GEORGIAN MATHEMATICAL JOURNAL: Vol. 5, No. 6, 1998, 545-564

ON THE NUMBER OF REPRESENTATIONS OF INTEGERS BY QUADRATIC FORMS IN TWELVE VARIABLES

G. LOMADZE

Abstract. A way of finding exact explicit formulas for the number of representations of positive integers by quadratic forms in 12 variables with integral coefficients is suggested.

Introduction

In [1] we derived a formula for the sum of singular series corresponding to the number of representations of positive integers by a positive primitive quadratic form

f = a_{1}(x_{1}^{2} + x_{2}^{2}) + a_{2}(x_{3}^{2} + x_{4}^{2}) + a_{3}(x_{5}^{2} + x_{6}^{2}) +
+ a_{4}(x_{7}^{2} + x_{8}^{2}) + a_{5}(x_{9}^{2} + x_{10}^{2}) + a_{6}(x_{11}^{2} + x_{12}^{2})  (1)

with integral coefficients a_{1}, a_{2}, …, a_{6}.

In [2, 3], several classes of entire modular forms of weight 6 are constructed for the congruence group Γ_{0}(4N) with Fourier coefficients in a simple arithmetical sense, which allows one sometimes to obtain explicit exact formulas for the number of representations of positive integers by quadratic forms of type (1).

In the present paper we have obtained such formulas for the forms

\begin{aligned}f_{1} & = x_{1}^{2} + ⋯ + x_{10}^{2} + 2(x_{11}^{2} + x_{12}^{2}), \\ f_{2} & = x_{1}^{2} + ⋯ + x_{6}^{2} + 2(x_{7}^{2} + ⋯ + x_{12}^{2}), \\ f_{3} & = x_{1}^{2} + x_{2}^{2} + 2(x_{3}^{2} + ⋯ + x_{12}^{2}), \\ f_{4} & = x_{1}^{2} + ⋯ + x_{8}^{2} + 2(x_{9}^{2} + x_{10}^{2}) + 4(x_{11}^{2} + x_{12}^{2}). \end{aligned}  (2)

545

1991 Mathematics Subject Classification. 11E20, 11E25.

Key words and phrases. Number of representations by quadratic forms, entire modular form, theta-functions with characteristics, singular series.

1072-947X/98/1100-0545$15.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1998 Plenum Publishing Corporation

546 G. LOMADZE

Namely, we prove the following formulas:

(I) r(n; f_{1}) = 4σ_{5}(u) − 8  ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4(x_{3}^{2}+x_{4}^{2})=2n \\ 2∤x_{1},2∤x_{2},x_{1}>0,x_{2}>0 \end{matrix}}x_{1}^{4} − 3x_{1}^{2}x_{2}^{2} if n ≡ 1 (mod 2),

= 128σ_{5}(u) + 56 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=2n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,··· ,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}

if n ≡ 2 (mod 4),

= \frac{72}{31} (2^{5(α−1)}55 + 7)σ_{5}(u) if n ≡ 0 (mod 4);

(II) r(n; f_{2}) = σ_{5}(u) − 6 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4(x_{3}^{2}+x_{4}^{2})=2n \\ 2∤x_{1},2∤x_{2},x_{1}>0,x_{2}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2} −

− ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=4n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,··· ,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}

if n ≡ 1 (mod 2),

= 32σ_{5}(u) + 40 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=2n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,··· ,x_{4}>0 \end{matrix}}(\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}

if n ≡ 2 (mod 4),

= \frac{24}{31} (2^{5(α−1)}41 + 21)σ_{5}(u) if n ≡ 0 (mod 4);

(III) r(n; f_{3}) = \frac{1}{4} σ_{5}(u) − 2 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4(x_{3}^{2}+x_{4}^{2})=2n \\ 2∤x_{1},2∤x_{2},x_{1}>0,x_{2}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2} −

− \frac{1}{4} ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=4n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,··· ,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}

if n ≡ 1 (mod 2),

= 8σ_{5}(u) + 16 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=2n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,··· ,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}

if n ≡ 2 (mod 4),

= \frac{24}{31} (2^{5α−4}5 + 21)σ_{5}(u) if n ≡ 0 (mod 4);

(IV) r(n; f_{4}) = σ_{5}(u) + 15 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=4n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,··· ,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}

if n ≡ 1 (mod 4),

= σ_{5}(u) + 15 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=4n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,··· ,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4} −

− 256 ∑_{\begin{matrix}2(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}=2n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,··· ,x_{4}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2} +

+ 8 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4(x_{3}^{2}+x_{4}^{2})=2n \\ 2∤x_{1},2∤x_{2},x_{1}>0,x_{2}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2} if n ≡ 3 (mod 4),

= 32σ_{5}(u) − 56 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4(x_{3}^{2}+x_{4}^{2})=n \\ 2∤x_{1},2∤x_{2},x_{1}>0,x_{2}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2} −

− 28 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=2n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,··· ,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}

if n ≡ 2 (mod 4),

= 1024σ_{5}(u) + 568 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,··· ,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}

if n ≡ 4 (mod 8),

= \frac{8}{21} (2^{5(α−2)}3967 + 63)σ_{5}(u) if n ≡ 0 (mod 8).

1. Some Known Results

1.1. In the present paper N, a, k, n, q, r denote positive integers; b, u, v are odd positive integers; p is a prime number; c, g, h, j, m, α, β, γ, δ are integers; i is the imaginary unit; z, τ are complex variables (Im τ > 0); e(z) = exp 2πiz; Q = e(τ); (\frac{h}{u}) is the generalized Jacobi symbol; ∑_{hmodq} and

\begin{matrix}∑ & ′ \\ hmodq \end{matrix} denote sums in which h runs through the complete and the reduced

residue system modulo q, respectively; σ_{5}(u) is the sum of the fifth powers of positive divisors of u.

548 G. LOMADZE

Let

ϑ_{gh}(z|τ; c, N) =
= ∑_{m≡c (mod N)}(−1)^{h(m−c)/N}e(\frac{1}{2N} (m + \frac{g}{2})^{2}τ)e((m + \frac{g}{2})z); (1.1)

hence

\frac{∂^{n}}{∂z^{n}} ϑ_{gh}(z|τ; c, N) = (πi)^{n} ∑_{m≡c (mod N)}(−1)^{h(m−c)/N}(2m + g)^{n} ×
× e(\frac{1}{2N} (m + \frac{g}{2})^{2}τ)e((m + \frac{g}{2})z).  (1.2)

Let

ϑ_{gh}(τ; c, N) = ϑ_{gh}(0|τ; c, N), ϑ_{gh}^{(n)}(τ; c, N) = \frac{∂^{n}}{∂z^{n}} ϑ_{gh}(z|τ; c, N)|_{z=0}. (1.3)

We have (see, e.g., [4], formulas (2.3), (2.5) on page 112)

\begin{aligned}ϑ_{g+2j,h}(τ; c, N) & = ϑ_{gh}(τ; c + j, N), \\ ϑ_{g+2j,h}^{(n)}(τ; c, N) & = ϑ_{gh}^{(n)}(τ; c + j, N); \end{aligned}  (1.4)
\begin{aligned}ϑ_{gh}(τ; c + N_{j}, N) & = (−1)^{hj}ϑ_{gh}(τ; c, N), \\ ϑ_{gh}^{(n)}(τ; c + N_{j}) & = (−1)^{hj}ϑ_{gh}(τ; c, N). \end{aligned}  (1.5)

In particular, according to (1.3), it follows from (1.1) that

ϑ_{gh}(τ; 0, N) = ∑_{m=−∞}^{∞}(−1)^{hm}Q^{(2Nm+g)^{2}/8N},  (1.6)
ϑ_{gh}^{(n)}(τ; 0, N) = (πi)^{n} ∑_{m=−∞}^{∞}(−1)^{hm}(2Nm + g)^{n}Q^{(2Nm+g)^{2}/8N}.  (1.7)

Expressions (1.6) and (1.7) yield

ϑ_{−g,h}(τ; 0, N) = ϑ_{gh}(τ; 0, N), ϑ_{−g,h}^{(n)}(τ; 0, N) = (−1)^{n}ϑ_{gh}(τ; 0, N). (1.8)

Throughout this paper, Δ = ∏_{k=1}^{6} a_{k}^{2} is the determinant of the quadratic form (1) and a is the least common multiple of its coefficients.

Denoting by r(n; f) the number of representations of n by the form (1), we get

∏_{k=1}^{6} ϑ_{00}^{2}(τ; 0, 2a_{k}) = 1 + ∑_{n=1}^{∞} r(n; f)Q^{n}.  (1.9)

Further, put

Θ(τ; f) = 1 + ∑_{n=1}^{∞} ρ(n; f)Q^{n},  (1.10)

where ρ(n; f) is the sum of singular series corresponding to the function r(n; f).

Finally, let

Γ_{0}(4N) = \{\frac{ατ + β}{γτ + δ} | αδ − βγ = 1, γ ≡ 0 (mod 4N)\}

(nonhomogeneous congruence subgroup modulo 4N).

1.2. For convenience we shall quote here some well-known results.

Lemma 1 (see, e.g., [4], Lemma 1, p. 114). The entire modular form F (τ) of weight r for the congruence group Γ_{0}(4N) is identically zero, if in the expansion into the series

F (τ) = ∑_{m=0}^{∞} C_{m}Q^{m}

we have C_{m} = 0 for all m ≤ \frac{r}{12} 4N ∏_{p|4N}(1 + \frac{1}{p}).

Lemma 2 ([5], Lemma 14, p. 18 and Remark to Lemma 18, p. 21).

The functions Θ(τ; f) and ∏_{k=1}^{6} ϑ_{00}^{2}(τ; 0, 2a_{k}) are entire modular forms of

weight 6 and character χ(δ) = (\frac{Δ}{|δ|}) for Γ_{0}(4a).

Lemma 3 ([2], Theorem 2, pp. 70, 73, and [3], Theorem 2, pp. 196, 197). For given N, the functions

Φ_{2}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =

= \{\frac{1}{N_{1}^{2}} ϑ_{g_{1}h_{1}}^{(4)}(τ; 0, 2N_{1})ϑ_{g_{2}h_{2}}(τ; 0, 2N_{2}) +

+\frac{1}{N_{2}^{2}} ϑ_{g_{2}h_{2}}^{(4)}(τ; 0, 2N_{2})ϑ_{g_{1}h_{1}}(τ; 0, 2N_{1}) −

−\frac{6}{N_{1}N_{2}} ϑ′′_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ′′_{g_{2}h_{2}}(τ; 0, 2N_{2})\} ∏_{k=3}^{4} ϑ_{g_{k}h_{k}}(τ; 0, 2N_{k}), (1.11)

Φ_{3}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =

= \{\frac{1}{N_{1}} ϑ′′_{g_{1}h_{1}}(τ; 0, 2N_{1})ϑ_{g_{2}h_{2}}(τ; 0, 2N_{2}) −

−\frac{1}{N_{2}} ϑ′′_{g_{2}h_{2}}(τ; 0, 2N_{2})ϑ_{g_{1}h_{1}}(τ; 0, 2N_{1})\} ∏_{k=3}^{4} ϑ′_{g_{k}h_{k}}(τ; 0, 2N_{k}), (1.12)
Φ_{4}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =
= ∏_{k=1}^{4} ϑ′_{g_{k}h_{k}}(τ; 0, 2N_{k}),  (1.13)

where

(a) 2 | g_{k}, N_{k} | N (k = 1, 2, 3, 4), 4 | N ∑_{k=1}^{4} \frac{h_{k}^{2}}{N_{k}}, 4 | ∑_{k=1}^{4} \frac{g_{k}^{2}}{4N_{k}};

(b) for all α and δ with αδ ≡ 1 (mod 4N),

(\frac{∏_{k=1}^{4} N_{k}}{|δ|} )Φ_{j}(τ; αg_{1}, …, αg_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) =

= (\frac{Δ}{|δ|})Φ_{j}(τ; g_{1}, …, g_{4}; h_{1}, …, h_{4}; 0, …, 0; N_{1}, …, N_{4}) (j = 2, 3, 4),

are entire modular forms of weight 6 and character χ(Δ) = (\frac{Δ}{|δ|}) for Γ_{0}(4N).

Remark. In what follows we shall write Φ_{2}(τ; g_{1}, …, g_{4}; N_{1}, …, N_{4}) instead of Φ_{2}(τ; g_{1}, …, g_{4}; 0, …, 0; N_{1}, …, N_{4}).

Lemma 4 ([1], Theorem, p. 383). Let n = 2^{α}uv, u = ∏_{p|n, p∤2Δ}p^{β}, v =

∏_{\begin{matrix}p|n, p|Δ \\ p>2 \end{matrix}} p^{β} (α ≥ 0, β ≥ 0). Then

ρ(n; f) = \frac{2^{5α+3}v^{5}}{Δ^{1/2}} χ_{2} ∏_{p|Δ, p>2} χ_{p} ∏_{p|Δ, p>2}(1 − p^{−6})^{−1}σ_{5}(u).

The values χ_{2} and χ_{p} are given in Lemmas 5 and 6 of [1].

2. Formulas for r(n; f)

2.1. We consider the quadratic forms f_{1}, f_{2}, and f_{3} from (2), whose determinants are respectively Δ = 2^{2}, Δ = 2^{6}, and Δ = 2^{10}.

Lemma 5. The functions

ψ(τ; f_{1}) = ϑ_{00}^{10}(τ; 0, 2)ϑ_{00}^{2}(τ; 0, 4) − Θ(τ; f_{1}) +
+ 2 \frac{1}{128π^{4}} X_{1}(τ) − \frac{7}{2} \frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4),  (2.1)
ψ(τ; f_{2}) = ϑ_{00}^{6}(τ; 0, 2)ϑ_{00}^{6}(τ; 0, 4) − Θ(τ; f_{2}) +

+ \frac{3}{2} \frac{1}{128π^{4}} X_{1}(τ) + \frac{1}{16} \frac{1}{16π^{4}} ϑ_{21}^{′4}(τ; 0, 2) −

− \frac{5}{2} \frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4),  (2.2)
ψ(τ; f_{3}) = ϑ_{00}^{2}(τ; 0, 2)ϑ_{00}^{10}(τ; 0, 4) − Θ(τ; f_{3}) +

+ \frac{1}{2} \frac{1}{128π^{4}} X_{1}(τ) + \frac{1}{64} \frac{1}{16π^{4}} ϑ_{21}^{′4}(τ; 0, 2) −

− \frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4),  (2.3)

where, for brevity, we put

X_{1}(τ) = Φ_{2}(τ; 4, 4, 0, 0; 2, 2, 2, 2),  (2.4)

are entire modular forms of weight 6 and character χ(δ) = 1 for Γ_{0}(8).

Proof. According to Lemma 2, the first two summands in (2.1)–(2.3) are entire modular forms of weight 6 and character χ(δ) = 1 for Γ_{0}(8).

In Lemma 3 put N = 2. It is then obvious that the other summands in (2.1)–(2.3) satisfy condition (a) of this lemma.

If αδ ≡ 1 (mod 8), then αδ ≡ 1 (mod 4), i.e.,

α ≡ ±1 (mod 4) and δ ≡ ±1 (mod 4), respectively.  (2.5)

It is easy to verify that the summands from (2.1)–(2.3) involve

(\frac{∏_{k=1}^{4} N_{k}}{|δ|} ) = 1 and (\frac{Δ}{|δ|}) = 1.  (2.6)

By (1.11), (1.4), (1.5), (2.5), and (1.8),

Φ_{2}(τ; 4α, 4α, 0, 0; 2, 2, 2, 2) =

= \{\frac{1}{2} ϑ_{4α,0}^{(4)}(τ; 0, 4)ϑ_{4α,0}(τ; 0, 4) − \frac{3}{2} ϑ_{4α,0}^{′′2}(τ; 0, 4)\}ϑ_{00}^{2}(τ; 0, 4) =

= \{\frac{1}{2} ϑ_{±4,0}^{(4)}(τ; 2(α ∓ 1), 4)ϑ_{±4,0}(τ; 2(α ∓ 1), 4) −

−\frac{3}{2} ϑ_{±4,0}^{′′2}(τ; 2(α ∓ 1), 4)\}ϑ_{00}^{2}(τ; 0, 4) =

= \{\frac{1}{2} ϑ_{±4,0}^{(4)}(τ; 0, 4)ϑ_{±4,0}(τ; 0, 4) − \frac{3}{2} ϑ_{±4,0}^{′′2}(τ; 0, 4)\}ϑ_{00}^{2}(τ; 0, 4) =

= \{\frac{1}{2} ϑ_{40}^{(4)}(τ; 0, 4)ϑ_{40}(τ; 0, 4) − \frac{3}{2} ϑ_{40}^{′′2}(τ; 0, 4)\}ϑ_{00}^{2}(τ; 0, 4) =

= Φ_{2}(τ; 4, 4, 0, 0; 2, 2, 2, 2).

Similarly, by (1.13), (1.4), (1.5), (2.5) and (1.8),

ϑ_{2α,1}^{′4}(τ; 0, 2) = ϑ_{±2,1}^{′4}(τ; α ∓ 1, 2) = ϑ_{±2,1}^{′4}(τ; 0, 2) = ϑ_{2,1}^{′4}(τ; 0, 2),
ϑ_{4α,1}^{′4}(τ; 0, 4) = ϑ_{±4,1}^{′4}(τ; 2(α∓)1, 4) = ϑ_{±4,1}^{′4}(τ; 0, 4) = ϑ_{4,1}^{′4}(τ; 0, 4)

Hence by (2.6) the functions X_{1}(τ), ϑ_{21}^{′4}(τ; 0, 2) and ϑ_{41}^{′4}(τ; 0, 4) satisfy also condition (b) from Lemma 3. Thus they are entire modular forms of weight 6 and character χ(δ) = 1 for Γ_{0}(8).

Theorem 1. ϑ_{00}^{10}(τ; 0, 2)ϑ_{00}^{2}(τ; 0, 4) =

= Θ(τ; f_{1}) − 2 \frac{1}{128π^{4}} X_{1}(τ) + \frac{7}{2} \frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4),  (2.7)

ϑ_{00}^{6}(τ; 0, 2)ϑ_{00}^{6}(τ; 0, 4) = Θ(τ; f_{2}) − \frac{3}{2} \frac{1}{128π^{4}} X_{1}(τ) −

−\frac{1}{16} \frac{1}{16π^{4}} ϑ_{21}^{′4}(τ; 0, 2) + \frac{5}{2} \frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4),  (2.8)

ϑ_{00}^{2}(τ; 0, 2)ϑ_{00}^{10}(τ; 0, 4) = Θ(τ; f_{3}) − \frac{1}{2} \frac{1}{128π^{4}} X_{1}(τ) −

−\frac{1}{64} \frac{1}{16π^{4}} ϑ_{21}^{′4}(τ; 0, 2) + \frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4).  (2.9)

Proof. By Lemma 1, the functions ψ(τ; f_{1}), ψ(τ; f_{2}), and ψ(τ; f_{3}) will be identically zero, if all coefficients by Q^{n} (n ≤ 6) in their expansion into the powers of Q are zero.

I. In Lemma 4, putting n = 2^{α}u, v = 1, Δ = 2^{2}, 2^{6}, and 2^{10}, we respectively obtain

ρ(n; f_{1}) = \frac{2^{5α+3}}{2} χ_{2}σ_{5}(u), ρ(n; f_{2}) = \frac{2^{5α+3}}{2^{3}} χ_{2}σ_{5}(u),

ρ(n; f_{3}) = \frac{2^{5α+3}}{2^{5}} χ_{2}σ_{5}(u).

In Lemma 5 from [1] we put γ_{6} = 1, γ_{5} = ⋯ = γ_{1} = 0, γ = 1, b_{1} = ⋯ = b_{6} = 1. Then

χ_{2} = 1 if α = 0, 1,
χ_{2} = 2^{−5α}{2^{5α} − 2(2^{5}(2^{5(α−2)} − 1)31^{−1} − 1)} =

= \frac{2 · 9}{2^{5α}31} (2^{5(α−1)}55 + 7) if α ≥ 2.

Thus

\begin{aligned}ρ(n; f_{1}) & = 4σ_{5}(u) if α = 0, \\ & = 128σ_{5}(u) if α = 1, \\ & = 72 · 31^{−1}(2^{5(α−1)}55 + 7)σ_{5}(u) if α ≥ 2. \end{aligned}  (2.10)

Calculating the values of ρ(n; f_{1}) by (2.10) for all n ≤ 6, by virtue of (1.10) we obtain

Θ(τ; f_{1}) =
= 1 + 4Q + 128Q^{2} + 976Q^{3} + 4104Q^{4} + 12504Q^{5} + 31232Q^{6} + · · · . (2.11)

Now in Lemma 5 from [1] put γ_{6} = γ_{5} = γ_{4} = 1, γ_{3} = γ_{2} = γ_{1} = 0, γ = 3, b_{1} = ⋯ = b_{6} = 1. Then

χ_{2} = 1 if α = 0, 1,

= 2^{−5α}{2^{5α} − 2^{3}(2^{5}(2^{5(α−2)} − 1)31^{−1} − 1)} =

= \frac{2^{3}3}{2^{5α}31} (2^{5(α−1)}41 + 21) if α ≥ 2.

Thus

\begin{aligned}ρ(n; f_{2}) & = σ_{5}(u) if α = 0, \\ & = 32σ_{5}(u) if α = 1, \\ & = 24 · 31^{−1}(2^{5(α−1)}41 + 21)σ_{5}(u) if α ≥ 2. \end{aligned}  (2.12)

Calculating the values of ρ(n; f_{2}) by (2.12) for all n ≤ 6, we get

Θ(τ; f_{2}) =
= 1 + Q + 32Q^{2} + 244Q^{3} + 1032Q^{4} + 3126Q^{5} + 7808Q^{6} + · · · . (2.13)

Finally, in Lemma 5 from [1] put γ_{6} = ⋯ = γ_{2} = 1, γ_{1} = 0, γ = 5, b_{1} = ⋯ = b_{6} = 1. Then

χ_{2} = 1 if α = 0, 1,

= 2^{−5α}{2^{5α} − 2^{5}(2^{5}(2^{5(α−2)} − 1)31^{−1} − 1)} =

= \frac{2^{5}3}{2^{5α}31} (2^{5α−4}5 + 21) if α ≥ 2.

Thus

\begin{aligned}ρ(n; f_{3}) & = \frac{1}{4} σ_{5}(u) if α = 0, \\ & = 8σ_{5}(u) if α = 1, \\ & = 24 · 31^{−1}(2^{5α−4}5 + 21)σ_{5}(u) if α ≥ 2. \end{aligned}  (2.14)

554 G. LOMADZE

Calculating the values of ρ(n; f_{3}) by (2.14) for all n ≤ 6, we get

Θ(τ; f_{3}) =

= 1 + \frac{1}{4} Q + 8Q^{2} + 61Q^{3} + 264Q^{4} + \frac{1563}{2} Q^{5} + 1952Q^{6} + · · · . (2.15)

II. It follows from (1.6) that

ϑ_{00}^{10}(τ; 0, 2)ϑ_{00}^{2}(τ; 0, 4) = 1 + 20Q + 184Q^{2} + 1040Q^{3} +

= 4104Q^{4} + 12344Q^{5} + 30560Q^{6} + · · · .  (2.16)

Equations (1.7) and (1.6) imply

ϑ_{40}^{(4)}(τ; 0, 4)ϑ_{40}(τ; 0, 4) =

= 256π^{4} ∑_{m=−∞}^{∞}(2m + 1)^{4}Q^{(2m+1)^{2}/2} ∑_{m=−∞}^{∞} Q^{(2m+1)^{2}/2} = (2.17)
= 256π^{4}(4Q)(1 + 82Q^{4} + 81Q^{8} + 62Q^{12} + ⋯),
ϑ_{40}^{′′2}(τ; 0, 4) = (16(πi)^{2} ∑_{m=−∞}^{∞}(2m + 1)^{2}Q^{(2m+1)^{2}/2})^{2} = (2.18)
= 256π^{4}(4Q)(1 + 18Q^{4} + 81Q^{8} + 50Q^{12} + ⋯),
ϑ_{00}^{2}(τ; 0, 4) = ( ∑_{m=−∞}^{∞} Q^{2m^{2}})^{2} =  (2.19)
= 1 + 4Q^{2} + 4Q^{4} + 4Q^{8} + 8Q^{10} + 4Q^{16} + · · · .

Thus, by (2.4) and (1.11),

X_{1}(τ) = \{\frac{1}{2} ϑ_{40}^{(4)}(τ; 0, 4)ϑ_{40}(τ; 0, 4) − \frac{3}{2} ϑ_{40}^{′′2}(τ; 0, 4)\}ϑ_{00}^{2}(τ; 0, 4) = (2.20)
= 128π^{4}(4Q){(1 + 82Q^{4} + 81Q^{8} + 626Q^{12} + ⋯) −
−3(1 + 18Q^{4} + 81Q^{8} + 50Q^{12} + ⋯)} ×  (2.20_{1})
×(1 + 4Q^{2} + 4Q^{4} + 4Q^{8} + 8Q^{10} + ⋯) =

=128π^{4}(4Q)( − 2− 8Q^{2}+ 20Q^{4}+ 112Q^{6}− 58Q^{8}− 664Q^{10}− 60Q^{12} + ⋯), i.e,

\frac{1}{128π^{4}} X_{1}(τ) = −8Q − 32Q^{3} + 80Q^{5} + 448Q^{7} −
− 232Q^{9} − 2656Q^{11} − 240Q^{13} + · · · .  (2.21)

From (1.7) we have

ϑ_{41}^{′4}(τ; 0, 4) = (4πi ∑_{m=−∞}^{∞}(−1)^{m}(2m + 1)Q^{(2m+1)^{2}/2})^{4} =  (2.22)
= 256π^{4}(16Q^{2})(1 − 3Q^{4} + 5Q^{12} + ⋯)^{4},

i.e.,

\frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4) = 16Q^{2} − 192Q^{6} + 864Q^{10} − 1408Q^{14} + · · · . (2.23)

From (1.6) we find that

ϑ_{00}^{6}(τ; 0, 2)ϑ_{00}^{6}(τ; 0, 4) = 1 + 12Q + 72Q^{2} + 304Q^{3} +
+ 1032Q^{4} + 2952Q^{5} + 7328Q^{6} + ⋯,  (2.24)

while (1.7) yields

ϑ_{21}^{′4}(τ; 0, 2) = (2πi ∑_{m=−∞}^{∞}(−1)^{m}(2m + 1)Q^{(2m+1)^{2}/4})^{4} =  (2.25)
= 16π^{4}(16Q)(1 − 3Q^{2} + 5Q^{6} − 7Q^{12} + ⋯)^{4},

i.e.,

\frac{1}{16π^{4}} ϑ_{21}^{′4}(τ; 0, 2) = 16Q − 192Q^{3} + 864Q^{5} − 1408Q^{7} −
− 1584Q^{9} + 8640Q^{11} − 6688Q^{13} + · · · .  (2.26)

It follows from (1.6) that

ϑ_{00}^{2}(τ; 0, 2)ϑ_{00}^{10}(τ; 0, 4) = 1 + 4Q + 24Q^{2} + 80Q^{3} +
+ 264Q^{4} + 728Q^{5} + 1760Q^{6} + ⋯  (2.27)

It is easy to verify that all coefficients of Q^{n} (n ≤ 6) in expansions of functions ψ(τ; f_{i}) (i = 1, 2, 3) into the powers of Q are zero. Thus identity (2.9) is proved.

Proof of formulas (I)–(III). Equating the coefficients by Q^{n} in both sides of identities (2.7)–(2.9), according to (1.9) and (1.10), we get

r(n; f_{1}) = ρ(n; f_{1}) − 2ν_{1}(n) + \frac{7}{2} ν̃_{1}(n),  (2.28)
r(n; f_{2}) = ρ(n; f_{2}) − \frac{3}{2} ν_{1}(n) − \frac{1}{16} ν_{11}(n) + \frac{5}{2} ν̃_{1}(n),  (2.29)
r(n; f_{3}) = ρ(n; f_{3}) − \frac{1}{2} ν_{1}(n) − \frac{1}{64} ν_{11}(n) + ν̃_{1}(n),  (2.30)

where ν_{1}(n), ν̃_{1}(n) and ν_{11}(n) denote the coefficients by Q^{n} in the expansion of the functions \frac{1}{128π^{4}} X_{1}(τ), \frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4), and \frac{1}{16π^{4}} ϑ_{21}^{′4}(τ; 0, 2) in powers of Q, respectively.

It follows from (2.20) and (2.17)–(2.19) that

\frac{1}{128π^{4}} X_{1}(τ) = ( ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{4}Q^{\{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}\}/2} −

−3 ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{2}(2m_{2} + 1)^{2}Q^{{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}}/2}) ×

× ∑_{m_{3},m_{4}=−∞}^{∞} Q^{2m_{3}^{2}+2m_{4}^{2}} = ∑_{n=1}^{∞} ( ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4(x_{3}^{2}+x_{4}^{2})=2n \\ 2∤x_{1},2∤x_{2} \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2})Q^{n},

i.e.,

ν_{1}(n) = 4 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4(x_{3}^{2}+x_{4}^{2})=2n \\ 2∤x_{1},2∤x_{2},x_{1}>0,x_{2}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2}.  (2.31)

Obviously, ν_{1}(n) ≠ 0 for n ≡ 1 (mod 2) only. From (2.22) we find that

\frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4) = ∑_{m_{1},...,m_{4}=−∞}^{∞}(−1)^{m_{1}+···+m_{4}}(2m_{1} + 1) ⋯ (2m_{4} + 1) ×

×Q^{{(2m_{1}+1)^{2}+···+(2m_{4}+1)^{2}}/2} =

= ∑_{n=1}^{∞} ( ∑_{\begin{matrix}x_{1}^{2}+···+x_{4}^{2}=2n \\ 2∤x_{1},...,2∤x_{4} \end{matrix}}(−1)^{(x_{1}−1)/2+···+(x_{4}−1)/2}x_{1}x_{2}x_{3}x_{4})Q^{n},

i.e.,

ν̃_{1}(n) = 16 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=2n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,...,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}. (2.32)

Obviously, ν̃_{1}(n) ≠ 0 for n ≡ 2 (mod 4) only. From (2.25) follows

\frac{1}{16π^{4}} ϑ_{21}^{′4}(τ; 0, 2) = ∑_{m_{1},...,m_{4}=−∞}^{∞}(−1)^{m_{1}+···+m_{4}}(2m_{1} + 1) ⋯ (2m_{4} + 1) ×

×Q^{{(2m_{1}+1)^{2}+···+(2m_{4}+1)^{2}}/4} =

= ∑_{n=1}^{∞} ( ∑_{\begin{matrix}x_{1}^{2}+···+x_{4}^{2}=4n \\ 2∤x_{1},...,2∤x_{4} \end{matrix}}(−1)^{(x_{1}−1)/2+···+(x_{4}−1)/2}x_{1}x_{2}x_{3}x_{4})Q^{n},

i.e,

ν_{11}(n) = 16 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=4n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,...,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}. (2.33)

Obviously, ν_{11}(n) ≠ 0 for n ≡ 1 (mod 2) only.

From (2.28), (2.10), (2.31), and (2.32) follows formula (I), from (2.29), (2.12), and (2.31)–(2.33) formula (II) and from (2.30), (2.14), and (2.31)– (2.33) formula (III).

2.2. Consider now the quadratic form

f_{4} = x_{1}^{2} + x_{2}^{2} + ⋯ + x_{8}^{2} + 2(x_{9}^{2} + x_{10}^{2}) + 4(x_{11}^{2} + x_{12}^{2})

whose determinant is Δ = 2^{6}.

Lemma 6. The function

ψ(τ; f_{4}) = ϑ_{00}^{8}(τ; 0, 2)ϑ_{00}^{2}(τ; 0, 4)ϑ_{00}^{2}(τ; 0, 8) − Θ(τ; f_{4}) −

− 2 \frac{1}{128π^{4}} X_{1}(τ) − \frac{15}{16} \frac{1}{16π^{4}} ϑ_{21}^{′4}(τ; 0, 2) + 16 \frac{1}{512π^{4}} X_{2}(τ) +

+ 2 \frac{1}{128π^{4}} X_{3}(τ) + \frac{7}{4} \frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4) +

+ 14 \frac{1}{512π^{4}} X_{4}(τ) − \frac{71}{2} \frac{1}{4096π^{4}} ϑ_{81}^{′4}(τ; 0, 8),

where

X_{1}(τ) = Φ_{2}(τ; 4, 4, 0, 0; 2, 2, 2, 2),  (2.34)
X_{2}(τ) = Φ_{2}(τ; 8, 8, 4, 4; 4, 4, 2, 2),  (2.35)
X_{3}(τ) = Φ_{2}(τ; 4, 4, 0, 0; 2, 2, 4, 4),  (2.36)
X_{4}(τ) = Φ_{2}(τ; 8, 8, 0, 0; 4, 4, 4, 4),  (2.37)

is entire modular form of weight 6 and character χ(δ) = 1 for Γ_{0}(16).

Proof. According to Lemma 2, the first two summands of the function ψ(τ; f_{4}) are entire modular forms of weight 6 and character χ(δ) = 1 for Γ_{0}(16).

In Lemma 3 put N = 4. Then it is obvious that the other summands of ψ(τ; f_{4}) satisfy condition (a) of this lemma.

If αδ ≡ 1 (mod 16), then αδ ≡ 1 (mod 4), i.e.,

α ≡ ±1 (mod 4) and δ ≡ ±1 (mod 4), respectively. (2.38)

It is easy to verify that all these summands involve

(\frac{∏_{k=1}^{4} N_{k}}{|δ|} ) = 1 and (\frac{Δ}{|δ|}) = 1.  (2.39)

Reasoning as in Lemma 5, by (1.11), (1.4), (1.5), (2.38) and (1.8), we get

Φ_{2}(τ; 8α, 8α, 4α, 4α; 4, 4, 2, 2) = Φ_{2}(τ; 8, 8, 4, 4; 4, 4, 2, 2),
Φ_{2}(τ; 4α, 4α, 0, 0; 2, 2, 4, 4) = Φ_{2}(τ; 4, 4, 0, 0; 2, 2, 4, 4),
Φ_{2}(τ; 8α, 8α, 0, 0; 4, 4, 4, 4) = Φ_{2}(τ; 8, 8, 0, 0; 4, 4, 4, 4),
ϑ_{8α,1}^{′4}(τ; 0, 8) = ϑ_{±8,1}^{′4}(τ; 4(α ∓ 1), 8) = ϑ_{±8,1}^{′4}(τ; 0, 8) = ϑ_{81}^{′4}(τ; 0, 8).

Similar relations for the other three summands of the function ψ(τ; f_{4}) have already been given in Lemma 5. Thus, according to (2.39), all these functions satisfy condition (b) from Lemma 3 as well. Consequently, they are entire modular forms of weight 6 and character χ(δ)=1 for Γ_{0}(16).

Theorem 2. ϑ_{00}^{8}(τ; 0, 2)ϑ_{00}^{2}(τ; 0, 4)ϑ_{00}^{2}(τ; 0, 8) =

= Θ(τ; f_{4}) + 2 \frac{1}{128π^{4}} X_{1}(τ) + \frac{15}{16} \frac{1}{16π^{4}} ϑ_{21}^{′4}(τ; 0, 2) −

−16 \frac{1}{512π^{4}} X_{2}(τ) − 2 \frac{1}{128π^{4}} X_{3}(τ) −

−\frac{7}{4} \frac{1}{256π^{4}} ϑ_{41}^{′4}(τ; 0, 4) − 14 \frac{1}{512π^{4}} X_{4}(τ) + \frac{71}{2} \frac{1}{4096π^{4}} ϑ_{81}^{′4}(τ; 0, 8). (2.40)

Proof. By Lemma 1, the function ψ(τ; f_{4}) will be identically zero if all coefficients by Q^{n} (n ≤ 12) in its expansion into the powers of Q are zero.

I. In Lemma 4 put n = 2^{α}u, v = 1, Δ = 2^{6}. Then ρ(n; f_{4}) = 2^{5α}χ_{2}σ_{5}(u). In Lemma 5 from [1] we put γ_{6} = 2, γ_{5} = 1, γ_{4} = ⋯ = γ_{1} = 0, γ = 3, b_{1} = ⋯ = b_{6} = 1. Then

χ_{2} = 1 if α = 0, 1, 2,

= 2^{−5α}{2^{5α} − 2^{3}(2^{5}(2^{5(α−3)} − 1)31^{−1} − 1)} =

= \frac{8}{2^{5α}31} (2^{5(α−2)}3967 + 63) if α ≥ 3.

ON THE NUMBER OF REPRESENTATIONS OF INTEGERS 559

Thus

ρ(n; f_{4}) = σ_{5}(u) if α = 0,

= 32σ_{5}(u) if α = 1, = 1024σ_{5}(u) if α = 2,

= \frac{8}{31} (2^{5(α−2)}3967 + 63)σ_{5}(u) if α ≥ 3.

(2.41)

Calculating the values of ρ(n; f_{4}) by (2.41) for all n ≤ 12, we get

Θ(τ; f_{4}) = 1 + Q + 32Q^{2} + 244Q^{3} + 1024Q^{4} + 3126Q^{5} +
+ 7808Q^{6} + 16808Q^{7} + 32776Q^{8} + 59293Q^{9} +
+ 100032Q^{10} + 161052Q^{11} + 249856Q^{12} + · · · .  (2.42)

II. From (1.6) follows

ϑ_{00}^{8}(τ; 0, 2)ϑ_{00}^{2}(τ; 0, 4)ϑ_{00}^{2}(τ; 0, 8) = 1 + 16Q + 116Q^{2} + +512Q^{3} + 1592Q^{4} + 3936Q^{5} + 8592Q^{6} + 17408Q^{7} + 32776Q^{8} +

+57808Q^{9} + 97400Q^{10} + 157184Q^{11} + 243040Q^{12} + · · · . (2.43)

Equations (1.7) and (1.6) imply

ϑ_{80}^{(4)}(τ; 0, 8)ϑ_{80}(τ; 0, 8) =

= 4096π^{4} ∑_{m=−∞}^{∞}(2m + 1)^{4}Q^{(2m+1)^{2}} ∑_{m=−∞}^{∞} Q^{(2m+1)^{2}} =  (2.44)
= 4096π^{4}(4Q^{2})(1 + 82Q^{8} + 81Q^{16} + ⋯),  (2.45)
ϑ_{80}^{′′2}(τ; 0, 8) = 4096π^{4}( ∑_{m=−∞}^{∞}(2m + 1)^{2}Q^{(2m+1)^{2}})^{2} =  (2.46)
= 4096π^{4}(4Q^{2})(1 + 18Q^{8} + 81Q^{16} + ⋯),  (2.47)
ϑ_{40}^{2}(τ; 0, 4) = ( ∑_{m=−∞}^{∞} Q^{(2m+1)^{2}/2})^{2} =  (2.48)
= 4Q(1 + 2Q^{4} + Q^{8} + 2Q^{12} + ⋯).  (2.49)

Thus, by (2.35), (1.11), (2.45), (2.47), and (2.49),

X_{2}(τ) = \{\frac{1}{8} ϑ_{80}^{(4)}(τ; 0, 8)ϑ_{80}(τ; 0, 8) − \frac{3}{8} ϑ_{80}^{′′2}(τ; 0, 8)\}ϑ_{40}^{2}(τ; 0, 4) = (2.50)
= 512π^{4}(4Q^{2}){(1 + 82Q^{8} + 81Q^{16} + ⋯) −

−3(1 + 18Q^{8} + 81Q^{16} + ⋯)} ×  (2.50_{1})
×(4Q)(1 + 2Q^{4} + Q^{8} + 2Q^{12} + ⋯) =
= 512π^{4}(16Q^{3})(−2 − 4Q^{4} + 26Q^{8} + 52Q^{12} + ⋯)

i.e.,

\frac{1}{512π^{4}} X_{2}(τ) = −32Q^{3} − 64Q^{7} + 416Q^{11} + 832Q^{15} + ⋯ (2.51)

From (1.6) we find that

ϑ_{00}^{2}(τ; 0, 8) = ( ∑_{m=−∞}^{∞} Q^{4m^{2}})^{2} =  (2.52)
= (1 + 2Q^{4} + 2Q^{16} + ⋯)^{2} = 1 + 4Q^{4} + 4Q^{8} + 4Q^{16} + ⋯ (2.53)

Thus, by (2.36), (1.11), (2.20_{1}), and (2.53),

X_{3}(τ) = \{\frac{1}{2} ϑ_{40}^{(4)}(τ; 0, 4)ϑ_{40}(τ; 0, 4) − \frac{3}{2} ϑ_{40}^{′′2}(τ; 0, 4)\}ϑ_{00}^{2}(τ; 0, 8) = (2.54)
= 128π^{4}(4Q){(1 + 82Q^{4} + 81Q^{8} + 626Q^{12} + ⋯) −
−3(1 + 18Q^{4} + 81Q^{8} + 50Q^{12} + ⋯)} ×
×(1 + 4Q^{4} + 4Q^{8} + 4Q^{16} + ⋯) =
= 128π^{4}(4Q)(−2 + 20Q^{4} − 58Q^{8} − 60Q^{12} + ⋯),

i.e.,

\frac{1}{128π^{4}} X_{3}(τ) = −8Q + 80Q^{5} − 232Q^{9} − 240Q^{13} + · · · . (2.55)

From (2.37), (1.11), (2.50_{1}), and (2.53) we obtain

X_{4}(τ) = \{\frac{1}{8} ϑ_{80}^{(4)}(τ; 0, 8)ϑ_{80}(τ; 0, 8) − \frac{3}{8} ϑ_{80}^{′′2}(τ; 0, 8)\}ϑ_{00}^{2}(τ; 0, 8) = (2.56)
= 512π^{4}(4Q^{2}){(1 + 82Q^{8} + 81Q^{16} + ⋯) −
−3(1 + 18Q^{8} + 81Q^{16} + ⋯)} ×
×(1 + 4Q^{4} + 4Q^{8} + 4Q^{16} + ⋯) =
= 512π^{4}(4Q^{2})(−2 − 8Q^{4} + 20Q^{8} + 112Q^{12} + ⋯),

i.e.,

\frac{1}{512π^{4}} X_{4}(τ) = −8Q^{2} − 32Q^{6} + 80Q^{10} + 448Q^{14} + · · · . (2.57)

From (1.7) we have

ϑ_{81}^{′4}(τ; 0, 8) = (8πi ∑_{m=−∞}^{∞}(−1)^{m}(2m + 1)^{(2m+1)^{2}})^{4} =  (2.58)
= 4096π^{4}(16Q^{4})(1 − 12Q^{8} + 54Q^{16} + ⋯)

i.e.,

\frac{1}{4096π^{4}} ϑ_{81}^{′4}(τ; 0, 8) = 16Q^{4} − 192Q^{12} + 864Q^{20} + · · · .  (2.59)

According to (2.43), (2.42), (2.21), (2.25), (2.51), (2.55), (2.23), (2.57), and (2.59), it is easy to verify that all coefficients of Q^{n} (n ≤ 12) in the expansion of the function ψ(τ; f_{4}) into the powers of Q are zero.

Thus identity (2.40) is proved.

Proof of formula (IV). Equating the coefficients by Q^{n} in both sides of identity (2.40), by (1.9) and (1.10) we obtain

r(n; f_{4}) = ρ(n; f_{4}) + 2ν_{1}(n) + \frac{15}{16} ν_{11}(n) − 16ν_{2}(n) −
− 2ν_{3}(n) − \frac{7}{4} ν̃_{1}(n) − 14ν̃_{2}(n) + \frac{71}{2} ν̃_{3}(n),  (2.60)

where ν_{1}(n), ν_{11}(n), ν_{2}(n), ν_{3}(n), ν̃_{1}(n), ν̃_{2}(n), and ν̃_{3}(n) denote the coefficients of Q^{n} in the expansion of the functions \frac{1}{128π^{4}} X_{1}(τ), \frac{1}{16π^{4}} ϑ_{21}^{′4}(τ; 0, 2), \frac{1}{512π^{4}} X_{2}(τ), \frac{1}{128π^{4}} X_{3}(τ), \frac{1}{256π^{4}} ϑ_{41}^{′4}(τ;0,4), \frac{1}{511π^{4}} X_{4}(τ), and \frac{1}{4096π^{4}} ϑ_{81}^{′4}(τ;0,8) into the powers of Q, respectively.

It follows from (2.50), (2.44), (2.46), and (2.49) that

\frac{1}{512π^{4}} X_{2}(τ) = ( ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{4}Q^{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}} −

− 3 ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{2}(2m_{2} + 1)^{2}Q^{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}}) ×

× ∑_{m_{3},m_{4}=−∞}^{∞} Q^{{(2m_{3}+1)^{2}+(2m_{4}+1)^{2}}/2} =

= ∑_{n=1}^{∞} ( ∑_{\begin{matrix}2(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}=2n \\ 2∤x_{1},...,2∤x_{4} \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2})Q^{n},

i.e.,

ν_{2}(n) = 16 ∑_{\begin{matrix}2(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}=2n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,...,x_{4}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2}.  (2.61)

Obviously, ν_{2}(n) ≠ 0 for n ≡ 3 (mod 4) only. From (2.54), (2.17), (2.18) and (2.52) we find that

\frac{1}{128π^{4}} X_{3}(τ) = ( ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{4}Q^{\{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}\}/2} −

−3 ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{2}(2m_{2} + 1)^{2}Q^{{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}}/2}) ×

× ∑_{m_{3},m_{4}=−∞}^{∞} Q^{4m_{3}^{2}+4m_{4}^{2}} = ∑_{n=1}^{∞} ( ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+8(x_{3}^{2}+x_{4}^{2})=2n \\ 2∤x_{1},2∤x_{2} \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2})Q^{n},

i.e.,

ν_{3}(n) = 4 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+8(x_{3}^{2}+x_{4}^{2})=2n \\ 2∤x_{1},2∤x_{2},x_{1}>0,x_{2}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2}.  (2.62)

Obviously, ν_{3}(n) ≠ 0 for n ≡ 1 (mod 4) only. From (2.56), (2.44), (2.46), and (2.52) follows

\frac{1}{512π^{4}} X_{4}(τ) = ( ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{4}Q^{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}} −

−3 ∑_{m_{1},m_{2}=−∞}^{∞}(2m_{1} + 1)^{2}(2m_{2} + 1)^{2}Q^{(2m_{1}+1)^{2}+(2m_{2}+1)^{2}}) ×

× ∑_{m_{3},m_{4}=−∞}^{∞} Q^{4m_{3}^{2}+4m_{4}^{2}} = ∑_{n=1}^{∞} ( ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4(x_{3}^{2}+x_{4}^{2})=n \\ 2∤x_{1},2∤x_{2} \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2})Q^{n},

i.e.,

ν̃_{2}(n) = 4 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4(x_{3}^{2}+x_{4}^{2})=n \\ 2∤x_{1},2∤x_{2},x_{1}>0,x_{2}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2}.  (2.63)

Obviously, ν̃_{2}(n) ≠ 0 for n ≡ 2 (mod 4) only.

From (2.58) we find that

\frac{1}{4096π^{4}} ϑ_{81}^{′4}(τ; 0, 8) = ∑_{m_{1},...,m_{4}=−∞}^{∞}(−1)^{∑_{k=1}^{4} m_{k}} ∏_{k=1}^{4}(2m_{k} + 1)Q^{∑_{k=1}^{4}(2m_{k}+1)^{2}} =

= ∑_{n=1}^{∞} ( ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=n \\ 2∤x_{1},...,2∤x_{4} \end{matrix}}(−1)^{∑_{k=1}^{4}(x_{k}−1)/2}x_{1}x_{2}x_{2}x_{4})Q^{n},

i.e.,

ν̃_{3}(n) = 16 ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=n \\ 2∤x_{1},...,2∤x_{4},x_{1}>0,...,x_{4}>0 \end{matrix}} (\frac{−1}{x_{1}x_{2}x_{3}x_{4}} )x_{1}x_{2}x_{3}x_{4}. (2.64)

Obviously, ν̃_{3}(n) ≠ 0 for n ≡ 4 (mod 8) only.

The values ν_{1}(n), ν̃_{1}(n), and ν_{11}(n) have already been defined by formulas (2.31), (2.32), and (2.33), respectively. Formula (IV) follows from (2.60), (2.41), (2.31)–(2.33), and (2.61)–(2.63). But in the case where n ≡ 1 (mod 4), (2.60) implies that

r(n; f_{4}) = ρ(n; f_{4}) + 2ν_{1}(n) + \frac{15}{16} ν_{11}(n) − 2ν_{3}(n),

since by (2.61), (2.32), (2.63), and (2.64)

ν_{2}(n) = ν̃_{1}(n) = ν̃_{2}(n) = ν̃_{3}(n) = 0.

However, one can verify that

∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4(x_{3}^{2}+x_{4}^{2})=2n \\ 2∤x_{1},2∤x_{2},x_{1}>0,x_{2}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2} = ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+8(x_{3}^{2}+x_{4}^{2})=2n \\ 2∤x_{1},2∤x_{2},x_{1}>0,x_{2}>0 \end{matrix}} x_{1}^{4} − 3x_{1}^{2}x_{2}^{2},

i.e., 2ν_{1}(n) − 2ν_{3}(n) = 0.

2.3. Notice finally that one can derive in a similar way formulas for r(n; f) for some other quadratic forms of type (1). In particular, one can derive the well-known Liouville formula obtained by him in 1864 for r(n; f), when f = x_{1}^{2} + x_{2}^{2} + ⋯ = x_{12}^{2}, if we introduce the function

ψ(τ; f) = ϑ_{00}^{12}(τ; 0, 2) − Θ(τ; f) −

− \frac{1}{512π^{4}} \{2ϑ_{00}^{(4)}(τ; 0, 2)ϑ_{00}(τ; 0, 2) − 6ϑ_{00}^{′′2}(τ; 0, 2)\}ϑ_{00}^{2}(τ; 0, 2).

References

1. G. Lomadze, Summation of singular series corresponding to representations of numbers by some quadratic forms in twelve variables. Georgian Math. J. 5(1998), No. 4, 367–384.

2. G. Lomadze, On some entire modular forms of weights 5 and 6 for the congruence group Γ_{0}(4N). Georgian Math. J. 1(1994), No. 1, 53–76.

3. G. Lomadze, Construction of entire modular forms of weights 5 and 6 for the congruence group Γ_{0}(4N). Georgian Math. J. 2(1995), No. 2, 189–199.

4. G. A. Lomadze, On the number of representations of integers by positive quadratic forms with six variables. (Russian) Trudy Tbilis. Math. Inst. Razmadze 45(1974), 111–133.

5. G. A. Lomadze, On the representations of numbers by certain quadratic forms with six variables. I. (Russian) Trudy Tbilis. Univ. Mat. Mekh. Astron. 117(1966), 7–43.

(Received 28.05.1996)

Author’s address: Faculty of Mechanics and Mathematics I. Javakhishvili Tbilisi State University 2, University St., Tbilisi 380043 Georgia