ON THE ITERATES OF THE SHIFTED EULER’S FUNCTION

Abstract

Let $\varphi $ be Euler’s function and fix an integer $k\ge 0$. We show that for every initial value $x_1\ge 1$, the sequence of positive integers $(x_n)_{n\ge 1}$ defined by $x_{n+1}=\varphi (x_n)+k$ for all $n\ge 1$ is eventually periodic. Similarly, for all initial values $x_1,x_2\ge 1$, the sequence of positive integers $(x_n)_{n\ge 1}$ defined by $x_{n+2}=\varphi (x_{n+1})+\varphi (x_n)+k$ for all $n\ge 1$ is eventually periodic, provided that k is even.

1 Introduction and main results

Let $\mathbb {N}$ be the set of positive integers and fix an arithmetic function $f: \mathbb {N}^d\to \mathbb {N}$ for some $d \in \mathbb {N}$ . Let $(x_n)_{n\ge 1}$ be a sequence of positive integers which satisfies the recurrence

(1.1) $$ \begin{align} x_{n+d}=f(x_n,\ldots,x_{n+d-2},x_{n+d-1}) \quad \text{for all }n \in \mathbb{N}, \end{align} $$

with starting values $x_1,\ldots ,x_d\in \mathbb {N}$ . In the case where $d=1$ , the sequence $(x_n)_{n\ge 1}$ is simply the orbit of $x_1$ with respect to f. The aim of this note is to study whether certain recurrence sequences $(x_n)_{n\ge 1}$ of the type (1.1) are eventually periodic independent of their starting values, that is, for all $x_1,\ldots ,x_d \in \mathbb {N}$ , there exists $T \in \mathbb {N}$ such that $x_n=x_{n+T}$ for all sufficiently large n.

We will frequently use the basic observation that a recurrence sequence $(x_n)_{n\ge 1}$ , as in (1.1), is eventually periodic if and only if it is bounded (see [5, page 45]).

We start with a simple result for functions f which are not too large.

Proposition 1.1. Let $f: \mathbb {N}^d \to \mathbb {N}$ be an arithmetical function, with $d \in \mathbb {N}$ , and suppose that there exists $C \in \mathbb {N}$ such that

(1.2) $$ \begin{align} f(n_1,\ldots,n_d) < \max\{n_1,\ldots,n_d\} \end{align} $$

for all $n_1,\ldots ,n_d \in \mathbb {N}$ with $n_i \ge C$ for some $i\in \{1,\ldots ,d\}$ . Let $(x_n)_{n\ge 1}$ be a recurrence sequence as in (1.1), with starting values $x_1,\ldots ,x_d \in \mathbb {N}$ . Then

(1.3) $$ \begin{align} \limsup_{n\to \infty} x_n \le \max \{f(n_1,\ldots,n_d): n_1,\ldots,n_d\le C-1\}. \end{align} $$

In particular, $(x_n)_{n\ge 1}$ is bounded above and hence eventually periodic.

Special instances of Proposition 1.1 in the one-dimensional case $d=1$ have been previously obtained in the literature. For example, Porges [15] considered the case where $f(n)$ is the sum of squares of the digits of n (see also [7, 11, 19]). Note that (1.2) holds if $f(n_1,\ldots ,n_d)=o(N)$ as $N=\max \{n_1,\ldots ,n_d\}\to \infty $ .

Of course, there exist other functions f which do not satisfy (1.2) and such that every sequence $(x_n)_{n\ge 1}$ , as in (1.1), is eventually periodic: as a trivial example, one can consider $f(n_1,\ldots ,n_d):=\max \{n_1,\ldots ,n_d\}$ for all $(n_1,\ldots ,n_d) \in \mathbb {N}^d$ . At the opposite extreme, if f is slightly bigger (for instance, $f(n_1,\ldots ,n_d):=\max \{n_1,\ldots ,n_d\}+1$ ), then there are no eventually periodic sequences $(x_n)_{n\ge 1}$ as in (1.1). This is the starting point for this work, which motivates the heuristic: if a function f satisfies (1.2) ‘on average’, then every sequence $(x_{n})_{n\ge 1}$ as in (1.1) should be eventually periodic, independent of its starting values.

We are going to confirm the above heuristic in two cases which involve (shifted iterates of) Euler’s function $\varphi $ (recall that $\varphi (n)$ is the number of integers in $\{1,\ldots ,n\}$ which are coprime with n). Our first main result is the following theorem.

Theorem 1.2. Fix an integer $k\ge 0$ and let $(x_n)_{n\ge 1}$ be the recurrence sequence defined by

$$ \begin{align*} x_{n+1}=\varphi(x_n)+k \quad \text{for all }n \in \mathbb{N}, \end{align*} $$

with starting value $x_1 \in \mathbb {N}$ . Then,

(1.4) $$ \begin{align} \sup_{n \in \mathbb{N}} x_n \le \max\{x_1,k^4\}+(k+1)^2. \end{align} $$

In particular, $(x_n)_{n\ge 1}$ is eventually periodic.

Note that the dependence of the upper bound (1.4) on $x_1$ cannot be removed: indeed, if $k:=x_1-\varphi (x_1)$ for some $x_1\in \mathbb {N}$ , then the sequence $(x_{n})_{n\ge 1}$ is constantly equal to $x_1$ .

The trivial case $k=0$ in Theorem 1.2 has been already considered in the literature from different viewpoints (and, of course, it follows by Proposition 1.1 since $\varphi (n)\le n-1$ for all $n\ge 2$ ). Indeed, given a starting value $x_1 \in \mathbb {N}$ , then $x_{n+1}=\varphi ^{(n)}(x_1)$ for all $n \in \mathbb {N}$ , where $\varphi ^{(m)}$ is the m-fold iteration of $\varphi $ . For instance, Pillai [12] showed that

$$ \begin{align*} \bigg\lfloor \frac{\log x_1-\log 2}{\log 3}\bigg\rfloor+1 \le N(x_1) \le \bigg\lfloor \frac{\log x_1}{\log 2}\bigg\rfloor+1 \quad \text{for all }x_1 \in \mathbb{N}, \end{align*} $$

where $N(x_1)$ is the minimal integer n for which $x_n=1$ (see also [17]) and it has been conjectured by Erdős et al. [4] that $N(x_1) \sim \alpha \log x_1$ as $x_1\to \infty $ , for some $\alpha \in \mathbb {R}$ . It is known that the understanding of the multiplicative structure of $\varphi $ and its iterates is, in some sense, equivalent to the study of the behaviour of the integers of the form $p-1$ , where p is a prime. See also [8, 10, 14, 18] for related work.

However, if $k\ge 1$ , then the function $f(n):=\varphi (n)+k$ does not satisfy (1.2): indeed, $\varphi (p)=p-1$ for all primes p, and hence $f(p)\ge p$ . However, it is well known that

(1.5) $$ \begin{align} \frac{1}{n}\sum_{i=1}^n \varphi(i)=\frac{3}{\pi^2}n+O(\log n) \quad \text{as }n\to \infty \end{align} $$

(see, for example, [2, Theorem 8.6]). Hence, approximating roughly $f(n)$ with $cn+k$ , where $c=3/\pi ^2 \in (0,1)$ , we expect that (1.2) holds ‘on average’, which is the heuristic behind Theorem 1.2.

Our second main result is the following theorem.

Theorem 1.3. Fix an even integer $k\ge 0$ and let $(x_n)_{n\ge 1}$ be the recurrence sequence defined by

$$ \begin{align*} x_{n+2}=\varphi(x_{n+1})+\varphi(x_n)+k \quad \text{for all }n \in \mathbb{N}, \end{align*} $$

with starting values $x_1,x_2 \in \mathbb {N}$ . Then,

(1.6) $$ \begin{align} \sup_{n\in \mathbb{N}} x_n \le 4^{X^{3^{k+1}}}, \quad \text{where } X:=\frac{3x_1+5x_2+7k}{2}. \end{align} $$

In particular, $(x_n)_{n\ge 1}$ is eventually periodic.

The heuristic supporting Theorem 1.3 is similar: thanks to (1.5), the value $f(n,m):=\varphi (n)+\varphi (m)+k$ can be roughly bounded above by $2c\max \{n,m\}+k$ , which is definitely smaller than $\max \{n,m\}$ since $2c=6/\pi ^2 <1$ .

We end with an open question to check if d is sufficiently large, then there exist starting values $x_1,\ldots ,x_d \in \mathbb {N}$ such that the sequence $(x_n)_{n\ge 1}$ defined as in (1.1) with

$$ \begin{align*} f(n_1,\ldots,n_d)=\varphi(n_1)+\cdots+\varphi(n_d) \end{align*} $$

is not eventually periodic. In a sense, this is related to the open question known as Lehmer’s totient problem [9], which asks about the existence of a composite $q\ge 2$ such that $\varphi (q)$ divides $q-1$ : indeed, if $r:=(q-1)/\varphi (q)$ , $d=r$ and $x_1=\cdots =x_d=q$ , then the sequence $(x_n)_{n\ge 1}$ would be constant.

Lastly, as suggested by the referee, it would be interesting to check whether the upper estimates of our main results are, in some sense, sufficiently sharp and whether some nontrivial lower bounds on the size of $(x_n)_{n\ge 1}$ could be obtained. We do not have an answer to these questions. However, we suspect that they are quite difficult. Indeed, in the setting of Theorem 1.2, one could try to obtain a cycle with period $(x_1,\ldots ,x_T)$ made by all primes in arithmetic progression but the last one (so that $x_1=x_{T+1}=\varphi (x_T)+k$ ). A possible interpretation may amount to asking whether such examples exist with comparable sizes of T and k.

2 Proofs

Proof of Proposition 1.1.

Let $\mathcal {Q}$ be the set of d-uples $(n_1,\ldots ,n_d) \in \mathbb {N}^d$ with $n_i\le C-1$ for all $i \in \{1,\ldots ,d\}$ . Suppose that $(n_1,\ldots ,n_d) \in \mathbb {N}\setminus \mathcal {Q}$ . Hence, we can pick the largest index $i \in \{1,\ldots ,d\}$ such that $n_i=\max \{n_1,\ldots ,n_d\}$ . In particular, $n_i\ge C$ . We claim that there exists $m \in \mathbb {N}$ such that $(n_{m+1}, \ldots , n_{m+d}) \in \mathcal {Q}$ . In addition, if m is the least such integer, then $\max \{n_j,\ldots ,n_{j+d-1}\}$ is decreasing for $j \in \{1,\ldots ,m\}$ . To this aim, suppose for the sake of contradiction that the claim does not hold. By the standing hypothesis (1.2), we get $n_{d+1}=f(n_1,\ldots ,n_d)< n_{i}$ . Repeating this reasoning, we obtain $n_{d+j}< \max \{n_j,\ldots ,n_{d+j-1}\}= n_{i}$ for all $j\in \{1,\ldots ,i\}$ . Hence, $\max \{n_{i+1},\ldots ,n_{i+d}\}\le n_{i}~-~1$ . Proceeding similarly, it follows that

$$ \begin{align*} \max\{n_{i+(k-1)d+1},\ldots,n_{i+kd}\}\le n_{i}-k \quad \text{for all }k\in \mathbb{N}. \end{align*} $$

However, if $k=n_{i}+1-C$ , then $(n_{i+(k-1)d+1},\ldots ,n_{i+kd}) \in \mathcal {Q}$ , which proves the claim.

To complete the proof, fix starting values $x_1,\ldots ,x_d \in \mathbb {N}$ . By the above claim and the finiteness of $\mathcal {Q}$ , it follows that the sequence $(x_n)_{n\ge 1}$ is bounded above by the constant

$$ \begin{align*} \max(\{x_1,\ldots,x_d\}\cup \{f(n_1,\ldots,n_d): (n_1,\ldots,n_d) \in \mathcal{Q}\}) \end{align*} $$

and that the upper limit in (1.3) holds.

Proof of Theorem 1.2.

First, let us suppose $k \le 1$ and fix a starting value $x_1 \in \mathbb {N}$ . Then,

$$ \begin{align*} x_{n+1}=\varphi(x_n)+k \le \max\{1,x_n-1\}+1=\max\{2,x_n\} \end{align*} $$

for all $n \in \mathbb {N}$ , with the consequence that $x_n \le \max \{x_1,2\}$ for all $n \in \mathbb {N}$ .

Suppose hereafter that $k \ge 2$ . Note that for all $n,m \in \mathbb {N}$ ,

(2.1) $$ \begin{align} \begin{split} x_{n+m}&\le \max\{x_{n+m-1}-1,1\}+k \\ &= \max\{x_{n+m-1}+k-1,k+1\} \le \max\{x_n+m(k-1),k+1\}. \end{split} \end{align} $$

Let us suppose for the sake of contradiction that $(x_n)_{n\ge 1}$ is not bounded above. Hence, there exists a minimal $r_1 \in \mathbb {N}$ such that $x_{r_1} \ge k^4$ (in particular, $x_{r_1}>4$ ).

Claim 2.1. There exists $i \in \{1,\ldots ,k\}$ such that $x_{r_1+i}<x_{r_1}$ .

Proof. Since $\varphi (n) \le n-\sqrt {n}$ whenever n is composite (by the fact that there exists a divisor of n which is at most $\sqrt {n}$ ), it follows that, if $x_{{r_1}+i-1}$ is composite for some $i \in \{1,\ldots ,k\}$ , then

$$ \begin{align*} x_{r_1+i}=\varphi(x_{r_1+i-1})+k \le x_{r_1+i-1}-\sqrt{x_{r_1+i-1}}+k. \end{align*} $$

Considering that the map $x\mapsto x-\sqrt {x}$ is increasing on $(4,\infty )$ and using (2.1), we obtain

$$ \begin{align*} \begin{split} x_{r_1+i} &\le x_{r_1}+(i-1)(k-1)-\sqrt{x_{r_1}+(i-1)(k-1)}+k\\ &\le x_{r_1}+(k-1)^2-\sqrt{k^4}+k < x_{r_1}. \end{split} \end{align*} $$

To conclude, we show that there exists some $i \in \{1,\ldots ,k\}$ for which $x_{{r_1}+i-1}$ is composite. Indeed, in the opposite case, these $x_{{r_1}+i-1}$ terms are all primes (and greater than k), and hence

$$ \begin{align*} x_{r_1+i-1}=x_{r_1}+(i-1)(k-1) \equiv x_{r_1}-(i-1) \bmod{k} \end{align*} $$

for all $i \in \{1,\ldots ,k\}$ . This is impossible, because there would exist $i \in \{1,\ldots ,k\}$ such that k divides $x_{{r_1}+i-1}$ .

At this point, (2.1) and Claim 2.1 imply that there exists a minimal $i_1 \in \{1,\ldots ,k\}$ such that $x_{r_1+i_1}< x_{r_1}$ ; and hence

(2.2) $$ \begin{align} \max\{x_1,\ldots,x_{r_1+i_1-1}\} \le x_{r_1}+(i_1-1)(k-1) < x_{r_1}+k^2. \end{align} $$

With the same reasoning, we can construct recursively sequences of positive integers $(r_n)$ and $(i_n)$ such that for all $n \in {\mathbb {N}}$ :

1. (i) $r_{n+1}$ is the minimal integer such that $r_{n+1}\ge r_n+i_n$ and $x_{r_{n+1}}\ge k^4$ ;

2. (ii) $i_{n+1}$ is the minimal integer in $\{1,\ldots ,k\}$ such that $x_{r_{n+1}+i_{n+1}}<x_{r_n}$ ; and hence,

   (2.3) $$ \begin{align} \max\{x_{r_n+i_n},\ldots,x_{r_{n+1}+i_{n+1}-1}\} \le x_{r_{n+1}}+(i_{n+1}-1)(k-1) < x_{r_{n+1}}+k^2. \end{align} $$

Lastly, note that

(2.4) $$ \begin{align} x_{r_{n}}=\varphi(x_{r_n-1})+k\le x_{r_n-1}+k \le k^4+k\le x_{r_1}+k \end{align} $$

for all $n \in \mathbb {N}$ . Using (2.2), (2.3) and (2.4), we conclude that

$$ \begin{align*} x_n < x_{r_1}+k^2+k\le \max\{x_1,k^4+k\}+k^2+k<\max\{x_1,k^4\}+(k+1)^2 \end{align*} $$

for all $n \in \mathbb {N}$ . This proves (1.4), concluding the proof.

Remark 2.2. A sketch of a shorter proof that the sequence $(x_n)_{n\ge 1}$ in Theorem 1.2 is eventually periodic goes as follows. Set $y_n:=\varphi (x_n)$ for all $n \in \mathbb {N}$ and note that $y_n\le C n$ with $C:=\max \{x_1,k\}$ . Hence, $\{y_1,\ldots ,y_n\}$ is contained in $V_n:=\varphi (\mathbb {N})\cap [1,Cn]$ . By a classical result of Pillai [13] (see also [6] and references therein), we have $|V_n|=o(n)$ as $n\to \infty $ , and hence there exist distinct $i,j \in \mathbb {N}$ with $y_i=y_j$ . This implies that $x_{i+1}=x_{j+1}$ ; therefore, $(x_n)_{n\ge 1}$ is eventually periodic. However, this does not lead to an effective upper bound as in (1.4).

In the proof of Theorem 1.3, we will need also the effective version of the third Mertens’ theorem given by Rosser and Schoenfeld [16] in 1962 (see also [1, 3]). As usual, hereafter, we reserve the letter p for primes.

Proposition 2.3. Let $\gamma :=\lim _n(\sum _{i\le n}1/i-\log n)=0.57721\ldots $ be the Euler– Mascheroni constant. Then the following inequality holds for all $x\ge 2{:}$

$$ \begin{align*} \frac{e^{-\gamma}}{\log x} \bigg(1- \frac{1}{\log^2 x} \bigg) < \prod_{p\le x} \bigg( 1 - \frac{1}{p} \bigg) < \dfrac{e^{-\gamma}}{\log x} \bigg(1+ \frac{1}{2\log^2 x} \bigg). \end{align*} $$

Proof. See [16, Theorem 7 and its Corollary].

Corollary 2.4. If $x\ge 6$ , then

$$ \begin{align*} \prod_ {x<p\le x^3} \bigg( 1 - \frac{1}{p} \bigg)<\frac{1}{2}. \end{align*} $$

Proof. Thanks to Proposition 2.3, for each $r \ge 2$ , there exists $c_r \in (-1,\tfrac 12)$ such that

$$ \begin{align*} \prod_ {p\le r} \bigg( 1 - \frac{1}{p} \bigg) =\frac{e^{-\gamma}}{\log x} \bigg(1+ \frac{c_r}{\log^2 x} \bigg). \end{align*} $$

At this point, fix $x\ge 6$ . It follows that

$$ \begin{align*} \begin{split} \prod_ {x<p\le x^3} \bigg( 1 - \frac{1}{p} \bigg) &=\frac {\prod_{p \leq x^3} ( 1 - {1}/{p} )} {\prod_{p \leq x} ( 1 - {1}/{p} )}\\ &=\frac{e^{-\gamma}/(\log x^3)}{e^{-\gamma}/\log x}\cdot \frac{(1+c_{x^3}/(\log x^3)^2)}{1+c_x/(\log x)^2}\\ & < \frac{1}{3}\cdot \frac{1+0.5/ (9\log^2 x)}{1-1/(\log^2 x)} < \frac{1}{2}. \end{split} \end{align*} $$

Indeed, the last inequality is equivalent to

$$ \begin{align*} \frac{2}{3} \bigg(1+\frac{1}{18 \log^2 x}\bigg)< 1-\frac{1}{\log^2 x}, \end{align*} $$

which holds if and only if

$$ \begin{align*} x> e^{\sqrt{3(1+{1}/{27})}}. \end{align*} $$

The conclusion follows since the value of the right-hand side above is smaller than $6$ .

Proof of Theorem 1.3.

If $\max \{x_1,x_2\} \le 2$ and $k=0$ , then $x_n=2$ for all $n\ge 3$ , so the claimed (1.6) holds since $4^{X^{3^{k+1}}}\ge 4> x_n$ for all $n \in \mathbb {N}$ .

Suppose now that $\max \{x_1,x_2\} \ge 3$ or $k \ge 1$ , and note that ${X= \tfrac 12(3x_1+5x_2+7k)\ge 6}$ . Then, $\min \{x_3,x_4\}\ge 3$ ; and hence, $x_n$ is even for all $n\ge 5$ . In addition, since $\max \{x_1,\ldots ,x_6\}\le 2X$ and

$$ \begin{align*} 4^{X^{3^{k+1}}} \ge 4^X=2^{2X}>2X, \end{align*} $$

it follows that the claimed inequality holds for all $n\le 6$ .

Let $(p_n)_{n\ge 1}$ be the increasing enumeration of the primes greater than X. Since $\prod _{i=1}^n(1-{1}/{p_i})$ converges to zero as $n\to \infty $ by Proposition 2.3, one can find integers $1 =: r_0 < r_1 < \cdots < r_{k+1}$ such that

(2.5) $$ \begin{align} \prod_{i=r_j}^{r_{j+1} - 1} \bigg(1 - \frac{1}{p_i}\bigg) < \frac{1}{2} <\prod_{i=r_j}^{r_{j+1} - 2} \bigg(1 - \frac{1}{p_i}\bigg) \quad \text{for all }j \in \{0, 1, \ldots, k\}. \end{align} $$

Define also

$$ \begin{align*} q_j:=\prod_{i=r_j}^{r_{j+1} - 1}p_i \quad \text{for all }j \in \{0, 1, \ldots, k\}. \end{align*} $$

Since $\{q_0,q_1,\ldots ,q_k\}$ are pairwise coprime, the Chinese remainder theorem yields the existence of some $y \in \mathbb {N}$ such that $y\equiv j\bmod {q_j}$ for all $j \in \{0,1,\ldots ,k\}$ . In particular,

(2.6) $$ \begin{align} y \ge q_0 \ge p_{r_0}>X. \end{align} $$

Let us suppose for the sake of contradiction that $(x_n)_{n\ge 1}$ is not bounded. Then, there exists a minimal $v \in \mathbb {N}$ such that $x_v \ge 2y$ . Since $\max \{x_1,\ldots ,x_6\} \le 2X$ , it follows by (2.9) that $v\ge 7$ . In particular, $x_{v-1}$ and $x_{v-2}$ are even.

Claim 2.5. $\max \{\varphi (x_{v-1}),\varphi (x_{v-2})\} < y-k$ .

Proof. If $x_{v-1}<2y-2k$ then, by the fact that $x_{v-1}$ is even, $\varphi (x_{v-1}) \le \tfrac 12x_{v-1}<y-k$ . Otherwise, recalling that v is the minimal integer such that $x_v \ge 2y$ , then $2y-2k \le x_{v-1}<2y$ . In addition, $x_{v-1}$ is even, so $x_{v-1}=2y-2j$ for some $j \in \{1,\ldots ,k\}$ . It follows that

$$ \begin{align*} \varphi(x_{v-1})=x_{v-1} \, {\prod_{p\;\! \mid \;\! x_{v-1}}} \bigg(1 - \frac{1}{p}\bigg) = (y-j) {\prod_{p\;\! \mid \;\! x_{v-1}}}^{\!\!\prime} \bigg(1 - \frac{1}{p}\bigg), \end{align*} $$

where the last product is extended over the odd prime divisors of $x_{v-1}$ . Since, by construction, we have $y\equiv j\bmod {q_j}$ , we obtain by (2.8) that

$$ \begin{align*} \varphi(x_{v-1}) \le (y-j) \prod_{i=r_j}^{r_{j+1} - 1} \bigg(1 - \frac{1}{p_i}\bigg)< \frac{y-j}{2} < y-k. \end{align*} $$

Note that the last inequality holds because $y>2k$ , thanks to (2.9).

The same argument can be repeated for $x_{v-2}$ .

We conclude by Claim 2.5 that

$$ \begin{align*} 2y \le x_v = \varphi(x_{v-1}) + \varphi(x_{v-2}) + k < 2(y-k)+k \le 2y, \end{align*} $$

which is contradiction. It follows that $x_n<2y$ for all $n \in \mathbb {N}$ .

To complete the proof, it will be enough to show that $2y\le 4^{X^{3^{k+1}}}$ . For this, define $X_n:=X^{3^n}$ for all $n\ge 0$ and note that, thanks to Corollary 2.4, we have

$$ \begin{align*} \prod_{X_n<p\le X_n^3}\bigg(1-\frac{1}{p}\bigg)<\frac{1}{2} \quad \text{for all }n\ge 0. \end{align*} $$

By the definition of $r_j$ , it follows that

$$ \begin{align*} r_j \le X_j \quad \text{for all }j \in \{0,1,\ldots,k+1\}. \end{align*} $$

Lastly, since $\prod _{p\le x}p<4^x$ for all $x\ge 1$ (see, for example, [2, Lemma 2.8]), we conclude that

$$ \begin{align*} 2y \le 2\prod_{j=0}^k q_j \le \prod_{i=1}^{r_{k+1}}p_i \le \prod_{p\le X_{k+1}}p\le 4^{X_{k+1}}. \end{align*} $$

Therefore, $x_n \le 4^{X_{k+1}}$ for all $n \in \mathbb {N}$ .