On the Sum of the Reciprocals of the Middle Prime Factors of an Integer

Vincent Ouellet Département de mathématiques et de statistique Université Laval Québec G1V 0A6 Canada

                  vincent.ouellet.7@ulaval.ca

Abstract

We consider the arithmetical function p^{(β)}(n) := p_{max(1,⌊βk⌋)} for a given fixed number β ∈ (0, 1), where p_{1} < p_{2} < ⋯ < p_{k} are the prime factors of n. We provide an estimate for the sum of the reciprocals of p^{(β)}(n) for n ≤ x, which improves and generalizes an earlier result of De Koninck and Luca.

1 Introduction

Given an integer n ≥ 2, let P (n) denote its largest prime factor and let P (1) = 1. At the end of the 1970’s and early 1980’s, many papers focused on estimating the global behavior of the sum of the reciprocals of P (n) for n ≤ x. For the highlights, see the papers of Erdős and Ivić [7] and [8]. The best estimate was obtained in 1986 by Erdős, Ivić, and Pomerance [5, Thm. 1], as they proved that

∑_{n≤x} \frac{1}{P (n)} = x ∫_{2}^{x} ρ (\frac{log x}{log t}) t^{−2}dt (1 + O (\sqrt{\frac{log_{2} x}{log x}})) (x → ∞) ,

where ρ (u) is the Dickman function and log_{k} x denotes the k-th iterate of log evaluated at x. Here and in what follows, we shall assume that the input x in such an expression is sufficiently large so that the iterated logarithms are real and positive. For any integer k ≥ 2, letting P_{k}(n) stand for the k-th largest prime factor with multiplicity of the integer n, De Koninck [2, Thm. 2] proved that there exists a constant c_{k} such that

∑_{\begin{matrix}n≤x \\ Ω(n)≥k \end{matrix}} \frac{1}{P_{k}(n)} = c_{k} \frac{x (log_{2} x)^{k−2}}{log x} (1 + O (\frac{1}{log_{2} x})) (x → ∞) ,

where Ω(n) stands for the number of prime factors of n counting multiplicities.

During the 1984 Oberwolfach Conference on Analytic Number Theory, Erdős asked De Koninck if he had thought of estimating the sum of the reciprocals of the middle prime factors of the positive integers n ≤ x. Given an integer n ≥ 2, write it as n = p_{1}^{a_{1}}p_{2}^{a_{2}} ⋯ p_{k}^{a_{k}}, where p_{1} < p_{2} < ⋯ < p_{k} are its distinct prime factors, and the a_{i} are positive integers. Denote the number of distinct prime factors of n by ω(n), so that ω (n) = k, and let p^{(m)}(n) := p⌊_{\frac{ω(n)+1}{2}}⌋ denote its middle prime factor. De Koninck and Luca [4] proved that, as x → ∞,

∑_{1<n≤x} \frac{1}{p^{(m)}(n)} = \frac{x}{log x} exp ((\sqrt{2} + o(1)) \sqrt{log_{2} x log_{3} x}) .  (1)

Expanding the main ideas of the proof of the upper bound given by De Koninck and Luca [4], our goal here is to improve and generalize equation (1). For an integer n ≥ 2 and a fixed real number β ∈ (0, 1), we denote by p^{(β)}(n) = p_{max(1,⌊βk⌋)} the β-positioned prime factor of n, where p_{1} < p_{2} < ⋯ < p_{k} are its prime factors. As De Koninck and Luca did with the middle prime factor, we obtain an estimate for the sum of the reciprocals of the β-positioned prime factors of the integers n ≤ x.

Theorem 1. There exist four constants α_{1}, α_{2}, α_{3}, and α_{4} such that, as x → ∞,

∑_{1<n≤x} \frac{1}{p^{(β)}(n)} = \frac{x}{log x} exp (\frac{(log_{2} x)^{1−β} (log_{3} x)^{β}}{β^{β} (1 − β)^{1−2β}} (G (x, β) + O (\frac{1}{log_{3}^{2} x}))) ,  (2)

where G (x, β) = 1 + α_{1} \frac{log_{4} x}{log_{3} x} + α_{2} \frac{1}{log_{3} x} + α_{3} \frac{log_{4}^{2} x}{log_{3}^{2} x} + α_{4} \frac{log_{4} x}{log_{3}^{2} x}. In particular,

α_{1} = \frac{−β (2 − β)}{1 − β},  α_{3} = \frac{2 − β}{2} α_{1},
α_{2} = β (log β − \frac{3 − 2β}{1 − β} log (1 − β) − \frac{1}{1 − β} ) ,  α_{4} = (2 − β) α_{2} − α_{1} + \frac{β}{1 − β} .

By setting β = 1/2, the following corollary shows that (2) is an improvement over equation (1).

Corollary 2. As x → ∞,

∑_{1<n≤x} \frac{1}{p^{(m)}(n)} = \frac{x}{log x} exp (\sqrt{2 log_{2} x log_{3} x} (1 + c_{1} \frac{log_{4} x}{log_{3} x} + \frac{c_{2}}{log_{3} x} + c_{3} \frac{log_{4}^{2} x}{log_{3}^{2} x} + c_{4} \frac{log_{4} x}{log_{3}^{2} x}))

× exp (O (\sqrt{\frac{log_{2} x}{log_{3}^{3} x}})) ,

where c_{1} = \frac{−3}{2}, c_{2} = \frac{3}{2} log 2 − 1, c_{3} = \frac{−9}{8}, and c_{4} = 1 + \frac{9}{4} log 2.

2 Preliminary results

Throughout this paper, p and q always stand for prime numbers, β ∈ (0, 1) is a fixed real number, and x is a large number. Our goal is to estimate

∑_{1<n≤x} \frac{1}{p^{(β)}(n)} = ∑_{p≤x} \frac{1}{p}# \{n ≤ x : p^{(β)}(n) = p\} = ∑_{p≤x} \frac{1}{p} ∑_{k≥1} #N_{p,k}(x),  (3)

where N_{p,k}(x) := {n ≤ x : p^{(β)}(n) = p, ω(n) = k}. Note that, given any x, the sum over k is

finite since the integer k must satisfy k ≤ ⌊\frac{log x}{log 2}⌋. Moreover, it is possible that #N_{p,k}(x) = 0 for some primes p and integers k. Hence, the integers k and prime numbers p are dependent. We shall see that the main contribution to equation (3) is reached when the prime numbers p and the integers k are in particular sets. Let

N_{1}(x) := {n ≤ x : Ω(n) > 10 log_{2} x};
N_{2}(x) := {n ≤ x : p_{β}(n) > log x};
N_{3}(x) := {n ≤ x : ω(n) ∈ {1, 2, …, M}};
N_{4}(x) := {n ≤ x} \ (N_{1}(x) ∪ N_{2}(x) ∪ N_{3}(x)) ,

where M := ⌈max (\frac{2}{β} , \frac{2}{1−β} )⌉. We first show that

∑_{n∈N_{i}(x)} \frac{1}{p^{(β)}(n)} ≪ \frac{x (log_{2} x)^{M−1}}{log x}  for i = 1, 2, 3.  (4)

By [9, Lemma 13], it follows that

∑_{n∈N_{1}(x)} \frac{1}{p^{(β)}(n)} ≪ #N_{1}(x) = ∑_{\begin{matrix}n≤x \\ Ω(n)>10 log_{2} x \end{matrix}} 1 ≪ x log x \frac{10 log_{2} x}{2^{10 log_{2} x}} ≪ \frac{x}{(log x)^{5}} .  (5)

For the integers n ∈ N_{2}(x), we have

∑_{n∈N_{2}(x)} \frac{1}{p^{(β)}(n)} ≤ ∑_{n≤x} \frac{1}{log x} ≤ \frac{x}{log x}.  (6)

Finally,

∑_{n∈N_{3}(x)} \frac{1}{p^{(β)}(n)} ≪ #N_{3}(x) ≪ \frac{x (log_{2} x)^{M−1}}{log x} ,  (7)

by the Hardy-Ramanujan inequality (see Lemma 4). Hence, combining the bounds (5), (6) and (7), the upper bound (4) follows.

For each integer n ∈ N_{p,k}(x) ∩ N_{4}(x), we write ω(n) = k = \frac{1}{β} k_{0} + δ, where k_{0} = ⌊βk⌋,

so that δ ∈ [0, \frac{1}{β} ) is fixed. Note that k ∈ [M + 1, 10 log_{2} x] and that p ∈ [2, log x]. Let us write n = ap^{α}b, where a ≥ 2, P (a) < p, ω(a) = k_{0} − 1, 1 ≤ α ≤ 10 log_{2} x, p(b) > p, and ω(b) = (\frac{1}{β} − 1) k_{0} + δ, where P (n) and p(n) denote respectively the largest and the smallest prime factors of n. It follows from the bounds (5) and (7) that

∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)} = ∑_{p∈[3,log x]} \frac{1}{p} ∑_{k∈[M+1,10 log_{2} x]} # (N_{p,k}(x) ∩ N_{4}(x)) + O (\frac{x (log_{2} x)^{M−1}}{log x} ) , (8)

where p ≥ 3 comes from the fact that a ≥ 2. Note that the β-positioned prime factors of some integers n that are in the sets N_{i}(x) for i = 1, 2, 3 are counted multiple times on the right-hand side of (8), but that their contribution is taken into consideration by the error term of equation (8). The objective is now to estimate the main term of equation (8). For this, three preliminary results will be useful.

Lemma 3 (Alladi [1], Theorem 6). Given a positive integer λ, let

ω_{λ}(x, y) := # {n ≤ x : p(n) ≥ y, ω(n) = λ}

and

g (s, y, z) := ∏_{p} (1 + \frac{z}{p^{s} − 1}) (1 − \frac{1}{p^{s}} )^{z} ∏_{p<y} (1 + \frac{z}{p^{s} − 1})^{−1}

for each z ∈ ℂ. Then, for any r > 0, in the range 2 ≤ y ≤ exp ((log x)^{2/5}), ℜ(s) > \frac{1}{2} and λ ≤ r log_{2} x, we have

ω_{λ} (x, y) = \frac{x}{log x} \frac{g (1, y, μ)}{Γ (1 + μ)} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} + O (\frac{x (log_{2} x)^{λ−1} (log y)^{−μ} (log_{2} y)^{2} λ}{(λ − 1)! log x (log_{2} x)^{2}}) (x → ∞) ,

where μ := \frac{λ−1}{log_{2} x}.

Lemma 3 and the following lemmas are used to estimate the sum over b for the integers n = ap^{α}b.

Lemma 4 (Hardy-Ramanujan inequality). For any integer λ ≥ 1, define

Π_{λ}(x) = # {n ≤ x : ω(n) = λ} .

There exist positive constants c and x_{0} such that, uniformly for 1 ≤ λ ≤ 10 log_{2} x,

Π_{λ}(x) ≤ c\frac{x}{log x} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!}

for all x > x_{0}.

The next lemma follows from the proof of Theorem 1 in Erdős and Tenenbaum [6]. It will be used to obtain an estimate for the sum over a of the integers n = ap^{α}b ≤ x.

Lemma 5 (Erdős and Tenenbaum). Let ϵ > 0. For every prime number p ≥ 5, define the function ρ = ρ (k_{0} − 1, p) as the unique solution to ∑_{q<p} \frac{ρ}{q−1+ρ} = k_{0}−1 for 1 ≤ k_{0}−1 ≤ π(p)−2,

and define the functions

w(t) = \begin{cases}Γ (t + 1) t^{−t}e^{t}, & if t > 0; \\ 1, & if t = 0; \end{cases}

and

F (z, p) = ∏_{q<p} (1 + \frac{z}{q − 1})

for z ∈ ℂ. Moreover, let G := {a ∈ ℕ : ω(a) = k_{0} − 1, P (a) < p}. Then, uniformly for 1 ≤ k_{0} ≤ p^{1−ϵ}, we have

∑_{a∈G} \frac{1}{a} = \frac{F (ρ, p)}{ρ^{k_{0}−1}w (k_{0} − 1)} (1 + O (R^{−1})) ,

where R = log (\frac{log p}{log(k_{0}+1)}) (1 + log^{+} (_{\frac{k_{0}}{log( \frac{log p}{log(k_{0}+1)} )}})) and log^{+} x := max (0, log x).

In the following lemmas, we study three functions that are used to estimate the main term of the right-hand side of equation (8).

Lemma 6. Let B > 1 and define the function f : (0, ∞) → (0, ∞) by f(t) = (\frac{eB}{t} )^{t}. The function f is concave and reaches its maximum when t = B.

Definition 7. For x > −\frac{1}{e}, we define the Lambert-W function as the inverse of the realvalued function h(y) = ye^{y}, which is defined for y > −1, so that W (x)e^{W (x)} = x.

In particular, one can easily show that the Lambert-W function goes to ∞ as x → ∞, that it is strictly increasing and that it goes to 0 as x → 0. Using these facts, the following lemmas can be proved. Lemma 8. As x → ∞,

W (x) = log x − log_{2} x + \frac{log_{2} x}{log x} + O ((\frac{log_{2} x}{log x})^{2}) .

Lemma 9. As x → 0,

W (x) = x + O (x^{2}) .

The third function will be useful in the evaluation of some sums. Lemma 10. Let D > 0 and C ∈ ℝ, and define the function g : (e^{C}, ∞) → (0, ∞) by g(t) = exp (\frac{D}{β} (log t − C)^{β} − t). The function g is concave and reaches its maximum when

log t = log t_{0} := (1 − β) W (\frac{D^{\frac{1}{1−β}}}{1 − β} exp (\frac{−C}{1 − β} )) + C,

where W stands for the Lambert-W function. Proof. Clearly,

g′(t) = 0 ⟺ \frac{D}{t (log t − C)^{1−β}} − 1 = 0 ⟺ D = t (log t − C)^{1−β}

⟺ D^{\frac{1}{1−β}} = t^{\frac{1}{1−β}} (log t − C) ⟺ D^{\frac{1}{1−β}} = exp (\frac{log t}{1 − β} ) (log t − C)

⟺ \frac{D^{\frac{1}{1−β}} exp (\frac{−C}{1−β} )}{1 − β} = exp (\frac{log t − C}{1 − β} ) (\frac{log t − C}{1 − β} ) .

Hence, by the definition of the Lambert-W function, it follows that this last equation is equivalent to \frac{log t−C}{1−β} = W (^{\frac{D^{\frac{1}{1−β}}}{1−β}} exp (\frac{−C}{1−β} )). Since t > e^{C}, this equation always has a unique solution.

3 Estimation of the main term

We will now consider the primes p belonging to the interval

I := [exp ((\frac{log_{2} x}{(log_{3} x)^{\frac{4}{1−β}} log_{4} x})^{1−β}) , exp ((log_{2} x)^{1−β} log_{3} x)]  (9)

and the positive integers k belonging to the interval

J := [\frac{1}{4} \frac{(log_{2} x)^{1−β} (log_{3} x)^{β}}{β^{β} (1 − β)^{1−3β}} , 2e\frac{(log_{2} x)^{1−β} (log_{3} x)^{β}}{β^{β} (1 − β)^{1−β}} ] .  (10)

We will show that the main contribution to the right-hand side of equation (8) comes from the primes p ∈ I and integers k ∈ J. Note that #N_{p,k}(x) ≠ 0 for any prime number p ∈ I and integer k ∈ J. Let

A = A (k, p) := {a ∈ ℕ : ω(a) = k_{0} − 1, P (a) < p and Ω(a) ≤ 10 log_{2} x}  (11)

and

B = B (k, p) := \{b ∈ ℕ : ω(b) = (\frac{1}{β} − 1) k_{0} + δ and p(b) > p\} ,  (12)

so that n = ap^{α}b ∈ N_{4}(x) for a ∈ A and b ∈ B. Hence, for p ∈ I and k ∈ J, we have from the upper bound (5) that

# (N_{p,k}(x) ∩ N_{4}(x)) = ∑_{\begin{matrix}a≤x \\ a∈A \end{matrix}} ∑_{α=1}^{⌊10 log_{2} x⌋} ∑_{\begin{matrix}b≤ \frac{x}{ap^{α}} \\ b∈B \end{matrix}} 1 + O (\frac{x}{(log x)^{5}} ) .  (13)

Thus, it follows from equation (8) that

∑_{\begin{matrix}n∈N_{4}(x) \\ p^{(β)}(n)∈I \\ ω(n)∈J \end{matrix}} \frac{1}{p^{(β)}(n)} = ∑_{p∈I} \frac{1}{p} ∑_{k∈J} ∑_{\begin{matrix}a≤x \\ a∈A \end{matrix}} ∑_{α=1}^{⌊10 log_{2} x⌋} ∑_{\begin{matrix}b≤ \frac{x}{ap^{α}} \\ b∈B \end{matrix}} 1 + O (\frac{x (log_{2} x)^{M−1}}{log x} ) .  (14)

It remains to estimate the sums in the right-hand side of (14). Let

#N′_{p,k}(x) := ∑_{\begin{matrix}a≤x \\ a∈A \end{matrix}} ∑_{α=1}^{⌊10 log_{2} x⌋} ∑_{\begin{matrix}b≤ \frac{x}{ap^{α}} \\ b∈B \end{matrix}} 1.

Since ap^{α} = x^{o(1)}, we obtain from Lemma 3 that

∑_{\begin{matrix}b≤ \frac{x}{ap^{α}} \\ b∈B \end{matrix}} 1 = v_{λ} (\frac{x}{ap^{α}} , p + 2) = \frac{x}{ap^{α} log x} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} (1 + o(1)) (x → ∞) ,

where λ = (\frac{1}{β} − 1) k_{0} + δ ≥ 2, because k ≥ M + 1. Hence, as x → ∞,

#N′_{p,k}(x) ∼ \frac{x}{p log x} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} ∑_{\begin{matrix}a≤x \\ a∈A \end{matrix}} \frac{1}{a} ∑_{α=1}^{⌊10 log_{2} x⌋} \frac{1}{p^{α−1}} ∼ \frac{x}{p log x} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} ∑_{a∈A} \frac{1}{a},  (15)

because a = x^{o(1)} for all a ∈ A. Moreover, we have ∑_{a∈A} \frac{1}{a} ∼ ∑_{a∈G} \frac{1}{a} as x → ∞, where the set G is defined as in Lemma 5. Indeed, we have

∑_{a∈G} \frac{1}{a} = ∑_{a∈A} \frac{1}{a} + ∑_{\begin{matrix}a∈G \\ Ω(a)>10 log_{2} x \end{matrix}} \frac{1}{a},

and Rankin’s method (see, for example, [4, Chap. 9]) shows that ∑_{\begin{matrix}n>x \\ P (n)<y \end{matrix}} \frac{1}{n} ≪ e^{−u/2} log y for

any x ≥ y ≥ 2, where u = \frac{log x}{log y}. Hence, it follows that

∑_{\begin{matrix}a∈G \\ Ω(a)>10 log_{2} x \end{matrix}} \frac{1}{a} ≤ ∑_{\begin{matrix}a>2^{10 log_{2} x} \\ P (a)<p \end{matrix}} \frac{1}{a} ≪ exp (−\frac{5 log_{2} x log 2}{log p}) log p.

We get from equation (15) that

#N′_{p,k}(x) ∼ \frac{x}{p log x} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} ∑_{a∈G} \frac{1}{a}

as x → ∞. By an explicit evaluation of ∑_{a∈G} \frac{1}{a} using Lemma 5, we obtain

#N′_{p,k}(x) = \frac{x}{p log x} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} (\frac{eA}{k_{0}})^{k_{0}} R,  (16)

where A = A(x, p) = log_{2} p − log_{4} x − log (1 − β) + \frac{log_{4} x}{log_{3} x} and

R = R(x) = exp (O (\begin{matrix}(log_{2} x) & 1−β \\ (log_{3} x) & 2−β \end{matrix} )) .

Indeed, following the proof of Erdős and Tenenbaum [6, Lemma 1], for any prime number p ∈ I and integer k ∈ J, we have

\frac{1}{ρ} = \frac{log_{2} p − log_{2} k_{0} + \frac{log_{3} p}{log k_{0}}}{k_{0}} + O (\frac{1}{k_{0} log k_{0}} )
= \frac{log_{2} p − log_{4} x − log (1 − β) + \frac{log_{4} x}{log_{3} x}}{k_{0}} + O (\frac{1}{k_{0} log k_{0}} ) ,  (17)

and, by Erdős and Tenenbaum [6, Lemma 2], we obtain F (ρ, p) = exp (k_{0} + O (\frac{ρ}{log ρ})).

Using the Stirling’s formula to estimate (λ − 1)!, it follows from equation (16) and the definition of λ in terms of k_{0} that

#N′_{p,k}(x) = \frac{x}{p log x} (\frac{e (log_{2} x)^{1−β} A^{β}}{(\frac{1}{β} − 1) k_{0}} )^{\frac{k_{0}}{β}} R,

and, since \frac{k_{0}}{β} = k + O(1), we obtain

#N′_{p,k}(x) = \frac{x}{p log x} (\frac{eB}{k} )^{k} R,  (18)

where B = B(x, p) = \frac{(log_{2} x)^{1−β}A^{β}}{β(\frac{1}{β} −1)^{1−β}}. Hence, combining equations (14) and (18), we have

∑_{\begin{matrix}n∈N_{4}(x) \\ p^{(β)}(n)∈I \\ ω(n)∈J \end{matrix}} \frac{1}{p^{(β)}(n)} = \frac{x}{log x}R ∑_{p∈I} \frac{1}{p^{2}} ∑_{k∈J} (\frac{eB}{k} )^{k} + O (\frac{x log_{2} x log_{3} x}{log x}) .

We can conclude from Lemma 6 that

∑_{\begin{matrix}n∈N_{4}(x) \\ p^{(β)}(n)∈I \\ ω(n)∈J \end{matrix}} \frac{1}{p^{(β)}(n)} = \frac{x}{log x}R ∑_{p∈I} \frac{1}{p^{2}} exp (B) + O (\frac{x log_{2} x log_{3} x}{log x}) .  (19)

To estimate the sum over p, we note that

∑_{p∈I} \frac{1}{p^{2}} exp (B) = ∑_{p∈I} \frac{1}{p} exp (B − log p) = ∑_{p∈I} \frac{1}{p}g (log p) ,

where g is the same function as the one defined in Lemma 10, C = log_{4} x+log (1 − β)− \frac{log_{4} x}{log_{3} x}

and D = (\frac{β log_{2} x}{1−β})^{1−β}. Moreover, from this same lemma, we have

∑_{p∈I} \frac{1}{p}g (log p) ≤ g (t_{0}) ∑_{p∈I} \frac{1}{p} ≪ g (t_{0}) log_{4} x,  (20)

where, in particular,

t_{0} = (\frac{β}{(1 − β)^{2}} \frac{log_{2} x}{log_{3} x})^{1−β} (1 + (2 − β) \frac{log_{4} x}{log_{3} x} + O (\frac{1}{log_{3} x})) .  (21)

Lemma 10 also provides a lower bound for the sum over p. Indeed, let p_{0} be the largest prime number in I such that log p_{0} ≤ t_{0}. In particular, by Bertrand’s postulate, one can conclude that the prime number p_{0} satisfies p_{0} ∈ (\frac{1}{2}e^{t_{0}}, e^{t_{0}}]. Thus, for every prime number

p ∈ (\frac{1}{log_{2} x}p_{0}, p_{0} log_{2} x] ⊂ I, there exist positive constants c_{1}, c_{2}, and c_{3} such that

g (log p) ≥ g (log p_{0} − c_{1} (log_{3} x)) ≥ g (t_{0} − c_{2} (log_{3} x)) ≥ g (t_{0}) exp (−c_{3} ((log_{3} x)^{2−β}))

for every x > e^{e^{e}}, so that

∑_{p∈I} \frac{1}{p}g (log p) ≥ g (t_{0}) exp (−c_{3} (log_{3} x)^{2−β}) ∑_{p∈[ \frac{p_{0}}{log_{2} x} ,p_{0} log_{2} x]} \frac{1}{p}

≫ g (t_{0}) exp (−c_{3} (log_{3} x)^{2−β}) (\frac{log_{3} x}{log_{2} x})^{1−β}

≥ g (t_{0}) exp ((log_{3} x)^{2−β}) .  (22)

Hence, from equation (19), the upper bound (20), and the lower bound (22), we have

∑_{\begin{matrix}n∈N_{4}(x) \\ p^{(β)}(n)∈I \\ ω(n)∈J \end{matrix}} \frac{1}{p^{(β)}(n)} = \frac{x}{log x}g (t_{0}) R.  (23)

Estimates (4), (8), and (23) allow us to write

∑_{n≤x} \frac{1}{p^{(β)}(n)} = \frac{x}{log x}g (t_{0}) R + E(x),  (24)

where the error term E(x) is defined by

E(x) = ∑_{\begin{matrix}n∈N_{4}(x) \\ p^{(β)}(n)∉I or ω(n)∉J \end{matrix}} \frac{1}{p^{(β)}(n)}.

In particular, an explicit evaluation of g (t_{0}) using equation (21) yields the main term on the right-hand side of (2). What is left to do is to obtain an upper bound for the error term.

4 Estimation of the error term

In this section, we show that the error term E(x) satisfies E(x) = o (\frac{x}{log x}g (t_{0})) as x → ∞. We can proceed as in the proof of the upper bound given by De Koninck and Luca [4]. First, we have from upper bound (4) and equation (8) that

∑_{n≤x} \frac{1}{p^{(β)}(n)}= ∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)} + O (\frac{x (log_{2} x)^{M−1}}{log x} )

= ∑_{p∈[3,log x]} \frac{1}{p} ∑_{k∈[M+1,10 log_{2} x]} # (N_{p,k}(x) ∩ N_{4}(x)) + O (\frac{x (log_{2} x)^{M−1}}{log x} ) .

(25)

Moreover, by equation (13), we have

# (N_{p,k}(x) ∩ N_{4}(x)) ≤ ∑_{\begin{matrix}a≤x \\ a∈A \end{matrix}} ∑_{α=1}^{⌊10 log_{2} x⌋} ∑_{\begin{matrix}b≤ \frac{x}{ap^{α}} \\ b∈B \end{matrix}} 1.

Hence, we get from Lemma 4 that

# (N_{p,k}(x) ∩ N_{4}(x)) ≪ \frac{x}{log x} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} ∑_{\begin{matrix}a≤x \\ a∈A \end{matrix}} \frac{1}{a} ∑_{α=1}^{⌊10 log_{2} x⌋} \frac{1}{p^{α}} ≪ \frac{x}{p log x} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} ∑_{a∈A} \frac{1}{a}. (26)

In light of the definition of the set G, upper bound (26) yields

# (N_{p,k}(x) ∩ N_{4}(x)) ≪ \frac{x}{p log x} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} ∑_{a∈G} \frac{1}{a}.  (27)

On the other hand, observe that ∑_{a∈G} \frac{1}{a} = 0 if p < p_{k_{0}} and that for p ≥ p_{k_{0}}, we have

∑_{a∈G} \frac{1}{a} ≪ ∑_{\begin{matrix}p_{1}<p_{2}<···<p_{k_{0}−1}<p \\ α_{i}≥1 \end{matrix}} \frac{1}{p_{1}^{α_{1}} ⋯ p_{k_{0}−1}^{α_{k_{0}−1}}} = ∑_{\begin{matrix}p_{1}<p \\ α_{1}≥1 \end{matrix}} \frac{1}{p_{1}^{α_{1}}} ∑_{\begin{matrix}p_{1}<p_{2}<p \\ α_{2}≥1 \end{matrix}} \frac{1}{p_{2}^{α_{2}}} ⋯ ∑_{\begin{matrix}p_{k_{0}−2}<p_{k_{0}−1}<p \\ α_{k_{0}−1}≥1 \end{matrix}} \frac{1}{p_{k_{0}−1}^{α_{k_{0}−1}}}

≪ ∑_{p_{1}<p} \frac{1}{p_{1} − 1} ∑_{p_{1}<p_{2}<p} \frac{1}{p_{2} − 1} ⋯ ∑_{p_{k_{0}−2}<p_{k_{0}−1}≤p} \frac{1}{p_{k_{0}−1} − 1}

≤ \frac{1}{(k_{0} − 1)!} (∑_{q≤p} \frac{1}{q − 1})^{k_{0}−1} ≪ \frac{1}{(k_{0} − 1)!} (log_{2} p + c)^{k_{0}−1}

for some positive constant c, where the inequality comes from Mertens’ estimate. Hence, it follows from upper bound (27) that

# (N_{p,k}(x) ∩ N_{4}(x)) ≪ \frac{x}{p log x} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} \frac{((1 + o(1)) log_{2} p)^{k_{0}−1}}{(k_{0} − 1)!} .  (28)

Using bound (28) in equation (25) yields

∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)} ≪ \frac{x}{log x} ∑_{p∈[3,log x]} \frac{1}{p^{2}} ∑_{k∈[M+1,10 log_{2} x]} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} \frac{(log_{2} p + c)^{k_{0}−1}}{(k_{0} − 1)!} + \frac{x (log_{2} x)^{M−1}}{log x} .

(29)

Observe that we can assume that k ≥ log_{3} x. Indeed, it follows from upper bound (29) that

∑_{\begin{matrix}n∈N_{4}(x) \\ ω(n)<log_{3} x \end{matrix}} \frac{1}{p^{β}(n)} ≪ \frac{x}{log x} (log_{2} x)^{log_{3} x} ((1 + o(1)) log_{2} x)^{log_{3} x} .  (30)

Hence, combining estimates (29) and (30), we obtain

∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)}≪ \frac{x}{log x} ∑_{p∈[3,log x]} \frac{1}{p^{2}} ∑_{k∈[log_{3} x,10 log_{2} x]} \frac{(log_{2} x)^{λ−1}}{(λ − 1)!} \frac{(log_{2} p + c)^{k_{0}−1}}{(k_{0} − 1)!}
+\frac{x}{log x} exp (2 (log_{3} x)^{2} + o (log_{3} x)) .

By applying Stirling’s formula to (λ − 1)!, we then have

∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)}≪\frac{x}{log x} ∑_{p∈[3,log x]} \frac{1}{p^{2}} ∑_{k∈[log_{3} x,10 log_{2} x]} \frac{1}{k} (\frac{e log_{2} x}{λ})^{λ−1} (\frac{e (log_{2} p + c)}{k_{0}})^{k_{0}−1}
+ \frac{x}{log x} exp (2 (log_{3} x)^{2} + o (log_{3} x))
≪\frac{x}{log x} ∑_{p∈[3,log x]} \frac{1}{p^{2}} ∑_{k∈[log_{3} x,10 log_{2} x]} (\frac{e log_{2} x}{λ})^{λ} (\frac{e (log_{2} p + c)}{k_{0}})^{k_{0}}
+ \frac{x}{log x} exp (2 (log_{3} x)^{2} + o (log_{3} x)) .

Expressing λ and k_{0} in terms of k, we get from this upper bound that

∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)}≪\frac{x log_{2} x}{log x} ∑_{p∈[3,log x]} \frac{1}{p^{2}} ∑_{k∈[log_{3} x,10 log_{2} x]} (\frac{e (log_{2} x)^{1−β} (log_{2} p + c)^{β}}{β^{β} (1 − β)^{1−β} k} )^{k}
+ \frac{x}{log x} exp (2 (log_{3} x)^{2} + o (log_{3} x)) .

(31)

Let K_{1} and K_{2} be the smallest and largest integers in the interval J respectively. Then, from Lemma 6, since

K_{1} = \frac{(1 − β)^{β}}{4} \frac{(log_{2} x)^{1−β} (log_{3} x)^{β}}{β^{β} (1 − β)^{1−2β}} + O(1)

is smaller than the maximum of the function (\frac{e(log_{2} x)^{1−β}(log_{3} x+c)^{β}}{β^{β}(1−β)^{1−β}t} )^{t}, we obtain that

∑_{k∈[log_{3} x,K_{1}]} (\frac{e (log_{2} x)^{1−β} (log_{2} p + c)^{β}}{β^{β} (1 − β)^{1−β} k} )^{k}≪K_{1} (\frac{e (log_{2} x)^{1−β} (log_{3} x + c)^{β}}{β^{β} (1 − β)^{1−β} K_{1}} )^{K_{1}}

≪ (log_{2} x)^{1−β} (log_{3} x)^{β} (\frac{4e}{(1 − β)^{2β}} )^{K_{1}} exp (o (K_{1})) .

Since (log_{2} x)^{1−β} (log_{3} x)^{β} ≪ exp (o (K_{1})), it follows that

∑_{k∈[log_{3} x,K_{1}]} (\frac{e (log_{2} x)^{1−β} (log_{2} p + c)^{β}}{β^{β} (1 − β)^{1−β} k} )^{k}≪ exp (K_{1} (1 + log 4 − 2β log (1 − β) + o(1)))

≪ exp ((\frac{3}{5} + o(1)) \frac{(log_{2} x)^{1−β} (log_{3} x)^{β}}{β^{β} (1 − β)^{1−2β}} ) , (32)

where the last inequality comes from the fact that

K_{1} (1 + log 4 − 2β log (1 − β))=\frac{1 + log 4 − 2β log (1 − β)}{4 (1 − β)^{−β}} \frac{(log_{2} x)^{1−β} (log_{3} x)^{β}}{β^{β} (1 − β)^{1−2β}} + O(1),

where the first fraction is strictly smaller than 3/5. Indeed, the function F (β) defined for β ∈ [0, 1) by

F (β) = \frac{1 + log 4 − 2β log (1 − β)}{4 (1 − β)^{−β}}

is strictly decreasing and F (0) = \frac{1+log 4}{4} < \frac{3}{5}. Hence, from upper bounds (31) and (32), we have that

∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)}≪\frac{x log_{2} x}{log x} ∑_{p∈[3,log x]} \frac{1}{p^{2}} ∑_{k∈[K_{1},10 log_{2} x]} (\frac{e (log_{2} x)^{1−β} (log_{2} p + c)^{β}}{β^{β} (1 − β)^{1−β} k} )^{k}

+ \frac{x}{log x} exp ((\frac{3}{5} + o(1)) \frac{(log_{2} x)^{1−β} (log_{3} x)^{β}}{β^{β} (1 − β)^{1−2β}} ) .

(33)

Similarly, from Lemma 6, since

K_{2} = \frac{2e (log_{2} x)^{1−β} (log_{3} x)^{β}}{β^{β} (1 − β)^{1−β}} + O(1)

is larger than the maximum of the function (\frac{e(log_{2} x)^{1−β}(log_{3} x+c)^{β}}{β^{β}(1−β)^{1−β}t} )^{t}, we obtain

∑_{k∈[K_{2},10 log_{2} x]} (\frac{e (log_{2} x)^{1−β} (log_{2} p + c)^{β}}{β^{β} (1 − β)^{1−β} k} )^{k}≪ ∑_{k∈[K_{2},10 log_{2} x]} (\frac{e (log_{2} x)^{1−β} (log_{3} x + c)^{β}}{β^{β} (1 − β)^{1−β} k} )^{k}

≪ log_{2} x (\frac{e (log_{2} x)^{1−β} (log_{3} x)^{β} (1 + o(1))}{β^{β} (1 − β)^{1−β} K_{2}})^{K_{2}}

≪ log_{2} x (\frac{1}{2} + o(1))^{K_{2}} ≪ log_{2} x.  (34)

It follows from estimates (33) and (34) that

∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)}≪\frac{x log_{2} x}{log x} ∑_{p∈[3,log x]} \frac{1}{p^{2}} ∑_{k∈J} (\frac{e (log_{2} x)^{1−β} (log_{2} p + c)^{β}}{β^{β} (1 − β)^{1−β} k} )^{k}

+ \frac{x}{log x} exp ((\frac{3}{5} + o(1)) \frac{(log_{2} x)^{1−β} (log_{3} x)^{β}}{β^{β} (1 − β)^{1−2β}} ) .

(35)

Hence, from Lemma 6, we can conclude that

∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)} ≪ \frac{x log_{2} x}{log x} ∑_{p∈[3,log x]} \frac{1}{p^{2}} exp (\frac{(log_{2} x)^{1−β} (log_{2} p + c)^{β}}{β^{β} (1 − β)^{1−β}} )

+ \frac{x}{log x} exp ((\frac{3}{5} + o(1)) \frac{(log_{2} x)^{1−β} (log_{3} x)^{β}}{β^{β} (1 − β)^{1−2β}} ) .

Let q_{1} and q_{2} be the smallest and largest prime numbers in the interval I respectively. Then, for p ≤ q_{1}, we have

log_{2} p ≤ log_{2} q_{1} = (1 − β) (log_{3} x − \frac{4}{1 − β} log_{4} x − log_{5} x (1 + o(1))) =: (1 − β) C(x), so that

\frac{x log_{2} x}{log x} ∑_{p∈[3,q_{1}]} \frac{1}{p^{2}} exp (\frac{(log_{2} x)^{1−β} (log_{2} p + c)^{β}}{β^{β} (1 − β)^{1−β}} ) ≪ \frac{x}{log x} exp (\frac{(log_{2} x)^{1−β} (C(x))^{β}}{β^{β} (1 − β)^{1−2β}} ) . (36)

Combining upper bounds (35) and (36), we obtain

∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)}≪\frac{x log_{2} x}{log x} ∑_{p∈[q_{1},log x]} \frac{1}{p^{2}} ∑_{k∈J} (\frac{e (log_{2} x)^{1−β} (log_{2} p + c)^{β}}{β^{β} (1 − β)^{1−β} k} )^{k}

+ \frac{x}{log x} exp (\frac{(log_{2} x)^{1−β} (log_{3} x − \frac{4}{1−β} log_{4} x + O (log_{5} x))^{β}}{β^{β} (1 − β)^{1−2β}} ) ,

(37) since the second term in the right-hand side of estimate (35) is smaller than the one of the bound (37). When the prime number p satisfies q_{2} ≤ p ≤ log x, we have that log_{2} p ≤ log_{3} x and that ∑_{p≥q_{2}} \frac{1}{p^{2}} ≪ \frac{1}{q_{2} log q_{2}} , where q_{2} = exp ((log_{2} x)^{1−β} log_{3} x) + O(1). Hence, as x → ∞,

∑_{p∈[q_{2},log x]} \frac{exp (\frac{(log_{2} x)^{1−β}(log_{2} p+c)^{β}}{β^{β}(1−β)^{1−β}} )}{p^{2}} ≪ exp (\frac{(log_{2} x)^{1−β}(log_{3} x)^{β}}{β^{β}(1−β)^{1−β}} (1 + o(1))) ; exp ((log_{2} x)^{1−β} log_{3} x) = o(1).  (38)

It follows from estimates (37) and (38) that

∑_{n∈N_{4}(x)} \frac{1}{p^{(β)}(n)}≪\frac{x log_{2} x}{log x} ∑_{p∈I} \frac{1}{p^{2}} ∑_{k∈J} (\frac{e (log_{2} x)^{1−β} (log_{2} p + c)^{β}}{β^{β} (1 − β)^{1−β} k} )^{k}

+ \frac{x}{log x} exp (\frac{(log_{2} x)^{1−β} (log_{3} x − \frac{4}{1−β} log_{4} x + O (log_{5} x))^{β}}{β^{β} (1 − β)^{1−2β}} ) .

(39)

Finally, in light of equation (24), since

g (t_{0}) = exp (\frac{(log_{2} x)^{1−β} (log_{3} x − \frac{2−β}{1−β} log_{4} x + O(1))^{β}}{β^{β} (1 − β)^{1−2β}} + O ((\frac{log_{2} x}{log_{3} x})^{1−β})) ,

it follows from upper bound (39) that

\frac{x}{log x} exp (\frac{(log_{2} x)^{1−β} (log_{3} x − \frac{4}{1−β} log_{4} x − log_{5} x (1 + o(1)))^{β}}{β^{β} (1 − β)^{1−2β}} ) = o (\frac{x}{log x}g (t_{0})) .

Hence, we can conclude that E(x) = o (\frac{x}{log x}g (t_{0})), which completes the proof of Theorem 1.

5 Final remarks

The error term R in Theorem 1 seems difficult to improve if one wants to obtain an explicit result, the reason being that our result comes directly from the estimation of ρ and F (ρ, p) provided in Erdős and Tenenbaum [6]. In fact, estimate (17) is the same as the one given by Erdős and Tenenbaum [6, Lemma 1], so that obtaining an explicit result better than the one given in Theorem 1 would require improving the Erdős and Tenenbaum estimates. However, it would still be interesting to obtain an estimate for #N_{p,k}(x) in a range wider than the one provided by the primes p ∈ I and integers k ∈ J.

6 Acknowledgments

The author wishes to thank J. M. De Koninck, N. Doyon, and the reviewers for their helpful remarks and support. This study has been funded in part by NSERC and FRQNT.

References

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[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] J. M. De Koninck and F. Luca, On the middle prime factor of an integer, J. Integer Sequences 16 (2013), Article 13.5.5.

[5] P. Erdős, A. Ivić, and C. Pomerance, On sums involving reciprocals of the largest prime factor of an integer, Glas. Mat. Ser. III 21 (1986), 283–300.

[6] P. Erdős and G. Tenenbaum, Sur les densités de certaines suites d’entiers, Proc. Lond. Math. Soc. (3) 59 (1989), 417–438.

[7] P. Erdős and A. Ivić, Estimates for sums involving the largest prime factor of an integer and certain related additive functions, Studia Sci. Math. Hungar. 15 (1980), 183–199.

[8] P. Erdős and A. Ivić, On sums involving reciprocals of certain arithmetical functions, Publ. Inst. Math. (Beograd) (N.S.) 32 (1982), 49–56.

[9] F. Luca and C. Pomerance, Irreducible radical extensions and Euler-function chains, in Combinatorial Number Theory, de Gruyter, 2007, pp. 351–361.

2010 Mathematics Subject Classification: Primary 11N37. Keywords: number of distinct prime factors, smooth number, middle prime factor.

Received September 14 2017; revised version received October 1 2017. Published in Journal of Integer Sequences, October 29 2017.

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