MODULAR FUNCTIONS, MAPLE AND ANDREWS’S 10TH PROBLEM

FRANK GARVAN

Abstract. We discuss how Andrews’s 10th problem for generalized Frobenius partitions can be attacked using modular functions and a little help from Maple. Both elementary and function theoretic methods are employed. A multiplicative analogue of the Hecke operator U_{q}^{∗} is constructed.

1. Introduction

Given a partition π we may form the corresponding Frobenius symbol as follows: Firstly, we draw the Ferrers diagram of the partition and circle the main diagonal; then we read the number of nodes in each node to the right of the diagonal and the number of nodes in each column below the main diagonal. In this way we obtain two equal length sequences of decreasing nonnegative integers giving the desired Frobenius symbol:

( \begin{matrix}a_{1} & a_{2} & a_{3} & ⋯ & a_{r} \\ b_{1} & b_{2} & b_{3} & ⋯ & b_{r} \end{matrix} ) .

This construction is illustrated below in Figure 1, using the partition π = 7 + 7 + 5 + 4 + 4 + 2 + 2 + 1.

The corresponding Frobenius symbol is

( \begin{matrix}6 & 5 & 2 & 0 \\ 7 & 5 & 2 & 1 \end{matrix} ) .

Observe the number being partitioned can be obtained by adding the sum of the entries to the length of the symbol:

32 = 4 + (6 + 5 + 2 + 0) + (7 + 5 + 2 + 1).

One nice feature of the Frobenius symbol is that conjugation corresponds to swapping the rows of the symbol. In [2], Andrews considered generalizations of the Frobenius symbol, which originated in the solution of certain problems in statistical mechanics [3]. A generalized

Fig. 1 Ferrers graph of π

Frobenius partition (or more simply an F -partition) of n has the form

( \begin{matrix}a_{1} & a_{2} & a_{3} & ⋯ & a_{r} \\ b_{1} & b_{2} & b_{3} & ⋯ & b_{r} \end{matrix} ) ,

where

a_{1} ≥ a_{2} ≥ ⋯ ≥ a_{r} ≥ 0,
b_{1} ≥ b_{2} ≥ ⋯ ≥ b_{r} ≥ 0,

and

n = r + ∑_{i=1}^{r} a_{i} + ∑_{i=1}^{r} b_{i}.

So a generalized Frobenius partition resembles an ordinary Frobenius symbol except repetition of parts is allowed.

Let k > 0. We define ϕ_{k}(n) to be the number of F -partitions of n in which entries on each row are repeated at most k times. According to Frobenius

ϕ_{1}(n) = p(n),

the number of unrestricted partitions of n. We define

Φ_{k}(q) := ∑_{n≥0} ϕ_{k}(n)q^{n}.

Example: There are five F -partitions of 3 with k = 2:

( \begin{matrix}2 \\ 0 \end{matrix} ) , ( \begin{matrix}0 \\ 2 \end{matrix} ) , ( \begin{matrix}1 \\ 1 \end{matrix} ) , ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) , ( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} ) ,

and

Φ_{2}(q) = 1 + q + 3q^{2} + 5q^{3} + 9q^{4} + · · · .

At the end of [2] Andrews proposed ten open problems. Many of these problems have been solved. See [9] and [4]. The final and most bewildering problem is given below.

Problem 10, [2]. Define integers a_{n} uniquely by

Φ_{4}(q) = ∏_{n=1}^{∞}(1 − q^{n})^{−a_{n}}.

Then

(1.1)  a_{10n−5} = 2.

for all n. All other a’s seem to become large in absolute value for large subscript.

In this paper we prove (1.1) and give some related conjectures (see § 6). In § 2 we show how (1.1) may be reduced to the identity

(1.2)  \frac{X(q)}{X(−q)} = ∏_{n=1}^{∞} [\frac{(1 − q^{10n−5})(1 + q^{2n−1})^{5}}{(1 − q^{2n−1})^{5}(1 + q^{10n−5})}]^{2} ,

where

(1.3)  X(q) := \frac{Φ(q^{1/5})Φ(ζq^{1/5}) ⋯ Φ(ζ^{4}q^{1/5})}{Φ(q)},

Φ_{4}(q) = Φ(q), and ζ = exp(2πi/5).

We prove (1.2) by rewriting it as an equation between two modular functions on Γ_{0}(40) (see equation (5.4)) and then checking that enough of their Fourier coefficients agree.

1.1. Notation. We use the standard notation

(a)_{∞} = (a; q)_{∞} = ∏_{n=1}^{∞}(1 − aq^{n−1}),

where |q| < 1. Let q := exp(2πiτ) (with ℑτ > 0). As usual the Dedekind eta function is defined as

(1.4)  η(τ) := exp(πiτ/12) ∏_{n=1}^{∞}(1 − exp(2πinτ))

= q^{1/24} ∏_{n=1}^{∞}(1 − q^{n}). (1.5)

2. Reduction to a q-identity

In this section we show that the main result (1.1) is equivalent to a q-series identity (2.5). We let

Φ(q) := Φ_{4}(q) = ∏_{n=1}^{∞} \frac{1}{(1 − q^{n})^{a_{n}}}

= ∏_{n≢0 (mod 5)} \frac{1}{(1 − q^{n})^{a_{n}}} ∏_{n=1}^{∞} \frac{1}{(1 − q^{5n})^{a_{5n}}}

Now we let ζ = exp(2πi/5) so that

Φ(q)Φ(ζq)Φ(ζ^{2}q)Φ(ζ^{3}q)Φ(ζ^{4}q)

= ∏_{n≢0 (mod 5)} \frac{1}{(1 − q^{5n})^{a_{n}}} ∏_{n=1}^{∞} \frac{1}{(1 − q^{5n})^{5a_{5n}}}

We find that

(2.1) X(q) := \frac{Φ(q^{1/5})Φ(ζq^{1/5}) ⋯ Φ(ζ^{4}q^{1/5})}{Φ(q)} = ∏_{n=1}^{∞} [\frac{(1 − q^{5n})}{(1 − q^{n})^{5}}]^{a_{5n}}

= ∏_{n=1}^{∞} [\frac{(1 − q^{10n})}{(1 − q^{2n})^{5}}]^{a_{10n}} ∏_{n=1}^{∞} [\frac{(1 − q^{10n−5})}{(1 − q^{2n−1})^{5}}]^{a_{10n−5}} . (2.2)

Now we let

d_{n} = a_{10n−5},

for all n ≥ 1. We have

(2.3) \begin{aligned}\frac{X(q)}{X(−q)} & = & ∏_{n=1}^{∞} [\frac{(1 − q^{10n−5})(1 + q^{2n−1})^{5}}{(1 − q^{2n−1})^{5}(1 + q^{10n−5})}]^{d_{n}} \\ & = & ∏_{n=1}^{∞} [\frac{(1 − q^{10n−5})^{2}}{(1 − q^{2n−1})^{10}} ]^{d_{n}}  ∏_{n=1}^{∞} [\frac{(1 − q^{4n−2})^{5}}{(1 − q^{20n−10})}]^{d_{n}} . \end{aligned}

Lemma 2.1 Let X(q) be defined as above. Then

(2.4)  a_{10n−5} = 2,

for all n, if and only if,

(2.5)  \frac{X(q)}{X(−q)} = ∏_{n=1}^{∞} [\frac{(1 − q^{10n−5})(1 + q^{2n−1})^{5}}{(1 − q^{2n−1})^{5}(1 + q^{10n−5})}]^{2}

Proof. We have already seen how (2.4) implies (2.5). Now we assume that (2.5) holds. We must show that

d_{n} = a_{10n−5} = 2,

for all n. By examining (2.3) we see that exponent of (1 − q^{10n−5}) in the product expansion of X(q)/X(−q) is

2d_{n} − 10d_{5n−2}.

Similarly, when k = 0, 1, 3 or 4 we see that the exponent of (1 − q^{2(5n−k)−1}) = (1 − q^{10n−2k−1}) is

−10d_{5n−k}.

Hence we find that (2.5) implies that

d_{n} = \{ \begin{matrix}2, & if n ≢ 3 (mod 5), \\ \frac{1}{5}(8 + d_{(n+2)/5}), & if n ≡ 3 (mod 5). \end{matrix}

The result follows by induction on n. □

3. A palatable form for the generating function

In this section we show how the generating function Φ_{4}(q) may be written in terms of theta functions. We let ζ = exp(2πi/5). From eq.

6 FRANK GARVAN

(1) in [2] we have

f(q) := (q)_{∞}^{4}Φ_{4}(q)

= ∑_{m_{1},m_{2},m_{3}=−∞}^{∞} ζ^{3m_{1}+2m_{2}+m_{3}}q^{m_{1}^{2}+m_{2}^{2}+m_{3}^{2}+m_{1}m_{2}+m_{2}m_{3}+m_{1}m_{3}}.

We observe that the exponent of q above can be written neatly as

m_{1}^{2} + m_{2}^{2} + m_{3}^{2} + m_{1}m_{2} + m_{2}m_{3} + m_{1}m_{3}

= \frac{1}{2}(m_{1} + m_{2})^{2} + \frac{1}{2}(m_{2} + m_{3})^{2} + \frac{1}{2}(m_{1} + m_{3})^{2}.

We change variables in the summation and find that

f(q) = ∑_{n_{1}+n_{2}+n_{3}≡0 (mod 2)} ζ^{2n_{1}+n_{3}}q^{\frac{1}{2} (n_{1}^{2}+n_{2}^{2}+n_{3}^{2})},

so that

(3.1)  2f(q^{2}) = ∑_{n_{1},n_{2},n_{3}} ζ^{2n_{1}+n_{3}}(−1)^{n_{1}+n_{2}+n_{3}}q^{(n_{1}^{2}+n_{2}^{2}+n_{3}^{2})}

+ ∑_{n_{1},n_{2},n_{3}} ζ^{2n_{1}+n_{3}}q^{(n_{1}^{2}+n_{2}^{2}+n_{3}^{2})}.

We need Jacobi’s triple product identity [1]

(3.2) ∑_{n=−∞}^{∞}(−1)^{n}z^{n}q^{n^{2}} = (zq; q^{2})_{∞}(q/z; q^{2})_{∞}(q^{2}; q^{2})_{∞}.

Thus we may write the first sum on the right side of (3.1) as

∑_{n_{1}}(−1)^{n_{1}}ζ^{2n_{1}}q^{n_{1}^{2}} ∑_{n_{2}}(−1)^{n_{2}}q^{n_{2}^{2}} ∑_{n_{3}}(−1)^{n_{3}}ζ^{n_{3}}q^{n_{3}^{2}}

= (ζ^{2}q; q^{2})_{∞}(q/ζ^{2}; q^{2})_{∞}(q^{2}; q^{2})_{∞}(q; q^{2})_{∞}^{2}(q^{2}; q^{2})_{∞}

(ζq; q^{2})_{∞}(q/ζ; q^{2})_{∞}(q^{2}; q^{2})_{∞}
= (q; q^{2})_{∞}^{2}(q^{5}; q^{10})_{∞}^{2}(q^{2}; q^{2})_{∞}^{3}.

Again via (3.2) we have

(q; q^{2})_{∞}(q^{4}; q^{4})_{∞} = (q; q^{4})_{∞}(q^{3}; q^{4})_{∞}(q^{4}; q^{4})_{∞}
= ∑_{n}(−1)^{n}q^{2n^{2}−n}
= ∑_{n} q^{8n^{2}−2n} − q ∑_{n} q^{8n^{2}+6n}.

Thus if we define

α(q) := ∑_{n} q^{4n^{2}+n}, β(q) := ∑_{n} q^{4n^{2}+3n};

then we find that

(q; q^{2})_{∞} = \frac{1}{(q^{4}; q^{4})_{∞}} [α(q^{2}) − qβ(q^{2})]

and after some simplification that

f(q) = (q)_{∞}^{4}Φ_{4}(q) = \frac{(q)_{∞}^{3}}{(q^{2}; q^{2})_{∞}(q^{10}; q^{10})_{∞}} \{α(q)α(q^{5}) + q^{3}β(q)β(q^{5})\} ,

and

(3.3)  Φ_{4}(q) = \frac{α(q)α(q^{5}) + q^{3}β(q)β(q^{5})}{(q)_{∞}(q^{2}; q^{2})_{∞}(q^{10}; q^{10})}.

As a side remark it turns out that

q^{3/8} {α(−q)α(−q^{5}) + q^{3}β(−q)β(−q^{5})} = η(4τ) η(5τ).

We omit the details.

4. Modular Functions

In this section we show how Φ_{4}(−q) can be written in terms of modular functions and set up the necessary results on modular functions to complete the proof.

4.1. Background theory. The necessary background theory of modular functions and modular forms may be found in [14], [16], [10] and [13]. Many of the results that we require are contained in [6], [7] and [8]. Let Γ(1) denote the full modular group and as usual let

Γ_{0}(N) := \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ(1) : c ≡ 0 (mod N)\} .

Let Γ be a subgroup of Γ(1) with finite index. For a modular function f on Γ and a cusp ζ the order of f (mod Γ) at ζ is denoted by ORD (f; ζ; Γ) and the invariant order of f at ζ is denoted by ord (f; ζ). We have

(4.1)  ORD (f; ζ; Γ) = κ(Γ; ζ) ord (f; ζ; Γ)

where κ(Γ; ζ) denotes the fan width of the cusp ζ (mod Γ), and

(4.2)  ord (f | A; ζ) = ord (f; Aζ),

for A ∈ Γ(1). Here we use the usual stroke operator notation

(f | A)(τ) := f(Aτ).

Any non-zero modular function f must satisfy the valence formula:

(4.3)  ∑_{s∈F} ORD (f; s; Γ) = 0,

where F is a fundamental set of Γ and the sum is taken over s with non-zero order so that this is a finite sum. For s ∈ H the order is interpreted in the usual sense. See [14] for more details. We shall use the valence formula to prove certain modular function identities. This is a standard technique.

4.2. Transformation Formulae. Following [15, p.6], [17] we define

θ_{01}(v, τ) := ∏_{n=1}^{∞}(1 − q^{n−1/2}e^{2πiv})(1 − q^{n−1/2}e^{−2πiv})(1 − q^{n}).

Here q = exp(2πiτ). For odd k we define

(4.4)  U_{k}(τ) := q^{k^{2}/16} \frac{θ_{01}(kτ, 8τ)}{θ_{01}(0, 8τ)}

= q^{k^{2}/16} ∏_{n=1}^{∞} \frac{(1 − q^{8n−4+k})(1 − q^{8n−4−k})}{(1 − q^{8n−4})^{2}}.

We remark that

(4.5)  α(−q) = ∑_{n}(−1)^{n}q^{4n^{2}+n} = θ_{01}(τ, 8τ),
(4.6)  β(−q) = ∑_{n}(−1)^{n}q^{4n^{2}+3n} = θ_{01}(3τ, 8τ).

Weber [17] has calculated transformation formulae for the thetafunctions θ_{ij} under the action of each element of the full modular group Γ. Using these formulae it is a simple matter to show that

(4.7)  U_{k}(τ) | A = exp(πik^{2}ab/8) U_{ka}(τ),

where A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(8).

We need to write Φ_{4}(q) in terms of a modular function. The function we need is given in the following lemma.

TOPICS IN NUMBER THEORY 9

Lemma 4.1 The function

(4.8)  F (τ) := U_{1}(τ)U_{3}(τ) [U_{1}(τ)U_{1}(5τ) − U_{3}(τ)U_{3}(5τ)]

is a modular function on the congruence subgroup Γ_{0}(40). Proof. Let A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(40). Then

A^{∗} = ( \begin{matrix}a & 5b \\ c/5 & d \end{matrix} ) ∈ Γ_{0}(8).

Hence

U_{1}(5τ) | A = (U_{1}(τ) | A^{∗}) (5τ) = exp(5abπi/8) U_{a}(5τ),

and

U_{1}(τ)U_{1}(5τ) | A = exp(3abπi/4) U_{a}(τ)U_{a}(5τ).

Similarly we find that

U_{3}(τ)U_{3}(5τ) | A = exp(3abπi/4) U_{3a}(τ)U_{3a}(5τ),

and

U_{1}(τ)U_{3}(τ) | A = exp(5abπi/4) U_{a}(τ)U_{3a}(τ).

Hence

F (τ) | A = U_{a}(τ)U_{3a}(τ) [U_{a}(τ)U_{a}(5τ) − U_{3a}(τ)U_{3a}(5τ)] .

From the definition (4.4) it is easy to show that

U_{k}(τ) = −U_{k+8}(τ), U_{k}(τ) = U_{−k}(τ),

for all integers k. The result

F (τ) | A = F (τ),

follows easily by considering the two cases a ≡ ±1 (mod 8) and a ≡ ±3 (mod 8). Thus F is a modular function on Γ_{0}(40). □

By (3.3), and using (4.4)–(4.6), (4.8) we have

(4.9)  Φ_{4}(−q) = q^{1/6} \frac{η^{2}(20τ)η^{7}(4τ)}{η(40τ)η(10τ)η^{5}(8τ)η^{3}(2τ)} F (τ).

4.3. Orders at cusps. We will need the following lemmas

Lemma 4.2, [8]. If (r, s) = 1, then the fan width of Γ_{0}(N) at \frac{r}{s} is

κ (Γ_{0}(N); \frac{r}{s}) = \frac{N}{(N, s^{2})}.

Lemma 4.3, [8]. If no square divides N other than 1 or 4, then a complete set of inequivalent cusps for Γ_{0}(N) is

\{\frac{1}{a} : a | n\} .

Lemma 4.4, [8]. If no square divides N other than 1 or 4, and if (r, s) = (r′, s′) = 1, then \frac{r}{s} and \frac{r′}{s′} are equivalent cusps modulo Γ_{0}(N) if and only if (s, N) = (s′, N).

Theorem 4.5 Suppose 4 | N and N has no other square divisors. If f(τ) is a modular function on Γ_{0}(N) then the function f(τ + \frac{1}{2}) is also a modular function on Γ_{0}(N). Further

(1)

(4.10)  ord (f(τ + \frac{1}{2}); 0) = \frac{1}{4}ord (f; \frac{1}{2}).

(2) If 4 | c then

(4.11)  ord (f(τ + \frac{1}{2}); \frac{1}{c}) = ord (f; \frac{1}{c}).

(3) If c ≡ 2 (mod 4) then

(4.12)  ord (f(τ + \frac{1}{2}); \frac{1}{c}) = 4 ord (f; \frac{2}{c}).

(4) If c is odd then

(4.13)  ord (f(τ + \frac{1}{2}); \frac{1}{c}) = \frac{1}{4} ord (f; \frac{1}{2c}).

Proof. Suppose 4 | N and N has no other square divisors, and suppose

f(τ) is a modular function on Γ_{0}(N). Let A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(N) so that 4 | c. Since ad − bc = 1, a ≡ d ≡ 1 (mod 2). We let

B = ( \begin{matrix}2 & 1 \\ 0 & 2 \end{matrix} ) and

C = BAB^{−1} = ( \begin{matrix}a + c/2 & \frac{1}{2}(d − a) + b − c/4 \\ c & d − c/2 \end{matrix} ) ,

so that C ∈ Γ_{0}(N) and BA = CB. We have

f(τ + \frac{1}{2}) | A = (f | B) | A = f | (BA)
= f | (CB) = (f | C) | B = f | B  (since f is a modular function on Γ_{0}(N))
= f(τ + \frac{1}{2}),

and f(τ + \frac{1}{2}) is a modular function on Γ_{0}(N).

We will use (4.2) with ζ = ∞ to prove (4.10)–(4.13). For a cusp

\frac{1}{c} we take A = ( \begin{matrix}1 & 1 \\ c & c + 1 \end{matrix} ) ∈ Γ(1). For such a fixed cusp we show

that there is a matrix D = ( \begin{matrix}g & h \\ 0 & 4/g \end{matrix} ) ∈ Γ(1), where g, h ∈ ℤ, g | 4

and such that C = BAD^{−1} ∈ Γ(1). In this case

C = ( \begin{matrix}\frac{1}{g}(2 + c) & \frac{g}{4}(3 + c) − \frac{h}{4}(2 + c) \\ \frac{2c}{g} & \frac{1}{2}((c + 1)g − ch) \end{matrix} ) ,

det(C) = 1 and BA = CD. Assuming C ∈ Γ(1) we have

f(τ + \frac{1}{2}) | A = (f | B) | A = (f | C) | D

so that

(4.14)  ord (f(τ + \frac{1}{2}); A∞) = \frac{g^{2}}{4} ord (f; C∞).

There are four cases. In each case it is a simple matter to verify that C ∈ Γ(1), and the corresponding desired result follows via Lemma 4.4 and equation (4.14). Case 1. 4 | c. We take g = 2, h = 1, and find that the cusp C∞ is equivalent to \frac{1}{c} and we have (4.11). Case 2. c ≡ 2 (mod 4). We take g = 4, h = 2, and find that the cusp C∞ is equivalent to \frac{1}{c/2} and we have (4.12). Case 3. c ≡ 1 (mod 4). We take g = 1, h = 4, and find that the cusp C∞ is equivalent to \frac{1}{2c} and we have (4.13). We also note that (4.10) follows in this case since 0 and 1 are equivalent cusps.

Case 4. c ≡ 3 (mod 4). We take g = 1, h = 2, and find that the cusp C∞ is equivalent to \frac{1}{2c} and we have the remaining case of (4.13).

This completes the proof. □

Let p be a prime and suppose p | N. Then the Hecke operator U_{p}^{∗} defined by

(4.15)  f | U_{p}^{∗} := ∑_{j=0}^{p−1} f (\frac{τ + j}{p} )

is well-known. The notation U_{p}^{∗} is due to Atkin and Lehner [5, p.138]. If p is a prime, p | N and f(τ) is a modular function on Γ_{0}(N) then f | U_{p}^{∗} is also a modular function on Γ_{0}(N). See [5, Lemma 6]. We need a multiplicative analog of this Hecke operator. For a modular function f(τ) and a prime p we define

(4.16)  f^{∗(p)}(τ) := ∏_{j=0}^{p−1} f (\frac{τ + j}{p} ) .

We do not use the “stroke” notation here since this is clearly not a linear operator. We use the notation ^{∗(p)} simply to indicate the dependence on p.

We have the following

Theorem 4.6 Suppose 4 | N, N has no other square divisors and p is a fixed odd prime divisor of N. If f(τ) is a modular function on Γ_{0}(N) then f^{∗(p)}(τ) is also a modular function on Γ_{0}(N). Further

(1) if p ∤ c then

(4.17)  ord (f^{∗(p)}; \frac{1}{c}) = p ord (f; \frac{1}{c}) + \frac{(p−1)}{p}ord (f; \frac{1}{pc}); and

(2) if p | c then

(4.18)  ord (f^{∗(p)}; \frac{1}{c}) = ord (f; \frac{1}{c}).

Proof. Suppose N, p satisfy the conditions of the theorem and f(τ) is a modular function on Γ_{0}(N). The proof that f^{∗(p)}(τ) is also a modular function on Γ_{0}(N) is completely analogous to the proof that the Hecke operator U_{p}^{∗} has the same property. See [5, Lemma 6].

By Lemma 4.3, we need only consider cusps of the form \frac{1}{c} where c | N. For 0 ≤ k ≤ p − 1 define

A_{k} = ( \begin{matrix}1 & k \\ 0 & p \end{matrix} )

and suppose c | N. Let

B = ( \begin{matrix}1 & 1 \\ c & c + 1 \end{matrix} ) .

For a = 1 or a = p we let

C = ( \begin{matrix}a & b \\ 0 & p/a \end{matrix} ) .

and

M = A_{k}BC^{−1} = ( \begin{matrix}\frac{1+kc}{a} & \frac{1}{p}((1 + kc + k)a − b(1 + kc)) \\ \frac{pc}{a} & a + ac − bc \end{matrix} ) ,

so that det(M) = 1. If M ∈ Γ(1) we have

f (\frac{τ+k}{p}) | B = f | A_{k}B = (f | M) | C.

Since C corresponds to a translation it follows that

(4.19) ord (f(\frac{τ+k}{p}); B∞) = ord (f(\frac{τ+k}{p}) | B; ∞)

= ord (f | MC; ∞) = \frac{a^{2}}{p} ord (f; M∞),

again assuming that M ∈ Γ(1). Case 1. p ∤ c. Suppose 0 ≤ k ≤ p − 1. There exists a unique value of k, such that kc ≡ −1 (mod p). We denote this k by k_{0}.

• Suppose k = k_{0}. Then we take a = p and b = 0 so that

ord (f(\frac{τ+k_{0}}{p}); \frac{1}{c}) = p ord (f; \frac{1}{c}),

since M∞ = \frac{(1+k_{0}c)/p}{c} is equivalent to the cusp \frac{1}{c} by Lemma 4.4.

• Suppose k ≠ k_{0}. Therefore p ∤ kc + 1 and we take a = 1. This time we choose b so that b(kc+1) ≡ 1+kc+k (mod p). Using (4.19) we find that

ord (f(\frac{τ+k}{p}); \frac{1}{c}) = \frac{1}{p} ord (f; \frac{1}{pc}),

since M∞ = \frac{1+k_{0}c}{pc} is equivalent to the cusp \frac{1}{pc} by Lemma 4.4. Hence we have

ord (f^{∗(p)}; \frac{1}{c}) = ∑_{k=0}^{p−1} ord (f(\frac{τ+k}{p}); \frac{1}{c}) = p ord (f; \frac{1}{c}) + \frac{(p−1)}{p} ord (f; \frac{1}{pc}),

which is (4.17). Case 2. p | c. Again we suppose 0 ≤ k ≤ p − 1. We take a = 1 and b = 1 + k and find

M = ( \begin{matrix}1 + kc & −k^{2}c/p \\ pc & 1 − kc \end{matrix} ) ∈ Γ(1).

Using (4.19) we find that

ord (f(\frac{τ+k}{p}); \frac{1}{c}) = \frac{1}{p} ord (f; \frac{1+kc}{pc}).

Now (1 + kc, pc) = (1, pc) = 1 and (pc, N) = (c, N) = c since p | c and p^{2} ∤ N. Therefore the two cusps \frac{1+kc}{pc} and \frac{1}{c} are equivalent by Lemma 4.4. For each k we have

ord (f(\frac{τ+k}{p}); \frac{1}{c}) = \frac{1}{p} ord (f; \frac{1}{c}),

and (4.18) follows. □

We will need results on the orders at cusps of eta-products and theta-functions. Newman [12] has found necessary and sufficient conditions under which an eta-product is a modular function on Γ_{0}(N). Let N > 0 be a fixed integer. Here an eta-product takes the form

(4.20)  f(τ) = ∏_{d|N} η(dτ)^{m_{d}},

where each d > 0 and m_{d} ∈ ℤ.

Theorem 4.7, [12]. The function f(τ) (given in (4.20)) is a modular function on Γ_{0}(N) if and only if

(1) ∑_{d|N} m_{d} = 0,

(2) ∑_{d|N} dm_{d} ≡ 0 (mod 24),

(3) ∑_{d|N} \frac{Nm_{d}}{d} ≡ 0 (mod 24), and

(4) ∏_{d|N} d^{|m_{d}|} is a square.

Ligozat [11] has computed the order of the eta-product f(τ) at the cusps of Γ_{0}(N).

Theorem 4.8, [11]. If the function f(τ) (given in (4.20)) is a modular function on Γ_{0}(N), then its order at the cusp s = \frac{b}{c} (assuming (b, c) = 1) is

(4.21)  ord (f(τ); s) = ∑_{d|N} \frac{(d, c)^{2}m_{d}}{24d} .

Biagoli [8] has computed the order of the theta function:

(4.22) f_{n,ρ}(τ) = q^{(n−2ρ)^{2}/(8n)} ∏_{m=1}^{∞}(1 − q^{mn−ρ})(1 − q^{mn−(ρ−n)})(1 − q^{mn}),

at any cusp. Here n, ρ ∈ ℤ with n ≥ 1 and n ∤ ρ.

Theorem 4.9, [8, Lemma 3.2, p.285]. The order at the cusp s = \frac{b}{c} (assuming (b, c) = 1) of the theta function f_{n,ρ}(τ) (defined above and assuming ρ ∤ n) is

(4.23)  ord (f_{n,ρ}(τ); s) = \frac{e^{2}}{2n} (\frac{bρ}{e} − [\frac{bρ}{e}] − \frac{1}{2})^{2} ,

where e = (n, c) and [ ] is the greatest integer function.

5. Proof of the main result

In section 2 we saw in Lemma 2.1 how the main result (2.4) is equivalent to the q-series identity (2.5). We now recast this identity in terms of modular functions and complete the proof.

Recall that F (τ) was defined in (4.8). We define

(5.1)  H(τ) := \frac{η^{6}(10τ)η^{10}(4τ)η^{20}(τ)}{η^{2}(20τ)η^{4}(5τ)η^{30}(2τ)}

Cusp s ord (F; s) ORD (F; s; Γ_{0}(40))
0  0  0
1/40  1  1
1/20  −1  −1
1/10  −1/2  −1
1/8  4/5  4
1/5  0  0
1/4  −3/5  −3
1/2  −1/5  −2

Table 1. Order of F (τ) at each inequivalent cusp of Γ_{0}(40)

and

(5.2)  Ξ(τ) := \frac{F^{∗}(τ)}{F (τ)}.

Using (1.4), (1.5), it is straightforward to show that H(τ) is the reciprocal of the function on the right side of (2.5), and is a modular function on Γ_{0}(40) by Theorem 4.7. From (4.9) we observe that the quotient of Φ(−q) and F (τ) is an even function of q. It follows that (2.5) is equivalent to the identity

(5.3)  \frac{Ξ(τ)}{Ξ(τ + \frac{1}{2})} = H(τ);

i.e.

(5.4)  F^{∗}(τ)F (τ + \frac{1}{2}) = H(τ) F (τ)F^{∗}(τ + \frac{1}{2}).

Thus we have been able to reduce our original problem to a modular function identity. Here F^{∗} = F^{∗(p)} with p = 5. Thus all the functions that appear in (5.4) are modular functions on Γ_{0}(40) by Theorems 4.5 and 4.6.

Using Lemma 4.1 and Theorem 4.9 we may calculate lower bounds for the order of F (τ) at each cusp of Γ_{0}(40). Using the techniques of [8] it is possible to calculate the exact order at each cusp with a little more work. We omit the details. These orders are given below in Table 1. A set of inequivalent cusps for Γ_{0}(40) is

\{0, \frac{1}{40}, \frac{1}{20}, \frac{1}{10}, \frac{1}{8}, \frac{1}{5}, \frac{1}{4}, \frac{1}{2}\} .

Cusp s ord (L; s) = ord (R; s) ORD
0  −1/20  −2
1/40  2  2
1/20  −2  −2
1/10  −1/2  −1
1/8  28/5  28
1/5  −1/8  −1
1/4  −22/5  −22
1/2  −7/5  −14

Table 2. Orders of L(τ) and R(τ) at each inequivalent cusp of Γ_{0}(40)

Observe that from Table 1 we see that

(5.5)  ∑_{s∈S} ORD (F; s; Γ_{0}(40)) = −2.

Here S is a set of inequivalent cusps for Γ_{0}(40). (See first column of Table 1). This means, by the valence formula (4.3), that F (τ) has two inequivalent zeros (counted with multiplicity) in the upper-half plane H. The author has found these zeros but this result and more general results will be left until a later paper.

We let L(τ) (resp. R(τ)) denote the modular function on the left (resp. right) side of (5.4). We will use the valence formula (4.3) to prove that L(τ) − R(τ) is identically zero. Since we know the exact order of F (τ) at each cusp we may compute the exact order of L(τ) and R(τ) using (4.1), and Theorems 4.5, 4.6 and 4.8. The results are given below in Table 2.

From the table we have

(5.6) ∑_{s∈S} ORD (L; s; Γ_{0}(40)) = ∑_{s∈S} ORD (R; s; Γ_{0}(40)) = −12,

and

(5.7)  ∑_{s∈S^{∗}} ORD (L − R; s; Γ_{0}(40)) ≥ −14.

Here S^{∗} = S \ {1/40}.

We note that ∞ is equivalent to the cusp 1/40. In view of the valence formula (4.3), our main result will follow if we can show

ORD (L − R; ∞; Γ_{0}(40)) ≥ 15.

We do this indirectly. Here it is easier to calculate a product expansion than a q-series expansion. Since κ(Γ_{0}(40), ∞) = 1, and ORD (F; ∞; Γ_{0}(40)) = ord (F; ∞) = ord (F^{∗}(τ + \frac{1}{2})) = 1 we need to show the q-expansion of both sides of (5.3) agree up through q^{13}. Now define a sequence of integers b_{n} by

(5.8)  ∏_{n=1}^{∞}(1 − q^{n})^{b_{n}} = q^{−3/8} (U_{1}(τ)U_{1}(5τ) − U_{3}(τ)U_{3}(5τ)) .

Using (4.8), the fact that

(5.9)  U_{1}(τ)U_{3}(τ) = q^{5/8} ∏_{n=1}^{∞} \frac{(1 − q^{2n−1})}{(1 − q^{8n−4})^{4}} ,

and an argument analogous to that used in the proof of Lemma 2.1, we need only show that

(5.10)  b_{10n−5} = 1,

for n = 1, 2, 3, 4, 5 and 6. We have been able to verify this using the symbolic algebra package MAPLE, thus proving our main result. In fact we have verified (5.10) for the first 20 values of n.

6. Conjectures

We ask whether the phenomenon of Andrews’s 10th Problem occurs in any other naturally occurring q-series. For fixed odd p we define two sequences a_{n} and b_{n} by

(6.1) ∏_{n=1}^{∞}(1 − q^{n})^{a_{n}} = q^{−(p+1)/16} (U_{1}(τ)U_{1}(pτ) + U_{3}(τ)U_{3}(pτ)) ,

and

(6.2) ∏_{n=1}^{∞}(1 − q^{n})^{b_{n}} = q^{−(p+1)/16} (U_{1}(τ)U_{1}(pτ) − U_{3}(τ)U_{3}(pτ)) .

Then we

Conjecture:

(1) If p ≡ 3 (mod 8) then a_{2p(2n+1)} = 0 for all n.

(2) If p ≡ 5 (mod 8) then a_{p(2n+1)} = b_{p(2n+1)} = 1 for all n.

(3) If p ≡ 7 (mod 8) then a_{2(2n+1)} = 0 for all n.

Andrews’s 10th Problem is equivalent to the case for b_{p(2n+1)} with p = 5. See (5.10). The conjectures for a_{n} and b_{n} have been checked for all odd p ≤ 91 and n ≤ 994, using Maple. For instance, p = 91 ≡ 3 (mod 8) and we have checked that for this p, a_{182} = a_{546} = a_{910} = 0. On the face of it, this is quite remarkable when we observe that the other terms seem to grow rapidly in absolute value. For example,

a_{909} = −92151123491267743803787143128
= −2^{3} · 11518890436408467975473392891.

Each of the conjectures corresponds to a modular function identity analogous to (5.4). Since there are infinitely many such conjectures the methods of this paper are not sufficient. An alternative approach is to show that functions on each side of a particular identity have the same order for each cusp and have the same zeros in the upper-half plane. This author has made some progress in this direction. More details and hopefully a proof of the conjectures will appear in a later paper.

References

[1] George E. Andrews. The theory of partitions. Addison-Wesley Publishing Co., Reading, Mass.-London-Amsterdam, 1976. Encyclopedia of Mathematics and its Applications, Vol. 2.

[2] George E. Andrews. Generalized Frobenius partitions. Mem. Amer. Math. Soc., 49(301):iv+44, 1984.

[3] George E. Andrews. Use and extension of Frobenius’ representation of partitions. In Enumeration and design (Waterloo, Ont., 1982), pages 51–65. Academic Press, Toronto, Ont., 1984.

[4] George E. Andrews. q-series: their development and application in analysis, number theory, combinatorics, physics, and computer algebra, volume 66 of CBMS Regional Conference Series in Mathematics. Published for the Conference Board of the Mathematical Sciences, Washington, D.C., 1986.

[5] A. O. L. Atkin and J. Lehner. Hecke operators on Γ_{0}(m). Math. Ann., 185:134–160, 1970.

[6] Bruce C. Berndt, Anthony J. Biagioli, and James M. Purtilo. Ramanujan’s “mixed” modular equations. J. Ramanujan Math. Soc., 1(1-2):46–70, 1986.

[7] Bruce C. Berndt, Anthony J. Biagioli, and James M. Purtilo. Ramanujan’s modular equations of “large” prime degree. J. Indian Math. Soc. (N.S.), 51:75–110 (1988), 1987.

[8] Anthony J. F. Biagioli. A proof of some identities of Ramanujan using modular forms. Glasgow Math. J., 31(3):271–295, 1989.

[9] F.G. Garvan. Generalizations of Dyson’s rank. PhD thesis, The Pennsylvania State University, 1986.

[10] Marvin I. Knopp. Modular functions in analytic number theory. Markham Publishing Co., Chicago, Ill., 1970.

[11] Gérard Ligozat. Courbes modulaires de genre 1. Société Mathématique de France, Paris, 1975. Bull. Soc. Math. France, Mém. 43, Supplément au Bull. Soc. Math. France Tome 103, no. 3.

[12] Morris Newman. Construction and application of a class of modular functions. II. Proc. London Math. Soc. (3), 9:373–387, 1959.

[13] Hans Rademacher. Topics in analytic number theory. Springer-Verlag, New York, 1973. Edited by E. Grosswald, J. Lehner and M. Newman, Die Grundlehren der mathematischen Wissenschaften, Band 169.

[14] Robert A. Rankin. Modular forms and functions. Cambridge University Press, Cambridge, 1977.

[15] Øystein Rødseth. Dissections of the generating functions of q(n) and q_{0}(n). Ȧrbok Univ. Bergen Mat.-Natur. Ser., 1969(12):12 pp. (1970).

[16] Bruno Schoeneberg. Elliptic modular functions: an introduction. Springer-Verlag, New York, 1974. Translated from the German by J. R. Smart and E. A. Schwandt, Die Grundlehren der mathematischen Wissenschaften, Band 203.

[17] Heinrich Weber. Lehrbuch der Algebra. Braunschweig, F. Vieweg und sohn,

1908. Vol. 3.

E-mail address: frank@math.ufl.edu

Department of Mathematics, University of Florida, Gainesville, FL 32611