TWO-DIVISIBILITY OF THE COEFFICIENTS OF CERTAIN WEAKLY HOLOMORPHIC MODULAR FORMS

DARRIN DOUD, PAUL JENKINS, AND JOHN LOPEZ

Abstract. We study a canonical basis for spaces of weakly holomorphic modular forms of weights 12, 16, 18, 20, 22, and 26 on the full modular group. We prove a relation between the Fourier coefficients of modular forms in this canonical basis and a generalized Ramanujan τ-function, and use this to prove that these Fourier coefficients are often highly divisible by 2.

1. Introduction

The unique cusp form Δ(z) of weight 12 for SL_{2}(ℤ) has a Fourier expansion given by

Δ(z) = q ∏_{n=1}^{∞}(1 − q^{n})^{24} = ∑_{n=1}^{∞} τ(n)q^{n},

where q = e^{2πiz}; the Fourier coefficients of Δ(z) are the values of the Ramanujan τ-function. Since Δ(z) is a Hecke eigenform, the function τ(n) is multiplicative, so that τ(mn) = τ(m)τ(n) for (m, n) = 1, and for a prime p and a positive integer a ≥ 2, we have

τ(p^{a}) = τ(p)τ(p^{a−1}) − p^{11}τ(p^{a−2}).

Congruences for the values of τ(n) have been studied by many authors, as detailed by Swinnerton-Dyer in [17, 18]. For instance, Ramanujan [15], [10, p. 165] showed that

τ(2n) ≡ 0 (mod 2),
τ(3n) ≡ 0 (mod 3),
τ(5n) ≡ 0 (mod 5),

and that

τ(7n), τ(7n + 3), τ(7n + 5), τ(7n + 6) ≡ 0 (mod 7).

More generally, he showed [5] that with σ_{k}(n) = ∑_{d|n} d^{k},

τ(2n + 1) ≡ σ_{11}(2n + 1) (mod 2^{8}),
τ(n) ≡ n^{2}σ_{7}(n) (mod 27),
τ(n) ≡ nσ_{9}(n) (mod 25),
τ(n) ≡ nσ_{3}(n) (mod 7).

Swinnerton-Dyer [17, 18] showed that even stronger forms of these congruences may be derived from the existence of certain ℓ-adic representations ρ_{ℓ} : G_{ℚ} → GL_{2}(ℤ_{ℓ}); the existence of such representations was conjectured by Serre and proved by Deligne. The idea is that when the image of ρ_{ℓ} is “small”, then ℓ-adic knowledge of the determinant of an element provides information about the trace and thus gives a congruence for τ(p) (mod ℓ). This representation can also be used in conjunction with the Chebotarev density theorem to show that τ(p) is usually nonzero. However, these methods for obtaining congruences generally depend on the fact that Δ(z) is a cusp form, and do not easily generalize to meromorphic modular forms.

Date: July 11, 2011.

Congruences similar to those described by Swinnerton-Dyer for τ(n) were proved by Lehner [13] for the coefficients of the weight 0 modular j-function

j(z) = \frac{E_{4}^{3}(z)}{Δ(z)} = q^{−1} + ∑_{n=0}^{∞} c(n)q^{n}.

Specifically, Lehner showed that

c(2n) ≡ 0 (mod 2^{11}),
c(3n) ≡ 0 (mod 3^{5}),
c(5n) ≡ 0 (mod 5^{2}),
c(7n) ≡ 0 (mod 7).

In fact, these divisibility properties of c(n) may be strengthened to obtain results modulo powers of these primes; Lehner proved that for odd n and for a ≥ 1,

c(2^{a}n) ≡ 0 (mod 2^{3a+8}),

with similar congruences for the primes 3, 5, and 7 [12, 13]. Kolberg [11] and Aas [1, 2] generalized these divisibility properties to congruences for c(n) modulo even higher powers of these primes.

It is natural to ask whether such divisibility or congruence results hold for the Fourier coefficients of other modular forms. For large families of weakly holomorphic modular forms with small weights, this question has recently been answered affirmatively.

A weakly holomorphic modular form for a subgroup Γ of finite index in SL_{2}(ℤ) is a function that is holomorphic on the upper half plane and satisfies the modular equation f(γz) = (cz + d)^{k}f(z) for some integer weight k and for all γ = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ, with poles of finite order at the cusps of Γ. We shall denote the space of holomorphic modular forms of weight k and level N by M_{k}(N), and the space of weakly holomorphic modular forms of weight k and level N by M_{k}^{!}(N). It is well known that M_{k}(N) is finite dimensional.

If we write k = 12ℓ + k′, with k′ ∈ {0, 4, 6, 8, 10, 14}, there is a canonical basis {f_{k,m} : m ≥ −ℓ} for M_{k}^{!}(1) consisting of forms with Fourier expansions

f_{k,m}(z) = q^{−m} + ∑_{n=ℓ+1}^{∞} a_{k}(m, n)q^{n}.

We define a_{k}(m, n) to be zero when either m or n is nonintegral, and when m < −ℓ or n < ℓ+1. The Fourier coefficients a_{k}(m, n) possess a remarkable duality property relating weights k and 2 − k, proven by Duke and the second author in [7]. Namely, for all even k and all m, n ∈ ℤ,

a_{k}(m, n) = −a_{2−k}(n, m).

Using this duality, the first and second authors [6] proved that for k ∈ {4, 6, 8, 10, 14} the coefficients a_{k}(m, n) are often divisible by high powers of 2, 3, 5, and 7. As an example, for odd m, n and a, b > 0,

a_{4}(2^{a}m, 2^{b}n) ≡ \begin{cases}0 (mod 2^{7}) & if a > b, \\ 0 (mod 2^{3(b−a)+7}) & if b > a. \end{cases}

An important part of the proof is the one-dimensionality of M_{k}(1) for these weights. Similar results were proved by Griffin in [9] for the weight k = 0.

In this paper, we will prove similar divisibility results for weakly holomorphic modular forms in weights k for which M_{k}(1) is two-dimensional. As is well known, these weights are k ∈ {12, 16, 18, 20, 22, 26}. To be precise, for each weight k, we define constants γ, ρ, χ, ν, η, and ω as follows:

γ = \begin{cases}3 & for k = 12, 16 or 20 \\ 4 & for k = 18 or 26 \\ 5 & for k = 22, \end{cases} ρ = \begin{cases}7 & for k = 12, 20 or 22 \\ 8 & for k = 16 or 18 \\ 9 & for k = 26, \end{cases}

χ = \begin{cases}7 & for k = 12, 20 or 22 \\ 9 & for k = 16 or 18 \\ 8 & for k = 26, \end{cases} ν = \begin{cases}15 & for k = 12 or 26 \\ 16 & for k = 16, 18, 20 or 22, \end{cases}

η = \begin{cases}12 & for k = 12 or 22, \\ 13 & for k = 16 or 20, \\ 10 & for k = 18 or 26, \end{cases} ω = \begin{cases}3 & for k = 18, 22, or 26 \\ 4 & for k = 12 or 20, \\ 6 & for k = 16 \end{cases}

Note that although these constants depend on the weight k, this dependence is omitted from the notation. We will prove the following theorem:

Theorem 1.1. Let m, n be odd positive integers, and let a, b be non-negative integers. For k ∈ {12, 16, 18, 20, 22, 26}, we have

a_{k}(2^{a}m, 2^{b}n) ≡ 0 \{ \begin{matrix}(mod 2^{γb}) & for a = 0,  (see Lemma 7.4) \\ (mod 2^{ν}) & for a > 0, b = 0,  (see Prop. 6.6) \\ (mod 2^{χ}) & for a > b ≥ 1,  (see Prop. 6.8) \\ (mod 2^{ρ+γ(b−a)}) & for b > a ≥ 1.  (see Prop. 7.9) \end{matrix}

In addition, a_{k}(−1, 2^{b}n) ≡ 0 (mod 2^{γb}) (see Lemma 3.3) and

a_{k}(0, 2^{b}n) ≡ 0 \{ \begin{matrix}(mod 2^{η}) & for b = 0,  (see Lemma 6.3), \\ (mod 2^{ω}) & for b > 0.  (see Lemma 6.4), \end{matrix}

Note that there are no results for a = b. In these cases, the best possible exponent appears to be 0, since the corresponding coefficients are often odd.

The remainder of the paper is structured as follows: In Sections 2, 3, and 4, we introduce notation for specific modular forms and operators that will be used throughout the paper, prove a relation between the coefficients a_{k}(m, n) and a generalized Ramanujan τ-function, and compute the orders of poles of certain modular forms at the cusp at 0. In Section 5, we review Kolberg’s method of two-dissections and use it to prove congruences for certain coefficients. In Sections 6 and 7, we prove Theorem 1.1 for the a > b and a < b cases, respectively.

The authors thank the referee for a careful and thoughtful review that greatly improved the paper. Additionally, the third author thanks the Brigham Young University Department of Mathematics for its generous support of his work on this project.

2. Modular forms and Operators

In this section we fix notation for certain operators on spaces of modular forms and for certain specific modular forms that will be used often.

2.1. Operators on modular forms. For p a prime, and f(z) = ∑_{n≥n_{0}} a_{n}q^{n} a weakly holomorphic modular form, we define the U_{p} operator by

(f|U_{p})(z) = ∑_{n≥n_{0}/p} a_{pn}q^{n}.

An alternative definition of f|U_{p} that we will find useful is [3, Thm. 4.5]

(f|U_{p})(z) = \frac{1}{p} ∑_{j=0}^{p−1} f (\frac{z + j}{p} ) .

If f has weight k and level N, then f|U_{p} is again a weakly holomorphic modular form of weight k and level N (if p|N) or level pN (if p ∤ N) [14, Prop. 2.22].

We also define the operators V_{p} by (f|V_{p})(z) = f(pz) and the Hecke operators T_{p} by (f|T_{p}) = (f|U_{p}) + p^{k−1}(f|V_{p}). Note that V_{p} takes modular forms of weight k and level N to forms of weight k and level Np, while T_{p} preserves both weight and level [14, Prop. 2.2, Thm. 4.5].

2.2. Holomorphic Modular Forms. Let

σ_{k}(n) = ∑_{d|n} d^{k}.

We define the Eisenstein series

E_{4}(z) = 1 + 240 ∑_{n=1}^{∞} σ_{3}(n)q^{n} and E_{6}(z) = 1 − 504 ∑_{n=1}^{∞} σ_{5}(n)q^{n},

and we note that these are modular forms of level 1 and weights 4 and 6 respectively. Next we define

S_{4}(z) = \frac{E_{4}(z) − E_{4}(2z)}{240}, S_{6}(z) = \frac{E_{6}(z) − E_{6}(2z)}{−504}

T_{4}(z) = \frac{2^{4}E_{4}(2z) − E_{4}(z)}{2^{4} − 1}, and T_{6}(z) = \frac{2^{6}E_{6}(2z) − E_{6}(z)}{2^{6} − 1}.

We note that each of these forms is a holomorphic modular form of level 2, and weight 4 or 6. The forms S_{4} and S_{6} vanish at infinity to order 1, while the forms T_{4} and T_{6} vanish at 0 to order 1. All of these forms have integer coefficients, and using the valence formula for level 2 modular forms [8], S_{4} and T_{4} are easily seen to be nonvanishing in the upper half plane.

For k ∈ {12, 16, 18, 20, 22, 26}, we have already noted that M_{k}(1) is two-dimensional. It is spanned by the standard Eisenstein series E_{k}(z) and a unique normalized cusp form that is an eigenform of all the Hecke operators. We will denote this eigenform by Δ_{k}(z), and its Fourier coefficients by τ_{k}(n). Hence,

Δ_{k}(z) = ∑_{n=1}^{∞} τ_{k}(n)q^{n}.

We note that Δ_{12} is the modular form Δ and τ_{12} is the Ramanujan τ function. Note that in addition to being the unique normalized cusp form in weight k, Δ_{k} is also the form denoted by us as f_{k,−1}. Hence, we may also write τ_{k}(n) = a_{k}(−1, n).

2.3. Weakly holomorphic modular functions of level 2. We note that Γ_{0}(2) has cusps at 0 and at ∞. Following [6], for k < 0 we define α_{k}(z) (resp. θ_{k}) to be the weakly holomorphic modular form of weight k and level 2 that is holomorphic at ∞ and has a pole of minimal possible order at 0 (resp. holomorphic at 0 and has a pole of minimal possible order at ∞). The fact that there is a unique form with these properties is discussed in [6]. The following fact about these functions is easily proved:

Theorem 2.1. Let k < 0 be even. If k ≡ 2 (mod 4), set i = (6 − k)/4. Then

α_{k}(z) = T_{6}(z)/T_{4}(z)^{i}, θ_{k}(z) = S_{6}(z)/S_{4}(z)^{i}

and

θ_{k}(−1/(2z)) = −2^{3−k}z^{k}α_{k}(z).

If k ≡ 0 (mod 4), set i = −k/4. Then

α_{k}(z) = 1/T_{4}(z)^{i}, θ_{k}(z) = 1/S_{4}(z)^{i}

and

θ_{k}(−1/(2z)) = 2^{−k}z^{k}α_{k}(z).

For each relevant value of k, we define μ_{k} to be the exponent such that θ_{k}(−1/(2z)) = ±2^{μ_{k}}z^{k}α_{k}(z). We have the following congruences on the coefficients of α_{k}, which define constants ξ_{k}.

Lemma 2.2. Let k ∈ {−10, −14, −16, −18, −20, −24}. Then α_{k} ≡ 1 (mod 2^{ξ_{k}}) with

ξ_{−10} = ξ_{−14} = ξ_{−18} = 3, ξ_{−20} = 4, ξ_{−24} = 5, ξ_{−16} = 6.

Proof. This follows from [6, Lemma 6.2]. □

Finally, following Apostol [3, pg. 87], we define

Φ(z) = \frac{Δ(2z)}{Δ(z)}

and ψ(z) = 1/Φ(z). We recall from [3] that both ψ and Φ are in M_{0}^{!}(2), both have integer Fourier coefficients, Φ has a simple zero at ∞ and a simple pole at 0, ψ has a simple pole at ∞ and a simple zero at 0, and

ψ (\frac{−1}{2z}) = 2^{12}Φ(z).

3. Hecke Operators

Using Hecke operators, we now prove a two useful relation between certain coefficients a_{k}(m, n) and the τ_{k} function. In order to state and prove these relations, we define the symbol

δ_{x,y} = \begin{cases}1 & if y|x, \\ 0 & if y ∤ x. \end{cases}

Lemma 3.1. Let k ∈ {12, 16, 18, 20, 22, 26}, let m, n be positive integers, and let p be a prime. Then

a_{k}(m/p, n) + a_{k}(m, p)τ_{k}(n) + p^{k−1}a_{k}(mp, n) = a_{k}(m, np) + p^{k−1}a_{k}(m, n/p).

Proof. Applying the Hecke operator T_{p} to the modular form f_{k,m}, we find

f_{k,m}|T_{p} = p^{k−1}q^{−mp} + δ_{m,p}q^{−m/p} + a_{k}(m, p)q + O(q^{2})
= p^{k−1}f_{k,mp} + δ_{m,p}f_{k,m/p} + a_{k}(m, p)Δ_{k},

The nth coefficient of this form is

p^{k−1}a_{k}(mp, n) + a_{k}(m/p, n) + a_{k}(m, p)τ_{k}(n).

By the definition of the Hecke operator, the nth coefficient of f_{k,m}|T_{p} is

a_{k}(m, np) + p^{k−1}a_{k}(m, n/p).

Hence, the lemma is proved. □

For the second relation, we apply the T_{p} operator twice.

Lemma 3.2. Let k ∈ {12, 16, 18, 20, 22, 26}, let m, n be positive integers, and let p be a prime. Then

p^{2k−2}a_{k}(mp^{2}, n)+(1 + δ_{m,p})p^{k−1}a_{k}(m, n) + a_{k}(m/p^{2}, n) + a_{k}(m, p^{2})τ_{k}(n)

= p^{2k−2}a_{k}(m, n/p^{2}) + (1 + δ_{n,p})p^{k−1}a_{k}(m, n) + a_{k}(m, np^{2}).

Proof. Applying T_{p} to f_{k,m}|T_{p}, we obtain

f_{k,m}|T_{p}|T_{p} = p^{2k−2}q^{−mp^{2}} + (1 + δ_{m,p})p^{k−1}q^{−m} + δ_{m,p^{2}}q^{−m/p^{2}} + a_{k}(m, p^{2})q + O(q^{2})
= p^{2k−2}f_{k,mp^{2}} + (1 + δ_{m,p})p^{k−1}f_{k,m} + δ_{m,p^{2}}f_{k,m/p^{2}} + a_{k}(m, p^{2})Δ_{k}.

Computing the coefficient of q^{n} in this expression gives the left-hand side of the desired equation; computing it directly from the definition of f_{k,m}|T_{p}|T_{p} gives the right hand side.

□

For the remainder of this paper, we will use these lemmas only in the case p = 2. We now prove a divisibility result for the coefficients τ_{k}. In addition to being

important in the remainder of the paper, under the identification τ_{k}(n) = a_{k}(−1, n) this result is part of Theorem 1.1.

Lemma 3.3. Let k ∈ {12, 16, 18, 20, 22, 26}. For a nonnegative integer b and a positive odd integer n, we have

τ_{k}(2^{b}n) ≡ 0 (mod 2^{γb}).

Proof. We note that Δ_{k} is the unique normalized cusp form of weight k, so that it is an eigenform of all the Hecke operators. Hence, we have ([4, Theorem 3]) that

τ_{k}(2^{b+1}n) = τ_{k}(2)τ_{k}(2^{b}n) − 2^{k−1}τ_{k}(2^{b−1}n).

The result then follows from the fact that τ_{k}(2) is divisible by 2^{γ} by induction on b, using that k − 1 > 2γ. □

Another way to write the result of this lemma is to note that a_{k}(−1, 2^{b}n) ≡ 0 (mod 2^{γb}) for b ≥ 0.

4. Operators acting on weakly holomorphic modular forms

The standard U_{p} and V_{p} operators take weakly holomorphic modular forms to weakly holomorphic modular forms, so they preserve modularity and holomorphicity in the upper half plane. In this section, we study the effects that the U_{p} and V_{p} operators have on poles of weakly holomorphic modular forms. Because the actions of U_{p} and V_{p} are easily described in terms of Fourier expansions, the effects on poles at ∞ are easily seen. Hence, we concentrate on studying the poles at 0. A basic result is the following.

Theorem 4.1. [6, Corollary 4.2] Let f be a weakly holomorphic modular form of even weight k and level 1, and let f_{p} = f|U_{p}. Then

p(pz)^{−k}f_{p}(−1/pz) = −f(z) + pf_{p}(pz) + p^{k}f(p^{2}z)

and p(pz)^{−k}f_{p}(−1/pz) is modular of weight k and level p.

We wish to extend this theorem to study f|U_{p^{2}}. Using a proof similar to that found in [6], we obtain

Theorem 4.2. Let f be a weakly holomorphic modular form of even weight k and level 1, and let f_{p^{2}} = f|U_{p^{2}} and f_{p} = f|U_{p}. Then

p(pz)^{−k}f_{p^{2}}(−1/pz) = −f_{p}(z) + pf_{p^{2}}(pz) − p^{k−1}f(pz) + p^{k}f_{p}(p^{2}z) + p^{2k−1}f(p^{3}z).

In addition, p(pz)^{−k}f_{p^{2}}(−1/pz) is modular of weight k and level p.

Proof. We proceed as in the proof of [6, Lemma 4.1]. For an integer j with 1 ≤ j ≤ p − 1, let j′ be the unique solution to jj′ ≡ 1 (mod p) with −(p − 1) < j′ < −1, and set b_{j} = \frac{jj′−1}{p}. We then write

pz^{−k}f_{p^{2}}(−1/z) = pz^{−k} \frac{1}{p} ∑_{j=0}^{p−1} f_{p} (\frac{−1/z + j}{p} )

= z^{−k} ∑_{j=0}^{p−1} f_{p} (\frac{jz − 1}{pz})

= z^{−k} ∑_{j=1}^{p−1} f_{p} ((\begin{matrix}j & b_{j} \\ p & j′ \end{matrix}) \frac{z − j′}{p} ) + z^{−k}f_{p} (\frac{−1}{pz})

= z^{−k} ∑_{j=1}^{p−1} (p (\frac{z − j′}{p} ) + j′)^{k} f_{p} (\frac{z − j′}{p} ) + z^{−k}f_{p} (\frac{−1}{pz})

= z^{−k} ∑_{j=1}^{p−1} z^{k}f_{p} (\frac{z − j′}{p} )

+ z^{−k} (\frac{(pz)^{k}}{p} (−f(z) + pf_{p}(pz) + p^{k}f(p^{2}z)))

= ∑_{j=0}^{p−1} f_{p} (\frac{z + j}{p} ) − f_{p}(z/p) + p^{k−1} (−f(z) + pf_{p}(pz) + p^{k}f(p^{2}z))

= pf_{p^{2}}(z) − f_{p}(z/p) − p^{k−1}f(z) + p^{k}f_{p}(pz) + p^{2k−1}f(p^{2}z)

The formula for p(pz)^{−k}f_{p^{2}}(−1/pz) follows by replacing z by pz in this equation. The statement about the level and the weight is immediate, once we notice that

−f_{p}(z) + pf_{p^{2}}(pz) − p^{k−1}f(pz) + p^{k}f_{p}(p^{2}z) + p^{2k−1}f(p^{3}z)

is equal to

p(f|T_{p}|T_{p}|V_{p}) − f|T_{p} − p^{k}f|V_{p}.
□

We also note the trivial identity below that allows us to find the pole at 0 of f|V_{p}, where f is a weakly holomorphic form of level 1.

Proposition 4.3. Let f be a weakly holomorphic modular form of weight k and level 1. Then

z^{−k}(f|V_{p})(−1/pz) = f(z).

5. Two-dissections

As a first step in proving the divisibility results on the a_{k}(m, n) we begin by reviewing Kolberg’s method of two-dissections [11]. We do this in order to prove a key congruence that will be used in the proof of Lemma 7.5.

Given a power series

f = ∑_{n∈ℤ} a(n)q^{n},

we define the two-dissection f = (f)_{0} + (f)_{1} by

(f)_{0} = f|U_{2}|V_{2} = ∑_{n∈ℤ} a(2n)q^{2n} and (f)_{1} = f − (f)_{0} = ∑_{n∈ℤ} a(2n + 1)q^{2n+1}.

Note that if f is modular of weight k and level N, then both (f)_{0} and (f)_{1} are modular of weight k and level 4N. This two-dissection allows us to separate the coefficients of even and odd powers of q in the expansion of f. If we wish to make a finer distinction than studying the coefficients of odd or even powers of q, we can (after dividing by q in the case of (f)_{1}) replace q^{2} by q in the expansion of (f)_{0} or (f)_{1} and two-dissect the resulting function. By two-dissecting a function repeatedly, we can isolate powers of q with exponent in a fixed residue class modulo a power of 2.

To allow us to easily compute two-dissections, we define (following Kolberg) the following functions:

φ(q) = ∏_{k=1}^{∞}(1 − q^{k}), Q(q) = ∏_{k=1}^{∞}(1 + q^{k}) = \frac{φ(q^{2})}{φ(q)},

R(q) = Q(q^{2}), S(q) = R(q^{2}), T (q) = S(q^{2}).

Note that although the expansion of Q contains odd powers of q, the expansions of R, S, and T contain only even powers of q. In addition, under the transformation taking q^{2} to q, we see that R maps to Q, S maps to R, and T maps to S.

Kolberg showed [11, p. 5], that we can write

Q^{2} = R(cos α + i sin α)

with cos(α) = R^{2}S^{3}T^{−2} and sin(α) = −2iqR^{2}ST^{2}. One then checks that (Q^{2})_{0} = R cos α and (Q^{2})_{1} = iR sin(α). We then compute the two-dissection of Q^{2k} for any integer value of k by setting

Q^{2k} = R^{k}(cos(kα) + i sin(kα))

and using trigonometric identities to compute cos(kα) and sin(kα) in terms of Q, R, S, and T. For instance

sin(2α) = 2 sin(α) cos(α) = −4iqR^{4}S^{4}.

Other basic identities used by Kolberg are

cos(2α) = R^{8}S^{−4} and sin(4α) = −8iqR^{12}.

We have written a computer program to compute two-dissections of q-series. Since our purpose is to compute congruences modulo powers of two, we are interested in computing two-dissections with few terms modulo given powers of two.

For odd dissections of powers of Q, the program needs to compute sin(kα) in terms of Kolberg’s functions. It does this by using Chebyshev polynomials to write sin(kα) as sin(α) times a polynomial in cos(α) when k is odd, as sin(2α) times a polynomial in cos(2α) if k is even, or as sin(4α) times a polynomial in cos(4α) = 1 − 2 sin^{2}(2α) if k is divisible by 4).

For even dissections of powers of Q, the program must compute cos(kα) in terms of Kolberg’s functions. To do this for odd values of k, it uses Chebyshev polynomials to write cos(kα) as a polynomial in cos(α). If k is divisible by 4, it computes k as a polynomial in sin^{2}(2α). Otherwise it uses the identity cos(kα) = 1 − 2 sin^{2}((k/2)α) to reduce the problem to computing a sine function, as described above.

To illustrate this, we demonstrate the two-dissection of Q^{16}. We note that we can write cos(8α) = 1 − 8 sin^{2}(2α) + 8 sin^{4}(2α), so we have

(Q^{16})_{0} = R^{8} cos(8α)
= R^{8}(1 − 8 sin^{2}(2α) + 8 sin^{4}(2α))
= R^{8} + 2^{7}qR^{16}S^{8} + 2^{11}q^{4}R^{24}S^{16},

where we have substituted −4iqR^{4}S^{4} for sin(2α). This shows that (Q^{16})_{0} is congruent to R^{8} modulo 2^{7}.

Similarly, we compute the odd two-dissection of Q^{16} using the identity

sin(8α) = 2 sin(4α) cos(4α) = 2 sin(4α)(1 − 2 sin^{2}(2α)).

We then substitute in for sin(4α) and sin(2α), obtaining

(Q^{16})_{1} = R^{8}i sin(8α) = R^{8}i(2 sin(4α)(1 − 2 sin^{2}(2α))) = 16qR^{20} + 512q^{3}R^{28}S^{8}.

This shows that (Q^{16})_{1} is congruent to 16qR^{20} modulo 2^{9}, and also that it is congruent to 0 modulo 2^{4}.

To dissect arbitrary functions written in terms of Q, R, S, T, φ(q^{2}), φ(q^{4}), and φ(q^{8}), we see that it suffices to be able to two-dissect powers of Q, since only Q contains terms with odd powers of q.

We note that the computer program uses no approximate techniques; in all cases, it manipulates trigonometric identities to obtain proven dissections. Using this program, we prove the following result:

Lemma 5.1. For k ∈ {12, 16, 18, 20, 22, 26} and m positive and odd, we have

a_{k}(2m, 4) ≡ 0 (mod 2^{ρ+γ}).

Proof. By duality, this is equivalent to showing that

a_{2−k}(4, 2m) ≡ 0 (mod 2^{ρ+γ})

for each k. In other words, we wish to show that a_{2−k}(4, j) ≡ 0 (mod 2^{ρ+γ}) for all j ≡ 2 (mod 4).

To prove these results, we set g(q^{2}) = (f_{2−k,4})_{0} = ∑_{n=0}^{∞} a_{2−k}(4, 2n)q^{2n}. Then

g(q) = ∑_{n=0}^{∞} a_{2−k}(4, 2n)q^{n}

and the odd two-dissection of g(q) is

(g(q))_{1} = ∑_{n=0}^{∞} a_{2−k}(4, 2(2n + 1))q^{2n+1} = ∑_{n=0}^{∞} a_{2−k}(4, 4n + 2)q^{2n+1},

so we see that we want (g(q))_{1} ≡ 0 (mod 2^{ρ+γ}).

To illustrate this, we will look at the k = 16 case. We are obtaining a result modulo 2^{ρ+γ} = 2^{11}. We use the facts that [11, 14]

Δ(z) = qφ(q)^{24}, E_{4}(z) = φ(q^{2})^{8}(Q^{−16} + 2^{8}qQ^{8}),
E_{6}(z) = φ(q^{2})^{12}Q^{−24} − 2^{5} · 3 · 5qφ(q^{2})^{12} − 2^{9} · 3 · 11q^{2}Q^{8}φ(q^{2})^{4}φ(q^{4})^{8}
+ 2^{13}q^{3}φ(q^{2})^{−12}φ(q^{4})^{24},

and

j(z) = E_{4}(z)^{3}/Δ(z) = q^{−1}Q^{−24} + 2^{8} · 3 + 2^{16} · 3qQ^{24} + 2^{24}q^{2}Q^{48},

to write f_{−14,4}(z) = E_{4}(z)E_{6}(z)Δ^{−2}(z)(j(z)^{2} − 1272j(z) + 192600) modulo 2^{12} as

\frac{φ(q^{2})^{8} + 40qφ(q^{2})^{8}Q^{24} + 3584q^{2}φ(q^{4})^{8}Q^{32} + 344q^{2}φ(q^{2})^{8}Q^{48} + 768q^{3}φ(q^{2})^{8}Q^{72}}{q^{4}Q^{40}φ(q^{2})^{36}} .

Note that to get equality, there would be many additional terms; we have eliminated all terms that are multiples of 2^{12}.

We now use the program to two-dissect f_{−14,4} by two-dissecting each of the powers of Q appearing in this expansion, obtaining

g(q^{2}) ≡ (f_{−14,4})_{0} ≡ \frac{2816R^{40}S^{8}}{φ(q^{4})^{28}} + \frac{R^{8}}{φ(q^{4})^{28}q^{4}} + \frac{3288R^{32}}{φ(q^{4})^{28}q^{2}} + \frac{800R^{16}S^{8}}{φ(q^{4})^{28}q^{2}} (mod 2^{12}).

Replacing q^{2} by q yields

g(q) ≡ \frac{2816Q^{40}R^{8}}{φ(q^{2})^{28}} + \frac{Q^{8}}{φ(q^{2})^{28}q^{2}} + \frac{3288Q^{32}}{φ(q^{2})^{28}q} + \frac{800Q^{16}R^{8}}{φ(q^{2})^{28}q} (mod 2^{12}).

Finally, we perform the odd dissection of g(q) to get the desired result:

(g(q))_{1} ≡ \frac{2048qR^{40}}{φ(q^{2})^{28}} (mod 2^{12}).

Therefore, a_{16}(2m, 4) is divisible by 2^{11}.

The proofs for the other weights follow similarly, by writing f_{2−k,4}(z) in terms of E_{4}, E_{6}, Δ, and j, and two-dissecting the resulting expression twice. In each case, we find that the coefficients of powers of q with exponent congruent to 2 modulo 4 are all divisible by 2^{ρ+γ}.

□
6. Coefficients a_{k}(2^{a}m, 2^{b}n) with a > b

6.1. Special bases for M_{k}(2). Writing holomorphic modular forms in terms of special basis elements will often help us to prove congruences on their coefficients (see Lemma 6.3). Standard dimension formulas [16, Proposition 6.1] yield the following formula for the dimension of M_{k}(2).

Theorem 6.1. For k ≥ 4 even,

dim(M_{k}(2)) = 1 + ⌊\frac{k}{4} ⌋ .

Hence, in order to find a basis for M_{k}(2), we need only find 1 + ⌊\frac{k}{4} ⌋ linearly independent modular forms in M_{k}(2). For k ∈ {12, 16, 18, 20, 22, 26} we do this as follows.

Lemma 6.2. For k ∈ {12, 16, 18, 20, 22, 26}, the modular forms listed in Table 1 form a basis of M_{k}(2). All of the forms in these bases have integer coefficients and leading coefficient 1, and the ith form in each basis vanishes to order i − 1 at ∞.

Proof. One easily checks that each of the listed forms is in the appropriate M_{k}(2), has integer coefficients, and vanishes to the indicated order at ∞. For a given k, each form has a different order of vanishing, which implies that they are linearly independent, and since there are 1 + ⌊\frac{k}{4} ⌋ of them, they form a basis. □

We now prove a result on the divisibility of the coefficients of the holomorphic modular form f_{k,0}.

k Basis Elements
12 E_{4}(2z)^{3}, Δ(z), Δ(2z), S_{4}(z)^{3}
16 E_{4}(2z)^{4}, Δ_{16}, Δ_{16}(2z), E_{4}(z)S_{4}(z)^{3}, S_{4}(z)^{4}
18 E_{4}(2z)^{3}E_{6}(2z), Δ_{18}, Δ_{18}(2z), S_{4}^{3}E_{6}, S_{4}^{3}S_{6}
20 E_{4}(2z)^{5}, Δ_{20}, Δ_{20}(2z), S_{4}^{3}E_{4}^{2}, S_{4}^{4}E_{4}, S_{4}^{5}
22 E_{4}(2z)^{4}E_{6}(2z), Δ_{22}, Δ_{22}(2z), S_{4}^{3}E_{4}E_{6}, S_{4}^{4}E_{6}, S_{4}^{4}S_{6}
26 E_{4}(2z)^{5}E_{6}(2z), Δ_{26}, Δ_{26}(2z), S_{4}^{3}E_{4}^{2}E_{6}, S_{4}^{4}E_{4}E_{6}, S_{4}^{5}E_{6}, S_{4}^{5}S_{6}

Table 1. Bases for M_{k}(2)

Lemma 6.3. For k ∈ {12, 16, 18, 20, 22, 26}, and for n odd and positive, a_{k}(0, n) ≡ 0 (mod 2^{η}).

Proof. Note that f_{k,0} ∈ M_{k}(2), hence it can be written in terms of the basis described in Lemma 6.2. For k = 12, we obtain

f_{12,0}(z) = E_{4}(2z)^{3} + 2^{8}765Δ(2z) + 2^{12}4095S_{4}(z)^{3}.

Note that the only form with nonzero odd coefficients is multiplied by 2^{12} so that all odd coefficients of f_{12,0} are divisible by 2^{12}. The proof for all other weights is similar–when we write f_{k,0} in terms of the basis, all forms with odd coefficients are multiplied by 2^{η}. □

Although it is not used anywhere else in the proof, we include the following result for completeness.

Lemma 6.4. For k ∈ {12, 16, 18, 20, 22, 26}, for n odd and positive, and for b > 0, a_{k}(0, 2^{b}n) ≡ 0 (mod 2^{ω})

Proof. Each f_{k,0} is a holomorphic modular form of level 1, with constant term 1, and with the coefficient of q equal to 0. We may thus write it as

f_{k,0} = E_{4}(z)^{r}E_{6}(z)^{s} − CΔ_{k}(z)

where 4r + 6s = k, 0 ≤ s ≤ 1, and C is the coefficient of q in E_{4}(z)^{a}E_{6}(z)^{b}. Using the facts that E_{4}(z) ≡ 1 (mod 2^{4}) and E_{6}(z) ≡ 1 (mod 2^{3}), one easily checks that E_{4}(z)^{r}E_{6}(z)^{s} ≡ 1 (mod 2^{ω}), and the lemma follows. □

Using the above result, we now prove the following two results–the first about coefficients of odd powers of q in specific forms, and the second generalizing this result to odd powers of q in all forms f_{k,m} with m even.

Lemma 6.5. For k ∈ {12, 16, 18, 20, 22, 26} and n positive and odd, a_{k}(2, n) ≡ 0 (mod 2^{ν}).

Proof. We examine the function G(z) = f_{k,2}(z) − (f_{k,1}|V_{2})(z) − a_{k}(2, 2)Δ_{k}(2z) ∈ M_{k}^{!}(2). This function is holomorphic at ∞ and in the upper half plane. Using Proposition 4.3, together with the modularity of f_{k,2} and Δ_{k}, we easily compute the expansion of G(z) at 0 to be

2(2z)^{−k}G(−1/(2z)) = 2q^{−4} − 2^{1−k}q^{−1} + O(1).

Hence, we may write

2(2z)^{−k}G(−1/(2z)) = ∑_{j=1}^{4} C_{j}ψ^{j+1}(z)Δ_{k}(z) + F′(z)

where F′(z) is holomorphic at ∞. From the expansion given above, we see that 2^{k−1}C_{j} ∈ ℤ for all j. Now replacing z by −1/(2z) and dividing by 2z^{k}, we obtain

G(z) = ∑_{j=1}^{4} \frac{1}{2}C_{j}2^{12(j+1)}Φ(z)^{j+1}2^{k}Δ_{k}(2z) + \frac{1}{2}z^{−k}F′(−1/(2z)).

We see from this that \frac{1}{2}z^{−k}F′(−1/(2z)) is a holomorphic modular function of weight k and level 2, which we will denote by F (z). We note that the coefficients \frac{1}{2}C_{j}2^{12j+12+k} are all divisible by 2 to the power of 12j+12+k−(k−1)−1 = 12j+12. Hence, F (z) actually has integer coefficients, and

G(z) ≡ F (z) (mod 2^{24}).

Since F (z) is a linear combination of the basis elements listed above, we see that if the first dim(M_{k}(2)) coefficients of G(z) are divisible by 2^{ν}, then the same is true for the first coefficients of F (z), and hence for all coefficients of F (z) and G(z). We have computed G(z) for each k ∈ {12, 16, 18, 20, 22, 26} and verified that all of the first dim(M_{k}(2)) coefficients are divisible by 2^{ν}. Hence, the lemma is proved. □

Proposition 6.6. For k ∈ {12, 16, 18, 20, 22, 26}, m, n positive and odd, and a > 0,

a_{k}(2^{a}m, n) ≡ 0 (mod 2^{ν}).

Proof. We will show that f_{2−k,n}|U_{2} ≡ a_{2−k}(n, 0) (mod 2^{ν}). This immediately implies that for all a > 0, the coefficient a_{2−k}(n, 2^{a}m) ≡ 0 (mod 2^{ν}), and we are done by duality.

Now f_{2−k,n}|U_{2} is holomorphic at ∞, and using Theorem 4.1, we see that it has a pole of order 4n at 0. In fact, choosing values of C_{j} to cancel out all negative powers of q, we see that

2(2z)^{−(2−k)}(f_{2−k,n}|U_{2})(−1/(2z)) = 2^{2−k}q^{−4n} + ⋯ = ∑_{j=0}^{4n} C_{j}ψ(z)^{j}θ_{2−k}(z),

where equality follows from the fact that there are no nonzero holomorphic modular forms of negative weight. Note that A_{j} = 2^{k−2}C_{j} is an integer.

We then have that

f_{2−k,n}|U_{2} = ± ∑_{j=0}^{4n} 2^{2−k}A_{j}2^{12j−1}2^{μ_{2−k}}α_{2−k}(z)Φ(z)^{j},

with the sign depending on k.

Reducing this modulo 2^{16}, we find that all but the first two terms vanish. We then find that for some integer C,

f_{2−k,n}|U_{2} ≡ a_{2−k}(n, 0)α_{2−k}(z) + Cα_{2−k}(z)Φ(z) (mod 2^{16}).

Since a_{2−k}(n, 0) ≡ 0 (mod 2^{η}) (by Lemma 6.3), and α_{2−k} ≡ 1 (mod 2^{ξ_{2−k}}), we see that C ≡ a_{2−k}(n, 2) (mod 2^{η+ξ_{2−k}}). Hence, as long as η + ξ_{2−k} ≥ ν, we are done, since a_{2−k}(n, 2) ≡ 0 (mod 2^{ν}) by Lemma 6.5 and duality. One checks easily for each value of k that η + ξ_{2−k} = ν. □

Next we prove a congruence on coefficients of q^{2} in certain forms, which will allow us to complete the proof of Proposition 6.8, which gives the desired divisibility for all a_{k}(2^{i}m, 2^{j}n) with i > j.

Lemma 6.7. Let k ∈ {12, 16, 18, 20, 22, 26}. For a > 1 and m positive and odd, we have a_{k}(2^{a}m, 2) ≡ 0 (mod 2^{χ}).

Proof. By duality, it is enough to show that a_{2−k}(2, 2^{a}m) ≡ 0 (mod 2^{χ}). We will do this by showing that f_{2−k,2}|U_{4} ≡ a_{2−k}(2, 0) (mod 2^{χ}). This will imply that all coefficients of q^{n} in f_{2−k,2} with 4|n and n > 0 will be divisible by 2^{χ}.

We begin by noting that by Theorem 4.2, f_{2−k,2}|U_{4} is a modular form of weight 2 − k and level 2, holomorphic at infinity and on the upper half plane, with a pole of order 16 at 0. This implies that it can be written as a linear combination

f_{2−k,2}|U_{4} = ∑_{i=0}^{16−δ} C_{i}Φ^{i}α_{2−k},

where δ is the order of the pole at 0 of α_{2−k}. We could use Theorem 4.2 to directly compute the coefficients C_{i}, but instead use the fact that the Φ^{i}α_{2−k} are linearly independent to compute the expansion directly (by cancelling out the constant term, then the q term, etc.). The expansion for k = 12 is then computed to be

f_{−10,2}|U_{4} = −2^{4}12285α_{−10} − 2^{13}560584703Φ^{1}α_{−10} − 2^{32}81073083Φ^{2}α_{−10}

− 2^{41}1530945507Φ^{3}α_{−10} − 2^{52}2152441109Φ^{4}α_{−10} − 2^{61}5180299059Φ^{5}α_{−10}

− 2^{72}1612138839Φ^{6}α_{−10} − 2^{81}1156390619Φ^{7}α_{−10} − 2^{93}63543825Φ^{8}α_{−10}

− 2^{102}17635779Φ^{9}α_{−10} − 2^{117}48367Φ^{10}α_{−10} − 2^{126}5199Φ^{11}α_{−10}
− 2^{138}39Φ^{12}α_{−10} − 2^{147}Φ^{13}α_{−10}.

All terms except the first vanish modulo 2^{7}, and, since α_{−10} ≡ 1 (mod 2^{3}), we see that

f_{−10,2}|U_{4} ≡ −2^{4}12285 (mod 2^{7}).

The proof for the other weights proceeds similarly, computing the expansion of f_{2−k,2}|U_{4} and noting that all terms except the first vanish modulo 2^{χ} and that the first term is a constant modulo 2^{χ}. □

Proposition 6.8. Let k ∈ {12, 16, 18, 20, 22, 26}. For a > b ≥ 0, and for m, n positive and odd, we have

a_{k}(2^{a}m, 2^{b}n) ≡ 0 (mod 2^{χ}).

Proof. We proceed by induction on b. The case b = 0 is just Proposition 6.6.

We assume (by way of induction) that the theorem is true for all b ≤ B. Assume that a > B + 1. Then by Lemma 3.1, with m and n replaced respectively by 2^{a}m and 2^{B}n, we obtain

a_{k}(2^{a−1}m, 2^{B}n) + a_{k}(2^{a}m, 2)τ_{k}(2^{B}n) + 2^{k−1}a_{k}(2^{a+1}m, 2^{B}n)
= a_{k}(2^{a}m, 2^{B+1}n) + 2^{k−1}a_{k}(2^{a}m, 2^{B−1}n).

Since k − 1 ≥ χ, this reduces to

a_{k}(2^{a−1}m, 2^{B}n) + a_{k}(2^{a}m, 2)τ_{k}(2^{B}n) = a_{k}(2^{a}m, 2^{B+1}n) (mod 2^{χ}).

However, by our induction hypothesis, a_{k}(2^{a−1}m, 2^{B}n) vanishes modulo 2^{χ}, and by Lemma 6.7, a_{k}(2^{a}m, 2) vanishes modulo 2^{χ}. Hence, the theorem is proven. □

7. Coefficients a_{k}(2^{a}m, 2^{b}n) with a < b

We now prove the theorem for the coefficients a_{k}(2^{a}m, 2^{b}n) for which a < b. We will prove this by induction on the difference b − a. To begin, we deal with the cases where b−a < 3 (each of which is proved by induction on a). In order to prove these cases, we first need the following divisibility result for coefficients of q^{2}.

Lemma 7.1. Let k ∈ {12, 16, 18, 20, 22, 26}. Then for odd m, a_{k}(m, 2) is divisible by 2^{γ}.

Proof. We note that v_{2}(a_{k}(m, 2)) = v_{2}(a_{2−k}(2, m)) by duality. For k ≠ 22, we note (by explicit computation and comparing poles) that f_{2−k,2}E_{4}^{3} = f_{2−k+12,2} + C_{k}f_{2−k+12,1}, with C_{12} = 744, C_{16} = 504, C_{18} = 1248, C_{20} = 264, and C_{26} = 768. Since E_{4}^{3} ≡ 1 (mod 2^{4}), we see that

v_{2}(a_{k}(m, 2)) = v_{2}(a_{2−k}(2, m))
≥ min(v_{2}(a_{2−k+12}(2, m)), 4, v_{2}(C_{k}))
= min(v_{2}(a_{k−12}(m, 2), 4, v_{2}(C_{k}))
= min(4, v_{2}(C_{k})),

where the last equality arises from the fact that v_{2}(a_{k−12}(m, 2)) ≥ 7 by [6, Theorem 1.3] and [9, Theorem 2.1]. Since min(4, v_{2}(C_{k})) = γ, we are done.

For k = 22, we do a similar computation, using that E_{4}^{4} ≡ 1 (mod 2^{5}):

f_{−20,2}E_{4}^{4} = f_{−4,2} + 1248f_{−4,1}.

Hence for m odd,

a_{22}(m, 2) = −a_{−20}(2, m) ≡ −a_{−4}(2, m) = a_{6}(m, 2) ≡ 0 (mod 2^{5})

by duality and [6, Theorem 1.3]. □

We now prove the cases b − a = 1 and b − a = 2.

Lemma 7.2. Let k ∈ {12, 16, 18, 20, 22, 26}. For a ≥ 0 and m, n > 0 odd, we have

a_{k}(2^{a}m, 2^{a+1}n) ≡ 0 (mod 2^{γ}).

Proof. Applying Lemma 3.1, we obtain

a_{k}(m, 2)τ_{k}(n) + 2^{k−1}a_{k}(2m, n) = a_{k}(m, 2n).

Now k − 1 > γ and by Lemma 7.1, a_{k}(m, 2) is divisible by 2^{γ}, so we see that a_{k}(m, 2n) ≡ 0 (mod 2^{γ}), and the Lemma is proved for a = 0.

We now assume that A > 0 and that the Lemma is true for a = A − 1. Applying Lemma 3.1 with m, n replaced by 2^{A}m and 2^{A}n,

a_{k}(2^{A−1}m, 2^{A}n) + a_{k}(2^{A}m, 2)τ_{k}(2^{A}n) ≡ a_{k}(2^{A}m, 2^{A+1}n) (mod 2^{γ}).

By Lemma 3.3, τ_{k}(2^{A}n) ≡ 0 (mod 2^{γ}), so by the induction hypothesis we see that a_{k}(2^{A}m, 2^{A+1}n) ≡ 0 (mod 2^{γ}). □

Lemma 7.3. For k ∈ {12, 16, 18, 20, 22, 26}, a ≥ 0 and m, n ∈ ℤ positive and odd,

a_{k}(2^{a}m, 2^{a+2}n) ≡ 0 (mod 2^{2γ}).

Proof. Replacing n by 2n in Lemma 3.1, we obtain

a_{k}(m, 4n) ≡ a_{k}(m, 2)τ_{k}(2n) (mod 2^{k−1}).

By Lemmas 3.3 and 7.1, and the fact that 2γ < k − 1, we see that a_{k}(m, 4n) ≡ 0 (mod 2^{2γ}).

Letting A > 0 and assuming the theorem for a = A − 1, we apply Lemma 3.1 with m, n replaced by 2^{A}m and 2^{A+1}n, to obtain

a_{k}(2^{A}m, 2^{A+2}n) ≡ a_{k}(2^{A−1}m, 2^{A+1}n) + a_{k}(2^{A}m, 2)τ_{k}(2^{A+1}n) (mod 2^{k−1}).

Lemma 3.3 and the induction hypothesis then imply that a_{k}(2^{A}m, 2^{A+2}n) ≡ 0 (mod 2^{2γ}). □

Next, we prove a weaker divisibility than our final result, from which we will derive the stronger divisibility that we need.

Lemma 7.4. Let k ∈ {12, 16, 18, 20, 22, 26}. For 0 ≤ a < b and m, n positive and odd, we have that

a_{k}(2^{a}m, 2^{b}n) ≡ 0 (mod 2^{γ(b−a)}).

Proof. Note that the theorem has already been proven for b − a = 1 (Lemma 7.2) and for b − a = 2 (Lemma 7.3). Assume, by way of induction, that the theorem is true for 0 < b − a < N. Then applying Lemma 3.1 with m and n replaced by 2^{a}m and 2^{a+N−1}n, we obtain

a_{k}(2^{a−1}m, 2^{a+N−1}n) + a_{k}(2^{a}m, 2)τ_{k}(2^{a+N−1}n) + 2^{k−1}a_{k}(2^{a+1}m, 2^{a+N−1}n)
= a_{k}(2^{a}m, 2^{a+N}n) + 2^{k−1}a_{k}(2^{a}m, 2^{a+N−2}n)

We note that a_{k}(2^{a}m, 2)τ_{k}(2^{a+N−1}n) ≡ 0 (mod 2^{γN}) (using Lemma 3.3 if a > 0, and using Lemmas 3.3 and 7.1 if a = 0). Further, by our induction hypothesis, 2^{k−1}a_{k}(2^{a+1}m, 2^{a+N−1}n) and 2^{k−1}a_{k}(2^{a}m, 2^{a+N−2}n) are both divisible by 2^{k−1+γ(N−2)}, which (since k − 1 > 2γ) is divisible by 2^{γN}. Hence, we see that

a_{k}(2^{a}m, 2^{a+N}n) ≡ a_{k}(2^{a−1}m, 2^{a+N−1}n) (mod 2^{γN}).

For a = 0, the right hand side of this congruence is 0. Hence, by a simple induction on a, the left hand side is 0 modulo 2^{γN} for all nonnegative a. This completes the induction. □

For coefficients a_{k}(2^{a}m, 2^{b}n) with b > a and a = 0, the congruence in Lemma 7.4 is our final result. However, for a > 0, we can derive a better congruence by bootstrapping from this result. The proof will again be by induction on b − a, but we need an additional divisibility result on the coefficients a_{k}(2m, 2^{b}n) with b ≥ 2 to complete the induction.

Lemma 7.5. Let k ∈ {12, 16, 18, 20, 22, 26}. For b ≥ 2 and m, n positive and odd,

a_{k}(2m, 2^{b}n) ≡ 0 (mod 2^{ρ+γ(b−1)}).

Proof. Applying Lemma 3.2 with m and n replaced by 2m and 2^{B−2}n, we obtain

2^{2k−2}a_{k}(8m, 2^{B−2}n) + (1 − δ_{2^{B−2}n,2})2^{k−1}a_{k}(2m, 2^{B−2}n) + a_{k}(2m, 4)τ_{k}(2^{B−2}n)
= 2^{2k−2}a_{k}(2m, 2^{B−4}n) + a_{k}(2m, 2^{B}n).

For B = 2, we note that ρ + (B − 1)γ = ρ + γ ≤ k − 1. For 3 ≤ B ≤ 6, we note that ρ + (B − 1)γ ≤ ρ + 5γ ≤ 2k − 2 and that (1 − δ_{2^{B−2}n,2}) = 0. In both cases, the formula above reduces to

a_{k}(2m, 2^{B}n) ≡ a_{k}(2m, 4)τ_{k}(2^{B−2}n) (mod 2^{ρ+(B−1)γ}).

Now, by Lemma 5.1, we know that a_{k}(2m, 4) is divisible by 2^{ρ+γ}, and by Lemma 3.3, we know that τ_{k}(2^{B−2}n) is divisible by 2^{γ(B−2)}. Hence, for 2 ≤ b ≤ 6 we have that a_{k}(2m, 2^{b}n) ≡ 0 (mod 2^{ρ+(b−1)γ}).

We will now assume that B > 6, and that the theorem is true for all b with 2 ≤ b < B. Now, since 2k − 2 > 4γ, the induction hypothesis implies that 2^{2k−2}a_{k}(2m, 2^{B−4}n) vanishes modulo 2^{ρ+(B−1)γ}. Lemma 7.4 and the fact that 2k − 2 > ρ + 4γ imply that 2^{2k−2}a_{k}(8m, 2^{B−2}n) vanishes modulo 2^{ρ+(B−1)γ}. By Lemmas 3.3 and 5.1, a_{k}(2m, 4)τ_{k}(2^{B−2}n) vanishes modulo 2^{ρ+(B−1)γ}. Hence, a_{k}(2m, 2^{B}n) ≡ 0 (mod 2^{ρ+(B−1)γ}), and by induction the theorem is true for all b ≥ 2. □

The proof for b − a = 1 proceeds similarly to the proof of Lemma 7.2.

Lemma 7.6. Let k ∈ {12, 16, 18, 20, 22, 26}. For a ≥ 1 and m, n positive and odd,

a_{k}(2^{a}m, 2^{a+1}n) ≡ 0 (mod 2^{ρ+γ}).

Proof. By Lemma 7.5 we have that a_{k}(2m, 4n) ≡ 0 (mod 2^{ρ+γ}). Hence, the theorem is true for a = 1.

Assume now that A > 1, and that the theorem is true for all 1 ≤ a < A. By Lemma 3.1 with m, n replaced by 2^{A}m and 2^{A}n, we find

a_{k}(2^{A}m, 2^{A+1}n) ≡ a_{k}(2^{A−1}m, 2^{A}n) + a_{k}(2^{A}m, 2)τ_{k}(2^{A}n) (mod 2^{k−1}).

We know that ρ + γ < k − 1, that for A > 1, 2^{2γ}|τ_{k}(2^{A}n) by Lemma 3.3, that 2^{χ}|a_{k}(2^{A}m, 2) by Lemma 6.7, that 2^{ρ+γ}|a_{k}(2^{A−1}m, 2^{A}n) by the induction hypothesis, and that χ + γ > ρ. Hence, a_{k}(2^{A}m, 2^{A−1}n) ≡ 0 (mod 2^{ρ+γ}), completing our induction. □

In order to complete the case where b−a = 2 (Lemma 7.8), we need the following result on coefficients for which a = b.

Lemma 7.7. Let k ∈ {12, 16, 18, 20, 22, 26}. For b ≥ 1 and m, n odd,

a_{k}(2^{b+1}m, 2^{b+1}n) ≡ a_{k}(2^{b}m, 2^{b}n) (mod 2^{χ+γ}).

Proof. Applying Lemma 3.1 with m, n replaced by 2^{b+1}m and 2^{b}n, we obtain

a_{k}(2^{b}m, 2^{b}n) + a_{k}(2^{b+1}m, 2)τ_{k}(2^{b}n) + 2^{k−1}a_{k}(2^{b+2}m, 2^{b}n)
= a_{k}(2^{b+1}m, 2^{b+1}n) + 2^{k−1}a_{k}(2^{b+1}m, 2^{b−1}n).

Since k − 1 > χ + γ, and by Lemma 3.3 and Proposition 6.8 we know that a_{k}(2^{b+1}m, 2)τ_{k}(2^{b}n) vanishes modulo 2^{χ+γ}, the result follows. □

Lemma 7.8. Let k ∈ {12, 16, 18, 20, 22, 26}. For a ≥ 1 and m, n positive and odd,

a_{k}(2^{a}m, 2^{a+2}n) ≡ 0 (mod 2^{ρ+2γ}).

Proof. By Lemma 7.5, we see that the lemma is true for a = 1. We will now assume that A > 1 and that the lemma is true for all a with 1 ≤ a < A. By Lemma 3.1 with m, n replaced by 2^{A}m and 2^{A+1}n we obtain

a_{k}(2^{A}m, 2^{A+2}n) = a_{k}(2^{A−1}m, 2^{A+1}n) + a_{k}(2^{A}m, 2)τ_{k}(2^{A+1}n)
+ 2^{k−1}(a_{k}(2^{A+1}m, 2^{A+1}n) − a_{k}(2^{A}m, 2^{A}n)).

By Lemma 7.7 we know that 2^{k−1}(a_{k}(2^{A+1}m, 2^{A+1}n) − a_{k}(2^{A}m, 2^{A}n)) vanishes modulo 2^{k−1+χ+γ}. Since k − 1 + χ + γ > ρ + 2γ, we see that this term vanishes modulo 2^{ρ+2γ}. By our induction hypothesis, a_{k}(2^{A−1}m, 2^{A+1}n) vanishes modulo 2^{ρ+2γ}. Finally, by Lemmas 6.7 and 3.3, we see that

a_{k}(2^{A}m, 2)τ_{k}(2^{A+1}n)

vanishes modulo 2^{ρ+2γ}, since χ + (A + 1)γ = (χ + γ) + Aγ ≥ ρ + 2γ. □

We now conclude by induction on b − a, using Lemmas 7.6 and 7.8 as base cases.

Proposition 7.9. Let k ∈ {12, 16, 18, 20, 22, 26}. For b > a ≥ 1 and m, n positive and odd, we have that

a_{k}(2^{a}m, 2^{b}n) ≡ 0 (mod 2^{ρ+γ(b−a)}).

Proof. The result is true for b − a = 1 and b − a = 2 by Lemmas 7.6 and 7.8. In addition, by Lemma 7.5 it is true for a = 1 and any b ≥ 2. We will assume that b − a = N > 2 with a > 1, and that the theorem is true for b − a < N.

Now, using Lemma 3.1 with m, n replaced by 2^{a}m and 2^{b−1}n, we obtain

a_{k}(2^{a−1}m, 2^{b−1}n) + a_{k}(2^{a}m, 2)τ_{k}(2^{b−1}n) + 2^{k−1}a_{k}(2^{a+1}m, 2^{b−1}n)
= a_{k}(2^{a}m, 2^{b}n) + 2^{k−1}a_{k}(2^{a}m, 2^{b−2}n).

By the induction hypothesis, both terms that are multiplied by 2^{k−1} are divisible by 2^{k−1}2^{ρ+γ(N−2)}. Because k − 1 > 2γ, both of these terms vanish modulo 2^{ρ+γN}.

Since a > 1, we see that a_{k}(2^{a}m, 2) is divisible by 2^{χ} by Lemma 6.7. In addition, τ_{k}(2^{b−1}n) is divisible by 2^{γ(b−1)}. Since χ + γ(b − 1) = γ(b − a) + γ(a − 1) + χ ≥ γ(b − a) + (γ + χ) ≥ ρ + γN, we see that

a_{k}(2^{a}m, 2^{b}n) ≡ a_{k}(2^{a−1}m, 2^{b−1}n) (mod 2^{ρ+γN}).

Since the theorem is true for a = 1 (by Lemma 7.5), a simple induction on a completes the proof. □

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Department of Mathematics, Brigham Young University, Provo, UT 84602

E-mail address: doud@math.byu.edu

Department of Mathematics, Brigham Young University, Provo, UT 84602

E-mail address: jenkins@math.byu.edu

Department of Mathematics, Brigham Young University, Provo, UT 84602

E-mail address: johnvlopez@gmail.com