On the least common multiple of random q-integers

Abstract

For every positive integer n and for every $\alpha \in [0, 1]$, let ${\mathcal {B}}(n, \alpha )$ denote the probabilistic model in which a random set ${\mathcal {A}} \subseteq \{1, \ldots , n\}$ is constructed by picking independently each element of $\{1, \ldots , n\}$ with probability $\alpha$. Cilleruelo, Rué, Šarka, and Zumalacárregui proved an almost sure asymptotic formula for the logarithm of the least common multiple of the elements of ${\mathcal {A}}$.Let q be an indeterminate and let $[k]_q := 1 + q + q^2 + \cdots + q^{k-1} \in {\mathbb {Z}}[q]$ be the q-analog of the positive integer k. We determine the expected value and the variance of $X := \deg {\text {lcm}}\!\big ([{\mathcal {A}}]_q\big )$, where $[{\mathcal {A}}]_q := \big \{[k]_q : k \in {\mathcal {A}}\big \}$. Then we prove an almost sure asymptotic formula for X, which is a q-analog of the result of Cilleruelo et al.

1 Introduction

A nice consequence of the Prime Number Theorem is the asymptotic formula

$$\begin{aligned} \log {\text {lcm}}(1, 2, \ldots , n) \sim n , \quad \text { as } n \rightarrow +\infty , \end{aligned}$$

(1)

where ${\text {lcm}}$ denotes the least common multiple. Indeed, precise estimates for $\log {\text {lcm}}(1, \ldots , n)$ are equivalent to the Prime Number Theorem with an error term. Thus, a natural generalization is to study estimates for $L_f(n) := \log {\text {lcm}}(f(1), \ldots , f(n))$, where f is a well-behaved function, for instance, a polynomial with integer coefficients. (We ignore terms equal to 0 in the ${\text {lcm}}$ and we set ${\text {lcm}}\varnothing := 1$.) When $f \in {\mathbb {Z}}[x]$ is a linear polynomial, the product of linear polynomials, or an irreducible quadratic polynomial, asymptotic formulas for $L_f(n)$ were proved by Bateman et al. [3], Hong et al. [10], and Cilleruelo [6], respectively. In particular, for $f(x) = x^2 + 1$, Rué et al. [15] determined a precise error term for the asymptotic formula. When f is an irreducible polynomial of degree $d \ge 3$, Cilleruelo [6] conjectured that $L_f(n) \sim (d - 1)\, n \log n$, as $n \rightarrow +\infty$, but this is still an open problem. However, bounds for $L_f(n)$ were proved by Maynard and Rudnick [13], and Sah [16]. Moreover, Rudnick and Zehavi [14] studied the growth of $L_f(n)$ along a shifted family of polynomials.

Another direction of research consists in considering the least common multiple of random sets of positive integers. For every positive integer n and every $\alpha \in [0, 1]$, let ${\mathcal {B}}(n, \alpha )$ denote the probabilistic model in which a random set ${\mathcal {A}} \subseteq \{1, \ldots , n\}$ is constructed by picking independently each element of $\{1, \ldots , n\}$ with probability $\alpha$. Cilleruelo et al. [9] studied the least common multiple of the elements of ${\mathcal {A}}$ and proved the following result (see [1] for a more precise version, and [4, 5, 7, 8, 12, 17,18,19] for other results of a similar flavor).

Theorem 1.1

Let ${\mathcal {A}}$ be a random set in ${\mathcal {B}}(n, \alpha )$. Then, as $\alpha n \rightarrow +\infty$, we have

$$\begin{aligned} \log {\text {lcm}}({\mathcal {A}}) \sim \frac{\alpha \log (1/\alpha )}{1 - \alpha } \cdot n , \end{aligned}$$

with probability $1 - o(1)$, where the factor involving $\alpha$ is meant to be equal to 1 for $\alpha = 1$.

Remark 1.1

In the deterministic case $\alpha = 1$, we have ${\mathcal {A}} = \{1, \ldots , n\}$ (surely) and Theorem 1.1 corresponds to (1).

Let q be an indeterminate. The q-analog of a positive integer k is defined by

$$\begin{aligned}{}[k]_q := 1 + q + q^2 + \cdots + q^{k - 1} \in {\mathbb {Z}}[q] . \end{aligned}$$

The q-analogs of many other mathematical objects (factorial, binomial coefficients, hypergeometric series, derivative, integral...) have been extensively studied, especially in Analysis and Combinatorics [2, 11]. For every set ${\mathcal {S}}$ of positive integers, let $[{\mathcal {S}}]_q := \big \{[k]_q : k \in {\mathcal {S}}\big \}$.

The aim of this paper is to study the least common multiple of the elements of $[{\mathcal {A}}]_q$ for a random set ${\mathcal {A}}$ in ${\mathcal {B}}(n, \alpha )$. Our main results are the following:

Theorem 1.2

Let ${\mathcal {A}}$ be a random set in ${\mathcal {B}}(n, \alpha )$ and put $X := \deg {\text {lcm}}\!\big ([{\mathcal {A}}]_q\big )$. Then, for every integer $n \ge 2$ and every $\alpha \in [0,1]$, we have

$$\begin{aligned} {\mathbb {E}}[X] = \frac{3}{\pi ^2} \cdot \frac{\alpha {\text {Li}}_2(1 - \alpha )}{1 - \alpha } \cdot n^2 + O\!\left( \alpha n (\log n)^2 \right) , \end{aligned}$$

(2)

where ${\text {Li}}_2(z) := \sum _{k=1}^\infty z^k / k^2$ is the dilogarithm and the factor involving $\alpha$ is meant to be equal to 1 when $\alpha = 1$. In particular,

$$\begin{aligned} {\mathbb {E}}[X] \sim \frac{3}{\pi ^2} \cdot \frac{\alpha {\text {Li}}_2(1 - \alpha )}{1 - \alpha } \cdot n^2 , \end{aligned}$$

as $n \rightarrow +\infty$, uniformly for $\alpha \in [0,1]$.

Theorem 1.3

Let ${\mathcal {A}}$ be a random set in ${\mathcal {B}}(n, \alpha )$ and put $X := \deg {\text {lcm}}\!\big ([{\mathcal {A}}]_q\big )$. Then there exists a function $\mathrm {v} : {(0,1)} \rightarrow {\mathbb {R}}^+$ such that, as $\alpha n / \big ((\log n)^3 (\log \log n)^2\big ) \rightarrow +\infty$, we have

$$\begin{aligned} {\mathbb {V}}[X] = (\mathrm {v}(\alpha ) + o(1)) \, n^3 . \end{aligned}$$

(3)

Moreover, the upper bound

$$\begin{aligned} {\mathbb {V}}[X] \ll \alpha n^3 , \end{aligned}$$

(4)

holds for every positive integer n and every $\alpha \in [0, 1]$.

As a consequence of Theorems 1.2 and 1.3, we obtain the following q-analog of Theorem 1.1.

Theorem 1.4

Let ${\mathcal {A}}$ be a random set in ${\mathcal {B}}(n, \alpha )$. Then, as $\alpha n \rightarrow +\infty$, we have

$$\begin{aligned} \deg {\text {lcm}}\!\big ([{\mathcal {A}}]_q\big ) \sim \frac{3}{\pi ^2} \cdot \frac{\alpha {\text {Li}}_2(1 - \alpha )}{1 - \alpha } \cdot n^2 , \end{aligned}$$

with probability $1 - o(1)$, where the factor involving $\alpha$ is meant to be equal to 1 for $\alpha = 1$.

Remark 1.2

In the deterministic case $\alpha = 1$, we have (see Lemma 4.1 below)

$$\begin{aligned} \deg {\text {lcm}}\!\big [\{1, 2, \ldots , n\}\big ]_q = \sum _{1 \,<\, d \,\le \, n} \varphi (d) , \end{aligned}$$

and Theorem 1.4 corresponds to the well-known asymptotic formula $\sum _{d \le n} \varphi (d) \sim \tfrac{3}{\pi ^2} n^2$ (Lemma 3.3 below) for the sum of the first values of the Euler function $\varphi$.

Remark 1.3

In Theorem 1.4 the condition $\alpha n \rightarrow +\infty$ is necessary. Indeed, if $\alpha n \le C$, for some constant $C > 0$, then

$$\begin{aligned} {\mathbb {P}}[{\mathcal {A}} = \varnothing ] = (1 - \alpha )^n \ge \left( 1 - \frac{C}{n}\right) ^n \rightarrow \mathrm {e}^C \end{aligned}$$

as $n \rightarrow +\infty$, and so no (nontrivial) asymptotic formula for $\deg {\text {lcm}}\!\big ([{\mathcal {A}}]_q\big )$ can hold with probability $1 - o(1)$.

We conclude this section with some possible questions for further research on this topic. Alsmeyer, Kabluchko, and Marynych [1, Corollary 1.5] proved that, for fixed $\alpha \in [0, 1]$ and for a random set ${\mathcal {A}}$ in ${\mathcal {B}}(n, \alpha )$, an appropriate normalization of the random variable $\log {\text {lcm}}({\mathcal {A}})$ converges in distribution to a standard normal random variable, as $n \rightarrow +\infty$. In light of Theorems 1.2 and 1.3, it is then natural to ask whether the random variable

$$\begin{aligned} \frac{\deg {\text {lcm}}\!\big ([{\mathcal {A}}]_q\big ) - \frac{3}{\pi ^2}\cdot \frac{\alpha {\text {Li}}_2(1 - \alpha )}{1 - \alpha }\cdot n^2 }{\sqrt{\mathrm {v}(\alpha )n^3}} \end{aligned}$$

converges in distribution to a normal random variable, or to some other random variable.

Another problem could be considering polynomial values, similarly to the results done in the context of integers, and studying ${\text {lcm}}\!\big ([f(1)]_q, \cdots , [f(n)]_q\big )$ for $f \in {\mathbb {Z}}[x]$ or, more generally, ${\text {lcm}}\!\big ([f(k)]_q : k \in {\mathcal {A}}\big )$ with ${\mathcal {A}}$ a random set in ${\mathcal {B}}(n, \alpha )$.

2 Notation

We employ the Landau–Bachmann “Big Oh” and “little oh” notations O and o, as well as the associated Vinogradov symbol $\ll$, with their usual meanings. Any dependence of the implied constants is explicitly stated or indicated with subscripts. For real random variables X and Y, depending on some parameters, we say that “$X \sim Y$ with probability $1 - o(1)$”, as the parameters tend to some limit, if for every $\varepsilon > 0$ we have ${\mathbb {P}}\big [\,|X - Y| > \varepsilon |Y|\,\big ] = o_\varepsilon (1)$, as the parameters tend to the limit. We let (a, b) and [a, b] denote the greatest common divisor and the least common multiple, respectively, of two integers a and b. As usual, we write $\varphi (n)$, $\mu (n)$, $\tau (n)$, and $\sigma (n)$, for the Euler totient function, the Möbius function, the number of divisors, and the sum of divisors, of a positive integer n, respectively.

3 Preliminaries

In this section we collect some preliminary results needed in later arguments.

Lemma 3.1

We have

$$\begin{aligned} \sum _{m \,\le \, x} \tau (m) \ll x \log x , \end{aligned}$$

for every $x \ge 2$.

Proof

See, e.g., [20, Part I, Theorem 3.2]. $\square$

Lemma 3.2

We have

$$\begin{aligned} \sum _{[e_1\!,\, e_2] \,>\, x} \frac{1}{e_1 e_2 [e_1, e_2]} \ll \frac{\log x}{x} , \end{aligned}$$

for every $x \ge 2$.

Proof

From Lemma 3.1 and partial summation, it follows that

$$\begin{aligned} \sum _{m \,>\, x} \frac{\tau (m)}{m^2}&= \left[ \frac{\sum _{m \le t} \tau (m)}{t^2}\right] _{t\,=\, x}^{+\infty } + 2\int _x^{+\infty } \frac{\sum _{m \le t} \tau (m)}{t^3}\,\mathrm {d} t \\&\ll \int _x^{+\infty } \frac{\log t}{t^2}\,\mathrm {d} t = \left[ -\frac{\log t + 1}{t}\right] _{t \,=\, x}^{+\infty } \ll \frac{\log x}{x} . \end{aligned}$$

Let $e := (e_1, e_2)$ and $e_i^\prime := e_i / e$ for $i=1,2$. Then we have

$$\begin{aligned} \sum _{[e_1\!,\, e_2] \,>\, x} \frac{1}{e_1 e_2 [e_1, e_2]}&\le \sum _{e \,\ge \, 1} \frac{1}{e^3} \sum _{e_1^\prime e_2^\prime \,>\, x / e} \frac{1}{(e_1^\prime e_2^\prime )^2} = \sum _{e \,\ge \, 1} \frac{1}{e^3} \sum _{m \,>\, x / e} \frac{\tau (m)}{m^2} \\&\ll \sum _{e \,\le \, x / 2} \frac{1}{e^3} \frac{\log (x/e)}{x/e} + \sum _{e \,>\, x / 2} \frac{1}{e^3} \ll \frac{\log x}{x} + \frac{1}{x^2} \ll \frac{\log x}{x} , \end{aligned}$$

as desired. $\square$

Let us define

$$\begin{aligned} \Phi (x) := \sum _{n \,\le \, x} \varphi (n) \quad \text { and }\quad \Phi (a_1, a_2; x) := \sum _{n \,\le \, x} \varphi (a_1 n)\, \varphi (a_2 n) , \end{aligned}$$

for every $x \ge 1$ and for all positive integers $a_1, a_2$.

Lemma 3.3

We have

$$\begin{aligned} \Phi (x) = \frac{3}{\pi ^2} \, x^2 + O(x \log x) , \end{aligned}$$

for every $x \ge 2$.

Proof

See, e.g., [20, Part I, Theorem 3.4]. $\square$

Lemma 3.4

We have

$$\begin{aligned} \Phi (a_1, a_2; x) = C_1(a_1, a_2) \, x^3 + O\big (\sigma (a_1)\,\sigma (a_2) \,x^2 (\log x)^2\big ) , \end{aligned}$$

(5)

for every $x \ge 2$, where

$$\begin{aligned} C_1(a_1, a_2) := \frac{a_1 a_2}{3}\sum _{d_1\!,\, d_2 \,\ge \, 1} \frac{\mu (d_1)\mu (d_2)}{d_1 d_2 \big [d_1 / (a_1, d_1), d_2 / (a_2, d_2)\big ]} \end{aligned}$$

(6)

and the series is absolutely convergent.

Proof

From the identity $\varphi (n) / n = \sum _{d \,\mid \;\;\!\!\! n} \mu (d) / d$, it follows that

$$\begin{aligned} \sum _{n \,\le \, x} \frac{\varphi (a_1 n)}{a_1 n} \,\frac{\varphi (a_2 n)}{a_2 n}&= \sum _{n \,\le \, x} \left( \sum _{d_1 \,\mid \, a_1 n} \frac{\mu (d_1)}{d_1} \sum _{d_2 \,\mid \, a_2 n} \frac{\mu (d_2)}{d_2} \right) \\&= \sum _{\begin{array}{c} d_1 \,\le \, a_1 x \\ d_2 \,\le \, a_2 x \end{array}} \frac{\mu (d_1)}{d_1} \, \frac{\mu (d_2)}{d_2}\, \#\big \{n \le x : d_1 \mid a_1 n \text { and } d_2 \mid a_2 n \big \} \\&= \sum _{\left[ \frac{d_1}{(a_1\!,\, d_1)},\, \frac{d_2}{(a_2,\, d_2)}\right] \,\le \, x} \frac{\mu (d_1)}{d_1} \, \frac{\mu (d_2)}{d_2} \left( \frac{x}{\big [d_1 / (a_1, d_1), d_2 / (a_2, d_2)\big ]} + O(1)\right) . \end{aligned}$$

Let $c_i := (a_i, d_i)$ and $e_i := d_i / c_i$, for $i=1,2$. On the one hand, we have

$$\begin{aligned} E_1 := \sum _{\left[ \frac{d_1}{(a_1\!,\, d_1)},\, \frac{d_2}{(a_2,\, d_2)}\right] \,\le \, x} \frac{1}{d_1 d_2} \le \sum _{c_1 \,\mid \, a_1} \frac{1}{c_1} \sum _{c_2 \,\mid \, a_2} \frac{1}{c_2} \sum _{e_1 \,\le \, x} \frac{1}{e_1} \sum _{e_2 \,\le \, x} \frac{1}{e_2}\ll \frac{\sigma (a_1)\,\sigma (a_2)}{a_1 a_2} \, (\log x)^2 . \end{aligned}$$

On the other hand, thanks to Lemma 3.2, we have

$$\begin{aligned} E_2&:= \sum _{\left[ \frac{d_1}{(a_1\!,\, d_1)},\, \frac{d_2}{(a_2,\, d_2)}\right] \,>\, x} \frac{1}{d_1 d_2 \big [d_1 / (a_1, d_1), d_2 / (a_2, d_2)\big ]} \\&\le \sum _{c_1 \,\mid \, a_1} \frac{1}{c_1} \sum _{c_2 \,\mid \, a_2} \frac{1}{c_2} \sum _{[e_1\!,\, e_2] \,>\, x} \frac{1}{e_1 e_2 [e_1, e_2]} \ll \frac{\sigma (a_1)\,\sigma (a_2)}{a_1 a_2} \,\frac{\log x}{x} , \end{aligned}$$

which, in particular, implies that the series

$$\begin{aligned} C_0(a_1, a_2) := \sum _{d_1\!,\, d_2 \,\ge \, 1} \frac{\mu (d_1)\mu (d_2)}{d_1 d_2 [d_1 / (a_1, d_1), d_2 / (a_2, d_2)]} \end{aligned}$$

is absolutely convergent. Therefore, we obtain

$$\begin{aligned} \sum _{n \,\le \, x} \frac{\varphi (a_1 n)}{a_1 n} \,\frac{\varphi (a_2 n)}{a_2 n}&= \big (C_0(a_1, a_2) + O(E_2)\big )\, x + O(E_1) \nonumber \\&= C_0(a_1, a_2) \, x + O\!\left( \frac{\sigma (a_1)\,\sigma (a_2)}{a_1 a_2} \,(\log x)^2\right) . \end{aligned}$$

(7)

Now (5) follows from (7) by partial summation and since $C_1(a_1, a_2) = \dfrac{a_1 a_2}{3}\,C_0(a_1, a_2)$. $\square$

Remark 3.1

The obvious bound $\varphi (m) \le m$ yields $C_1(a_1, a_2) \le \dfrac{a_1 a_2}{3}$ (which is not so obvious from (6)).

We end this section with an easy observation that will be useful later.

Remark 3.2

It holds $1 - (1 - x)^k \le k x$, for all $x \in [0, 1]$ and for all integers $k \ge 0$.

4 Proofs

Henceforth, let ${\mathcal {A}}$ be a random set in ${\mathcal {B}}(n, \alpha )$, let $[{\mathcal {A}}]_q$ be its q-analog, and put $L := {\text {lcm}}\!\big ([{\mathcal {A}}]_q\big )$ and $X := \deg L$. For every positive integer d, let us define

$$\begin{aligned} I_{{\mathcal {A}}}(d) := {\left\{ \begin{array}{ll} 1 &{} \text { if } d \mid k \text { for some } k \in {\mathcal {A}}; \\ 0 &{} \text { otherwise.} \end{array}\right. } \end{aligned}$$

The following lemma gives a formula for X in terms of $I_{{\mathcal {A}}}$ and the Euler function.

Lemma 4.1

We have

$$\begin{aligned} X = \sum _{1 \,<\, d \,\le \, n} \varphi (d)\, I_{{\mathcal {A}}}(d) . \end{aligned}$$

(8)

Proof

For every positive integer k, it holds

$$\begin{aligned}{}[k]_q = \frac{q^k - 1}{q - 1} = \prod _{\begin{array}{c} d \,\mid \!\; k \\ d \,>\, 1 \end{array}} \Phi _d(q) , \end{aligned}$$

where $\Phi _d(q)$ is the dth cyclotomic polynomials. Since, as it is well known, every cyclotomic polynomial is irreducible over ${\mathbb {Q}}$, it follows that L is the product of the polynomials $\Phi _d(q)$ such that $d > 1$ and $d \mid k$ for some $k \in {\mathcal {A}}$. Finally, the equality $\deg \!\big (\Phi _d(q)\big ) = \varphi (d)$ and the definition of $I_{{\mathcal {A}}}$ yield (8). $\square$

Let $\beta := 1 - \alpha$. The next lemma provides two expected values involving $I_{{\mathcal {A}}}$.

Lemma 4.2

For all positive integers $d, d_1, d_2$, we have

$$\begin{aligned} {\mathbb {E}}\big [I_{\mathcal {A}}(d)\big ] = 1 - \beta ^{\lfloor n / d\rfloor } \end{aligned}$$

(9)

and

$$\begin{aligned} {\mathbb {E}}\big [I_{\mathcal {A}}(d_1)I_{\mathcal {A}}(d_2)\big ] = 1 - \beta ^{\lfloor n / d_1 \rfloor } - \beta ^{\lfloor n / d_2 \rfloor } + \beta ^{\lfloor n / d_1 \rfloor + \lfloor n / d_2 \rfloor - \lfloor n / [d_1\!,\, d_2] \rfloor } . \end{aligned}$$

Proof

On the one hand, by the definition of $I_{{\mathcal {A}}}$, we have

$$\begin{aligned} {\mathbb {E}}\big [I_{{\mathcal {A}}}(d)\big ] = {\mathbb {P}}\big [\exists k \in {\mathcal {A}} : d \mid k\big ] = 1 - {\mathbb {P}}\left[ \bigwedge _{m \,\le \, \lfloor n / d\rfloor } (dm \notin {\mathcal {A}})\right] = 1 - \beta ^{\lfloor n / d \rfloor } , \end{aligned}$$

which is (9). On the other hand, by linearity of the expectation and by (9), we have

$$\begin{aligned} {\mathbb {E}}\big [I_{\mathcal {A}}(d_1)I_{\mathcal {A}}(d_2)\big ]&= {\mathbb {E}}\big [I_{\mathcal {A}}(d_1) + I_{\mathcal {A}}(d_2) - 1 + \big (1 - I_{\mathcal {A}}(d_1)\big )\big (1 - I_{\mathcal {A}}(d_2)\big )\big ] \\&= {\mathbb {E}}\big [I_{\mathcal {A}}(d_1)\big ] + {\mathbb {E}}\big [I_{\mathcal {A}}(d_2)\big ] - 1 + {\mathbb {E}}\big [\big (1 - I_{\mathcal {A}}(d_1)\big )\big (1 - I_{\mathcal {A}}(d_2)\big )\big ] \\&= 1 - \beta ^{\lfloor n / d_1 \rfloor } - \beta ^{\lfloor n / d_2 \rfloor } + {\mathbb {E}}\big [\big (1 - I_{\mathcal {A}}(d_1)\big )\big (1 - I_{\mathcal {A}}(d_2)\big )\big ] , \end{aligned}$$

where the last expected value can be computed as

$$\begin{aligned} {\mathbb {E}}\big [\big (1 - I_{\mathcal {A}}(d_1)\big ) \big (1 - I_{\mathcal {A}}(d_2)\big )\big ]&= {\mathbb {P}}\big [\forall k \in {\mathcal {A}} : d_1 \not \mid k \text { and } d_2 \not \mid k\big ] \\&= {\mathbb {P}}\left[ \bigwedge _{\begin{array}{c} k \,\le \, n \\ d_1 \,\mid \, k \text { or } d_2 \,\mid \, k \end{array}}(k \notin {\mathcal {A}})\right] = \beta ^{\lfloor n / d_1 \rfloor + \lfloor n / d_2 \rfloor - \lfloor n / [d_1\!,\, d_2] \rfloor } , \end{aligned}$$

and second claim follows. $\square$

We are ready to compute the expected value of X.

Proof of Theorem 1.2

From Lemmas 4.1 and 4.2, it follows that

$$\begin{aligned} {\mathbb {E}}[X] = \sum _{1 \,<\, d \,\le \, n} \varphi (d)\, {\mathbb {E}}\big [I_{{\mathcal {A}}}(d)\big ] = \sum _{1 \,<\, d \,\le \, n} \varphi (d) \big (1 - \beta ^{\lfloor n / d \rfloor }\big ) . \end{aligned}$$

(10)

Moreover, since $\lfloor n / d \rfloor = j$ if and only if $n / (j + 1) < d \le n / j$, we get that

$$\begin{aligned} \sum _{d \,\le \, n} \varphi (d) \big (1 - \beta ^{\lfloor n / d \rfloor }\big )&= \sum _{j \,\le \, n} (1 - \beta ^j) \sum _{n / (j + 1) \,<\, d \,\le \, n / j} \varphi (d) \nonumber \\&= \sum _{j \,\le \, n} (1 - \beta ^j) \!\left( \Phi \!\left( \frac{n}{j}\right) - \Phi \!\left( \frac{n}{j + 1}\right) \right) \nonumber \\&= \alpha \sum _{j \,\le \, n} \beta ^{j - 1} \Phi \!\left( \frac{n}{j}\right) \nonumber \\&= \frac{3}{\pi ^2} \cdot \alpha \sum _{j \,\le \, n} \frac{\beta ^{j-1}}{j^2} \cdot n^2 + O\!\left( \alpha \sum _{j \,\le \, n} \frac{n}{j}\log \!\left( \frac{n}{j}\right) \right) \nonumber \\&= \frac{3}{\pi ^2} \cdot \frac{\alpha {\text {Li}}_2(1 - \alpha )}{1 - \alpha } \cdot n^2 + O \big (\alpha n (\log n)^2\big ) , \end{aligned}$$

(11)

where we used Lemma 3.3. Putting together (10) and (11), and noting that, by Remark 3.2, the addend of (11) corresponding to $d = 1$ is $1 - \beta ^n = O(\alpha n)$, we get (2). The proof is complete. $\square$

Now we consider the variance of X.

Proof of Theorem 1.3

From Lemmas 4.1 and 4.2, it follows that

$$\begin{aligned} {\mathbb {V}}[X]&= {\mathbb {E}}\big [X^2\big ] - {\mathbb {E}}[X]^2 \nonumber \\&= \sum _{1 \,<\, d_1\!,\, d_2 \,\le \, n} \varphi (d_1)\,\varphi (d_2) \Big ({\mathbb {E}}\big [I_{{\mathcal {A}}}(d_1)\, I_{{\mathcal {A}}}(d_2)\big ] - {\mathbb {E}}\big [I_{{\mathcal {A}}}(d_1)\big ]\,{\mathbb {E}}\big [I_{{\mathcal {A}}}(d_2)\big ]\Big ) \nonumber \\&= \sum _{1 \,<\, d_1\!,\, d_2 \,\le \, n} \varphi (d_1)\,\varphi (d_2) \, \beta ^{\lfloor n / d_1 \rfloor + \lfloor n / d_2 \rfloor - \lfloor n / [d_1, d_2] \rfloor } \big (1 - \beta ^{\lfloor n / [d_1, d_2] \rfloor } \big ) . \end{aligned}$$

(12)

Let us define

$$\begin{aligned} V_n(\alpha ) := \frac{1}{n^3}\sum _{d_1\!,\, d_2 \,\le \, n} \varphi (d_1)\,\varphi (d_2) \, \beta ^{\lfloor n / d_1 \rfloor + \lfloor n / d_2 \rfloor - \lfloor n / [d_1, d_2] \rfloor } \big (1 - \beta ^{\lfloor n / [d_1, d_2] \rfloor } \big ) . \end{aligned}$$

Clearly, we have

$$\begin{aligned} V_n(\alpha ) - \frac{{\mathbb {V}}[X]}{n^3} \ll \frac{1}{n^3}\sum _{d \,\le \, n} \varphi (d) \, \beta ^{n} \big (1 - \beta ^{\lfloor n / d \rfloor } \big ) \le \frac{1}{n^3}\sum _{d \,\le \, n} d \ll \frac{1}{n}. \end{aligned}$$

Hence, in order to prove (3), it suffices to show that $V_n(\alpha ) = \mathrm {v}(\alpha ) + o(1)$.

For all vectors $\varvec{a} := (a_1, a_2)$ and $\varvec{j} := (j_1, j_2, j_3)$ with components in the set of positive integers, define the quantities

$$\begin{aligned} \rho _1(\varvec{a}, \varvec{j}) := \max \!\left( \frac{1}{a_1(j_1 + 1)}, \frac{1}{a_2(j_2 + 1)}, \frac{1}{a_1 a_2 (j_3 + 1)}\right) \end{aligned}$$

and

$$\begin{aligned} \rho _2(\varvec{a}, \varvec{j}) := \min \!\left( \frac{1}{a_1 j_1}, \frac{1}{a_2 j_2}, \frac{1}{a_1 a_2 j_3}\right) . \end{aligned}$$

Let $d := (d_1, d_2)$ and $a_i := d_i / d$ for $i=1,2$. Then the equalities

$$\begin{aligned} j_1 = \left\lfloor \frac{n}{d_1}\right\rfloor , \quad j_2 = \left\lfloor \frac{n}{d_2}\right\rfloor , \quad j_3 = \left\lfloor \frac{n}{[d_1, d_2]}\right\rfloor , \end{aligned}$$

are equivalent to

$$\begin{aligned} j_1 \le \frac{n}{a_1 d}< j_1 + 1 , \quad j_2 \le \frac{n}{a_2 d}< j_2 + 1 , \quad j_3 \le \frac{n}{a_1 a_2 d} < j_3 + 1 , \end{aligned}$$

which in turn are equivalent to

$$\begin{aligned} \frac{n}{a_1 (j_1 + 1)}< d \le \frac{n}{a_1 j_1} , \quad \frac{n}{a_2 (j_2 + 1)}< d \le \frac{n}{a_2 j_2} , \quad \frac{n}{a_1 a_2 (j_3 + 1)} < d \le \frac{n}{a_1 a_2 j_3} , \end{aligned}$$

that is,

$$\begin{aligned} \rho _1(\varvec{a}, \varvec{j})\, n < d \le \rho _2(\varvec{a}, \varvec{j})\, n . \end{aligned}$$

Therefore, letting

$$\begin{aligned} {\mathcal {S}}_n := \big \{(\varvec{a}, \varvec{j}) \in {\mathbb {N}}^5 : (a_1, a_2) = 1,\; \exists d \in {\mathbb {N}} \text { s.t. }\! \rho _1(\varvec{a}, \varvec{j})\, n < d \le \rho _2(\varvec{a}, \varvec{j})\, n \big \} \end{aligned}$$

and

$$\begin{aligned} S(\varvec{a}, \varvec{j}; n) := \frac{1}{n^3} \sum _{\rho _1(\varvec{a},\, \varvec{j})\, n \,<\, d \,\le \, \rho _2(\varvec{a},\, \varvec{j})\, n} \varphi (a_1 d) \, \varphi (a_2 d) , \end{aligned}$$

we have

$$\begin{aligned} V_n(\alpha ) = \sum _{(\varvec{a},\, \varvec{j}) \,\in \, {\mathcal {S}}_n} \beta ^{j_1 + j_2 - j_3} (1 - \beta ^{j_3}) \,S(\varvec{a}, \varvec{j}; n) . \end{aligned}$$

Now let us define

$$\begin{aligned} \mathrm {v}(\alpha ) := \sum _{(\varvec{a},\, \varvec{j}) \,\in \, {\mathcal {S}}_\infty } \beta ^{j_1 + j_2 - j_3} (1 - \beta ^{j_3}) \, D(\varvec{a}, \varvec{j}) , \end{aligned}$$

(13)

where

$$\begin{aligned} {\mathcal {S}}_\infty := \bigcup _{m \,\ge \, 1} {\mathcal {S}}_m = \big \{(\varvec{a}, \varvec{j}) \in {\mathbb {N}}^5 : (a_1, a_2) = 1,\, \rho _1(\varvec{a}, \varvec{j}) < \rho _2(\varvec{a}, \varvec{j}) \big \} \end{aligned}$$

and

$$\begin{aligned} D(\varvec{a}, \varvec{j}) := C_1(a_1, a_2) \big (\rho _2(\varvec{a},\, \varvec{j})^3 - \rho _1(\varvec{a},\, \varvec{j})^3\big ) , \end{aligned}$$

recalling that $C_1(a_1, a_2)$ is defined by (6). The convergence of series (13) follows easily from Remark 3.1, $\rho _2(\varvec{a}, \varvec{j}) \le 1 / (a_1 a_2 j_3)$, and the fact that $\min (j_1, j_2) \ge j_3$ for all $(\varvec{a}, \varvec{j}) \in {\mathcal {S}}_\infty$.

Thanks to Lemma 3.4, for each $(\varvec{a}, \varvec{j}) \in {\mathcal {S}}_n$ we have

$$\begin{aligned} S(\varvec{a}, \varvec{j}; n) = D(\varvec{a}, \varvec{j}) + O\!\left( \sigma (a_1)\,\sigma (a_2) \,\rho _2(\varvec{a}, \varvec{j})^2 \cdot \frac{(\log n)^2}{n}\right) . \end{aligned}$$

Consequently, we get that

$$\begin{aligned} V_n(\alpha ) = \mathrm {v}(\alpha ) - \Sigma _1 + O\!\left( \Sigma _2 \cdot \frac{(\log n)^2}{n}\right) , \end{aligned}$$

(14)

where

$$\begin{aligned} \Sigma _1 := \sum _{(\varvec{a},\, \varvec{j}) \,\in \, {\mathcal {S}}_\infty \!\setminus {\mathcal {S}}_n} \beta ^{j_1 + j_2 - j_3} (1 - \beta ^{j_3}) \, D(\varvec{a}, \varvec{j}) \end{aligned}$$

and

$$\begin{aligned} \Sigma _2 := \sum _{(\varvec{a},\, \varvec{j}) \,\in \, {\mathcal {S}}_n} \beta ^{j_1 + j_2 - j_3} (1 - \beta ^{j_3}) \, \sigma (a_1)\,\sigma (a_2) \,\rho _2(\varvec{a}, \varvec{j})^2 . \end{aligned}$$

Now we have to bound both $\Sigma _1$ and $\Sigma _2$.

If $(\varvec{a}, \varvec{j}) \in {\mathcal {S}}_\infty \setminus {\mathcal {S}}_n$ then $\big (\rho _2(\varvec{a}, \varvec{j}) - \rho _1(\varvec{a}, \varvec{j})\big ) n < 1$ and consequently, also by Remark 3.1,

$$\begin{aligned} D(\varvec{a}, \varvec{j}) \ll a_1 a_2 \big (\rho _2^3 - \rho _1^3\big )= & {} a_1 a_2 \big (\rho _1^2 + \rho _1 \rho _2 + \rho _2^2\big )(\rho _2 - \rho _1) \ll \frac{a_1 a_2 \rho _2^2}{n}\nonumber \\\le & {} \frac{1}{a_1 a_2 j_3^2 n} , \end{aligned}$$

(15)

where, for brevity, we wrote $\rho _i := \rho _i(\varvec{a}, \varvec{j})$ for $i=1,2$.

If $(\varvec{a}, \varvec{j}) \in {\mathcal {S}}_\infty$ then, as we already noticed, $\min (j_1, j_2) \ge j_3$ and, moreover,

$$\begin{aligned} \frac{j_2}{j_3 + 1}< a_1< \frac{j_2 + 1}{j_3} \quad \text { and }\quad \frac{j_1}{j_3 + 1}< a_2 < \frac{j_1 + 1}{j_3} . \end{aligned}$$

Hence, we have

$$\begin{aligned} \sum _{(\varvec{a},\, \varvec{j}) \,\in \, {\mathcal {S}}_\infty } \frac{\beta ^{j_1 + j_2 - j_3} (1 - \beta ^{j_3})}{a_1 a_2 j_3^2}&\le \sum _{j_3 \,\ge \, 1} \frac{1 - \beta ^{j_3}}{j_3^2} \sum _{j_1,\, j_2 \,\ge \, j_3} \beta ^{j_1 + j_2 - j_3} \sum _{\begin{array}{c} j_2 / (j_3 + 1) \,<\, a_1 \,<\, (j_2 + 1) / j_3 \\ j_1 / (j_3 + 1) \,<\, a_2 \,<\, (j_1 + 1) / j_3 \end{array}} \frac{1}{a_1 a_2} \nonumber \\&\ll \sum _{j_3 \,\ge \, 1} \frac{1 - \beta ^{j_3}}{j_3^2} \sum _{j_1,\, j_2 \,\ge \, j_3} \beta ^{j_1 + j_2 - j_3} = \frac{1}{\alpha ^2}\sum _{j \,\ge \, 1} \frac{(1 - \beta ^j)\beta ^j}{j^2} \nonumber \\&\le \frac{1}{\alpha } \sum _{j \,\le \, 1 / \alpha } \frac{1}{j} + \frac{1}{\alpha ^2}\sum _{j \,>\, 1 / \alpha } \frac{1}{j^2} \ll \frac{\log (1 / \alpha ) + 1}{\alpha } , \end{aligned}$$

(16)

where we used the inequality $1 - \beta ^j \le \alpha j$, which follows from Remark 3.2.

On the one hand, from (15) and (16) it follows that

$$\begin{aligned} \Sigma _1 \ll \frac{\log (1 / \alpha ) + 1}{\alpha n} = o(1) , \end{aligned}$$

(17)

as $\alpha n / \!\big ((\log n)^3 (\log \log n)^2\big ) \rightarrow +\infty$ (actually, $\alpha n / \!\log n \rightarrow +\infty$ is sufficient).

On the other hand, from $\rho _2(\varvec{a}, \varvec{j}) \le 1 / (a_1 a_2 j_3)$, (16), and the bound $\sigma (m) \ll m \log \log m$ (see, e.g., [20, Part I, Theorem 5.7]) it follows that

$$\begin{aligned} \Sigma _2\le & {} \!\!\!\! \sum _{(\varvec{a},\, \varvec{j}) \,\in \, {\mathcal {S}}_n}\!\!\!\! \frac{\beta ^{j_1 + j_2 - j_3} (1 - \beta ^{j_3})}{a_1 a_2 j_3^2} \cdot \frac{\sigma (a_1)\,\sigma (a_2)}{a_1 a_2} \ll \frac{(\log (1/\alpha ) + 1) (\log \log n)^2\!}{\alpha }\nonumber \\= & {} o\!\left( \frac{n}{(\log n)^2}\right) , \end{aligned}$$

(18)

as $\alpha n / \big ((\log n)^3 (\log \log n)^2\big ) \rightarrow +\infty$.

At this point, putting together (14), (17), and (18), we obtain $V_n(\alpha ) = \mathrm {v}(\alpha ) + o(1)$, as desired. The proof of (3) is complete.

It remains only to prove the upper bound (4). From (12) it follows that

$$\begin{aligned} {\mathbb {V}}[X]&\le \sum _{[d_1\!,\, d_2] \,\le \, n} \varphi (d_1)\,\varphi (d_2) \, \beta ^{\lfloor n / d_1 \rfloor + \lfloor n / d_2 \rfloor - \lfloor n / [d_1, d_2] \rfloor } \big (1 - \beta ^{\lfloor n / [d_1, d_2] \rfloor } \big ) \\&\le \sum _{[d_1\!,\, d_2] \,\le \, n} d_1 d_2 \cdot \frac{\alpha n}{[d_1, d_2]} = \alpha n \sum _{[d_1\!,\, d_2] \,\le \, n} (d_1, d_2) \le \alpha n \sum _{d \,\le \, n} d \sum _{a_1 a_2 \,\le \, n / d} 1 \\&= \alpha n \sum _{d \,\le \, n} d \sum _{m \,\le \, n / d} \tau (m) \ll \alpha n^2 \sum _{d \,\le \, n} \log \!\left( \frac{n}{d}\right) = \alpha n^2 \big (n \log n - \log (n!)\big ) < \alpha n^3 , \end{aligned}$$

where we used Remark 3.2, Lemma 3.1, and the bound $n! > (n / \mathrm {e})^n$. Thus (4) is proved. $\square$

Proof of Theorem 1.4

By Chebyshev’s inequality, Theorems 1.2 and 1.3, we have

$$\begin{aligned} {\mathbb {P}}\big [\,|X - {\mathbb {E}}[X]| > \varepsilon \!\; {\mathbb {E}}[X] \,\big ] \le \frac{{\mathbb {V}}[X]}{\big (\varepsilon {\mathbb {E}}[X]\big )^2} \ll \frac{\alpha n^3}{(\varepsilon \alpha n)^2} = \frac{1}{\varepsilon ^2 \alpha n} = o_\varepsilon (1) , \end{aligned}$$

as $\alpha n \rightarrow +\infty$. Hence, using again Theorem 1.2, we get

$$\begin{aligned} X \sim \frac{3}{\pi ^2} \cdot \frac{\alpha {\text {Li}}_2(1 - \alpha )}{1 - \alpha } \cdot n^2 , \end{aligned}$$

with probability $1 - o(1)$, as $\alpha n \rightarrow +\infty$. $\square$