GEORGIAN MATHEMATICAL JOURNAL: Vol. 5, No. 4, 1998, 367-384

SUMMATION OF SINGULAR SERIES CORRESPONDING

TO REPRESENTATIONS OF NUMBERS BY SOME QUADRATIC FORMS IN TWELVE VARIABLES

G. LOMADZE

Abstract. Formulas for calculating the sum of singular series corresponding to the number of representations of integers by some quadratic forms in 12 variables with integral coefficient are derived.

Introduction

In this paper a formula is derived for the sum of the singular series corresponding to the number of representations of positive integers by positive primitive quadratic forms

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_{5}^{2} + x_{10}^{2}) + a_{6}(x_{11}^{2} + x_{12}^{2})  (1)

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

In our next paper a way to find explicit exact formulas for the number of representations of positive integers by the quadratic forms of type (1) will be suggested.

1. preliminaries

1.1. In this paper a, k, n, q, λ denote positive integers; b, m, u, v are odd positive integers; p is a prime number; α, β, γ, ν, l are non-negative integers,; h, j, x, y are integers; i is an imaginary unit; ∑_{hmodq}and^{∑′}_{hmodq} denote respectively sums in which h runs a complete and a reduced residue system modulo q; (\frac{h}{u}) is the generalized Jacobi symbol; σ_{5}(u) is the sum of the fifth powers of positive divisors of u; e(z) = e^{2πiz} for arbitrary complex number z; Δ is the determinant of the form (1).

367

1991 Mathematics Subject Classification. 11E99, 11P99.

Key words and phrases. Quadratic form, Gaussian sum, Ramanujan’s sum, singular series, Riemann ζ-function.

1072-947X/98/0700-0367$15.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1998 Plenum Publishing Corporation

368 G. LOMADZE

Let

S(h, q) = ∑_{jmodq} e(\frac{hj^{2}}{q} ) (Gaussian sum);  (1.1)
c(h, q) =\begin{matrix}∑ & ′ \\ jmodq \end{matrix} e(\frac{hj}{q} ) =\begin{matrix}∑ & ′ \\ jmodq \end{matrix} e( − \frac{hj}{q} ) (Ramanujan’s sum);  (1.2)
ρ(n; f) = \frac{π^{6}}{5!Δ^{1/2}} n^{5} ∑_{q=1}^{∞} A(q) (singular series of the problem),  (1.3)

where

A(q) = q^{−12}\begin{matrix}∑ & ′ \\ hmodq \end{matrix} e( − \frac{hn}{q}) ∏_{k=1}^{6} S^{2}(a_{k}h, q).  (1.4)

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

Lemma 1. If (h, q) = 1, then S(kh, kq) = kS(h, q).

Lemma 2 ([1], p. 13, Lemma 6). If (h, q) = 1, then

S^{2}(h, q) = (\frac{−1}{q})q for q ≡ 1 (mod 2),
= 2i^{h}q for q ≡ 0 (mod 4),
= 0 for q ≡ 2 (mod 4).

Lemma 3 ([1], p. 177, formula (20)). Let q = p^{λ} and p^{ν} || h. Then

c(h, q) = 0 for ν < λ − 1,
= −p^{λ−1} for ν = λ − 1,
= p^{λ−1}(p − 1) for ν > λ − 1.

Lemma 4. Let

χ_{p} = 1 + A(p) + A(p^{2}) + · · · .  (1.5)

Then

∑_{q=1}^{∞} A(q) = ∏_{p} χ_{p}.

SUMMATION OF SINGULAR SERIES 369

2. The summation of the singular series ρ(n; f) Lemma 5. Let n = 2^{α}m, a_{k} = 2^{γ_{k}}b_{k} (k = 1, 2, …, 6), γ_{6} ≥ γ_{5} ≥ γ_{4} ≥

γ_{3} ≥ γ_{2} ≥ γ_{1} = 0, γ = ∑_{k=2}^{6} γ_{k}, (b_{1}, b_{2}, …, b_{6}) = 1, b = [b_{1}, b_{2}, …, b_{6}]. Then

χ_{2} = 1 + (−1)^{(b_{1}−m)/2} for 0 ≤ α ≤ γ_{2} − 2;

= 1 for α = γ_{2} − 1, α = γ_{2} < γ_{3}, γ_{2} = γ_{3} ≤ α = γ_{4} − 1, γ_{2} + 1 = γ_{3} ≤ α = γ_{4} − 1, γ_{2} = γ_{3} ≤ α = γ_{4} < γ_{5}, γ_{2} + 1 = γ_{3} ≤ α = γ_{4} < γ_{5}, γ_{4} = γ_{5} ≤ α = γ_{6} − 1, γ_{4} + 1 = γ_{5} ≤ α = γ_{6} − 1, γ_{4} = γ_{5} ≤ α = γ_{6}, γ_{4} + 1 = γ_{5} ≤ α = γ_{6}

(in the last four conditions γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1);

χ_{2} = 2^{−α}{2^{α} + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}}(2^{α−γ_{2}} − 3)} for γ_{2} + 1 ≤ α < γ_{3},

= 2^{−2(α+1)}{2^{2(α+1)} + (−1)^{(∑_{k=1}^{3} b_{k}−m)/2}2^{γ_{2}+γ_{3}}}

for γ_{2} = γ_{3} ≤ α ≤ γ_{4} − 2, γ_{2} + 1 = γ_{3} ≤ α ≤ γ_{4} − 2,

= 2^{−2(α+1)}{2^{2(α+1)} + 2^{γ_{2}}((−1)^{(b_{1}+b_{2})/2}(2^{2(α+1)−γ_{2}} − 2^{2α−γ_{3}+3}) +

+ (−1)^{(∑_{k=1}^{3} b_{k}−m)/2} · 2^{γ_{3}})} for γ_{2} + 2 ≤ γ_{3} ≤ α = γ_{4} − 2,

= 2^{−γ_{3}}{2^{γ_{3}} + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}+1}(2^{γ_{3}−γ_{2}−1} − 1)}

for γ_{2} + 2 ≤ γ_{3} ≤ α = γ_{4} − 1, γ_{2} + 2 ≤ γ_{3} ≤ α = γ_{4} < γ_{5},

= 2^{−3α}{2^{3α} + (−1)^{(b_{1}+b_{2})/2} · 2^{3α+γ_{2}−γ_{3}+1}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{3}(2^{3(α−γ_{4}−1)} − 1)7^{−1} − 1)}

for γ_{4} + 1 ≤ α < γ_{5}, but γ_{3} ≥ γ_{2} + 2,

= 2^{−3α}{2^{3α} + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{3}(2^{3(α−γ_{4}−1)} − 1)7^{−1} − 1)}

for γ_{4} + 1 ≤ α < γ_{5}, but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1,

= 2^{−4(α+1)}{2^{4(α+1)} + (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}}}

370 G. LOMADZE

for γ_{4} = γ_{5} ≤ α ≤ γ_{6} − 2, γ_{4} + 1 = γ_{5} ≤ α ≤ γ_{6} − 2, but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1,

χ_{2} = 2^{−4(α+1)}{2^{4(α+1)} + (−1)^{(b_{1}+b_{k})/2} · 2^{4α+5+γ_{2}−γ_{3}}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}}} for γ_{4} = γ_{5} ≤ α ≤ γ_{6} − 2,

γ_{4} + 1 = γ_{5} ≤ α ≤ γ_{6} − 2, but γ_{3} ≥ γ_{2} + 2,

= 2^{−(γ_{3}−γ_{2}−1)}{2^{γ_{3}−γ_{2}−1} + (−1)^{(b_{1}+b_{2})/2}(2^{γ_{3}−γ_{2}−1} − 1)}

for γ_{4} = γ_{5} ≤ α = γ_{6} − 1, γ_{4} + 1 = γ_{5} ≤ α = γ_{6} − 1,

γ_{4} = γ_{5} ≤ α = γ_{6}, γ_{4} + 1 = γ_{5} ≤ α = γ_{6}, but γ_{3} ≥ γ_{2} + 2,

= 2^{−4(α+1)}{2^{4(α+1)} + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{4(α+1)−3γ_{4}} +

− 2^{4(α+1)−3(γ_{5}−1)})7^{−1} + (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}}}

for γ_{4} + 2 ≤ γ_{5} ≤ α ≤ γ_{6} − 2, but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1,

= 2^{−4(α+1)}{2^{4(α+1)} + (−1)^{(b_{1}+b_{2})/2} · 2^{4α+5+γ_{2}−γ_{3}}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{α+7}(2^{3(α−γ_{4}−1)} − 2^{3(α−γ_{5})})7^{−1}) +

+ (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}}} for γ_{4} + 2 ≤ γ_{5} ≤ α ≤ γ_{6} − 2,

but γ_{3} ≥ γ_{2} + 2,

= 2^{−4(α+1)}{2^{4(α+1)} + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}2^{α+7} ×

× (2^{3(α−γ_{4}−1)} − 2^{3(α−γ_{5}−1)})7^{−1}} for γ_{4} + 2 ≤ γ_{5} ≤ α = γ_{6} − 1,

γ_{4} + 2 ≤ γ_{5} ≤ α = γ_{6}, but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1,

= 2^{−3(γ_{5}−1)}{2^{3(γ_{5}−1)} + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−γ_{3}+3γ_{5}−2}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{3(γ_{5}−γ_{4}−1)} − 1)7^{−1}}

SUMMATION OF SINGULAR SERIES 371

for γ_{4} + 2 ≤ γ_{5} ≤ α = γ_{6}, γ_{4} + 2 ≤ γ_{5} ≤ α = γ_{6} − 1, but γ_{3} ≥ γ_{2} + 2,

χ_{2} = 2^{−5α}{2^{5α} + (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ}(2^{5}(2^{5(α−γ_{6}−1)} − 1)31^{−1} − 1)}

for α ≥ γ_{6} + 1, but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1 and γ_{5} = γ_{4} or γ_{5} ≤ γ_{4} + 1,

= 2^{−5α}{2^{5α} + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{5α+γ−4γ_{5}−γ_{6}+3}(2^{3(γ_{5}−γ_{4}−1)} − 1)7^{−1} +

+ (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ}(2^{5}(2^{5(α−γ_{6}+1)} − 1)31^{−1} − 1)} for α ≥ γ_{6} + 1,

but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1 and γ_{5} ≥ γ_{4} + 2,

= 2^{−5α}{2^{5α} + (−1)^{(b_{1}+b_{2})/2} · 2^{5α+γ_{2}−γ_{3}+1}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ}(2^{5}(2^{5(α−γ_{6}−1)} − 1)31^{−1} − 1)} for α ≥ γ_{6} + 1,

but γ_{3} ≥ γ_{2} + 2 and γ_{5} = γ_{4} or γ_{5} = γ_{4} + 1,

= 2^{−5α}{2^{5α} + (−1)^{(b_{1}+b_{2})/2} · 2^{5α+γ_{2}−γ_{3}+1}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{5α+γ−4γ_{5}−γ_{6}+3}(2^{3(γ_{5}−γ_{4}−1)} − 1)7^{−1} +

+ (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ}(2^{5}(2^{5(α−γ_{6}+1)} − 1)31^{−1} − 1)}

for α ≥ γ_{6} + 1, but γ_{3} ≥ γ_{2} + 2 and γ_{5} ≥ γ_{4} + 2.

Proof. I. If in (1.4) we put q = 2^{λ} and then instead of h introduce a new letter of summation y defined by the congruence h ≡ by (mod 2^{λ}), then we get

A(2^{λ}) = 2^{−12λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(−2^{α−λ}mby) ∏_{k=1}^{6} S^{2}(2^{γ_{k}}b_{k}by, 2^{λ}).  (2.1)

From (2.1), according to Lemmas 1, 2 and 3 it follows that:

(1) for λ = γ_{k} + 1, as S^{2}(2^{γ_{k}}b_{k}by, 2^{γ_{k}+1}) = 0 (k = 1, 2, …, 6),

A(2^{λ}) = 0;  (2.2)

372 G. LOMADZE

(2) for 2 ≤ λ ≤ γ_{2}

A(2^{λ}) = 2^{−12λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(−2^{α−λ}mby)(2i^{b_{i}by} · 2^{λ})2^{10λ} =

= 2^{1−λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(\frac{b_{1}by}{4} − \frac{2^{α}mby}{2^{λ}} ) =

= 2^{1−λ}e(\frac{b_{1}b − 2^{α−λ+2}mb}{4}) ∑_{y=0}^{2^{λ−1}−1} e(\frac{(2^{λ−2}b_{1} − 2^{α}m)by}{2^{λ−1}} ) =

= \begin{cases}e(\frac{b_{1}b − 2^{α−λ+2}mb}{4}) & if 2^{λ−1} | (2^{λ−2}b_{1} − 2^{α}m)b, \\ 0 & if 2^{λ−1} ∤ (2^{λ−2}b_{1} − 2^{α}m)b, \end{cases}

i.e.,

A(2^{λ}) = \begin{cases}(−1)^{(b_{1}−m)/2} & if λ = α + 2, \\ 0 & if λ ≠ α + 2; \end{cases} \begin{matrix}(2.3) \\ (2.3_{1}) \end{matrix}

(3) for γ_{2} + 2 ≤ λ ≤ γ_{3}

A(2^{λ}) = 2^{−12λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(−2^{α−λ}mby)(2i^{b_{i}by} · 2^{λ})2^{2γ_{2}}(2i^{b_{2}by} · 2^{λ−γ_{2}})2^{8λ} =

= (−1)^{(b_{1}+b_{2})/2} · 2^{−2λ+γ_{2}+2}c(2^{α}mb, 2^{λ}) =

= \begin{cases}(−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−(λ−1)} & if λ < α + 1, \\ (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−α} & if λ = α + 1, \\ 0 & if λ > α + 1; \end{cases} \begin{matrix}(2.4) \\ (2.4_{1}) \\ (2.4_{2}) \end{matrix}

(4) for γ_{3} + 2 ≤ λ ≤ γ_{4}

A(2^{λ}) = 2^{−12λ}\begin{matrix}∑ & ′ \\ ymod2^{λ} \end{matrix} e(−2^{α−λ}mby)(2i^{b_{i}by} · 2^{λ}) ×

× 2^{γ_{2}}(2i^{b_{2}by} · 2^{λ−γ_{2}})2^{γ_{3}}(2i^{b_{3}by} · 2^{λ−γ_{3}})2^{6λ} =

= 2^{−3λ+γ_{2}+γ_{3}+3}e(\frac{(b_{1} + b_{2} + b_{3})b}{4} − \frac{2^{α}mb}{2^{λ}}) ×

× ∑_{y=0}^{2^{λ−1}−1} e(\frac{(2^{λ−2}(b_{1} + b_{2} + b_{3}) − 2^{α}m)by}{2^{λ−1}} ) =

= \begin{cases}(−1)^{(b_{1}+b_{2}+b_{3}−m)/2} · 2^{γ_{2}+γ_{3}−2α−2} & if λ = α + 2, \\ 0 & if λ ≠ α + 2; \end{cases} \begin{matrix}(2.5) \\ (2.5_{1}) \end{matrix}

(5) for γ_{4} + 2 ≤ λ ≤ γ_{5}, similarly as in (3),

A(2^{λ}) = \begin{cases}(−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}−3(λ−1)} & if λ < α + 1, \\ −(−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}−3α} & if λ = α + 1, \\ 0 & if λ > α + 1; \end{cases} \begin{matrix}(2.6) \\ (2.6_{1}) \\ (2.6_{2}) \end{matrix}

(6) for γ_{5} + 2 ≤ λ ≤ γ_{6}, similarly as in (4),

A(2^{λ}) = \begin{cases}(−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}−4α−4} & if λ = α + 2, \\ 0 & if λ ≠ α + 2; \end{cases} \begin{matrix}(2.7) \\ (2.7_{1}) \end{matrix}

(7) for λ ≥ γ_{6} + 2, similarly as in (3) and in (5),

A(2^{λ}) = \begin{cases}(−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ−5(λ−1)} & if λ < α + 1, \\ −(−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ−5α} & if λ = α + 1, \\ 0 & if λ > α + 1; \end{cases} \begin{matrix}(2.8) \\ (2.8_{1}) \\ (2.8_{2}) \end{matrix}

II. According to (1.5) and (2.2), we have

χ_{2} = 1 + ∑_{λ=2}^{γ_{2}} A(2^{λ}) + ∑_{λ=γ_{2}+2}^{γ_{3}} A(2^{λ}) + ∑_{λ=γ_{3}+2}^{γ_{4}} A(2^{λ}) +

+ ∑_{λ=γ_{4}+2}^{γ_{5}} A(2^{λ}) + ∑_{λ=γ_{5}+2}^{γ_{6}} A(2^{λ}) + ∑_{λ=γ_{6}+2}^{∞} A(2^{λ}).  (2.9)

Consider the following cases:

(1) Let 0 ≤ α ≤ γ_{2} − 2. Then from (2.9), (2.3), (2.3_{1}), (2.4_{2}), (2.5_{1}), (2.6_{2}), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1 + ∑_{λ=2}^{γ_{2}} A(2^{λ}) = 1 + (−1)^{(b_{1}−m)/2}.

(2) Let α = γ_{2} − 1, or α = γ_{2} < γ_{3} or γ_{2} = γ_{3} ≤ α = γ_{4} − 1 or γ_{2} + 1 = γ_{3} ≤ α = γ_{4} − 1 or γ_{2} = γ_{3} ≤ α = γ_{4} < γ_{5} or γ_{2} + 1 = γ_{3} ≤ α = γ_{4} < γ_{5} or γ_{4} = γ_{5} ≤ α = γ_{6} − 1 or γ_{4} + 1 = γ_{5} ≤ α = γ_{6} − 1, or γ_{4} = γ_{5} ≤ α = γ_{6} or γ_{4} + 1 = γ_{5} ≤ α = γ_{6} (in the last four conditions γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1). Then from (2.9), (2.3_{1}), (2.4_{2}), (2.5_{1}), (2.6_{2}), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1.

(3) Let γ_{2} + 1 ≤ α < γ_{3}. Then from (2.9), (2.3_{1}), (2.4)–(2.4_{2}), (2.5_{1}), (2.6_{2}), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}} ∑_{λ=γ_{2}+2}^{α} 2^{−(λ−1)} − (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−α} =

= 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}}(2^{−γ_{2}−1} − 2^{−α})2 − (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−α} =

= 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}}(2^{−γ_{2}} − 3 · 2^{−α}).

(4) Let γ_{2} = γ_{3} ≤ α ≤ γ_{4} − 2 or γ_{2} + 1 = γ_{3} ≤ α ≤ γ_{4} − 2. Then from (2.9), (2.3_{1}), (2.4_{2}), (2.5), (2.5_{1}), (2.6_{2}), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2}+b_{3}−m)/2} · 2^{γ_{2}+γ_{3}−2α−2}.

(5) Let γ_{2} + 2 ≤ γ_{3} ≤ α ≤ γ_{4} − 2. Then from (2.9), (2.3_{1}), (2.4), (2.5), (2.5_{1}), (2.6_{2}), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}} ∑_{λ=γ_{2}+2}^{γ_{3}} 2^{−(λ−1)} +

+ (−1)^{(b_{1}+b_{2}+b_{3}−m)/2} · 2^{γ_{2}+γ_{3}−2α−2} =

= 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}}(2^{−γ_{2}} − 2^{−γ_{3}+1}) +

+ (−1)^{(∑_{k=1}^{3} b_{k}−m)/2} · 2^{γ_{2}+γ_{3}−2α−2}.

(6) Let γ_{2} + 2 ≤ γ_{3} ≤ α = γ_{4} − 1 or γ_{2} + 2 ≤ γ_{3} ≤ α = γ_{4} < γ_{5}. Then from (2.9), (2.3_{1}), (2.4), (2.4_{2}), (2.5_{1}), (2.6_{2}), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}}(2^{−γ_{2}} − 2^{−γ_{3}+1}).

(7) Let γ_{4} + 1 ≤ α < γ_{5}, but γ_{3} ≥ γ_{2} + 2. Then from (2.9), (2.3_{1}), (2.4), (2.5_{1}), (2.6), (2.6_{1}), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}} ∑_{λ=γ_{2}+2}^{γ_{3}} 2^{−(λ−1)} +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}} ∑_{λ=γ_{4}+2}^{α} 2^{−3(λ−1)} − (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}−3α} =

= 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}}(2^{−γ_{2}} − 2^{−γ_{3}+1}) +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{−3γ_{4}−3} − 2^{−3α})2^{3} · 7^{−1} +

− (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}−3α} =

= 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−γ_{3}+1}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{−3α+3}(2^{3(α−γ_{4}−1)} − 1)7^{−1} − 2^{−3α}).

(8) Let γ_{4} + 1 ≤ α < γ_{5}, but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1. Then from (2.9), (2.3_{1}), (2.4_{2}), (2.5_{1}), (2.6), (2.6_{1}), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}} ∑_{λ=γ_{4}+2}^{α} 2^{−3(λ−1)} − (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}−3α}.

(9) Letγ_{4} = γ_{5} ≤ α ≤ γ_{6} − 2 or γ_{4} + 1 = γ_{5} ≤ α ≤ γ_{6} − 2, but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1. Then from (2.9), (2.3_{1}), (2.4_{2}), (2.5_{1}), (2.6_{2}), (2.7), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}−4α−4}.

(10) Let γ_{4} = γ_{5} ≤ α ≤ γ_{6} − 2 or γ_{4} + 1 = γ_{5} ≤ α ≤ γ_{6} − 2, but γ_{3} ≥ γ_{2} + 2. Then from (2.9), (2.3_{1}), (2.4), (2.5_{1}), (2.6_{2}), (2.7), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}} ∑_{λ=γ_{2}+2}^{γ_{3}} 2^{−(λ−1)} +

+ (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}−4α−4} =

= 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−γ_{3}+1}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}−4α−4}.

(11) Let γ_{4} = γ_{5} ≤ α = γ_{6} − 1 or γ_{4} + 1 = γ_{5} ≤ α = γ_{6} − 1or γ_{4} = γ_{5} ≤ α = γ_{6} or γ_{4} + 1 = γ_{5} ≤ α = γ_{6}, but γ_{3} ≥ γ_{2} + 2. Then from (2.9), (2.3_{1}), (2.4), (2.5_{1}), (2.6_{2}), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}} ∑_{λ=γ_{2}+2}^{γ_{3}} 2^{−(λ−1)} =
= 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−γ_{3}+1}(2^{γ_{3}−γ_{2}−1} − 1).

(12) Let γ_{4} + 2 ≤ γ_{5} ≤ α ≤ γ_{6} − 2, but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1. Then from (2.9), (2.3_{1}), (2.4_{2}), (2.5_{1}), (2.6), (2.7), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}} ∑_{λ=γ_{4}+2}^{γ_{5}} 2^{−3(λ−1)} +

+ (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}−4α−4} =

= 1 + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{−3γ_{4}−3} − 2^{−3γ_{5}})2^{3} · 7^{−1} +

+ (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}−4α−4}.

(13) Let γ_{4} + 2 ≤ γ_{5} ≤ α ≤ γ_{6} − 2, but γ_{3} ≥ γ_{2} + 2. Then from (2.9), (2.3_{1}), (2.4), (2.5_{1}), (2.6), (2.7), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}} ∑_{λ=γ_{2}+2}^{γ_{3}} 2^{−(λ−1)} +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}} ∑_{λ=γ_{4}+2}^{γ_{5}} 2^{−3(λ−1)} +

+ (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}−4α−4} =

= 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−γ_{3}+1}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{−3γ_{4}} − 2^{−3γ_{5}+3})7^{−1} +

+ (−1)^{(∑_{k=1}^{5} b_{k}−m)/2} · 2^{γ−γ_{6}−4α−4}.

(14) Let γ_{4} + 2 ≤ γ_{5} ≤ α = γ_{6} − 1 or γ_{4} + 2 ≤ γ_{5} ≤ α = γ_{6}, but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1. Then from (2.9), (2.3_{1}), (2.4_{2}), (2.5_{1}), (2.6), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{−3γ_{4}} − 2^{−3γ_{5}+3})7^{−1}.

(15) Let γ_{4} + 2 ≤ γ_{5} ≤ α = γ_{6} − 1 or γ_{4} + 2 ≤ γ_{5} ≤ α = γ_{6}, but γ_{3} ≥ γ_{2} + 2. Then from (2.9), (2.3_{1}), (2.4), (2.5_{1}), (2.6), (2.7_{1}), and (2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−γ_{3}+1}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{−3γ_{4}} − 2^{−3γ_{5}+3})7^{−1}.

(16) Let α ≥ γ_{6}+1, but γ_{3} = γ_{2} or γ_{3} = γ_{2}+1 and γ_{5} = γ_{4} or γ_{5} = γ_{4}+1. Then from (2.9), (2.3_{1}), (2.4_{2}), (2.5_{1}), (2.6_{2}), (2.7_{1}), (2.8)–(2.8_{2}) we get

χ_{2} = 1 + (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ} ∑_{λ=γ_{6}+2}^{α} 2^{−5(λ−1)} − (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ−5α} =

= 1 + (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ}(2^{−5γ_{6}−5} − 2^{−5α})2^{5} · 31^{−1} +

− (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ−5α} =

= 1 + (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ}(2^{−5(α−1)}(2^{5(α−γ_{6}−1)} − 1)31^{−1} − 2^{−5α}).

(17) Let α ≥ γ_{6} + 1, but γ_{3} = γ_{2} or γ_{3} = γ_{2} + 1 and γ_{5} ≥ γ_{4} + 2. Then from (2.9), (2.3_{1}), (2.4_{2}), (2.5_{1}), (2.6), (2.7_{1}), (2.8)–(2.8_{2}) we get

χ_{2} = 1 + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{∑_{k=2}^{4} γ_{k}}(2^{−3γ_{4}} − 2^{−3γ_{5}+3})7^{−1} +

+ (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ}((2^{−5γ_{6}−5} − 2^{−5α})2^{5} · 31^{−1} − 2^{−5α}) =

= 1 + (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ−4γ_{5}−γ_{6}+3}(2^{3(γ_{5}−γ_{4}−1)} − 1)7^{−1} +

+ (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ}(2^{−5(α−1)}(2^{5(α−γ_{6}−1)} − 1)31^{−1} − 2^{−5α}).

(18) Let α ≥ γ_{6} + 1, but γ_{3} ≥ γ_{2} + 2 and γ_{5} = γ_{4} or γ_{5} = γ_{4} + 1. Then from (2.9), (2.3_{1}), (2.4), (2.5_{1}), (2.6_{2}), (2.7_{1}), (2.8)–(2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−γ_{3}+1}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ}(2^{−5(α−1)}(2^{5(α−γ_{6}−1)} − 1)31^{−1} − 2^{−5α}).

(19) Let α ≥ γ_{6} + 1, but γ_{3} ≥ γ_{2} + 2 and γ_{5} ≥ γ_{4} + 2. Then from (2.9), (2.3_{1}), (2.4), (2.5_{1}), (2.6), (2.7_{1}), (2.8)–(2.8_{2}) we get

χ_{2} = 1 + (−1)^{(b_{1}+b_{2})/2} · 2^{γ_{2}−γ_{3}+1}(2^{γ_{3}−γ_{2}−1} − 1) +

+ (−1)^{(∑_{k=1}^{4} b_{k})/2} · 2^{γ−4γ_{5}−γ_{6}+3}(2^{3(γ_{5}−γ_{4}−1)} − 1)7^{−1} +

+ (−1)^{(∑_{k=1}^{6} b_{k})/2} · 2^{γ}(2^{−5(α−1)}(2^{5(α−γ_{6}−1)} − 1)31^{−1} − 2^{−5α}).

From all that was said above the formulas for χ_{2} follow.

Lemma 6. Let p > 2, p^{β} || n, p^{l_{k}} || a_{k} (k = 1, 2, …, 6). Let the values of l_{k} taken in decreasing order be l_{6} ≥ l_{5} ≥ l_{4} ≥ l_{3} ≥ l_{2} ≥ l_{1} = 0, l = ∑_{k=1}^{6} l_{k}; η(l_{2}) = 1 if 2 | l_{2} and η(l_{2}) = 0 if 2 ∤ l_{2}. Then

χ_{p} = p^{−1}(p − 1)(β + 1) for l_{2} ≥ β + 1, p ≡ 1 (mod 4),

= p^{−1}(p + 1) for l_{2} ≥ β + 1, p ≡ 3 (mod 4), 2 | β, = 0 for l_{2} ≥ β + 1, p ≡ 3 (mod 4), 2 ∤ β;

χ_{p} = p^{−(β−l_{2}+2)}{p^{β−l_{2}+1}((p + 1) + (p − 1)l_{2}) − (p + 1)}
for l_{2} ≤ β < l_{3}, p ≡ 1 (mod 4),

= −p^{−(β−l_{2}+2)}(p + 1){p^{β−l_{2}+1}η(l_{2}) − (−1)^{l_{2}}}

for l_{2} ≤ β < l_{3}, p ≡ 3 (mod 4);
χ_{p} = p^{−(2β−l_{2}−l_{3}+3)}{p^{2(β+1)−l_{2}−l_{3}}((p + 1) + (p − 1)l_{2}) +

+ p^{2}((p^{2(β−l_{3})} − 1)(p + 1)^{−1} − p^{2(β−l_{3})}) − 1}

for l_{3} ≤ β < l_{4}, p ≡ 1 (mod 4),
χ_{p} = p^{−(2β−l_{2}−l_{3}+3)}{p^{2β−l_{2}−l_{3}+2}(p + 1)η(l_{2}) − (−1)^{l_{2}}p^{2} ×

× (p^{2(β−l_{3})} + (p^{2(β−l_{3})} − (−1)^{β+l_{3}})(p − 1)(p^{2} + 1)^{−1}) + (−1)^{β+l_{2}+l_{3}}}

for l_{3} ≤ β < l_{4}, p ≡ 3 (mod 4);

χ_{p} = p^{−(3β−l_{2}−l_{3}−l_{4}+4)}{p^{3(β+1)−l_{2}−l_{3}−l_{4}}((p + 1) + (p − 1)l_{2}) +

+ p^{3(β−l_{4}+1)}((p^{2(l_{4}−l_{3})} − 1)(p + 1)^{−1} − p^{2(l_{4}−l_{3})}) +

+ p^{3}(p^{3(β−l_{4})} − 1)(p − 1)(p^{3} − 1)^{−1} − 1}

for l_{4} ≤ β < l_{5}, p ≡ 1 (mod 4),
χ_{p} = p^{−(3β−l_{2}−l_{3}−l_{4}+4)}{p^{3β−l_{2}−l_{3}−l_{4}+3}(p + 1)η(l_{2}) − (−1)^{l_{2}}p^{3(β−l_{4}+1)} ×

× (p^{2(l_{4}−l_{3})} + ((−1)^{l_{3}−l_{4}+1}p^{2(l_{4}−l_{3})} − 1)(p − 1)(p^{2} + 1)^{−1}) +

+ (−1)^{l_{2}+l_{3}+l_{4}}(p^{3}(p^{3(β−l_{4})} − 1)(p − 1)(p^{3} − 1)^{−1} − 1}

for l_{4} ≤ β < l_{5}, p ≡ 3 (mod 4);
χ_{p} = p^{−(4β−l+l_{6}+5)}{p^{4(β+1)−l+l_{6}}((p + 1) + (p − 1)l_{2}) +

+ p^{4(β−l_{5}+1)}(p^{3(l_{5}−l_{4})}((p^{2(l_{4}−l_{3})} − 1)(p + 1)^{−1} − p^{3l_{5}−2l_{3}−l_{4}}) +

+ (p^{3(l_{5}−l_{4})} − 1)(p − 1)(p^{3} − 1)^{−1}) +

+ (p^{4(β−l_{5})} − 1)p^{4}(p − 1)(p^{4} − 1)^{−1} − 1}

for l_{5} ≤ β < l_{6}, p ≡ 1 (mod 4),
χ_{p} = p^{−(4β−l+l_{6}+5)}{p^{4β−l+l_{6}+4}(p + 1)η(l_{2}) − (−1)^{l_{2}}p^{4(β+1)−3l_{4}−l_{5}} ×

× (p^{2(l_{4}−l_{3})} + (p^{2(l_{4}−l_{3})} − (−1)^{l_{3}+l_{4}})(p − 1)(p^{2} + 1)^{−1}) +

+ (−1)^{l_{2}+l_{3}+l_{4}}p^{4(β+1−l_{5})}(p^{3(l_{5}−l_{4})} − 1)(p − 1)(p^{3} − 1)^{−1} +

+ (−1)^{l−l_{6}}(((−1)^{l_{5}+1}p^{4(β−l_{5})} + (−1)^{β})(p − 1)p^{4}(p^{4} + 1)^{−1} + (−1)^{β})}

for l_{5} ≤ β < l_{6}, p ≡ 3 (mod 4);
χ_{p} = p^{−(5β−l+6)}{p^{5(β+1)−l}((p + 1) + (p − 1)l_{2}) +

+ p^{5(β+1)−4l_{5}−l_{6}}(p^{3(l_{5}−l_{4})}((p^{2(l_{4}−l_{3})} − 1)(p + 1)^{−1} − p^{2(l_{4}−l_{3})}) +

+ (p^{3(l_{5}−l_{4})} − 1)(p − 1)(p^{3} − 1)^{−1}) +

+ p^{5}(p − 1)(p^{5(β−l_{6})}(p^{4(l_{6}−l_{5})} − 1)(p^{4} + 1)^{−1} +

+ (p^{5(β−l_{6})} − 1)(p^{5} − 1)^{−1}) − 1} for β ≥ l_{6}, p ≡ 1 (mod 4),

χ_{p} = p^{−(5β−l+6)}{p^{5β−l+5}(p + 1)η(l_{2}) − (−1)^{l_{2}}p^{5(β+1)−l+l_{2}+l_{3}−2l_{4}} ×

× (p^{2(l_{4}−l_{3})} + (p^{2(l_{4}−l_{3})} − (−1)^{l_{3}−l_{4}})(p − 1)(p^{2} + 1)^{−1}) +

+ (−1)^{l_{2}+l_{3}+l_{4}}p^{5(β_{+}1)−4l_{5}−l_{6}}(p^{3(l_{5}−l_{4})} − 1)(p − 1)(p^{3} − 1)^{−1} +

− (−1)^{l}p^{5}(p − 1)(p^{5(β−l_{6})}((−1)^{l_{5}−l_{6}}p^{4(l_{6}−l_{5})} − 1)(p^{4} + 1)^{−1} +

− (p^{5(β−l_{6})} − 1)(p^{5} − 1)^{−1}) − (−1)^{l}} for β ≥ l_{6}, p ≡ 3 (mod 4).

Proof. I. In (1.4) put 2 ∤ q, q = (q, a_{k})q_{k} (k = 1, 2, …, 6). Then from Lemmas 1 and 2 we get

A(q) = q^{−12}\begin{matrix}∑ & ′ \\ hmodq \end{matrix} e( − \frac{hn}{q}) ∏_{k=1}^{6}(q, a_{k})^{2}S^{2}(\frac{a_{k}}{(q, a_{k})} h, q_{k}) =

= q^{−6}\begin{matrix}∑ & ′ \\ hmodq \end{matrix} e( − \frac{hn}{q}) ∏_{k=1}^{6}(q, a_{k})(\frac{−1}{q_{k}}).

Putting q = p^{λ} and taking into account that (a_{1}, …, a_{6}) = 1, it follows that

A(p^{λ}) = p^{−6λ}\begin{matrix}∑ & ′ \\ hmodp^{λ} \end{matrix} e( − \frac{hn}{p^{λ}})p^{∑_{k=2}^{6} min(λ,l_{k})}(\frac{−1}{p})^{∑_{k=2}^{6} min(λ,l_{k})} =

= (\frac{−1}{p})^{∑_{k=2}^{6} min(λ,l_{k})}p^{∑_{k=2}^{6} min(λ,l_{k})} · p^{−6λ}c(n, p^{λ}).  (2.10)

From (2.10) amd Lemma 3 it follows that

(1) for λ ≤ l_{2}

A(p^{λ}) = (\frac{−1}{p})^{λ}(1 − p^{−1}) if λ < β + 1,  (2.11)
= −(\frac{−1}{p})^{β+1}p^{−1} if λ = β + 1;  (2.11_{1})

(2) for l_{2} < λ ≤ l_{3}

A(p^{λ}) = (\frac{−1}{p})^{l_{2}}p^{l_{2}−λ}(1 − p^{−1}) if λ < β + 1,  (2.12)
= −(\frac{−1}{p})^{l_{2}}p^{l_{2}−β−2} if λ = β + 1;  (2.12_{1})

SUMMATION OF SINGULAR SERIES 381

(3) for l_{3} < λ ≤ l_{4}

A(p^{λ}) = (\frac{−1}{p})^{l_{2}+l_{3}+λ}p^{l_{2}+l_{3}−2λ}(1 − p^{−1}) if λ < β + 1,  (2.13)
= −(\frac{−1}{p})^{l_{2}+l_{3}+β+1}p^{l_{2}+l_{3}−2β−3} if λ = β + 1;  (2.13_{1})

(4) for l_{4} < λ ≤ l_{5}

A(p^{λ}) = (\frac{−1}{p})^{l_{2}+l_{3}+l_{4}}p^{l_{2}+l_{3}+l_{4}−3λ}(1 − p^{−1}) if λ < β + 1,  (2.14)
= −(\frac{−1}{p})^{l_{2}+l_{3}+l_{4}}p^{l_{2}+l_{3}+l_{4}−3β−4} if λ = β + 1;  (2.14_{1})

(5) for l_{5} < λ ≤ l_{6}

A(p^{λ}) = (\frac{−1}{p})^{l−l_{6}+λ}p^{l−l_{6}−4λ}(1 − p^{−1}) if λ < β + 1,  (2.15)
= −(\frac{−1}{p})^{l−l_{6}+β+1}p^{l−l_{6}−4β−5} if λ = β + 1;  (2.15_{1})

(6) for λ ≥ l_{6}

A(p^{λ}) = (\frac{−1}{p})^{l}p^{l−5λ}(1 − p^{−1}) if λ < β + 1,  (2.16)
= −(\frac{−1}{p})^{l}p^{l−5β−6} if λ = β + 1.  (2.16_{1})

In all the above-mentioned cases

A(p^{λ}) = 0 if λ > β + 1.  (2.17)

II. According to (1.5) we have

χ_{p} = 1 + ∑_{λ=1}^{l_{2}} A(p^{λ}) + ∑_{λ=l_{2}+1}^{l_{3}} A(p^{λ}) + ∑_{λ=l_{3}+1}^{l_{4}} A(p^{λ}) +
+ ∑_{λ=l_{4}+1}^{l_{5}} A(p^{λ}) + ∑_{λ=l_{5}+1}^{l_{6}} A(p^{λ}) + ∑_{λ=l_{6}+1}^{∞} A(p^{λ}).  (2.18)

Also it is obvious that

1 + (1 − p^{−1}) ∑_{λ=1}^{l_{2}}(−1)^{λ} = \begin{cases}1 & if 2 | l_{2}, \\ p^{−1} & if 2 ∤ l_{2}. \end{cases}  (2.19)

Consider the following cases:

(1) Let l_{2} ≥ β + 1. Then from (2.18), (2.11), (2.11_{1}), and (2.17) we get

χ_{p} = 1 + (1 − p^{−1}) ∑_{λ=1}^{β} (\frac{−1}{p})^{λ} − (\frac{−1}{p})^{β+1}p^{−1}.

(2) Let l_{2} ≤ β < l_{3}. Then from (2.18), (2.11), (2.12), (2.12_{1}), and (2.17) we get

χ_{p} = 1 + (1 − p^{−1}) ∑_{λ=1}^{l_{2}} (\frac{−1}{p})^{λ} +

+ (1 − p^{−1})(\frac{−1}{p})^{l_{2}} ∑_{λ=l_{2}+1}^{β} p^{l_{2}−λ} − (\frac{−1}{p})^{l_{2}}p^{l_{2}−β−2}.

(3) Let l_{3} ≤ β < l_{4}. Then from (2.18), (2.11), (2.12), (2.13), (2.13_{1}), and (2.17) we get

χ_{p} = 1 + (1 − p^{−1}) ∑_{λ=1}^{l_{2}} (\frac{−1}{p})^{λ} + (1 − p^{−1})(\frac{−1}{p})^{l_{2}} ∑_{λ=l_{2}+1}^{l_{3}} l^{l_{2}−λ} +

+ (1 − p^{−1})(\frac{−1}{p})^{l_{2}} ∑_{λ=l_{3}+1}^{β} (\frac{−1}{p})^{λ+l_{3}}p^{l_{2}+l_{3}−2λ} +

− (\frac{−1}{p})^{l_{2}+l_{3}+β+1}p^{l_{2}+l_{3}−2β−3}.

(4) Let l_{4} ≤ β < L_{5}. Then from (2.18), (2.11), (2.12), (2.13), (2.14), (2.14_{1}), and (2.17) we get

χ_{p} = 1 + (1 − p^{−1}) ∑_{λ=1}^{l_{2}} (\frac{−1}{p})^{λ} + (1 − p^{−1})(\frac{−1}{p})^{l_{2}} ∑_{λ=l_{2}+1}^{l_{3}} p^{l_{2}−λ} +

+ (1 − p^{−1})(\frac{−1}{p})^{l_{2}+l_{3}} ∑_{λ=l_{3}+1}^{l_{4}} (\frac{−1}{p})^{λ}p^{l_{2}+l_{3}−2λ} + (1 − p^{−1}) ×

× (\frac{−1}{p})^{l_{2}+l_{3}+l_{4}} ∑_{λ=l_{4}+1}^{β} p^{l_{2}+l_{3}+l_{4}−3λ} − (\frac{−1}{p})^{l_{2}+l_{3}+l_{4}}p^{l_{2}+l_{3}+l_{4}−3β−4}.

(5) Let l_{5} ≤ β < l_{6}. Then from (2.18), (2.11)–(2.15), (2.15_{1}), and (2.17) we get

χ_{p} = 1 + (1 − p^{−1}) ∑_{λ=1}^{l_{2}} (\frac{−1}{p})^{λ} + (1 − p^{−1})(\frac{−1}{p})^{l_{2}} ∑_{λ=l_{2}+1}^{l_{3}} p^{l_{2}−λ} +

+ (1 − p^{−1})(\frac{−1}{p})^{l_{2}+l_{3}} ∑_{λ=l_{3}+1}^{l_{4}} (\frac{−1}{p})^{λ}p^{l_{2}+l_{3}−2λ} +

+ (1 − p^{−1})(\frac{−1}{p})^{l_{2}+l_{3}+l_{4}} ∑_{λ=l_{4}+1}^{l_{5}} p^{l_{2}+l_{3}+l_{4}−3λ} + (1 − p^{−1})(\frac{−1}{p})^{l−l_{6}} ×

× ∑_{λ=l_{5}+1}^{β} (\frac{−1}{p})^{λ}p^{l−l_{6}−4λ} − (\frac{−1}{p})^{l−l_{6}+β+1}p^{l−l_{6}−4β−5}.

(6) Let β ≥l_{6}. Then from (2.18), (2.11)–(2.16), (2.16_{1}), and (2.17) we get

χ_{p} = 1 + (1 − p^{−1}) ∑_{λ=1}^{l_{2}} (\frac{−1}{p})^{λ} + (1 − p^{−1})(\frac{−1}{p})^{l_{2}} ∑_{λ=l_{2}+1}^{l_{3}} p^{l_{2}−λ} +

+ (1 − p^{−1})(\frac{−1}{p})^{l_{2}+l_{3}} ∑_{λ=l_{3}+1}^{l_{4}} (\frac{−1}{p})^{λ}p^{l_{2}+l_{3}−2λ} +

+ (1 − p^{−1})(\frac{−1}{p})^{l_{2}+l_{3}+l_{4}} ∑_{λ=l_{4}+1}^{l_{5}} p^{l_{2}+l_{3}+l_{4}−3λ} +

+ (1 − p^{−1})(\frac{−1}{p})^{l−l_{6}} ∑_{λ=l_{5}+1}^{l_{6}} (\frac{−1}{p})^{λ}p^{l−l_{6}−4λ} +

+ (1 − p^{−1})(\frac{−1}{p})^{l} ∑_{λ=l_{6}+1}^{β} p^{l−5λ} − (\frac{−1}{p})^{l}p^{l−5β−6}.  (2.20)

From the above expressions for χ_{p} one can get the corresponding formulas stated in the lemma in the cases where p ≡ 1 (mod 4) and p ≡ 3 (mod 4).

Theorem. Let n = 2^{α}m = 2^{α}uv, u = ∏_{p|n,p∤2Δ} p^{β}, v = ∏_{\begin{matrix}p|n \\ p|Δ,p>2 \end{matrix}} p^{β}. Then

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

Proof. Let p > 2, p^{β} || n, p ∤ Δ (i.e., l = 0). Then in (2.20), putting l = 0, we get

χ_{p} = 1 + (1 − p^{−1}) ∑_{λ=1}^{β} p^{−5λ} − p^{−5β−6} =

= 1 + ∑_{λ=1}^{β} p^{−5λ} − p^{−6} ∑_{λ=1}^{β} p^{−5(λ−1)} − p^{−5β−6} =

= ∑_{λ=0}^{β} p^{−5λ} − p^{−6} ∑_{λ=0}^{β} p^{−5λ} = (1 − p^{−6}) ∑_{d|p^{β}} d^{−5}.  (2.21)

For p ∤ Δn, i.e., for β = 0, from (2.21) we get

χ_{p} = 1 − p^{−6}.  (2.22)

Thus from Lemma 4, (2.21), and (2.22) it follows that

∑_{q=1}^{∞} A(q) = χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{\begin{matrix}p∤Δ \\ p>2 \end{matrix}} χ_{p} = χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{\begin{matrix}p|n \\ p∤2Δ \end{matrix}} χ_{p} ∏_{\begin{matrix}p∤Δn \\ p>2 \end{matrix}} χ_{p} =

= χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{\begin{matrix}p|n \\ p∤2Δ \end{matrix}} ((1 − p^{−6}) ∑_{d|p^{β}} d^{−5}) ∏_{\begin{matrix}p∤Δn \\ p>2 \end{matrix}}(1 − p^{−6}) =

= χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{p>2}(1 − p^{−6}) ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}}(1 − p^{−6})^{−1} ∑_{d|u} d^{−5} =

= \frac{2^{6} · 3 · 5}{π^{6}} χ_{2} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}} χ_{p} ∏_{\begin{matrix}p|Δ \\ p>2 \end{matrix}}(1 − p^{−6})^{−1} \frac{1}{u^{5}} σ_{5}(u),  (2.23)

as it is well known that

∏_{p>2}(1 − p^{−6}) = (1 − 2^{−6})^{−1}ζ^{−1}(6) = \frac{2^{6} · 3 · 5}{π^{6}}.

Thus the theorem follows from (1.3) and (2.23).

References

1. A. Walfisz, Gitterpunkte in mehrdimensionalen Kugeln. Panstwowe Wydawnictwo Naukowe, Warszawa, 1957.

(Received 03.05.1996)

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