod {12}$ . For each $v \in \{1,2\}$ , we define the modular form

$$ \begin{align*} h_v(z) = \sum_{n \equiv 11 \!\!\pmod{12}} t_v(n) q^n \in M_{50}(\Gamma_0(144 \cdot 23),\chi_{23}). \end{align*} $$

Next, let

$$ \begin{align*} F(z) := \frac{h_1(z)-h_2(z)}{2} = \sum_{n \equiv 11 \!\!\pmod{12}} \alpha_n q^n \end{align*} $$

and

$$ \begin{align*} G(z) := \frac{h_1(z)+h_2(z)}{2\sqrt{6}} = \sum_{n \equiv 11 \!\!\pmod{12}} \beta_n q^n. \end{align*} $$

Theorem 3.4. Suppose n is a positive integer with $11 \nmid n$ . Then

$$ \begin{align*} b_{23}(n) \equiv \bigg( \frac{n}{11} \bigg) (\alpha_{12n+11} + 9 \beta_{12n+11}) \pmod{11}. \end{align*} $$

Proof. Note first that if $H(z)$ is the modular form in Proposition 2.1, then our result is equivalent to the integral power series congruence

$$ \begin{align*} H(z) \equiv F(z) + 9G(z) \pmod{11}. \end{align*} $$

Recall the weight ten normalised Eisenstein series

$$ \begin{align*} E_{10}(z) = 1 - 264 \sum_{n=1}^{\infty} \sigma_9(n)q^n \in M_{10}(\Gamma_0(1)), \end{align*} $$

where $\sigma _9(n) = \sum _{d \mid n} d^9$ . Since $F(z)+9G(z) \in M_{50}( \Gamma _0(144 \cdot 23),\chi _{23})$ ,

$$ \begin{align*} (F(z)+9G(z)) E_{10}(z)^{128} \in M_{1330}( \Gamma_0(144 \cdot 23),\chi_{23}), \end{align*} $$

and hence both $(F(z)+9G(z)) E_{10}(z)^{128}$ and $H(z)$ lie in the space

$$ \begin{align*} M_{1330}( \Gamma_0(144 \cdot 11^3 \cdot 23),\chi_{23}). \end{align*} $$

Since $E_{10}(z) \equiv 1 \pmod {11}$ , on checking that the coefficients of the Fourier expansions of $H(z)$ and $F(z)+9G(z)$ agree modulo $11$ as far as the Sturm bound [8], our proof is complete.

Remark 3.5. Theorems 3.1 and 3.4, when used in conjunction with (2.4), (2.5), the value of $s_1(23)$ and the fact that $s_1(p)=0$ for odd primes p with $\big(\frac{-69}{p}\big)=-1$ , allow one to evaluate $b_{23}(n)$ modulo $11$ for any n not divisible by $11$ .

To illustrate, first let $n=197$ . Since $12n+11=5^3 \cdot 19$ , we begin by noting that

$$ \begin{align*} 5=F_{5,2,14}(-1,0) \quad \mathrm{and} \quad 19=F_{7,2,10}(-1,-1). \end{align*} $$

By (3.3) and (3.2),

$$ \begin{align*} s_1(5) = i \sqrt{10-3 \sqrt{6}} \cdot \rho(-5)(-1)E(-1,0) = - i \sqrt{10-3 \sqrt{6}} \cdot E(-1,0) \end{align*} $$

and

$$ \begin{align*} s_1(19) = \sqrt{14-5 \sqrt{6}} \cdot \rho(-8 - \theta) (1 + \sqrt{6}) D(-1,-1) = \sqrt{14-5 \sqrt{6}} \cdot (-1 - \sqrt{6}) D(-1,-1). \end{align*} $$

Then (2.5) yields

$$ \begin{align*} s_1(5^2) = (-10+3 \sqrt{6}) E(-1,0)^2 + 5^{49} \end{align*} $$

and

$$ \begin{align*} s_1(5^3) = -i \sqrt{10-3 \sqrt{6}} \cdot [ (-10+3 \sqrt{6}) E(-1,0)^3 + 2 \cdot 5^{49}E(-1,0)], \end{align*} $$

and so (2.4) allows us to conclude that $s_1(5^3 \cdot 19)$ is equal to

$$ \begin{align*} ( 4 - \sqrt{6} ) D(-1,-1) [(-10+3 \sqrt{6} ) E(-1,0)^3 + 2 \cdot 5^{49}E(-1,0) ] \sqrt{-23}. \end{align*} $$

Since $D(-1,-1) \equiv 19^4 \equiv 4 \pmod {11}$ , $E(-1,0) \equiv 5^4 \equiv 9 \pmod {11}$ and $5^{49} \equiv 9 \pmod {11}$ , it follows that

$$ \begin{align*} t_1( 5^3 \cdot 19 ) \equiv 4 + \sqrt{6} \pmod{11}. \end{align*} $$

Theorem 3.4 then yields

$$ \begin{align*} b_{23}(197) \equiv \bigg( \frac{197}{11} \bigg) (4 + 9 \cdot 1) \equiv 9 \pmod{11}, \end{align*} $$

which one may verify using the exact value $b_{23}(197)=2626664703430$ .

Now let $n=102$ . Since $12n+11=5 \cdot 13 \cdot 19$ , we proceed by noting that $13=F_{6,6,13}(0,1)$ , which gives

$$ \begin{align*} s_1(13) = t_1(13) \equiv \rho (16 + \theta) (1) (13^4) \sqrt{6} \equiv 6 \sqrt{6} \pmod{11}. \end{align*} $$

Combining this with our values of $s_1(5)$ and $s_1(19)$ , we can calculate

$$ \begin{align*} t_1(5 \cdot 13 \cdot 19) \equiv 2 + 6 \sqrt{6} \pmod{11}, \end{align*} $$

and hence Theorem 3.4 implies that

$$ \begin{align*} b_{23}(102) \equiv \bigg( \frac{102}{11} \bigg) (2 + 9 \cdot 6) \equiv 1 \pmod{11}, \end{align*} $$

which can be checked against $b_{23}(102)= 226889906$ .