Exponential Sums Involving the k-th Largest Prime Factor Function

Jean-Marie De Koninck Département de Mathématiques

Université Laval Québec G1V 0A6

Canada

 jmdk@mat.ulaval.ca

Imre Kátai

Computer Algebra Department

Eötvös Loránd University

1117 Budapest Pázmány Péter Sétány I/C

Hungary

katai@compalg.inf.elte.hu

Dedicated to Jean-Paul Allouche on the occasion of his 60^{th} birthday

Abstract

Letting P_{k}(n) stand for the k-th largest prime factor of n ≥ 2 and given an irrational number α and a multiplicative function f such that |f(n)| = 1 for all positive integers n, we prove that ∑_{n≤x} f(n) exp{2πiαP_{k}(n)} = o(x) as x → ∞.

1 Introduction

In 1954, Vinogradov [7] showed that, given any irrational number α, if p_{1} < p_{2} < ⋯ stands for the sequence of primes, then

∑_{n≤x} e(αp_{n}) = o(x) as x → ∞,  (1)

where we used the standard notation e(z) = exp{2πiz}. In light of the well known Weyl criteria (see the book of Kuipers and Niederreiter [5]), statement (1) is equivalent to asserting that the sequence αp_{n}, n = 1, 2, …, is uniformly distributed mod 1.

In 2005, Banks, Harman and Shparlinski [1] proved that for every irrational number α,

∑_{n≤x} e(αP (n)) = o(x) as x → ∞,  (2)

where P (n) stands for the largest prime factor of the integer n ≥ 2 with P (1) = 1.

Let M denote the set of all complex valued multiplicative arithmetical functions and let M_{1} be those f ∈ M for which |f(n)| = 1 for all positive integers n. In [2], we generalized (2) by showing that for any irrational number α and any function f ∈ M_{1}, we have ∑_{n≤x} f(n)e(αP (n)) = o(x) as x → ∞.

Let ω(n) stand for the number of distinct prime divisors of n ≥ 2 with ω(1) = 0. Given an integer k ≥ 1, for each integer n ≥ 2, we let P_{k}(n) stand for the k-th largest prime factor of n if ω(n) ≥ k, while we set P_{k}(n) = 1 if ω(n) ≤ k − 1. Thus, if n = p_{1}^{α_{1}}p_{2}^{α_{2}} ⋯ p_{s}^{α_{s}} stands for the prime factorization of n, where p_{1} < p_{2} < ⋯ < p_{s}, then

P_{1}(n) = P (n) = p_{s},  P_{2}(n) = p_{s−1},  P_{3}(n) = p_{s−2}, …

In this paper, we prove that, given any integer k ≥ 2 and any irrational number α, then

∑_{n≤x} f(n)e(αP_{k}(n)) = o(x) as x → ∞.

2 Main result

Theorem 1. Given an integer k ≥ 2 and an irrational number α, let f ∈ M_{1} and consider the sum

S_{f}(x) = ∑_{n≤x} f(n)e(αP_{k}(n)).

Then

S_{f}(x) = o(x)  as x → ∞.  (3)

3 Notation and preliminary results

We say that a function L : ℝ^{+} → ℝ^{+} is slowly oscillating if lim_{y→∞} L(cy)/L(y) for each real number c > 0.

In 1968, Halász [4] established the following result.

Lemma 2 (Halász’s theorem). Let f be a complex-valued multiplicative arithmetical function such that |f(n)| ≤ 1 for all positive integers n. The following two statements hold:

(a) If there exists a real number τ_{0} for which the series

∑_{p} \frac{1 − ℜ(f(p)/p^{iτ_{0}})}{p}

is convergent, then, as x → ∞,

∑_{n≤x} f(n) = x · \frac{x^{iτ_{0}}}{1 + ıτ_{0}} ∏_{p≤x} (1 − \frac{1}{p}) (1 + ∑_{r=1}^{∞} \frac{f(p^{r})}{p^{r(1+iτ_{0})}} ) + o(x).

(b) If the series

∑_{p} \frac{1 − ℜ(f(p)/p^{iτ})}{p}

is divergent for every real number τ, then

∑_{n≤x} f(n) = o(x) as x → ∞.

Proof. For a proof, see the book of Schwarz and Spilker ([6, Thm. 3.1]).

Fix an integer k ≥ 2 and for each real number τ, let

R_{τ}(x) := ∑_{n≤x} f(n)n^{iτ}e(αP_{k}(n)).

We then have the following result. Lemma 3. Let τ_{1}, τ_{2} ∈ ℝ. Then, as x → ∞,

(a) R_{τ_{1}}(x) = o(x) ⟺ (b) R_{τ_{2}}(x) = o(x). (4)

Proof. It is clear that (a) holds if and only if, given any ε > 0,

\frac{1}{εx} ∑_{x≤n≤(1+ε)x} f(n)n^{iτ_{1}}e(αP_{k}(n)) → 0 as x → ∞,

while (b) holds if and only if, given any ε > 0,

\frac{1}{εx} ∑_{x≤n≤(1+ε)x} f(n)n^{iτ_{2}}e(αP_{k}(n)) → 0 as x → ∞.

But since each n ∈ [x, (1 + ε)x] can be written as n = x + δx for some 0 ≤ δ ≤ ε, we have

n^{iτ_{2}} = (x + δx)^{iτ_{2}} = x^{iτ_{2}} (1 + δ)^{iτ_{2}} = x^{iτ_{2}}(1 + O(ε)),

and similarly

n^{iτ_{1}} = x^{iτ_{1}}(1 + O(ε)).

It follows that (a) and (b) are equivalent, thus proving (4).

Lemma 4. For all 2 ≤ y ≤ x, let Ψ(x, y) := #{n ≤ x : P (n) ≤ y}. Then,

(a) As x → ∞,

Ψ(x, y) = (1 + o(1))ρ(u) x,

where u = log x/ log y and ρ(u) is the Dickman function defined by the initial condition ρ(u) = 1 for 0 ≤ u ≤ 1 and thereafter as the continuous solution of the differential equation with shift differences

uρ′(u) + ρ(u − 1) = 0 (u > 1).

(b) For all 2 ≤ y ≤ x, Ψ(x, y) ≪ x exp \{−\frac{1}{2} \frac{log x}{log y}\}.

Proof. Proofs of these results can be found in the book of De Koninck and Luca ([3], pages 134 and 138).

Lemma 5. Given an arbitrary irrational number α, set

S_{1}(x) = ∑_{n≤x} e(αP_{k}(n)).

Then

S_{1}(x) = o(x) as x → ∞.

Proof. Let ε > 0 be a small number. It is easy to see that in the sum representing S_{1}(x), we may drop three types of integers n ≤ x, namely (i) those for which ω(n) ≤ k + 1, (ii) those for which P_{k+1}(n) ≤ x^{ε} and finally (iii) those for which p^{2}|n for some prime p ≥ P_{k}(n), the reason being that the number of these exceptional n’s is O(εx). So, let us write the remaining integers n ≤ x as

n = νp_{k}p_{k−1} ⋯ p_{1}, where x^{ε} < P (ν) < p_{k} < p_{k−1} < ⋯ < p_{1}

and set

Q_{k} = p_{k}p_{k−1} ⋯ p_{1} (< x^{1−ε}).

Using this set up, we may write

S_{1}(x) = ∑_{{x^{ε}<p_{k}<···<p_{1} \atop Q_{k}<x^{1−ε}}} e(αp_{k})Ψ (\frac{x}{Q_{k}} , p_{k}) + O(εx).

Let

T_{1}(x) = ∑_{{x^{ε}<p_{k}<···<p_{1} \atop Q_{k}<x^{1−ε}}} e(αp_{k})Ψ (\frac{x}{Q_{k}} , p_{k}) ,

so that

S_{1}(x) = T_{1}(x) + O(εx),  (5)

Now, observe that, using Lemma 4 and the fact that Q_{k} = p_{k}Q_{k−1}, we have

Ψ (\frac{x}{Q_{k}} , p_{k}) = \frac{x}{Q_{k}} ρ (\frac{log x − log Q_{k}}{log p_{k}} ) + o (\frac{x}{Q_{k}} )
= \frac{x}{Q_{k}} ρ (\frac{log x − log Q_{k−1}}{log p_{k}} − 1) + o (\frac{x}{Q_{k}} ) .

Substituting this last identity in (5), we get

T_{1}(x) = ∑_{{x^{ε}<p_{k}<···<p_{1} \atop p_{k}Q_{k−1}<x^{1−ε}}} e(αp_{k})\frac{x}{Q_{k}} ρ (\frac{log x − log Q_{k−1}}{log p_{k}} − 1) + o(x)
= ∑_{{x^{ε}<p_{k}<···<p_{1} \atop p_{k}Q_{k−1}<x^{1−ε}}} \frac{x}{Q_{k−1}} ∑_{x^{ε}<p_{k}<min(p_{k−1}, \frac{x^{1−ε}}{Q_{k−1}} )} \frac{e(αp_{k})}{p_{k}}ρ (\frac{log x − log Q_{k−1}}{log p_{k}} − 1)
+ o(x).  (6)

Setting t(p_{k−1}, …, p_{1}) := min (p_{k−1}, \frac{x^{1−ε}}{Q_{k−1}} ), we now subdivide the above inner sum into two separate sums, depending if

t(p_{k−1}, …, p_{1}) ≤ 2x^{ε} or t(p_{k−1}, …, p_{1}) > 2x^{ε},

and thus we write T_{1}(x) = T′_{1}(x) + T′′_{1}(x).

On the one hand, using the fact that ∑_{x^{ε}<p_{k}≤2x^{ε}} \frac{1}{p_{k}} ≪ \frac{1}{ε log x}, we obtain

|T′_{1}(x)| ≪ \frac{x}{ε log x} ( ∑_{x^{ε}<p_{k}≤2x^{ε}} \frac{1}{p_{k}} )^{k} ≪_{ε} \frac{x}{log x}.  (7)

On the other hand, using the Vinogradov theorem (see (1)) and the continuity of the ρ function, we obtain that, as x → ∞,

|T′′_{1}(x)| ≤ ∑_{x^{ε}<p_{k}<t(p_{k−1},...,p_{1})} \frac{e(αp_{k})}{p_{k}}ρ (\frac{log x − log Q_{k−1}}{log p_{k}} − 1)
= o (  ∑_{x^{ε}<p_{k}<t(p_{k−1},...,p_{1})} \frac{1}{p_{k}} ρ (\frac{log x − log Q_{k−1}}{log p_{k}} − 1))
= o(x).  (8)

Substituting (7) and (8) in (6), and thus in light of (5) completes the proof of Lemma 5.

4 Proof of Theorem 1

Let us first assume (case (b) of Halász’s theorem) that

∑_{p} \frac{1 − ℜ(f(p)p^{iτ})}{p} = ∞ for all τ ∈ ℝ.

Let us set

E(x) = ∑_{n≤x} f(n) and E(x|y) = ∑_{{n≤x \atop P (n)≤y}} f(n).

It follows from Halász’s theorem (Lemma 2) that E(x) = o(x) as x → ∞, in which case there exists a positive decreasing function δ(x) which tends to 0 as x → ∞ and for which we have

|E(x)| ≤ xδ(x).  (9)

Let ε > 0 be a fixed small number and choose y satisfying x^{ε} ≤ y ≤ x. Further set Π_{y} := ∏_{y≤p≤x} p. We then have

E(x|y) = ∑_{n≤x} f(n) ∑_{d|(n,Π_{y})} μ(d) = ∑_{d|Π_{y}} μ(d) ∑_{md≤x} f(md)

= ∑_{d|Π_{y}} μ(d)f(d) ∑_{m≤x/d} f(m) + O ( ∑_{dm≤x ; {d|Πy \atop (d,m)>1}} 1)

= ∑_{d|Π_{y}} μ(d)f(d)E(x/d) + O (x ∑_{p>y} \frac{1}{p^{2}} ) .

Consequently, uniformly for x^{ε} ≤ y ≤ x, and in light of (9), we have

|E(x|y)| ≤ x ∑_{d|Π_{y}} \frac{δ(x/d)}{d} + O (\frac{x}{y}) .  (10)

In light of (10), in order to show that

E(x|y) = o(x)  as x → ∞,  (11)

we only need to show that

T_{0} := ∑_{d|Π_{y}} \frac{δ(x/d)}{d} = o(1) as x → ∞.  (12)

We split the above sum in two parts as follows:

T_{0} = ∑_{{d|Πy \atop d≤x/ log x}} \frac{δ(x/d)}{d} + ∑_{{d|Πy \atop x/ log x<d≤x}} \frac{δ(x/d)}{d}

≤ δ(log x) ∑_{{d|Πy \atop d≤x/ log x}} \frac{1}{d} + c ∑_{{d|Πy \atop x/ log x<d≤x}} \frac{1}{d}

= δ(log x)T_{1} + cT_{2},  (13)

where c is some positive constant. On the one hand, we have

T_{1} ≤ ∏_{x^{ε}≤p≤x} (1 + \frac{1}{p}) ≪ exp \{ ∑_{x^{ε}≤p≤x} \frac{1}{p}\} ≪ \frac{1}{ε}.  (14)

On the other hand, setting U_{0} = x/ log x and letting j_{0} be the smallest positive integer satisfying 2^{j_{0}+1}U_{0} > x, we have

T_{2} ≤ ∑_{j=0}^{j_{0}} \frac{1}{2^{j}U_{0}} ∑_{{2^{j} U_{0}≤d<2^{j+1}U_{0} \atop p(d)>x^{ε}}} 1

≤ ∑_{j=1}^{j_{0}} ∏_{p<x^{ε}} (1 − \frac{1}{p}) ≪ \frac{j_{0}}{log x} ≪ \frac{log log x}{log x}.  (15)

Combining (14) and (15), we immediately obtain (12), from which (11) follows.

On the other hand,

Ψ(x, y) ≥_{ε} x for x^{ε} ≤ y ≤ x.  (16)

Combining (11) and (16), we get that

lim_{x→∞} max_{x^{ε}≤y≤x} \frac{|E(x|y)|}{Ψ(x, y)} = 0.  (17)

Given a positive integer k and a positive integer n, it will be convenient to write

Q_{k}(n) = Q_{k} = P_{k}(n)P_{k−1}(n) ⋯ P_{1}(n).

Then, write

S_{f}(x) = ∑_{{n≤x \atop P_{k}(n)≤x^{ε}}} f(n)e(αP_{k}(n)) + ∑_{{n≤x \atop P_{k}(n)>x^{ε}}} f(n)e(αP_{k}(n)) = S′_{f}(x) + S′′_{f}(x),  (18)

say.

First, observe that it is an easy consequence of the Turán-Kubilius inequality that

∑_{n≤x} ( ∑_{{p|n \atop x^{ε}<p≤x}} 1 − ∑_{x^{ε}<p≤x} \frac{1}{p})^{2} ≪ x ∑_{x^{ε}<p≤x} \frac{1}{p} ≪ x log(1/ε),

from which it follows that

∑_{{n≤x \atop P_{k}(n)≤x^{ε}}} (k − log(1/ε))^{2} ≪ x log(1/ε).

Using this, we conclude that

|S′_{f}(x)| ≤ #\{n ≤ x : P_{k}(n) ≤ x^{ε}\} ≪ \frac{x}{log(1/ε)}.  (19)

Similarly, we can say that

#\{n ≤ x : P_{k+1}(n) ≤ x^{ε}\} ≪ \frac{x}{log(1/ε)}.

This implies that Q_{k}(n) ≤ x^{1−ε} for all but O (\frac{x}{log(1/ε)}) integers n ≤ x.

This means that

|S′′_{f}(x)| ≤ |e(αP_{k})f(Q_{k})E(x/Q_{k}|P_{k})| + O (\frac{x}{log(1/ε)}) .  (20)

Using (17), we obtain that the summation on the right-hand side of (20) is

o (x ∑_{{p_{k}···p_{1}≤x \atop x^{ε}<p_{k}<···<p_{1}}} \frac{1}{p_{k} ⋯ p_{1}} ) = o (x\frac{1}{k!} ( ∑_{x^{ε}<p≤x} \frac{1}{p})^{k}) = o (x (log \frac{1}{ε})^{k}) ,

implying that

S′′_{f}(x) = o(x) as x → ∞.  (21)

Substituting (19) and (21) in (18), we obtain (3).

It remains to consider case (a) of Halász’s theorem (Lemma 2), that is when there exists a real number τ_{0} for which the series

∑_{p} \frac{1 − ℜ(f(p)/p^{iτ_{0}})}{p}

is convergent. In light of Lemma 3 we can assume that τ_{0} = 0, that is that

∑_{p} \frac{1 − ℜ(f(p))}{p} < ∞.  (22)

For each prime power p^{a}, let us write f(p^{a}) = exp\{iu(p^{a})\} where u(p^{a}) ∈ [−\frac{π}{2}, π). It follows that

∑_{p} \frac{u^{2}(p)}{p} < ∞.

Now let D be a large number and define the multiplicative functions f_{D} and g_{D} on prime powers p^{a} by

f_{D}(p^{a}) = \begin{cases}f(p^{a}), & if p ≤ D; \\ 1, & if p > D; \end{cases} and g_{D}(p^{a}) = \begin{cases}1, & if p ≤ D; \\ f(p^{a}), & if p > D. \end{cases}

Then define the arithmetical function t(n) implicitly by the relation f_{D}(n) = ∑_{δ|n} t(δ). Since one easily sees that t(p) = 0 if p > D, it follows that the above summation runs over only those divisors δ for which P (δ) ≤ D.

Further define

a_{D}(x) := ∑_{D<p≤x} \frac{u(p)}{p}.

Using the Turán-Kubilius inequality, we obtain that

∑_{n≤x} (∑_{{p^{a}∥n \atop p>D}} u(p^{a}) − a_{D}(x))^{2} ≪ x ∑_{p>D} \frac{u^{2}(p^{a})}{p^{a}} ≪ xη_{D}^{2},

say, where η_{D} → 0 as D → ∞.

It follows from this that

∑_{n≤x} |f(n) − f_{D}(n)e^{ia_{D}(x)}|^{2} ≪ η_{D}^{2}x,

and therefore that

∑_{n≤x} |f(n) − f_{D}(n)e^{ia_{D}(x)}| ≪ η_{D}x.

We may conclude from this that

S_{f}(x) = e^{−ia_{D}(x)}A_{D}(x) + O(η_{D}x),

where

A_{D}(x) := ∑_{n≤x} f_{D}(n)e(αP_{k}(n)).

For each integer δ ≥ 1, let

B_{δ}(y) = ∑_{m≤y} e(αP_{k}(δm)).

With this definition, we may write

A_{D}(x) = ∑_{{δ≤x \atop P (δ)≤D}} t(δ)B_{δ} (\frac{x}{δ}) .  (23)

Now if P_{k}(δm) ≠ P_{k}(m), then either ω(m) ≤ k − 1 or P_{k}(m) ≤ D. Thus

|B_{δ} (\frac{x}{δ}) − B_{1} (\frac{x}{δ})| ≤ ∑_{{m≤x/δ \atop ω(m)≤k−1}} 1 + ∑_{{Qν≤x/δ \atop ω(Q)≤k−1, P (ν)≤D}} 1 = U_{1}(x) + U_{2}(x),  (24)

say. Write

U_{1}(x) = ∑_{δ≤\sqrt{x}} ∗ + ∑_{\sqrt{x}<δ≤x} ∗ = U′_{1}(x) + U′′_{1}(x),  (25)

say. Then, it is clear that

U′′_{1}(x) ≤ ∑_{m≤\sqrt{x}} 1 ≤ \sqrt{x}.  (26)

On the other hand, using the Hardy-Ramanujan inequality (see, for instance, [3, Theorem 10.1]), it follows that there exist two absolute positive constants c_{1} and c_{2} such that

U′_{1}(x) ≤ \frac{c_{1}x}{δ log x} \frac{(log log x + c_{2})^{k−2}}{(k − 2)!} .  (27)

On the other hand,

U_{2}(x) ≤ U′_{2}(x) + U′′_{2}(x),  (28)

where in U′_{2}(x), we sum over those Q ≤ \sqrt{x/δ}, while in U′′_{2}(x), we sum over those ν ≤ \sqrt{x/δ}. To estimate U′_{2}(x), we proceed as follows. First, using Lemma 4 (b), we get

U′_{2}(x) ≤ ∑_{{Q≤\sqrt{x/δ} \atop ω(Q)≤k−1}} ∑_{{ν≤x/δQ \atop P (ν)≤D}} 1 ≪ ∑_{{Q≤\sqrt{x/δ} \atop ω(Q)≤k−1}} \frac{x}{δQ} exp \{−\frac{1}{2} \frac{log(x/δQ)}{log D}\} .  (29)

Since \frac{x}{δQ} ≥ (\frac{x}{δ})^{1/4} ≥ x^{1/8}, it follows from (29) that

U′_{2}(x) ≪ ∑_{{Q≤\sqrt{x/δ} \atop ω(Q)≤k−1}} \frac{x}{δQ} exp \{−\frac{1}{16} \frac{log x}{log D}\} .  (30)

Since

∑_{{Q≤x \atop ω(Q)≤k−1}} \frac{1}{Q} ≪ (log log x)^{k−1},

it follows from (30) that, given any positive number K,

U′_{2}(x) ≪_{D} \frac{x}{δ}(log x)^{−K}.  (31)

On the other hand, setting π_{k}(x) := #{n ≤ x : ω(n) = k} and again using the Hardy-Ramanujan inequality, it follows that

U′′_{2}(x) ≤ ∑_{{ν≤\sqrt{x/δ} \atop P (ν)≤D}} ∑_{{Q≤x/δν \atop ω(Q)≤k−1}} 1

≤ (k − 1) ∑_{{ν≤\sqrt{x/δ} \atop P (ν)≤D}} π_{k−1} (\frac{x}{δν} )

≤ c_{1} ∑_{P (ν)≤D} \frac{kx}{δν} · \frac{1}{log x} \frac{(log log x + c_{2})^{k−2}}{(k − 2)!}

≤ \frac{c_{1}x}{δ log x}(log log x + c_{2})^{k−2} ∏_{p≤D} (1 − \frac{1}{p})^{−1}

≪ \frac{x}{δ} \frac{log D}{log x}(log log x)^{k−2}.  (32)

Substituting (26) and (27) in (25), and then using (31) and (32) in (28), we obtain from (24) that

max_{δ≤\sqrt{x}} \frac{1}{x/δ} |B_{δ} (\frac{x}{δ}) − B_{1} (\frac{x}{δ})| ≪ \frac{1}{\sqrt{log x}},

say. It follows from this last estimate and (23) that for some positive constant c_{3}

|A_{D}(x)| ≤ x ∑_{{\sqrt{x}<δ<x \atop P (δ)≤D}} \frac{|t(δ)|}{δ} + ∑_{{δ≤\sqrt{x} \atop P (δ)≤D}} |t(δ)| |B_{1} (\frac{x}{δ})| + \frac{c_{3}}{\sqrt{log x}} ∑_{δ≤\sqrt{x}} |t(δ)|\frac{x}{δ}

= xW_{1}(x) + W_{2}(x) + \frac{c_{3}x}{\sqrt{log x}}W_{3}(x),  (33)

say.

Since

W_{3}(x) ≤ ∏_{p≤D} (1 + \frac{|t(p)|}{p} + \frac{|t(p^{2})|}{p^{2}} + ⋯)

and since |t(p^{a})| = |f(p^{a}) − f(p^{a−1})| ≤ 2, it follows that

W_{3}(x) ≤ c(log D)^{2}.  (34)

Using Lemma 5, we obtain that, as x → ∞,

W_{2}(x) = o (xW_{3}(x)) = o (x(log D)^{2}) .  (35)

In order to estimate W_{1}(x), let us first find an upper bound for

κ(v) := ∑_{{v≤δ≤2v \atop P (δ)≤D}} t(δ) for \sqrt{x} ≤ v ≤ x.

We have

κ(v) ≤ 2 ∑_{{k≤\sqrt{2v} \atop P (k)≤D}} ∑_{{ℓ∈[v/k,2v/k] \atop P (ℓ)≤D}} 1 ≤ 2 ∑_{{k≤\sqrt{2v} \atop P (k)≤D}} Ψ (\frac{2v}{k} , D) .  (36)

Since \frac{2v}{k} ≥ \sqrt{2v} ≥ \sqrt{x}, it follows that, given any arbitrary large number R > 0,

Ψ (\frac{2v}{k} , D) ≤ \frac{2vc}{k}(log x)^{−R}.  (37)

Let v_{0} = \sqrt{x} and, for each integer j ≥ 1, let v_{j} = 2^{j}\sqrt{x}. Letting j_{0} be the smallest positive integer such that v_{j_{0}} ≥ x, so that j_{0} = O(log x), we obtain, using (37) in (36), that

W_{1}(x) ≤ ∑_{j=0}^{j_{0}} \frac{κ(v_{j})}{v_{j}} ≪ \frac{j_{0} + 1}{(log x)^{R}} .  (38)

Substituting (34), (35) and (38) in (33), we obtain that

A_{D}(x) = o(x) as x → ∞,

thus completing the proof of Theorem 1.

5 Acknowledgments

Research of the first author was supported by a grant from NSERC and that of the second author by ELTE IK.

References

[1] W. D. Banks, G. Harman and I. E. Shparlinski, Distributional properties of the largest prime factor, Michigan Math. J. 53 (2005), 665–681.

[2] J. M. De Koninck and I. Kátai, Exponential sums involving the largest prime factor function, Acta Arith. 146 (2011), 233–245.

[3] J. M. De Koninck and F. Luca, Analytic Number Theory: Exploring the Anatomy of Integers, Graduate Studies in Mathematics, Vol. 134, American Mathematical Society, 2012.

[4] G. Halász, Über die Mittelwerte multiplikativen zahlentheoretischer Funktionen, Acta Math. Acad. Scient. Hungaricae 19 (1968), 365–404.

[5] L. Kuipers and H. Niederreiter, Uniform Distribution of Sequences, John Wiley & Sons, 1974.

[6] W. Schwarz and J. Spilker, Arithmetical Functions, London Mathematical Society Lecture Note Series, Vol. 184, Cambridge University Press, 1994.

[7] I. M. Vinogradov, The Method of Trigonometrical Sums in the Theory of Numbers, translated from the Russian, revised and annotated by K. F. Roth and Anne Davenport. Reprint of the 1954 translation. Dover Publications, 2004.

2010 Mathematics Subject Classification: Primary 11L07; Secondary 11N37. Keywords: exponential sum, largest prime factor.

Received October 7 2012; revised version received October 27 2012. Published in Journal of Integer Sequences, March 2 2013.

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