An Asymptotic Formula for Short Sums of Gcd-Sum Functions

Olivier Bordellès 2, allée de la Combe 43000 Aiguilhe France

                      borde43@wanadoo.fr

Abstract In this note, we prove an asymptotic formula for sums, over short segments, of the composition of the gcd and arithmetic functions belonging to certain classes. Two examples are also given.

1 Introduction and main results

There has been renewed interest in gcd-sum functions over the last few years. We refer to [4, 6] for a survey. In [1], it is shown that

∑_{n≤x} (∑_{i=1}^{n} f(gcd(i, n))) = \frac{x^{2}F (2)}{2ζ(2)} + O \{x ∏_{p≤x} (1 + ∑_{l=1}^{∞} \frac{|f (p^{l}) − f (p^{l−1})|}{p^{l}}) + x(log x)^{α}\}

under the sole hypothesis

∑_{n≤x} |(f ⋆ μ)(n)| ≪ x(log x)^{α}  (1)

where α ≥ 0, F (s) is the Dirichlet series of the arithmetic function f and the Dirichlet convolution product f ⋆ g of f and g is defined by (f ⋆ g) (n) = ∑_{d|n} f(d)g (n/d).

It can easily be deduced that, if f satisfies (1), then for large real numbers x, y such that x^{1/2} ≤ y ≤ x, we have

∑_{x<n≤x+y} (∑_{i=1}^{n} f(gcd(i, n))) = y (2x + y) \frac{F (2)}{2ζ(2)}

+ O \{x ∏_{p≤x} (1 + ∑_{l=1}^{∞} \frac{|f (p^{l}) − f (p^{l−1})|}{p^{l}}) + x(log x)^{α}\} .

The aim of the present work is to study similar short sums when f lies into a wider class of arithmetic functions. More precisely, we consider the class of real-valued multiplicative functions satisfying

∑_{n≤x} \frac{|(f ⋆ μ)(n)|}{n} ≪ x^{a} (0 ≤ a ≤ \frac{1}{2})  (2)

∑_{n≤x} \frac{(f ⋆ μ)(n)^{2}}{n^{2}} ≪ x^{b} (0 ≤ b ≤ 1)  (3)

for all x ≥ 1. We will prove the following asymptotic formula.

Theorem 1. Let f be a real-valued multiplicative function satisfying (2) and (3) with Dirichlet series F (s). Then for x^{1/3} ≤ y ≤ x, we have

∑_{x<n≤x+y} (∑_{i=1}^{n} f(gcd(i, n))) = y (2x + y) \frac{F (2)}{2ζ(2)} + O (x^{1+a/2+b/4}y^{1/4}(log x)^{1/4}) .

Note that this result is non-trivial whenever x^{(4+2a+b)/7}(log x)^{1/7} ≤ y ≤ x.

Let k ≥ 2 be a fixed integer and s_{k} be the characteristic function of the set of k-full numbers. Denote Id : n ⟼ n and consider the multiplicative function Id ·s_{k}. Similarly, let M_{k}(n) be the maximal k-full divisor of n (see [5] for instance). It can easily be seen that both functions Id ·s_{k} and M_{k} satisfy the assumptions (2) and (3) with a = b = \frac{1}{k}, and hence do not satisfy (1). Theorem 1 implies the following result.

Corollary 2. Let k ≥ 2 be an integer. For x^{1/3} ≤ y ≤ x, we have

∑_{x<n≤x+y} ( ∑_{\begin{matrix}i=1 \\ gcd(i,n) k-full \end{matrix}}^{n} gcd(i, n)) = \frac{y (2x + y)}{2ζ(2)} ∏_{p} (1 + \frac{1}{p^{k−1}(p − 1)})+ O (x^{1+ \frac{3}{4k}} y^{1/4}(log x)^{1/4})

and

∑_{x<n≤x+y} (∑_{i=1}^{n} M_{k}(gcd(i, n))) = \frac{C_{k} y (2x + y)}{2ζ(2)} + O (x^{1+ \frac{3}{4k}} y^{1/4}(log x)^{1/4})

where

C_{k} := ∏_{p} (1 + \frac{p^{2k} + p^{k+1}(p + 1) − p^{2}}{p^{2k}(p^{2} − 1)} ) .

2 Notation

In what follows, 1 ≤ y ≤ x are large real numbers and N ≥ 1 is a large integer. Let f be a real-valued multiplicative function satisfying (2) and (3), and let g := f ⋆ μ be the Eratosthenes transform of f. According to the usual practice, for all x ∈ ℝ, ⌊x⌋ is the integer part of x and ∥x∥ is the distance from x to its nearest integer. φ is the Euler totient function and μ is the Möbius function.

3 Proof of Theorem 1

First, note that by (2) and Abel summation, we have for all z ≥ 1

∑_{n>z} \frac{|g(n)|}{n^{2}} ≪ z^{−1+a}.  (4)

Next, using Cesáro’s identity [3]

∑_{i=1}^{n} f(gcd(i, n)) = (f ⋆ φ)(n)

and using φ = μ ⋆ Id, we get

∑_{x<n≤x+y} (∑_{i=1}^{n} f(gcd(i, n))) = ∑_{x<n≤x+y} (g ⋆ Id) (n)
= ∑_{d≤x+y} g(d) ∑_{x/d<k≤(x+y)/d} k
= \frac{1}{2} ∑_{d≤x+y} g(d) (⌊\frac{x + y}{d} ⌋ − ⌊\frac{x}{d}⌋) (⌊\frac{x + y}{d} ⌋ + ⌊\frac{x}{d}⌋ + 1)
= \frac{1}{2} ∑_{d≤2y} g(d) (⌊\frac{x + y}{d} ⌋ − ⌊\frac{x}{d}⌋) (⌊\frac{x + y}{d} ⌋ + ⌊\frac{x}{d}⌋ + 1)

+ ∑_{2y<d≤x+y} g(d) ⌊\frac{x + y}{d} ⌋ (⌊\frac{x + y}{d} ⌋ − ⌊\frac{x}{d}⌋)

:= Σ_{1} + Σ_{2}.

3.1 The sum Σ_{1}

Since

\frac{1}{2} (⌊\frac{x + y}{d} ⌋ − ⌊\frac{x}{d}⌋) (⌊\frac{x + y}{d} ⌋ + ⌊\frac{x}{d}⌋ + 1) = \frac{y}{2d^{2}} (2x + y) + O (\frac{x}{d})

we get using (2) and (4)

Σ_{1} = \frac{y}{2} (2x + y) ∑_{d≤2y} \frac{g(d)}{d^{2}} + O (x ∑_{d≤2y} \frac{|g(d)|}{d})
= \frac{y}{2} (2x + y) ∑_{d=1}^{∞} \frac{g(d)}{d^{2}} − \frac{y}{2} (2x + y) ∑_{d>2y} \frac{g(d)}{d^{2}} + O (xy^{a})
= y (2x + y) \frac{F (2)}{2ζ(2)} + O (xy^{a}) .

3.2 The sum Σ_{2}

Lemma 3. If x^{1/3} ≤ y ≤ x, then

∑_{2y<n≤x+y} (⌊\frac{x + y}{n} ⌋ − ⌊\frac{x}{n}⌋) ≪ y log x.

Proof. If N ≥ 1 is a large integer, δ ∈ (0, \frac{1}{2}) and if h : [N, 2N] ⟶ ℝ is any function, then let R(h, N, δ) be the number of integers n ∈ [N, 2N] such that ∥h(n)∥ < δ. The second derivative test for R(h, N, δ) (see [2, Corollaire]) states that, if h ∈ C^{2}[N, 2N] such that there exists T ≥ 1 such that |h′(x)| ≍ T N^{−1} and |h′′(x)| ≍ T N^{−2}, then

R(h, N, δ) ≪ (NT)^{1/3} + Nδ + (T δ)^{1/2}.

Now using this estimate with T = xN^{−1} and δ = yN^{−1}, we get

∑_{2y<n≤x+y} (⌊\frac{x + y}{n} ⌋ − ⌊\frac{x}{n}⌋) = ( ∑_{2y<n≤x} + ∑_{x<n≤x+y}) (⌊\frac{x + y}{n} ⌋ − ⌊\frac{x}{n}⌋)

≪ max_{2y<N≤x} ∑_{N<n≤2N} (⌊\frac{x + y}{n} ⌋ − ⌊\frac{x}{n}⌋) log x + y
≪ max_{2y<N≤x} R (\frac{x}{n}, N, \frac{y}{N} ) log x + y
≪ max_{2y<N≤x} (x^{1/3} + y + (xy)^{1/2}N^{−1}) log x + y
≪ (x^{1/3} + y + (xy^{−1})^{1/2}) log x + y ≪ y log x

since y ≥ x^{1/3}.

Now using the Cauchy-Schwarz inequality and Lemma 3, we get

|Σ_{2}| ≤ 2x ∑_{2y<d≤x+y} \frac{|g(d)|}{d} (⌊\frac{x + y}{d} ⌋ − ⌊\frac{x}{d}⌋)

≪ x ( ∑_{2y<d≤x+y} \frac{g(d)^{2}}{d^{2}} )^{1/2} ( ∑_{2y<d≤x+y} (⌊\frac{x + y}{d} ⌋ − ⌊\frac{x}{d}⌋))^{1/2}

≪ xy^{1/2} ( ∑_{2y<d≤x+y} \frac{g(d)^{2}}{d^{2}} )^{1/2} (log x)^{1/2}

whenever x^{1/3} ≤ y ≤ x, and the hypothesis (3) provides

|Σ_{2}| ≪ x^{1+b/2}y^{1/2}(log x)^{1/2}.

We also have trivially using (2)

|Σ_{2}| ≪ x ∑_{2y<d≤x+y} \frac{|g(d)|}{d} ≪ x^{1+a}

and hence

|Σ_{2}| ≪ min (x^{1+a}, x^{1+b/2}y^{1/2}(log x)^{1/2}) ≪ x^{1+a/2+b/4}y^{1/4}(log x)^{1/4}.

We conclude the proof with the following remarks: if 0 ≤ a ≤ \frac{1}{4}, then

xy^{a} ≤ xy^{1/4} ≤ x^{1+a/2+b/4}y^{1/4}.

If \frac{1}{4} < a ≤ \frac{1}{2}, then x^{1+a/2+b/4}y^{1/4} ≥ xy^{a} as soon as y ≤ x^{\frac{2a+b}{4a−1}} which is trivially true since in that case we have

\frac{2a + b}{4a − 1} ≥ b + 1 ≥ 1

since b ≥ 0.

4 Acknowledgments

I am indebted to the referee for his careful reading of the manuscript and some valuable suggestions and corrections he made.

References

[1] O. Bordellès, The composition of the gcd and certain arithmetic functions, J. Integer Sequence 10 (2010), Article 10.7.1.

[2] M. Branton and P. Sargos, Points entiers au voisinage d’une courbe plane à très faible courbure, Bull. Sci. Math. 118 (1994), 15–28.

[3] E. Cesáro, Étude moyenne du plus grand commun diviseur de deux nombres, Ann. Mat. Pura Appl. 13 (1885), 235–250.

[4] P. Haukkanen, On a gcd-sum function, Aequationes Math. 76 (2008), 168–178.

[5] D. Suryanarayana and P. Subrahmanyam, The maximal k-full divisor of an integer, Indian J. Pure Appl. Math. 12 (1981), 175–190.

[6] L. Tóth, A survey of gcd-sum functions, J. Integer Seq. 12 (2010), Article 10.8.1.

2010 Mathematics Subject Classification: Primary 11A25; Secondary 11N37. Keywords: Short sum, composition, Dirichlet convolution, asymptotic result.

(Concerned with sequence \underline{A018804}.)

Received April 13 2012. Revised versions received June 14 2012; July 15 2012. Published in Journal of Integer Sequences, July 17 2012.

Return to Journal of Integer Sequences home page.