CONGRUENCES FOR ANDREWS’ SPT-FUNCTION MODULO POWERS OF 5, 7 AND 13

F. G. GARVAN

Dedicated to my friend and mentor Michael D. Hirschhorn on the occasion of his 63rd birthday

Abstract. Congruences are found modulo powers of 5, 7 and 13 for Andrews’ smallest parts partition function spt(n). These congruences are reminiscent of Ramanujan’s partition congruences modulo powers of 5, 7 and 11. Recently, Ono proved explicit Ramanujan-type congruences for spt(n) modulo ℓ for all primes ℓ ≥ 5 which were conjectured earlier by the author. We extend Ono’s method to handle the powers of 5, 7 and 13 congruences. We need the theory of weak Maass forms as well as certain classical modular equations for the Dedekind eta-function.

1. Introduction

Andrews [3] 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. He also proved some surprising congruences mod 5, 7 and 13. Namely, he showed that

(1.1)  spt(n) = np(n) − \frac{1}{2}N_{2}(n),

where N_{2}(n) is the second rank moment function [4] and p(n) is the number of partitions of n, and he proved that

(1.2)  spt(5n + 4) ≡ 0 (mod 5),
(1.3)  spt(7n + 5) ≡ 0 (mod 7),
(1.4)  spt(13n + 6) ≡ 0 (mod 13).

Bringmann [9] studied analytic, arithmetic and asymptotic 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 [16], Folsom and Ono [14] and Ono [22]. In particular, Ono [22] proved that if (\frac{1−24n}{ℓ}) = 1 then

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

Date: December 14, 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 September 20, 2010. Submitted to Trans. Math. Soc. November 20, 2010. Accepted by Trans. Math. Soc. November 27, 2010. Final version completed December 14, 2010.

for any prime ℓ ≥ 5. This amazing result was originally conjectured by the author^{1}. Earlier special cases were observed by Tina Garrett [17] and her students.

We prove some suprising congruences for spt(n) modulo powers of 5, 7 and 13. For a, b, c ≥ 3,

(1.6)  spt(5^{a}n + δ_{a}) + 5 spt(5^{a−2}n + δ_{a−2}) ≡ 0 (mod 5^{2a−3}),
(1.7)  spt(7^{b}n + λ_{b}) + 7 spt(7^{b−2}n + λ_{b−2}) ≡ 0 (mod 7^{⌊ \frac{1}{2} (3b−2)⌋}),
(1.8)  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. This together with (1.2)–(1.4) implies that

(1.9)  spt(5^{a}n + δ_{a}) ≡ 0 (mod 5^{⌊ \frac{a+1}{2} ⌋}),
(1.10)  spt(7^{b}n + λ_{b}) ≡ 0 (mod 7^{⌊ \frac{b+1}{2} ⌋}),
(1.11)  spt(13^{c}n + γ_{c}) ≡ 0 (mod 13^{⌊ \frac{c+1}{2} ⌋}),

for a, b, c ≥ 1. These congruences are reminiscent of Ramanujan’s partition congruences for powers of 5, 7 and 11:

(1.12)  p(5^{a}n + δ_{a}) ≡ 0 (mod 5^{a}),
(1.13)  p(7^{b}n + λ_{b}) ≡ 0 (mod 7^{⌊ \frac{b+2}{2} ⌋}),
(1.14)  p(11^{c}n + φ_{c}) ≡ 0 (mod 11^{c}),

for all a, b, c ≥ 1. Here φ_{c} is the reciprocal of 24 mod 11^{c}. The congruences mod powers of 5 and 7 were proved by Watson [25], although many of the details had been worked out earlier by Ramanujan in an unpublished manuscript. The powers of 11 congruence was proved by Atkin [7].

Following Ono [22], we define

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

for n ≥ 0, and define

(1.16)  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 [9] showed that α(24z) is the holomorphic part of a weight \frac{3}{2} weak Maass form. Using this observation and the idea of using the weight \frac{3}{2} Hecke operator T (ℓ^{2}) to annihilate the nonholomorphic part enabled Ono [22] to prove the general congruence (1.5). We use a similar idea. Instead of a Hecke operator we use Atkin’s U(ℓ) operator to annihilate the nonholomorphic part.

We show that

(1.17)  a(5^{a}n + δ_{a}) + 5 a(5^{a−2}n + δ_{a−2}) ≡ 0 (mod 5^{⌊ \frac{1}{2} (5a−7)⌋}),
(1.18)  a(7^{b}n + λ_{b}) + 7 a(7^{b−2}n + λ_{b−2}) ≡ 0 (mod 7^{⌊ \frac{1}{2} (3b−2)⌋}),
(1.19)  a(13^{c}n + γ_{c}) − 13 a(13^{c−2}n + γ_{c−2}) ≡ 0 (mod 13^{c−1}),

1The congruence (1.5) was first conjectured by the author in a Colloquium given at the University of Newcastle, Australia on July 17, 2008.

for all a, b, c ≥ 3. We note that (1.17) is a stronger congruence than (1.6). The congruences (1.6)–(1.7) follow from (1.17)–(1.18) and Ramanujan’s partition congruences for powers of 5 and 7 that were first proved by Watson [25]. The congruence (1.8) follows easily from (1.19).

Let ℓ ≥ 5 be prime. In Section 2 we use results of Bringmann [9] to show how Atkin’s U(ℓ) operator can be used to annihilate the nonholomorphic part of the weight \frac{3}{2} weak Maass form that corresponds to the function α(24z), and prove that the function

(1.20)  α_{ℓ}(z) := ∑_{n=0}^{∞} (a(ℓn − \frac{1}{24}(ℓ^{2} − 1)) − χ_{12}(ℓ) ℓ a (\frac{n}{ℓ})) q^{n− \frac{ℓ}{24}}

is a weakly holomorphic weight \frac{3}{2} modular form on Γ_{0}(ℓ). Here χ_{12} is the character given below in (2.2), and we note a(n) = 0 if n is not a nonnegative integer. We determine the multiplier of this form and exact information about the orders at cusps. See Theorem 2.2. This enables us to prove identities such as (1.21)

α_{5}(z) = ∑_{n=0}^{∞} (a(5n − 1) + 5 a (\frac{n}{5})) q^{n− \frac{5}{24}} = \frac{5}{4} \frac{(5E_{2}(5z) − E_{2}(z))}{η(5z)} (125\frac{η(5z)^{6}}{η(z)^{6}} − 1) ,

where E_{2}(z) is the usual quasimodular Eisenstein series of weight 2, and η(z) is the Dedekind eta-function. We then use Watson’s [25] and Atkin’s [8] method of modular equations to prove the congruences (1.17)–(1.19). These details are carried out in Section 3. In Section 4 we improve some results in [16] and [10] on spt(ℓn − \frac{1}{24}(ℓ^{2} − 1)) and N_{2}(ℓn − \frac{1}{24}(ℓ^{2} − 1)) modulo ℓ.

Since this paper was first written Ahlgren, Bringmann and Lovejoy [2] have generalized Ono’s congruence (1.5) to higher powers of ℓ. They have also obtained analogous results for other spt-like functions which were studied by Bringmann, Lovejoy and Osburn [11], [12]. We state their theorem for the spt-function. Suppose ℓ ≥ 5 is prime and m ≥ 1. Then Ahlgren, Bringmann and Lovejoy have shown the following two congruences.

(i) If (\frac{−23−24n}{ℓ}) = 1 then

(1.22)  spt(ℓ^{2m}n + d_{ℓ,2m}) ≡ 0 (mod ℓ^{m}).

(ii) If n ≥ 0 then

(1.23)  spt(ℓ^{2m+1}n + d_{ℓ,2m+1}) ≡ χ_{12}(ℓ) spt(ℓ^{2m−1}n + d_{ℓ,2m−1}) (mod ℓ^{m}).

Here d_{ℓ,a} is the least positive integer such that 24d_{ℓ,a} ≡ 1 (mod ℓ^{a}). The congruences (1.22)–(1.23) are truly amazing results. For the cases ℓ = 5, 7 and 13 these congruences follow from (1.9)–(1.11). Our congruences (1.6)–(1.8) do not follow from Ahlgren, Bringmann and Lovejoy’s results. Ahlgren, Bringmann and Lovejoy’s proof of (1.22)–(1.23) is an extension of Ono’s proof of (1.5). Their proofs utilize the Hecke operators T (ℓ^{2m}). Our results and proofs are different in that they involve the Atkin operators U(ℓ^{a}).

2. The Atkin operator U_{ℓ}^{∗}

In this section we prove that the function α_{ℓ}(z), which is defined in (1.20) is a weakly holomorphic weight \frac{3}{2} modular form on Γ_{0}(ℓ) when ℓ ≥ 5 is prime. The proof uses results of Bringmann [9] and the idea of using the Atkin operator U_{ℓ} to annihilate the nonholomorphic part of a certain weak Maass form.

Following Bringmann [9] and Ono [22] we define

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

where η(z) := q^{\frac{1}{24}} ∏_{n=1}^{∞}(1−q^{n}) is the Dedekind eta-function and α(z) is defined in (1.16). Then M(z) is a weight \frac{3}{2} harmonic Maass form on Γ_{0}(576) with Nebentypus χ_{12} where

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

Let

(2.3)  N (z) = −\frac{3i}{π\sqrt{2}} ∫_{−\overline{z}}^{i∞} \frac{η(24τ) dτ}{(−i(τ + z))^{\frac{3}{2}}} = \frac{3}{π\sqrt{2}} ∫_{y}^{∞} \frac{η(24(−x + it)) dt}{(y + t)^{3/2}},

where z = x + iy, y > 0, so that

(2.4)  M(z) = α(24z) + N (z).

We define

(2.5)  A(z) := M (\frac{z}{24}) .

The following theorem follows in a straightforward way from the work of Bringmann [9]. Theorem 2.1.

A (\frac{az + b}{cz + d}) = \frac{(cz + d)^{3/2}}{ν_{η}(A)} A(z),

where A = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ SL_{2}(ℤ), and ν_{η}(A) is the eta-multiplier.

Remark. When defining z^{3/2} we use the principal branch; i.e. for z = re^{iθ}, r > 0, −π ≤ θ < π, we take z^{3/2} = r^{3/2}e^{3iθ/2}.

Proof. We note that

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

where E_{2}(z) = 1 − 24 ∑_{n=1}^{∞} σ(n)q^{n} is a quasi-modular form that satisfies

(2.7)  E_{2} (\frac{az + b}{cz + d}) = (cz + d)^{2}E_{2}(z) − \frac{6iz}{π} c(cz + d).

Using (2.7) and Corollary 4.3 and Lemma 4.4 in [9],

M (−\frac{1}{z}) = \frac{−(−iz)^{3/2}}{48\sqrt{6}} M (\frac{z}{576}) ,

and hence

A (−\frac{1}{z}) = −(−iz)^{3/2}A(z) = e^{πi/4}z^{3/2}A(z).

Therefore,

A(Sz) = \frac{z^{3/2}}{ν_{η}(S)}A(z),

where S = (\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix}). From (1.16), (2.3) and (2.4)

M(z + \frac{1}{24}) = e^{−πi/12}M(z),
N (z + \frac{1}{24}) = e^{−πi/12}N (z),
A(z + 1) = e^{−πi/12}A(z),
A(T z) = \frac{1}{ν_{η}(T)}A(z),

where T = (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}). Since S, T generate SL_{2}(ℤ) the result follows. □

In what follows ℓ ≥ 5 is prime. We let d_{ℓ} denote the least nonnegative residue of the reciprocal of 24 mod ℓ so that 24d_{ℓ} ≡ 1 (mod ℓ). We define

(2.8)  r_{ℓ} := \frac{24d_{ℓ} − 1}{ℓ},  r_{ℓ}^{∗} := \frac{24d_{ℓ} + ℓ^{2} − 1}{24ℓ},  s_{ℓ} := \frac{(ℓ^{2} − 1)}{24}.

so that

(2.9)  α_{ℓ}(z) := ∑_{n=−r_{ℓ}^{∗}}^{∞} (a(ℓn + d_{ℓ}) − χ_{12}(ℓ) ℓ a (\frac{n + r_{ℓ}^{∗}}{ℓ} )) q^{n+ \frac{r_{ℓ}}{24}}

= ∑_{n=0}^{∞} (a(ℓn − s_{ℓ}) − χ_{12}(ℓ) ℓ a (\frac{n}{ℓ})) q^{n− \frac{ℓ}{24}} .

For a function G(z) we define the Atkin-type operator U_{ℓ}^{∗} by

(2.10)  U_{ℓ}^{∗}(G) := \frac{1}{ℓ} ∑_{k=0}^{ℓ−1} G (\frac{z + 24k}{ℓ} ) ,

so that

α_{ℓ}(z) = U_{ℓ}^{∗}(α) − χ_{12}(ℓ) ℓ α(ℓz).

The usual Atkin operator U_{ℓ} is defined by

(2.11)  U_{ℓ}(G) := \frac{1}{ℓ} ∑_{k=0}^{ℓ−1} G (\frac{z + k}{ℓ} ) .

We need U_{ℓ}^{∗} since α(z) has fractional powers of q, and we note that

U_{ℓ}^{∗}(G) = U_{ℓ}(G^{∗})(z/24),

where G^{∗}(z) = G(24z). For a congruence subgroup Γ we let M_{k}(Γ) denote the space of entire modular forms of weight k with respect to the group Γ, and we let M_{k}(Γ, χ) denote the space of entire modular forms of weight k and character χ with respect to the group Γ. Then we have Theorem 2.2. If ℓ ≥ 5 is prime, then

(2.12)  G_{ℓ}(z) := α_{ℓ}(z) \frac{η^{2ℓ}(z)}{η(ℓz)} ∈ M_{ℓ+1}(Γ_{0}(ℓ)).

In other words, the function G_{ℓ}(z) is an entire modular form of weight ℓ + 1 with respect to the group Γ_{0}(ℓ).

Proof. We assume ℓ ≥ 5 is prime. We divide the proof into four parts: (i) U_{ℓ}^{∗}(A) − ℓ χ_{12}(ℓ) A(ℓz) = α_{ℓ}(z) and G_{ℓ}(z) is holomorphic for ℑ(z) > 0.

(ii) G_{ℓ}(Az) = (cz + d)^{ℓ+1}G_{ℓ}(z) for all A = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ Γ_{0}(ℓ).

(iii) G_{ℓ}(z) is holomorphic at i∞.

(iv) G_{ℓ}(z) is holomorphic at the cusp 0. Part (i). It is well-known (and an easy exercise) to show that

(2.13)  U_{ℓ}(η(24z)) = χ_{12}(ℓ) η(24ℓz).

Using (2.3) and (2.13) we easily find that

U_{ℓ}(N (z)) = ℓ χ_{12}(ℓ) N (ℓz).

It follows that

U_{ℓ}(M) − ℓ χ_{12}(ℓ) M(ℓz)

is holomorphic for ℑ(z) > 0. By replacing z by \frac{z}{24} we see that

U_{ℓ}^{∗}(A) − ℓ χ_{12}(ℓ) A(ℓz) = U_{ℓ}^{∗}(α) − ℓ χ_{12}(ℓ) α(ℓz) = α_{ℓ}(z)

and it is clear that G_{ℓ}(z) is holomorphic for ℑ(z) > 0.

Part (ii). Now let A = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ Γ_{0}(ℓ). We must show that

G_{ℓ}(Az) = (cz + d)^{ℓ+1}G_{ℓ}(z).

Since it is well-known that

(\frac{η^{ℓ}(z)}{η(ℓz)})^{2} ∈ M_{ℓ−1}(Γ_{0}(ℓ)),

it suffices to show that

α_{ℓ}(Az) η(ℓAz) = (cz + d)^{2}α_{ℓ}(z)η(ℓz).

We need to show that

(2.14)  f_{ℓ}(Az) = (cz + d)^{2}f_{ℓ}(z),
(2.15)  g_{ℓ}(Az) = (cz + d)^{2}g_{ℓ}(z),

where

f_{ℓ}(z) = U_{ℓ}^{∗}(A) η(ℓz), g_{ℓ}(z) = A(ℓz) η(ℓz).

Let

A^{∗} = (\begin{matrix}a & ℓb \\ c/ℓ & d \end{matrix}) .

Then A^{∗} ∈ SL_{2}(ℤ) and (2.15) follows from Theorem 2.1 and the fact that

A(ℓAz) η(ℓAz) = A(A^{∗}ℓz) η(A^{∗}ℓz).

Now,

f_{ℓ}(z) = U_{ℓ}^{∗}(A)η(ℓz) = U_{ℓ}^{∗}(A(z)η(ℓ^{2}z)).

ANDREWS’ SPT-FUNCTION MODULO POWERS OF 5, 7 AND 13 7

We define

(2.16)  F_{ℓ}(z) := A(z)η(ℓ^{2}z) = A(z)η(z) \frac{η(ℓ^{2}z)}{η(z)}.

Using Theorem 2.1 and the fact that \frac{η(ℓ^{2}z)}{η(z)} is a modular function on Γ_{0}(ℓ^{2}) we have

F_{ℓ}(Cz) = (c_{1}z + d_{1})^{2}F_{ℓ}(z),

for C = (\begin{matrix}a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix}) ∈ Γ_{0}(ℓ^{2}).

Now for 0 ≤ k ≤ ℓ − 1, let

B_{k} = (\begin{matrix}1 & 24k \\ 0 & ℓ \end{matrix})

so that

f_{ℓ}(z) = U_{ℓ}^{∗}(F_{ℓ}(z)) = \frac{1}{ℓ} ∑_{k=0}^{ℓ−1} F_{ℓ}(B_{k}z).

Since A ∈ Γ_{0}(ℓ), (a, ℓ) = 1 and we can choose unique 0 ≤ k^{∗} ≤ ℓ − 1 such that

24ak^{∗} ≡ b + 24kd (mod ℓ).

Then

B_{k}A = A_{k}^{∗}B_{k^{∗}},

where A_{k}^{∗} ∈ Γ_{0}(ℓ^{2}). We have

f_{ℓ}(Az) = \frac{1}{ℓ} ∑_{k=0}^{ℓ−1} F_{ℓ}(B_{k}Az) = \frac{1}{ℓ} ∑_{k^{∗}=0}^{ℓ−1} F_{ℓ}(A_{k}^{∗}B_{k^{∗}}z) = \frac{(cz + d)^{2}}{ℓ} ∑_{k^{∗}=0}^{ℓ−1} F_{ℓ}(B_{k^{∗}}z) = (cz+d)^{2}f_{ℓ}(z),

which is (2.14). Part (iii). First we note that r_{ℓ}^{∗} is a positive integer. We have

G_{ℓ}(z) = α_{ℓ}(z) \frac{η^{2ℓ}(z)}{η(ℓz)} = ∑_{n=−r_{ℓ}^{∗}} (a(ℓn + d_{ℓ}) − χ_{12}(ℓ) ℓ a (\frac{n + r_{ℓ}^{∗}}{ℓ} )) q^{n+r_{ℓ}∗} \frac{E(q)^{2ℓ}}{E(q^{ℓ})}

where

E(q) = ∏_{n=1}^{∞}(1 − q^{n}).

We see that G_{ℓ}(z) is holomorphic at i∞. Part (iv). We need to find G_{ℓ} (\frac{−1}{ℓz}).

U_{ℓ}^{∗}(A) = \frac{1}{ℓ} ∑_{k=0}^{ℓ−1} A (\frac{z + 24k}{ℓ} ) = \frac{1}{ℓ}A (\frac{z}{ℓ} )+\frac{1}{ℓ} ∑_{k=1}^{ℓ−1} A (\frac{z + 24k}{ℓ} ) = \frac{1}{ℓ}A (\frac{z}{ℓ} )+\frac{1}{ℓ} ∑_{k=1}^{ℓ−1} A(B_{k}z).

For each 1 ≤ k ≤ ℓ − 1 choose 1 ≤ k^{∗} ≤ ℓ − 1 such that 576kk^{∗} ≡ −1 (mod ℓ). Then

B_{k}S = C_{k}B_{k^{∗}},

where

C_{k} = (\begin{matrix}24k & \frac{−1−576kk^{∗}}{ℓ} \\ ℓ & −24k^{∗} \end{matrix} ) ∈ Γ_{0}(ℓ).

Then

A(B_{k}Sz) = A(C_{k}B_{k^{∗}}z) = z^{3/2}(\frac{−24k^{∗}}{ℓ})e^{πiℓ/4}A(B_{k^{∗}}z),

by Theorem 2.1 since

ν_{η}(C_{k}) = (\frac{−24k^{∗}}{ℓ})e^{−πiℓ/4},

by [20, p.51]. Define

S_{ℓ} = (\begin{matrix}0 & −1 \\ ℓ & 0 \end{matrix}) .

By Theorem 2.1,

A (\frac{1}{ℓ}S_{ℓ}z) = e^{πi/4}(zℓ^{2})^{3/2}A(ℓ^{2}z),
A (ℓS_{ℓ}z) = e^{πi/4}z^{3/2}A(z).

.

Hence, if we define

(2.17)  H_{ℓ}(z) := U_{ℓ}^{∗}(A) − ℓχ_{12}(ℓ) A(ℓz),

then

H_{ℓ}(S_{ℓ}z)

= ℓz^{3/2}e^{πi/4} (ℓ A(ℓ^{2}z) + \frac{1}{\sqrt{ℓ}}e^{πi(ℓ−1)/4} ∑_{k=1}^{ℓ−1} (\frac{−24k}{ℓ})A (z + \frac{24k}{ℓ} ) − χ_{12}(ℓ)A(z)) .

Replacing z by 24z gives

H_{ℓ}(S_{ℓ}24z)

= ℓ(24z)^{3/2}e^{πi/4} (ℓ M(ℓ^{2}z) + \frac{1}{\sqrt{ℓ}}χ_{12}(ℓ)ϵ_{ℓ}^{3} ∑_{k=1}^{ℓ−1} (\frac{−k}{ℓ})M (z + \frac{k}{ℓ} ) − χ_{12}(ℓ)M(z)) ,

since

e^{πi(ℓ−1)/4}(\frac{24}{ℓ}) = χ_{12}(ℓ)ϵ_{ℓ}^{3}.

Here

ϵ_{ℓ} = \begin{cases}1 & if ℓ ≡ 1 (mod 4), \\ i & if ℓ ≡ 3 (mod 4). \end{cases}

By [24, p.451] we have

H_{ℓ}(S_{ℓ}24z) = ℓ(24z)^{3/2}e^{πi/4} (M|T (ℓ^{2}) − χ_{12}(ℓ)M(z) − U_{ℓ^{2}}(M)) ,

= ℓ(24z)^{3/2}e^{πi/4} ((M|T (ℓ^{2}) − χ_{12}(ℓ)(1 + ℓ)M(z)) − (U_{ℓ^{2}}(M) − ℓχ_{12}(ℓ)M(z))) ,

where T (ℓ^{2}) is the Hecke operator which acts on harmonic Maass forms of weight \frac{3}{2}, and was used by Ono [22]. When the form is meromorphic it corresponds to the usual Hecke operator as described by Shimura [24]. Ono [22] showed that function

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

is a weakly holomorphic modular form. In fact, he showed that

(2.18)  F_{ℓ}(z) := η(z)^{ℓ^{2}}M_{ℓ}(z/24)

is a weight (l^{2} + 3)/2 entire modular form on SL_{2}(ℤ). See [22, Theorem 2.2]. We also note that the function

U_{ℓ^{2}}(M) − ℓχ_{12}(ℓ)M(z) = U_{ℓ} (U_{ℓ}(M) − ℓχ_{12}(ℓ)M(ℓz))

is holomorphic for ℑ(z) > 0 by the remarks in Part (i). Thus we find that

G_{ℓ} (\frac{−1}{ℓz}) (2.19)

= −(izℓ)^{ℓ+1} \frac{E(q^{ℓ})^{2ℓ}}{E(q)} ∑_{n=−s_{ℓ}}^{∞} (χ_{12}(ℓ)a(n) ((\frac{1 − 24n}{ℓ}) − 1) + ℓa (\frac{n + s_{ℓ}}{ℓ^{2}} )) q^{n+2s_{ℓ}},

where s_{ℓ} = \frac{ℓ^{2}−1}{24}. It follows that G_{ℓ}(z) is holomorphic at the cusp 0. □

Since G_{ℓ}(z) ∈ M_{ℓ+1}(Γ_{0}(ℓ)), the function z^{−ℓ−1}G_{ℓ} (\frac{−1}{ℓz}) ∈ M_{ℓ+1}(Γ_{0}(ℓ)) by [5, Lemma 1]. Thus if we define

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

then the proof of Part (iv) of Theorem 2.2 yields Corollary 2.3. If ℓ ≥ 5 is prime, then

(2.21)  J_{ℓ}(z) := β_{ℓ}(z) \frac{η^{2ℓ}(ℓz)}{η(z)} ∈ M_{ℓ+1}(ℓ).

We illustrate the case ℓ = 5. For ℓ prime we define

(2.22)  E_{2,ℓ}(z) := \frac{1}{ℓ − 1} (ℓE_{2}(ℓz) − E_{2}(z)) .

It is well-known that E_{2,ℓ}(z) ∈ M_{2}(Γ_{0}(ℓ)). We also note that E_{2,ℓ}(z) has integral coefficients when ℓ = 5, 7 or 13. By [19, Theorem 3.8] dim M_{6}(Γ_{0}(5)) = 3, and it can be shown that

\{E_{2,5}(z)\frac{η(5z)^{10}}{η(z)^{2}} , E_{2,5}(z)η(5z)^{4}η(z)^{4}, E_{2,5}(z)\frac{η(z)^{10}}{η(5z)^{2}} \}

is a basis. We find that

G_{5}(z) = 5 E_{2,5}(z) (125 η(5z)^{4}η(z)^{4} − \frac{η(z)^{10}}{η(5z)^{2}} ) ,

and

J_{5}(z) = 5 E_{2,5}(z) (\frac{η(5z)^{10}}{η(z)^{2}} − η(5z)^{4}η(z)^{4}) .

Thus

(2.23)  ∑_{n=0}^{∞} (a(5n − 1) + 5 a (\frac{n}{5})) q^{n− \frac{5}{24}} = 5 \frac{E_{2,5}(z)}{η(5z)} (125 \frac{η(5z)^{6}}{η(z)^{6}} − 1) ,

and (2.24)

∑_{n=−1}^{∞} (−a(n) ((\frac{1 − 24n}{5}) − 1) + 5 a (\frac{n + 1}{25})) q^{n− \frac{1}{24}} = 5 \frac{E_{2,5}(z)}{η(z)} (1 − \frac{η(z)^{6}}{η(5z)^{6}} ) .

3. The Congruences

In this section we derive explicit formulas for the generating functions of

(3.1)  a(ℓ^{a}n + d_{ℓ,a}) − χ_{12}(ℓ) ℓ a(ℓ^{a−2}n + d_{ℓ,a−2}),

when ℓ = 5, 7, and 13. As before, d_{ℓ,a} is the least positive integer such that 24d_{ℓ,a} ≡ 1 (mod ℓ^{a}). The presentation of the identities is analogous to those of the partition function as given by Hirschhorn and Hunt [18] and the author [15]. In each case we start by using Theorem 2.2 to find identities for α_{ℓ}(z). This basically gives the initial case a = 1. Then we use Watson’s [25] and Atkin’s [8] method of modular equations to do the induction step and study the arithmetic properties of the coefficients in these identities. The main congruences (1.6)-(1.8) then follow in a straightforward way.

3.1. The SPT-function modulo powers of 5. Theorem 3.1. If a ≥ 1 then

(3.2) ∑_{n=0}^{∞} (a(5^{2a−1}n − t_{a}) + 5 a(5^{2a−3}n − t_{a−1})) q^{n− \frac{5}{24}} = \frac{E_{2,5}(z)}{η(5z)} ∑_{i≥0} x_{2a−1,i}Y^{i},

(3.3)  ∑_{n=0}^{∞} (a(5^{2a}n − t_{a}) + 5 a(5^{2a−2}n − t_{a−1})) q^{n− \frac{1}{24}} = \frac{E_{2,5}(z)}{η(z)} ∑_{i≥0} x_{2a,i}Y^{i},

where

t_{a} = \frac{1}{24}(5^{2a} − 1), Y (z) = \frac{η(5z)^{6}}{η(z)^{6}} ,

x⃗_{1} = (x_{1,0}, x_{1,1}, ⋯) = (−5, 5^{4}, 0, 0, 0, ⋯),

and for a ≥ 1

(3.4)  x⃗_{a+1} = \begin{cases}x⃗_{a}A, & a odd, \\ x⃗_{a}B, & a even. \end{cases}

Here A = (a_{i,j})_{i≥0,j≥0} and B = (b_{i,j})_{i≥0,j≥0} are defined by

(3.5)  a_{i,j} = m_{6i,i+j},  b_{i,j} = m_{6i+1,i+j},

where the matrix M = (m_{i,j})_{i,j≥0} is defined as follows: The first five rows of M are

(\begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & ⋯ \\ 0 & 5^{3} & 0 & 0 & 0 & 0 & ⋯ \\ 0 & 4 · 5^{2} & 5^{5} & 0 & 0 & 0 & ⋯ \\ 0 & 9 · 5 & 9 · 5^{4} & 5^{7} & 0 & 0 & ⋯ \\ 0 & 2 · 5 & 44 · 5^{3} & 14 · 5^{6} & 5^{9} & 0 & ⋯ \end{matrix})

and for i ≥ 5, m_{i,0} = 0 and for j ≥ 1,

(3.6) m_{i,j} = 25 m_{i−1,j−1} + 25 m_{i−2,j−1} + 15 m_{i−3,j−1} + 5 m_{i−4,j−1} + m_{i−5,j−1}.

Lemma 3.2. If n is a positive integer then there are integers c_{m} (⌈\frac{n}{5}⌉ ≤ m ≤ n) such that

U_{5}(E_{2,5}Z^{n}) = E_{2,5} ∑_{m=⌈ \frac{n}{5} ⌉}^{n} c_{m}Y^{m},

where

(3.7)  Z(z) = \frac{η(25z)}{η(z)},  Y (z) = \frac{η(5z)^{6}}{η(z)^{6}} .

Proof. We need the following dimension formulas which follow from [13] and [19, Theorem 3.8]. For k even,

dim M_{k}(Γ_{0}(5)) = 2 ⌊\frac{k}{4} ⌋ + 1,
dim M_{k}(Γ_{0}(5), (\frac{·}{5})) = k − 2 ⌊\frac{k}{4} ⌋ .

Let n be a positive integer. Then

U_{5}(E_{2,5}Z^{n}) = U_{5} (E_{2,5}(z) (\frac{η(5z)^{5}}{η(z)} )^{n} (\frac{η(25z)}{η(5z)^{5}})^{n}) = U_{5} (E_{2,5}(z) (\frac{η(5z)^{5}}{η(z)} )^{n}) (\frac{η(5z)}{η(z)^{5}})^{n} .

When n is even the function

E_{2,5}(z) (\frac{η(5z)^{5}}{η(z)} )^{n}

belongs to the space M_{2n+2}(Γ_{0}(5)), which has as a basis

{E_{2,5}(z)η(z)^{5n−6m}η(5z)^{6m−n} , 0 ≤ m ≤ n} .

This follows from the dimension formula. We note that

ord (E_{2,5}(z)η(z)^{5n−6m}η(5z)^{6m−n}; i∞) = m.

The operator U_{5} preserves the space M_{2n+2}(Γ_{0}(5)). It follows that there are integers c_{m} (⌈\frac{n}{5}⌉ ≤ m ≤ n) such that

U_{5}(E_{2,5}Z^{n}) = E_{2,5}(z) ∑_{m=⌈ \frac{n}{5} ⌉}^{n} c_{m} η(z)^{5n−6m}η(5z)^{6m−n} (\frac{η(5z)}{η(z)^{5}})^{n} = E_{2,5}(z) ∑_{m=⌈ \frac{n}{5} ⌉}^{n} c_{m}Y^{m}.

When n is odd the proof is similar except this time one needs to work in the space M_{2n+2}(Γ_{0}(5), (\frac{·}{5})). □

Corollary 3.3.

(3.8)  U_{5}(E_{2,5}) = E_{2,5}
(3.9)  U_{5}(E_{2,5}Z) = 5^{3} E_{2,5}Y
(3.10)  U_{5}(E_{2,5}Z^{2}) = 5^{2} E_{2,5}(4Y + 5^{3}Y^{2})
(3.11)  U_{5}(E_{2,5}Z^{3}) = 5 E_{2,5}(9Y + 9 · 5^{3}Y^{2} + 5^{6}Y^{3})
(3.12)  U_{5}(E_{2,5}Z^{4}) = 5 E_{2,5}(2Y + 44 · 5^{2}Y^{2} + 14 · 5^{5}Y^{3} + 5^{8}Y^{4}).

Proof. Equation (3.8) is elementary. It also follows from the fact that dim M_{2}(Γ_{0}(5)) =

1. Equations (3.9)–(3.12) follow from Lemma 3.2 and straightforward calculation. □

We need the 5th order modular equation that was used by Watson to prove Ramanujan’s partition congruences for powers of 5.

(3.13)  Z^{5} = (25Z^{4} + 25Z^{3} + 15Z^{2} + 5Z + 1) Y (5z).

12 F. G. GARVAN

Lemma 3.4. For i ≥ 0

U_{5}(E_{2,5}Z^{i}) = E_{2,5}(z) ∑_{j=⌈ \frac{i}{5} ⌉}^{i} m_{i,j}Y^{j},

where Z = Z(z), Y = Y (z) are defined in (3.7), and the m_{i,j} are defined in Theorem 3.1. Proof. The result holds for 0 ≤ i ≤ 4 by Corollary 3.3. By (3.13) we have

U_{5}(E_{2,5}Z^{i}) = (25U_{5}(E_{2,5}Z^{i−1}) + 25U_{5}(E_{2,5}Z^{i−2}) + 15U_{5}(E_{2,5}Z^{i−3}) + 5U_{5}(E_{2,5}Z^{i−4})
+U_{5}(E_{2,5}Z^{i−5})) Y (z),

for i ≥ 5. The result follows by induction on i using the recurrence (3.6). □ Lemma 3.5. For i ≥ 0,

(3.14)  U_{5}(E_{2,5}Y^{i}) = E_{2,5}(z) ∑_{j=⌈ \frac{i}{5} ⌉}^{5i} a_{i,j}Y^{j},
(3.15)  U_{5}(E_{2,5}ZY^{i}) = E_{2,5}(z) ∑_{j=⌈ \frac{i+1}{5} ⌉}^{5i+1} b_{i,j}Y^{j},

where the a_{i,j}, b_{i,j} are defined in (3.5). Proof. Suppose i ≥ 0. By Lemma 3.4

U_{5}(E_{2,5}Y^{i}) = U_{5}(E_{2,5}Z^{6i}Y (5z)^{−i}) = Y^{−i}U_{5}(E_{2,5}Z^{6i})

= Y^{−i}E_{2,5}(z) ∑_{j=⌈ \frac{6i}{5} ⌉}^{6i} m_{6i,j}Y^{j}

= E_{2,5}(z) ∑_{j=⌈ \frac{i}{5} ⌉}^{5i} m_{6i,i+j}Y^{j} = E_{2,5}(z) ∑_{j=⌈ \frac{i}{5} ⌉}^{5i} a_{i,j}Y^{j},

which is (3.14). Similarly

U_{5}(E_{2,5}ZY^{i}) = U_{5}(E_{2,5}Z^{6i+1}Y (5z)^{−i}) = Y^{−i}U_{5}(E_{2,5}Z^{6i+1})

= Y^{−i}E_{2,5}(z) ∑_{j=⌈ \frac{6i+1}{5} ⌉}^{6i+1} m_{6i+1,j}Y^{j}

= E_{2,5}(z) ∑_{j=⌈ \frac{i+1}{5} ⌉}^{5i+1} m_{6i+1,i+j}Y^{j} = E_{2,5}(z) ∑_{j=⌈ \frac{i+1}{5} ⌉}^{5i+1} b_{i,j}Y^{j},

which is (3.15). □ Proof of Theorem 3.1. We proceed by induction. The case a = 1 of (3.2) is (2.23). We now suppose a ≥ 1 is fixed and (3.2) holds. Thus

E(q^{5}) ∑_{n=0}^{∞} (a(5^{2a−1}n − t_{a}) + 5 a(5^{2a−3}n − t_{a−1})) q^{n} = E_{2,5}(z) ∑_{i≥0} x_{2a−1,i}Y^{i}.

We now apply the U_{5} operator to both sides and use Lemma 3.5.

E(q) ∑_{n=0}^{∞} (a(5^{2a}n − t_{a}) + 5 a(5^{2a−2}n − t_{a−1})) q^{n} = ∑_{i≥0} x_{2a−1,i}U_{5}(E_{2,5}(z)Y^{i})

= E_{2,5}(z) ∑_{i≥0} x_{2a−1,i} ∑_{j≥0} a_{i,j}Y^{j} = E_{2,5}(z) ∑_{j≥0} (∑_{i≥0} x_{2a−1,i}a_{i,j}) Y^{j}

= E_{2,5}(z) ∑_{j≥0} x_{2a,j}Y^{j}.

We obtain (3.3) by dividing both sides by η(z).

Now again suppose a is fixed and (3.3) holds. Multiplying both sides by η(25z) gives

E(q^{25}) ∑_{n=0}^{∞} (a(5^{2a}n − t_{a}) + 5 a(5^{2a−2}n − t_{a−1})) q^{n+1} = E_{2,5}(z) ∑_{i≥0} x_{2a,i}ZY^{i}.

We apply the U_{5} operator to both sides.

E(q^{5}) ∑_{n=0}^{∞} (a(5^{2a}(5n − 1) − t_{a}) + 5 a(5^{2a−2}(5n − 1) − t_{a−1})) q^{n}

= ∑_{i≥0} x_{2a,i}U_{5}(E_{2,5}(z)ZY^{i}).

Using Lemma 3.5 and the fact that t_{a+1} = 5^{2a} + t_{a} we have

E(q^{5}) ∑_{n=0}^{∞} (a(5^{2a+1}n − t_{a+1}) + 5 a(5^{2a−1}n − t_{a})) q^{n} = E_{2,5}(z) ∑_{i≥0} x_{2a,i} ∑_{j≥0} b_{i,j}Y^{j}

= E_{2,5}(z) ∑_{j≥0} (∑_{i≥0} x_{2a,i}b_{i,j}) Y^{j} = E_{2,5}(z) ∑_{j≥0} x_{2a+1,j}Y^{j}.

We obtain (3.2) with a replaced by a + 1 after dividing both sides by η(5z). This completes the proof of the theorem. □

Throughout this section we will make repeated use of the following lemma which we leave as an exercise. Lemma 3.6. Suppose x, y, n ∈ ℤ and n > 0. Then

(3.16)  ⌊\frac{x}{n}⌋ + ⌊\frac{y}{n}⌋ ≥ ⌊\frac{x + y − n + 1}{n}⌋ .

For any prime ℓ we let π(n) = π_{ℓ}(n) denote the exact power of ℓ that divides n. Then we have

Lemma 3.7.

π_{5}(m_{i,j}) ≥ ⌊\frac{1}{2}(5j − i + 1)⌋,

where the matrix M = (m_{i,j})_{i,j≥0} is defined in Theorem 3.1.

Proof. First we verify the result for 0 ≤ i ≤ 4. The result is easily proven for i ≥ 5 using the recurrence (3.6). □

14 F. G. GARVAN

Corollary 3.8.

π_{5}(a_{i,j}) ≥ ⌊\frac{1}{2}(5j − i + 1)⌋, π_{5}(b_{i,j}) ≥ ⌊\frac{1}{2}(5j − i)⌋,

where the a_{i,j}, b_{i,j} are defined by (3.5). Lemma 3.9. For b ≥ 2, and j ≥ 1,

(3.17)  π_{5}(x_{2b−1,j}) ≥ 5b − 6 + max(0, ⌊\frac{1}{2}(5j − 7)⌋),
(3.18)  π_{5}(x_{2b,j}) ≥ 5b − 4 + ⌊\frac{1}{2}(5j − 4)⌋.

Proof. A calculation gives

x⃗_{3} = (x_{3,0}, x_{3,1}, x_{3,2}, ⋯)
= (0, 669303124 · 5^{4}, 3328977476 · 5^{11}, 366098988268 · 5^{14},
201318006648837 · 5^{15}, 1618593700646527 · 5^{18}, 6370852555263938 · 5^{21},

2900024541422883 · 5^{25}, 4237895677971369 · 5^{28}, 21327793208615511 · 5^{30}, 15532659183030861 · 5^{33}, 8481639849706179 · 5^{36}, 3564573506915806 · 5^{39},

1175454967692313 · 5^{42}, 1542192101361916 · 5^{44}, 325171329708596 · 5^{47},
55431641829564 · 5^{50}, 1532152033009 · 5^{54}, 171561318777 · 5^{57},
77490966671 · 5^{59}, 5598792206 · 5^{62}, 318906274 · 5^{65},
2799863 · 5^{69}, 91379 · 5^{72}, 10439 · 5^{74}, 149 · 5^{77},
5^{80}, 0, ⋯),
π_{5}(x⃗_{3}) = (∞, 4, 11, 14, 15, 18, 21, 25, 28, 30, 33, 36, 39, 42, 44, 47, 50, 54, 57, 59, 62,
65, 69, 72, 74, 77, 80, ∞, ∞, ⋯),

and (3.17) holds for b = 2. Now suppose b ≥ 2 is fixed and (3.17) holds. By (3.4)

x_{2b,j} = ∑_{i≥1} x_{2b−1,i}a_{i,j}.

Then using Corollary 3.8 π_{5}(x_{2b,1}) ≥ min(\{5b−4\}∪\{5b−6+⌊\frac{1}{2}(5i−7)⌋+⌊(\frac{1}{2}(6−i)⌋ : 2 ≤ i ≤ 5\}) = 5b−4, and (3.18) holds for j = 1. Suppose j ≥ 2. Then

π_{5}(x_{2b,j}) ≥ min_{1≤i≤5j}(π_{5}(x_{2b−1,i}) + π_{5}(a_{i,j}))

≥ min_{2≤i≤5j}(π_{5}(x_{2b−1,1}) + π_{5}(a_{1,j}), (π_{5}(x_{2b−1,i}) + π_{5}(a_{i,j}))

≥ min(\{5b − 6 + ⌊\frac{1}{2}(5j)⌋\} ∪ \{5b − 6 + ⌊\frac{1}{2}(5i − 7)⌋) + ⌊\frac{1}{2}(5j − i + 1)⌋ : 2 ≤ i ≤ 5j\}). Now

5b − 6 + ⌊\frac{1}{2}(5j)⌋ = 5b − 4 + ⌊\frac{1}{2}(5j − 4)⌋.

If 2 ≤ i ≤ 5j, then using Lemma 3.6 we have

5b − 6 + ⌊\frac{1}{2}(5i − 7)⌋) + ⌊\frac{1}{2}(5j − i + 1)⌋ ≥ 5b − 6 + ⌊\frac{1}{2}(5j + 4i − 7)⌋

≥ 5b − 6 + ⌊\frac{1}{2}(5j + 1)⌋ = 5b − 4 + ⌊\frac{1}{2}(5j − 3)⌋

and (3.18) holds. Now suppose b ≥ 2 is fixed and (3.18) holds. By (3.4)

x_{2b+1,j} = ∑_{i≥1} x_{2b,i}b_{i,j}.

ANDREWS’ SPT-FUNCTION MODULO POWERS OF 5, 7 AND 13 15

We observe that π_{5}(b_{1,1}) = π_{5}(500) = 3. Then using Corollary 3.8 π_{5}(x_{2b+1,1}) ≥ min(\{5b−1\}∪\{5b−4+⌊\frac{1}{2}(5i−4)+⌊(\frac{1}{2}(5−i)⌋ : 2 ≤ i ≤ 4\}) = 5b−1, and (3.17) holds for j = 1 with b replaced by b + 1. Suppose j ≥ 2. Then

π_{5}(x_{2b+1,j}) ≥ min_{1≤i≤5j−1}(π_{5}(x_{2b,i}) + π_{5}(b_{i,j}))

≥ min_{2≤i≤5j−1}(π_{5}(x_{2b,1}) + π_{5}(b_{1,j}), (π_{5}(x_{2b,i}) + π_{5}(b_{i,j}))

≥ min(\{5b − 4 + ⌊\frac{1}{2}(5j − 1)⌋\} ∪ \{5b − 4 + ⌊\frac{1}{2}(5i − 4)⌋) + ⌊\frac{1}{2}(5j − i)⌋ : 2 ≤ i ≤ 5j − 1\}). Now

5b − 4 + ⌊\frac{1}{2}(5j − 1)⌋ = 5b − 1 + ⌊\frac{1}{2}(5j − 7)⌋.

If 2 ≤ i ≤ 5j − 1, then again using Lemma 3.6 we have

5b − 4 + ⌊\frac{1}{2}(5i − 4)⌋) + ⌊\frac{1}{2}(5j − i)⌋ ≥ 5b − 4 + ⌊\frac{1}{2}(5j + 4i − 5)⌋

≥ 5b − 4 + ⌊\frac{1}{2}(5j + 3)⌋ = 5b − 1 + ⌊\frac{1}{2}(5j − 3)⌋

and (3.17) holds with b replaced by b + 1. Lemma 3.9 follows by induction. □ Corollary 3.10. For b ≥ 2,

(3.19)  a(5^{2b−1}n + δ_{2b+1}) + 5 a(5^{2b−3}n + δ_{2b−3}) ≡ 0 (mod 5^{5b−6}),
(3.20)  a(5^{2b}n + δ_{2b}) + 5 a(5^{2b−2}n + δ_{2b−2}) ≡ 0 (mod 5^{5b−4}).

For a ≥ 1,

(3.21)  spt(5^{a+2}n + δ_{a+2}) + 5 spt(5^{a}n + δ_{a}) ≡ 0 (mod 5^{2a+1}),
(3.22)  spt(5^{a}n + δ_{a}) ≡ 0 (mod 5^{⌊ \frac{a+1}{2} ⌋}).

Proof. The congruences (3.19)–(3.20) follow from Theorem 3.1 and Lemma 3.9. Let

dp(n) = (24n − 1)p(n).

Then

(3.23)  dp(5^{a}n + δ_{a}) ≡ 0 (mod 5^{2a}),

by (1.12). The congruence (3.21) follows from (3.19)–(3.20), and (3.23). Andrews’ congruence (1.2) implies that (3.22) holds for a = 1, 2. The general result follows by induction using (3.21). □

We note that when a = 0 there is a stronger congruence than (3.21). We prove that

(3.24)  spt(25n − 1) + 5 spt(n) ≡ 0 (mod 25).

We have calculated

x⃗_{2} = (x_{2,0}, x_{2,1}, x_{2,2}, ⋯)
= (−5^{1}, 63 · 5^{6}, 104 · 5^{9}, 189 · 5^{11}, 24 · 5^{14}, 5^{17}, 0, ⋯).

16 F. G. GARVAN

Thus

∑_{n=0}^{∞} (a(25n − 1) + 5 a(n)) q^{n− \frac{1}{24}} (3.25)

= 5\frac{E_{2,5}(z)}{η(z)} (−1 + 63 · 5^{5} \frac{η^{6}(5z)}{η^{6}(z)} + 104 · 5^{8} \frac{η^{12}(5z)}{η^{12}(z)} + 189 · 5^{10} \frac{η^{18}(5z)}{η^{18}(z)}

+24 · 5^{13} \frac{η^{24}(5z)}{η^{24}(z)} + 5^{16} \frac{η^{30}(5z)}{η^{30}(z)}) ,

and

∑_{n=0}^{∞} (a(25n − 1) + 5 a(n)) q^{n− \frac{1}{24}} ≡ 20\frac{E_{2}(z)}{η(z)} (mod 25).

But from (2.6) we see that

∑_{n=0}^{∞} (dp(25n − 1) + 5 dp(n)) q^{n− \frac{1}{24}} ≡ 20\frac{E_{2}(z)}{η(z)} (mod 25),

and

12 ∑_{n=0}^{∞} (spt(25n − 1) + 5 spt(n)) q^{n− \frac{1}{24}}

= ∑_{n=0}^{∞} (a(25n − 1) + 5 a(n)) q^{n− \frac{1}{24}} − ∑_{n=0}^{∞} (dp(25n − 1) + 5 dp(n)) q^{n− \frac{1}{24}}

≡ 0 (mod 25), which gives (3.24). 3.2. The SPT-function modulo powers of 7. Theorem 3.11. If a ≥ 1 then

(3.26) ∑_{n=0}^{∞} (a(7^{2a−1}n − u_{a}) + 7 a(7^{2a−3}n − u_{a−1})) q^{n− \frac{7}{24}} = \frac{E_{2,7}(z)}{η(7z)} ∑_{i≥0} x_{2a−1,i}Y^{i},
(3.27) ∑_{n=0}^{∞} (a(7^{2a}n − u_{a}) + 7 a(7^{2a−2}n − u_{a−1})) q^{n− \frac{1}{24}} = \frac{E_{2,7}(z)}{η(z)} ∑_{i≥0} x_{2a,i}Y^{i},

where

u_{a} = \frac{1}{24}(7^{2a} − 1), Y (z) = \frac{η(7z)^{4}}{η(z)^{4}} ,

x⃗_{1} = (x_{1,0}, x_{1,1}, ⋯) = (−7, 3 · 7^{3}, 7^{5}, 0, 0, ⋯),

and for a ≥ 1

x⃗_{a+1} = \begin{cases}x⃗_{a}A, & a odd, \\ x⃗_{a}B, & a even. \end{cases}

Here A = (a_{i,j})_{i≥0,j≥0} and B = (b_{i,j})_{i≥0,j≥0} are defined by

(3.28)  a_{i,j} = m_{4i,i+j},  b_{i,j} = m_{4i+1,i+j},

where the matrix M = (m_{i,j})_{i,j≥0} is defined as follows: The first seven rows of M are defined so that

U_{7}(E_{2,7}Z^{i}) = ∑_{j=⌈ \frac{2i}{7} ⌉}^{2i} m_{i,j}Y^{j} (0 ≤ i ≤ 6),

where

Z(z) = \frac{η(49z)}{η(z)}.

and for i ≥ 7, m_{i,0} = 0, m_{i,1} = 0, and for j ≥ 2,

m_{i,j} = 49 m_{i−1,j−1} + 35 m_{i−2,j−1} + 7 m_{i−3,j−1} + 343 m_{i−1,j−2} + 343 m_{i−2,j−2} (3.29)

+ 147 m_{i−3,j−2} + 49 m_{i−4,j−2} + 21 m_{i−5,j−2} + 7 m_{i−6,j−2} + m_{i−7,j−2}.

The proof of the following lemma is analogous to that of Lemma 3.2. Lemma 3.12. If n is a positive integer then there are integers c_{m} (⌈\frac{2n}{7}⌉ ≤ m ≤ 2n) such that

U_{7}(E_{2,7}Z^{n}) = E_{2,7} ∑_{m=⌈ \frac{2n}{7} ⌉}^{2n} c_{m}Y^{m},

where

(3.30)  Z(z) = Z_{7}(z) = \frac{η(49z)}{η(z)},  Y (z) = \frac{η(7z)^{4}}{η(z)^{4}} .

Corollary 3.13.

U_{7}(E_{2,7}) = E_{2,7} (3.31)
U_{7}(E_{2,7}Z) = 7^{2}E_{2,7}(3 Y + 7^{2} Y^{2}) (3.32)
U_{7}(E_{2,7}Z^{2}) = 7E_{2,7}(10 Y + 27 · 7^{2} Y^{2} + 10 · 7^{4} Y^{3} + 7^{6} Y^{4}) (3.33)
U_{7}(E_{2,7}Z^{3}) = 7E_{2,7}(Y + 190 · 7 Y^{2} + 255 · 7^{3} Y^{3} + 104 · 7^{5} Y^{4} + 17 · 7^{7} Y^{5} + 7^{9} Y^{6}) (3.34)
U_{7}(E_{2,7}Z^{4}) = 7^{2}E_{2,7}(82 Y^{2} + 352 · 7^{2} Y^{3} + 2535 · 7^{3} Y^{4} + 1088 · 7^{5} Y^{5} + 230 · 7^{7} Y^{6} (3.35)
+ 24 · 7^{9} Y^{7} + 7^{11} Y^{8})
U_{7}(E_{2,7}Z^{5}) = 7E_{2,7}(114 Y^{2} + 253 · 7^{3} Y^{3} + 4169 · 7^{4} Y^{4} + 3699 · 7^{6} Y^{5} + 11495 · 7^{7} Y^{6} (3.36)
+ 2852 · 7^{9} Y^{7} + 405 · 7^{11} Y^{8} + 31 · 7^{13} Y^{9} + 7^{15} Y^{10})
U_{7}(E_{2,7}Z^{6}) = 7E_{2,7}(9 Y^{2} + 736 · 7^{2} Y^{3} + 27970 · 7^{3} Y^{4} + 6808 · 7^{6} Y^{5} + 38475 · 7^{7} Y^{6} (3.37)
+ 17490 · 7^{9} Y^{7} + 33930 · 7^{10} Y^{8} + 5890 · 7^{12} Y^{9} + 629 · 7^{14} Y^{10}
+ 38 · 7^{16} Y^{11} + 7^{18} Y^{12})

We need the 7th order modular equation that was used by Watson to prove Ramanujan’s partition congruences for powers of 7.

(3.38)  Z^{7} = (1 + 7 Z + 21 Z^{2} + 49 Z^{3} + 147 Z^{4} + 343 Z^{5} + 343 Z^{6}) Y (7z)^{2}
+ (7 Z^{4} + 35 Z^{5} + 49 Z^{6}) Y (7z).

Lemma 3.14. For i ≥ 0

U_{7}(E_{2,7}Z^{i}) = E_{2,7}(z) ∑_{j=⌈ \frac{2i}{7} ⌉}^{2i} m_{i,j}Y^{j},

where Z = Z(z), Y = Y (z) are defined in (3.30), and the m_{i,j} are defined in Theorem 3.11. Lemma 3.15. For i ≥ 0,

(3.39)  U_{7}(E_{2,7}Y^{i}) = E_{2,7}(z) ∑_{j=⌈ \frac{i}{7} ⌉}^{7i} a_{i,j}Y^{j},

(3.40)  U_{7}(E_{2,7}ZY^{i}) = E_{2,7}(z) ∑_{j=⌈ \frac{i+2}{7} ⌉}^{7i+2} b_{i,j}Y^{j}

where the a_{i,j}, b_{i,j} are defined in (3.28).

Let π_{7}(n) denote the exact power of 7 dividing n. Then we have Lemma 3.16.

π_{7}(m_{i,j}) ≥ ⌊\frac{1}{4}(7j − 2i + 3)⌋,

where the matrix M = (m_{i,j})_{i,j≥0} is defined in Theorem 3.11. Corollary 3.17.

π_{7}(a_{i,j}) ≥ ⌊\frac{1}{4}(7j − i + 3)⌋, π_{7}(b_{i,j}) ≥ ⌊\frac{1}{4}(7j − i + 1)⌋,

where the a_{i,j}, b_{i,j} are defined by (3.28). Lemma 3.18. For b ≥ 2, and j ≥ 1,

(3.41)  π_{7}(x_{2b−1,j}) ≥ 3b − 3 + ⌊\frac{1}{4}(7j − 4)⌋.
(3.42)  π_{7}(x_{2b,j}) ≥ 3b − 1 + ⌊\frac{1}{4}(7j − 6)⌋.

Corollary 3.19. For b ≥ 2,

(3.43)  a(7^{2b−1}n + λ_{2b+1}) + 7 · a(7^{2b−3}n + λ_{2b−3}) ≡ 0 (mod 7^{3b−3}),
(3.44)  a(7^{2b}n + λ_{2b}) + 7 · a(7^{2b−2}n + λ_{2b−2}) ≡ 0 (mod 7^{3b−1}).

For a ≥ 1,

(3.45)  spt(7^{a+2}n + λ_{a+2}) + 7 · spt(7^{a}n + λ_{a}) ≡ 0 (mod 7^{⌊ \frac{1}{2} (3a+4)⌋}),
(3.46)  spt(7^{a}n + λ_{a}) ≡ 0 (mod 7^{⌊ \frac{a+1}{2} ⌋}).

We note that (3.45) also holds for a = 0 taking λ_{0} = 1. The proof of the congruence

(3.47)  spt(49n − 2) + 7 · spt(n) ≡ 0 (mod 49).

is analogous to the proof of (3.24).

3.3. The SPT-function modulo powers of 13. Theorem 3.20. If a ≥ 1 then

∑_{n=0}^{∞} (a(13^{2a−1}n − v_{a}) − 13 a(13^{2a−3}n − v_{a−1})) q^{n− \frac{13}{24}} = \frac{E_{2,13}(z)}{η(13z)} ∑_{i≥0} x_{2a−1,i}Y^{i}, (3.48)
∑_{n=0}^{∞} (a(13^{2a}n − v_{a}) − 13 a(13^{2a−2}n − v_{a−1})) q^{n− \frac{1}{24}} = \frac{E_{2,13}(z)}{η(z)} ∑_{i≥0} x_{2a,i}Y^{i}, (3.49)

where

v_{a} = \frac{1}{24}(13^{2a} − 1), Y (z) = \frac{η(13z)^{2}}{η(z)^{2}} ,

x⃗_{1} = (x_{1,0}, x_{1,1}, ⋯)

= (13, 11 · 13^{2}, 108 · 13^{3}, 190 · 13^{4}, 140 · 13^{5}, 54 · 13^{6}, 11 · 13^{7}, 13^{8}, 0, 0, 0, ⋯), and for a ≥ 1

x⃗_{a+1} = \begin{cases}x⃗_{a}A, & a odd, \\ x⃗_{a}B, & a even. \end{cases}

Here A = (a_{i,j})_{i≥0,j≥0} and B = (b_{i,j})_{i≥0,j≥0} are defined by

(3.50) a_{i,j} = m_{2i,i+j}, b_{i,j} = m_{2i+1,i+j},

where the matrix M = (m_{i,j})_{i≥−12,j≥−6} is defined as follows: The first 13 rows of M are

(\begin{matrix}0 & 0 & 0 & 0 & 0 & 0 & 13^{6} & 0 & 0 & ⋯ \\ 0 & 82 · 13 & 456 · 13^{2} & 360 · 13^{3} & 126 · 13^{4} & 18 · 13^{5} & 13^{6} & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & 0 & 0 & 13^{5} & 0 & 0 & ⋯ \\ 0 & 0 & 18 · 13 & −36 · 13^{2} & −40 · 13^{3} & −14 · 13^{4} & −13^{5} & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & 0 & 0 & 13^{4} & 0 & 0 & ⋯ \\ 0 & 0 & 0 & −14 · 13 & −12 · 13^{2} & 0 & 13^{4} & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & 0 & 0 & 13^{3} & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & 4 · 13 & 6 · 13^{2} & 13^{3} & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & 0 & 0 & 13^{2} & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & 0 & 0 & −13^{2} & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & 0 & 0 & 13 & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & 0 & 0 & −13 & 0 & 0 & ⋯ \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & ⋯ \end{matrix})

and for m_{k,ℓ} = 0 for k ≥ 1 and −6 ≤ ℓ ≤ 0; and for i ≥ 1 and j ≥ 1,

(3.51)  m_{i,j} = ∑_{r=1}^{13} ∑_{s=⌊ \frac{1}{2} (r+2)⌋}^{7} ψ_{r,s}m_{i−r,j−s},

where Ψ = (ψ_{r,s})_{1≤r≤13,1≤s≤7} is the matrix (3.52)

Ψ = (\begin{matrix}11 · 13 & 36 · 13^{2} & 38 · 13^{3} & 20 · 13^{4} & 6 · 13^{5} & 13^{6} & 13^{6} \\ 0 & −204 · 13 & −346 · 13^{2} & −222 · 13^{3} & −74 · 13^{4} & −13^{6} & −13^{6} \\ 0 & 36 · 13 & 126 · 13^{2} & 102 · 13^{3} & 38 · 13^{4} & 7 · 13^{5} & 7 · 13^{5} \\ 0 & 0 & −346 · 13 & −422 · 13^{2} & −184 · 13^{3} & −37 · 13^{4} & −3 · 13^{5} \\ 0 & 0 & 38 · 13 & 102 · 13^{2} & 56 · 13^{3} & 13^{5} & 15 · 13^{4} \\ 0 & 0 & 0 & −222 · 13 & −184 · 13^{2} & −51 · 13^{3} & −5 · 13^{4} \\ 0 & 0 & 0 & 20 · 13 & 38 · 13^{2} & 13^{4} & 19 · 13^{3} \\ 0 & 0 & 0 & 0 & −74 · 13 & −37 · 13^{2} & −5 · 13^{3} \\ 0 & 0 & 0 & 0 & 6 · 13 & 7 · 13^{2} & 15 · 13^{2} \\ 0 & 0 & 0 & 0 & 0 & −13^{2} & −3 · 13^{2} \\ 0 & 0 & 0 & 0 & 0 & 13 & 7 · 13 \\ 0 & 0 & 0 & 0 & 0 & 0 & −13 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{matrix}) .

The proof of the following lemma is analogous to that of Lemma 3.2.

Lemma 3.21. If n is a positive integer then there are integers c_{m} (⌈\frac{7n}{13}⌉ ≤ m ≤ 7n) such that

U_{13}(E_{2,13}Z^{n}) = E_{2,13} ∑_{m=⌈ \frac{7n}{13} ⌉}^{7n} c_{m}Y^{m},

where

(3.53)  Z(z) = Z_{13}(z) = \frac{η(169z)}{η(z)},  Y (z) = \frac{η(13z)^{2}}{η(z)^{2}} .

We need a version for Lemma 3.21 when n is negative.

Lemma 3.22. If n is a nonnegative integer then there are integers c_{m} (−6n ≤ m ≤ n − ⌈\frac{6n}{13}⌉) such that

U_{13}(E_{2,13}Z^{−n}) = E_{2,13} ∑_{m=−6n}^{n−⌈ \frac{6n}{13} ⌉} c_{m}Y^{−m}.

Proof. The proof is analogous to Lemma 3.21. The main difference is that we write

U_{13}(E_{2,13}Z^{−n}) = U_{13} (E_{2,13}(z) (η(z)η^{11}(13z))^{n}) (η^{11}(z)η(13z))^{−n} ,

and use the fact that E_{2,13}(z) (η(z)η^{11}(13z))^{n} ∈ M_{2+6n}(Γ_{0}(13), (\frac{·}{13})^{n}). □

Corollary 3.23.

U_{13}(E_{2,13}) = E_{2,13}
U_{13}(E_{2,13}Z^{−1}) = −13 E_{2,13}
U_{13}(E_{2,13}Z^{−2}) = 13 E_{2,13}
U_{13}(E_{2,13}Z^{−3}) = −13^{2} E_{2,13}
U_{13}(E_{2,13}Z^{−4}) = 13^{2} E_{2,13}
U_{13}(E_{2,13}Z^{−5}) = 13 E_{2,13}(4 Y^{−2} + 6 · 13 Y^{−1} + 13^{2})
U_{13}(E_{2,13}Z^{−6}) = 13^{3} E_{2,13}
U_{13}(E_{2,13}Z^{−7}) = 13 E_{2,13}(−14 Y^{−3} − 12 · 13 Y^{−2} + 13^{3})
U_{13}(E_{2,13}Z^{−8}) = 13^{4} E_{2,13}
U_{13}(E_{2,13}Z^{−9}) = 13 E_{2,13}(18 Y^{−4} − 36 · 13 Y^{−3} − 40 · 13^{2} Y^{−2} − 14 · 13^{3} Y^{−1} − 13^{4})
U_{13}(E_{2,13}Z^{−10}) = 13^{5} E_{2,13}
U_{13}(E_{2,13}Z^{−11}) = 13 E_{2,13}(82 Y^{−5} + 456 · 13 Y^{−4} + 360 · 13^{2} Y^{−3} + 126 · 13^{3} Y^{−2}
+ 18 · 13^{4} Y^{−1} + 13^{5})
U_{13}(E_{2,13}Z^{−12}) = 13^{6} E_{2,13}

We need the 13th order modular equation that was used by Atkin and O’Brien [6] to study properties of p(n) modulo powers of 13. Lehner [21] derived this equation earlier.

(3.54)  Z^{13}(z) = ∑_{r=1}^{13} ∑_{s=⌊ \frac{1}{2} (r+2)⌋}^{7} ψ_{r,s}Y^{s}(13z) Z^{13−r}(z),

where the matrix Ψ = (ψ_{i,j}) is given in (3.52), and Y (z), Z(z) are given in (3.53). The modular equation and the matrix Ψ are given explicitly in Appendix C in [6] Lemma 3.24. For i ≥ 0

U_{13}(E_{2,13}Z^{i}) = E_{2,13}(z) ∑_{j=⌈ \frac{7i}{13} ⌉}^{7i} m_{i,j}Y^{j},

where Z = Z(z), Y = Y (z) are defined in (3.53), and the m_{i,j} are defined in Theorem 3.20. Lemma 3.25. For i ≥ 0,

(3.55)  U_{13}(E_{2,13}Y^{i}) = E_{2,13}(z) ∑_{j=⌈ \frac{i}{13} ⌉}^{13i} a_{i,j}Y^{j},

(3.56)  U_{13}(E_{2,13}ZY^{i}) = E_{2,13}(z) ∑_{j=⌈ \frac{i+7}{13} ⌉}^{13i+7} b_{i,j}Y^{j}

where the a_{i,j}, b_{i,j} are defined in (3.50).

Let π_{13}(n) denote the exact power of 13 dividing n. Then we have Lemma 3.26. For i, j ≥ 0,

(3.57)  π_{13}(m_{i,j}) ≥ ⌊\frac{1}{14}(13j − 7i + 13)⌋,

where the matrix M = (m_{i,j}) is defined in Theorem 3.20.

Proof. As noted in [6] we observe that

(3.58)  π_{13}(ψ_{r,s}) ≥ ⌊\frac{1}{14}(13s − 7r + 13)⌋,

for all 1 ≤ r ≤ 13 and 1 ≤ s ≤ 7. We verify the result for 0 ≤ i ≤ 12 by direct computation using the recurrence (3.51). We use (3.58), the recurrence (3.51) and Lemma 3.6 to prove the general result by induction. □

Corollary 3.27.

π_{13}(a_{i,j}) ≥ ⌊\frac{1}{14}(13j − i + 13)⌋, π_{13}(b_{i,j}) ≥ ⌊\frac{1}{14}(13j − i + 6)⌋,

where the a_{i,j}, b_{i,j} are defined by (3.50).

We provide more complete details for the proof of the following lemma since congruences for the spt-function modulo 13 are stronger than those for the partition function. Lemma 3.28.

(3.59)  π_{13}(x_{2,0}) = 1,
(3.60)  π_{13}(x_{2,j}) ≥ 3 + ⌊\frac{1}{14}(13j)⌋  for j ≥ 1
(3.61)  π_{13}(x_{2b−1,j}) ≥ 2b − 2 + ⌊\frac{1}{14}(13j − 10)⌋  for b ≥ 2, and j ≥ 1
(3.62)  π_{13}(x_{2b,j}) ≥ 2b − 1 + ⌊\frac{1}{14}(13j)⌋  for b ≥ 2, and j ≥ 1.

Proof. We have calculated x⃗_{2} and verified (3.59)–(3.60). We note that x_{2,j} = 0 for j > 91. Now,

x_{3,j} = ∑_{i≥0} x_{2,i}b_{i,j},

and we note that x_{3,0} = 0. We have

π_{13}(x_{2,0}b_{0,j}) = 1 + π_{13}(b_{0,j}) ≥ 2 + ⌊\frac{1}{14}(13j − 8)⌋

by Corollary 3.27. For i ≥ 1

π_{13}(x_{2,i}b_{i,j}) = π_{13}(x_{2,i}) + π_{13}(b_{i,j}) ≥ 3 + ⌊\frac{1}{14}(13i)⌋ + ⌊\frac{1}{14}(13j − i + 6)⌋
≥ 3 + ⌊\frac{1}{14}(13j + 12i − 7)⌋ ≥ 2 + ⌊\frac{1}{14}(13j − 9)⌋,

again by Corollary 3.27. It follows that

π_{13}(x_{3,j}) ≥ 2 + ⌊\frac{1}{14}(13j − 9)⌋,

and (3.61) holds for b = 2. Now supposed b ≥ 2 is fixed and that (3.61) holds. We have

x_{2b,j} = ∑_{i≥1} x_{2b−1,i}a_{i,j}.

Now

π_{13}(x_{2b−1,1}a_{1,j}) = π_{13}(x_{2b−1,1})+π_{13}(a_{1,j}) ≥ 2b−2+π_{13}(a_{1,j}) ≥ 2b−1+⌊\frac{1}{14}(13j)⌋,

by a direct calculation noting that a_{1,j} = 0 for j > 13. For i ≥ 2

π_{13}(x_{2b−1,i}a_{i,j}) = π_{13}(x_{2b−1,i}) + π_{13}(a_{i,j}) ≥ 2b − 2 + ⌊\frac{1}{14}(13i − 10)⌋ + ⌊\frac{1}{14}(13j − i + 13)⌋
≥ 2b − 2 + ⌊\frac{1}{14}(13j + 12i − 10)⌋ ≥ 2b − 1 + ⌊\frac{1}{14}(13j)⌋,

again by Corollary 3.27. It follows that

π_{13}(x_{2b,j}) ≥ 2b − 1 + ⌊\frac{1}{14}(13j)⌋,

and (3.62) holds. For i ≥ 1 Again suppose b ≥ 2 is fixed, and that (3.62) holds. We have

x_{2b+1,j} = ∑_{i≥1} x_{2b,i}b_{i,j}.

For i ≥ 1

π_{13}(x_{2b,i}b_{i,j}) = π_{13}(x_{2b,i}) + π_{13}(b_{i,j}) ≥ 2b − 1 + ⌊\frac{1}{14}(13i)⌋ + ⌊\frac{1}{14}(13j − i + 6)⌋
≥ 2b − 1 + ⌊\frac{1}{14}(13j + 12i − 8)⌋ ≥ 2b + ⌊\frac{1}{14}(13j − 10)⌋,

again by Corollary 3.27. It follows that

π_{13}(x_{2b+1,j}) ≥ 2b + ⌊\frac{1}{14}(13j − 10)⌋,

and (3.61) holds with b replaced by b + 1. Lemma 3.28 follows by induction. □

Corollary 3.29. For c ≥ 3,

(3.63)  a(13^{c}n + γ_{c}) − 13 · a(13^{c−2}n + γ_{c−2}) ≡ 0 (mod 13^{c−1}).

For a ≥ 1,

(3.64)  spt(13^{a+2}n + γ_{a+2}) − 13 · spt(13^{a}n + γ_{a}) ≡ 0 (mod 13^{a+1}),
(3.65)  spt(13^{a}n + γ_{a}) ≡ 0 (mod 13^{⌊ \frac{a+1}{2} ⌋}).

We note that (3.63) holds when c = 2 by taking γ_{0} = 1. Also when a = 0 the congruence (3.64) has a stronger form. The proof of the congruence

(3.66)  spt(169n − 7) − 13 · spt(n) ≡ 0 (mod 169).

is analogous to the proof of (3.24).

4. The spt-function modulo ℓ

In this section we improve on results in [16] and [10] for the spt-function and the second moment rank function modulo ℓ. We let

J_{ℓ}(z) = ∑_{n=s_{ℓ}}^{∞} j_{ℓ}(n)q^{n},

where J_{ℓ}(z) is defined in (2.21), and define

(4.1)  K_{ℓ}(z) := G_{ℓ}(z) + (−1)^{\frac{1}{2} (ℓ−1)} ℓ ∑_{n=⌈ \frac{s_{ℓ}}{ℓ} ⌉}^{∞} j_{ℓ}(ℓn)q^{n},

where G_{ℓ}(z) is defined in (2.12). Then we have

Theorem 4.1. If ℓ ≥ 5 is prime, then K_{ℓ}(z) is an entire modular form of weight (ℓ + 1) on the full modular group SL_{2}(ℤ).

Proof. Suppose ℓ ≥ 5 is prime. We utilize Serre’s [23, pp.223–224] results on the trace of a modular form on Γ_{0}(ℓ). By Theorem 2.2 we know that G_{ℓ}(z) is an entire modular form of weight (ℓ + 1) on Γ_{0}(ℓ). By [23, Lemma 7],

(4.2)  Tr(G_{ℓ}) = G_{ℓ} + ℓ^{1− \frac{1}{2} (ℓ+1)} G_{ℓ} | W | U

is an entire modular form of weight (ℓ + 1) on SL_{2}(ℤ). See [23, pp.223–224] for definition of W, U and the notation used. From (2.19) we find that

(4.3)  G_{ℓ} | W = (−1)^{\frac{1}{2} (ℓ−1)} ℓ^{\frac{1}{2} (ℓ+1)} J_{ℓ}.

From (4.1), (4.2) and (4.3) we see that

K_{ℓ} = Tr(G_{ℓ})

is an entire modular form of weight (ℓ + 1) on SL_{2}(ℤ). □

We observed special cases of the following Corollary in [16, Theorem 6.1]. Corollary 4.2. Suppose ℓ ≥ 5 is prime. Then

(4.4)  ∑_{n=⌈ \frac{ℓ}{24} ⌉}^{∞} spt(ℓn − s_{ℓ})q^{n− \frac{ℓ}{24}} ≡ η(z)^{r_{ℓ}} L_{ℓ}(z) (mod ℓ)

for some integral entire modular form L_{ℓ}(z) on the full modular group of weight ℓ + 1 − 12⌈\frac{ℓ}{24}⌉, and where r_{ℓ} and s_{ℓ} are defined in (2.8).

Proof. Suppose ℓ ≥ 5 is prime. Since

(24n − 1) p(n) ≡ 0 (mod ℓ),

for 24n ≡ 1 (mod ℓ), and using Theorem 4.1 we have

\frac{η(z)^{2ℓ}}{η(ℓz)} ∑_{n=0}^{∞} a(ℓn − s_{ℓ})q^{n− \frac{ℓ}{24}} ≡ P_{ℓ}(z) (mod ℓ),

for some integral P_{ℓ}(z) ∈ M_{ℓ+1}(Γ(1)). We note that

spt(ℓn − s_{ℓ}) ≠ 0

implies that ℓn − s_{ℓ} ≥ 1 and n ≥ ⌈\frac{ℓ}{24}⌉. It follows that

\frac{η(z)^{2ℓ}}{η(ℓz)} ∑_{n=⌈ \frac{ℓ}{24} ⌉}^{∞} spt(ℓn − s_{ℓ})q^{n− \frac{ℓ}{24}} ≡ Δ(z)^{c} L_{ℓ}(z) (mod ℓ),

where Δ(z) is Ramanujan’s function

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

c = ⌈\frac{ℓ}{24}⌉ and L_{ℓ}(z) is some integral modular form in M_{ℓ+1−12c}(Γ(1)). Thus

∑_{n=⌈ \frac{ℓ}{24} ⌉}^{∞} spt(ℓn − s_{ℓ})q^{n− \frac{ℓ}{24}} ≡ η(z)^{24c−ℓ} L_{ℓ}(z) (mod ℓ),

and the result follows since

r_{ℓ} = 24c − ℓ.
□

We conclude the paper by improving a result in [10] for the second rank moment function. From (1.1)

(4.6)  N_{2}(n) = 2n p(n) − 2 spt(n).

We note that the analog of Corollary 4.2 holds for the partition function p(n) except the weight is 2 less. See either [16, Theorem 3.4] or [1, Theorem3]. This together with Corollary 4.2 and (4.6) implies Corollary 4.3. Suppose ℓ ≥ 5 is prime. Then

(4.7)  ∑_{n=⌈ \frac{ℓ}{24} ⌉}^{∞} N_{2}(ℓn − s_{ℓ})q^{n− \frac{ℓ}{24}} ≡ η(z)^{r_{ℓ}} (Q_{ℓ}(z) + L_{ℓ}(z)) (mod ℓ)

for some integral entire modular forms Q_{ℓ}(z) and L_{ℓ}(z) on the full modular group of weights k and k + 2 respectively where k = ℓ − 1 − 12⌈\frac{ℓ}{24}⌉.

We illustrate Theorem 4.1 and Corollaries 4.2 and 4.3 in the case ℓ = 17. We find that

K_{17}(z) = G_{17}(z) + 17 ∑_{n=1}^{∞} j_{17}(17n)q^{n} = −17 E_{6}(z)^{3} − 26148 Δ(z) E_{6}(z),

∑_{n=0}^{∞} spt(17n + 5)q^{n+ \frac{7}{24}} ≡ 14 η(z)^{7} E_{6}(z) (mod 17),

and

∑_{n=0}^{∞} N_{2}(17n + 5)q^{n+ \frac{7}{24}} ≡ η(z)^{7} (2 E_{4}(z) + 6 E_{6}(z)) (mod 17).

Here E_{4}(z) and E_{6}(z) are the usual Eisenstein series

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

where σ_{k}(n) = ∑_{d|n} q^{k}. Acknowledgements I would like to thank Gerard Garvan, Zachary Kent, Karl Mahlburg and Ken Ono for their comments and suggestions.

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

E-mail address: fgarvan@ufl.edu
URL: http://www.math.ufl.edu/ fgarvan