Georgian Mathematical Journal Volume 8 (2001), Number 1, 111-127

ON THE NUMBER OF REPRESENTATIONS OF POSITIVE INTEGERS BY THE QUADRATIC FORM x_{1}^{2} + ⋯ + x_{8}^{2} + 4x_{9}^{2}

G. LOMADZE

Abstract. An explicit exact (non asymptotic) formula is derived for the number of representations of positive integers by the quadratic form x_{1}^{2} + ⋯+x_{8}^{2}+4x_{9}^{2}. The way by which this formula is derived, gives us a possibility to develop a method of finding the so-called Liouville type formulas for the number of representations of positive integers by positive diagonal quadratic forms in nine variables with integral coefficients.

2000 Mathematics Subject Classification: 11E20, 11E25.

Key words and phrases: Entire modular form, congruence subgroup, theta-functions with characteristics.

In [6], entire modular forms of weight 9/2 on the congruence subgroup Γ_{0}(4N) are constructed. The Fourier coefficients of these modular forms have a simple arithmetical sense. This allows one to get the Liouville type formulas for the number of representations of positive integers by positive diagonal quadratic forms in nine variables with integral coefficients.

Let r(n; f) denote the number of representations of a positive integer n by the positive quadratic form f = a_{1}x_{1}^{2} + a_{2}x_{2}^{2} + ⋯ + a_{9}x_{9}^{2}.

1. Preliminaries

1.1. In this paper N, a, d, n, q, r, s denote positive integers; u, v are odd positive integers; ω is a square free integer; p is a prime number; k, ℓ are nonnegative integers; c, g, h, j, m, x, α, β, γ, δ are integers; z, τ are complex variables (Im τ > 0); e(z) = exp(2πiz); Q = e(τ); (\frac{h}{u}) is the generalized Jacobi symbol; η(γ) = 1 if γ ≥ 0 and η(γ) = −1 if γ < 0; a denotes the least common multiple of the coefficients of the quadratic form and Δ is its determinant. Further, ∑_{hmodq} and \begin{matrix}∑ & ′ \\ hmodq \end{matrix} denote respectively summation with respect to the

complete and the reduced residue system modulo q; ρ(n; f) is a sum of the singular series corresponding to r(n; f).

Since

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)

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

112 G. LOMADZE

(theta-function with characteristics g, h), we have

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

Suppose

\begin{aligned}ϑ_{gh}(τ; c, N) & = ϑ_{gh}(0 | τ; c, N), \\ ϑ′_{gh}(τ; c, N) & = \frac{∂}{∂z} ϑ_{gh}(z | τ; c, N)|_{z=0}. \end{aligned}  (1.3)

It is known (see, e.g., [5], p.112) that

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

From (1.1) and (1.2), in particular according to (1.3), it follows that

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

From (1.6) and (1.7) we obtain

ϑ_{−g,h}(τ; 0, N) = ϑ_{gh}(τ; 0, N), ϑ′_{−g,h}(τ; 0, N) = −ϑ′_{gh}(τ; 0, N). (1.8)

It is clear that

∏_{k=1}^{9} ϑ_{00}(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)

It is known that

ρ(n; f) = \frac{π^{9/2}n^{7/2}}{Γ(9/2)Δ^{1/2}} ∑_{q=1}^{∞} A(q),  (1.11)

where Γ(·) is the gamma-function,

A(q) = q^{−9/2}\begin{matrix}∑ & ′ \\ h mod q \end{matrix} e( − \frac{hn}{q}) ∏_{k=1}^{9} S(a_{k}h, q)  (1.12)

and S(a_{k}h, q) is the Gaussian sum.

1.2. For the convenience we quote some known results as the following lemmas.

Lemma 1 ([2], p. 811, the end of the p. 954). The entire modular form F (τ) of weight r on the congruence subgroup Γ_{0}(4N) is identically zero if in its expansion in the series

F (τ) = C_{0} + ∑_{n=1}^{∞} C_{n}Q^{n},

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

Lemma 2 ([6], Theorem on p. 64). For given N ≥ a, the function

ψ(τ; f) = ∏_{k=1}^{9} ϑ_{00}(τ; 0, 2a_{k}) − θ(τ; f) − λ ∏_{k=1}^{3} ϑ′_{g_{k}h_{k}}(τ; 0, 2N_{k}),

where λ is an arbitrary constant, is an entire modular form of weight 9/2, and the multiplier system

v(M) = i^{η(γ)(sgn δ−1)/2} · i^{(|δ|−1)^{2}/4}(\frac{βΔ sgn δ}{|δ|} )

on Γ_{0}(4N) (M = (\begin{matrix}α & β \\ γ & δ \end{matrix}) is a substitution matrix from Γ_{0}(4N)) if the following

conditions hold:

1) 2 | g_{k}, N_{k} | N (k = 1, 2, 3), a | N,

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

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

sgn δ(\frac{N_{1}N_{2}N_{3}}{|δ|} ) ∏_{k=1}^{3} ϑ′_{αg_{k},h_{k}}(τ; 0, 2N_{k}) = (\frac{−Δ}{|δ|}) ∏_{k=1}^{3} ϑ′_{g_{k}h_{k}}(τ; 0, 2N_{k}).

Lemma 3 (see, e.g., [4], p. 14, Lemma 10). Let

χ_{p} = 1 + A(p) + A(p^{2}) + ⋯;

then for s > 4

∑_{q=1}^{∞} A(q) = ∏_{p} χ_{p}.

Lemma 4 ([1], Theorem 2, p. 531). Let n be a positive integer and f be a positive definite quadratic form in s > 4 variables with integral coefficients, Δ

114 G. LOMADZE

its determinant. If 2 ∤ s, 2^{k} || n, 2^{k′} || Δ, Δn = 2^{k+k′}uv = r^{2}ω, p^{ℓ} || n, p^{ℓ′} || Δ,

u = ∏_{\begin{matrix}p|n \\ p∤2Δ \end{matrix}}p^{ℓ} = r_{1}^{2}ω_{1}, v = ∏_{\begin{matrix}p|Δn, p|Δ \\ p>2 \end{matrix}} p^{ℓ+ℓ′} = r_{2}^{2}ω_{2}, then

∏_{p} χ_{p} = \frac{(s − 1)!r_{1}^{2−s}}{2^{s−2}π^{s−1}|B_{s−1}|} χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{p|2Δ}(1 − p^{1−s})^{−1}L(\frac{s − 1}{2}, (−1)^{(s−1)/2}ω)

× ∏_{\begin{matrix}p|r_{2} \\ p>2 \end{matrix}} (1 − (\frac{(−1)^{(s−1)/2}ω}{p} )p^{(1−s)/2})

× ∑_{d|r_{1}} d^{s−2} ∏_{p|d} (1 − (\frac{(−1)^{(s−1)/2}ω}{p} )p^{(1−s)/2}),

where L is the Dirichlet series and B_{s−1} is the Bernoulli number.

If in Lemma 4 we put s = 9, then by (1.11) and Lemma 3 we get

ρ(n; f) = \frac{π^{9/2}n^{7/2}}{Γ(9/2)Δ^{1/2}} \frac{8!r_{1}^{−7}}{2^{7}π^{8}|B_{8}|} χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{p|2Δ}(1 − p^{−8})^{−1}L(4, ω)

× ∏_{\begin{matrix}p|r_{2} \\ p>2 \end{matrix}} (1 − (\frac{ω}{p} )p^{−4}) ∑_{d|r_{1}} d^{7} ∏_{p|d} (1 − (\frac{ω}{p} )p^{−4})

= \frac{n^{7/2} · 48 · 30}{π^{4}Δ^{1/2}r_{1}^{7}} χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{p|2Δ}(1 − p^{−8})^{−1}L(4, ω)

× ∏_{\begin{matrix}p|r_{2} \\ p>2 \end{matrix}} (1 − (\frac{ω}{p} )p^{−4}) ∑_{d|r_{1}} d^{7} ∏_{p|d} (1 − (\frac{ω}{p} )p^{−4}).  (1.13)

Lemma 5 ([3], p. 298). The sum of Dirichlet series L(4, ω) is equal to

L(4, 1) = \frac{π^{4}}{2^{5} · 3} , L(4, 2) = \frac{11π^{4}}{2^{8} · 3\sqrt{2}} ,

L(4, ω) = −2π^{4}ω^{−1/2} ∑_{0<h≤ω/2} (\frac{h}{ω} ) (\frac{h^{2}}{2^{2}ω^{2}} − \frac{h^{3}}{3ω^{3}} )

if ω > 1, ω ≡ 1 (mod 4),

L(4, ω) = 2π^{4}ω^{−1/2}\{ ∑_{0<h≤ω/4} (\frac{h}{ω} ) (\frac{h}{2^{4}ω} − \frac{h^{3}}{3ω^{3}} )

− ∑_{ω/4<h≤ω/2} (\frac{h}{ω} ) (\frac{1}{2^{5} · 3} − \frac{3h}{2^{4}ω} + \frac{h^{2}}{2ω^{2}} − \frac{h^{3}}{3ω^{3}} )\}

if ω ≡ 3 (mod 4),

REPRESENTATIONS OF INTEGERS BY THE QUADRATIC FORM 115

L(4, ω) = 2π^{4}ω^{−1/2}\{ ∑_{0<h≤ω/16} (\frac{h}{ω/2}) (\frac{11}{2^{8} · 3} − \frac{h^{2}}{ω^{2}} )

+ ∑_{ω/16<h≤3ω/16} (\frac{h}{ω/2}) (\frac{5}{2^{7} · 3} + \frac{h}{2^{4}ω} − \frac{2h^{2}}{ω^{2}} + \frac{2^{4}}{3} \frac{h^{3}}{ω^{3}} )

+ ∑_{3ω/16<h≤ω/4} (\frac{h}{ω/2}) (\frac{37}{2^{8} · 3} − \frac{h}{2ω} + \frac{h^{2}}{ω^{2}} )\}

if ω > 2, ω ≡ 2 (mod 8),

L(4, ω) = 2π^{4}ω^{−1/2}\{ ∑_{0<h≤ω/16} (\frac{h}{ω/2}) (\frac{3h}{2^{4}ω} − \frac{2^{4}h^{3}}{3ω^{3}} )

− ∑_{ω/16<h≤3ω/16} (\frac{h}{ω/2}) (\frac{1}{2^{8} · 3} + \frac{h}{2^{2}ω} + \frac{h^{2}}{ω^{2}} )

− ∑_{3ω/16<h≤ω/4} (\frac{h}{ω/2}) (\frac{7}{2^{6} · 3} − \frac{13h}{2^{4}ω} + \frac{2^{2}h^{2}}{ω^{2}} − \frac{2^{4}h^{3}}{3ω^{3}} )\}

if ω ≡ 6 (mod 8).

In particular, using this lemma, by tedious calculations we get:

L(4, 3) = \frac{23π^{4}}{2^{4} · 3^{4}\sqrt{3}} , L(4, 5) = \frac{17π^{4}}{2 · 3 · 5^{3}\sqrt{5}} , L(4, 6) = \frac{29π^{4}}{2^{7} · 3^{2}\sqrt{6}} ,

L(4, 7) = \frac{113π^{4}}{3 · 4 · 7^{3}\sqrt{7}} , L(4, 10) = \frac{1577π^{4}}{2^{7} · 3 · 5^{3}\sqrt{10}} ,

L(4, 11) = \frac{2153π^{4}}{2^{4} · 3 · 11^{3}\sqrt{11}} , L(4, 13) = \frac{493π^{4}}{2 · 3 · 13^{3}\sqrt{13}} ,

L(4, 14) = \frac{2503π^{4}}{2^{6} · 3 · 7^{3}\sqrt{14}} , L(4, 15) = \frac{179π^{4}}{20 · 15^{2}\sqrt{15}} ,

L(4, 17) = \frac{205π^{4}}{17^{3}\sqrt{17}} , L(4, 19) = \frac{14933π^{4}}{48 · 19^{3}\sqrt{19}} ,

L(4, 21) = \frac{187π^{4}}{9 · 21^{2}\sqrt{21}} , L(4, 22) = \frac{24889π^{4}}{2^{7} · 3 · 11^{3}\sqrt{22}} ,

L(4, 23) = \frac{7093π^{4}}{2^{2} · 3 · 23^{3}\sqrt{23}} , L(4, 26) = \frac{43679π^{4}}{2^{7} · 3 · 13^{2}\sqrt{26}} ,

L(4, 29) = \frac{2669π^{4}}{2 · 29^{3}\sqrt{29}} , L(4, 30) = \frac{36451π^{4}}{2^{6} · 3^{4} · 5^{3}\sqrt{30}} ,

L(4, 31) = \frac{10357π^{4}}{6 · 31^{3}\sqrt{31}} , L(4, 33) = \frac{2115π^{4}}{33^{3}\sqrt{33}} ,

L(4, 34) = \frac{57241π^{4}}{2^{6} · 3 · 17^{3}\sqrt{34}} , L(4, 35) = \frac{61733π^{4}}{2^{3} · 3 · 35^{3}\sqrt{35}} .

2. Formulas for r(n; f) if f = x_{1}^{2} + x_{2}^{2} + ⋯ + x_{8}^{2} + 4x_{9}^{2}

Lemma 6. The function

ψ(τ; f) = ϑ_{00}^{8}(τ; 0, 2)ϑ_{00}(τ; 0, 8) − θ(τ; f)

− \frac{16 · 9}{17} \frac{1}{512(πi)^{3}} ϑ_{80}^{′3}(τ; 0, 24)

− \frac{64 · 27}{17} \frac{1}{2048(πi)^{3}} ϑ_{16,0}^{′2}(τ; 0, 24)ϑ′_{80}(τ; 0, 24)

− \frac{32 · 21}{17} \frac{1}{1024(πi)^{3}} ϑ_{80}^{′2}(τ; 0, 24)ϑ′_{16,0}(τ; 0, 24)

− \frac{896}{17} \frac{1}{4096(πi)^{3}} ϑ_{16,0}^{′3}(τ; 0, 24)  (2.1)

is an entire modular form of weight 9/2 and the multiplier system

v(M) = i^{η(γ)(sgn δ−1)/2} · i^{(|δ|−1)^{2}/4}(\frac{4β sgn δ}{|δ|} )

on Γ_{0}(48) (M = (\begin{matrix}α & β \\ γ & δ \end{matrix}) is a substitution matrix from Γ_{0}(48)).

Proof. In Lemma 2 put a = Δ = 4; g_{1} = g_{2} = g_{3} = 8; g_{1} = g_{2} = 16, g_{3} = 8; g_{1} = g_{2} = 8, g_{3} = 16; g_{1} = g_{2} = g_{3} = 16; h_{1} = h_{2} = h_{3} = 0; N_{1} = N_{2} = N_{3} = N = 12. It is easy to verify that the last four summands in (2.1) satisfy the conditions 1) and 2) of this Lemma. Further,

(\frac{N_{1}N_{2}N_{3}}{|δ|} ) = (\frac{12^{3}}{|δ|} ) = (\frac{3}{|δ|}), (\frac{−Δ}{|δ|}) = (\frac{−4}{|δ|}) = (\frac{−1}{|δ|}).  (2.2)

If αδ ≡ 1 (mod 48), then αδ ≡ 1 (mod 24), whence, in particular,

α ≡ ±1 (mod 24), then respectively δ ≡ ±1 (mod 24).  (2.3)

From (2.3) it follows that if α ≡ ±1 (mod 4) and α ≡ ±1 (mod 3), then respectively

δ ≡ ±1 (mod 4) and δ ≡ ±1 (mod 3).  (2.4)

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

ϑ_{8α,0}^{′3}(τ; 0, 24) = ϑ_{±8+8(α∓1),0}^{′3}(τ; 0, 24) = ϑ_{±8,0}^{′3}(τ; 4(α ∓ 1), 24)
= ϑ_{±8,0}^{′3}(τ; 0, 24) = \begin{cases}ϑ_{80}^{′3}(τ; 0, 24) & for α ≡ 1 (mod 24), \\ −ϑ_{80}^{′3}(τ; 0, 24) & for α ≡ −1 (mod 24). \end{cases}  (2.5)

Similarly,

ϑ_{16α,0}^{′2}(τ; 0, 24)ϑ′_{8α,0}(τ; 0, 24)
= \begin{cases}ϑ_{16,0}^{′2}(τ; 0, 24)ϑ′_{80}(τ; 0, 24) & for α ≡ 1 (mod 24), \\ −ϑ_{16,0}^{′2}(τ; 0, 24)ϑ′_{80}(τ; 0, 24) & for α ≡ −1 (mod 24), \end{cases}  (2.6)
ϑ_{8α,0}^{′2}(τ; 0, 24)ϑ′_{16α,0}(τ; 0, 24)
= \begin{cases}ϑ_{80}^{′2}(τ; 0, 24)ϑ′_{16,0}(τ; 0, 24) & for α ≡ 1 (mod 24), \\ −ϑ_{80}^{′2}(τ; 0, 24)ϑ′_{16,0}(τ; 0, 24) & for α ≡ −1 (mod 24), \end{cases}  (2.7)

ϑ_{16α,0}^{′3}(τ; 0, 24) = \begin{cases}ϑ_{16,0}^{′3}(τ; 0, 24) & for α ≡ 1 (mod 24), \\ −ϑ_{16,0}^{′3}(τ; 0, 24) & for α ≡ −1 (mod 24). \end{cases}  (2.8)

From (2.2) it follows that

sgn δ(\frac{N_{1}N_{2}N_{3}}{|δ|} ) = sgn δ(\frac{3}{|δ|}) = \begin{cases}(\frac{3}{δ}) & if δ > 0, \\ −(\frac{3}{−δ} ) & if δ < 0. \end{cases}  (2.9)

If in (2.9) we have δ > 0 and δ ≡ 1 (mod 24), i.e., δ ≡ 1 (mod 4) and δ ≡ 1 (mod 3), then (\frac{3}{δ}) = (\frac{δ}{3}) = (\frac{1}{3}) = 1 and (\frac{−1}{δ}) = 1. Thus if δ > 0, then sgn δ(\frac{3}{|δ|}) = (\frac{−1}{δ}) = (\frac{−1}{|δ|}). But if in (2.9) we have δ < 0 and δ ≡ 1 (mod 24), i.e., δ ≡ 1 (mod 4) and δ ≡ 1 (mod 3), then, as −δ ≡ −1 (mod 4), we get (\frac{3}{−δ}) = −(\frac{−δ}{3}) = −(\frac{−1}{3}) = 1 and (\frac{−1}{−δ}) = −1. Thus if δ < 0, then sgn δ(\frac{3}{|δ|}) = −(\frac{3}{−δ}) = (\frac{−1}{−δ}) = (\frac{−1}{|δ|}).

Now suppose that in (2.9) we have δ ≡ −1 (mod 24); hence δ ≡ −1 (mod 4) and δ ≡ −1 (mod 3). Then if δ > 0, we get (\frac{3}{δ}) = −(\frac{δ}{3}) = 1 and (\frac{−1}{δ}) = −1, i.e., sgn δ(\frac{3}{|δ|}) = −(\frac{−1}{δ}) = −(\frac{−1}{|δ|}). But if δ < 0, then (\frac{3}{−δ}) = (\frac{−δ}{3}) = 1 and (\frac{−1}{−δ}) = 1, i.e., sgn δ(\frac{3}{−δ}) = −(\frac{3}{−δ}) = −(\frac{−1}{−δ}) = −(\frac{−1}{|δ|}).

Thus

sgn δ(\frac{N_{1}N_{2}N_{3}}{|δ|} ) = \begin{cases}(\frac{−1}{|δ|}) & for δ ≡ 1 (mod 24), \\ −(\frac{−1}{|δ|}) & for δ ≡ −1 (mod 24). \end{cases}  (2.10)

Since in the investigated quadratic form Δ = 4, we have (\frac{−1}{|δ|}) = (\frac{−4}{|δ|}), and according to (2.3), formula (2.10) can be rewritten as

sgn δ(\frac{N_{1}N_{2}N_{3}}{|δ|} ) = \begin{cases}(\frac{−4}{|δ|}) & for δ ≡ 1 (mod 24), \\ & i.e., also for α ≡ 1 (mod 24), \\ −(\frac{−4}{|δ|}) & for δ ≡ −1 (mod 24), \\ & i.e., also for α ≡ −1 (mod 24). \end{cases} (2.11)

Hence by (2.11) and (2.5)–(2.8), for all α and δ with αδ ≡ 1 (mod 24) we have:

1) sgn δ(\frac{N_{1}N_{2}N_{3}}{|δ|} )ϑ_{8α,0}^{′3}(τ; 0, 24) = (\frac{−4}{|δ|})ϑ_{80}^{′3}(τ; 0, 24),

2) sgn δ(\frac{N_{1}N_{2}N_{3}}{|δ|} )ϑ_{16α,0}^{′2}(τ; 0, 24)ϑ′_{8α,0}(τ; 0, 24)= (\frac{−4}{|δ|})ϑ_{16,0}^{′2}(τ; 0, 24)ϑ′_{80}(τ; 0, 24),

3) sgn δ(\frac{N_{1}N_{2}N_{3}}{|δ|} )ϑ_{8α,0}^{′2}(τ; 0, 24)ϑ′_{16α,0}(τ; 0, 24)= (\frac{−4}{|δ|})ϑ_{80}^{′2}(τ; 0, 24)ϑ′_{16,0}(τ; 0, 24),

4) sgn δ(\frac{N_{1}N_{2}N_{3}}{|δ|} )ϑ_{16α,0}^{′3}(τ; 0, 24) = (\frac{−4}{|δ|})ϑ_{16,0}^{′3}(τ; 0, 24). Thus the condition 3) of Lemma 2 also is satisfied.

Theorem 1. The following indentity takes place:

ϑ_{00}^{8}(τ; 0, 2)ϑ_{00}(τ; 0, 8) = θ(τ; f) + \frac{16 · 9}{17} \frac{1}{512(πi)^{3}} ϑ_{80}^{′3}(τ; 0, 24)

+ \frac{64 · 27}{17} \frac{1}{2048(πi)^{3}} ϑ_{16,0}^{′2}(τ; 0, 24)ϑ′_{80}(τ; 0, 24)

+ \frac{32 · 21}{17} \frac{1}{1024(πi)^{3}} ϑ_{80}^{′2}(τ; 0, 24)ϑ′_{16,0}(τ; 0, 24)

+ \frac{896}{17} \frac{1}{4096(πi)^{3}} ϑ_{16,0}^{′3}(τ; 0, 24).

Proof. According to Lemma 1, the function ψ(τ; f) will be identically zero if all coefficients of Q^{n} (n ≤ 36) in its expansion in powers of Q are zero.

If in (1.13) we put Δ = 4, i.e., v = 1, r_{2} = 1, and n = 2^{k}u, then

ρ(n; f) = \frac{2^{7k/2}u^{7/2}}{π^{4} · 2r_{1}^{7}} · 48 · 30χ_{2} \frac{256}{255} L(4, ω) ∑_{d|r_{1}} d^{7} ∏_{p|d} (1 − (\frac{ω}{p} )p^{−4})

= \frac{2^{7k/2}u^{7/2}}{17π^{4} · r_{1}^{7}} · 12288χ_{2}L(4, ω) ∑_{d|r_{1}} d^{7} ∏_{p|d} (1 − (\frac{ω}{p} )p^{−4}).  (2.12)

A method of finding the values of χ_{2} in the general case for an arbitrary quadratic form is developed in [7]. In the case of the quadratic form f = x_{1}^{2} + x_{2}^{2} + ⋯ + x_{8}^{2} + 4x_{9}^{2}, we get

χ_{2} = 1 if k = 0,

= \frac{7}{8} if k = 1,

= \frac{143 − 255 · 2^{−7k/2−3}}{127} if k ≥ 2, 2 | k, u ≡ 3 (mod 4),

= \frac{143 + (127 · (−1)^{(u^{2}−1)/8} − 8)2^{−7k/2−6}}{127}

if k ≥ 2, 2 | k, u ≡ 1 (mod 4),

= \frac{143 − 255 · 2^{−(7k−1)/2}}{127} if k ≥ 3, 2 ∤ k.

(2.13)

In particular,

χ_{2} = \frac{9224 + (−1)^{(u^{2}−1)/8}}{8192} if k = 2, u ≡ 1 (mod 4),
= \frac{1151}{1024} if k = 2, u ≡ 3 (mod 4) and if k = 3.

(2.14)

Our aim now is to find the values of ρ(n; f) for all 1 ≤ n ≤ 36. In the table below the values of χ_{2} are obtained by virtue of (2.13) and (2.14), the values of L(4, ω) are calculated by Lemma 5 and are placed after this lemma, and the values of ρ(n; f) are calculated by (2.12).

\begin{matrix}n & u & r_{1} & Δn & r & ω & χ_{2} & L(4, ω) & ρ(n; f) \\ 1 & 1 & 1 & 4 · 1 & 2 & 1 & 1 & \frac{π^{4}}{2^{5} · 3} & \frac{128}{17} \\ 3 & 3 & 1 & 4 · 3 & 2 & 3 & 1 & \frac{23π^{4}}{2^{4} · 3^{4}\sqrt{3}} & \frac{256 · 23}{17} \\ 5 & 5 & 1 & 4 · 5 & 2 & 5 & 1 & \frac{17π^{4}}{2 · 3 · 5^{3}\sqrt{5}} & 2048 \\ 7 & 7 & 1 & 4 · 7 & 2 & 7 & 1 & \frac{113π^{4}}{2^{2} · 3 · 7^{3}\sqrt{7}} & \frac{1024 · 113}{17} \\ 9 & 9 & 3 & 4 · 9 & 6 & 1 & 1 & \frac{π^{4}}{2^{5} · 3} & \frac{128 · 2161}{17} \\ 11 & 11 & 1 & 4 · 11 & 2 & 11 & 1 & \frac{2153π^{4}}{2^{4} · 3 · 11^{3}\sqrt{11}} & \frac{256 · 2153}{17} \\ 13 & 13 & 1 & 4 · 13 & 2 & 13 & 1 & \frac{493π^{4}}{2 · 3 · 13^{3}\sqrt{13}} & 59392 \\ 15 & 15 & 1 & 4 · 15 & 2 & 15 & 1 & \frac{179π^{4}}{20 · 15^{2}\sqrt{15}} & \frac{1024 · 1611}{17} \\ 17 & 17 & 1 & 4 · 17 & 2 & 17 & 1 & \frac{205π^{4}}{17^{3}\sqrt{17}} & \frac{4096 · 615}{17} \\ 19 & 19 & 1 & 4 · 19 & 2 & 19 & 1 & \frac{14933π^{4}}{48 · 19^{3}\sqrt{19}} & \frac{256 · 14933}{17} \\ 21 & 21 & 1 & 4 · 21 & 2 & 21 & 1 & \frac{187π^{4}}{9 · 21^{2}\sqrt{21}} & 315392 \\ 23 & 23 & 1 & 4 · 23 & 2 & 23 & 1 & \frac{7093π^{4}}{2^{2} · 3 · 23^{3}\sqrt{23}} & \frac{1024 · 7093}{17} \end{matrix}

\begin{matrix}n & u & r_{1} & Δn & r & ω & χ_{2} & L(4, ω) & ρ(n; f) \\ 25 & 25 & 5 & 4 · 25 & 10 & 1 & 1 & \frac{π^{4}}{2^{5} · 3} & \frac{128 · 78001}{17} \\ 27 & 27 & 3 & 4 · 27 & 6 & 3 & 1 & \frac{23π^{4}}{2^{4} · 3^{4}\sqrt{3}} & \frac{1024 · 12581}{17} \\ 29 & 29 & 1 & 4 · 29 & 2 & 29 & 1 & \frac{2669π^{4}}{2 · 29^{3}\sqrt{29}} & 964608 \\ 31 & 31 & 1 & 4 · 31 & 2 & 31 & 1 & \frac{10357π^{4}}{6 · 31^{3}\sqrt{31}} & \frac{2048 · 10357}{17} \\ 33 & 33 & 1 & 4 · 33 & 2 & 33 & 1 & \frac{2115π^{4}}{33^{3}\sqrt{33}} & \frac{4096 · 6345}{17} \\ 35 & 35 & 1 & 4 · 35 & 2 & 35 & 1 & \frac{61733π^{4}}{8 · 3 · 35^{3}\sqrt{35}} & \frac{512 · 61733}{17} \\ 2 & 1 & 1 & 8 · 1 & 2 & 2 & \frac{7}{8} & \frac{11π^{4}}{2^{8} · 3\sqrt{2}} & \frac{16 · 77}{17} \\ 6 & 3 & 1 & 8 · 3 & 2 & 6 & \frac{7}{8} & \frac{29π^{4}}{2^{7} · 3^{2}\sqrt{6}} & \frac{32 · 1827}{17} \\ 10 & 5 & 1 & 8 · 5 & 2 & 10 & \frac{7}{8} & \frac{1577π^{4}}{2^{7} · 3 · 5^{3}\sqrt{10}} & \frac{32 · 11039}{17} \\ 14 & 7 & 1 & 8 · 7 & 2 & 14 & \frac{7}{8} & \frac{2503π^{4}}{2^{6} · 3 · 7^{3}\sqrt{14}} & \frac{64 · 17521}{17} \\ 18 & 9 & 3 & 8 · 9 & 6 & 2 & \frac{7}{8} & \frac{11π^{4}}{2^{8} · 3\sqrt{2}} & \frac{16 · 170555}{17} \\ 22 & 11 & 1 & 8 · 11 & 2 & 22 & \frac{7}{8} & \frac{24889π^{4}}{2^{7} · 3 · 11^{3}\sqrt{22}} & \frac{32 · 7 · 24889}{17} \\ 26 & 13 & 1 & 8 · 13 & 2 & 26 & \frac{7}{8} & \frac{43679π^{4}}{2^{7} · 3 · 13^{3}\sqrt{26}} & \frac{32 · 305753}{17} \\ 30 & 15 & 1 & 8 · 15 & 2 & 30 & \frac{7}{8} & \frac{36451π^{4}}{2^{6} · 3^{4} · 5^{3}\sqrt{30}} & \frac{64 · 255157}{17} \\ 34 & 17 & 1 & 8 · 17 & 2 & 34 & \frac{7}{8} & \frac{57241π^{4}}{2^{6} · 3 · 17^{3}\sqrt{34}} & \frac{64 · 400687}{17} \\ 4 & 1 & 1 & 16 · 1 & 4 & 1 & \frac{9225}{8192} & \frac{π^{4}}{2^{5} · 3} & \frac{2 · 9225}{17} \end{matrix}

REPRESENTATIONS OF INTEGERS BY THE QUADRATIC FORM 121

\begin{matrix}n & u & r_{1} & Δn & r & ω & χ_{2} & L(4, ω) & ρ(n; f) \\ 8 & 1 & 1 & 32 · 1 & 4 & 2 & \frac{1151}{1024} & \frac{11π^{4}}{2^{8} · 3\sqrt{2}} & \frac{16 · 12661}{17} \\ 12 & 3 & 1 & 16 · 3 & 4 & 3 & \frac{1151}{1024} & \frac{23π^{4}}{2^{4} · 3^{4}\sqrt{3}} & \frac{32 · 26473}{17} \\ 16 & 1 & 1 & 64 · 1 & 8 & 1 & \frac{1180681}{2^{20}} & \frac{π^{4}}{2^{5} · 3} & \frac{2 · 1180681}{17} \\ 20 & 5 & 1 & 16 · 5 & 4 & 5 & \frac{9223}{8192} & \frac{17π^{4}}{2 · 3 · 5^{3}\sqrt{5}} & 295136 \\ 24 & 3 & 1 & 32 · 3 & 4 & 6 & \frac{1151}{1024} & \frac{29π^{4}}{2^{7} · 3^{2}\sqrt{6}} & \frac{32 · 300411}{17} \\ 28 & 7 & 1 & 16 · 7 & 4 & 7 & \frac{1151}{1024} & \frac{113π^{4}}{3 · 4 · 7^{3}\sqrt{7}} & \frac{128 · 130063}{17} \\ 32 & 1 & 1 & 128 · 1 & 8 & 2 & \frac{18743041}{2^{17} · 127} & \frac{11π^{4}}{2^{8} · 3\sqrt{2}} & \frac{16 · 11 · 147583}{17} \\ 36 & 9 & 3 & 16 · 9 & 12 & 1 & \frac{9225}{8192} & \frac{π^{4}}{2^{5} · 3} & \frac{2 · 9225 · 2161}{17} \end{matrix}

Thus according to (1.10) we get

θ(τ; f) = 1 + \frac{128}{17} Q + \frac{16 · 77}{17} Q^{2} + \frac{256 · 23}{17} Q^{3} + \frac{2 · 9225}{17} Q^{4} + 2048Q^{5}

+ \frac{32 · 1827}{17} Q^{6} + \frac{1024 · 113}{17} Q^{7} + \frac{16 · 12661}{17} Q^{8} + \frac{128 · 2161}{17} Q^{9}

+ \frac{32 · 11039}{17} Q^{10} + \frac{256 · 2153}{17} Q^{11} + \frac{32 · 26473}{17} Q^{12} + 59392Q^{13}

+ \frac{64 · 17521}{17} Q^{14} + \frac{1024 · 1611}{17} Q^{15} + \frac{2 · 1180681}{17} Q^{16}

+ \frac{4096 · 615}{17} Q^{17} + \frac{16 · 170555}{17} Q^{18} + \frac{256 · 14933}{17} Q^{19}

+ 295136Q^{20} + 315392Q^{21} + \frac{32 · 7 · 24889}{17} Q^{22}

+ \frac{1024 · 7093}{17} Q^{23} + \frac{32 · 300411}{17} Q^{24} + \frac{128 · 78001}{17} Q^{25}

+ \frac{32 · 305753}{17} Q^{26} + \frac{1024 · 12581}{17} Q^{27} + \frac{128 · 130063}{17} Q^{28}

+ 964608Q^{29} + \frac{64 · 255157}{17} Q^{30} + \frac{2048 · 10357}{17} Q^{31}

+ \frac{16 · 11 · 147583}{17} Q^{32} + \frac{4096 · 6345}{17} Q^{33} + \frac{64 · 400687}{17} Q^{34}

+ \frac{512 · 61733}{17} Q^{35} + \frac{2 · 9225 · 2161}{17} Q^{36} + · · · .  (2.15)

Using (1.6), we have

ϑ_{00}(τ; 0, 8) = ∑_{m=−∞}^{∞} Q^{(16m)^{2}/64} = ∑_{m=−∞}^{∞} Q^{4m^{2}}
= 1 + 2Q^{4} + 2Q^{16} + 2Q^{36} + ⋯  (2.16)

and

ϑ_{00}(τ; 0, 2) = ∑_{m=−∞}^{∞} Q^{(4m)^{2}/16} = ∑_{m=−∞}^{∞} Q^{m^{2}}
= 1 + 2Q + 2Q^{4} + 2Q^{9} + 2Q^{16} + 2Q^{25} + 2Q^{36} + · · · .  (2.17)

From (2.17) it follows that

ϑ_{00}^{2}(τ; 0, 2) = 1 + 4Q + 4Q^{2} + 4Q^{4} + 8Q^{5} + 4Q^{8} + 4Q^{9} + 8Q^{10}

+ 8Q^{13} + 4Q^{16} + 8Q^{17} + 4Q^{18} + 8Q^{20} + 12Q^{25} + 8Q^{26}

+ 8Q^{29} + 4Q^{32} + 8Q^{34} + 4Q^{36} + · · · .

It is not difficult to verify, but by very tedious calculations, that

ϑ_{00}^{4}(τ; 0, 2) = 1 + 8Q + 24Q^{2} + 32Q^{3} + 24Q^{4} + 48Q^{5} + 96Q^{6} + 64Q^{7}

+ 24Q^{8} + 104Q^{9} + 144Q^{10} + 96Q^{11} + 96Q^{12} + 112Q^{13}

+ 192Q^{14} + 192Q^{15} + 24Q^{16} + 144Q^{17} + 312Q^{18} + 160Q^{19}

+ 144Q^{20} + 256Q^{21} + 288Q^{22} + 192Q^{23} + 96Q^{24} + 248Q^{25}

+ 336Q^{26} + 320Q^{27} + 192Q^{28} + 240Q^{29} + 576Q^{30} + 256Q^{31}

+ 24Q^{32} + 384Q^{33} + 432Q^{34} + 384Q^{35} + 312Q^{36} + ⋯

and

ϑ_{00}^{8}(τ; 0, 2) = 1 + 16Q + 112Q^{2} + 448Q^{3} + 1136Q^{4} + 2016Q^{5}

+ 3136Q^{6} + 5504Q^{7} + 9328Q^{8} + 12112Q^{9} + 14112Q^{10}

+ 21312Q^{11} + 31808Q^{12} + 35168Q^{13} + 38528Q^{14}

+ 56448Q^{15} + 74864Q^{16} + 78624Q^{17} + 84784Q^{18} + 109760Q^{19}

+ 143136Q^{20} + 154112Q^{21} + 149184Q^{22} + 194688Q^{23}

+ 261184Q^{24} + 252016Q^{25} + 246176Q^{26} + 327040Q^{27}

+ 390784Q^{28} + 390240Q^{29} + 395136Q^{30} + 476672Q^{31}

+ 599152Q^{32} + 596736Q^{33} + 550368Q^{34}

+ 693504Q^{35} + 859952Q^{36} + · · · .  (2.18)

REPRESENTATIONS OF INTEGERS BY THE QUADRATIC FORM 123

From (2.18) and (2.16), again by tedious calculations, we get

ϑ_{00}^{8}(τ; 0, 2)ϑ_{00}(τ; 0, 8)

= 1 + 16Q + 112Q^{2} + 448Q^{3} + 1138Q^{4} + 2048Q^{5} + 3360Q^{6} +6400Q^{7} + 11600Q^{8} + 16144Q^{9} + 20384Q^{10} + 32320Q^{11}

+50464Q^{12} + 59392Q^{13} + 66752Q^{14} + 99072Q^{15} +138482Q^{16} + 148992Q^{17} + 162064Q^{18} + 223552Q^{19} +295136Q^{20} + 315392Q^{21} + 325024Q^{22} + 425216Q^{23} +566112Q^{24} + 584464Q^{25} + 572768Q^{26} + 759040Q^{27} +976768Q^{28} + 964608Q^{29} + 964544Q^{30} + 1243648Q^{31}

+1530448Q^{32} + 1534464Q^{33} + 1510208Q^{34}

+1866368Q^{35} + 2344530Q^{36} + · · · .  (2.19)

From (1.7) it follows that

1) ϑ_{80}^{′3}(τ; 0, 24) = (8πi ∑_{m=−∞}^{∞}(6m + 1)Q^{(6m+1)^{2}/3})^{3} =

= (δπi)^{3}Q(1 − 15Q^{8} + 96Q^{16} − 335Q^{24} + 672Q^{32} − ⋯),  (2.20)

i.e.,

\frac{1}{512(πi)^{3}} ϑ_{80}^{′3}(τ; 0, 24) = Q − 15Q^{9} + 96Q^{17} − 335Q^{25} + 672Q^{33} + · · · . (2.21)

2) ϑ_{16,0}^{′2}(τ; 0, 24)ϑ′_{80}(τ; 0, 24) = (16πi ∑_{m=−∞}^{∞}(3m + 1)Q^{4(3m+1)^{2}/3})^{2}

×(8πi ∑_{m=−∞}^{∞}(6m + 1)Q^{(6m+1)^{2}/3})

= 2048(πi)^{3}Q^{3}(1 − 4Q^{4} − Q^{8} + 20Q^{12} − 13Q^{16}

−20Q^{20} + 12Q^{24} − 40Q^{28} + 70Q^{32} + 76Q^{36} + ⋯),  (2.22)

i.e.,

\frac{1}{2048(πi)^{3}} ϑ_{16,0}^{′2}(τ; 0, 24)ϑ′_{80}(τ; 0, 24) = Q^{3} − 4Q^{7} − Q^{11} + 20Q^{15}

−13Q^{19} − 20Q^{23} + 12Q^{27} − 40Q^{31} + 70Q^{35} + · · · .  (2.23)

3) ϑ_{80}^{′2}(τ; 0, 24)ϑ′_{16,0}(τ; 0, 24) = (8πi ∑_{m=−∞}^{∞}(6m + 1)Q^{(6m+1)^{2}/3})^{2}

×(16πi ∑_{m=−∞}^{∞}(3m + 1)Q^{4(3m+1)^{2}/3})

= 1024(πi)^{3}Q^{2}(1 − 2Q^{4} − 10Q^{8} + 20Q^{12} + 39Q^{16} − 74Q^{20}
−70Q^{24} + 100Q^{28} + 44Q^{32} + 58Q^{36} + ⋯),  (2.24)

i.e.,

\frac{1}{1024(πi)^{3}} ϑ_{80}^{′2}(τ; 0, 24)ϑ′_{16,0}(τ; 0, 24) = Q^{2} − 2Q^{6} − 10Q^{10} + 20Q^{14}

+39Q^{18} − 74Q^{22} − 70Q^{26} + 100Q^{30} + 44Q^{34} + · · · .  (2.25)

4) ϑ_{16,0}^{′3}(τ; 0, 24) = (16πi ∑_{m=−∞}^{∞}(3m + 1)Q^{4(3m+1)^{2}/3})^{2}

= (16πi)^{3}Q^{4}(1 − 6Q^{4} + 12Q^{8} − 8Q^{12} + 12Q^{20} − 48Q^{24}
+48Q^{28} − 15Q^{32} + 60Q^{36} + ⋯),  (2.26)

i.e.,

\frac{1}{4096(πi)^{3}} ϑ_{16,0}^{′3}(τ; 0, 24) = Q^{4} − 6Q^{8} + 12Q^{12} − 8Q^{16}

+12Q^{24} − 48Q^{28} + 48Q^{32} − 15Q^{36} + · · · .  (2.27)

By (2.1), (2.19), (2.15), (2.21), (2.23), (2.25) and (2.27), one can verify that all coefficients of Q^{n} (n ≤ 36) in the expansion of ψ(τ; f) in powers of Q are zero. Thus, according to Lemma 1, the theorem is proved.

Theorem 2. Let n = 2^{k}u, 4n = r^{2}ω, r_{1}^{2} | u. Then

r(n; f) = \frac{12288u^{7/2}}{17π^{4}r_{1}^{7}} L(4, ω) ∑_{d|r_{1}} d^{7} ∏_{p|d} (1 − (\frac{ω}{p} )p^{−4})

+ \frac{16 · 9}{17} ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=3n \\ x_{1}≡x_{2}≡x_{3}≡1 (mod 6) \end{matrix}} x_{1}x_{2}x_{3} if n ≡ 1 (mod 4),

= \frac{12288u^{7/2}}{17π^{4}r_{1}^{7}} L(4, ω) ∑_{d|r_{1}} d^{7} ∏_{p|d} (1 − (\frac{ω}{p} )p^{−4})

+ \frac{64 · 27}{17} ∑_{\begin{matrix}4(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}=3n \\ x_{1}≡x_{2}≡1 (mod 3), x_{3}≡1 (mod 6) \end{matrix}} x_{1}x_{2}x_{3} if n ≡ 3 (mod 4),

= \frac{12288 · 2^{7k/2}u^{7/2}}{17π^{4}r_{1}^{7}} χ_{2}L(4, ω) ∑_{d|r_{1}} d^{7} ∏_{p|d} (1 − (\frac{ω}{p} )p^{−4})

+ \frac{32 · 21}{17} ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4x_{3}^{2}=3n \\ x_{1}≡x_{2}≡1 (mod 6), x_{3}≡1 (mod 3) \end{matrix}} x_{1}x_{2}x_{3} if n ≡ 2 (mod 4),

r(n; f) = \frac{12288 · 2^{7k/2}u^{7/2}}{17π^{4}r_{1}^{7}} χ_{2}L(4, ω) ∑_{d|r_{1}} d^{7} ∏_{p|d} (1 − (\frac{ω}{p} )p^{−4})

+ \frac{896}{17} ∑_{\begin{matrix}4(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})=3n \\ x_{1}≡x_{2}≡x_{3}≡1 (mod 3) \end{matrix}} x_{1}x_{2}x_{3} if n ≡ 0 (mod 4).

The values of χ_{2} can be calculated by formulas (2.13) and (2.14), the values of L(4, ω) by Lemma 5.

Remark. If p^{2} ∤ n (p > 2), i.e., r_{1} = 1, then in all above formulas

∑_{d|r_{1}} d^{7} ∏_{p|d} (1 − (\frac{ω}{p} )p^{−4}) = 1.

Proof. Equating the coefficients of Q^{n} in both sides of the identity from Theorem 1, by (1.9) and (1.10) we get

r(n; f) = ρ(n; f) + \frac{16 · 9}{17} w_{1}(n) + \frac{64 · 27}{17} w_{2}(n)

+ \frac{32 · 21}{17} w_{3}(n) + \frac{896}{17} w_{4}(n),

where w_{1}(n), w_{2}(n), w_{3}(n) and w_{4}(n) respectively denote the coefficients of Q^{n} in the expansions of the functions

\frac{1}{512(πi)^{3}} ϑ_{80}^{′3}(τ; 0, 24), \frac{1}{2048(πi)^{3}} ϑ_{16,0}^{′2}(τ; 0, 24)ϑ′_{80}(τ; 0, 24),

\frac{1}{1024(πi)^{3}} ϑ_{80}^{′2}(τ; 0, 24)ϑ′_{16,0}(τ; 0, 24), \frac{1}{4096(πi)^{3}} ϑ_{16,0}^{′3}(τ; 0, 24)

in powers of Q.

The values of ρ(n; f) are given in (2.12).

Further,

a) from (2.20) it follows that

\frac{1}{512(πi)^{3}} ϑ_{80}^{′3}(τ; 0, 24)

= ∑_{m_{1},m_{2},m_{3}=−∞}^{∞}(6m_{1} + 1)(6m_{2} + 1)(6m_{3} + 1)Q^{((6m_{1}+1)^{2}+(6m_{2}+1)^{2}+(6m_{3}+1)^{2})/3}

= ∑_{n=1}^{∞} ( ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=3n \\ x_{1}≡x_{2}≡x_{3}≡1 (mod 6) \end{matrix}} x_{1}x_{2}x_{3})Q^{n},

i.e.,

w_{1}(n) = ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=3n \\ x_{1}≡x_{2}≡x_{3}≡1 (mod 6) \end{matrix}} x_{1}x_{2}x_{3};

it is obvious that w_{1}(n) = 0 for n ≡ 0 (mod 2) and n = 3 (mod 4). b) from (2.22) it follows that

\frac{1}{2048(πi)^{3}} ϑ_{16,0}^{′2}(τ; 0, 24)ϑ′_{80}(τ; 0, 24)

= ∑_{m_{1},m_{2},m_{3}=−∞}^{∞}(3m_{1} + 1)(3m_{2} + 1)(6m_{3} + 1)Q^{(4(3m_{1}+1)^{2}+4(3m_{2}+1)^{2}+(6m_{3}+1)^{2})/3}

= ∑_{n=1}^{∞} ( ∑_{\begin{matrix}4(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}=3n \\ x_{1}≡x_{2}≡1 (mod 3), x_{3}≡1 (mod 6) \end{matrix}} x_{1}x_{2}x_{3})Q^{n},

i.e.,

w_{2}(n) = ∑_{\begin{matrix}4(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}=3n \\ x_{1}≡x_{2}≡1 (mod 3), x_{3}≡1 (mod 6) \end{matrix}} x_{1}x_{2}x_{3};

it is obvious that w_{2}(n) = 0 for n ≡ 0 (mod 2) and n ≡ 1 (mod 4). c) from (2.24) it follows that

\frac{1}{1024(πi)^{3}} ϑ_{80}^{′2}(τ; 0, 24)ϑ′_{16,0}(τ; 0, 24)

= ∑_{m_{1},m_{2},m_{3}=−∞}^{∞}(6m_{1} + 1)(6m_{2} + 1)(3m_{3} + 1)Q^{((6m_{1}+1)^{2}+(6m_{2}+1)^{2}+4(3m_{3}+1)^{2})/3}

= ∑_{n=1}^{∞} ( ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4x_{3}^{2}=3n \\ x_{1}≡x_{2}≡1 (mod 6), x_{3}≡1 (mod 3) \end{matrix}} x_{1}x_{2}x_{3})Q^{n},

i.e.,

w_{3}(n) = ∑_{\begin{matrix}x_{1}^{2}+x_{2}^{2}+4x_{3}^{2}=3n \\ x_{1}≡x_{2}≡1 (mod 6), x_{3}≡1 (mod 3) \end{matrix}} x_{1}x_{2}x_{3};

it is obvious that w_{3}(n) = 0 for n ≡ 1 (mod 2) and n ≡ 0 (mod 4). d) from (2.26) it follows that

\frac{1}{4096(πi)^{3}} ϑ_{16,0}^{′3}(τ; 0, 24)

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

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

i.e.,

w_{4}(n) = ∑_{\begin{matrix}4(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})=3n \\ x_{1}≡x_{2}≡x_{3}≡1 (mod 3) \end{matrix}} x_{1}x_{2}x_{3};

it is obvious that w_{4}(n) = 0 for n ≡ 1 (mod 2) and n ≡ 2 (mod 4).

The author expresses his gratitude to G. Gogishvili for providing formulas (2.13) and (2.14).

References

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3. G. A. Lomadze, On the representation of numbers by odd number of squares. (Russian) Trudy Tbiliss. Math. Inst. Razmadze 17(1949), 281–314.

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(Received 01.06.2000)

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