On an Open Problem of Tóth

Deyu Zhang School of Mathematical Sciences Shandong Normal University Jinan 250014 Shandong P. R. China

                  zdy 78@yahoo.com.cn

Wenguang Zhai Department of Mathematics China University of Mining and Technology Beijing, 100083 P. R. China

                  zhaiwg@hotmail.com

Abstract

In a recent paper, Tóth mentioned that it is an open problem to give the asymptotic formula for ∑_{n≤x} P^{k}(n), where P (n) is the well-known gcd-sum function and k ≥ 2 is a fixed integer. In this paper, we use the analytic properties of the Dirichlet divisor function to obtain the asymptotic formula for it.

1 Introduction

In 1933, Pillai [8] introduced the gcd-sum function

P (n) = ∑_{k=1}^{n} gcd(k, n).  (1)

By grouping the terms according to the values of gcd(k, n) we have

P (n) = ∑_{d|n} dϕ(n/d) = n ∑_{d|n} \frac{ϕ(d)}{d},  (2)

where ϕ is Euler’s function. Many authors have studied the properties of P (n), see [1, 2, 3, 4, 5, 8, 9, 10]; it is Sloane’s sequence \underline{A018804}. Chidambaraswamy and Sitaramachandrarao [5] showed that, given an arbitrary ϵ > 0,

∑_{n≤x} P (n) = e_{1}x^{2} log x + e_{2}x^{2} + O(x^{1+θ+ϵ}),

where θ is the constant appearing in the error term of the Dirichlet divisor problem, e_{1}, e_{2} are certain constants.

It follows from (2) that the arithmetic mean of gcd(1, n), …, gcd(n, n) is given by

A(n) = \frac{P (n)}{n} = ∑_{d|n} \frac{ϕ(d)}{d}.  (3)

Tóth [9] showed that

∑_{n≤x} A^{2}(n) = x(C_{1} log^{3} x + C_{2} log^{2} x + C_{3} log x + C_{4}) + O(x^{1/2+ϵ}),  (4)

where C_{1}, C_{2}, C_{3}, C_{4} are computable constants. Furthermore, he listed some open problems concerning the gcd-sum function, one of which is to derive the asymptotic formula for ∑_{n≤x} P^{k}(n), where k ≥ 2 is a fixed integer.

In this paper, we use the analytic method to get the asymptotic formula for ∑_{n≤x} A^{k}(n).

Theorem 1. Let k ≥ 2 be a fixed integer. Then

∑_{n≤x} A^{k}(n) = xQ_{2^{k}−1}(log x) + O(x^{β_{k}+ϵ}),  (5)

where Q_{2^{k}−1}(t) is a polynomial of degree 2^{k} − 1 in t and

β_{2} = \frac{1}{2}, β_{3} = \frac{5}{8}, β_{4} = \frac{7}{9}, β_{5} = \frac{31}{36}, β_{6} = \frac{207}{224}, β_{k} = 1 − 2^{− \frac{2}{3} k}/50, k ≥ 7.

Corollary 2. Let k ≥ 2 be a fixed integer. Then

∑_{n≤x} P^{k}(n) = x^{2}Q′_{2^{k}−1}(log x) + O(x^{1+β_{k}+ϵ}),  (6)

where Q′_{2^{k}−1}(t) is a polynomial of degree 2^{k} − 1 in t.

Theorem 3. Let k ≥ 2 be a fixed integer and

E_{k}(x) = ∑_{n≤x} A^{k}(n) − xQ_{2^{k}−1}(log x).

Then for k = 3, 4, 5, we have

∫_{1}^{U} E_{k}(x)dx ≪ U^{1+δ_{k}+ϵ},

where

δ_{3} = 1/2, δ_{4} = 0.6030739, δ_{5} = 0.773114.

2 Preliminary Lemmas

Lemma 4. Let s be a complex number with Re(s) > 1. Then

∑_{n=1}^{∞} \frac{A^{k}(n)}{n^{s}} = ζ^{2^{k}}(s)G_{k}(s),

where G_{k}(s) = ∑_{n=1}^{∞} \frac{g(n)}{n^{s}} is a Dirichlet series which is absolutely convergent for Re(s) > 1/2. Proof. Recall that A(n) is multiplicative function. Then it follows from the Euler product representation that for Re(s) > 1,

F (s) := ∑_{n=1}^{∞} \frac{A^{k}(n)}{n^{s}} = ∏_{p} (1 + ∑_{α=1}^{∞} \frac{A^{k}(p^{α})}{p^{αs}}) .  (7)

where

A(p^{α}) = 1 + \frac{ϕ(p)}{p} + ⋯ + \frac{ϕ(p^{α})}{p^{α}} = 1 + α − \frac{α}{p}, α ≥ 1.

Thus we have

1 + ∑_{α=1}^{∞} \frac{A^{k}(p^{α})}{p^{αs}} = 1 + ∑_{α=1}^{∞} \frac{(1 + α − \frac{α}{p})^{k}}{p^{αs}}
= 1 + \frac{2^{k}}{p^{s}} − \frac{k · 2^{k−1}}{p^{s+1}} + … + (−1)^{k} \frac{1}{p^{s+k}}

+\frac{3^{k}}{p^{2s}} − \frac{2k · 3^{k−1}}{p^{2s+1}} + … + (−1)^{k} \frac{2^{k}}{p^{2s+k}}

+\frac{4^{k}}{p^{3s}} − \frac{3k · 4^{k−1}}{p^{3s+1}} + … + (−1)^{k} \frac{3^{k}}{p^{3s+k}}

+ … =: x_{k}(s).

We substitute the above formula to the formula (7) to get

∑_{n=1}^{∞} \frac{A^{k}(n)}{n^{s}} = ∏_{p} x_{k}(s) = ζ^{2^{k}}(s) · ∏_{p}(1 − \frac{1}{p^{s}})^{2^{k}} · x_{k}(s) =: ζ^{2^{k}}(s)G_{k}(s),

where

G_{k}(s) = ∏_{p}(1 − \frac{1}{p^{s}})^{2^{k}} · x_{k}(s)

= ∏_{p} (1 − \frac{k · 2^{k−1}}{p^{s+1}} + … + (−1)^{k} \frac{1}{p^{s+k}}

+\frac{3^{k} − 4^{k} + \binom{2^{k}}{2}}{p^{2s}} + \frac{k · 2^{2k−1} − 2k · 3^{k−1}}{p^{2s+1}} + …

+\frac{4^{k} − 6^{k} + \binom{2^{k}}{2}2^{k} − \binom{2^{k}}{3}}{p^{3s}} + \frac{2k · 2^{k}3^{k−1} − 3k · 4^{k−1} − k · 2^{k−1} · \binom{2^{k}}{2}}{p^{3s+1}} + …).

From the above formula, it is easy to see that G_{k}(s) can be expanded to a Dirichlet series ∑_{n=1}^{∞} \frac{g(n)}{n^{s}}, which is absolutely convergent for Re(s) > 1/2.

Lemma 5. Suppose 1/2 ≤ σ ≤ 1, then

ζ(σ + it) ≪ (|t| + 2)^{\frac{1−σ}{3}} log(|t| + 2).  (8)

Proof. We define the function μ(σ) for each σ as the infimum of number c ≥ 0 such that ζ(σ + it) ≪ t^{c}, or alternatively as

μ(σ) = lim sup_{t→∞} \frac{log |ζ(σ + it)|}{log t}.

Then μ(σ) is continuous, nonincreasing and for σ_{1} ≤ σ ≤ σ_{2},

μ(σ) ≤ μ(σ_{1})\frac{σ_{2} − σ}{σ_{2} − σ_{1}} + μ(σ_{2})\frac{σ − σ_{1}}{σ_{2} − σ_{1}} .

By the well-known estimates

ζ(1/2 + it) ≪ t^{1/6}, ζ(1 + it) ≪ log t,

we can easily get the formula (8). Lemma 6. If ζ(s) = χ(s)ζ(1 − s), then the estimate

χ(s) ≪ (|t| + 2)^{1/2−σ}

holds uniformly for 0 ≤ σ ≤ 1.

Proof. Using standard properties of the gamma-function one may write the functional equation of ζ(s) as

ζ(s) = χ(s)ζ(1 − s), χ(s) = (2π)^{s}/(2Γ(s) cos(πs/2)).

From Stirling’s formula

|Γ(s)| = (2π)^{\frac{1}{2}} |t|^{σ− \frac{1}{2}} e^{− \frac{π|t|}{2}} (1 + O(|t|^{−1})) (|t| ≥ t_{0}),

it follows that

\begin{aligned}χ(s) & = (\frac{2π}{|t|} )^{σ+i|t|− \frac{1}{2}} e^{i(|t|+ \frac{π}{4} )} (1 + O(|t|^{−1})) \\ & ≪ (|t| + 2)^{1/2−σ}. \end{aligned}

3 Proofs of Theorem 1 and Corollary 2

Recall that the generalized divisor function

d_{k}(n) = ∑_{n=n_{1}···n_{k}} 1,

and its Dirichlet series is

ζ^{k}(s) = ∑_{n=1}^{∞} \frac{d_{k}(n)}{n^{s}}.

From [7, Theorem 13.2 and 13.3] it follows that

∑_{n≤x} d_{2^{k}}(n) = xP_{2^{k}−1}(log x) + O(x^{β_{k}+ϵ}),  (9)

where P_{2^{k}−1}(t) is a polynomial of degree 2^{k} − 1 in t, and

β_{2} = \frac{1}{2}, β_{3} = \frac{5}{8}, β_{4} = \frac{7}{9}, β_{5} = \frac{31}{36}, β_{6} = \frac{207}{224}, β_{k} = 1 − 2^{− \frac{2}{3} k}/50, k ≥ 7.

Then by Lemma 4, we have that

∑_{n≤x} A^{k}(n) = ∑_{mℓ≤x} d_{2^{k}}(m)g(ℓ) = ∑_{ℓ≤x} g(ℓ) ∑_{m≤x/ℓ} d_{2^{k}}(m),

and formula (9) applied to the inner sum gives

∑_{n≤x} A^{k}(n) = ∑_{ℓ≤x} g(ℓ) \{\frac{x}{ℓ}P_{2^{k}−1} (log(\frac{x}{ℓ})) + O ((\frac{x}{ℓ})^{β_{k}+ϵ})\}

= xQ_{2^{k}−1}(log x) + O(x^{β_{k}+ϵ}),

if we notice from Lemma 4 that the infinite series ∑_{ℓ=1}^{∞} \frac{g(ℓ)}{ℓ} and ∑_{ℓ=1}^{∞} \frac{g(ℓ) log^{k} ℓ}{ℓ} are absolutely convergent, and

∑_{ℓ≤x} |g(ℓ)| ≪ x^{1/2+ϵ}.

From the definitions of P (n) and Abel’s summation formula, we can easily get

∑_{n≤x} P^{k}(n) = x^{2}Q′_{2^{k}−1}(log x) + O(x^{1+β_{k}+ϵ}),

where Q′_{2^{k}−1}(t) is a polynomial of degree 2^{k} − 1 in t.

4 Proof of Theorem 3

It suffices to prove that

∫_{U}^{2U} E_{k}(x)dx ≪ U^{1+δ_{k}+ϵ},  (10)

where

δ_{3} = 1/2, δ_{4} = 0.6030739, δ_{5} = 0.773114.

By Perron’s formula (see for example, [6, Chapter 5]), we have for T ≤ x ≤ 2T that

∑_{n≤x} A^{k}(n) = \frac{1}{2πi} ∫_{1+ε−iT}^{1+ε+iT} ζ^{2^{k}}(s)G_{k}(s)\frac{x^{s}}{s} ds + O(T^{ε}).

Then we move the integration to the parallel segment with Re(s) = 1 − ε to get

E_{k}(x) = \frac{1}{2πi} ∫_{1−ε−iT}^{1−ε+iT} ζ^{2^{k}}(s)G_{k}(s)\frac{x^{s}}{s} ds + O(T^{ε}).

So

∫_{U}^{2U} E_{k}(x)dx = \frac{1}{2πi} ∫_{1−ε−iT}^{1−ε+iT} \frac{ζ^{2^{k}}(s)G_{k}(s)}{s} (∫_{U}^{2U} x^{s}dx) ds + O(U^{1+ε})
= \frac{1}{2πi} ∫_{1−ε−iT}^{1−ε+iT} \frac{ζ^{2^{k}}(s)G_{k}(s)(2^{s+1} − 1)U^{s+1}}{s(s + 1)} ds + O(U^{1+ε}).  (11)

Moving the integral line in the last integral of (11) to σ = c, where \frac{1}{2} < c < 1, we have

∫_{U}^{2U} E_{k}(x)dx ≪ U^{1+c} ∫_{c−iT}^{c+iT} \frac{|ζ(s)|^{2^{k}}}{|s(s + 1)|}ds ≪ U^{1+c} ∫_{1}^{T} \frac{|ζ(c + it)|^{2^{k}}}{T^{2}} ds,  (12)

if we notice that G_{k}(s) is absolutely convergent in Re(s) > \frac{1}{2}.

For the case k = 3, it follows from [7, Theorem 8.3] that

∫_{1}^{T} |ζ(\frac{1}{2} + it)|^{8} dt ≪ T^{\frac{3}{2}} .

On taking c = 1/2 + ε in (12), we have

∫_{U}^{2U} E_{3}(x)dx ≪ U^{3/2}.  (13)

For the case k = 4, from [7, Theorem 8.3] and [7, Theorem 8.4], it follows that

∫_{1}^{T} |ζ(\frac{1}{2} + it)|^{16} dt ≪ T^{1+ \frac{350}{216}}

and

∫_{1}^{T} |ζ(σ + it)|^{16} dt ≪ T^{1+ε},

where σ satisfies

\frac{12408}{4537 − 4890σ} = 16,

which gives σ = 0.7692229. By [7, Lemma 8.3], we have

∫_{1}^{T} |ζ(0.6030739 + it)|^{16} dt ≪ T^{2}.

In the formula (12), we take c = 0.6030739 to get

∫_{U}^{2U} E_{4}(x)dx ≪ U^{1+0.6030739}.  (14)

Similarly, we can get δ_{k}, k ≥ 5. For example, δ_{5} = 0.773114.

5 Acknowledgments

The authors express their gratitude to the referee for a careful reading of the manuscript and many valuable suggestions, which highly improve the quality of this paper.

Deyu Zhang is supported by National Natural Science Foundation of China (Grant No. 11001154) and Shandong Province Natural Science Foundation (ZR2010AQ009). Wenguang Zhai is supported by National Natural Science Foundation of China (Grant No. 11171344) and Natural Science Foundation of Beijing (No. 1112010).

References

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[8] S. S. Pillai, On an arithmetic function, J. Annamalai Univ.2 (1933), 243–248.

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2010 Mathematics Subject Classification: Primary 11N37. Keywords: gcd-sum function, Dirichlet divisor problem, asymptotic formula.

(Concerned with sequence \underline{A018804}.)

Received January 10 2013; revised versions received March 6 2013; June 14 2013. Published in Journal of Integer Sequences, July 25 2013.

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