The Number of Prime Factors on Average in Certain Integer Sequences

Rafael Jakimczuk División Matemática

Universidad Nacional de Luján Buenos Aires Argentina

jakimczu@mail.unlu.edu.ar

Matilde Lalín

Département de mathématiques et de statistique

Université de Montréal CP 6128, succ. Centre-ville

Montréal, QC H3C 3J7

Canada

mlalin@dms.umontreal.ca

Abstract

Let Ω(n) denote the total number of primes in the factorization of the integer n. We obtain asymptotic formulae for the sums ∑_{n≤x} Ω(n), where n runs over the squarefree numbers, the square-full numbers, the perfect powers, and generalizations of these families of numbers.

1 Introduction

Let us consider the prime factorization of a positive integral number n, namely

n = q_{1}^{s_{1}} ⋯ q_{r}^{s_{r}},  (1)

where the q_{i} (i = 1, …, r) are its distinct prime factors and the s_{i} (i = 1, …, r) are their respective multiplicities. Let Ω(n) be the total number of prime factors in the factorization of n, that is, Ω(n) = s_{1} + ⋯ + s_{r}. The following formula is well-known (see, for example, [2, Thm. 430, p. 355] and [1, Sect. 1.4.4]):

∑_{n≤x} Ω(n) = x log log x + B_{2}x + O (\frac{x}{log x})  (x ≥ 2),  (2)

where

B_{2} = B_{1} + ∑_{p} \frac{1}{p(p − 1)},

B_{1} is the Mertens constant given by

B_{1} = γ + ∑_{p} (log (1 − \frac{1}{p}) + \frac{1}{p}) ,

and

γ = lim_{n→∞} (∑_{k=1}^{n} \frac{1}{k} − log n) ≈ 0.57721566 …

is the Euler-Mascheroni constant.

A positive integer n is square-free if and only if its prime factorization has no factors with an exponent larger than one, that is, n = q_{1} ⋯ q_{r}, where the q_{i} (i = 1, …, r) are distinct. Let S_{2} denote the set of square-free numbers, and let Q_{2}(x) denote the cardinality of the set of square-free numbers not exceeding x. It is well-known [2, Thm. 333, p. 269] that these numbers have density \frac{1}{ζ(2)} = \frac{6}{π^{2}} , where ζ(s) denotes the Riemann zeta function, that is,

Q_{2}(x) = \frac{x}{ζ(2)} + O (x^{1/2}) = \frac{6}{π^{2}} x + O (x^{1/2}) .  (3)

More generally, given a positive integer n with prime factorization n = q_{1}^{s_{1}} ⋯ q_{r}^{s_{r}} as in (1), we shall say that n is h-free if s_{i} ≤ h − 1 (i = 1, …, r). In particular, if h = 2 we obtain the square-free numbers. Let S_{h} denote the set of h-free numbers, and let Q_{h}(x) be the cardinality of the set of h-free numbers not exceeding x. It is well-known that these numbers have density \frac{1}{ζ(h)}, that is,

Q_{h}(x) = \frac{x}{ζ(h)} + O (x^{1/h}) ,  (h ≥ 2).  (4)

In this note we show a proof for the following statement.

Theorem 1. Let h ≥ 2 be an arbitrary fixed integer, and let S_{h} be the set of h-free numbers. Then

∑_{\begin{matrix}n≤x \\ n∈S_{h} \end{matrix}} Ω(n) = \frac{1}{ζ(h)}x log log x + O(x).

Let h ≥ 2 be an arbitrary fixed integer. A positive integer is said to be h-full if all the factors in its prime factorization have exponent greater than or equal to h. That is, the number n = q_{1}^{s_{1}} ⋯ q_{r}^{s_{r}} is h-full if s_{i} ≥ h (i = 1, …, r). If h = 2 these numbers are called square-full or powerful. Let N_{h} denote the set of h-full numbers and let A_{h}(x) denote the cardinality of the set of h-full numbers not exceeding x. The number A_{h}(x) has been estimated by Ivić and Shiu [3] with the following asymptotic formula:

A_{h}(x) = γ_{0,h}x^{1/h} + γ_{1,h}x^{1/(h+1)} + ⋯ + γ_{h−1,h}x^{1/(2h−1)} + Δ_{h}(x),

where γ_{0,h}, γ_{1,h} …, γ_{h−1,h} are certain computable constants and Δ_{h}(x) ≪ x^{ρ} with ρ ≤ \frac{263}{2052}. (In fact, ρ can be much smaller, depending on h.)

The constant for the main term in the above asymptotics can be precisely described by the following [5]:

γ_{0,h} = \frac{6}{π^{2}} ∑_{n=1}^{∞} \frac{1}{σ(u(n))n^{1/h}} = ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) ,

where σ(n) denotes the sum of divisors of n and u(n) is the greatest square-free integer that divides n. Thus, if n = q_{1}^{s_{1}} ⋯ q_{r}^{s_{r}}, then σ(u(n)) = (q_{1} + 1) ⋯ (q_{r} + 1).

In this note we prove the following result.

Theorem 2. Let h ≥ 2 be an arbitrary fixed integer, and let N_{h} be the set of h-full numbers. Then

∑_{\begin{matrix}n≤x \\ n∈N_{h} \end{matrix}} Ω(n) = hγ_{0,h}x^{1/h} log log x

+ (h(B_{2} − log h) + ∑_{p} \frac{(h + 1)p^{1+1/h} − hp − 2hp^{2/h} + (2h − 1)p^{1/h}}{(p − 1) (p^{1/h} − 1) (p^{1+1/h} + p^{1/h} − p)} ) γ_{0,h}x^{1/h}

+ O_{h} (\frac{x^{1/h}}{\sqrt{log x}}) .

Setting h = 2 in the previous result, we immediately obtain the following corollary.

Corollary 3. We have

∑_{\begin{matrix}n≤x \\ n∈N_{2} \end{matrix}} Ω(n) = 2\frac{ζ(3/2)}{ζ(3)}x^{1/2} log log x

+ (2(B_{2} − log 2) + ∑_{p} \frac{3}{p^{3/2} + 1}) \frac{ζ(3/2)}{ζ(3)}x^{1/2} + O (\frac{x^{1/2}}{\sqrt{log x}}) .

A perfect power is a number of the form n^{m}, where m and n are positive integers such that m ≥ 2. Therefore, all perfect powers are square-full numbers. Let P denote the set of perfect powers and let N(x) be the cardinality of the set of perfect powers not exceeding x. The following formula is well-known [4, Thm. 5]

N(x) = x^{1/2} + f(x)x^{1/3},

where lim_{x⟶∞} f(x) = 1.

We prove the following result.

Theorem 4. Let P be the set of perfect powers. Then

∑_{\begin{matrix}n≤x \\ n∈P \end{matrix}} Ω(n) = 2x^{1/2} log log x + 2(B_{2} − log 2)x^{1/2} + O (\frac{x^{1/2}}{log x}) ,

and therefore,

∑_{\begin{matrix}n≤x \\ n∈P \end{matrix}} Ω(n) ∼ 2N(x) log log x.

The proofs of Theorems 1, 2, and 4 are detailed in Sections 2, 3, and 4 respectively.

2 Proof of Theorem 1

We start by considering the following Euler product (converging at Re(s) > 1), which gives the generating Dirichlet series for the h-free numbers:

\frac{ζ(s)}{ζ(hs)} = ∏_{p} (\frac{1 − \frac{1}{p^{hs}}}{1 − \frac{1}{p^{s}}} ) = ∏_{p} (1 + \frac{1}{p^{s}} + ⋯ + \frac{1}{p^{(h−1)s}} ) = ∑_{n∈S_{h}} \frac{1}{n^{s}} .

We will incorporate the coefficient Ω(n) by introducing it as the exponent of an extra variable, which we will later differentiate. This preserves the additive structure of Ω(n).

∏_{p} (1 + \frac{z}{p^{s}} + ⋯ + \frac{z^{h−1}}{p^{(h−1)s}} ) = ∑_{n∈S_{h}} \frac{z^{Ω(n)}}{n^{s}} .  (5)

We follow the Selberg-Delange method (see, for example, [6, Sect. 7.4] and [1, Sect. 1.4.4]). We will work with the Euler product

F (s, z) : = ∏_{p} (1 − \frac{1}{p^{s}} )^{z} (1 + \frac{z}{p^{s}} + ⋯ + \frac{z^{h−1}}{p^{(h−1)s}} )

= ∏_{p} (1 − \frac{1}{p^{s}} )^{z} (\frac{1 − \frac{z^{h}}{p^{hs}}}{1 − \frac{z}{p^{s}}} ) =: ∏_{p} H(p, s, z).  (6)

We claim that the product defining F (s, z) converges absolutely and uniformly over compact subsets of Re(s) > 1/2, |z| < \sqrt{2}. Indeed, taking the logarithm of a single factor of (6), we have

log H(p, s, z) = − log (1 − \frac{z}{p^{s}} ) + log (1 − \frac{z^{h}}{p^{hs}} ) + z log (1 − \frac{1}{p^{s}} )

= ∑_{k=1}^{∞} \frac{z^{k}}{kp^{sk}} − ∑_{k=1}^{∞} \frac{z^{hk}}{kp^{shk}} − z ∑_{k=1}^{∞} \frac{1}{kp^{sk}}

= ∑_{k=2}^{∞}(z^{k} − z)\frac{1}{kp^{sk}} − ∑_{\begin{matrix}k=1 \\ h|k \end{matrix}}^{∞} \frac{hz^{k}}{kp^{sk}} .  (7)

The sums in (7) converge for Re(s) > 1/2 and |z| < \sqrt{2}. More precisely, we have

log H(p, s, z) ≪_{h} ∑_{k=2}^{∞} \frac{max(1, |z|^{k})}{p^{k Re(s)}} = max (\frac{\frac{1}{p^{2 Re(s)}}}{1 − \frac{1}{p^{Re(s)}}} , \frac{\frac{|z|^{2}}{p^{2 Re(s)}}}{1 − \frac{|z|}{p^{Re(s)}}} ) ≪_{h,z} \frac{1}{p^{2 Re(s)}} ,

where we have used that the geometric series converges because |z|/p^{Re(s)} < 1. Summing over all the primes p, the above is bounded by ζ(2 Re(s)) and is therefore finite for Re(s) > 1/2, |z| < \sqrt{2}. Thus we deduce that log F (s, z) converges absolutely and uniformly over compact subsets of Re(s) > 1/2, |z| < \sqrt{2}, and the same is true for F (s, z).

Back to equation (5), we have

∑_{n∈S_{h}} \frac{z^{Ω(n)}}{n^{s}} = ∏_{p} (1 + \frac{z}{p^{s}} + ⋯ + \frac{z^{h−1}}{p^{(h−1)s}} ) = ζ(s)^{z}F (s, z),

and this converges for Re(s) > 1, |z| < \sqrt{2}.

Now we apply Theorem 7.18 from [6] and obtain that

∑_{\begin{matrix}n≤x \\ n∈S_{h} \end{matrix}} z^{Ω(n)} = \frac{F (1, z)}{Γ(z)}x log^{z−1} x + O (x(log x)^{Re(z)−2}) .  (8)

Differentiating both sides of (8) with respect to z, for z close to 1, we obtain

∑_{\begin{matrix}n≤x \\ n∈S_{h} \end{matrix}} Ω(n)z^{Ω(n)−1} = (\frac{F (1, z)}{Γ(z)})′ x log^{z−1} x + \frac{F (1, z)}{Γ(z)}x(log log x)(log^{z−1} x) + O (x) , (9)

where we have bounded the error term by using the mean value theorem and the fact that (log log x)(log^{Re(z)−2} x) ≪ 1 when z is near 1. Evaluating (9) at z = 1 gives

∑_{\begin{matrix}n≤x \\ n∈S_{h} \end{matrix}} Ω(n) = \frac{F (1, 1)}{Γ(1)}x(log log x) + O (x) ,

where the first term in (9) has been absorbed into the O(x) error upon evaluation. Since

F (1, 1) = ∏_{p} (1 − \frac{1}{p^{h}} ) = \frac{1}{ζ(h)},

and Γ(1) = 1, we obtain the desired result.

3 Proof of Theorem 2

Let n be an h-full number. We proceed to separate the primes in the factorization of n according to the congruence modulo h of their exponent. In other words, we write

n = \underbrace{q_{1,0}^{ht_{1,0}} ⋯ q_{ℓ_{0},0}^{ht_{ℓ_{0},0}}}_{exp≡0( mod h)} \underbrace{q_{1,1}^{ht_{1,1}+1} ⋯ q_{ℓ_{1},1}^{ht_{ℓ_{1},1}+1}}_{exp≡1( mod h)} ⋯ \underbrace{q_{1,h−1}^{ht_{1,h−1}+h−1} ⋯ q_{ℓ_{h−1},h−1}^{ht_{ℓ_{h−1},h−1}+h−1}}_{exp≡h−1( mod h)} .

By setting

m = q_{1,0}^{t_{1,0}} ⋯ q_{ℓ_{0},0}^{t_{ℓ_{0},0}}q_{1,1}^{t_{1,1}−1} ⋯ q_{ℓ_{1},1}^{t_{ℓ_{1},1}−1} ⋯ q_{1,h−1}^{t_{1,h−1}−1} ⋯ q_{ℓ_{h−1},h−1}^{t_{ℓ_{h−1},h−1}−1},
r_{j} = q_{1,j} ⋯ q_{ℓ_{j},j},  (j = 1, …, h − 1),

we can finally write

n = m^{h}r_{1}^{h+1} ⋯ r_{h−1}^{2h−1},

and we remark that the r_{j} are square-free and coprime in pairs.

After the above observation, the sum under consideration becomes

∑_{\begin{matrix}n≤x \\ n∈N_{h} \end{matrix}} Ω(n) = ∑_{m^{h}r_{1}^{h+1}···r_{h−1}^{2h−1}≤x} μ^{2}(r_{1} ⋯ r_{h−1})Ω(m^{h}r_{1}^{h+1} ⋯ r_{h−1}^{2h−1}).  (10)

Since Ω is totally additive and the r_{j} are square-free,

Ω(m^{h}r_{1}^{h+1} ⋯ r_{h−1}^{2h−1}) = hΩ(m) + (h + 1)Ω(r_{1}) + ⋯ + (2h − 1)Ω(r_{h−1})
= hΩ(m) + (h + 1)ω(r_{1}) + ⋯ + (2h − 1)ω(r_{h−1}),

where ω(n) denotes the number of distinct prime factors in the factorization of n (if n = q_{1}^{s_{1}} ⋯ q_{r}^{s_{r}}, then ω(n) = r).

Using this in (10), we obtain

∑_{\begin{matrix}n≤x \\ n∈N_{h} \end{matrix}} Ω(n) = h ∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x} μ^{2}(r_{1} ⋯ r_{h−1}) ∑_{m^{h}≤x/r_{1}^{h+1}···r_{h−1}^{2h−1}} Ω(m)

+ (h + 1) ∑_{r_{1}^{h+1}≤x} ω(r_{1}) ∑_{m^{h}r_{2}^{h+2}···r_{h−1}^{2h−1}≤x/r_{1}^{h+1}} μ^{2}(r_{1} ⋯ r_{h−1})

+ ⋯

+ (2h − 1) ∑_{r_{h−1}^{2h−1}≤x} ω(r_{h−1}) ∑_{m^{h}r_{1}^{h+1}···r_{h−2}^{2h−2}≤x/r_{h−1}^{2h−1}} μ^{2}(r_{1} ⋯ r_{h−1})

= hT_{0}(x) + (h + 1)T_{1}(x) + ⋯ + (2h − 1)T_{h−1}(x).  (11)

Eventually we will see that T_{0}(x) gives the main term, while the T_{j}(x) give secondary terms. In order to continue we need some auxiliary results.

Lemma 5. We have the following estimate

∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x} 1 ≪_{h} x^{1/(h+1)}.  (12)

Proof of Lemma 5. We expand the sum in (12) as a combination of iterated sums, and we evaluate the successive terms starting from the innermost sum.

∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x} 1

= ∑_{r_{1}≤x^{1/(h+1)}} ∑_{r_{2}≤_{\frac{x^{1/(h+2)}}{r{(h+1)/(h+2) \atop 1}}}} ⋯ ∑_{r_{h−2}≤_{\frac{x^{1/(2h−2)}}{r{(h+1)/(2h−2) \atop 1}···r{(2h−3)/(2h−2) \atop h−3}}}} ∑_{r_{h−1}≤_{\frac{x^{1/(2h−1)}}{r{(h+1)/(2h−1) \atop 1}···r{(2h−2)/(2h−1) \atop h−2}}}} 1

≪_{h} ∑_{r_{1}≤x^{1/(h+1)}} ∑_{r_{2}≤_{\frac{x^{1/(h+2)}}{r{(h+1)/(h+2) \atop 1}}}} ⋯ ∑_{r_{h−2}≤_{\frac{x^{1/(2h−2)}}{r{(h+1)/(2h−2) \atop 1}···r{(2h−3)/(2h−2) \atop h−3}}}} \frac{x^{1/(2h−1)}}{r_{1}^{(h+1)/(2h−1)} ⋯ r_{h−2}^{(2h−2)/(2h−1)}} .

In order to evaluate the innermost sum above, we approximate it with an integral

∑_{r_{h−2}≤_{\frac{x^{1/(2h−2)}}{r{(h+1)/(2h−2) \atop 1}···r{(2h−3)/(2h−2) \atop h−3}}}} \frac{x^{1/(2h−1)}}{r_{1}^{(h+1)/(2h−1)} ⋯ r_{h−2}^{(2h−2)/(2h−1)}}

≪_{h} ∫0^{\frac{x^{1/(2h−2)}}{r{(h+1)/(2h−2) \atop 1}···r{(2h−3)/(2h−2) \atop h−3}}} \frac{x^{1/(2h−1)}t^{− \frac{2h−2}{2h−1}}}{r_{1}^{(h+1)/(2h−1)} ⋯ r_{h−3}^{(2h−3)/(2h−1)}} dt

≪_{h} \frac{x^{1/(2h−1)}t^{\frac{1}{2h−1}}}{r_{1}^{(h+1)/(2h−1)} ⋯ r_{h−3}^{(2h−3)/(2h−1)}} |_{t=0}^{t=_{\frac{x^{1/(2h−2)}}{r{(h+1)/(2h−2) \atop 1}···r{(2h−3)/(2h−2) \atop h−3}}}}

≪_{h} \frac{x^{1/(2h−2)}}{r_{1}^{(h+1)/(2h−2)} ⋯ r_{h−3}^{(2h−3)/(2h−2)}} .

Continuing in this way, we obtain

∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x} 1

≪_{h} ∑_{r_{1}≤x^{1/(h+1)}} ∑_{r_{2}≤_{\frac{x^{1/(h+2)}}{r{(h+1)/(h+2) \atop 1}}}} ⋯ ∑_{r_{h−3}≤_{\frac{x^{1/(2h−3)}}{r{(h+1)/(2h−3) \atop 1}···r{(2h−4)/(2h−3) \atop h−4}}}} \frac{x^{1/(2h−2)}}{r_{1}^{(h+1)/(2h−2)} ⋯ r_{h−3}^{(2h−3)/(2h−2)}}

⋮

≪_{h} x^{1/(h+1)}, as desired. Lemma 6. We have the following estimate

∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x} \frac{μ^{2}(r_{1} ⋯ r_{h−1})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} = ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) + O_{h} (x^{−1/(h(h+1))}) .

Proof of Lemma 6. First notice that we can approximate the sum up to x with the full sum as follows

∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x} \frac{μ^{2}(r_{1} ⋯ r_{h−1})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}}

= ∑_{r_{1},...,r_{h−1}} \frac{μ^{2}(r_{1} ⋯ r_{h−1})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}}

+ O ( ∑_{r_{1}≥x^{1/(h+1)}} \frac{1}{r_{1}^{(h+1)/h}} ∑_{r_{2}} \frac{1}{r_{2}^{(h+2)/h}} ⋯ ∑_{r_{h−2}} \frac{1}{r_{h−2}^{(2h−2)/h}} ∑_{r_{h−1}} \frac{1}{r_{h−1}^{(2h−1)/h}}

+ ∑_{r_{1}≤x^{1/(h+1)}} \frac{1}{r_{1}^{(h+1)/h}} ∑_{r_{2}≥_{\frac{x^{1/(h+2)}}{r{(h+1)/(h+2) \atop 1}}}} \frac{1}{r_{2}^{(h+2)/h}} ⋯ ∑_{r_{h−2}} \frac{1}{r_{h−2}^{(2h−2)/h}} ∑_{r_{h−1}} \frac{1}{r_{h−1}^{(2h−1)/h}}

+ ⋯

+ ∑_{r_{1}≤x^{1/(h+1)}} \frac{1}{r_{1}^{(h+1)/h}} ∑_{r_{2}≤_{\frac{x^{1/(h+2)}}{r{(h+1)/(h+2) \atop 1}}}} \frac{1}{r_{2}^{(h+2)/h}} ⋯ ∑_{r_{h−2}≤_{\frac{x^{1/(2h−1)}}{r{(h+1)/(2h−2) \atop 1}···r{(2h−3)/(2h−2) \atop h−3}}}} \frac{1}{r_{h−2}^{(2h−2)/h}}

× ∑_{r_{h−1}≥_{\frac{x^{1/(2h−1)}}{r{(h+1)/(2h−1) \atop 1}···r{(2h−2)/(2h−1) \atop h−2}}}} \frac{1}{r_{h−1}^{(2h−1)/h}} ) .  (13)

We proceed to express the main term in (13) as an Euler product. This gives

∑_{r_{1},...,r_{h−1}} \frac{μ^{2}(r_{1} ⋯ r_{h−1})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} = ∏_{p} (1 + \frac{1}{p^{(h+1)/h}} + \frac{1}{p^{(h+2)/h}} + ⋯ + \frac{1}{p^{(2h−1)/h}} )

= ∏_{p} (1 + \frac{1}{p} \frac{p^{−1} − p^{−1/h}}{p^{−1/h} − 1} )

= ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) .  (14)

Now we treat the error term in (13). The inner sums with no condition over r or with a less-or-equal-than condition can be bounded by a constant, since

∑_{r=1}^{∞} \frac{1}{r^{(h+j)/h}} = ζ (\frac{h + j}{h} ) .  (15)

Therefore, the error term from (13) is

O_{h} (∑_{j=1}^{h−1} ∑_{r≥x^{1/(h+j)}} \frac{1}{r^{(h+j)/h}} ) = O_{h} (∑_{j=1}^{h−1} ∫_{x^{1/(h+j)}}^{∞} \frac{dt}{t^{(h+j)/h}} ) = O_{h} (x^{−1/(h(h+1))}) .

Combining the above with (14) in (13) yields the desired result.

We can now return to the proof of Theorem 2. First we proceed to calculate T_{0}(x) in (11). Using that Ω(m) = Ω(1) = 1 when x/2 < r_{1}^{h+1} ⋯ r_{h−1}^{2h−1} ≤ x and estimate (2) when r_{1}^{h+1} ⋯ r_{h−1}^{2h−1} ≤ x/2, we obtain

T_{0}(x) = ∑_{x/2<r_{1}^{h+1}···r_{h−1}^{2h−1}≤x} μ^{2}(r_{1} ⋯ r_{h−1})

+ ∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x/2} μ^{2}(r_{1} ⋯ r_{h−1}) (\frac{x^{1/h}}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} log log (\frac{x^{1/h}}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} )

+B_{2} \frac{x^{1/h}}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} ) + O_{h} (  ∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x/2} \frac{\frac{x^{1/h}}{r_{1}^{(h+1)/h}···r_{h−1}^{(2h−1)/h}}}{log (\frac{x^{1/h}}{r_{1}^{(h+1)/h}···r_{h−1}^{(2h−1)/h}} )}) .  (16)

Notice that the first term in the sum above is ≪_{h} x^{1/(h+1)} by Lemma 5. To bound the big-O error term, we fix 0 < ε < \frac{1}{h+1}. Since the function \frac{t}{log t} is strictly increasing for t > e we have, for ε > \frac{1}{log(y/r)} and 1 < r,

\frac{\frac{y}{r}}{log (\frac{y}{r})} = \frac{ε\frac{y^{ε}}{r^{ε}}}{log (\frac{y^{ε}}{r^{ε}} )} · \frac{y^{1−ε}}{r^{1−ε}} ≤ ε\frac{y^{ε}}{log (y^{ε})} · \frac{y^{1−ε}}{r^{1−ε}} = \frac{y}{log y} · \frac{1}{r^{1−ε}} .  (17)

On the one hand, we limit the sum so that \frac{1}{log(y/r)} ≤ \frac{1}{h+2} and therefore we can take \frac{1}{h+2} <

ε < \frac{1}{h+1}. Taking y = x^{1/h} and r = r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}, we have

∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x/e^{h(h+2)}} \frac{\frac{x^{1/h}}{r_{1}^{(h+1)/h}···r_{h−1}^{(2h−1)/h}}}{log (\frac{x^{1/h}}{r_{1}^{(h+1)/h}···r_{h−1}^{(2h−1)/h}} )}

≤ \frac{x^{1/h}}{log (x^{1/h})} ∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x/e^{h(h+2)}} \frac{1}{r_{1}^{(1−ε)(h+1)/h} ⋯ r_{h−1}^{(1−ε)(2h−1)/h}}

≪_{h} \frac{x^{1/h}}{log x},  (18)

where the final sum was bounded using a similar idea to the one we used in (15), since \frac{(1−ε)(h+j)}{h} > 1.

On the other hand,

∑_{x/e^{h(h+2)}<r_{1}^{h+1}···r_{h−1}^{2h−1}≤x/2} \frac{\frac{x^{1/h}}{r_{1}^{(h+1)/h}···r_{h−1}^{(2h−1)/h}}}{log (\frac{x^{1/h}}{r_{1}^{(h+1)/h}···r_{h−1}^{(2h−1)/h}} )} ≪_{h} ∑_{x/e^{h(h+2)}<r_{1}^{h+1}···r_{h−1}^{2h−1}≤x/2} \frac{he^{h+2}}{log 2}

≪_{h} x^{1/(h+1)}  (19)

by Lemma 5.

Now we study the main term in (16). By writing

log log (\frac{y}{r} ) = log log y + log (1 − \frac{log r}{log y} ) ,

and applying this to log log (\frac{x^{1/h}}{r_{1}^{(h+1)/h}···r_{h−1}^{(2h−1)/h}} ) we are led to consider

M : = ∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x/2} \frac{x^{1/h}}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} |log (1 − \begin{matrix}log & (r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}) \\ log & (x^{1/h}) \end{matrix})| .

We use that for 0 ≤ t < 1, we have

− log(1 − t) ≤ \frac{t}{\sqrt{1 − t}},

which can be proven by considering the derivative of log(1 − t) + \frac{t}{\sqrt{1−t}}. This gives

M ≪ ∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x/2} \frac{x^{1/h}}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} · log (r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}) ; log (x^{1/h}) \sqrt{1 − ^{\begin{matrix}log & (r_{1}^{(h+1)/h}···r_{h−1}^{(2h−1)/h}) \\ log & (x^{1/h}) \end{matrix}}}

≪\frac{x^{1/h}}{log x} ∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x/2} \frac{log (r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}\sqrt{1 − ^{\begin{matrix}log & ( \frac{x^{1/h}}{2^{1/h}} ) \\ log & (x^{1/h}) \end{matrix}}}} .

Taking advantage of the fact that

\sqrt{1 − \begin{matrix}log & (\frac{x^{1/h}}{2^{1/h}} ) \\ log & (x^{1/h}) \end{matrix}} = \sqrt{\frac{log 2}{log x}},

and applying considerations similar to (15), we can simplify the bound to

M ≪_{h} \frac{x^{1/h}}{\sqrt{log x}}.  (20)

By applying Lemma 6 in (16) and by incorporating the error terms (18) and (20) we obtain

T_{0}(x) = ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) x^{1/h} log log x

+ (B_{2} − log h) ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) x^{1/h} + O_{h} (\frac{x^{1/h}}{\sqrt{log x}}) .  (21)

We proceed similarly to calculate the T_{j}(x) in (11).

T_{j}(x) = ∑_{r_{j}^{h+j}≤x} ω(r_{j}) ∑_{r_{1}^{h+1}···r̂_{j}^{h+j}···r_{h−1}^{2h−1}≤x/r_{j}^{h+j}} μ^{2}(r_{1} ⋯ r_{h−1}) ∑_{m^{h}≤_{\frac{x}{r{h+1 \atop 1}···r{2h−1 \atop h−1}}}} 1

= ∑_{r_{j}^{h+j}≤x} ω(r_{j}) ∑_{r_{1}^{h+1}···r̂_{j}^{h+j}···r_{h−1}^{2h−1}≤x/r_{j}^{h+j}} μ^{2}(r_{1} ⋯ r_{h−1}) ⌊\frac{x^{1/h}}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} ⌋ ,

where the hat symbol indicates that the corresponding factor has been excluded. We write

⌊\frac{x^{1/h}}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} ⌋ = \frac{x^{1/h}}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} + O(1)

and use the trivial bound

ω(n) ≤ \frac{log n}{log 2}  (n ≥ 2)  (22)

to obtain

T_{j}(x) = x^{1/h} ∑_{r_{j}^{h+j}≤x} ω(r_{j}) ∑_{r_{1}^{h+1}···r̂_{j}^{h+j}···r_{h−1}^{2h−1}≤x/r_{j}^{h+j}} \frac{μ^{2}(r_{1} ⋯ r_{h−1})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}}

+ O (log x ∑_{r_{1}^{h+1}r_{2}^{h+2}···r_{h−1}^{2h−1}≤x} 1) .  (23)

For the main term in (23), we complete the sum and proceed similarly as in the proof of Lemma 6. The difference here is that we have to work with sums of the form

∑_{r=1}^{∞} \frac{ω(r)}{r^{(h+j)/h}} and ∑_{r≥x^{1/(h+j)}} \frac{ω(r)}{r^{(h+j)/h}} .

We can bound these sums using the fact that given any arbitrarily small ε > 0, it follows from inequality (22) that ω(r) ≪ r^{ε} provided r is sufficiently large. This results in

∑_{r_{1}^{h+1}···r_{h−1}^{2h−1}≤x} \frac{μ^{2}(r_{1} ⋯ r_{h−1})ω(r_{j})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} = ∑_{r_{1},...,r_{h−1}} \frac{μ^{2}(r_{1} ⋯ r_{h−1})ω(r_{j})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} + O_{h} (x^{−1/(h(h+1))+ε}) .

For the error term in (23) we can use Lemma 5. Combining the above with (12) in (23), we obtain

T_{j}(x) = x^{1/h} ∑_{r_{1},...,r_{h−1}} \frac{μ^{2}(r_{1} ⋯ r_{h−1})ω(r_{j})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}} + O_{h} (x^{1/(h+1)} log x) .  (24)

We aim at simplifying the main term in the above expression as much as possible. In order to do this, consider the following expression for a generating series with its corresponding Euler product:

G_{j}(z) := ∑_{r_{1},...,r_{h−1}} \frac{μ^{2}(r_{1} ⋯ r_{h−1})z^{ω(r_{j})}}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}}

= ∏_{p} (1 + \frac{1}{p^{(h+1)/h}} + ⋯ + \frac{z}{p^{(h+j)/h}} + ⋯ + \frac{1}{p^{(2h−1)/h}} ) .

The function G_{j}(z) is holomorphic, since the product converges independently of z. For us, it suffices that G_{j}(z) be holomorphic in a neighborhood of z = 1. Notice that

G′_{j}(1) = ∑_{r_{1},...,r_{h−1}} \frac{μ^{2}(r_{1} ⋯ r_{h−1})ω(r_{j})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}}.

In order to compute G′_{j}(1) from the Euler product, we consider the logarithmic derivative of G_{j}(z) and obtain

\frac{G′_{j}(1)}{G_{j}(1)} = ∑_{p} \frac{p^{−(h+j)/h}}{1 + \frac{1}{p^{(h+1)/h}} + ⋯ + \frac{1}{p^{(2h−1)/h}}}

= ∑_{p} \frac{p^{1+(1−j)/h} − p^{1−j/h}}{(p − 1) (p^{1+1/h} + p^{1/h} − p)}.

Multiplying by G_{j}(1) and using (14), we obtain

G′_{j}(1) = ∑_{r_{1},...,r_{h−1}} \frac{μ^{2}(r_{1} ⋯ r_{h−1})ω(r_{j})}{r_{1}^{(h+1)/h} ⋯ r_{h−1}^{(2h−1)/h}}

= ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) ∑_{p} \frac{p^{1+(1−j)/h} − p^{1−j/h}}{(p − 1) (p^{1+1/h} + p^{1/h} − p)}.

Replacing the above in the main term of (24), and replacing this result, as well as (21), in (11), we get

∑_{\begin{matrix}n≤x \\ n∈N_{h} \end{matrix}} Ω(n) = h ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) x^{1/h} log log x

+ (h(B_{2} − log h) + ∑_{p} \frac{∑_{j=1}^{h−1}(h + j) (p^{1+(1−j)/h} − p^{1−j/h})}{(p − 1) (p^{1+1/h} + p^{1/h} − p)})

× ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) x^{1/h} + O_{h} (\frac{x^{1/h}}{\sqrt{log x}})

= h ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) x^{1/h} log log x

+ (h(B_{2} − log h) + ∑_{p} \frac{p (p^{1/h} − 1) ∑_{j=1}^{h−1}(h + j)p^{−j/h}}{(p − 1) (p^{1+1/h} + p^{1/h} − p)} )

× ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) x^{1/h} + O_{h} (\frac{x^{1/h}}{\sqrt{log x}}) .

We use that

∑_{j=1}^{h−1}(h + j)x^{j} = \frac{(2h − 1)x^{h+1} − 2hx^{h} − hx^{2} + (h + 1)x}{(x − 1)^{2}}

in order to obtain

∑_{\begin{matrix}n≤x \\ n∈N_{h} \end{matrix}} Ω(n) = h ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) x^{1/h} log log x

+ (h(B_{2} − log h) + ∑_{p} \frac{(h + 1)p^{1+1/h} − hp − 2hp^{2/h} + (2h − 1)p^{1/h}}{(p − 1) (p^{1/h} − 1) (p^{1+1/h} + p^{1/h} − p)} )

× ∏_{p} (1 + \frac{p − p^{1/h}}{p^{2} (p^{1/h} − 1)}) x^{1/h} + O_{h} (\frac{x^{1/h}}{\sqrt{log x}}) .

This concludes the proof of Theorem 2.

4 Proof of Theorem 4

We first split the sum into a term involving squares and another term involving higher powers, as we expect that the squares are the only contributors to the main term.

∑_{\begin{matrix}n≤x \\ n∈P \end{matrix}} Ω(n) = ∑_{n^{2}≤x} Ω(n^{2}) + ∑_{\begin{matrix}n^{k}≤x \\ k≥3 \end{matrix}} Ω(n^{k})

= 2 ∑_{n≤x^{1/2}} Ω(n) + ∑_{\begin{matrix}n^{k}≤x \\ k≥3 \end{matrix}} Ω(n^{k})

= S_{1}(x) + S_{2}(x).  (25)

We can estimate the contribution from the squares by using (2),

S_{1}(x) = 2x^{1/2} log log x^{1/2} + 2B_{2}x^{1/2} + O (\frac{x^{1/2}}{log x})
= 2x^{1/2} log log x + 2(B_{2} − log 2)x^{1/2} + O (\frac{x^{1/2}}{log x}) .  (26)

Using (22), we obtain the following bound for the contribution of the higher powers:

S_{2}(x) = ∑_{\begin{matrix}n^{k}≤x \\ k≥3 \end{matrix}} Ω(n^{k}) ≤ ∑_{\begin{matrix}n^{k}≤x \\ k≥3 \end{matrix}} \frac{log n^{k}}{log 2} ≤ \frac{log x}{log 2}x^{1/3}.

Combining this estimate with equation (26) in (25) gives the desired result.

5 Acknowledgments

The authors would like to thank the anonymous referee who provided extensive arguments and constructive remarks on an earlier version of this article, as well as helpful suggestions for the final version. The first author is grateful to Universidad Nacional de Luján. The second author is partially supported by the Natural Sciences and Engineering Research Council of Canada (Discovery Grant 355412-2013) and the Fonds de recherche du Québec—Nature et technologies (Projets de recherche en équipe 256442 and 300951).

References

[1] S. R. Finch, Mathematical Constants II, Encyclopedia of Mathematics and its Applications, Vol. 169, Cambridge University Press, 2019.

[2] G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, sixth ed., Oxford University Press, 2008.

[3] A. Ivić and P. Shiu, The distribution of powerful integers, Illinois J. Math. 26 (1982), 576–590.

[4] R. Jakimczuk, On the distribution of perfect powers, J. Integer Sequences 14 (2011), Article 11.8.5.

[5] R. Jakimczuk, The kernel of powerful numbers, Int. Math. Forum 12 (2017), 721–730.

[6] H. L. Montgomery and R. C. Vaughan, Multiplicative Number Theory. I. Classical Theory, Cambridge Studies in Advanced Mathematics, Vol. 97, Cambridge University Press, Cambridge, 2007.

2020 Mathematics Subject Classification: Primary 11N37; Secondary 11A25, 11B99. Keywords: Ω(n), integer sequence, h-free number, h-full number, perfect power.

(Concerned with sequences \underline{A005117}, \underline{A001694}, \underline{A004709}, and \underline{A036966}.)

Received October 29 2021; revised versions received November 17 2021; January 17 2022. Published in Journal of Integer Sequences, January 30 2022.

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