Long and Short Sums of a Twisted Divisor Function

Olivier Bordellès 2, allée de la Combe 43000 Aiguilhe France

                  borde43@wanadoo.fr

Abstract

Let q > 2 be a prime number and define λ_{q} := (\frac{τ}{q} ) where τ(n) is the number of divisors of n and (\frac{·}{q} ) is the Legendre symbol. When τ(n) is a quadratic residue modulo q, then the convolution (λ_{q} ⋆ 1) (n) could be close to the number of divisors of n. The aim of this work is to compare the mean value of the function λ_{q} ⋆ 1 to the well known average order of τ. A bound for short sums in the case q = 5 is also given, using profound results from the theory of integer points close to certain smooth curves.

1 Introduction and main result

If Ω(n) stands for the number of total prime factors of n and λ = (−1)^{Ω} is the Liouville function, then

L(s, λ) = \frac{ζ(2s)}{ζ(s)} (σ > 1) .

This implies the convolution identity

∑_{n⩽x} (λ ⋆ 1) (n) = ⌊x^{1/2}⌋ ,

where, as usual, F ⋆ G is the Dirichlet convolution product of the arithmetic functions F and G given by

(F ⋆ G)(n) := ∑_{d|n} F (d)G(n/d).

Define λ_{3} := (\frac{τ}{3}) where τ(n) is the number of divisors of n and (\frac{·}{3}) is the Legendre symbol. Then from Proposition 3 below

L(s, λ_{3}) = \frac{ζ(3s)}{ζ(s)} (σ > 1) ,

implying the convolution identity

∑_{n⩽x} (λ_{3} ⋆ 1) (n) = ⌊x^{1/3}⌋ .

Now let q > 2 be a prime number and define λ_{q} := (\frac{τ}{q} ) where (\frac{·}{q}) is the Legendre symbol. Our main aim is to investigate the sum

∑_{n⩽x} (λ_{q} ⋆ 1) (n).

When τ(n) is a quadratic residue modulo q, one may wonder if (λ_{q} ⋆ 1) (n) has a high probability to be equal to the number of divisors of n. Note that this function is multiplicative, and, for any prime p, (λ_{q} ⋆ 1) (p) = 1 + (\frac{2}{q}). Consequently, when 2 is a quadratic residue modulo q, then (λ_{q} ⋆ 1) (n) = τ(n) for all squarefree numbers n. On the other hand, when 2 is a quadratic nonresidue modulo q, then (λ_{q} ⋆ 1) (n) = 0 unless n = 1 or n is squarefull, so that ∑_{n⩽x} (λ_{q} ⋆ 1) (n) ≪ x^{1/2} in this case. It then could be interesting to compare this sum to the average order of the function τ, i.e.,

∑_{n⩽x} τ(n) = x (log x + 2γ − 1) + O (x^{θ+ε}) ,  (1)

where

\frac{1}{4} ⩽ θ ⩽ \frac{131}{416},  (2)

the left-hand side being established by Hardy [5], the right-hand side being the best estimate to date due to Huxley [6].

To state our first main result, some specific notation are needed. For any prime q > 3, let c_{q}, respectively d_{q}, be least positive integer m ∈ \{1, …, q − 2\} for which (\frac{m}{q}) ≠ (\frac{m+1}{q}),

respectively (\frac{m}{q}) ≠ − (\frac{m+1}{q}). Note that c_{q} and d_{q} are well-defined, since it is known from

[3, p. 75-76] that the number of m for which (\frac{m}{q}) = (\frac{m+1}{q}) and (\frac{m}{q}) = − (\frac{m+1}{q}) are respectively \frac{1}{2}(q − 3) and \frac{1}{2}(q − 1). Hence there is at least \frac{1}{2}(q − 3) integers m for which

(\frac{m}{q}) ≠ ± (\frac{m+1}{q}). For convenience, set d_{3} = 3.

As usual in number theory, we adopt Riemann’s notation s = σ + it ∈ ℂ and ζ is the Riemann zeta function, and define the Euler products

P_{q}(s) := ∏_{p} (1 + ∑_{m=c_{q}}^{q−1} ((\frac{m + 1}{q}) − (\frac{m}{q})) \frac{1}{p^{ms}} ) (σ > \frac{1}{c_{q}} ) ,

and

R_{q}(s) := ∏_{p} (1 + ∑_{m=3}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} ) (σ > \frac{1}{3}) .

Theorem 1. Let q > 3 be a prime number.

(a) If q ≡ ±1 (mod 8)

∑_{n⩽x} (λ_{q} ⋆ 1) (n) = xζ(q)P_{q}(1) (log x + 2γ − 1 + q \frac{ζ′}{ζ} (q) + \frac{P′_{q}}{P_{q}} (1))+O_{q,ε} (x^{max(1/c_{q},θ)+ε}) ,

where θ is defined in (1) and (2).

(b) If q ≡ ±11 (mod 24)

∑_{n⩽x} (λ_{q} ⋆ 1) (n) = x^{1/2}ζ (\frac{q}{2}) R_{q} (\frac{1}{2}) + O_{q,ε} (x^{1/3+ε}) .

(c) If q ≡ ±5 (mod 24), there exists c > 0 such that

∑_{n⩽x} (λ_{q} ⋆ 1) (n) ≪_{q} x^{1/2}e^{−c(log x^{1/4})^{3/5}(log log x^{1/4})^{−1/5}}.

Furthermore, if the Riemann hypothesis is true, then for x sufficiently large

∑_{n⩽x} (λ_{q} ⋆ 1) (n) ≪_{q,ε} x^{1/4}e^{(log^{\sqrt{x}})^{1/2}(log log \sqrt{x})^{5/2+ε}}.

Example 2.

∑_{n⩽x} (λ_{7} ⋆ 1) (n) ≐ 0.454 x (log x + 2γ + 0.784) + O_{ε} (x^{1/2+ε}) .

∑_{n⩽x} (λ_{23} ⋆ 1) (n) ≐ 0.899 x (log x + 2γ − 0.678) + O_{ε} (x^{131/416+ε}) .

∑_{n⩽x} (λ_{13} ⋆ 1) (n) ≐ 1.969 x^{1/2} + O_{ε} (x^{1/3+ε}) .

∑_{n⩽x} (λ_{5} ⋆ 1) (n) ≪ x^{1/2}e^{−c(log x^{1/4})^{3/5}(log log x^{1/4})^{−1/5}}.

2 Notation

In what follows, x ⩾ e^{4} is a large real number, ε ∈ (0, \frac{1}{4}) is a small real number which does not need to be the same at each occurrence, q always denotes an odd prime number, (\frac{·}{q}) is the Legendre symbol and

λ_{q} := (\frac{τ}{q} ) ,

where τ(n) := ∑_{d|n} 1. Also, 1 is the constant arithmetic function equal to 1.

For any arithmetic functions F and G, L(s, F) is the Dirichlet series of F, and F^{−1} is the Dirichlet convolution inverse of F. If r ∈ ℤ_{⩾2}, then

a_{r}(n) := \begin{cases}1, & if n = m^{r}; \\ 0, & otherwise. \end{cases}

For some c > 0, set

δ_{c}(x) := e^{−c(log x)^{3/5}(log log x)^{−1/5}} and ω(x) := e^{(log x)^{1/2}(log log x)^{5/2+ε}}.

Finally, let M(x) and L(x) be respectively the Mertens function and the summatory function of the Liouville function, i.e.

M(x) := ∑_{n⩽x} μ(n) and L(x) := ∑_{n⩽x} λ(n).

3 The Dirichlet series of λ_{q}

Proposition 3. Let q ⩾ 3 be a prime number. For any s ∈ ℂ such that σ > 1 ▷ If q ≡ ±1 (mod 8)

L (s, λ_{q}) = ζ(qs)ζ(s) ∏_{p} (1 + ∑_{m=c_{q}}^{q−1} ((\frac{m + 1}{q}) − (\frac{m}{q})) \frac{1}{p^{ms}} ) .

▷ If q ≡ ±3 (mod 8)

L (s, λ_{q}) = \frac{ζ(qs)ζ(2s)}{ζ(s)} ∏_{p} (1 + ∑_{m=d_{q}}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} ) .

Proof. Set χ_{q} := (\frac{·}{q}) for convenience. From [8, Lemma 2.1], we have

L (s, λ_{q}) = ∏_{p} (1 + ∑_{α=1}^{∞} \frac{χ_{q}(α + 1)}{p^{sα}}) = ∏_{p} (1 + p^{s} ∑_{α=2}^{∞} \frac{χ_{q}(α)}{p^{sα}})

= ∏_{p} (1 + p^{s} ((1 − \frac{1}{p^{qs}} )^{−1} ∑_{m=1}^{q−1} (\frac{m}{q}) \frac{1}{p^{ms}} − p^{−s}))

= ∏_{p} ((1 − \frac{1}{p^{qs}} )^{−1} ∑_{m=1}^{q−1} (\frac{m}{q}) \frac{1}{p^{(m−1)s}} )

= ζ(qs) ∏_{p} (1 + ∑_{m=2}^{q−1} (\frac{m}{q}) \frac{1}{p^{(m−1)s}} ) .

If q ≡ ±1 (mod 8), then (\frac{2}{q}) = 1 and

L (s, λ_{q}) = ζ(qs)ζ(s) ∏_{p} (1 − \frac{1}{p^{s}} + (1 − \frac{1}{p^{s}} ) ∑_{m=2}^{q−1} (\frac{m}{q}) \frac{1}{p^{(m−1)s}} ) ,

where

(1 − \frac{1}{p^{s}} ) ∑_{m=2}^{q−1} (\frac{m}{q}) \frac{1}{p^{(m−1)s}} = ∑_{m=2}^{q−1} (\frac{m}{q}) (\frac{1}{p^{(m−1)s}} − \frac{1}{p^{ms}} )

= ∑_{m=1}^{q−2} (\frac{m + 1}{q}) \frac{1}{p^{ms}} − ∑_{m=2}^{q−1} (\frac{m}{q}) \frac{1}{p^{ms}}

= (\frac{2}{q}) \frac{1}{p^{s}} + ∑_{m=2}^{q−1} ((\frac{m + 1}{q}) − (\frac{m}{q})) \frac{1}{p^{ms}} − (\frac{q}{q} ) \frac{1}{p^{(q−1)s}}

= ∑_{m=2}^{q−1} ((\frac{m + 1}{q}) − (\frac{m}{q})) \frac{1}{p^{ms}} + \frac{1}{p^{s}} .

Similarly, if q ≡ ±3 (mod 8), then (\frac{2}{q}) = −1 and

L (s, λ_{q}) = \frac{ζ(qs)ζ(2s)}{ζ(s)} ∏_{p} (1 + \frac{1}{p^{s}} + (1 + \frac{1}{p^{s}} ) ∑_{m=2}^{q−1} (\frac{m}{q}) \frac{1}{p^{(m−1)s}} ) ,

where

(1 + \frac{1}{p^{s}} ) ∑_{m=2}^{q−1} (\frac{m}{q}) \frac{1}{p^{(m−1)s}} = ∑_{m=2}^{q−1} (\frac{m}{q}) (\frac{1}{p^{(m−1)s}} + \frac{1}{p^{ms}} )
= ∑_{m=1}^{q−2} (\frac{m + 1}{q}) \frac{1}{p^{ms}} + ∑_{m=2}^{q−1} (\frac{m}{q}) \frac{1}{p^{ms}}
= (\frac{2}{q}) \frac{1}{p^{s}} + ∑_{m=2}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} − (\frac{q}{q} ) \frac{1}{p^{(q−1)s}}
= ∑_{m=2}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} − \frac{1}{p^{s}} .

We achieve the proof noting that, if q ≡ ±1 (mod 24), then (\frac{3}{q}) − (\frac{2}{q}) = (\frac{4}{q}) − (\frac{3}{q}) = 0 and, similarly, if q ≡ ±11 (mod 24), then (\frac{3}{q}) + (\frac{2}{q}) = 0 whereas (\frac{4}{q}) + (\frac{3}{q}) = 2.

4 Proof of Theorem 1

4.1 The case q ≡ ±1 (mod 8)

For σ > 1, we set

G_{q}(s) = ζ(qs) ∏_{p} (1 + ∑_{m=c_{q}}^{q−1} ((\frac{m + 1}{q}) − (\frac{m}{q})) \frac{1}{p^{ms}} ) = ζ(qs)P_{q}(s) := ∑_{n=1}^{∞} \frac{g_{q}(n)}{n^{s}}.

This Dirichlet series is absolutely convergent in the half-plane σ > \frac{1}{c_{q}}, so that

∑_{n⩽x} |g_{q}(n)| ≪_{q,ε} x^{1/c_{q}+ε}.

By partial summation, we infer

∑_{n⩽x} \frac{g_{q}(n)}{n} = ζ(q)P_{q}(1) + O (x^{−1+1/c_{q}+ε}) ,
∑_{n⩽x} \frac{g_{q}(n)}{n} log \frac{x}{n} = ζ(q)P_{q}(1) log x + qP_{q}(1)ζ′(q) + P′_{q}(1)ζ(q) + O (x^{−1+1/c_{q}+ε}) .

From Proposition 3, λ_{q} ⋆ 1 = g_{q} ⋆ τ. Consequently

∑_{n⩽x} (λ_{q} ⋆ 1) (n) = ∑_{d⩽x} g_{q}(d) ∑_{k⩽x/d} τ(k)
= ∑_{d⩽x} g_{q}(d) (\frac{x}{d} log \frac{x}{d} + (2γ − 1)\frac{x}{d} + O ((\frac{x}{d})^{θ+ε}))
= x (ζ(q)P_{q}(1) log x + qP_{q}(1)ζ′(q) + P′_{q}(1)ζ(q) + (2γ − 1)ζ(q)P_{q}(1))
+ O (x^{max(1/c_{q},θ)+ε}) ,

where θ is defined in (1) and where we used

x^{−ε} ∑_{d⩽x} \frac{|g_{q}(d)|}{d^{θ}} ≪ \begin{cases}x^{1/c_{q}−θ}, & if c_{q}^{−1} ⩾ θ; \\ 1, & otherwise. \end{cases}

4.2 The case q ≡ ±11 (mod 24)

For σ > 1, we set

H_{q}(s) = ζ(qs) ∏_{p} (1 + ∑_{m=3}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} ) = ζ(qs)R_{q}(s) := ∑_{n=1}^{∞} \frac{h_{q}(n)}{n^{s}}.

Since q > 5, this Dirichlet series is absolutely convergent in the half-plane σ > \frac{1}{3}, so that

∑_{n⩽x} |h_{q}(n)| ≪_{q,ε} x^{1/3+ε}.

From Proposition 3, λ_{q} ⋆ 1 = h_{q} ⋆ a_{2}, hence

∑_{n⩽x} (λ_{q} ⋆ 1) (n) = ∑_{d⩽x} h_{q}(d) ⌊\sqrt{\frac{x}{d}}⌋
= x^{1/2} ∑_{d⩽x} \frac{h_{q}(d)}{\sqrt{d}} + O (x^{1/3+ε})
= x^{1/2}H_{q} (\frac{1}{2}) + O (x^{1/3+ε}) .

4.3 The case q ≡ ±5 (mod 24)

In this case, it is necessary to rewrite L(s, λ_{q}) in the following shape.

Lemma 4. Assume q ≡ ±5 (mod 24). For any σ > 1, L(s, λ_{q}) = \frac{K_{q}(s)}{ζ(s)ζ(2s)} with

K_{q}(s) := \begin{cases}ζ(5s), & if q = 5; \\ ζ(4s)L_{q}(s), & if q ≡ ±19, ±29 (mod 120); \\ \frac{L_{q}(s)}{ζ(4s)}, & if q ≡ ±43, ±53 (mod 120); \end{cases}

where

L_{q}(s) := ζ(qs) ∏_{p} (1 + \frac{2 (p^{2s} + p^{s} + 1)}{p^{7s} − p^{5s}} + \frac{p^{2s} + 1}{p^{2s} − 1} ∑_{m=6}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} )

and

L_{q}(s) := ζ(qs) ∏_{p} (1 − \frac{2p^{2s} − 1}{(p^{2s} − 1)^{3} (p^{2s} + 1)}

+ \frac{p^{8s}}{(p^{2s} − 1)^{3} (p^{2s} + 1)} ∑_{m=6}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} ) .

The Dirichlet series L_{q} is absolutely convergent in the half-plane σ > \frac{1}{5}, and the Dirichlet series L_{q} is absolutely convergent in the half-plane σ > \frac{1}{6}.

Proof. From Proposition 3, we immediately get

L(s, λ_{5}) = \frac{ζ(5s)}{ζ(s)ζ(2s)}.  (3)

Now suppose q > 5 and q ≡ ±5 (mod 24). In this case, (\frac{3}{q}) + (\frac{2}{q}) = −2 and (\frac{4}{q}) + (\frac{3}{q}) = 0 so that we may write by Proposition 3

L (s, λ_{q}) = \frac{ζ(qs)ζ(2s)}{ζ(s)} ∏_{p} (1 − \frac{2}{p^{2s}} + ∑_{m=4}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} )

= \frac{K_{q}(s)}{ζ(s)ζ(2s)}

where

K_{q}(s) := ζ(qs) ∏_{p} (1 − \frac{1}{(p^{2s} − 1)^{2}} + \frac{p^{4s}}{(p^{2s} − 1)^{2}} ∑_{m=4}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} ) .

Assume q ≡ ±19, ±29 (mod 120). Then

(\frac{5}{q}) + (\frac{4}{q}) = (\frac{6}{q}) + (\frac{5}{q}) = 2.

K_{q}(s) can therefore be written as

K_{q}(s) = ζ(qs) ∏_{p} (1 + \frac{p^{s} + 2}{p^{s} (p^{2s} − 1)^{2}} + \frac{p^{4s}}{(p^{2s} − 1)^{2}} ∑_{m=6}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} )
= ζ(qs)ζ(4s) ∏_{p} (1 + \frac{2 (p^{2s} + p^{s} + 1)}{p^{7s} − p^{5s}} + \frac{p^{2s} + 1}{p^{2s} − 1} ∑_{m=6}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} )
= ζ(4s)L_{q}(s).

Similarly, if q ≡ ±43, ±53 (mod 120), then

(\frac{5}{q}) + (\frac{4}{q}) = (\frac{6}{q}) + (\frac{5}{q}) = 0.

Hence

K_{q}(s) := ζ(qs) ∏_{p} (1 − \frac{1}{(p^{2s} − 1)^{2}} + \frac{p^{4s}}{(p^{2s} − 1)^{2}} ∑_{m=6}^{q−1} ((\frac{m + 1}{q}) + (\frac{m}{q})) \frac{1}{p^{ms}} )
= \frac{L_{q}(s)}{ζ(4s)}.

The proof is complete.

We now are in a position to prove Theorem 1 in the case q ≡ ±5 (mod 24).

Assume first that q ≡ ±19, ±29 (mod 120) and let ℓ_{q}(n) be the n-th coefficient of the Dirichlet series L_{q}(s). From Lemma 4, λ_{q} ⋆ 1 = ℓ_{q} ⋆ a_{4} ⋆ a_{2}^{−1} and therefore

∑_{n⩽x} (λ_{q} ⋆ 1) (n) = ∑_{d⩽x} ℓ_{q}(d) ∑_{m⩽(x/d)^{1/4}} M (\frac{1}{m^{2}} \sqrt{\frac{x}{d}}) = ∑_{d⩽x} ℓ_{q}(d)L (\sqrt{\frac{x}{d}}) .

Since L(z) ≪ zδ_{c}(z) for some c > 0

∑_{n⩽x} (λ_{q} ⋆ 1) (n) ≪ x^{1/2} ∑_{d⩽x} \frac{|ℓ_{q}(d)|}{\sqrt{d}} δ_{c} (\sqrt{\frac{x}{d}})
≪ x^{1/2} ( ∑_{d⩽\sqrt{x}} + ∑_{\sqrt{x}<d⩽x}) \frac{|ℓ_{q}(d)|}{\sqrt{d}} δ_{c} (\sqrt{\frac{x}{d}})
≪ x^{1/2}δ_{c} (x^{1/4}) + x^{1/2} ∑_{d>\sqrt{x}} \frac{|ℓ_{q}(d)|}{\sqrt{d}}.

The Dirichlet series L_{q}(s) := ∑_{n=1}^{∞} ℓ_{q}(n)n^{−s} is absolutely convergent in the half-plane σ > \frac{1}{5}, consequently

∑_{d⩽z} |ℓ_{q}(d)| ≪_{q,ε} z^{1/5+ε}

and by partial summation

∑_{d>z} \frac{|ℓ_{q}(d)|}{\sqrt{d}} ≪_{q,ε} z^{−3/10+ε}.

We infer that

∑_{n⩽x} (λ_{q} ⋆ 1) (n) ≪ x^{1/2}δ_{c} (x^{1/4}) + x^{7/20+ε} ≪ x^{1/2}δ_{c} (x^{1/4}) .

Now suppose that the Riemann hypothesis is true. By [1], which is a refinement of [9], we know that M(z) ≪_{ε} z^{1/2} ω(z). The method of [9, 1] may be adapted to the function L yielding

L(z) ≪_{ε} z^{1/2} ω(z) log z.

Observe that, for any a ⩾ 2, ε > 0 and z ⩾ e^{e^{e}}

log z exp (\sqrt{log z} (log log z)^{a}) ⩽ exp (\sqrt{log z} (log log z)^{a+ε})

so that L(z) ≪_{ε} z^{1/2} ω(z) and hence

∑_{n⩽x} (λ_{q} ⋆ 1) (n) ≪ x^{1/4} ∑_{d⩽x} \frac{|ℓ_{q}(d)|}{d^{1/4}} ω (\sqrt{\frac{x}{d}}) ≪ x^{1/4}ω (\sqrt{x}) ∑_{d⩽x} \frac{|ℓ_{q}(d)|}{d^{1/4}} ≪ x^{1/4}ω (\sqrt{x})

completing the proof in that case. The case q = 5 is similar but simpler since λ_{5}⋆1 = a_{5}⋆a_{2}^{−1} by (3).

Finally, when q ≡ ±43, ±53 (mod 120), we proceed as above. Let ν_{q}(n) be the n-th coefficient of the Dirichlet series L_{q}(s). Then λ_{q} ⋆ 1 = ν_{q} ⋆ a_{4}^{−1} ⋆ a_{2}^{−1} from Lemma 4, so that

∑_{n⩽x} (λ_{q} ⋆ 1) (n) = ∑_{d⩽x} ν_{q}(d) ∑_{m⩽(x/d)^{1/4}} μ(m)M (\frac{1}{m^{2}} \sqrt{\frac{x}{d}})

and estimating trivially yields

∑_{n⩽x} (λ_{q} ⋆ 1) (n) ≪ x^{1/2} ∑_{d⩽x} \frac{|ν_{q}(d)|}{\sqrt{d}} ∑_{m⩽(x/d)^{1/4}} \frac{1}{m^{2}} δ_{c} (\frac{1}{m^{2}} \sqrt{\frac{x}{d}})

and we complete the proof as in the previous case.

Remark 5. Let us stress that a bound of the shape

∑_{n⩽x} (λ_{q} ⋆ 1) (n) ≪ x^{1/4+ε}

for all x sufficiently large and small ε > 0, is a necessary and sufficient condition for the Riemann hypothesis. Indeed, if this estimate holds, then by partial summation the series ∑_{n=1}^{∞} (λ_{q} ⋆ 1) (n)n^{−s} is absolutely convergent in the half-plane σ > \frac{1}{4}. Consequently, the function K_{q}(s)ζ(2s)^{−1} is analytic in this half-plane. In particular, ζ(2s) does not vanish in this half-plane, implying the Riemann hypothesis, proving the necessary condition, the sufficiency being established above.

5 A short interval result for the case q = 5

5.1 Introduction

This section deals with sums of the shape

∑_{x<n⩽x+y} (λ_{5} ⋆ 1) (n)

where x^{ε} ⩽ y ⩽ x. From Theorem 1

∑_{x<n⩽x+y} (λ_{5} ⋆ 1) (n) ≪ x^{1/2}e^{−c(log x^{1/4})^{3/5}(log log x^{1/4})^{−1/5}}

and if the Riemann hypothesis is true, then

∑_{x<n⩽x+y} (λ_{5} ⋆ 1) (n) ≪_{ε} x^{1/4}e^{(log^{\sqrt{x}})^{1/2}(log log \sqrt{x})^{5/2+ε}}.

The purpose is to improve significantly upon these estimates when y = o(x), by using fine results belonging to the theory of integer points near a suitably chosen smooth curve. To this end, we need the following additional notation. Let δ ∈ (0, \frac{1}{4}), N ∈ ℤ_{⩾1} large, f : [N, 2N] ⟶ ℝ be any map, and define R(f, N, δ) to be the number of elements of the set of integers n ∈ [N, 2N] such that ∥f(n)∥ < δ, where ∥x∥ is the distance from x to its nearest integer. Note that the trivial bound is given by

∑_{x<n⩽x+y} (λ_{5} ⋆ 1) (n) ≪ ∑_{x<n⩽x+y} τ(n) ≪ y log x.

5.2 Tools from the theory

In what follows, N ∈ ℤ_{⩾1} is large and δ ∈ (0, \frac{1}{4}). The first result is [7, Theorem 5] with k = 5. See also [2, Theorem 5.23 (iv)].

Lemma 6 (5th derivative test). Let f ∈ C^{5} [N, 2N] such that there exist λ_{4} > 0 and λ_{5} > 0 satisfying λ_{4} = Nλ_{5} and, for any x ∈ [N, 2N]

|f^{(4)}(x)| ≍ λ_{4} and |f^{(5)}(x)| ≍ λ_{5}.

Then

R(f, N, δ) ≪ Nλ_{5}^{1/15} + Nδ^{1/6} + (δλ_{4}^{−1})^{1/4} + 1.

Remark 7. The basic result of the theory is the following first derivative test (see [2, Theorem 5.6]): Let f ∈ C^{1} [N, 2N] such that there exist λ_{1} > 0 such that |f′(x)| ≍ λ_{1}. Then

R(f, N, δ) ≪ Nλ_{1} + Nδ + δλ_{1}^{−1} + 1.  (4)

This result is essentially a consequence of the mean value theorem.

The second tool is [4, Theorem 7] with k = 3.

Lemma 8. Let s ∈ ℚ^{∗} \ {±2, ±1} and X > 0 such that N ⩽ X^{1/s}. Then there exists a constant c_{3} := c_{3}(s) ∈ (0, \frac{1}{4}) depending only on s such that, if

N^{2}δ ⩽ c_{3}  (5)

then

R (\frac{X}{n^{s}} , N, δ) ≪ (XN^{3−s})^{1/7} + δ (XN^{59−s})^{1/21} .

Our last result relates the short sum of λ_{5} ⋆ 1 to a problem of counting integer points near a smooth curve.

Lemma 9. Let 1 ⩽ y ⩽ x. Then

∑_{x<n⩽x+y} (λ_{5} ⋆ 1) (n) ≪ max_{(16y^{2}x^{−1})^{1/5}<N⩽(2x)^{1/5}}R (\sqrt{\frac{x}{n^{5}}} , N, \frac{y}{\sqrt{N^{5}x}}) log x + yx^{−1/2} + x^{−1/5}y^{2/5}.

Proof. Using (3), we get

∑_{n⩽x} (λ_{5} ⋆ 1) (n) = ∑_{d⩽\sqrt{x}} μ(d) ⌊(\frac{x}{d^{2}} )^{1/5}⌋

so that

∑_{x<n⩽x+y} (λ_{5} ⋆ 1) (n) = ∑_{d⩽\sqrt{x}} μ(d) (⌊(\frac{x + y}{d^{2}} )^{1/5}⌋ − ⌊(\frac{x}{d^{2}} )^{1/5}⌋) + ∑_{\sqrt{x}<d⩽\sqrt{x+y}} μ(d)
≪ ∑_{d⩽\sqrt{x}} (⌊(\frac{x + y}{d^{2}} )^{1/5}⌋ − ⌊(\frac{x}{d^{2}} )^{1/5}⌋) + yx^{−1/2}
≪ ∑_{d⩽\sqrt{x}} ∑_{x<d^{2}n^{5}⩽x+y} 1 + yx^{−1/2}
≪ ∑_{n⩽(2x)^{1/5}} ∑_{(\frac{x}{n^{5}} )^{1/2}<d⩽(\frac{x+y}{n^{5}})^{1/2}} 1 + yx^{−1/2}
≪ ∑_{(16y^{2}x^{−1})^{1/5}<n⩽(2x)^{1/5}} (⌊\sqrt{\frac{x + y}{n^{5}}} ⌋ − ⌊\sqrt{\frac{x}{n^{5}}} ⌋) + x^{−1/5}y^{2/5} + yx^{−1/2}

and for any integers N ∈ ](16y^{2}x^{−1})^{1/5} , (2x)^{1/5}] and n ∈ [N, 2N]

\sqrt{\frac{x + y}{n^{5}}} − \sqrt{\frac{x}{n^{5}}} < \frac{y}{\sqrt{N^{5}x}} < \frac{1}{4}

so that the sum does not exceed

≪ max_{(16y^{2}x^{−1})^{1/5}<N⩽(2x)^{1/5}}R (\sqrt{\frac{x}{n^{5}}} , N, \frac{y}{\sqrt{N^{5}x}}) log x + x^{−1/5}y^{2/5} + yx^{−1/2}

as asserted.

5.3 The main result

Theorem 10. Assume y ⩽ c_{3} x^{11/20} where c_{3} := c_{3} (\frac{5}{2}) is given in (5). Then

∑_{x<n⩽x+y} (λ_{5} ⋆ 1) (n) ≪ (x^{1/12} + yx^{−4/9}) log x.

Furthermore, if y ⩽ c_{3} x^{19/36}

∑_{x<n⩽x+y} (λ_{5} ⋆ 1) (n) ≪ x^{1/12} log x.

Proof. We split the first term in Lemma 9 into three parts, according to the ranges (16y^{2}x^{−1})^{1/5} < N ⩽ 2x^{1/10}, 2x^{1/10} < N ⩽ 2x^{1/6} and 2x^{1/6} < N ⩽ (2x)^{1/5}.

In the first case, we use Lemma 6 with λ_{4} = (xN^{−13})^{1/2} and λ_{5} = (xN^{−15})^{1/2} which yields

max_{(16y^{2}x^{−1})^{1/5}<N⩽2x^{1/10}} R (\sqrt{\frac{x}{n^{5}}} , N, \frac{y}{\sqrt{N^{5}x}}) ≪ x^{1/12} + x^{−1/40}y^{1/6} + x^{−3/20}y^{1/4}.

For the second range, we use Lemma 8 with X = x^{1/2}, s = \frac{5}{2} and δ = y (N^{5}x)^{−1/2}. Notice that the conditions N > 2x^{1/10} and y ⩽ c_{3} x^{11/20} ensure that δ < \frac{1}{4} and N^{2}δ ⩽ c_{3}. We get

max_{2x^{1/10}<N⩽2x^{1/6}} R (\sqrt{\frac{x}{n^{5}}} , N, \frac{y}{\sqrt{N^{5}x}}) ≪ x^{1/12} + yx^{−4/9}.

The last range is easily treated with (4), giving

max_{2x^{1/6}<N⩽(2x)^{1/5}} R (\sqrt{\frac{x}{n^{5}}} , N, \frac{y}{\sqrt{N^{5}x}}) ≪ x^{1/12} + yx^{−3/4}.

Using Lemma 9, we finally get

∑_{x<n⩽x+y} (λ_{5} ⋆ 1) (n) ≪ (x^{1/12} + x^{−1/40}y^{1/6} + x^{−3/20}y^{1/4} + yx^{−4/9}) log x + x^{−1/5}y^{2/5}

and note that x^{−1/40}y^{1/6}+x^{−3/20}y^{1/4}+x^{−1/5}y^{2/5} ≪ x^{1/12} as soon as y ⩽ x^{13/20}. This completes the proof of the first estimate, the second one being obvious.

6 Acknowledgments

The author deeply thanks Prof. Kannan Soundararajan for the help he gave him to adapt his result to the function L(x), and Benoit Cloitre for bringing this problem to his attention. The author gratefully acknowledges the anonymous referee for some corrections and remarks that have significantly improved the paper.

References

[1] M. Balazard and A. de Roton, Notes de lecture de l’article “Partial sums of the Möbius function” de Kannan Soundararajan, preprint, 2008,

   https://arxiv.org/abs/0810.3587.

[2] O. Bordellès, Arithmetic Tales, Springer, 2012.

[3] H. Davenport, The Higher Arithmetic, 5th edition, Cambridge University Press, 1982.

[4] M. Filaseta and O. Trifonov, The distribution of fractional parts with applications to gap results in number theory, Proc. London Math. Soc. 73 (1996), 241–278.

[5] G. H. Hardy, On Dirichlet’s divisor problem, Proc. London Math. Soc. 15 (1916), 1–25.

[6] M. N. Huxley, Exponential sums and lattice points III, Proc. London Math. Soc. 87 (2003), 591–609.

[7] M. N. Huxley and P. Sargos, Points entiers au voisinage d’une courbe plane de classe C^{n}, II, Functiones et Approximatio 35 (2006), 91–115.

[8] R. K. Muthumalai, Note on Legendre symbols connecting with certain infinite series, Notes on Number Theory and Discrete Mathematics 19 (2013), 77–83.

[9] K. Soundararajan, Partial sums of the Möbius function, J. Reine Angew. Math. 631 (2009), 141–152.

2010 Mathematics Subject Classification: Primary 11N37; Secondary 11A25, 11M41. Keywords: Number of divisors, Legendre symbol, mean value, Riemann hypothesis.

(Concerned with sequences \underline{A000005}, \underline{A008836}, and \underline{A091337}.)

Received January 14 2017; revised versions received June 1 2017; June 26 2017. Published in Journal of Integer Sequences, July 1 2017.

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