quadratic twists and the coefficients of weakly holomorphic...

QUADRATIC TWISTS AND THE COEFFICIENTS OF WEAKLYHOLOMORPHIC MODULAR FORMS

STEPHANIE TRENEER

Abstract. Linear congruences have been shown to exist for the partition function and forseveral other arithmetic functions that are generated by weakly holomorphic modular forms.In previous work, the author showed that such congruences exist for the Fourier coefficientsof any weakly holomorphic modular form in a large class. In this paper, quadratic twistsare used to produce new congruences for the coefficients of these forms. To demonstrate ourmain theorems, we give new congruences for overpartitions, and extend a result of Pennistonregarding congruences for `-regular partitions.

1. Introduction and Statement of Results

Suppose that k and N ≥ 1 are integers, with 4 | N . Let M k2(Γ1(N)) (resp. Mk(Γ1(N))),

denote the space of weakly holomorphic half-integral weight (resp. integral weight) modularforms for the congruence subgroup Γ1(N). These are meromorphic functions on the complexupper half plane H whose poles (if any) are restricted to the cusps of Γ1(N). Several authorshave recently studied the Fourier coefficients of weakly holomorphic modular forms in order todeduce congruence properties for certain arithmetic functions, most notably for the partitionfunction.

Let p(n) be the number of nonincreasing sequences of positive integers that sum to n.Ono [27] and Ahlgren and Ono [2, 4] greatly extended earlier work of Ramanujan, Watson[36], Atkin [7, 8] and others (see [28, 5] for references), by applying the theory of modularforms and Galois representations to the study of congruences for p(n). They showed that ifM is a positive integer coprime to 6, then there exist infinitely many non-nested arithmeticprogressions An+B such that

(1.1) p(An+B) ≡ 0 (mod M)

for all n ≥ 0. In contrast, Ono [26] showed that in the case M = 2, any arithmeticprogression r (mod t) contains infinitely many integers m with p(m) ≡ 0 (mod 2), butalso contains infinitely many integers n with p(n) 6≡ 0 (mod 2) provided there is at least onesuch n. Quantitative forms of these results were given by Serre (in an appendix to a paperof Nicolas, Ruzsa and Sarkozy [25]) and Ahlgren [1]. Similarly, Ahlgren and Ono [5] haveconjectured that in any arithmetic progression r (mod t) there are infinitely many integers nwith p(n) 6≡ 0 (mod 3). A result in this direction has been proved by Boylan [9].

Like p(n), many other interesting arithmetic functions are generated by weakly holo-morphic modular forms, so it is reasonable to investigate congruence properties for thesefunctions as well. Adapting their methods for p(n), Ahlgren and Ono [6] established con-gruences of type (1.1) for traces of singular moduli. Other results obtained in a similar wayinclude congruences for the number of partitions of n into distinct parts due to Lovejoy [21]

2000 Mathematics Subject Classification. Primary: 11F37; Secondary: 11F33.

1

2 STEPHANIE TRENEER

and Ahlgren-Lovejoy [3], Swisher’s congruences for the Andrews-Stanley partition function[34], Mahlburg’s congruences for the crank function [24], and Penniston’s congruences for`-regular partitions [30]. In [35], the author showed that these results fit into a generalframework, by proving that infinitely many linear congruences exist for the coefficients ofany weakly holomorphic modular form of any weight, level or character.

Work of Deligne and Serre [13, 14] involving modular Galois representations provides amethod for showing the existence of congruences for the coefficients of integral weight cuspforms. This coupled with Shimura’s correspondence leads to congruences for the coefficientsof half-integral weight cusp forms. The challenge, then, is to relate half-integral weightweakly holomorphic modular forms to half-integral weight cusp forms in a way that pre-serves important properties of the original modular forms. There are two approaches inthe literature for building cusp forms from weakly holomorphic forms, one which uses theU -operator [27, 2] and the other which uses a quadratic twist [4, 6]. In [35, Theorem 1.1],we used the U -operator approach to prove the following general theorem.

Theorem 1.1. Suppose that p is an odd prime, and that k and m are integers with kodd. Let N be a positive integer with 4 | N and (N, p) = 1, and let χ be a Dirichletcharacter modulo N . Let K be a number field with ring of integers OK, and suppose that

f(z) =∑a(n)qn ∈M k

2(Γ0(N), χ)∩OK((q)). If m is sufficiently large, then for each positive

integer j, a positive proportion of the primes Q ≡ −1 (modNpj) have the property that

a(Q3pmn) ≡ 0 (mod pj)

for all n coprime to Qp.

Remark 1. It is straightforward to extend Theorem 1.1 to weakly holomorphic forms forΓ1(N). We also include the analogous theorem for integral weight weakly holomorphicmodular forms in [35].

In this paper, we investigate the congruences which may be established using the quadratictwist method. In order for this method to produce congruences modulo any power of a primep, the cusp expansions of a weakly holomorphic modular form f must satisfy a certain set ofhypotheses which we refer to collectively as condition C, and which depends upon a numberεp ∈ {±1}. If the hypotheses are satisfied, we say that “f satisfies condition C for p.” Weshow in Theorem 2.1 below that every f satisfies condition C for a positive proportion ofprimes p. Because the statement of condition C is somewhat technical, we will delay givingit until §2.

We now state our main theorem for congruences using the quadratic twist method.

Theorem 1.2. Suppose that p is an odd prime, and that k is an odd integer. Let N be apositive integer with 4 | N and (N, p) = 1. Let K be a number field with ring of integers OK,

and suppose that f(z) =∑a(n)qn ∈ M k

2(Γ1(N)) ∩ OK((q)). Let f satisfy condition C for

p. Then for each positive integer j, a positive proportion of the primes Q ≡ −1 (mod Npj)have the property that

a(Q3n) ≡ 0 (mod pj)

for all n coprime to Q such that(−np

)= −εp.

QUADRATIC TWISTS 3

Remark 2. Although condition C limits the set of associated prime moduli for a given formf , Theorem 1.2 produces congruences for the coefficients a(n) with n in any of p−1

2residue

classes modulo p, whereas with Theorem 1.1, n must be 0 modulo p.

Remark 3. Theorems 1.1 and 1.2 may be applied to the holomorphic parts of harmonicweak Maass forms whose non-holomorphic part is the period integral of a single variabletheta function. Indeed, Brigmann and Ono [10] apply Theorem 1.1 to a particular family ofMaass forms to get congruences for rank generating functions. A more general extension ofTheorem 1.1 to Maass forms is given by Garthwaite and Penniston [15].

We have an analog of Theorem 1.2 for integer weight modular forms. If f(z) ∈Mk(Γ1(N)),we say that f(z) satisfies condition C for p if it does so when viewed as a half-integral weightmodular form.

Theorem 1.3. Suppose that p is an odd prime, and that k is an integer. Let N be a positiveinteger with (N, p) = 1. Let K be a number field with ring of integers OK, and suppose thatf(z) =

∑a(n)qn ∈ Mk(Γ1(N)) ∩ OK((q)). Let f satisfy condition C for p. Then for each

positive integer j,

(i) a(n) ≡ 0 (mod pj) for almost all n with(np

)= −εp, and

(ii) a positive proportion of the primes Q ≡ −1 (modNpj) have the property that

a(Qn) ≡ 0 (mod pj)

for all n coprime to Q such that(−np

)= −εp.

As with Theorem 1.1, Theorems 1.2 and 1.3 apply to a wide class of weakly holomorphicmodular forms which can be built from Dedekind’s eta-function,

(1.2) η(z) := q124

∞∏n=1

(1− qn).

An eta-quotient is a function of the form

(1.3) f(z) =∏δ|N

ηrδ(δz)

where the rδ are integers and N ≥ 1. We will require in addition that

(1.4)∑δ|N

rδδ ≡ 0 (mod 24),

to ensure that f(z) has a Fourier expansion in integer powers of q. We have the followingresult for eta-quotients.

Corollary 1.4. Suppose p is an odd prime and N is an integer with (N, p) = 1. Let f(z) =∑δ|N

ηrδ(δz) satisfy (1.4), and suppose that f(z) has the Fourier expansion f(z) =∑a(n)qn.

Let f(z) satisfy condition C for p, and set k :=∑

δ|N rδ.

(a) If k is odd, then for each positive integer j, a positive proportion of the primesQ ≡ −1 (mod Npj) have


for all (n,Q) = 1 with(−np

)= −εp.

4 STEPHANIE TRENEER

(b) If k is even, then for each positive integer j,

(i) a(n) ≡ 0 (mod pj) for almost all n with(np

)= −εp, and

(ii) a positive proportion of the primes Q ≡ −1 (mod Npj) have


for all (n,Q) = 1 with(−np

)= −εp.

In particular, Corollary 1.4 gives congruences for a wide class of partition functions whosegenerating functions are eta-quotients. As an example, we consider the partition functionp(n).

Example 1. The generating function

f(z) :=1

η(24z)=

∑n≡−1 (mod 24)

p

(n+ 1

24

)qn

for p(n) is a weakly holomorphic modular form of weight −1/2 for Γ1(576). It can be easilyshown that f(z) satisfies condition C for each ` ≥ 5 with ε` =

(−1`

). Therefore by Corollary

1.4(a), for each j ≥ 1, a positive proportion of the primes Q ≡ −1 (mod 576`j) have theproperty that

(1.5) p

(Q3n+ 1

24

)≡ 0 (mod `j)

for all (n,Q) = 1 with Q3n ≡ −1 (mod 24) and(−n`

)= −ε`. In fact, by making a slight

modification of our general argument, we can show that (1.5) also holds for (n,Q) = 1with Q3n ≡ −1 (mod 24) and ` | n. Taken together, this duplicates all of the congruencesguaranteed by Ahlgren and Ono in [4].

An overpartition of n is a partition in which the first occurrence of a number may beoverlined. We let p(n) denote the number of overpartitions of n. For example, 2 + 1 + 1,2 + 1 + 1, 2 + 1 + 1 and 2 + 1 + 1 are distinct overpartitions of 4, and one easily checksthat p(4) = 14. For a more detailed account of overpartitions, see for example Corteel andLovejoy [12], Corteel and Hitczenko [11], and Lovejoy [23, 22]. The generating function forp(n) is given by

f(z) =∞∑n=0

p(n)qn =∞∏n=1

1 + qn

1− qn=η(2z)

η2(z)∈M− 1

2(Γ0(16)).

Applying Corollary 1.4(a) to f(z) yields the following result.

Proposition 1.5. For each j ≥ 1, a positive proportion of the primes Q ≡ −1 (mod 16`j)have the property that

p(Q3n) ≡ 0 (mod `j)

for each (n,Q) = 1 with(−n`

)= −

(−1p

).

As a final application, we are able to extend Penniston’s results regarding the distributionof the number of `-regular partitions of n [30].

QUADRATIC TWISTS 5

For an integer ` > 1, an `-regular partition is one in which no part is divisible by `. Wedenote the number of `-regular partitions of an integer n by b`(n). The generating functionfor b`(n) is

∞∑n=0

b`(n)qn =∞∏n=1

1− q`n

1− qn=

∞∏n=1

(1 + qn + q2n + · · ·+ q(`−1)n).

Rewriting, we can express the generating function as an eta-quotient:

η(24`z)

η(24z)=

∑n≡`−1 (mod 24)

b`

(n− `+ 1

24

)qn ∈M0(Γ1(576`)).

Penniston proves the following in [30].

Theorem 1.6. (Penniston, [30, Theorem 1.1]) Suppose that ` and p are distinct primes with3 ≤ ` ≤ 23 and p ≥ 5, and that j is a positive integer. Then

lim infX→∞

#{1 ≤ n ≤ X | b`(n) ≡ 0 (mod pj)}/X ≥

{p−1p

if p | `− 1, andp+12p

if p - `− 1.

Remark 4. Since the number of partitions into odd parts is the same as the number ofpartitions into distinct parts, the same result was shown for ` = 2 by Ahlgren and Lovejoy[3].

Using Corollary 1.4, we extend Theorem 1.6 to the following result.

Theorem 1.7. Let ` ≥ 3 and p ≥ 5 be distinct primes, and let j be a positive integer.

Suppose that for each 0 ≤ n < `−124

of the form n = k(3k±1)2

with p - (24n + 1 − `), thequadratic symbols (

24n+ 1− `

p

)each take on the same value. Then

lim infX→∞

#{1 ≤ n ≤ X | b`(n) ≡ 0 (mod pj)}/X ≥

p−1p

if ` ≤ 23 and p | `− 1,p+12p

if ` ≤ 23 and p - `− 1,p−12p

+ 1pm0

otherwise,

where m0 = min{m ≥ 1 : pm > 4(`− 1)}.

Remark 5. Note that when 3 ≤ ` ≤ 23, we get exactly Theorem 1.6.

Finally, we prove a more general statement about congruences for weakly holomorphicmodular forms. We extend Theorems 1.2 and 1.3 to show that simultaneous congruencesexist for any finite set of weakly holomorphic modular forms for congruence subgroups ofthe form Γ1(N), even when the modular forms have different levels and weights.

Theorem 1.8. Fix a positive integer r. For each 1 ≤ i ≤ r, suppose that ki is an integerand that Ni is a positive integer with 4 | Ni. Let f1(z), f2(z), . . . , fr(z) be weakly holomorphicmodular forms with

fi(z) =∑

ai(n)qn ∈M ki2

(Γ1(Ni))

6 STEPHANIE TRENEER

for each i, and suppose that all of the coefficients of the fi(z) are contained in the ring ofintegers OK of some number field K. For each 1 ≤ i ≤ r, let pi be a prime with pi - Ni,and suppose that each fi(z) satisfies condition C for pi. Define

t(i) :=

{1 if ki is even,

3 if ki is odd.

Then for each choice of integers j1, . . . , jr ≥ 1, a positive proportion of the primes Q ≡ −1(mod lcm(N1, . . . , Nr, p

j11 , . . . , p

jrr )) have the property that

ai(Qt(i)n) ≡ 0 (mod pjii )

for each 1 ≤ i ≤ r and each (n,Q) = 1 satisfying the conditions{(

−npi

)= −εpi : 1 ≤ i ≤ r

}.

In the next section, we recall some facts that we will need, and define condition C. In §3,we prove a key technical result which we then use to establish Theorems 1.2 and 1.3. Section4 concerns the application of our main theorems to eta-quotients. We prove Corollary 1.4and explore its consequences for p(n), then prove Proposition 1.5 and Theorem 1.7. Finally,in §5, we prove Theorem 1.8.

2. Preliminaries

We begin by reviewing some facts about half-integral weight modular forms. For morecomplete details, see for example [19, 33, 29].

Let G be the four-fold cover of GL+2 (Q) defined by

(2.1) G :=

{(α, φ(z)) : α =

(a bc d

)∈ GL+

2 (Q), φ2(z) =±(cz + d)

detα

},

which is a group under the operation

(α, φ(z))(β, ψ(z)) = (αβ, φ(βz)ψ(z)).

We denote by G′ the restriction of G to those (α, φ(z)) with α ∈ SL2(Z).Suppose that k is an integer, (α, φ(z)) ∈ G, and f(z) is a meromorphic function defined

on the complex upper half plane H. Then the slash operator is given by

(2.2) f(z)| k2(α, φ(z)) := φ(z)−kf(αz).

For γ =

(a bc d

)∈ Γ0(4) and z ∈ H, let

(2.3) j(γ, z) :=( cd

)ε−1d

√cz + d,

where(cd

)is the extended Jacobi symbol and

εd :=

{1 if d ≡ 1 (mod 4),

i if d ≡ 3 (mod 4),

and set γ := (γ, j(γ, z)). If Γ′ 6 Γ0(4) is a congruence subgroup, then write Γ′ := {γ : γ ∈ Γ′}.Let k be an integer. A weakly holomorphic modular form of weight k

2for Γ′ is a function

f(z) which is holomorphic on H and meromorphic at the cusps, and which satisfies

f | k2γ = f

QUADRATIC TWISTS 7

for all γ ∈ Γ′. We say that f is holomorphic if it is holomorphic at the cusps, and a cuspform if it vanishes at each cusp. The spaces of weakly holomorphic, holomorphic, and cusp

forms are denoted by M k2(Γ′), M k

2(Γ′), and S k

2(Γ′), respectively.

From now on, let f(z) ∈ M k2(Γ1(N)). Then f(z) has a Fourier expansion at ∞ of the

form

f(z) =∑

a(n)qn

where q := e2πiz. A twist of f(z) by a Dirichlet character χ modulo m is given by

f(z)⊗ χ =∑

χ(n)a(n)qn ∈M k2( ˜Γ1(Nm2)).

Suppose that v is an integer and p is an odd prime. Let

(2.4) τv,p =

((1 −v/p0 1

), 1

)∈ G.

If χtrivp is the trivial character modulo p , then using (2.2), it is straightforward to see that

(2.5) f(z)⊗ χtrivp = f(z)− 1

p

p−1∑v=0

f(z)| k2τv,p.

Recall that our goal is to build a cusp form from f . We do so by using a quadratic twistto remove the terms in the Fourier expansions for f which have non-positive exponents. Foran odd prime p, let εp ∈ {±1}, and define

fp := f ⊗ χtrivp − εpf ⊗

(•p

)= 2

∑(n

p )=−εp

a(n)qn.

Now if the non-holomorphic part of f is supported only on terms with exponents n satisfying(np

)= εp or 0, then fp vanishes at infinity. This is the inspiration for condition C. In order

for fp to be a cusp form, however, it must vanish at every cusp, so the full statement ofcondition C requires similar behavior at every cusp.

To consider the Fourier expansion of f(z) at any cusp s ∈ Q ∪ {∞}, we associate to s anelement ψs = (α, φ(z)) ∈ G′, with α∞ = s. Then

(2.6) f(z)| k2ψs =

∑aψs(n)q

n+ rs4

hs

where hs | N and rs ∈ {0, 1, 2, 3} [19, pp. 181–182]. The integer hs depends only onthe Γ1(N)-equivalence class of s, and rs depends only on hs and k. If a different elementξs = (β, θ(z)) ∈ G′ were chosen with β∞ = s, then f | k

2ξs and f | k

2ψs would differ only

by multiplication by a root of unity. Hence the Fourier expansion of f(z) at s is given byf(z)| k

2ψs up to a root of unity.

Suppose that f(z) is an eta-quotient, as in (1.3), and that s = ac, where (a, c) = 1. Then

in (2.6), hs = N(c2,N)

, and the order of vanishing of f(z) at s is

(2.7) ords(f) =N

24(c2, N)

∑δ|N

(c, δ)2

δrδ.

This can be calculated using (2.2), or can be found in [20].

8 STEPHANIE TRENEER

Definition. We say that f satisfies condition C for a prime p if there exists εp ∈ {±1} suchthat for each ψs, the following is true for the Fourier expansion in (2.6): for all n < 0 withp - (4n+ rs) and aψs(n) 6= 0, we have(

4n+ rsp

)= εp

(hsp

).

Remark 6. In the case s = ∞ we have hs = 1 and rs = 0, so condition C requires that(np

)= εp for all n < 0 with p - n and a(n) 6= 0.

Remark 7. By the discussion above, condition C is well-defined. Further, it is sufficient tocheck the condition for one element from each Γ1(N)-equivalence class of cusps, because ifs′ = γs for γ ∈ Γ1(N), then γα∞ = s′, and f | k

2γψs = f | k

2ψs.

Remark 8. It is possible for f(z) to trivially satisfy condition C for a prime p if for eachcusp s, there is no n < 0 with aψs(n) 6= 0 and p - (4n + rs). In Corollary 3.8 in the nextsection, we show that in this special case, a broader range of congruences may be obtainedthan those given in Theorems 1.2 and 1.3.

Example 2. To illustrate condition C, consider the eta-quotient

1

η48(z)= q−2 + 48q−1 + 1244 + · · · ∈ M−24(SL2(Z)).

Since ∞ is the only cusp of SL2(Z), condition C requires only that(−2p

)=(−1p

), so 1

η(48z)

satisfies condition C for all primes p ≡ 1, 7 (mod 8).

The next result shows that every weakly holomorphic modular form of integral or half-integral weight on Γ1(N) satisfies condition C for a positive proportion of primes p.

Theorem 2.1. Let k and N be integers with N ≥ 1 and 4 | N . Suppose that f(z) =∑a(n)qn ∈ M k

2(Γ1(N)). Then a positive proportion of primes p have the property that f

satisfies condition C for p.

Proof. Let S be a complete set of representatives for the equivalence classes of cusps underthe action of Γ1(N). For each s ∈ S, let hs, rs and ψs be as in the definition of condition C,and recall that hs | N . Let Ms := {m ∈ Z<0 : aψs(m) 6= 0}. By Remark 7, the set Ms doesnot depend on our choice of representative s. If p is prime and a is an integer, then p ≡ 1

(mod 4a) implies that(ap

)= 1. Set M :=

∏s∈S

∏m∈Ms

(4m + rs), and M ′ :=∏

p|MN

p. Then for

any prime p ≡ 1 (mod 4M ′), we have (hsp

)= 1

for each s ∈ S, and (4m+ rs

p

)= 1

for each m ∈ Ms and s ∈ S. Then f satisfies condition C for each p ≡ 1 (mod 4M ′).By Dirichlet’s theorem on primes in arithmetic progressions, the density of such primes is1/φ(4M ′). This is a lower bound for the density of primes p for which f satisfies conditionC. This completes the proof of Theorem 2.1. �

QUADRATIC TWISTS 9

We define the quadratic Gauss sum

(2.8) ga :=

p−1∑t=0

(t

p

)e2πiat/p,

and set g := g1. Properties of Gauss sums may be found in [18], including the facts

(2.9) ga =

(a

p

)g,

and

(2.10) g2 =

(−1

p

)p,

which we will use later.We will also need the following identity which relates Gauss sums with quadratic twists.

Let(•p

)be the quadratic character modulo p. Then

(2.11) f(z)⊗(•p

)=g

p

p−1∑v=0

(v

p

)f(z)| k

2τv,p.

Finally, we review some facts about Hecke operators. The integral weight Hecke operatorTk,N(p) preserves Mk(Γ1(N)). If p ≡ −1 (mod N), its effect on the Fourier expansion off(z) =

∑a(n)qn ∈Mk(Γ1(N)) is given by

(2.12) f(z)|Tk,N(p) =∑[

a(pn) + pk−1a

(n

p

)]qn.

Similarly, the half-integral weight Hecke operator T k2,N(p2) preserves the space M k

2(Γ1(N)).

If p ≡ −1 (mod N), then

(2.13) f(z)|T k2,N(p2) =

∑[a(p2n) +

((−1)

k−12 n

p

)p

k−32 a(n) + pk−2a

(n

p2

)]qn.

3. Proof of Main Theorems

The following result is critical for proving Theorem 1.2.

Theorem 3.1. Suppose that p is an odd prime, and that k is an integer. Let N be a positiveinteger with 4 | N and (N, p) = 1. Let K be a number field with ring of integers OK, and

suppose that f(z) =∑a(n)qn ∈ M k

2(Γ1(N)) ∩ OK((q)). Let f satisfy condition C for p.

Then for every positive integer j, there exists an integer β ≥ j − 1 and a cusp form

gp,j(z) ∈ S k2+

pβ(p2−1)2

( ˜Γ1(Np2)) ∩ OK [[q]],

with the property that

gp,j(z) ≡∑

(np )=−εp

a(n)qn (mod pj).

To prove Theorem 3.1, we need the following technical result.

10 STEPHANIE TRENEER

Proposition 3.2. Suppose that p is an odd prime, and that k is an integer. Let N be a

positive integer with 4 | N and (N, p) = 1. Let f(z) =∑a(n)qn ∈ M k

2(Γ1(N)). Suppose

that µ ∈ {±1,±i}, and that ξ :=

((a bcp2 d

), µ√cp2z + d

)∈ G′ with ac > 0. Let

f(z)| k2ξ =

∑n≥nξ

aξ(n)qn+ r

4h .

Then (f(z)⊗

(•p

))∣∣∣∣k2

ξ =

(h

p

)∑n≥nξ

(4n+ r

p

)aξ(n)q

n+ r4

h .

Proof. Let g be the Gauss sum defined in (2.8). Then recalling (2.11), we have

(3.1)

(f(z)⊗

(•p

))∣∣∣∣k2

ξ =g

p

p−1∑v=0

(v

p

)f(z)| k

2τv,pξ.

For each v, we can choose a positive integer sv so that

(3.2) 4Nsv ≡ a−1vd (mod p),

and set

(3.3) wv := 4Nsv.

Define

(3.4) δv :=

(1− awvcp+ vwvc

2p2 a2wv−avdp

− acvwv + bvcp

−wvc2p3 1 + awvcp

),

and

(3.5) βv :=(δv,√−wvc2p3z + 1 + awvcp

).

A computation in the group G (2.1) shows that

(3.6) τv,pξ = βvξτwv ,p.

Then combining (3.6)and (3.1) yields(f(z)⊗

(•p

))∣∣∣∣k2

ξ =g

p

p−1∑v=0

(v

p

)f(z)| k

2βvξτwv ,p.

The next lemma shows that βv ∈ Γ1(N).

Lemma 3.3. Let βv be defined as in (3.5). Then βv ∈ Γ1(N).

Proof. By the definitions of wv in (3.3) and δv in (3.4), it is clear that δv ∈ Γ1(N). It remainsto show that

j(δv, z) =√−wvc2p3z + 1 + wvacp.

By definition (2.3),

j(δv, z) =

(−wvc2p3

1 + wvacp

)ε−11+wvacp

√−wvc2p3z + 1 + wvacp.

QUADRATIC TWISTS 11

Since 4 | wv, we have ε1+wvacp = 1. Now

(−wvc2p3

1 + wvacp

)=

(−wvp

1 + wvacp

).

Write wv = 2er, where r is odd. Then using properties of Jacobi symbols,

(−wvp

1 + wvacp

)=

(−1

1 + wvacp

)(2

1 + wvacp

)e(rp

1 + wvacp

)= 1,

since wvacp ≡ 0 (mod 8). This proves Lemma 3.3. �

Returning to the proof of Proposition 3.2, we see by Lemma 3.3 that

(f(z)⊗

(•p

))∣∣∣∣k2

ξ =g

p

p−1∑v=0

(v

p

)f(z)| k

2ξτwv ,p

=g

p

p−1∑v=0

(v

p

)∑n≥nξ

aξ(n)qn+ r

4h

∣∣∣∣∣∣k2

τwv ,p

=g

p

p−1∑v=0

(v

p

)∑n≥nξ

aξ(n) exp

(2πi(z − wv/p)(4n+ r)

4h

)

=g

p

∑n≥nξ

aξ(n)qn+ r

4h

p−1∑v=0

(v

p

)exp

(−2πiwv(4n+ r)

4hp

).(3.7)

Now recalling the definition of sv (3.2), we see that since 4, a, d, and N are coprime to p, thesv = wv/4N run through the residue classes modulo p as v does. Also, v ≡ wvad

−1 (mod p).Then writing N = hh′, we have

p−1∑v=0

(v

p

)exp

(−2πiwv(4n+ r)

4hp

)=

p−1∑v=0

(wvad

−1

p

)exp

(2πiwvp

(−4n− r

4h

))

=

(ad−1

p

) p−1∑v=0

(4Nsvp

)exp

(2πi(4Nsv)

p

(−4n− r

4h

))

=

(Nad−1

p

) p−1∑v=0

(svp

)exp

(2πisvp

(−h′(4n+ r))

)=

(Nad−1

p

)(−h′(4n+ r)

p

)g.(3.8)


The last line follows from (2.9). Combining (3.7) and (3.8), and using (2.10), we have(f(z)⊗

(•p

))∣∣∣∣k2

ξ =g

p

∑n≥nξ

aξ(n)qn+ r

4h

(Nad−1

p

)(−h′(4n+ r)

p

)g

=

(−1

p

)(Nad−1

p

)∑n≥nξ

(−h′(4n+ r)

p

)aξ(n)q

n+ r4

h

=

(Nad−1h′

p

)∑n≥nξ

(4n+ r

p

)aξ(n)q

n+ r4

h .(3.9)

Now we use the facts that ad ≡ 1 (mod p), that h′ ≡ h−1N (mod p), and that if b is coprime

to p, then(b−1

p

)=(bp

), to see that

(3.10)

(Nad−1h′

p

)=

(N2ad−1h−1

p

)=

(adh

p

)=

(h

p

).

Therefore by (3.9) and (3.10), we have(f(z)⊗

(•p

))∣∣∣∣k2

ξ =

(h

p

)∑n≥nξ

(4n+ r

p

)aξ(n)q

n+ r4

h .

This concludes the proof of Proposition 3.2. �

Proposition 3.4. Suppose that p is an odd prime, and k is an integer. Let N be a positive

integer with 4 | N and (N, p) = 1. Let f(z) =∑a(n)qn ∈ M k

2(Γ1(N)) satisfy condition C

for p with εp. Define

fp(z) := f(z)⊗ χtrivp − εpf(z)⊗(•p

)∈M k

2( ˜Γ1(Np2)).

Then fp vanishes at each cusp acp2

.

Proof. Assume without loss of generality that ac > 0, and let

ξ :=

((a bcp2 d

), µ√cp2z + d

)∈ G′

with µ ∈ {±1,±i}. Then by (2.5), we have

(3.11) fp(z)| k2ξ = f(z)| k

2ξ − 1

p

p−1∑v=0

f(z)| k2τv,pξ − εpf(z)⊗

(•p

)| k2ξ.

Recall the definitions of sv (3.2), wv (3.3) and βv (3.5). Using (3.6) and Lemma 3.3, we have

1

p

p−1∑v=0

f(z)| k2τv,pξ =

1

p

p−1∑v=0

f(z)| k2ξτwv ,p

=1

p

p−1∑v=0

∑n≥nξ

aξ(n)qn+ r

4h

| k2τwv ,p

QUADRATIC TWISTS 13

=1

p

p−1∑v=0

∑n≥nξ

aξ(n) exp

(2πi(z − wv/p)

4h(4n+ r)

)

=1

p

∑n≥nξ

aξ(n)qn+ r

4h

p−1∑v=0

exp

(−2πisv

p(h′(4n+ r))

)=

∑h′(4n+r)≡0 (mod p)

aξ(n)qn+ r

4h

=∑

4n+r≡0 (mod p)

aξ(n)qn+ r

4h .(3.12)

The last line follows since p - h′.Now by (3.11), (3.12) and Proposition 3.2, we have

(3.13) fp(z)| k2ξ =

∑4n+r 6≡0 (mod p)

aξ(n)qn+ r

4h − εp

(h

p

)∑n≥nξ

(4n+ r

p

)aξ(n)q

n+ r4

h .

Therefore, fp(z) vanishes at ξ if for each n < 0 with p - (4n+ r) and aξ(n) 6= 0, we have(4n+ r

p

)= εp

(h

p

).

This requirement is met since f satisfies condition C for p. This concludes the proof ofProposition 3.4. �

Now for each odd prime p, we define the eta-quotient

(3.14) Fp(z) :=

ηp2

(z)η(p2z)

∈M p2−12

(Γ0(p2)) if p ≥ 5,

η27(z)η3(9z)

∈M12(Γ0(9)) if p = 3.

Using (2.7), we see that Fp(z) vanishes at every cusp ac

with p2 - c. By the definition (1.2) ofη(z) (1.2), it is clear that Fp(z) ≡ 1 (mod p), and an easy induction argument shows that

Fp(z)ps−1 ≡ 1 (mod ps) for any integer s ≥ 1.

Proof of Theorem 3.1. Let f be as in the hypothesis of Theorem 3.1, and fix j ≥ 1. Thecusps a

chave either p2 | c or p2 - c. If β ≥ j − 1 is sufficiently large, then

fp(z) · Fp(z)pβ ≡ fp(z) (mod pj)

vanishes at all cusps where p2 - c, and by Proposition 3.4, it vanishes at all cusps with p2 | c.Set

(3.15) gp,j(z) :=1

2fp(z) · Fp(z)p

β ∈ S k2+

pβ(p2−1)2

( ˜Γ1(Np2)) ∩ OK [[q]].

Then since

fp(z) =∑p-n

a(n)qn − εp∑(

n

p

)a(n)qn = 2

∑(n

p )=−εp

a(n)qn,


we have

(3.16) gp,j(z) ≡∑

(np )=−εp

a(n)qn (mod pj).

Combining (3.15) and (3.16) proves Theorem 3.1. �

The following Proposition gives a straightforward modification of a result of Serre [32, Ex.6.4].

Proposition 3.5. Let k > 1 and N ∈ N, and let F be an algebraic number field with ringof integers OF . Suppose that m is an ideal of OF with norm M . Then a positive proportionof the primes p ≡ −1 (mod MN) have the property that

f(z)|Tk,N(p) ≡ 0 (mod m)

for each f(z) ∈ Sk(Γ1(N)) ∩ OF [[q]].

Sketch of proof . Let Sk(Γ1(N))OF /m be the finite set of reductions modulo m of the formsin Sk(Γ1(N)) ∩ OK [[q]]. Each g(z) ∈ Sk(Γ1(N))OF /m satisfies a congruence of the form

g(z) ≡∑h

αg(h)h(δz) (mod m)

where the h run over all newforms of level dividing N , the δ | N , and the αg(h) are algebraic.We can choose a nonzero integer C so that each Cαg(h) is an algebraic integer. Let K bea finite extension of F such that the coefficients of each h and each Cαg(h) lies in OK , andset m′ := Cm. Then consider the representation

ρ = χ⊕⊕h

ρh

where χ is the cyclotomic character moduloMN and each ρh : Gal(Q/Q

)→ GL2 (OK/m

′OK)is the Galois representation associated to the newform h such that

Tr(ρ(Frobp)) ≡ ah(p) (mod m′)

for each prime p - NM ′, where ah(p) is the Hecke eigenvalue of Tk,N(p) for h (see [13, 14]

for details). Now let c ∈ Gal(Q/Q

)be complex conjugation. Then each ρh(c) is a matrix of

trace zero, and the Chebotarev density theorem guarantees that the Frobenius elements of apositive proportion of primes p lie in the same conjugacy class as c. Further, all such p mustsatisfy p ≡ −1 (mod MN), since χ(c) = −1 in Z/MNZ. This gives the desired result. �

The next result, due to Ahlgren and Ono [4], follows from Proposition 3.5 by usingShimura’s correspondence.

Proposition 3.6. (Ahlgren-Ono, [4, Lemma 3.1]) Let N ≥ 1 and k > 3, and let K be a num-

ber field with ring of integers OK. Suppose that f(z) =∑∞

n=1 a(n)qn ∈ S k2(Γ1(N))∩ OK [[q]],

and m is an ideal of OK with norm M . Then a positive proportion of the primes p ≡ −1(mod MN) have the property that

f(z)|T k2,N(p2) ≡ 0 (mod M).

QUADRATIC TWISTS 15

We are now able to prove Theorem 1.2.

Proof of Theorem 1.2. Let f and p be as in the hypotheses of Theorem 1.2, fix an integerj ≥ 1, and let

gp,j(z) ≡∑

(np )=−εp

a(n)qn (mod pj)

be the cusp form guaranteed by Theorem 3.1. Set κ := k + pβ(p2 − 1). By Proposition 3.6,a positive proportion of the primes Q ≡ −1 (modNpj) have the property that

gp,j(z)|Tκ2,Np2(Q

2) ≡ 0 (mod pj).

If we write gp,j(z) =∑∞

n=1 b(n)qn, then

(3.17) gp,j(z)|Tκ2,Np2(Q

2)

=∞∑n=1

(b(Q2n) +

((−1)

κ−12 n

Q

)Q

κ−32 b(n) +Qκ−2b

(n

Q2

))qn ≡ 0 (mod pj).

Replacing n by Qn in (3.17), we have

(3.18)∞∑n=1

(b(Q3n) +

((−1)

κ−12 Qn

Q

)Q

κ−32 b(Qn) +Qκ−2b

(Qn

Q2

))qQn

≡ 0 (mod pj).

If (Q, n) = 1, then the coefficient of qQn in (3.18) is just b(Q3n). So

a(Q3n) ≡ b(Q3n) ≡ 0 (mod pj)

for all n coprime to Q with(Q3np

)= −εp. Since Q ≡ −1 (mod p), this completes the proof

of Theorem 1.2. �

For the integral weight analog, we will require the following result of Serre.

Proposition 3.7. (Serre, [31, Corollaire du Theoreme 1]) Let

f(z) =∞∑n=0

cnqnM , M ≥ 1,

be a modular form of integral weight k ≥ 1 on a congruence subgroup of SL2(Z), and supposethat the coefficients cn lie in the ring of integers of an algebraic number field K. Then forany integer m ≥ 1,

cn ≡ 0 (mod m)

for almost all n.

Proof of Theorem 1.3. Let f and p be as in the hypotheses of Theorem 1.3, and fix j ≥ 1.First, if 4 - N , we may consider f(z) as a modular form on Γ1(4N), so we will assume that

4 | N , and view f(z) as a member of M 2k2(Γ1(N)) = Mk(Γ1(N)). Then by Theorem 3.1,


there exists a cusp form gp,j of positive integral weight with algebraic integer coefficientssuch that

(3.19) gp,j(z) ≡∑

(np )=−εp

a(n)qn (mod pj).

The first assertion of Theorem 1.3 now follows from (3.19) and Proposition 3.7.For the second assertion, it follows from (3.19) and Proposition 3.5 that if gp,j(z) has

weight κ, then a positive proportion of the primes Q ≡ −1 (mod Npj) have the propertythat

(3.20) gp,j(z)|Tκ,Np2(Q) ≡ 0 (mod pj).

If we write gp,j(z) =∑∞

n=1 b(n)qn, then (2.12) and (3.20) imply that

(3.21) gp,j(z)|Tκ,Np2(Q) =∞∑n=1

(b(Qn) +Qκ−1b

(n

Q

))qn ≡ 0 (mod pj).

If (Q,n) = 1, then the coefficient of qn in (3.21) is just b(Qn). Therefore

a(Qn) ≡ b(Qn) ≡ 0 (mod pj)

for all n coprime to Q with(Qnp

)= −εp. This concludes the proof of Theorem 1.3. �

We now record our results in the special case where a weakly holomorphic modular formf(z) trivially satisfies condition C, as in Remark 8.

Corollary 3.8. Suppose that p is an odd prime, and that k is an integer. Let N be a positiveinteger with 4 | N and (N, p) = 1. Let K be a number field with ring of integers OK, and

suppose that f(z) =∑a(n)qn ∈M k

2(Γ1(N)) ∩ OK((q)). Let f trivially satisfy condition C

for p.

(a) If k is odd, then for each positive integer j, a positive proportion of the primes Q ≡ −1(mod Npj) have the property that


for all n coprime to Qp.(b) If k is even, then for each positive integer j,

(i) a(n) ≡ 0 (mod pj) for almost all n not divisible by p, and(ii) a positive proportion of the primes Q ≡ −1 (mod Npj) have


for all n coprime to Qp.

Proof. We set

fp(z) := f(z)⊗ χtrivp .

Defining ξ as in Proposition 3.4, and following the first part of the proof of that result, wesee that

fp(z)| k2ξ =

∑4n+r 6≡0 (mod p)

a(n)qn+ r

4h .

QUADRATIC TWISTS 17

Then since condition C is trivially satisfied, fp(z)| k2ξ vanishes at each cusp a

cwith p2 | c.

Now for j ≥ 1, we set

gp,j(z) := fp(z) · Fp(z)pβ

for sufficiently large β, so that the cusp form gp,j(z) satisfies

gp,j(z) ≡∑p-n

a(n)qn (mod pj).

Now by proceeding as in the proofs of Theorems 1.2 and 1.3, we reach the desired result. �

4. Eta-quotients

We now prove Corollary 1.4.

Proof of Corollary 1.4. Let f(z) satisfy the hypotheses of Corollary 1.4. Since η(z) isholomorphic and non-vanishing on H, then f(z) is also holomorphic on H. Since η(z) hasinteger coefficients, so does f(z). We can replace N by a power of N , if necessary, so thatf(z) satisfies

(4.1) N∑δ|N

rδδ≡ 0 (mod 24).

Then f(z) is a weakly holomorphic modular form of weight k2

with level N [16]. We maynow apply either Theorem 1.2 or Theorem 1.3, depending on the parity of k, to completethe proof of Corollary 1.4. �

Next, we prove Proposition 1.5, which gives congruences for the number of overpartitionsof n.

Proof of Proposition 1.5. Let ` be an odd prime. By (2.7), f(z) = η(2z)η2(z)

is holomorphic at

the cusps ac

with c even. If c is odd, then the expansion of f at ac

has the form

C · q−1hs

+O(1),

where hs = 16(c2,16)

. Therefore, as in Example 1, condition C holds for f(z) for each odd prime

` with ε` =(−1`

). The result now follows from Corollary 1.4(a). �

Finally, we prove Theorem 1.7.

Proof of Theorem 1.7. Fix a prime ` ≥ 3, and let

f`(z) :=η(24`z)

η(24z).

We must first explicitly determine condition C for f`(z). Let s = ac

be a cusp, and choose

ψs = (α, φ(z)) ∈ G′ with α =

(a bc d

). Using (2.7), we find that if ` | c (resp. if 9 | c when

` = 3) then f`(z) vanishes at s, so we may assume that ` - c (resp. assume that 9 - c when


` = 3). We will use the following two useful facts. Let γ ∈ SL2(Z). Then for any choice of(γ, φ(z)) ∈ G′,

(4.2) η(z)| 12(γ, φ(z)) = εγ · η(z)

for some root of unity εγ. Secondly, for any positive integer δ, there exists a γ′ ∈ SL2(Z) andintegers A, B and D, with A and D positive, such that

(4.3)

(δ 00 1

)γ = γ′

(A B0 D

).

Using (4.2) and (4.3) together with (2.2) and (1.2), we find that

(4.4) f`(z)|0ψs = C · q1−`hs

∞∏n=1

(1− (ζ`q24hs

)n)

(1− (ζq24`hs

)n)

where C is a nonzero constant, ζ and ζ` are roots of unity depending on s and `, andhs = 576`

(c2,576`). Expanding the infinite product in (4.4), we see that the denominator does not

contribute to the terms with powers of qhs less than 24`. In fact,

C · q1−`hs

∞∏n=1

(1− (ζ`q24hs

)n)

(1− (ζq24`hs

)n)= C · q1−`

hs

(`−1∑n=0

c(n)ζn` q24nhs

+O(q24`hs

))

= C ·`−1∑n=0

c(n)ζn` q24n+1−`hs

+O(q23`+1hs

),

where

(4.5) c(n) =n∑r=1

(−1)rp(n; r),

and p(n; r) is the number of partitions of n into r distinct parts. Note that c(n) does notdepend on ` or on s. Hence f`(z) satisfies condition C for a prime p ≥ 5 if there exists anεp ∈ {±1} such that for each 0 ≤ n < `−1

24with c(n) 6= 0 and p - (24n+ 1− `), we have

(4.6)

(24n+ 1− `

p

)= εp

(hsp

)= εp

(`

p

)for all cusps s = a

cwith ` - c. The last equality in (4.6) follows from the fact that for each

such s, we have hs = d2` for some integer d.Now it is a result of Euler [17, pp. 285–287] that

c(n) =

{(−1)k if n = k(3k±1)

2,

0 otherwise.

Therefore if ` ≥ 23 and p ≥ 5 satisfy the hypotheses of the theorem, then by setting n = 0(if p - ` − 1) or n = 1 (if p | ` − 1) in equation (4.6), and solving for εp, we see that f`(z)satisfies condition C for p with

(4.7) εp :=

(`−`2p

)if p - `− 1, and(

25`−`2p

)if p | `− 1.

QUADRATIC TWISTS 19

Then by Corollary 1.4(b)(i), for each j ≥ 1, b`(n) ≡ 0 (mod pj) for almost all n with(np

)= −εp. If m ≥ m0, then by the integral weight analog of Theorem 1.1 [35, Theorem 1.2],

b`(pmn) ≡ 0 (mod pj) for almost all (n, p) = 1. Together, such n comprise p−1

2arithmetic

progressions modulo p, and one progression modulo pm0 , so that

lim infX→∞

#{1 ≤ n ≤ X | b`(n) ≡ 0 (mod pj)}/X ≥ p− 1

2p+

1

pm0.

Now we consider ` ≤ 23. In this case, for each cusp s,

f`(z)|0ψs = q1−`hs

+O(q).

If p | `− 1, then f`(z) trivially satisfies condition C for p, so by Corollary 3.8,

lim infX→∞

#{1 ≤ n ≤ X | b`(n) ≡ 0 (mod pj)}/X ≥ p− 1

p.

If p - `− 1, then set εp :=(`−`2p

)and define

f`,p(z) := f`(z)− εpf`(z)⊗(•p

)=∑p|n

a(n)qn + 2∑

(np )=−εp

a(n)qn.

(Note that we do not twist f`(z) by the trivial character modulo p.) Using arguments as inthe proof of Proposition 3.4, f`,p(z) vanishes at each cusp a

cwith p2 | c. Then for each j ≥ 1,

recalling (3.14), we have a cusp form

f`,p(z) · Fp(z)pβ ≡ f`,p(z) (mod pj)

for sufficiently large β ≥ j − 1. Applying Proposition 3.7 to this family of cusp forms, we

see that b`(n) ≡ 0 (mod pj) for almost all n with(np

)= −εp or p | n. Therefore

lim infX→∞

#{1 ≤ n ≤ X|b`(n) ≡ 0 (mod pj)}/X ≥ p+ 1

2p.

This concludes the proof of Theorem 1.7. �

5. Simultaneous congruences

To proof Theorem 1.8, we need the following generalizations of Propositions 3.5 and 3.6.

Proposition 5.1. Fix an integer r ≥ 1. Associate to each 1 ≤ i ≤ r, the integers kiand Ni ≥ 1. Suppose that F is a number field with ring of integers OF , and let mi be anideal of OF with norm Mi for each i. Then a positive proportion of the primes p ≡ −1(mod lcm(M1, . . . ,Mr, N1, . . . , Nr)) have the property that

f(z)|Tki,Ni(p) ≡ 0 (mod mi)

for every 1 ≤ i ≤ r and each f(z) ∈ Ski(Γ1(Ni)) ∩ OF [[q]].

Sketch of proof. For each i, let Ski(Γ1(Ni))OF /mi

be the finite set of reductions modulo mi ofthe forms in Ski

(Γ1(Ni)) ∩ OK [[q]]. Each gi(z) ∈ Ski(Γ1(Ni))OF /mi

satisfies a congruence ofthe form

gi(z) ≡∑hi

αgi(hi)hi(δz) (mod mi)


where the hi run over all newforms of level dividing Ni, the δ | Ni, and the αgi(hi) are

algebraic. We can choose a nonzero integer C so that each Cαgi(hi) is an algebraic integer.

Let K be a finite extension of F such that the coefficients of each hi and each Cαgi(hi) lies

in OK for all i, and set m′i := Cmi. Then consider the representation

ρ = χ⊕r⊕i=1

⊕hi

ρhi,

where χ is the cyclotomic character modulo lcm(M1,M2, . . . ,Mr, N1, N2, . . . , Nr) and

ρhi: Gal

(Q/Q

)→ GL2 (OK/m

′iOK)

is the modulo m′ Galois representation associated to the newform hi. The result now followsas in the proof of Proposition 3.5. �

Proposition 5.2. Fix a positive integer r. For each 1 ≤ i ≤ r, let ki be an integer and letNi be a positive integer with 4 | Ni. Let K be an algebraic number field, and suppose thatf1(z), f2(z), . . . , fr(z) are cusp forms with each

fi(z) =∑

ai(n)qn ∈ S ki2

(Γ1(Ni)) ∩ OK [[q]].

For each i, let mi ⊂ OK be an ideal with norm Mi, and suppose that each ki > 3. Thena positive proportion of the primes p ≡ −1 (mod lcm(M1, . . . ,Mr, N1, . . . , Nr)) have theproperty that

fi(z)|T ki2,Ni

(p2) ≡ 0 (mod mi)

if ki is odd, andfi(z)|T ki

2,Ni

(p) ≡ 0 (mod mi)

if ki is even, for each 1 ≤ i ≤ r.

Proof. Proposition 5.2 follows from Proposition 5.1 by applying the Shimura lift to those fiof half-integral weight, in the manner of Proposition 3.6. �

Now we show that Theorem 1.8 follows from Theorem 5.2.

Proof of Theorem 1.8. For each i, Theorem 3.1 guarantees cusp forms

gi,j(z) ∈ Ski2

+pβii

(p2i−1)

2

( ˜Γ1(Nip2i )) ∩ OK [[q]],

with the property that

gi,j(z) ≡∑

“npi

”=−εpi

ai(n)qn (mod pji ),

for each j ≥ 1. Then by Theorem 5.2, for each r-tuple (j1, . . . , jr) with ji ≥ 1, a positiveproportion of the primes Q ≡ −1 (mod N1 · · ·Nrp

j11 · · · pjrr ) have the property that

gi,ji(z)|Tκi2,Nip2i

(Q2) ≡ 0 (mod pjii ),

if ki is odd, or

gi,ji(z)|Tκi2,Nip2i

(Q) ≡ 0 (mod pjii ),

QUADRATIC TWISTS 21

if ki is even, where κi := ki + pβi

i (p2i − 1), for each 1 ≤ i ≤ r. Using (2.13) (resp. (2.12)) if

ki is odd (resp. even), it follows as in the proof of Theorem 1.2 (resp. Theorem 1.3) that

ai(Qt(i)n) ≡ 0 (mod pjii )

for each 1 ≤ i ≤ r and each (n,Q) = 1 satisfying the conditions{(

−npi

)= −εpi : 1 ≤ i ≤ r

}.

�

References

[1] S. Ahlgren, Distribution of parity of the partition function in arithmetic progressions, Indag. Math.(N.S.) 10 (1999), no. 2, 173–181.

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Department of Mathematics, Dartmouth College, Hanover, New Hampshire 03755E-mail address: [email protected]

quadratic twists and the coefficients of weakly holomorphic...

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