On the Truncated Kernel Function

Jean-Marie De Koninck Département de Mathématiques et de Statistique Université Laval Québec G1V 0A6 Canada

                  jmdk@mat.ulaval.ca

Ismaïla Diouf Département de Mathématiques et d’Informatique FST - Université Cheikh Anta DIOP BP 5005, Dakar-Fann Senegal

                 isma.diouf@gmail.com

Nicolas Doyon Département de Mathématiques et de Statistique Université Laval Québec G1V 0A6 Canada

                nicodoyon77@hotmail.com

Abstract

We study properties of the truncated kernel function γ_{2} defined on integers n ≥ 2 by γ_{2}(n) = γ(n)/P (n), where γ(n) = ∏_{p|n} p is the well-known kernel function and P (n) is the largest prime factor of n. In particular, we show that the maximal order of γ_{2}(n) for n ≤ x is (1 + o(1))x/ log x as x → ∞ and that ∑_{n≤x} 1/γ_{2}(n) = (1 + o(1))ηx/ log x, where η = ζ(2)ζ(3)/ζ(6). We further show that, given any positive real number u < 1, lim_{x→∞} \frac{1}{x}#\{n ≤ x : γ_{2}(n) < x^{u}\} = lim_{x→∞} \frac{1}{x}#\{n ≤ x : n/P (n) < x^{u}\} = 1 − ρ(1/(1 − u)), where ρ is the Dickman function. We also show that n/P (n) can very often be much larger than γ_{2}(n), namely by proving that, given any c ∈ [1, ξ), where ξ is the unique solution to ξ log 2 = log(1 + ξ) + ξ log(1 + 1/ξ), then

#{n ≤ x : γ_{2}(n) ≥ n/(c log n)} = o (#{n ≤ x : n/P (n) ≥ n/(c log n)}) (x → ∞).

1 Introduction

Let γ(n) = ∏_{p|n} p for n ≥ 2 and γ(1) = 1 be the traditional kernel function, also at times called the largest squarefree divisor function. This arithmetic function has been extensively studied. For instance, it was shown by Cohen [4] that

∑_{n≤x} γ(n) = c_{0} x^{2} + O(x^{3/2} log x),  (1)

where

c_{0} = \frac{1}{2} ∏_{p} (1 − \frac{1}{p(p + 1)}) ≈ 0.352.  (2)

It was later shown by De Koninck and Sitaramachandrarao [5] that, given any positive integer k,

∑_{n≤x} \frac{1}{n log γ(n)} = log log x + ∑_{m=0}^{k−1} \frac{b_{m}}{log^{m} x} + O (\frac{1}{log^{k} x}) ,

where the b_{m}’s are computable constants.

The kernel function is at the heart of several problems in number theory, one of which is the famous abc−conjecture (stated below). Although simple in appearance, the kernel function remains intriguing and in some aspects very mysterious.

As an example of its intriguing character, let us mention the search for an asymptotic formula for the sum of its reciprocal values up to a given number x, namely the sum

I(x) := ∑_{n⩽x} \frac{1}{γ(n)}.

At first, since γ(n) = n when n is squarefree and since this occurs for approximately 6x/π^{2} of all integers up to x, one might think that I(x) would be near ∑_{n⩽x}1/n ∼ log x. But, in

1962, N. G. de Bruijn [1] proved that, as x → ∞,

log I(x) = (1 + o(1))\sqrt{\frac{8 log x}{loglog x}},

the asymptotic behavior of I(x) itself remaining at that time a total mystery. Three years later, W. Schwarz [11] came up with the unexpected formula

I(x) = (1 + o(1))\frac{1}{2^{1/4}\sqrt{4π}} (\frac{loglog x}{log x})^{1/4} Q(x) (x → ∞),

where

Q(x) = min_{0<σ<∞} x^{σ}κ(σ) with κ(σ) := ∑_{n=1}^{∞} \frac{1}{n^{σ}γ(n)} = ∏_{p} (1 + \frac{1}{p(p^{σ} − 1)}) ,

and was thus able to prove the asymptotic formula

I(x) = (1 + o(1))\frac{1}{2^{1/4}\sqrt{4π}} (\frac{loglog x}{log x})^{1/4} exp\{−R(log x)\} (x → ∞),

where, by setting φ(σ) := log κ(σ) and Φ(σ) := φ′(σ), the function R(u) is defined by the relation

−R(u) = φ(Φ^{−1}(−u)) + uΦ^{−1}(−u).

The study of the kernel function can be seen as a way to investigate the multiplicative structure of integers. Now, the quantity γ(n) is easy to grasp when it is fairly large compared with n, say in the neighborhood of n/k for a fixed k ≥ 1. However, when γ(n) is very small compared with n, say when γ(n) ≤ \sqrt{n} or even worst when γ(n) ≤ log n, then its behavior depends on the small prime factors of n (since P (n) is small) and is therefore very hard to grasp, thus for instance explaining the difficulty in obtaining the above asymptotic value of I(x). This motivates us to study the truncated kernel function γ_{2} defined by γ_{2}(1) = 1 and for each integer n ≥ 2 by γ_{2}(n) = γ(n)/P (n), where P (n) stands for the largest prime factor of n (with P (1) = 1). This new function is very similar to γ(n) in many respects but it also has the advantage of being simpler to investigate, essentially because the distribution of its small values no longer depends on P (n) and is therefore much easier to understand.

We will study the global and local behavior of γ_{2}(n) and also compare it with that of γ(n) and of n/P (n). Although γ_{2}(n) is not a multiplicative function, in general its behavior is more easily understood than that of γ(n), as is the case when comparing the corresponding sums of their reciprocal values. However, as we will see, it turns out to be quite the opposite in many aspects.

2 Main results

Our first two theorems provide information about the global behavior of γ_{2}(n).

Theorem 1. With c_{0} defined as in (2), we have

∑_{n≤x} γ_{2}(n) = (1 + o(1))2c_{0}x^{2}δ(x)  (x → ∞),  (3)

where

δ(x) = ∫_{2}^{x} \frac{1}{t^{2}} ρ (\frac{log x}{log t}) dt  (4)

and ρ is the well-known Dickman function defined as the (unique) continuous solution to the differential-difference equation

uρ′(u) = −ρ(u − 1) (u > 1)

satisfying the initial condition ρ(u) = 1 for 0 ≤ u ≤ 1.

Theorem 2. For all x ≥ 2,

∑_{n≤x} \frac{1}{γ_{2}(n)} = c_{2} \frac{x}{log x} + O (\frac{x log log x}{log^{2} x}) ,  (5)

where c_{2} = \frac{ζ(2)ζ(3)}{ζ(6)} ≈ 1.9436 (here ζ stands for the Riemann Zeta Function).

The next theorem allows us to study the distribution of γ_{2}(n), in particular by showing that it behaves like the function n/P (n) almost everywhere.

Theorem 3. Given any positive real number u < 1,

lim_{x→∞} \frac{1}{x}#\{n ≤ x : γ_{2}(n) < x^{u}\} = lim_{x→∞} \frac{1}{x}#\{n ≤ x : n/P (n) < x^{u}\} = 1 − ρ (\frac{1}{1 − u}) .

Given a real number h ≥ 1, let us set

Ψ_{h}(x, y) = #{n ≤ x : P (n) ≤ y, γ(n) < n/h}.

Observe that γ_{2}(n) can be written as the product of two functions, namely γ_{2}(n) = \frac{γ(n)}{n} · \frac{n}{P (n)}.

The next result shows that γ_{2}(n) is the product of two basic functions which are statistically independent almost everywhere.

Theorem 4. Given real numbers u > 0 and h ≥ 1, then

Ψ_{h}(x, x^{1/u}) = (1 + o(1))x D(h) ρ(u)  (x → ∞),  (6)

where

D(h) := lim_{x→∞} \frac{1}{x}#\{n ≤ x : γ(n) < n/h\},

a number whose existence is established in Lemma 13.

While γ_{2}(n) and n/P (n) are independent almost everywhere, this is not true for extremal values. Indeed, our next theorem shows that the maximal value of γ_{2}(n) is smaller than the maximal value of n/P (n), while Theorem 6 shows that n/P (n) takes on large values more often than γ_{2}(n).

Theorem 5. As x → ∞,

max_{n≤x} γ_{2}(n) = (1 + o(1))\frac{x}{log x}.  (7)

Theorem 6. Let ξ be the unique solution of the equation

ξ log 2 = log(1 + ξ) + ξ log(1 + 1/ξ),

so that ξ ≈ 3.403. Fix any number c ∈ [1, ξ) and set

A(x) := # \{n ≤ x : γ_{2}(n) ≥ \frac{n}{c log n}\} ,
B(x) := # \{n ≤ x : \frac{n}{P (n)} ≥ \frac{n}{c log n}\} .

Then,

A(x) = o(B(x)) (x → ∞).

The following theorem is essentially the counterpart to the last two theorems, that is that there are many more integers n with a small γ_{2}(n) than there are integers n with a small P (n).

Theorem 7. Given a fixed positive integer k,

#\{n ≤ x : n/P (n) ≤ k\} = C(k)\frac{x}{log x} + O (\frac{x}{log^{2} x}) ,

where C(k) := ∑_{j=1}^{k} \frac{1}{j}, and

#\{n ≤ x : γ_{2}(n) ≤ k\} = E(k)\frac{x}{log x} + O (\frac{x}{log^{2} x}) ,

where E(k) := ∑_{{m≥1 \atop γ(m)≤k}} \frac{1}{m} = ∑_{j=1}^{k} \frac{μ^{2}(j)}{j} ∏_{p|j} (1 − \frac{1}{p})^{−1} = ∑_{j=1}^{k} \frac{μ^{2}(j)}{ϕ(j)}.

3 Preliminary results and notations

We say that a function λ : [1, +∞[→ [0, +∞[ is slowly increasing if, for each constant c > 0,

lim_{x→∞} λ(cx)/λ(x) = 1.

Proposition 8. Let f be a non negative multiplicative function for which there exist a positive real number k and a slowly increasing function λ such that, as x → ∞,

∑_{n≤x} f(n) = (1 + o(1)) x (log^{k−1} x) λ(log x),

and such that for all real numbers u > 1, as y → ∞,

∑_{y<p<y^{u}} \frac{f(p)}{p} = (1 + o(1)) k log u.

Then, as x = y^{u} → ∞,

∑_{{n⩽x \atop P (n)⩽y}} f(n) = (1 + o(1)) Γ(k) u^{1−k}ρ_{k}(u) ∑_{n⩽x}f(n),

uniformly for u bounded, where ρ_{k}(u) is the continuous solution to the differential difference equation with delayed argument, defined by

\begin{aligned}ρ_{k}(u) & = & 0 & (u ⩽ 0), \\ ρ_{k}(u) & = & \frac{u^{k−1}}{Γ(k)} & (0 < u ⩽ 1), \\ uρ′_{k}(u) & = & (k − 1)ρ_{k}(u) − kρ_{k}(u − 1) & (u > 1), \end{aligned}

where Γ stands for the Gamma function.

Proof. This result is due to de Bruijn and van Lint [3]. Proposition 9. As x → ∞,

∑_{n⩽x} \frac{1}{P (n)} = (1 + O (\sqrt{\frac{loglog x}{log x}})) xδ(x),  (8)
∑_{n⩽x} \frac{μ^{2}(n)}{P (n)} = (\frac{6}{π^{2}} + O (\sqrt{\frac{loglog x}{log x}})) ∑_{n⩽x} \frac{1}{P (n)},  (9)
∑_{{n⩽x \atop P (n)^{2}|n}} \frac{1}{P (n)} = x exp \{−\sqrt{4 log x loglog x} (1 + O (\frac{logloglog x}{loglog x}))\} = xδ(x)^{\sqrt{2}+o(1)}, (10)

where δ(x) is the function defined in (4). Proof. A proof of (8) was established by Erdős, Ivić, and Pomerance [6], while (9) can be found in Ivić [9], and (10) in Ivić and Pomerance [10]. Remark 10. Using the estimate

ρ(u) = exp{−u(log u + loglog u − 1 + o(1))} (u → ∞)  (11)

(see the book of Tenenbaum [12]), one can show that the function δ(x) defined in (4) is slowly increasing and satisfies

δ(x) = exp \{−(1 + o(1))\sqrt{2 log x loglog x}\} (x → ∞),  (12)

so that

δ(x) = L_{0}(x)^{−1+o(1)} (x → ∞),

where

L_{0}(x) := exp\{\sqrt{2 log x log log x}\},  (13)

Lemma 11. As x → ∞,

∫_{1}^{x} t ρ (\frac{log t}{log y} ) dt = ∫_{y}^{x} t ρ (\frac{log t}{log y} ) dt + O(y) = O(x),

uniformly for 2 ⩽ y ⩽ x. Proof. We first estimate the maximum value of

g(t) := t ρ (\frac{log t}{log y} )  (14)

for y ⩽ t ⩽ x and fixed y ∈ [2, x]. For this, consider h(t) := log g(t) and solve h′(t) = 0 for t.

In view of (11), we have

h′(t) = \frac{d}{dt} (−\frac{log t}{log y} (log log t − log log y + log log log t − log log log y − 1) + log t)

so that

log log t − log log y = log y,

in which case,

t = y^{y}.

Substituting this value of t in (14), we get in view of (11),

g(t) = ρ(y)y^{y} ≪ e^{−y log y}y^{y} = 1,

which completes the proof of the lemma. Lemma 12. Uniformly for x ≥ y ≥ 2,

log Ψ(x, y) = Z \{1 + O (\frac{1}{log y} + \frac{1}{log log x})\} ,

where

Z = \frac{log x}{log y} log (1 + \frac{y}{log x}) + \frac{y}{log y} log (1 + \frac{log x}{y}) .

Proof. This result is due to de Bruijn [2]. Lemma 13. Given any real number h ≥ 1, the limit

D(h) := lim_{x→∞} \frac{1}{x}#\{n ≤ x : γ(n) < n/h\}  (15)

exists and moreover, as h becomes large,

D(h) ≪ \frac{1}{\sqrt{h}}.  (16)

Proof. Writing each positive integer n as n = st, where s is powerful and t squarefree with (s, t) = 1, we have that the condition \frac{γ(n)}{n} < \frac{1}{h} is equivalent to \frac{γ(s)}{s} < \frac{1}{h}. In light of these facts, we have that

∑_{{n≤x \atop \frac{γ(n)}{n} < \frac{1}{h}}} 1 = ∑_{s≤x \frac{γ(s)}{s} < \frac{1}{h} ; s powerful} ∑_{{t≤x/s \atop (t,s)=1}} μ^{2}(t) = (1 + o(1))\frac{6}{π^{2}} x ∑_{s≤x \frac{γ(s)}{s} < \frac{1}{h} ; s powerful} \frac{1}{s ∏_{p|s} (1 + \frac{1}{p})}.  (17)

Since ∑_{s powerful ; \frac{γ(s)}{s} < \frac{1}{h}} \frac{1}{s ∏_{p|s} (1 + \frac{1}{p})} is clearly a convergent sequence, we can define D(h) as

D(h) = ∑_{s powerful ; \frac{γ(s)}{s} < \frac{1}{h}} \frac{1}{s ∏_{p|s} (1 + \frac{1}{p})}.

Now, recall the following result proved by S. Golomb [7] in 1970:

S(y) := #\{n ≤ y : n powerful\} = (c_{3} + o(1))\sqrt{y} (y → ∞),

where c_{3} = ζ(3/2)/ζ(2) ≈ 2.173. Finally, in order to prove (16), we only need to observe that, since the condition \frac{γ(s)}{s} < \frac{1}{h} certainly implies that s > \frac{s}{γ(s)} > h, we have

∑_{s>x s powerful ; \frac{γ(s)}{s} < \frac{1}{h}} \frac{1}{s} < ∑_{s>h ; s powerful} \frac{1}{s} ≪ ∫_{h}^{∞} \frac{1}{t}d S(t) = ∫_{h}^{∞} \frac{S(t)}{t} dt + ∫_{h}^{∞} \frac{S(t)}{t^{2}} dt ≪ \frac{1}{\sqrt{h}}.

4 The proof of the theorems

4.1 The proof of Theorem 1

We first write

S_{2}(x) := ∑_{n≤x} γ_{2}(n) = ∑_{{n≤x \atop P (n)∥n}} \frac{γ(n)}{P (n)} + ∑_{{n≤x \atop P (n)^{2}|n}} \frac{γ(n)}{P (n)} = Σ_{1} + Σ_{2},

say. Since it follows from (10) of Proposition 9 that

Σ_{2} ⩽ ∑_{{n⩽x \atop P (n)^{2}|n}} \frac{n}{P (n)} ⩽ x ∑_{{n⩽x \atop P (n)^{2}|n}} \frac{1}{P (n)} ≪ x^{2} δ(x)^{\sqrt{2}+o(1)} (x → ∞),  (18)

we only need to estimate Σ_{1}.

We first observe that the true order of Σ_{1} is x^{2} δ(x) and in fact that

\frac{Σ_{1}}{(x^{2}δ(x))/2} ∈ [\frac{6}{π^{2}} , 1]

since it is easily shown that, as x → ∞,

(1 + o(1))\frac{6}{π^{2}} \frac{x^{2}}{2} δ(x) ⩽ ∑_{{n⩽x \atop P (n)∥n}} \frac{γ(n)}{P (n)} ⩽ (1 + o(1))\frac{x^{2}}{2} δ(x).  (19)

Indeed, on the one hand,

∑_{{n≤x \atop P (n)∥n}} \frac{γ(n)}{P (n)} ⩽ ∑_{{n⩽x \atop P (n)∥n}} \frac{n}{P (n)} = (1 + o(1))\frac{x^{2}}{2} δ(x) (x → ∞),

by way of (8) and partial summation. On the other hand, using the trivial observation γ(n) ⩾ μ^{2}(n) n valid for all n ⩾ 1, we have

∑_{{n≤x \atop P (n)∥n}} \frac{γ(n)}{P (n)} ⩾ ∑_{{n⩽x \atop P (n)∥n}} \frac{μ^{2}(n) n}{P (n)} = ∑_{n⩽x} \frac{μ^{2}(n)n}{P (n)} = (1 + o(1))\frac{6}{π^{2}} \frac{x^{2}}{2} δ(x) (x → ∞),

where first we used (9) and partial summation and thereafter estimate (8) of Proposition 9. In order to estimate Σ_{1}, we shall first prove that

G(x, y) := ∑_{{n≤x \atop P (n)⩽y}} γ(n) = (1 + o(1)) c_{1}x^{2} ρ(u) (x → ∞),  (20)

where u = \frac{log x}{log y} and ρ is the Dickman function.

Let f(n) := γ(n)/n. First, it is an easy matter to derive from (1) that

∑_{n≤x} \frac{γ(n)}{n} = c_{1}x + O(x^{1/2} log x),  (21)

where c_{1} = 2c_{0}.

On the other hand, using Mertens’ formula, we have that

∑_{y<p<y^{u}} \frac{f(p)}{p} = ∑_{y<p<y^{u}} \frac{1}{p} = log u + O(1) (y → ∞).  (22)

Hence, it follows from (22) and (21) that f satisfies the conditions of Proposition 8 with k = 1, yielding

∑_{{n≤x \atop P (n)≤y}} \frac{γ(n)}{n} = (1 + o(1)) c_{1} ρ(u) x (x → ∞).  (23)

Using partial summation and Lemma 11, we get, as x → ∞,

∑_{{n≤x \atop P (n)≤y}} γ(n) = ∑_{{n⩽x \atop P (n)≤y}} \frac{γ(n)}{n}n

= (1 + o(1)) c_{1}ρ(u)x^{2} − c_{1}(1 + o(1)) ∫_{1}^{x} ρ (\frac{log t}{log y} ) t dt
= (1 + o(1)) c_{1}ρ(u)x^{2} + O(x),

which proves (20).

Getting back to the definition of Σ_{1}, we obtain

Σ_{1} = ∑_{p⩽x} ∑_{{mp⩽x \atop P (m)<p}} \frac{γ(m) p}{p} = ∑_{p⩽x} ∑_{{m⩽x/p \atop P (m)<p}} γ(m) = ∑_{p⩽x} ∑_{{m⩽x/p \atop P (m)⩽p}} γ(m) − ∑_{p⩽x} ∑_{{m⩽x/p \atop P (m)=p}} γ(m)

= ∑_{p⩽x}G (\frac{x}{p}, p) − ∑_{{mp⩽x \atop P (mp)^{2}|mp}} \frac{γ(mp)}{P (mp)}

= T_{1} − T_{2},  (24)

where, by (20),

T_{1} = (1 + o(1))c_{1}x^{2} ∑_{p⩽x} \frac{1}{p^{2}} ρ (\frac{log x}{log p} − 1) (x → ∞)

while, by (18),

T_{2} ≪ x^{2}δ(x)^{\sqrt{2}+o(1)} (x → ∞).

Following an argument used by Ivić and Pomerance [10], we obtain that

T_{1} = c_{1}(1 + o(1)) x^{2} ∫_{2}^{x} \frac{1}{t^{2} log t}ρ (\frac{log x}{log t} − 1) dt = c_{1}(1 + o(1)) x^{2}δ(x) (x → ∞), (25)

while clearly

T_{2} = o (x^{2}δ(x))  (x → ∞).  (26)

Substituting (25) and (26) in (24), (3) follows.

4.2 The proof of Theorem 2

Let K be a fixed large integer. Then,

∑_{n≤x} \frac{1}{γ_{2}(n)} = ∑_{k≤K} \frac{μ^{2}(k)}{k} ∑_{{n≤x \atop γ_{2}(n)=k}} 1 + ∑_{K<k≤x} \frac{μ^{2}(k)}{k} ∑_{{n≤x \atop γ_{2}(n)=k}} 1 = S_{1}(x; K) + S_{2}(x; K),  (27)

say.

Using the fact that, for any squarefree integer k,

∑_{{n≥1 \atop γ_{2}(n)=k}} \frac{1}{n} = \frac{1}{k} ∏_{p|k} (1 + \frac{1}{p} + \frac{1}{p^{2}} + …) = \frac{1}{ϕ(k)},

we easily obtain that

S_{2}(x; K) ≤ ∑_{K<k≤x} \frac{μ^{2}(k)}{k} ∑_{{n≤x \atop γ_{2}(n)=k}} \frac{x}{n} ≤ x ∑_{k>K} \frac{μ^{2}(k)}{kϕ(k)} < x ∑_{k>K} \frac{1}{k^{3/2}} < \frac{x}{K^{1/2}} ,  (28)

where we used the trivial inequality ϕ(k) > k^{1/2} valid for all k ≥ 7. On the other hand, in light of estimate (47), which is proved in section 4.7, we get

S_{1}(x; K) = ∑_{k≤K} \frac{μ^{2}(k)}{k} ( ∑_{{n≤x \atop γ_{2}(n)≤k}} 1 − ∑_{{n≤x \atop γ_{2}(n)≤k−1}} 1)

= ∑_{k≤K} \frac{μ^{2}(k)}{k} ((E(k) − E(k − 1))\frac{x}{log x} + O (\frac{x}{log^{2} x}) + O(π(k)))

= (∑_{k≤K} \frac{μ^{2}(k)}{kϕ(k)}) \frac{x}{log x} + O (\frac{x log K}{log^{2} x} ) + O (\frac{K}{log K} ) .  (29)

Again using the fact that ϕ(k) < k^{1/2} for all k ≥ 7, we have

∑_{k≤K} \frac{μ^{2}(k)}{kϕ(k)} = ∑_{k=1}^{∞} \frac{μ^{2}(k)}{kϕ(k)} − ∑_{k>K} \frac{μ^{2}(k)}{kϕ(k)} = c_{2} + O (\frac{1}{\sqrt{K}} ) .  (30)

Choosing K = log^{4} x and using (28), (29) and (30) in (27), estimate (5) follows, thereby completing the proof of Theorem 2.

4.3 The proof of Theorem 3

Since, for each integer n ≥ 2, we have γ_{2}(n) = \frac{γ(n)}{P (n)} ≤ \frac{n}{P (n)}, it follows that

∑_{{n≤x \atop γ_{2}(n)<x^{u}}} 1 ≥ ∑_{{n≤x \atop \frac{n}{P (n)} <x^{u}}} 1 = ∑_{{n≤x \atop \frac{P (n)}{n} > \frac{1}{x^{u}}}} 1

= [x] − ∑_{{n≤x \atop P (n)≤n/x^{u}}} 1 ≥ [x] − ∑_{{n≤x \atop P (n)≤x^{1−u}}} 1

= x (1 − ρ (\frac{1}{1 − u}) + o(1))  (x → ∞).  (31)

On the other hand, let ε > 0 be an arbitrarily small number and choose k large enough so that, using estimate (16) of Lemma 13, we can claim that

lim_{x→∞} \frac{1}{x}#\{n ≤ x : γ(n)/n < 1/k\} < ε.  (32)

Then, using (32), provided x is large enough, we have

∑_{{n≤x \atop γ_{2}(n)<x^{u}}} 1 = ∑_{n≤x ; {γ_{2}(n)<x^{u} \atop \frac{γ(n)}{n} < \frac{1}{k}}} 1 + ∑_{n≤x ; {γ_{2}(n)<x^{u} \atop \frac{γ(n)}{n} ≥ \frac{1}{k}}} 1 < εx + ∑_{{n≤x \atop \frac{n}{P (n)} <kx^{u}}} 1,  (33)

where the last sum was obtained using the fact that

\frac{n}{kP (n)} ≤ \frac{γ(n)}{P (n)} = γ_{2}(n) < x^{u}.

Hence, from (33), we conclude that

∑_{{n≤x \atop γ_{2}(n)<x^{u}}} 1 < εx + [x] − ∑_{{n≤x \atop P (n)≤x^{1−u}}} 1 = x (1 − ρ (\frac{1}{1 − u}) + o(1))  (x → ∞).  (34)

Combining estimates(31) and (34) completes the proof of Theorem 3.

4.4 The proof of Theorem 4

First, we can assume that P (n)∥n since we know from Proposition 9 that the number of positive integers n ≤ x such that P^{2}(n)|n is no larger than xδ(x)^{\sqrt{2}+1} = o(x).

We will use the same pattern of proof as the one that Granville [8] used to prove that ψ(x, x^{1/u}) = (1 + o(1))xρ(u) as x → ∞.

First, in light of the definition of D(h) given in (15), it is clear that

Ψ_{h}(x, x^{1/u}) = (1 + o(1))xD(h)ρ(u) for all 0 < u ≤ 1 (x → ∞).  (35)

Let us now consider the case where u ∈ [1, 2]. Then, with the assumption P (n)∥n, we have

∑_{n≤x ; {P (n)≤x^{1/u} \atop \frac{γ(n)}{n} < \frac{1}{h}}} 1 = ∑_{{n≤x \atop \frac{γ(n)}{n} < \frac{1}{h}}} 1 − ∑_{n≤x ; {x^{1/u}<P (n)≤x \atop \frac{γ(n)}{n} < \frac{1}{h}}} 1 = Σ_{1} − Σ_{2},  (36)

say.

Again, it follows from the definition of D(h) that

Σ_{1} = (1 + o(1))D(h)x (x → ∞).  (37)

On the other hand, writing n = mp with P (m) < p, then, because u ∈ [1, 2], the condition P (m) < p is automatically satisfied, while the assumption P (n)∥n guarantees that the condition \frac{γ(n)}{n} < \frac{1}{h} is equivalent to the condition \frac{γ(m)}{m} < \frac{1}{h}, implying that

Σ_{2} = ∑_{x^{1/u}<p≤x} ∑_{m≤ \frac{x}{p} ; {P (m)≤p \atop \frac{γ(m)}{m} < \frac{1}{h}}} 1 = ∑_{x^{1/u}<p≤x} ∑_{{m≤ \frac{x}{p} \atop \frac{γ(m)}{m} < \frac{1}{h}}} 1

= ∑_{x^{1/u}<p≤x} \frac{x}{p}D(h) = (1 + o(1))xD(h) log u (x → ∞).  (38)

Gathering (37) and (38) in (36), we obtain that, in the case u ∈ [1, 2],

Ψ_{h}(x, x^{1/u}) = (1 + o(1))x D(h) (1 − log u) (x → ∞).  (39)

Hence, in light of (35) and (39), we have thus proved that

Ψ_{h}(x, x^{1/u}) = (1 + o(1))x D(h) ρ(u) (x → ∞)  (40)

holds for 0 ≤ u ≤ 2 with

ρ(u) = \begin{cases}1, & if 0 < u ≤ 1, \\ 1 − log u, & if 1 ≤ u ≤ 2. \end{cases}

We now use induction. Assuming that (40) holds for all u ∈ (0, N], we will now prove that it must also hold for all u ∈ [N, N + 1]. For this, we will need the Buchstab identity (trivially generalized to Ψ_{h}(x, y)), that is

Ψ_{h}(x, y) = 1 + ∑_{p≤y} ∑_{mp≤x ; {P (m)<p \atop \frac{γ(m)}{m} < \frac{1}{h}}} 1 = 1 + ∑_{p≤y} Ψ_{h} (\frac{x}{p}, p) .  (41)

Using (41), we have

Ψ_{h}(x, x^{1/N}) = 1 + ∑_{p≤x^{1/N}} Ψ (\frac{x}{p}, p) ,

Ψ_{h}(x, x^{1/u}) = 1 + ∑_{p≤x^{1/u}} Ψ (\frac{x}{p}, p) .

Subtracting these two equations, we obtain

Ψ_{h}(x, x^{1/u}) = Ψ_{h}(x, x^{1/N}) − ∑_{x^{1/u}<p≤x^{1/N}} Ψ_{h} (\frac{x}{p}, p) .  (42)

Now, since p > x^{1/u} and u ∈ [N, N + 1], we have

\frac{log(x/p)}{log p} = \frac{log x}{log p} − 1 < \frac{log x}{log(x^{1/u})} − 1 = u − 1 ≤ N,

implying that (40) holds, say with u′ = \frac{log(x/p)}{log p} instead of u, thus allowing us to replace (42) by

Ψ_{h}(x, x^{1/u}) = (1 + o(1))xD(h)ρ(N) − (1 + o(1))D(h) ∑_{x^{1/u}<p≤x^{1/N}} \frac{x}{p} ρ (\frac{log x}{log p} − 1)

= (1 + o(1))xD(h)ρ(N) − (1 + o(1))xD(h) ∫_{x^{1/u}}^{x^{1/N}} ρ (\frac{log x}{log v} − 1) \frac{d θ(v)}{v log v}
= (1 + o(1))xD(h) (ρ(N) − ∫_{N}^{u} \frac{ρ(t − 1)}{t} dt)
= (1 + o(1))xD(h)ρ(u) (x → ∞),

(where we used the prime number theorem in the form θ(v) = ∑_{p≤v} log p = (1 + o(1))v as v → ∞), thus showing that (40) also holds for u ∈ [N, N + 1] and thus completing the induction argument.

4.5 The proof of Theorem 5

We first show that the bound is achieved for n = ∏_{p≤log x} p. Indeed, it follows from the prime number theorem that, as x → ∞,

γ_{2}(n) = \frac{1}{max_{p≤log x} p} × ∏_{p≤log x} p = \frac{1}{max_{p≤log x} p} × e^{(1+o(1)) log x} = (1 + o(1))\frac{x}{log x}.

On the other hand, this last expression is indeed an upper bound for γ_{2}(n). To prove this, first assume that P (n) > log(n/ log n). Then, in this case, as x → ∞,

γ_{2}(n) ≤ \frac{n}{P (n)} ≤ \frac{n}{log n − log log n} = \frac{n}{log n} (1 + O (\frac{log log n}{log n})) ≤ (1 + o(1))\frac{x}{log x}.

If, on the contrary, P (n) ≤ log(n/ log n), we have, by the prime number theorem,

γ_{2}(n) < ∏_{p≤P (n)} p ≤ ∏_{p≤log n−log log n} p = (1 + o(1))\frac{n}{log n} ≤ (1 + o(1))\frac{x}{log x}  (x → ∞).

This completes the proof of Theorem 5.

4.6 The proof of Theorem 6

Let us write each integer n ≥ 2 as n = st, where s is squarefull and t is squarefree with (s, t) = 1. On the one hand,

\frac{γ_{2}(n)}{n} ≥ \frac{1}{c log n} ⟺ \frac{sP (st)}{γ(s)} ≤ c log n.

But this last inequality implies that

P (st) ≤ \frac{sP (st)}{γ(s)} ≤ c log n ≤ c log x.  (43)

Since γ_{2}(n) ≤ γ(n) ≤ n/\sqrt{s}, we get that s ≤ (k log x)^{2}, which combined with (43) implies that P (t) ≤ c log x.

Therefore, as x → ∞,

A(x) ≤ #{t ≤ x : μ^{2}(t) = 1, P (t) ≤ c log x}
≤ 2^{π(c log x)} = exp \{(1 + o(1))c log 2\frac{log x}{log log x}\} ,  (44)

where again we made use of the prime number theorem.

We will now obtain a lower bound for B(x). Given any small δ > 0, we have, using Lemma 12, that as x → ∞,

B(x) = #{n ≤ x : P (n) ≤ c log n}
≥ #{x^{1−δ} < n ≤ x : P (n) ≤ c log n}
≥ #{x^{1−δ} < n ≤ x : P (n) ≤ (1 − δ)c log x}
= Ψ(x, (1 − δ)c log x) − Ψ(x^{1−δ}, (1 − δ)c log x)
= (1 + o(1))Ψ(x, (1 − δ)c log x)  (45)
= (1 + o(1)) exp Z
= exp \{(1 + o(1))\frac{log x}{log log x} (log(1 + (1 − δ)c) + (1 − δ)c log (1 + \frac{1}{c(1 − δ)}))\} .

Since δ can be taken arbitrarily small, it follows from (45) that, as x → ∞,

B(x) ≥ exp \{(1 + o(1))\frac{log x}{log log x} (log(1 + c) + c log (1 + \frac{1}{c}))\} .  (46)

Finally, by comparing (44) with (46) and observing that for c < ξ, we have

c log 2 < log(1 + c) + c log (1 + \frac{1}{c}) ,

the proof of Theorem 6 is complete.

4.7 The proof of Theorem 7

We first evaluate S_{1} = #{n ≤ x : n/P (n) ≤ k}. Writing each positive integer n ≤ x as n = mp with P (m) ≤ p, we have

S_{1} = ∑_{m≤k} ∑_{k<p≤x/m} 1
= ∑_{m≤k} ∑_{p≤x/m} 1 + O(1)
= ∑_{m≤k} π(x/m) + O(1)

= ∑_{m≤k} (\frac{x}{m log(x/m)} + O (\frac{x/m}{log^{2}(x/m)}))

= \frac{x}{log x} ∑_{m=1}^{k} \frac{1}{m} + O (\frac{x}{log^{2} x}) ,

where we used the prime number theorem, thus proving our first assertion.

Now let S_{2} = #{n ≤ x : γ_{2}(n) ≤ k}. Again, writing each positive integer n ≤ x as n = mp with P (m) ≤ p, we have, using Proposition 9,

S_{2} = ∑_{n≤x ; {P (n)^{2}|n \atop γ_{2}(n)≤k}} 1 + ∑_{m≤x ; {p≤x/m \atop P (m)<p, γ(m)≤k}} 1

= O (\frac{x}{exp\{(1 + o(1))2\sqrt{log x log log x}\}}) + ∑_{{m≤x \atop γ(m)≤k}} ∑_{P (m)<p≤x/m} 1

= O (\frac{x}{log^{2} x}) + ∑_{{m≤x \atop γ(m)≤k}} π(x/m) + O (π(k))

= ∑_{{m≤x \atop γ(m)≤k}} \frac{x/m}{log(x/m)} + O (\frac{x}{log^{2} x}) + O(π(k))

= \frac{x}{log x} ∑_{{m≥1 \atop γ(m)≤k}} \frac{1}{m} + O (\frac{x}{log^{2} x}) + O(π(k))

= E(k)\frac{x}{log x} + O (\frac{x}{log^{2} x}) + O(π(k)),  (47)

as we wanted to prove.

5 Final remarks

One could certainly extend the above results on γ_{2}(n) to the more general case γ_{k}(n), with k ≥ 3, defined by γ_{k}(n) = γ_{k−1}(n)/P (γ_{k−1}(n)). For instance, using the above arguments, one can easily show that

∑_{n≤x} \frac{1}{γ_{k}(n)} = (c_{k} + o(1))x\frac{(log log x)^{k−2}}{log x} (x → ∞)

and that

max_{n≤x} γ_{k}(n) = (1 + o(1))\frac{x}{log^{k−1} x} (x → ∞).

On the other hand, an interesting problem would be to investigate what kind of results one could obtain if k is replaced by a function of n, for instance, by choosing k = ⌊\frac{ω(n)}{2}⌋.

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2010 Mathematics Subject Classification: Primary 11N37; Secondary 11A25. Keywords: kernel function, arithmetic function.

Received November 24 2011; revised version received January 28 2012. Published in Journal of Integer Sequences, February 5 2012.

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