2 Main results

To illustrate our main results, we define an invariant property satisfied by both (1.2) and (1.5). Let $k\ge 1$ be an integer. Let $a: \mathbb {N}^k \to \mathbb {C}$ be a bounded function. We say that a is of invariant average under multiplication if the mean value of $a(n_1,\ldots ,n_k)$ exists and satisfies

(2.1) $$ \begin{align} \lim_{N\to\infty}\frac1{N^k}\sum_{1\leq n_1,\ldots,n_k \leq N} a(dn_1,\ldots,dn_k) = \lim_{N\to\infty}\frac1{N^k}\sum_{1\leq n_1,\ldots,n_k \leq N} a(n_1,\ldots,n_k) \end{align} $$

for all $d\in \mathbb {N}$ . This definition was introduced for the case $k=1$ in [Reference Li, Wang, Wang and Yi9, Reference Wang11]. Let $(X, \nu , T)$ be a uniquely ergodic topological dynamical system. For $k=1$ , by (1.1), the Liouville function $\lambda (n)=(-1)^{\Omega (n)}$ is of invariant average $0$ , and by (1.2), the sequence $\{h(T^{\Omega (n)}x)\}_{n\in \mathbb {N}}$ for each $x\in X$ is of invariant average $\int _X h \, d\nu $ . In general, by [Reference Bergelson and Richter2, Corollary 1.16], the sequence $\{h(T^{\Omega (mn+r)}x)\}_{n\in \mathbb {N}}$ for each $x\in X$ is of invariant average $\int _X h \, d\nu $ for any integers $m, r$ with $m\ge 1$ and $0\leq r\leq m-1$ . For $k=2$ , by (1.5), the function $(m,n)\mapsto h(T^{\Omega (m^2+n^2)}x)$ for each $x\in X$ is also of invariant average $\int _X h \, d\nu $ . Our first theorem shows that, for any $k\ge 2$ , the function

$$ \begin{align*}(n_1,n_2,\ldots,n_k)\mapsto h(T^{\Omega({n_1}/{\gcd(n_1,n_2,\ldots,n_k)})}x)\end{align*} $$

is of invariant average $\int _X h \, d\nu $ for each $x\in X$ .

Theorem 2.1. Let $k\ge 2$ and let $(X, \nu , T)$ be uniquely ergodic. Then, for any $h\in C(X)$ and every $x\in X$ ,

(2.2) $$ \begin{align} \lim_{N\to\infty}\frac1{N^k}\sum_{1\leq n_1,\ldots,n_k \leq N } h(T^{\Omega({n_1}/{\gcd(n_1,n_2,\ldots,n_k)})}x) = \int_X h \,d\nu. \end{align} $$

Next, we state our disjointness results on functions of invariant average under multiplication. For any two-bounded function $a, b: \mathbb {N}^k \to \mathbb {C}$ , we call a and b asymptotically uncorrelated if

$$ \begin{align*} \lim_{N\to\infty} \bigg|\frac1{N^k} & \sum_{1\leq n_1,\ldots,n_k \leq N} a(n_1,\ldots,n_k) \overline{b(n_1,\ldots,n_k)} \nonumber \\ &- \bigg(\frac1{N^k}\sum_{1\leq n_1,\ldots,n_k \leq N} a(n_1,\ldots,n_k) \bigg)\bigg(\frac1{N^k}\sum_{1\leq n_1,\ldots,n_k \leq N} \overline{b(n_1,\ldots,n_k)} \bigg) \bigg|=0. \end{align*} $$

For example, by (1.4), the sequences capturing the behaviours of the Erdős–Kac theorem in (1.3) and the Bergelson–Richter theorem in (1.2) are asymptotically uncorrelated. It is well known that the probability of randomly choosing two integers which are co-prime is $6/\pi ^2$ . By (1.6), sequences capturing the behaviours of co-prime integers and the Donoso–Le–Moreira–Sun theorem in (1.5) are asymptotically uncorrelated. Our next theorem shows that if an arithmetic function is of a certain moderate growth, then it is asymptotically uncorrelated to the bounded functions of invariant average under multiplication.

Theorem 2.2. Let $k\ge 1$ be an integer. Let $f,g: \mathbb {N}\to \mathbb {C}$ be two arithmetic functions satisfying $f(n)=\sum _{d\mid n} g(d)$ . Suppose $a: \mathbb {N}^k \to \mathbb {C}$ is a bounded function of invariant average $\alpha $ under multiplication. If the series $\sum _{d=1}^\infty ( {|g(d)|}/{d^k})$ converges, then

(2.3) $$ \begin{align} \lim_{N\to\infty}\frac1{N^k}\sum_{1\leq n_1,\ldots,n_k \leq N} f(\gcd(n_1,\ldots,n_k))a(n_1,\ldots,n_k) = \alpha \sum_{d=1}^\infty \frac{g(d)}{d^k}. \end{align} $$

Theorem 2.2 gives us a unified form of some results related to (1.2) or (1.5). Let $\delta (n)$ be the function defined by $\delta (1)=1$ and $\delta (n)=0$ for $n\neq 1$ . For $k=1$ , if we take $g(d)=\delta (d)$ and $a(n)=h(T^{\Omega (n)}x)$ , then $f(n)=1$ for all $n\ge 1$ and (1.2) is recovered from (2.3). If we take $g(d)=\mu (m)$ for any square $d=m^2\geq 1$ and $g(d)=0$ for any nonsquare d, where $\mu (m)$ is the Möbius function, then by $\mu ^2(n)=\sum _{d^2\mid n}\mu (d)$ , we have $f(n)=\mu ^2(n)$ and obtain [Reference Bergelson and Richter2, Corollary 1.8]:

$$ \begin{align*} \lim_{N\to\infty}\frac1N\sum_{\substack{1\leq n \leq N \\ n\,\text{square-free}}} h(T^{\Omega(n)}x)=\frac{6}{\pi^2}\int_X h \,d\nu \end{align*} $$

for any $h\in C(X)$ and every $x\in X$ . For $k=2$ , if we take $f(n)=1$ for all $n\ge 1$ and $a(n)=h(T^{\Omega (m^2+n^2)}x)$ , we recover (1.5) from (2.3). If we take $g(d)=\mu (d)$ for all $d\ge 1$ and $a(n)=h(T^{\Omega (m^2+n^2)}x)$ , then $f(n)=\delta (n)$ , and (1.6) follows from (1.5) and (2.3).

For $k\ge 2$ , if we take $g(d)=\mu (d)$ for all $d\ge 1$ , $f(n)=\delta (n)$ and $a(n_1,n_2,\ldots ,n_k)=h(T^{\Omega ({n_1}/{\gcd (n_1,n_2,\ldots ,n_k)})}x)$ , then by (2.2) and (2.3), we obtain the following corollary.

Corollary 2.3. Let $k\ge 2$ . Let $(X, \nu , T)$ be a uniquely ergodic topological dynamical system. Then, for any $h\in C(X)$ and $x\in X$ ,

$$ \begin{align*} \lim_{N\to\infty}\frac1{N^k}\sum_{\substack{1\leq n_1,\ldots,n_k \leq N \\ \gcd(n_1,n_2,\ldots,n_k)=1}} h(T^{\Omega(n_1)}x) = \frac{1}{\zeta(k)}\int_X h \,d\nu, \end{align*} $$

where $\zeta (z)=\sum _{n=1}^\infty 1/{n^z}$ is the classical Riemann zeta function for $\operatorname {Re}(z)>1$ .

We give some other examples of arithmetic functions that are asymptotically uncorrelated to the orbits in (1.2) and (1.5). For example, using Theorem 2.2 with

$$ \begin{align*} \frac{\varphi(n)}{n} = \sum_{d\mid n}\frac{\mu(d)}{d}\quad \text{and} \quad\frac{n}{\varphi(n)} =\sum_{d\mid n} \frac{\mu^2(d)}{\varphi(d)}, \end{align*} $$

we have

$$ \begin{align*} \lim_{N\to\infty}\frac1N\sum_{n=1}^N \frac{\varphi(n)}{n} h(T^{\Omega(n)}x) = \frac{6}{\pi^2}\int_X h \,d\nu \end{align*} $$

and

$$ \begin{align*} \lim_{N\to\infty}\frac1N\sum_{n=1}^N \frac{n}{\varphi(n)} h(T^{\Omega(n)}x) = \frac{\zeta(2)\zeta(3)}{\zeta(6)} \int_X h \,d\nu. \end{align*} $$

For the divisor function $\tau (n)= \sum _{d\mid n} 1$ ,

$$ \begin{align*} \lim_{N\to\infty}\frac1{N^2}\sum_{1\leq m, n \leq N} \tau(\gcd(m,n)) h(T^{\Omega(m^2+n^2)}x) = \frac{\pi^2}{6} \int_X h \,d\nu, \end{align*} $$

and for the r-free numbers,

(2.4) $$ \begin{align} \lim_{N\to\infty}\frac1{N^2}\sum_{\substack{1\leq m, n \leq N \\ \gcd(m,n) \text{ is } r\text{-free}}} h(T^{\Omega(m^2+n^2)}x) = \frac1{\zeta(2r)} \int_X h \,d\nu. \end{align} $$

Here, if the greatest common divisor of m and n is r-free, then integers m and n are said to be relatively r-prime. In [Reference Benkoski1], Benkoski proved that the probability that two integers are relatively r-prime is $1/\zeta (2r)$ .

One can verify that the arithmetic functions $({\varphi (n)}/{n})^a$ , $({n}/{\varphi (n)})^a$ , ${\kappa (n)}/{n}$ and $q_r^\ast (n)$ also satisfy the assumption of Theorem 2.2 for any $a>0$ , $r\ge 2$ and $k\ge 1$ , where $\kappa (n)=\prod _{p\mid n} p$ and $q_r^\ast (n)$ is the characteristic function of the r-skew numbers defined by Cohen in [Reference Cohen5].

To end of this section, we give another application of Theorem 2.2 to the result (1.5) of Donoso, Le, Moreira and Sun. Let $B>0$ be a positive number. For an integer n, if n has no prime divisor larger than B, we say that n is B-friable. In [Reference Cheon and Kim3], Cheon and Kim proved that the probability of the greatest common divisor of k products of r positive integers being B-friable is $\prod _{p>B}[1-\{1-(1-1/p)^r\}^k]$ for $k\ge 2$ and $r\ge 1$ . For $k=2$ and $r=1$ , the probability of the greatest common divisor of two positive integers being B-friable is $\prod _{p>B}(1-{1}/{p^2})$ .

Corollary 2.4. Let $B>0$ . Let $(X, \nu , T)$ be uniquely ergodic. Then, for any $h\in C(X)$ and $x\in X$ ,

$$ \begin{align*} \lim_{N\to\infty}\frac1{N^2}\sum_{\substack{1\leq m, n \leq N \\ \gcd(m,n) \text{ is } B\text{-friable}}} h(T^{\Omega(m^2+n^2)}x) = \prod_{p>B}\bigg(1-\frac{1}{p^2}\bigg) \int_X h \,d\nu. \end{align*} $$

In the following sections, we use some elementary techniques in analytic number theory to prove Theorems 2.1 and 2.2, and Corollary 2.4.

3 Proof of Theorem 2.1

The proof of Theorem 2.1 is based on a general mathematical principle. For the sake of simplicity, we introduce the big $\mathbb {E}$ notation that is frequently used in ergodic theory. If S is a finite nonempty set, we define the average

$$ \begin{align*} \underset{n\in S}{{{\mathbb{E}}}^{~}}\,a(n):=\frac1{|S|}\sum_{n\in S}a(n) \end{align*} $$

for any function $a: S\to \mathbb {C}$ on S.

Theorem 3.1. Let $k\ge 2$ and let $a: \mathbb {N}\to \mathbb {C}$ be a bounded arithmetic function of invariant average $\alpha $ under multiplication. Then,

(3.1) $$ \begin{align} \lim_{N\to\infty}\frac1{N^k}\sum_{1\leq n_1,\ldots,n_k \leq N} a \bigg(\frac{n_1}{\gcd(n_1,n_2,\ldots,n_k)}\bigg) = \alpha. \end{align} $$

Proof. Set

$$ \begin{align*}A := \sum_{1\leq n_1,\ldots,n_k \leq N} a\bigg(\frac{n_1}{\gcd(n_1,n_2,\ldots,n_k)}\bigg).\end{align*} $$

Recall that $\delta (n)$ is the function defined by $\delta (1)=1$ and $\delta (n)=0$ for $n\neq 1$ . Using $\delta (n)=\sum _{d\mid n} \mu (d)$ , we can write A as a multiple summation on $a(n)$ :

(3.2) $$ \begin{align} A & = \sum_{d \leq N} \sum_{\substack{n_1,\ldots,n_k \leq N \\ \gcd(n_1,\ldots,n_k)=d}} a\bigg(\frac{n_1}{d} \bigg) = \sum_{d \leq N} \sum_{\substack{n_1,\ldots,n_k \leq {N}/d \\ \gcd(n_1,\ldots,n_k)=1}} a(n_1) \nonumber\\ &= \sum_{d \leq N} \sum_{n_1,\ldots,n_k \leq {N}/d} a(n_1) \sum_{l \,\mid\, \gcd(n_1,\ldots,n_k)}\mu(l) = \sum_{d \leq N} \sum_{l\leq {N}/d} \mu(l) \sum_{n_1,\ldots,n_k \leq {N}/{dl}} a(ln_1) \nonumber \\ &= \sum_{d \leq N} \sum_{l\leq {N}/d} \mu(l) \lfloor{N}/{dl}\rfloor^{k-1}\sum_{n_1\leq {N}/{dl}} a(ln_1). \sum_{d \leq N} \sum_{l\leq {N}/d} \mu(l) \lfloor{N}/{dl}\rfloor^{k}\underset{n\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln). \end{align} $$

Since $a(n)$ is bounded, we have $\mathbb {E}_{n\leq {N}/{dl}} a(ln) =O(1)$ . Then, for any $D\leq N^{1/4}$ ,

$$ \begin{align*} \bigg|\!\sum_{D<d \leq N} \sum_{l\leq {N}/d} \mu(l) \bigg\lfloor\frac{N}{dl}\bigg\rfloor^{k}\underset{n_1\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln_1) \bigg|\ll N^k \sum_{d>D} \frac{1}{d^k}\sum_{l=1}^\infty \frac{1}{l^k} \ll \frac{N^k}{D^{k-1}}. \end{align*} $$

Similarly, for any $L\leq N^{1/4}$ ,

$$ \begin{align*} \bigg|\!\sum_{d \leq D} \sum_{L<l\leq {N}/d} \mu(l) \bigg\lfloor\frac{N}{dl}\bigg\rfloor^{k}\underset{n_1\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln_1) \bigg|\ll N^k \sum_{d=1}^\infty \frac{1}{d^k}\sum_{l>L} \frac{1}{l^k} \ll \frac{N^k}{L^{k-1}}. \end{align*} $$

It follows by (3.2) that

(3.3) $$ \begin{align} A=\sum_{d \leq D} \sum_{l\leq L} \mu(l) \bigg\lfloor\frac{N}{dl}\bigg\rfloor^{k}\underset{n\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln) + O\bigg( \frac{N^k}{D^{k-1}}\bigg)+ O\bigg( \frac{N^k}{L^{k-1}}\bigg). \end{align} $$

For the first summation in (3.3),

(3.4) $$ \begin{align} \sum_{d \leq D} \sum_{l\leq L} \mu(l) \bigg\lfloor\frac{N}{dl}\bigg\rfloor^{k}\underset{n\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln) &= \sum_{d \leq D} \sum_{l\leq L} \mu(l) \bigg(\frac{N}{dl} +O(1)\bigg)^{k}\underset{n\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln) \nonumber\\ &= \sum_{d \leq D} \sum_{l\leq L} \mu(l) \bigg( \bigg(\frac{N}{dl}\bigg)^k + O\bigg(\frac{N}{dl}\bigg)^{k-1} \bigg)\underset{n\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln) \nonumber \\ &= N^k \sum_{d \leq D} \frac{1}{d^k} \sum_{l\leq L} \frac{\mu(l)}{l^k} \underset{n\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln) + O\bigg( N^{k-1} \sum_{d \leq D} \frac1{d^{k-1}}\sum_{l\leq L} \frac{1}{l^{k-1}} \bigg) \nonumber\\ &= N^k \sum_{d \leq D} \frac{1}{d^k} \sum_{l\leq L} \frac{\mu(l)}{l^k} \underset{n\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln) + O(N^{k-1}(\log D)(\log L)). \end{align} $$

In the last line of (3.4), we have used the estimate $\sum _{d \leq D} 1/{d^{k-1}} \leq \sum _{d \leq D} 1/{d}\ll \log D$ since $k\ge 2$ . Combining (3.3) and (3.4) yields

(3.5) $$ \begin{align} A=N^k \sum_{d \leq D} \frac{1}{d^k} \sum_{l\leq L} \frac{\mu(l)}{l^k} \underset{n\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln) + O(N^{k-1}(\log D)(\log L)) + O\bigg( \frac{N^k}{D^{k-1}}\bigg)+ O\bigg( \frac{N^k}{L^{k-1}}\bigg). \end{align} $$

Dividing both sides of (3.5) by $N^k$ ,

(3.6) $$ \begin{align} &\frac1{N^k}\sum_{1\leq n_1,\ldots,n_k \leq N} a \bigg(\frac{n_1}{\gcd(n_1,n_2,\ldots,n_k)}\bigg) \nonumber\\ &\quad=\sum_{d \leq D} \frac{1}{d^k} \sum_{l\leq L} \frac{\mu(l)}{l^k} \underset{n\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln) + O\bigg(\frac{(\log D)(\log L)}{N}\bigg) + O\bigg( \frac{1}{D^{k-1}}\bigg)+ O\bigg( \frac{1}{L^{k-1}}\bigg) \end{align} $$

for any $D,L$ with $1\leq D\leq N^{1/4}$ and $1\leq L \leq N^{1/4}$ .

Fix D and L first. Since $a(n)$ is of invariant average $\alpha $ under multiplication,

$$ \begin{align*} \lim_{N\to\infty} \underset{n\leq {N}/{dl}}{{{\mathbb{E}}}^{~}}\, a(ln) = \alpha \end{align*} $$

for $d\leq D$ and $l\leq L$ . Taking $N\to \infty $ in (3.6),

(3.7) $$ \begin{align} \lim_{N\to\infty} \frac1{N^k}\sum_{1\leq n_1,\ldots,n_k \leq N} a \bigg(\frac{n_1}{\gcd(n_1,n_2,\ldots,n_k)}\bigg) & = \alpha \sum_{d \leq D} \frac{1}{d^k} \sum_{l\leq L} \frac{\mu(l)}{l^k} + O\bigg( \frac{1}{D^{k-1}}\bigg)+ O\bigg( \frac{1}{L^{k-1}}\bigg). \end{align} $$

Now, taking $D\to \infty $ and $L\to \infty $ in (3.7), and using

$$ \begin{align*} \sum_{d=1}^\infty \frac{1}{d^k}=\zeta(k) \quad\text{and}\quad \sum_{l=1}^\infty \frac{\mu(l)}{l^k} = \frac1{\zeta(k)}, \end{align*} $$

we obtain (3.1), as desired.

Proof of Theorem 2.1

By (1.2), the sequence $\{h(T^{\Omega (n)}x)\}_{n\in \mathbb {N}}$ is of invariant average $\int _X h \, d\nu $ for each $x\in X$ . Theorem 2.1 follows by taking $a(n)=h(T^{\Omega (n)}x)$ in Theorem 3.1. Observe that $\gcd (dn_1,dn_2,\ldots ,dn_k)=d\cdot \gcd (n_1,n_2,\ldots ,n_k)$ for all integers $n_1,n_2,\ldots ,n_k, d\in \mathbb {N}$ . Therefore, the function

$$ \begin{align*}(n_1,n_2,\ldots,n_k)\mapsto h(T^{\Omega({n_1}/{\gcd(n_1,n_2,\ldots,n_k)})}x)\end{align*} $$

for each $x\in X$ is of invariant average $\int _X h \, d\nu $ under multiplication.

4 Proof of Theorem 2.2

We use the method in the proof of Theorem 3.1 to prove Theorem 2.2. We first write the summation on the left-hand side of (2.3) as a multiple summation on $a(dn_1, \ldots , dn_k)$ over integers $d, n_1,\ldots , n_k$ . Then, we split this summation up into two parts according to the size of the modulus d and estimate them.

Proof of Theorem 2.2

To prove Theorem 2.2, similar to the proof of Theorem 3.1, we estimate the summation on the left-hand side of (2.3) first. By the assumption that $f(n)=\sum _{d\mid n} g(d)$ , we have $f(\gcd (n_1,\ldots ,n_k))=\sum _{d\mid \gcd (n_1,\ldots ,n_k)} g(d)$ . So, we can write the left summation of (2.3) as a double summation on $a(n_1,\ldots , n_k)$ :

(4.1) $$ \begin{align} \sum_{1\leq n_1,\ldots,n_k \leq N} f(\gcd(n_1,\ldots,n_k))a(n_1,\ldots,n_k) & = \sum_{1\leq n_1,\ldots,n_k \leq N} a(n_1,\ldots,n_k)\sum_{d\mid \gcd(n_1,\ldots,n_k)} g(d) \nonumber\\ & = \sum_{\substack{1\leq n_1,\ldots,n_k \leq N \\ d\mid n_1,\ldots, d\mid n_k}} g(d)a(n_1,\ldots,n_k)\nonumber\\ & = \sum_{1\leq d \leq N} g(d) \sum_{\substack{1\leq n_1,\ldots,n_k \leq N \\ d\mid n_1, \ldots, d\mid n_k}} a(n_1,\ldots,n_k)\nonumber\\ & = \sum_{1\leq d \leq N} g(d) \sum_{ 1\leq n_1,\ldots,n_k \leq N/d } a(dn_1,\ldots,dn_k) \nonumber\\ &= \sum_{1\leq d \leq N} g(d) \bigg\lfloor \frac{N}d \bigg\rfloor^k \underset{ 1\leq n_1,\ldots,n_k \leq N/d }{{{\mathbb{E}}}^{~}}\, a(dn_1,\ldots,dn_k). \end{align} $$

We divide this summation into two parts. Choose D with $1\leq D \leq N$ . Since $\mathbb {E}_{ 1\leq n_1,\ldots ,n_k \leq N/d } a(dn_1,\ldots ,dn_k)= O(1)$ ,

(4.2) $$ \begin{align} \bigg|\!\sum_{D < d \leq N} g(d) \bigg\lfloor \frac{N}d \bigg\rfloor^k \underset{ 1\leq n_1,\ldots,n_k \leq N/d }{{{\mathbb{E}}}^{~}}\, a(dn_1,\ldots,dn_k) \bigg|\ll N^k \sum_{d>D} \frac{|g(d)|}{d^k}. \end{align} $$

For the summation over $1\leq d \leq D$ ,

(4.3) $$ \begin{align} & \sum_{1\leq d \leq D} g(d) \bigg\lfloor \frac{N}d \bigg\rfloor^k \underset{ 1\leq n_1,\ldots,n_k \leq N/d }{{{\mathbb{E}}}^{~}}\, a(dn_1,\ldots,dn_k) \nonumber\\ &\quad = \sum_{1\leq d \leq D} g(d) \bigg( \frac{N}d +O(1)\bigg)^k \underset{ 1\leq n_1,\ldots,n_k \leq N/d }{{{\mathbb{E}}}^{~}}\, a(dn_1,\ldots,dn_k)\nonumber\\ &\quad = \sum_{1\leq d \leq D} g(d) \bigg( \frac{N^k}{d^k} +O\bigg(\frac{N^{k-1}}{d^{k-1}}\bigg)\bigg)\underset{ 1\leq n_1,\ldots,n_k \leq N/d }{{{\mathbb{E}}}^{~}}\, a(dn_1,\ldots,dn_k)\nonumber\\ &\quad = N^k\sum_{1\leq d \leq D} \frac{g(d)}{d^k} \underset{ 1\leq n_1,\ldots,n_k \leq N/d }{{{\mathbb{E}}}^{~}}\, a(dn_1,\ldots,dn_k)+O\bigg(N^{k-1}\sum_{1\leq d \leq D} \frac{|g(d)|}{d^{k-1}}\bigg). \end{align} $$

Combining (4.1)–(4.3),

(4.4) $$ \begin{align} & \frac1{N^k}\sum_{1\leq n_1,\ldots,n_k \leq N} f(\gcd(n_1,\ldots,n_k))a(n_1,\ldots,n_k) \nonumber \\ & \quad =\sum_{1\leq d \leq D} \frac{g(d)}{d^k} \underset{ 1\leq n_1,\ldots,n_k \leq N/d }{{{\mathbb{E}}}^{~}}\, a(dn_1,\ldots,dn_k)+O\bigg(\frac1N\sum_{1\leq d \leq D} \frac{|g(d)|}{d^{k-1}}\bigg) + O\bigg(\sum_{d>D} \frac{|g(d)|}{d^k}\bigg). \end{align} $$

Since $a(n_1,\ldots ,n_k) $ is of invariant average $\alpha $ under multiplication, for any $d\leq D$ ,

$$ \begin{align*} \lim_{N\to\infty} \underset{ 1\leq n_1,\ldots,n_k \leq N/d }{{{\mathbb{E}}}^{~}}\, a(dn_1,\ldots,dn_k) =\alpha. \end{align*} $$

Then, (2.3) follows immediately by taking $N\to \infty $ first and then $D\to \infty $ in (4.4). This completes the proof of Theorem 2.2.

5 Proof of Corollary 2.4

To prove Corollary 2.4, we prove a general result.

Theorem 5.1. Let $(X, \nu , T)$ be uniquely ergodic and let S be an arbitrary subset of $\mathbb {N}$ . Then, for any $h\in C(X)$ and $x\in X$ ,

(5.1) $$ \begin{align} \lim_{N\to\infty}\frac1{N^2}\sum_{\substack{1\leq m, n \leq N \\ \gcd(m,n) \in S}} h(T^{\Omega(m^2+n^2)}x) = \frac{\zeta_S(2)}{\zeta(2)}\int_X h \,d\nu, \end{align} $$

where

$$ \begin{align*}\zeta_S(z)=\sum_{\substack{n=1 \\ n\in S}}^\infty \frac{1}{n^z} \quad\text{for } \operatorname{Re}(z)>1.\end{align*} $$

Proof. We define $\delta _S(n)$ by

$$ \begin{align*} \delta_S(n)= \begin{cases} 1 & \text{if } n\in S,\\ 0 & \text{if } n\notin S. \end{cases} \end{align*} $$

Let $g(n) =\sum _{d\mid n} \delta _S(d) \mu ({n}/{d})$ . By the Möbius inversion formula, $\delta _S(n)=\sum _{d\mid n}g(d)$ . Note that

$$ \begin{align*}|g(n)| \leq \sum_{d\mid n} \bigg|\delta_S(d) \mu\bigg(\dfrac{n}{d}\bigg)\bigg| \leq \tau(n) \ll n^\varepsilon\end{align*} $$

for any $\varepsilon>0$ . Here, $\tau (n)=\sum _{d\mid n}1$ is the number of positive divisors of n. This implies that $\sum _{d=1}^\infty {|g(d)|}/{d^2}$ converges. By Theorem 2.2 and (1.5),

(5.2) $$ \begin{align} \lim_{N\to\infty}\frac1{N^2}\sum_{\substack{1\leq m, n \leq N \\ \gcd(m,n) \in S}} h(T^{\Omega(m^2+n^2)}x) = \bigg(\sum_{d=1}^\infty\frac{g(d)}{d^2}\bigg)\int_X h \,d\nu. \end{align} $$

By [Reference Cohen4, (1.2)],

$$ \begin{align*} \sum_{d=1}^\infty\frac{g(d)}{d^2}=\frac{\zeta_S(2)}{\zeta(2)}. \end{align*} $$

Thus, (5.1) follows by (5.2) immediately.

Proof of Corollary 2.4

Let S be the set of all B-friable positive integers in Theorem 5.1. Then,

$$ \begin{align*} \zeta_S(2)=\sum_{\substack{n=1 \\ n \text{ is } B\text{-friable}}}^\infty\frac{1}{n^2} = \prod_{p \leq B} \bigg(1+\frac1{p^2}+\frac1{p^4}+\cdots\bigg) = \prod_{p \leq B} \bigg(1-\frac1{p^2}\bigg)^{-1} \end{align*} $$

and

$$ \begin{align*} \frac{\zeta_S(2)}{\zeta(2)} =\frac{{\prod_{p \leq B} \bigg(1-\dfrac1{p^2}\bigg)^{-1}}}{{\prod_{p} \bigg(1-\dfrac1{p^2}\bigg)^{-1}}} = \prod_{p>B}\bigg(1-\frac{1}{p^2}\bigg). \end{align*} $$

Thus, Corollary 2.4 follows by (5.1) of Theorem 5.1.

Remark 5.2. If we take $S=\{1\}$ in Theorem 5.1, then we obtain (1.6) from (5.1). If we take S to be the set of r-free integers, then by (5.1),

(5.3) $$ \begin{align} \lim_{N\to\infty}\frac1{N^2}\sum_{\substack{1\leq m, n \leq N \\ \gcd(m,n) \text{ is } r\text{-free}}} h(T^{\Omega(m^2+n^2)}x) = \frac{\zeta_S(2)}{\zeta(2)} \int_X h \,d\nu. \end{align} $$

By the well-known fact that

$$ \begin{align*} \sum_{d^r\mid n}\mu(d) = \begin{cases} 1 & \text{if } n \text{ is } r\text{-free},\\ 0 & \text{otherwise}, \end{cases} \end{align*} $$

we have

$$ \begin{align*}\bigg(\sum_{n=1}^\infty \frac{\mu(n)}{n^{2r}} \bigg)\bigg(\sum_{n=1}^\infty \frac{1}{n^2}\bigg) = \sum_{\substack{ n=1\\ n \text{ is } r\text{-free}}} ^\infty \frac{1}{n^2}. \end{align*} $$

That is, $\zeta _S(2)={\zeta (2)}/{\zeta (2r)}.$ Then, (2.4) follows by (5.3).