CONGRUENCES FOR ANDREWS’ SPT-FUNCTION MODULO 32760 AND EXTENSION OF ATKIN’S HECKE-TYPE PARTITION CONGRUENCES

F. G. GARVAN

Dedicated to the memory of A.J. (Alf) van der Poorten, my former teacher

Abstract. New congruences are found for Andrews’ smallest parts partition function spt(n). The generating function for spt(n) is related to the holomorphic part α(24z) of a certain weak Maass form M(z) of weight \frac{3}{2} . We show that a normalized form of the generating function for spt(n) is an eigenform modulo 72 for the Hecke operators T (ℓ^{2}) for primes ℓ > 3, and an eigenform modulo p for p = 5, 7 or 13 provided that (ℓ, 6p) = 1. The result for the modulus 3 was observed earlier by the author and considered by Ono and Folsom. Similar congruences for higher powers of p (namely 5^{6}, 7^{4} and 13^{2}) occur for the coefficients of the function α(z). Analogous results for the partition function were found by Atkin in 1966. Our results depend on the recent result of Ono that M_{ℓ}(z/24) is a weakly holomorphic modular form of weight \frac{3}{2} for the full modular group where

M_{ℓ}(z) = M(z)|T (ℓ^{2}) − (\frac{3}{ℓ})(1 + ℓ)M(z).

1. Introduction

Andrews [1] defined the function spt(n) as the number of smallest parts in the partitions of n. He related this function to the second rank moment and proved some surprising congruences mod 5, 7 and 13. Rank and crank moments were introduced by A. O. L. Atkin and the author [2]. Bringmann [6] studied analytic, asymptotic and congruence properties of the generating function for the second rank moment as a quasi-weak Maass form. Further congruence properties of Andrews’ spt-function were found by the author [10], [11], Folsom and Ono [8] and Ono [12]. In particular, Ono [12] proved that if (\frac{1−24n}{ℓ}) = 1 then

(1.1)  spt(ℓ^{2}n − \frac{1}{24}(ℓ^{2} − 1)) ≡ 0 (mod ℓ),

for any prime ℓ ≥ 5. This amazing result was originally conjectured by the author^{(i)}. Earlier special cases were observed by Tina Garrett [9] and her students. Recently the author [11] has proved the following congruences for powers of 5, 7 and 13. For a, b, c ≥ 3,

(1.2)  spt(5^{a}n + δ_{a}) + 5 spt(5^{a−2}n + δ_{a−2}) ≡ 0 (mod 5^{2a−3}),
(1.3)  spt(7^{b}n + λ_{b}) + 7 spt(7^{b−2}n + λ_{b−2}) ≡ 0 (mod 7^{⌊ \frac{1}{2} (3b−2)⌋}),

Date: November 8, 2010. 1991 Mathematics Subject Classification. Primary 11P83, 11F33, 11F37; Secondary 11P82, 05A15, 05A17. Key words and phrases. Andrews’s spt-function, weak Maass forms, congruences, partitions, modular forms. The author was supported in part by NSA Grant H98230-09-1-0051. The first draft of this paper was written

October 25, 2010.

(i)The congruence (1.1) was first conjectured by the author in a Colloquium given at the University of Newcastle, Australia on July 17, 2008.

(1.4)  spt(13^{c}n + γ_{c}) − 13 spt(13^{c−2}n + γ_{c−2}) ≡ 0 (mod 13^{c−1}),

where δ_{a}, λ_{b} and γ_{c} are the least nonnegative residues of the reciprocals of 24 mod 5^{a}, 7^{b} and 13^{c} respectively.

As in [12], [11] we define

(1.5)  a(n) := 12spt(n) + (24n − 1)p(n),

for n ≥ 0, and define

(1.6)  α(z) := ∑_{n≥0} a(n)q^{n− \frac{1}{24}} ,

where as usual q = exp(2πiz) and ℑ(z) > 0. We note that spt(0) = 0 and p(0) = 1. Bringmann [6] showed that α(24z) is the holomorphic part of the weight \frac{3}{2} weak Maass form M(z) on Γ_{0}(576) with Nebentypus χ_{12} where

(1.7)  M(z) := α(24z) − \frac{3i}{π\sqrt{2}} ∫_{−\overline{z}}^{i∞} \frac{η(24τ) dτ}{(−i(τ + z))^{\frac{3}{2}}} ,

η(z) := q^{\frac{1}{24}} ∏_{n=1}^{∞}(1 − q^{n}) is the Dedekind eta-function, the function α(z) is defined in (1.6), and

(1.8)  χ_{12}(n) = \begin{cases}1 & if n ≡ ±1 (mod 12), \\ −1 & if n ≡ ±5 (mod 12), \\ 0 & otherwise. \end{cases}

Ono [12] showed that for ℓ ≥ 5 prime, the operator

(1.9)  T (ℓ^{2}) − χ_{12}(ℓ)ℓ(1 + ℓ)

annihilates the nonholomorphic part of M(z), and the function M_{ℓ}(z/24) is a weakly holomorphic modular form of weight \frac{3}{2} for the full modular group where

(1.10)  M_{ℓ}(z) = M(z)|T (ℓ^{2}) − χ_{12}(ℓ)(1 + ℓ)M(z) = α(24z)|T (ℓ^{2}) − χ_{12}(ℓ)(1 + ℓ)α(24z).

In fact he obtained Theorem 1.1 (Ono [12]). If ℓ ≥ 5 is prime then the function

(1.11)  M_{ℓ}(z/24) η(z)^{ℓ^{2}}

is an entire modular form of weight \frac{1}{2}(ℓ^{2} + 3) for the full modular group Γ(1).

Applying this theorem Ono obtained

(1.12)  M_{ℓ}(z) ≡ 0 (mod ℓ).

The congruence (1.1) then follows easily.

Folsom and Ono [8] sketched the proof of the following Theorem 1.2 (Folsom and Ono). If ℓ ≥ 5 is prime then

(1.13) spt(ℓ^{2}n − s_{ℓ}) + χ_{12}(ℓ)(\frac{1 − 24n}{ℓ})spt(n) + ℓ spt (\frac{n + s_{ℓ}}{ℓ^{2}} ) ≡ χ_{12}(ℓ) (1 + ℓ) spt(n) (mod 3),

where

(1.14)  s_{ℓ} = \frac{1}{24}(ℓ^{2} − 1).

This result was observed earlier by the author. In this paper we prove a much stronger result.

Theorem 1.3. (i) If ℓ ≥ 5 is prime then

(1.15) spt(ℓ^{2}n − s_{ℓ}) + χ_{12}(ℓ)(\frac{1 − 24n}{ℓ})spt(n) + ℓ spt (\frac{n + s_{ℓ}}{ℓ^{2}} ) ≡ χ_{12}(ℓ) (1 + ℓ) spt(n) (mod 72).

(ii) If ℓ ≥ 5 is prime, t = 5, 7 or 13 and ℓ ≠ t then

(1.16) spt(ℓ^{2}n − s_{ℓ}) + χ_{12}(ℓ)(\frac{1 − 24n}{ℓ})spt(n) + ℓ spt (\frac{n + s_{ℓ}}{ℓ^{2}} ) ≡ χ_{12}(ℓ) (1 + ℓ) spt(n) (mod t).

Of course this implies the Corollary 1.4. If ℓ is prime and ℓ ∉ {2, 3, 5, 7, 13} then

(1.17) spt(ℓ^{2}n−s_{ℓ})+χ_{12}(ℓ)(\frac{1 − 24n}{ℓ})spt(n)+ℓ spt (\frac{n + s_{ℓ}}{ℓ^{2}} ) ≡ χ_{12}(ℓ) (1+ℓ) spt(n) (mod 32760).

This congruence modulo 32760 = 2^{3} · 3^{2} · 5 · 7 · 13 is the congruence referred in the title of this paper.

In 1966, Atkin [4] found a similar congruence for the partition function. Theorem 1.5 (Atkin). Let t = 5, 7, or 13, and c = 6, 4, or 2 respectively. Suppose ℓ ≥ 5 is prime and ℓ ≠ t. If (\frac{1−24n}{t}) = −1, then

(1.18)  ℓ^{3} p(ℓ^{2}n − s_{ℓ}) + ℓχ_{12}(ℓ)(\frac{1 − 24n}{ℓ})p(n) + p (\frac{n + s_{ℓ}}{ℓ^{2}} ) ≡ γ_{t} p(n) (mod t^{c}),

where γ_{t} is an integral constant independent of n.

We find that there is a corresponding result for the function a(n) defined in (1.5). Theorem 1.6. Let t = 5, 7, or 13, and c = 6, 4, or 2 respectively. Suppose ℓ ≥ 5 is prime and ℓ ≠ t. If (\frac{1−24n}{t}) = −1, then

(1.19)  a(ℓ^{2}n − s_{ℓ}) + χ_{12}(ℓ)(\frac{1 − 24n}{ℓ})a(n) + ℓ a (\frac{n + s_{ℓ}}{ℓ^{2}} ) ≡ χ_{12}(ℓ) (1 + ℓ) a(n) (mod t^{c}).

In Section 2 we prove Theorem 1.3. The method involves reviewing the action of weight −\frac{1}{2} Hecke operators T (ℓ^{2}) on the function η(z)^{−1} and doing a careful study of the action of weight \frac{3}{2} Hecke operators on the function \frac{d}{dz} η(z)^{−1} modulo 5, 7, 13, 27 and 32. In Section 3 we prove Theorem 1.6. The method involves extending Atkin’s [4] on modular functions to weight two modular forms on Γ_{0}(t) for t = 5, 7 and 13. The proof of both Theorems 1.3 and 1.6 depend on Ono’s Theorem 1.1.

2. Proof of Theorem 1.3

In this section we prove Theorem 1.3. Atkin [3] showed essentially that applying certain weight −\frac{1}{2} Hecke operators T (ℓ^{2}) to the function η(z)^{−1} produces a function with the same multiplier system as η(z)^{−1} and thus η(z) times this function is a certain polynomial (depending on ℓ) of Klein’s modular invariant j(z). We review Ono’s [13] recent explicit form for these polynomials. Although our proof does not depend on Ono’s result it is quite useful for computational purposes. The action of the corresponding weight \frac{3}{2} Hecke operators on \frac{d}{dz} η(z)^{−1} can be given in terms of the same polynomials. See Theorem 2.3 below. To finish the proof of the theorem we need to make a careful study of the action of these operators modulo 5, 7, 13, 27 and 32.

For ℓ ≥ 5 prime we define

(2.1)  Z_{ℓ}(z) = ∑_{n=−s_{ℓ}}^{∞} (ℓ^{3} p(ℓ^{2}n − s_{ℓ}) + ℓχ_{12}(ℓ)(\frac{1 − 24n}{ℓ})p(n) + p (\frac{n + s_{ℓ}}{ℓ^{2}} )) q^{n− \frac{1}{24}} .

Proposition 2.1 (Atkin [4]). The function Z_{ℓ}(z) η(z) is a modular function on the full modular group Γ(1).

It follows that Z_{ℓ}(z) η(z) is a polynomial in j(z), where j(z) is Klein’s modular invariant

(2.2)  j(z) := \frac{E_{4}(z)^{2}}{Δ(z)} = q^{−1} + 744 + 196884q + ⋯,

E_{2}(z), E_{4}(z), E_{6}(z) are the usual Eisenstein series (2.3)

E_{2}(z) := 1 − 24 ∑_{n=1}^{∞} σ_{1}(n)q^{n},  E_{4}(z) := 1 + 240 ∑_{n=1}^{∞} σ_{3}(n)q^{n},  E_{6}(z) := 1 − 504 ∑_{n=1}^{∞} σ_{5}(n)q^{n},

σ_{k}(n) = ∑_{d|n} q^{k}, and Δ(z) is Ramanujan’s function

(2.4)  Δ(z) := η(z)^{24} = q ∏_{n=1}^{∞}(1 − q^{n})^{24}.

In a recent paper, Ono [13] has found a nice formula for this polynomial. We define

(2.5)  E(q) := ∏_{n=1}^{∞}(1 − q^{n}) = q^{− \frac{1}{24}} η(z),

and a sequence of polynomials A_{m}(x) ∈ ℤ[x] by

(2.6)  ∑_{m=0}^{∞} A_{m}(x)q^{m} = E(q) \frac{E_{4}(z)^{2}E_{6}(z)}{Δ(z)} \frac{1}{j(z) − x}
= 1 + (x − 745)q + (x^{2} − 1489x + 160511)q^{2} + · · · .

Theorem 2.2 (Ono [13]). For ℓ ≥ 5 prime

(2.7)  Z_{ℓ}(z) η(z) = ℓ χ_{12}(ℓ) + A_{s_{ℓ}}(j(z)),

where Z_{ℓ}(z) is given in (2.1), and s_{ℓ} is given in (1.14).

We define a sequence of polynomials C_{ℓ}(x) ∈ ℤ[x] by

(2.8)  C_{ℓ}(x) := ℓ χ_{12}(ℓ) + A_{s_{ℓ}}(x),

= ∑_{n=0}^{s_{ℓ}} c_{n,ℓ}x^{n},

so that

(2.9)  Z_{ℓ}(z) η(z) = C_{ℓ}(j(z)).

We define

(2.10)  d(n) := (24n − 1) p(n),

so that

(2.11)  ∑_{n=0}^{∞} d(n)q^{24n−1} = q \frac{d}{dq} \frac{1}{η(24z)} = −\frac{E_{2}(24z)}{η(24z)},

and

(2.12)  a(n) = 12spt(n) + d(n).

ANDREWS’ SPT-FUNCTION MODULO 32760 5

For ℓ ≥ 5 prime we define

(2.13) Ξ_{ℓ}(z) = ∑_{n=−s_{ℓ}}^{∞} (d(ℓ^{2}n − s_{ℓ}) + χ_{12}(ℓ) ((\frac{1 − 24n}{ℓ}) − 1 − ℓ) d(n) + ℓ d (\frac{n + s_{ℓ}}{ℓ^{2}} )) q^{n− \frac{1}{24}} .

We then have the following analogue of Theorem 2.2. Theorem 2.3. For ℓ ≥ 5 prime we have

(2.14)  ℓ Ξ_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}} = − ∑_{n=0}^{s_{ℓ}} c_{n,ℓ} E_{4}(z)^{3n−1} Δ(z)^{s_{ℓ}−n} (24nE_{6}(z) + E_{4}(z)E_{2}(z))
+ χ_{12}(ℓ)ℓ(1 + ℓ)E_{2}(z) Δ(z)^{s_{ℓ}},

where the coefficients c_{n,ℓ} are defined by (2.6) and (2.8). Proof. Suppose ℓ ≥ 5 is prime. In equation (2.9) we replace z by 24z, apply the operator q \frac{d}{dq} and replace z by \frac{1}{24}z to obtain

(2.15)  ℓ Ξ_{ℓ}(z) η(z) = 24C′_{ℓ}(j(z)) q \frac{d}{dq} (j(z)) + (χ_{12}(ℓ)ℓ(1 + ℓ) − C_{ℓ}(j(z)) E_{2}(z)

The result then follows easily from the identities (2.16)

j(z) Δ(z) = E_{4}(z)^{3}, q \frac{d}{dq} (Δ(z)) = Δ(z) E_{2}(z), and q \frac{d}{dq} (j(z)) Δ(z) = −E_{4}(z)^{2}E_{6}(z),

which we leave as an easy exercise. □

We are now ready to prove Theorem 1.3. A standard calculation gives the following congruences.

(2.17) E_{4}(z)^{3} − 720 Δ(z) ≡ 1 (mod 65520), and  E_{2}(z) ≡ E_{4}(z)^{2}E_{6}(z) (mod 65520).

We now use (2.17) to reduce (2.15) modulo 65520.

ℓ Ξ_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}} (2.18)

≡ − ∑_{n=0}^{s_{ℓ}} c_{n,ℓ} E_{4}(z)^{3n−1} Δ(z)^{s_{ℓ}−n} (24nE_{6}(z)(E_{4}(z)^{3} − 720 Δ(z)) + E_{4}(z)^{3}E_{6}(z))

+ χ_{12}(ℓ)ℓ(1 + ℓ) E_{4}(z)^{2}E_{6}(z) Δ(z)^{s_{ℓ}} (mod 65520)

≡ − ∑_{n=0}^{s_{ℓ}}(24n + 1)c_{n,ℓ} E_{4}(z)^{3n+2} E_{6}(z) Δ(z)^{s_{ℓ}−n}

+ ∑_{n=0}^{s_{ℓ}} 720 · 24nc_{n,ℓ} E_{4}(z)^{3n−1} E_{6}(z) Δ(z)^{s_{ℓ}−n+1} + χ_{12}(ℓ)ℓ(1 + ℓ) E_{4}(z)^{2}E_{6}(z) Δ(z)^{s_{ℓ}} (mod 65520)

≡ (720 c_{1,ℓ} − c_{0,ℓ} + χ_{12}(ℓ)ℓ(1 + ℓ)) E_{4}(z)^{2} E_{6}(z) Δ(z)^{s_{ℓ}}

+ ∑_{n=1}^{s_{ℓ}−1} (720 · 24(n + 1)c_{n+1,ℓ} − (24n + 1)c_{n,ℓ}) E_{4}(z)^{3n+2} E_{6}(z) Δ(z)^{s_{ℓ}−n}

− (24s_{ℓ} + 1)c_{s_{ℓ}}E_{4}(z)^{3s_{ℓ}+2} E_{6}(z) (mod 65520).

We define

(2.19) A_{ℓ}(z) := ∑_{n=−s_{ℓ}}^{∞} (a(ℓ^{2}n − s_{ℓ}) + χ_{12}(ℓ) ((\frac{1 − 24n}{ℓ}) − 1 − ℓ) a(n) + ℓ a (\frac{n + s_{ℓ}}{ℓ^{2}} )) q^{n− \frac{1}{24}}

and (2.20)

S_{ℓ}(z) := ∑_{n=1}^{∞} (spt(ℓ^{2}n − s_{ℓ}) + χ_{12}(ℓ) ((\frac{1 − 24n}{ℓ}) − 1 − ℓ) spt(n) + ℓ spt (\frac{n + s_{ℓ}}{ℓ^{2}} )) q^{n− \frac{1}{24}} ,

so that

(2.21)  A_{ℓ}(z) = 12 S_{ℓ}(z) + Ξ_{ℓ}(z) = M_{ℓ}(z/24).

By Theorem 1.1 and equation (1.10) we see that the function

(2.22)  ℓA_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}} ∈ M_{\frac{1}{2} (ℓ^{2}+3)}(Γ(1)),

the space of entire modular forms of weight \frac{1}{2}(ℓ^{2} + 3) on Γ(1). Since \frac{1}{2}(ℓ^{2} + 3) = 2 + 12s_{ℓ} the set

(2.23)  {E_{4}(z)^{3n−1} E_{6}(z) Δ(z)^{s_{ℓ}−n} : 1 ≤ n ≤ s_{ℓ}}

is a basis. Hence there are integers b_{n,ℓ} (1 ≤ n ≤ s_{ℓ}) such that

(2.24)  A_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}} = ∑_{n=1}^{s_{ℓ}} b_{n,ℓ}E_{4}(z)^{3n−1} E_{6}(z) Δ(z)^{s_{ℓ}−n}.

Using (2.17) we find that

(2.25)  A_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}} ≡ −720b_{1,ℓ} E_{4}(z)^{2} E_{6}(z) Δ(z)^{s_{ℓ}}

+ ∑_{n=1}^{s_{ℓ}−1}(b_{n,ℓ} − 720b_{n+1,ℓ}) E_{4}(z)^{3n+2} E_{6}(z) Δ(z)^{s_{ℓ}−n}

+ b_{s_{ℓ},ℓ}E_{4}(z)^{3s_{ℓ}+2} E_{6}(z) (mod 65520).

By (2.18), (2.21) and (2.24) we deduce that there are integers a_{n,ℓ} (0 ≤ n ≤ s_{ℓ}) such that

(2.26)  12 ℓ S_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}} ≡ ∑_{n=0}^{s_{ℓ}} a_{n,ℓ} E_{4}(z)^{3n+2} E_{6}(z) Δ(z)^{s_{ℓ}−n} (mod 65520).

It follows that

(2.27)  12 ℓ S_{ℓ}(z) ≡ 0 (mod 65520),

since

(2.28)  ord_{i∞} (12 ℓ S_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}}) = s_{ℓ} + 1,
0 ≤ ord_{i∞} (E_{4}(z)^{3n+2} E_{6}(z) Δ(z)^{s_{ℓ}−n}) ≤ s_{ℓ},
E_{4}(z)^{3n+2} E_{6}(z) Δ(z)^{s_{ℓ}−n} = q^{s_{ℓ}−n} + ⋯,

for 0 ≤ n ≤ s_{ℓ} and all functions have integral coefficients. Since 65550 = 2^{4} · 3^{2} · 5 · 7 · 13, the congruence (2.27) implies Part (ii) of Theorem 1.3. To prove Part (i) we need to work a little harder. We note that the congruence (2.27) does imply

(2.29)  S_{ℓ}(z) ≡ 0 (mod 12).

We need to show this congruence actually holds modulo 72.

First we show the congruence holds modulo 8 by studying Ξ_{ℓ}(z) modulo 32. We need the congruences,

(2.30)  E_{2}(z) ≡ E_{4}(z) E_{6}(z) + 16Δ(z) (mod 32), and  E_{4}(z)^{2} ≡ 1 (mod 32),

which are routine to prove. We proceed as in the proof of (2.18) to find that

ℓ Ξ_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}} (2.31)

≡ (χ_{12}(ℓ)ℓ(1 + ℓ) − c_{0,ℓ} − 16c_{1,ℓ}) E_{2}(z) Δ(z)^{s_{ℓ}}

− ∑_{n=1}^{s_{ℓ}−1} ((24n + 1)c_{n,ℓ} + 16c_{n+1,ℓ}) E_{4}(z)^{3n−1} E_{6}(z) Δ(z)^{s_{ℓ}−n}

− (24s_{ℓ} + 1)c_{s_{ℓ}}E_{4}(z)^{3s_{ℓ}−1} E_{6}(z) (mod 32).

By (2.31), (2.21) and (2.24) we deduce that there are integers a′_{n,ℓ} (0 ≤ n ≤ s_{ℓ}) such that

(2.32) 12 ℓ S_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}} ≡ ∑_{n=1}^{s_{ℓ}} a′_{n,ℓ} E_{4}(z)^{3n−1} E_{6}(z) Δ(z)^{s_{ℓ}−n} + a′_{0,ℓ} E_{2}(z) Δ(z)^{s_{ℓ}} (mod 32).

Arguing as before, it follows that

(2.33)  12S_{ℓ}(z) ≡ 0 (mod 32), and  S_{ℓ}(z) ≡ 0 (mod 8).

To complete the proof, we need to study Ξ_{ℓ}(z) modulo 27. We need the congruences,

(2.34)  E_{2}(z) ≡ E_{4}(z)^{5} + 18Δ(z) (mod 27) and  E_{6}(z) ≡ E_{4}(z)^{6} (mod 27),

which are routine to prove. We proceed as in the proof of (2.18) and (2.31) to find that

ℓ Ξ_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}} (2.35)

≡ (χ_{12}(ℓ)ℓ(1 + ℓ) − c_{0,ℓ} − 18c_{1,ℓ}) E_{2}(z) Δ(z)^{s_{ℓ}}

− ∑_{n=1}^{s_{ℓ}−1} ((24n + 1)c_{n,ℓ} + 18c_{n+1,ℓ}) E_{4}(z)^{3n−1} E_{6}(z) Δ(z)^{s_{ℓ}−n}

− (24s_{ℓ} + 1)c_{s_{ℓ}}E_{4}(z)^{3s_{ℓ}−1} E_{6}(z) (mod 27).

By (2.35), (2.21) and (2.24) we deduce that there are integers a′′_{n,ℓ} (0 ≤ n ≤ s_{ℓ}) such that

(2.36) 12 ℓ S_{ℓ}(z) η(z) Δ(z)^{s_{ℓ}} ≡ ∑_{n=1}^{s_{ℓ}} a′′_{n,ℓ} E_{4}(z)^{3n−1} E_{6}(z) Δ(z)^{s_{ℓ}−n} + a′′_{0,ℓ} E_{2}(z) Δ(z)^{s_{ℓ}} (mod 27).

Arguing as before, it follows that

(2.37)  12S_{ℓ}(z) ≡ 0 (mod 27), and  S_{ℓ}(z) ≡ 0 (mod 9).

The congruences (2.33) and (2.37) give (1.15) and this completes the proof of Theorem 1.3.

3. Proof of Theorem 1.6

In this section we prove Theorem 1.6. Atkin [4] proved Theorem 1.5 by constructing certain special modular functions on Γ_{0}(t) and Γ_{0}(t^{2}) for t = 5, 7 and 13. We attack the problem by extending Atkin’s results to the corresponding weight 2 case.

Let GL_{2}^{+}(ℝ) denote the group of all real 2 × 2 matrices with positive determinant. GL_{2}^{+}(ℝ) acts on the complex upper half plane H by linear fractional transformations. We define the slash operator for modular forms of integer weight. Let k ∈ ℤ. For a function f : H ⟶ ℂ and

L = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ GL_{2}^{+}(ℝ) we define

(3.1)  f(z) |_{k} L = f |_{k} L = f | L = (det L)^{\frac{k}{2}} (cz + d)^{−k}f(Lz).

Let Γ′ ⊂ Γ(1) (a subgroup of finite index). We say f(z) is a weakly holomorphic modular form of weight k on Γ′ if f(z) is holomorphic on the upper half plane H, f(z) |_{k} L = f(z) for all L in Γ′, and f(z) has at most polar singularities in the local variables at the cusps of the fundamental region of Γ′. We say f(z) is a weakly holomorphic modular function if it is a weakly holomorphic modular form of weight 0. We say f(z) is an entire modular form of weight k on Γ′ if it is a weakly holomorphic modular form that is holomorphic at the cusps of the fundamental region of Γ′. We denote the space of entire modular forms of weight k on Γ′ by M_{k}(Γ′).

Suppose that t ≥ 5 is prime. We need

W_{t} = W = (\begin{matrix}0 & −1 \\ t & 0 \end{matrix}) , R = (\begin{matrix}1 & 0 \\ −1 & 1 \end{matrix}) , V_{a} = (\begin{matrix}a & λ \\ t & a′ \end{matrix}) , B_{t} = (\begin{matrix}t & 0 \\ 0 & 1 \end{matrix}) ,

T_{b,t} = (\begin{matrix}1 & b \\ 0 & 1 \end{matrix}) , Q_{b,t} = (\begin{matrix}1/t & b/t \\ 0 & 1 \end{matrix}) ,

where for 1 ≤ a ≤ t − 1, a′ is uniquely defined by 1 ≤ a′ ≤ t − 1, and a′a − λt = 1. We have

(3.2)  B_{t} R^{at} = W_{t} V_{a} T_{−a′/t}
(3.3)  R^{at} W_{t} = W_{t^{2}} Q_{a,t}.

We define

(3.4)  Φ_{t}(z) = Φ(z) := \frac{η(z)}{η(t^{2}z)}.

Then Φ_{t}(z) is a modular function of Γ_{0}(t),

(3.5)  Φ_{t}(z) | W_{t^{2}} = t Φ_{t}(z)^{−1}  ([4, (24)]),

and

(3.6)  Φ_{t}(z) | R^{at} = \sqrt{t} e^{πi(t−1)/4} e^{−πia′t/12} (\frac{a′}{t})\frac{η(z)}{η(z − a′/t)}  ([4, (25)]).

Although E_{2}(z) is not a modular form, it well-known that

(3.7)  E_{2,t}(z) := \frac{1}{t − 1} (t E_{2}(tz) − E_{2}(z)) ,

is an entire modular form of weight 2 on Γ_{0}(t) and

(3.8)  E_{2,t}(z) | W_{t} = −E_{2,t}(z).

Proposition 3.1. Suppose t ≥ 5 is prime, K(z) is a weakly holomorphic modular function on Γ_{0}(t), and

(3.9)  S(z) = E_{2,t}(tz) K^{∗}(tz)\frac{η(z)}{η(t^{2}z)} − χ_{12}(t) η(z) ∑_{n=m}^{∞} (\frac{1 − 24n}{t}) β_{t}(n)q^{n− \frac{1}{24}} ,

where

(3.10)  E_{2,t}(z) \frac{K(z)}{η(z)} = ∑_{n=m}^{∞} β_{t}(n)q^{n− \frac{1}{24}} ,

and

(3.11)  K^{∗}(z) = K(z) | W_{t}.

Then S(z) is a weakly holomorphic modular form of weight 2 on Γ_{0}(t).

Proof. Suppose t ≥ 5 is prime and K(z), K^{∗}(z), S(z) are defined as in the statement of the proposition. The function

(3.12)  H(z) := E_{2,t}(tz) Φ_{t}(z) K^{∗}(tz)

is a weakly holomorphic modular form of weight 2 on Γ_{0}(t^{2}). As in [4, Lemma1] the function

(3.13)  S_{1}(z) := ∑_{a=0}^{t−1} H(z) | R^{at}

is a weakly holomorphic modular form of weight 2 on Γ_{0}(t). Utilizing (3.2), (3.6), (3.8) and the evaluation of a quadratic Gauss sum [5, (1.7)] we find that (3.14)

S_{1}(z) = E_{2,t}(tz) K^{∗}(tz)\frac{η(z)}{η(t^{2}z)} − \frac{1}{\sqrt{t}} e^{πi(t−1)/4} η(z) ∑_{a′=1}^{t−1} e^{−πia′t/12}(\frac{a′}{t}) E_{2,t}(z − a′/t) \frac{K(z − a′/t)}{η(z − a′/t)}

Here we have also used the fact that

(3.15)  E_{2,t}(z) | R^{at} = −\frac{1}{t} E_{2,t}(z − a′/t),

where a a′ ≡ 1 (mod t). Hence

S_{1}(z) = S(z).

This gives the result. □

We illustrate Proposition 3.1 with two examples:

(3.16)  S(z) = E_{2,5}(z) (\frac{η(z)}{η(5z)})^{6}  (K(z) = 1 and t = 5)

and

(3.17)  S(z) = E_{2,7}(z) ((\frac{η(z)}{η(7z)})^{8} + 3 (\frac{η(z)}{η(7z)})^{4})  (K(z) = 1 and t = 7).

Corollary 3.2. Suppose t ≥ 5 is prime and S(z), K(z) and the sequence β_{t}(n) are defined as in Proposition 3.1. Then

(3.18)  S(z) | W_{t} = −η(tz) ∑_{tn−s_{t}≥m} β_{t}(tn − s_{t})q^{n− \frac{t}{24}} .

Proof. The result follows easily from (3.3), (3.5) and (3.8). □

We illustrate the corollary by applying W to both sides of the equations (3.16)–(3.17):

∑_{n=1}^{∞} β_{5}(5n − 1)q^{n− \frac{5}{24}} = 5^{3} \frac{E_{2,5}(z)}{η(5z)} (\frac{η(5z)}{η(z)})^{6}  (K(z) = 1 and t = 5) (3.19)

and

∑_{n=1}^{∞} β_{7}(7n − 2)q^{n− \frac{7}{24}} = 7^{2} \frac{E_{2,7}(z)}{η(7z)} (3 (\frac{η(7z)}{η(z)})^{4} + 7^{2} (\frac{η(7z)}{η(z)})^{8})  (K(z) = 1 and t = 7). (3.20)

For t and K(z) as in Proposition 3.1 we define

(3.21)  Ψ_{t,K}(z) = Ψ_{t}(z) = E_{2,t}(tz) \frac{K^{∗}(tz)}{η(t^{2}z)} − χ_{12}(t) ∑_{n=m}^{∞} (\frac{1 − 24m}{t}) β_{t}(n)q^{n− \frac{1}{24}}

− ∑_{t^{2}n−s_{t}≥m} β_{t}(t^{2}n − s_{t})q^{n− \frac{1}{24}} ,

where K^{∗}(z) and the sequence β_{t}(n) is defined in (3.10)–(3.11). We have the following analogue of 2.1. Corollary 3.3. The function Ψ_{t,K}(z) η(z) is a weakly holomorphic modular form of weight 2 on the full modular group Γ(1).

Proof. Let S(z) be defined as in (3.9), so that S(z) is a weakly holomorphic modular form of weight 2 on Γ_{0}(t). By [14, Lemma 7], the function

(3.22)  S(z) + S(z) | W_{t} | U

is a modular form of weight 2 on Γ(1). Here U = U_{t} is the Atkin operator

(3.23)  g(z) | U_{t} = \frac{1}{t} ∑_{a=0}^{t−1} g (\frac{z + a}{t}) .

The result then follows from applying the U-operator to equation (3.18). □

We illustrate the K(z) = 1 case of Corollary 3.3 with two examples:

(3.24)  Ψ_{5}(z) = \frac{E_{4}(z)^{2} E_{6}(z)}{η(z)^{25}}

and

(3.25)  Ψ_{7}(z) = \frac{1}{η(z)^{49}} (E_{4}(z)^{5} E_{6}(z) − 745E_{4}(z)^{2} E_{6}(z) Δ(z)) .

We need a weight 2 analogue of [4, Lemma 3]. For t = 5, 7 or 13 the genus of Γ_{0}(t) is zero, and a Hauptmodul is

(3.26)  G_{t}(z) := (\frac{η(z)}{η(tz)})^{24/(t−1)} .

This function satisfies

(3.27)  G_{t} (\frac{−1}{tz}) = t^{12/(t−1)}G_{t}(z)^{−1}.

Proposition 3.4. Suppose t = 5, 7 or 13, and let m be any negative integer such that 24m ≢ 1 (mod t). Suppose constants k_{j} (1 ≤ j ≤ −m) are chosen so that

(3.28)  β_{t}(n) = 0,  for m + 1 ≤ n ≤ −1,

where

(3.29)  K(z) = G_{t}(z)^{−m} + ∑_{k=1}^{−m−1} k_{j}G_{t}(z)^{j}

and

(3.30)  ∑_{n=m}^{∞} β_{t}(n)q^{n− \frac{1}{24}} = E_{2,t}(z)\frac{K(z)}{η(z)}.

Then

(3.31)  β_{t}(n) = 0,  for (\frac{1 − 24n}{t}) = −(\frac{1 − 24m}{t}).

Proof. Suppose t = 5, 7 or 13, and m is a negative integer such that 24m ≢ 1 (mod t). Suppose K(z) is chosen so that (3.28) holds. Let S(z) be defined as in (3.9), and define

(3.32)  B(z) := S(z) + χ_{12}(t) (\frac{1 − 24m}{t}) E_{2,t}(z) K(z),

so that

(3.33)  B(z) | W_{t} = S^{∗}(z) − χ_{12}(t) (\frac{1 − 24m}{t}) E_{2,t}(z) K^{∗}(z),

where

(3.34)  S^{∗}(z) = S(z) | W_{t}.

Since 24m ≢ 1 (mod t), we see that

(3.35)  ord_{0}(S(z)) = ord_{i∞}(S^{∗}(z)) > 0.

From (3.27) and (3.29) we see that

(3.36)  ord_{0}(K(z)) = ord_{i∞}(K^{∗}(z)) > 0

and hence

(3.37)  ord_{0}(B(z)) > 0.

Now

(3.38)  ord_{i∞} (E_{2,t}(tz) K^{∗}(tz)\frac{η(z)}{η(t^{2}z)}) ≥ t − \frac{1}{24}(t^{2} − 1) > 0,

for t = 5, 7, 13. By construction the coefficient of q^{m} in B(z) is zero and so (3.28), (3.38) imply that

(3.39)  ord_{i∞}(B(z)) ≥ 0.

Therefore B(z) is an entire modular form of weight 2 and hence a multiple of E_{2,t}(z) since there are no nontrivial cusp forms of weight 2 on Γ_{0}(t) for t = 5, 7 or 13 by [7]. This implies that B(z) is identically zero by (3.37). Hence

(3.40) \frac{B(z)}{E(q)} = q^{−s_{t}}E_{2,t}(tz) K^{∗}(tz)\frac{1}{E(q^{t^{2}})} − χ_{12}(t) ∑_{n=m}^{∞} ((\frac{1 − 24n}{t}) − (\frac{1 − 24m}{t})) β_{t}(n)q^{n} = 0.

Since −24s_{t} − 1 ≡ 0 (mod t), this implies that β_{t}(n) = 0 whenever (\frac{1−24n}{t}) = −(\frac{1−24m}{t}). □

We illustrate Proposition 3.4 with two examples:

∑_{n=−2}^{∞} β_{5}(n)q^{n} = \frac{E_{2,5}(z)}{E(q)} (G_{5}(z)^{2} + 5 G_{5}(z)) (3.41)

= q^{−2} + 1 − 379 q^{3} + 625 q^{4} + 869 q^{5} − 20125 q^{8} + 23125 q^{9} + 25636 q^{10} − 329236 q^{13} + · · · .

In this example, t = 5 and m = −2, and we see that β_{5}(n) = 0 for n ≡ 1, 2 (mod 5). In our second example, t = 7 and m = −1.

∑_{n=−1}^{∞} β_{7}(n)q^{n} = \frac{E_{2,7}(z)}{E(q)} G_{7}(z) (3.42)

= q^{−1} + 1 − 15 q^{2} + 49 q^{5} − 24 q^{6} + 88 q^{7} − 311 q^{9} + 392 q^{12} − 182 q^{13} + 811 q^{14} − 1886 q^{16} + · · · . (3.43)

In this example we see that β_{7}(n) = 0 for n ≡ 1, 3, 4 (mod 7).

The function

\frac{E_{4}(z)^{2} E_{6}(z)}{Δ(z)} = \frac{E_{6}(z)}{E_{4}(z)} j(z) (3.44)

= q^{−1} − 196884 q − 42987520 q^{2} − 2592899910 q^{3} − 80983425024 q^{4} − 1666013203000 q^{5} + ⋯

is a modular form of weight 2 on Γ(1). As a modular form on Γ_{0}(t) it has a simple pole at i∞ and a pole of order t at z = 0. When t = 5, 7 or 13, it is straightforward to show that there are integers a_{j,t} (−1 ≤ j ≤ t) such that

(3.45)  \frac{E_{6}(z)}{E_{4}(z)} j(z) = E_{2,t}(z) ∑_{j=−1}^{t} a_{j,t}G_{t}(z)^{j}.

For example,

\frac{E_{6}(z)}{E_{4}(z)} j(z) = E_{2,5}(z) (G_{5}(z) − 3^{2} · 5^{5} · 7 G_{5}(z)^{−1} − ·2^{3} · 5^{8} · 13 G_{5}(z)^{−2} − ·3^{3} · 5^{10} · 7 G_{5}(z)^{−3} (3.46)
−3 · 2^{3} · 5^{13} G_{5}(z)^{−4} − 5^{16} G_{5}(z)^{−5}) .

Reducing (3.45) mod t^{c} we obtain a weight 2 analogue of [4, Lemma 4]. Lemma 3.5. We have

(3.47)  \frac{E_{6}(z)}{E_{4}(z)} j(z) ≡ E_{2,5}(z) (G_{5}(z) + 2 · 31 · 5^{5} G_{5}(z)^{−1}) (mod 5^{8}),
(3.48)  \frac{E_{6}(z)}{E_{4}(z)} j(z) ≡ E_{2,7}(z) G_{7}(z) (mod 7^{4}),
(3.49)  \frac{E_{6}(z)}{E_{4}(z)} j(z) ≡ E_{2,13}(z) G_{13}(z) (mod 13^{2}).

We also need [4, Lemma 4]. Lemma 3.6 (Atkin [4]). [Atkin [4]] We have

(3.50)  j(z) ≡ G_{5}(z) + 750 + 3^{2} · 7 · 5^{5} G_{5}(z)^{−1} (mod 5^{8}),
(3.51)  j(z) ≡ G_{7}(z) + 748 (mod 7^{4}),
(3.52)  j(z) ≡ G_{13}(z) + 70 (mod 13^{2}).

Remark 3.7. In equation (3.50) we have corrected a misprint in [4, Lemma 4].

To handle the (t, c) = (5, 6) case of Theorem 1.6 we will need Lemma 3.8.

(3.53)  \frac{E_{6}(z)}{E_{4}(z)} j(z) ≡ E_{2,5}(z) (G_{5}(z) + 2 · 5^{5} G_{5}(z)^{−1}) (mod 5^{6}),
(3.54)  \frac{E_{6}(z)}{E_{4}(z)} j(z)^{2} ≡ E_{2,5}(z) (2 · 3 · 5^{3} G_{5}(z) + G_{5}(z)^{2}) (mod 5^{6}),

and

(3.55)  \frac{E_{6}(z)}{E_{4}(z)} j(z)^{a} ≡ E_{2,5}(z) (ε_{1,a} G_{5}(z)^{a−2} + ε_{2,a} G_{5}(z)^{a−1} + G_{5}(z)^{a}) (mod 5^{6}),

for a ≥ 3, where ε_{1,a}, ε_{2,a} are integers satisfying

(3.56)  ε_{1,a} ≡ 0 (mod 5^{5}) and  ε_{2,a} ≡ 0 (mod 5^{3}).

Proof. The result can be proved from Lemmas 3.5 and 3.6, some calculation and an easy induction argument. □

We need bases for M_{2+12s_{ℓ}}(Γ_{0}(t)) for t = 5, 7, 13. The following result follows from [7] and by checking the modular forms involved are holomorphic at the cusps i∞ and 0. Lemma 3.9. Suppose t = 5, 7 or 13 and ℓ > 3 is prime. Then

(3.57)  dim M_{2+12s_{ℓ}}(Γ_{0}(t)) = 1 + (1 + t) s_{ℓ},

and the set

(3.58)  {E_{2,t}(z) Δ(z)^{s_{ℓ}} G_{t}(z)^{a} : −ts_{ℓ} ≤ a ≤ s_{ℓ}}

is a basis for M_{2+12s_{ℓ}}(Γ_{0}(t)).

We are now ready to prove Theorem 1.6. We have two cases:

Case 1. In the first case we assume that (t, c) = (5, 5), (7, 4) or (13, 2). Suppose ℓ > 3 is prime and ℓ ≠ t. By (2.24) we have

(3.59)  A_{ℓ}(z) = ∑_{n=1}^{s_{ℓ}} b_{n,ℓ} \frac{E_{6}(z)}{E_{4}(z)} \frac{j(z)^{n}}{η(z)} .

By Theorem 1.1, equation (2.21) and Lemma 3.9 we have

(3.60)  A_{ℓ}(z) = ∑_{n=−ts_{ℓ}}^{s_{ℓ}} d_{n,ℓ} \frac{E_{2,t}(z)}{η(z)} G_{t}(z)^{n},

for some integers d_{n,ℓ} (−ts_{ℓ} ≤ n ≤ s_{ℓ}). Now let

(3.61)  K(z) = ∑_{n=1}^{s_{ℓ}} d_{n,ℓ}G_{t}(z)^{n}.

By using Lemmas 3.5 and 3.6 to reduce equation (3.59) modulo t^{c} and comparing the result with (3.60) we deduce that

(3.62)  A_{ℓ}(z) ≡ E_{2,t}(z) \frac{K(z)}{η(z)} (mod t^{c})

and that

(3.63)  d_{n,ℓ} ≡ 0 (mod t^{c}),

for −ts_{ℓ} ≤ n ≤ 0. By examining (3.60) we see that

(3.64)  A_{ℓ}(z) = E_{2,t}(z) \frac{K(z)}{η(z)} + O(q^{− \frac{1}{24}} ).

So if we let

(3.65)  E_{2,t}(z) \frac{K(z)}{η(z)} = ∑_{n=−s_{ℓ}}^{∞} β_{t,ℓ}(n)q^{n− \frac{1}{24}} , ,

then (3.62) may be rewritten as

(3.66)  a(ℓ^{2}n − s_{ℓ}) + χ_{12}(ℓ) ((\frac{1 − 24n}{ℓ}) − 1 − ℓ) a(n) + ℓ a (\frac{n + s_{ℓ}}{ℓ^{2}} ) ≡ β_{t,ℓ}(n) (mod t^{c})

and from (3.64) we have

(3.67)  a(ℓ^{2}n − s_{ℓ}) + χ_{12}(ℓ) ((\frac{1 − 24n}{ℓ}) − 1 − ℓ) a(n) + ℓ a (\frac{n + s_{ℓ}}{ℓ^{2}} ) = β_{t,ℓ}(n)

for −s_{ℓ} ≤ n ≤ −1. Equation (3.67) implies that

(3.68)  β_{t,ℓ}(−s_{ℓ}) = −ℓ

and

(3.69)  β_{t,ℓ}(n) = 0,

for −s_{ℓ} ≤ n ≤ −1. We can now apply Proposition 3.4 with m = −s_{ℓ} since 1 − 24m = ℓ^{2} and t ≠ ℓ. Hence

(3.70)  β_{t,ℓ}(n) = 0, provided (\frac{1 − 24n}{t}) = −1.

This gives Theorem 1.6 when (t, c) = (5, 5), (7, 4) or (13, 2) by (3.66).

Case 2. We consider the remaining case (t, c) = (5, 6) and assume ℓ > 5 is prime. We proceed as in Case 1. This time when we use Lemma 3.8 to reduce (2.24) modulo 5^{6} we see that the only extra term occurs when n = 1. We find that with K(z) as before we have

(3.71)  A_{ℓ}(z) ≡ E_{2,t}(z) \frac{K(z)}{η(z)} + b_{1,5} · 2 · 5^{5} \frac{E_{2,5}(z)}{η(z)} G_{5}(z)^{−1} (mod 5^{6}).

All that remains is to show that

(3.72)  b_{1,5} ≡ 0 (mod 5),

since then (3.62) actually holds modulo 5^{6} and the rest of the proof proceeds as in Case 1. Since E_{4}(z) ≡ 1 (mod 5) we may reduce (2.18) modulo 5 to obtain

(3.73)  ℓ Ξ_{ℓ} ≡ (χ_{12}(ℓ)ℓ(1 + ℓ) − c_{0,ℓ}) \frac{E_{6}(z)}{E_{4}(z)} \frac{1}{η(z)} − ∑_{n=1}^{s_{ℓ}}(24n + 1)c_{n,ℓ} \frac{E_{6}(z)}{E_{4}(z)} \frac{j(z)^{n}}{η(z)} (mod 5).

But

(3.74)  S_{ℓ}(z) ≡ 0 (mod 5),

by Theorem 1.3 (ii). Hence

(3.75) ℓA_{ℓ}(z) ≡ (χ_{12}(ℓ)ℓ(1 + ℓ) − c_{0,ℓ}) \frac{E_{6}(z)}{E_{4}(z)} \frac{1}{η(z)} − ∑_{n=1}^{s_{ℓ}}(24n + 1)c_{n,ℓ} \frac{E_{6}(z)}{E_{4}(z)} \frac{j(z)^{n}}{η(z)} (mod 5)

and we see that b_{1,5} the coefficient of \frac{E_{6}(z)}{E_{4}(z)} \frac{j(z)}{η(z)} is divisible by 5 as required. This completes the proof of Theorem 1.6.

We close by illustrating Theorem 1.6 when t = 5 and ℓ = 7. In this case the theorem predicts that

a(49n − 2) − (\frac{1 − 24n}{7}) a(n) + 7a (\frac{n + 2}{49}) ≡ −8 a(n) (mod 5^{6}),

when n ≡ 1, 2 (mod 5). When n = 1 this says

149077845 ≡ −280 (mod 5^{6}),

which is easy to check. Acknowledgements I would like to thank Ken Ono for sending me preprints of his recent work [12], [13].

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11. F. G. Garvan, Congruences for Andrews’ spt-function modulo powers of 5, 7 and 13, in preparation.

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URL: http://mathcs.emory.edu/~ono/publications-cv/pdfs/132.pdf

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14. J.-P. Serre, Formes modulaires et fonctions zêta p-adiques in “Modular functions of one variable, III”, (Proc. Internat. Summer School, Univ. Antwerp, 1972), pp. 191–268, Lecture Notes in Math., Vol. 350, Springer, Berlin, 1973.

URL: http://dx.doi.org/10.1007/978-3-540-37802-0_4

Department of Mathematics, University of Florida, Gainesville, Florida 32611-8105

E-mail address: fgarvan@ufl.edu