$\boldsymbol {p}$-ADIC QUOTIENT SETS: LINEAR RECURRENCE SEQUENCES

Abstract

Let $(x_n)_{n\geq 0}$ be a linear recurrence of order $k\geq 2$ satisfying $x_n=a_1x_{n-1}+a_2x_{n-2}+\cdots +a_kx_{n-k}$ for all integers $n\geq k$, where $a_1,\ldots ,a_k,x_0,\ldots , x_{k-1}\in \mathbb {Z},$ with $a_k\neq 0$. Sanna [‘The quotient set of k-generalised Fibonacci numbers is dense in $\mathbb {Q}_p$’, Bull. Aust. Math. Soc. 96(1) (2017), 24–29] posed the question of classifying primes p for which the quotient set of $(x_n)_{n\geq 0}$ is dense in $\mathbb {Q}_p$. We find a sufficient condition for denseness of the quotient set of the kth-order linear recurrence $(x_n)_{n\geq 0}$ satisfying $ x_{n}=a_1x_{n-1}+a_2x_{n-2}+\cdots +a_kx_{n-k}$ for all integers $n\geq k$ with initial values $x_0=\cdots =x_{k-2}=0,x_{k-1}=1$, where $a_1,\ldots ,a_k\in \mathbb {Z}$ and $a_k=1$. We show that, given a prime p, there are infinitely many recurrence sequences of order $k\geq 2$ whose quotient sets are not dense in $\mathbb {Q}_p$. We also study the quotient sets of linear recurrence sequences with coefficients in certain arithmetic and geometric progressions.

1 Introduction and statement of results

For a set of integers A, the set $R(A)=\{a/b:a,b\in A, b\neq 0\}$ is called the ratio set or quotient set of A. Several authors have studied the denseness of ratio sets of different subsets of $\mathbb {N}$ in the positive real numbers (see [3, 5–7, 15, 16–20, 24, 25, 29, 30]). An analogous study has also been done for algebraic number fields (see [12, 28]).

For a prime p, let $\mathbb {Q}_p$ denote the field of p-adic numbers. The denseness of ratio sets in $\mathbb {Q}_p$ has been studied by several authors (see [1, 2, 10, 13, 14, 21–23, 27]). Let $(F_n)_{n\geq 0}$ be the sequence of Fibonacci numbers, defined by $F_0=0$ , $F_1=1$ and $F_n=F_{n-1}+F_{n-2}$ for all integers $n\geq 2$ . In [14], Garcia and Luca showed that the ratio set of Fibonacci numbers is dense in $\mathbb {Q}_p$ for all primes p. Later, Sanna [27, Theorem 1.2] showed that, for any $k\geq 2$ and any prime p, the ratio set of the k-generalised Fibonacci numbers is dense in $\mathbb {Q}_p$ . Sanna remarked that his result could be extended to other linear recurrences over the integers. However, he used some specific properties of the k-generalised Fibonacci numbers in the proof. Therefore, he asked the following question.

Question 1.1 [27, Question 1.3].

Let $(S_n)_{n\geq 0}$ be a linear recurrence of order $k\geq 2$ satisfying $S_n=a_1S_{n-1}+a_2S_{n-2}+\cdots +a_kS_{n-k}$ for all integers $n\geq k$ , where $a_1,\ldots ,a_k,S_0,\ldots ,S_{k-1}\in \mathbb {Z},$ with $a_k\neq 0.$ For which prime numbers p is the quotient set of $(S_n)_{n\geq 0}$ dense in $\mathbb {Q}_p?$

In [13], Garcia et al. studied the quotient sets of certain second-order recurrences: given two fixed integers r and s, let $(a_n)_{n\geq 0}$ be defined by $a_{n}=ra_{n-1}+sa_{n-2}$ for $n\geq 2$ with initial values $a_0=0$ and $a_1=1$ , and let $(b_n)_{n\geq 0}$ be defined by $b_{n}=rb_{n-1}+sb_{n-2}$ for $n\geq 2$ with initial values $b_0=2$ and $b_1=r$ .

Theorem 1.2 [13, Theorem 5.2].

With the notation as above, let $A=\{a_n: n\geq 0\}$ and $B=\{b_n: n\geq 0\}$ .

1. (a) If $p\mid s$ and $p\nmid r$ , then $R(A)$ is not dense in $\mathbb {Q}_p$ .

2. (b) If $p\nmid s$ , then $R(A)$ is dense in $\mathbb {Q}_p$ .

3. (c) For all odd primes p, $R(B)$ is dense in $\mathbb {Q}_p$ if and only if there exists a positive integer n such that $p\mid b_n$ .

We study ratio sets of some other linear recurrences over the set of integers. Our results give some answers to Question 1.1. Our first result gives a sufficient condition for the denseness of the ratio sets of certain kth-order recurrence sequences. Finding a general solution to Question 1.1 seems to be a difficult problem. Hence, in Theorem 1.3, we consider kth-order recurrence sequences for which $a_k=1$ and with initial values $x_0=\cdots =x_{k-2}=0$ , $x_{k-1}=1$ . Recall that a Pisot number is a positive algebraic integer greater than $1$ all of whose conjugate elements have absolute value less than $1$ .

Theorem 1.3. Let $(x_n)_{n\geq 0}$ be a $kth$ -order linear recurrence satisfying

$$ \begin{align*} x_{n}=a_1x_{n-1}+a_2x_{n-2}+\cdots+a_{k-1}x_{n-k+1}+x_{n-k} \end{align*} $$

for all integers $n\geq k$ with initial values $x_0=x_1=\cdots =x_{k-2}=0, x_{k-1}=1$ and $a_1,\ldots ,a_{k-1}\in \mathbb {Z}$ . Suppose that the characteristic polynomial of the recurrence sequence has a root $\pm \alpha $ , where $\alpha $ is a Pisot number. If p is a prime such that the characteristic polynomial of the recurrence sequence is irreducible in $\mathbb {Q}_p$ , then the quotient set of $(x_n)_{n\geq 0}$ is dense in $\mathbb {Q}_p$ .

If we take $k=3$ in Theorem 1.3, then we have the following corollary.

Corollary 1.4. Let $(x_n)_{n\geq 0}$ be a third-order linear recurrence satisfying

$$ \begin{align*} x_{n}=ax_{n-1}+bx_{n-2}+x_{n-3} \end{align*} $$

for all integers $n\geq 3$ with initial values $x_0=x_1=0, x_2=1$ , where the integers a and b are such that $(a+b)(b-a-2)<0$ . If p is a prime such that the characteristic polynomial of the recurrence sequence is irreducible in $\mathbb {Q}_p$ , then the quotient set of $(x_n)_{n\geq 0}$ is dense in $\mathbb {Q}_p$ .

We discuss two examples as applications of Corollary 1.4.

Example 1.5. For $a\in \mathbb {N}$ , let $\ell $ be an odd positive integer less than $2a$ . Let $(x_n)_{n\geq 0}$ be a linear recurrence satisfying $x_{n}=ax_{n-1}+(a-\ell )x_{n-2}+x_{n-3}$ for all integers $n\geq 3$ with initial values $x_0=x_1=0, x_2=1$ . Then a and $b:=a-\ell $ satisfy $(a+b)(b-a-2) <0$ . The characteristic polynomial $p(x)=x^3-ax^2-(a-\ell )x-1$ is irreducible in $\mathbb {Q}_2$ because $p(0)\neq 0$ and $p(1)=-2a+\ell \not \equiv 0\pmod {2}$ . Therefore, by Theorem 1.3, $R((x_n)_{n\geq 0})$ is dense in $\mathbb {Q}_2$ .

Example 1.6. For $a\in \mathbb {N}$ such that $3\nmid a$ , let $\ell $ be an odd positive integer less than $2a$ and such that $3\mid \ell $ . Let $(x_n)_{n\geq 0}$ be a linear recurrence satisfying $x_{n}=ax_{n-1}+(a-\ell )x_{n-2}+ x_{n-3}$ for all integers $n\geq 3$ with initial values $x_0=x_1=0, x_2=1$ . Then a and $b=a-\ell $ satisfy $(a+b)(b-a-2)<0$ . The characteristic polynomial $p(x)=x^3-ax^2-(a-\ell ) x-1$ is irreducible in $\mathbb {Q}_3$ because $p(0)\neq 0$ , $p(1)=-2a+\ell \not \equiv 0\pmod {3}$ and $p(2)=-6a+2\ell +7\not \equiv 0\pmod {3}$ . Therefore, by Theorem 1.3, $R((x_n)_{n\geq 0})$ is dense in $\mathbb {Q}_3$ .

Next, we consider recurrence sequences whose nth term depends on all the previous $n-1$ terms and obtain the following results.

Theorem 1.7. Let $(x_n)_{n\geq 0}$ be a linear recurrence satisfying

$$ \begin{align*} x_{n}=x_{n-1}+2x_{n-2}+\cdots+(n-1)x_1+nx_0 \end{align*} $$

for all integers $n\geq 1$ with initial value $x_0=1$ . Then the quotient set of $(x_n)_{n\geq 0}$ is dense in $\mathbb {Q}_p$ for all primes p.

The recurrence relation given in Theorem 1.7 generates a subsequence of the Fibonacci sequence.

Theorem 1.8. Let $(x_n)_{n\geq 0}$ be a linear recurrence satisfying

$$ \begin{align*} x_{n}=ax_{n-1}+arx_{n-2}+\cdots+ar^{n-1}x_0 \end{align*} $$

for all integers $n\geq 1$ , with $x_0, a,r\in \mathbb {Z}$ . Then the quotient set of $(x_n)_{n\geq 0}$ is not dense in $\mathbb {Q}_p$ for all primes p.

In Theorem 1.2, Garcia et al. studied second-order recurrence relations with specific initial values. In the following result, we consider a particular second-order recurrence sequence with arbitrary initial values $x_0$ and $x_1$ in the set of integers.

Theorem 1.9. Let $(x_n)_{n\geq 0}$ be a second-order linear recurrence satisfying $x_{n}=2ax_{n-1}-a^2x_{n-2}$ for all integers $n\geq 2$ , where $a,x_0,x_1\in \mathbb {Z}$ . Then the quotient set of $(x_n)_{n\geq 0}$ is dense in $\mathbb {Q}_p$ for all primes p satisfying $p\nmid a(x_1-ax_0)$ .

For a prime p, let $\nu _p$ denote the p-adic valuation. The following theorem gives a set of linear recurrence sequences of order k whose ratio sets are not dense in $\mathbb {Q}_p.$

Theorem 1.10. Let $(x_n)_{n\geq 0}$ be a linear recurrence of order $k\geq 2$ satisfying

$$ \begin{align*} x_{n}=a_1x_{n-1}+\cdots+a_kx_{n-k} \end{align*} $$

for all integers $n\geq k$ , where $x_0,\ldots ,x_{k-1}, a_1,\ldots ,a_k\in \mathbb {Z}$ . If p is a prime such that $p\nmid a_k$ and $\min \{\nu _p(a_j):1\leq j<k\}>\max \{\nu _p(x_m)-\nu _p(x_n):0\leq m,n<k\}$ , then the quotient set of $(x_n)_{n\geq 0}$ is not dense in $\mathbb {Q}_p$ .

The next example is an application of Theorem 1.10. Given a prime p, this example gives infinitely many recurrence sequences of order $k\geq 2$ whose quotient sets are not dense in $\mathbb {Q}_p$ .

Example 1.11. Let $(x_n)_{n\geq 0}$ be a linear recurrence of order $k\geq 2$ satisfying

$$ \begin{align*} x_{n}=a_1x_{n-1}+\cdots+a_kx_{n-k} \end{align*} $$

for all integers $n\geq k$ , where $x_0=x_1=\cdots =x_{k-1}=1$ and $a_1,\ldots ,a_k\in \mathbb {Z}$ . If p is a prime such that $p\mid a_j,1\leq j\leq k-1$ , and $p\nmid a_k$ , then by Theorem 1.10, the quotient set of $(x_n)_{n\geq 0}$ is not dense in $\mathbb {Q}_p$ .

2 Preliminaries

Let p be a prime and r be a nonzero rational number. Then r has a unique representation of the form $r= \pm p^k a/b$ , where $k\in \mathbb {Z}, a, b \in \mathbb {N}$ and $\gcd (a,p)= \gcd (p,b)=\gcd (a,b)=1$ . The p-adic valuation of r is $\nu _p(r)=k$ and its p-adic absolute value is $\|r\|_p=p^{-k}$ . By convention, $\nu _p(0)=\infty $ and $\|0\|_p=0$ . The p-adic metric on $\mathbb {Q}$ is $d(x,y)=\|x-y\|_p$ . The field $\mathbb {Q}_p$ of p-adic numbers is the completion of $\mathbb {Q}$ with respect to the p-adic metric. The p-adic absolute value can be extended to a finite normal extension field K over $\mathbb {Q}_p$ of degree n. For $\alpha \in K$ , define $\|\alpha \|_p$ as the nth root of the determinant of the matrix of the linear transformation from the vector space K over $\mathbb {Q}_p$ to itself defined by $x\mapsto \alpha x$ for all $x\in K$ . Also, define $\nu _p(\alpha )$ as the unique rational number satisfying $\|\alpha \|_p=p^{-\nu _p(\alpha )}$ .

The following results will be used in the proofs of our theorems.

Lemma 2.1 [13, Lemma 2.1].

If S is dense in $\mathbb {Q}_p$ , then for each finite value of the p-adic valuation, there is an element of S with that valuation.

Lemma 2.2 [13, Lemma 2.3].

Let $A\subset \mathbb {N}$ .

1. (1) If A is p-adically dense in $\mathbb {N}$ , then $R(A)$ is dense in $\mathbb {Q}_p$ .

2. (2) If $R(A)$ is p-adically dense in $\mathbb {N}$ , then $R(A)$ is dense in $\mathbb {Q}_p$ .

Theorem 2.3 [4, Theorem 1].

Let $\alpha _1,\ldots ,\alpha _n$ be units in $\Omega _p$ , the completion of the algebraic closure of $\mathbb {Q}_p,$ which are algebraic over the rationals $\mathbb {Q}$ and whose p-adic logarithms are linearly independent over $\mathbb {Q}$ . These logarithms are then linearly independent over the algebraic closure of $\mathbb {Q}$ in $\Omega _p$ .

3 Proof of the theorems

Proof of Theorem 1.3.

Let $p(x)=x^k-a_1x^{k-1}-a_2x^{k-2}-\cdots -a_{k-1}x-1$ be the characteristic polynomial of the recurrence. Let $\alpha _1,\ldots ,\alpha _{k}$ be the k distinct roots of the characteristic polynomial in its splitting field, say, K over $\mathbb {Q}_p$ . The generating function of the sequence is

$$ \begin{align*} t(x)=\frac{x^{k-1}}{1-a_1x-a_2x^2-\cdots-x^k}=\sum_{i=1}^k\frac{1}{q(\alpha_i)}\sum_{n=0}^{\infty}\alpha_i^nx^n, \end{align*} $$

where $q(x):=p'(x)$ , the derivative of the polynomial $p(x)$ . Hence, the nth term of the sequence is given by

$$ \begin{align*} x_n=\sum_{i=1}^{k}\frac{1}{q(\alpha_i)}\alpha_i^n,\quad n\geq0. \end{align*} $$

Since $p(0)=-1$ , the roots of $p(x)$ are units in the ring formed by elements in K with p-adic absolute value less than or equal to $1$ . Following Sanna’s proof of [27, Theorem 1.2], we can choose an even $t\in \mathbb {N}$ such that the function

$$ \begin{align*} G(z):=\sum_{i=1}^{k}\frac{1}{q(\alpha_i)}\exp_p(z\log_p(\alpha_i^t)) \end{align*} $$

is analytic over $\mathbb {Z}_p$ and the Taylor series of $G(z)$ around $0$ converges for all $z\in \mathbb {Z}_p$ . Also, note that $x_{nt}=G(n)$ for $n\geq 0$ .

We now use a variant of the following lemma which gives the multiplicative independence of any $k-1$ roots among the k roots $\alpha _1,\ldots ,\alpha _k$ of the characteristic polynomial $x^k-x^{k-1}-\cdots -x-1$ of the k-generalised Fibonacci sequence in the field of complex numbers.

Lemma 3.1 [11, Lemma 1].

With the notation above, each set of $k-1$ different roots $\alpha _1,\ldots ,\alpha _{k-1}$ is multiplicatively independent, that is, $\alpha _1^{e_1}\cdots \alpha _{k-1}^{e_{k-1}}=1$ for some integers $e_1,\ldots ,e_{k-1}$ if and only if $e_1=\cdots =e_{k-1}=0$ .

Let $\sigma(\alpha_1) = \pm \alpha$ , where $\alpha$ is a Pisot number with absolute value greater than 1, the other roots, $\sigma (\alpha _2),\ldots ,\sigma (\alpha _k)$ , having absolute values less than $1$ , where $\sigma $ is an isomorphism from $\mathbb {Q}(\alpha _1,\ldots ,\alpha _k)$ to the splitting field of $p(x)$ over $\mathbb {Q}$ in the field of complex numbers. Therefore, the proof of Lemma 3.1 holds true for the roots of $p(x)$ , which are $\sigma (\alpha _1),\ldots , \sigma (\alpha _k)$ , since $\log |\sigma (\alpha _1)|$ is positive and $\log |\sigma (\alpha _2)|,\ldots ,\log |\sigma (\alpha _k)|$ are negative. Hence, $\sigma (\alpha _1),\ldots ,\sigma (\alpha _{k-1}$ ) are multiplicatively independent, implying that $\alpha _1^t,\ldots , \alpha _{k-1}^t$ are multiplicatively independent. Thus, $\log _p(\alpha _1^t),\ldots ,\log _p(\alpha _{k-1}^t)$ are linearly independent over $\mathbb {Z}$ and hence linearly independent over the algebraic numbers by Theorem 2.3.

Suppose $G'(0)=\sum _{i=0}^{k}({1}/{q(\alpha _i)})\log _p(\alpha _i^t)=0$ . Since $\log _p(\alpha _k^t)=-\log _p(\alpha _1^t)-\cdots -\log _p(\alpha _k^t)$ as the product of the roots is $-1$ and t is even, we obtain

$$ \begin{align*} \sum_{i=1}^{k-1} \bigg(\frac{1}{q(\alpha_i)}-\frac{1}{q(\alpha_k)}\bigg)\log_p(\alpha_i^t)=0. \end{align*} $$

By linear independence of $\log _p(\alpha _1^t),\ldots ,\log _p(\alpha _{k-1}^t)$ , we have ${1}/{q(\alpha _1)}=\cdots = {1}/{q(\alpha _k)}=c$ , for some p-adic number c. This gives k distinct roots $\alpha _1,\ldots ,\alpha _k$ of the ( $k-1$ )-degree polynomial $q(x)-{1}/{c}$ , which is not possible.Therefore, $G'(0)\neq 0$ . Since

$$ \begin{align*} G(z)=\sum_{j=0}^\infty\frac{G^{(\,j)}(0)}{j!}z^{\,j} \end{align*} $$

converges at $z=1$ , it follows that $\|{G^{(\,j)}(0)}/{j!}\|_p\rightarrow 0$ . Hence, there exists an integer $\ell $ such that $\nu _p(G^{(\,j)}(0)/j!)\geq -\ell $ for all j. Thus, we obtain $G(mp^h)=G'(0)mp^h+d$ where $\nu _p(d)\geq 2h-\ell $ for all $m,h\geq 0$ . Also, $G(0)=0$ for $h>h_0:=\ell +\nu _p(G'(0))$ and hence

$$ \begin{align*} \nu_p\bigg( \frac{G(mp^h)}{G(p^h)}-m\bigg)\geq h-h_0. \end{align*} $$

This yields

$$ \begin{align*} \lim_{h\rightarrow\infty}\bigg\lVert\frac{G(mp^h)}{G(p^h)}-m\bigg\rVert_p=0, \end{align*} $$

and hence $R(G(n)_{n\geq 0})$ is p-adically dense in $\mathbb {N}$ . Since $x_{nt}=G(n),n\geq 0$ , we find that $R((x_n)_{n\geq 0})$ is also p-adically dense in $\mathbb {N}$ . Therefore, by Lemma 2.2, $R((x_n)_{n\geq 0})$ is dense in $\mathbb {Q}_p$ .

Proof of Corollary 1.4.

Since $p(1)p(-1){\kern-1pt}={\kern-1pt}(-a{\kern-1pt}-{\kern-1pt}b)(b{\kern-1pt}-{\kern-1pt}a{\kern-1pt}-{\kern-1pt}2){\kern-1pt}>{\kern-1pt}0$ and $p(0){\kern-1pt}=-1$ , by continuity of the polynomial function in $\mathbb {R}$ , $p(x)$ has one real root with absolute value greater than 1 and two other roots with absolute values less than $1$ . Hence, the characteristic polynomial has a root $\pm \alpha$ , where $\alpha$ is a Pisot number, and the corollary follows from Theorem 1.3.

We need the following result to prove Theorem 1.7.

Corollary 3.2 [9, Corollary 2.2].

The linear recurrence relation $x_{n+1}=x_n+2x_{n-1}+\cdots +nx_1+(n+1)x_0,n\geq 0$ , with the initial data $x_0=1$ has the solution

$$ \begin{align*}x_n=\frac{1}{\sqrt{5}}\bigg({\bigg(\frac{3+\sqrt{5}}{2}\bigg)}^n-{\bigg(\frac{3-\sqrt{5}}{2}\bigg)}^n\bigg),\quad n\geq 1.\end{align*} $$

Proof of Theorem 1.7.

By Corollary 3.2, for $n\geq 1$ ,

$$ \begin{align*} x_n=\frac{1}{\sqrt{5}}\bigg({\bigg(\frac{3+\sqrt{5}}{2}\bigg)}^n-{\bigg(\frac{3-\sqrt{5}}{2}\bigg)}^n\bigg) =\frac{\alpha^{2n}-\beta^{2n}}{\sqrt{5}} =F_{2n}, \end{align*} $$

where $\alpha ={(1+\sqrt {5})}/{2}, \beta ={(1-\sqrt {5})}/{2}$ and $F_n$ denotes the nth Fibonacci number which is obtained by the Binet formula. From [14], the ratio set of the Fibonacci numbers is dense in $\mathbb {Q}_p$ for all primes p. Therefore, by Lemma 2.1, $\nu _p(F_n)$ is not bounded. Hence, for any $j\in \mathbb {N}$ , there exists $F_m$ such that $\nu _p(F_m)\geq j$ , that is, ${(\alpha ^{m}-\beta ^{m})}/{\sqrt {5}}\equiv 0 \pmod {p^{\,j}}$ which gives $\alpha ^m\equiv \beta ^m\pmod {p^{\,j}}$ . This yields

$$ \begin{align*} \alpha^{2mp^{\,j-1}(p-1)}=(\alpha^m\alpha^m)^{p^{\,j-1}(p-1)}\equiv(\alpha^m\beta^m)^{p^{\,j-1}(p-1)}\pmod{p^{\,j}}. \end{align*} $$

Since $\alpha \beta =-1$ , by using Euler’s theorem, we find that

$$ \begin{align*} \alpha^{2mp^{\,j-1}(p-1)}\equiv(\alpha^m\beta^m)^{p^{\,j-1}(p-1)}\equiv1\pmod{p^{\,j}}. \end{align*} $$

This gives $\alpha ^{2k}\equiv \beta ^{2k}\equiv 1\pmod {p^{\,j}}$ , where $k=mp^{\,j-1}(p-1)$ . Hence,

$$ \begin{align*} \frac{x_{kn}}{x_{k}}=\frac{F_{2kn}}{F_{2k}}=\frac{(\alpha^{2k})^n-(\beta^{2k})^n}{\alpha^{2k}-\beta^{2k}}=(\alpha^{2k})^{(n-1)}+(\alpha^{2k})^{n-2}\beta^{2k}+\cdots+(\beta^{2k})^{n-1}, \end{align*} $$

which is congruent to n modulo $p^{\,j}$ . Since, for $n\in \mathbb {N}$ , there exists $k\in \mathbb {N}$ such that $\|{x_{kn}}/{x_k}-n\|_p\leq p^{-j}$ , $R((x_n)_{n\geq 0})$ is p-adically dense in $\mathbb {N}$ . Therefore, by Lemma 2.2, $R((x_n)_{n\geq 0})$ is dense in $\mathbb {Q}_p$ .

We need the following results to prove Theorem 1.8.

Theorem 3.3 [9, Theorem 3.1].

The numbers $x_n$ are solutions of the linear recurrence relation with constant coefficients in geometric progression $x_{n+1}=ax_n+aqx_{n-1}+\cdots +aq^{n-1}x_1+aq^nx_0,n\geq 0$ , with initial data $x_0$ , if and only if they form the geometric progression given by the formula $x_n=ax_0(a+q)^{n-1},n\geq 1$ .

Lemma 3.4 [13, Lemma 2.2].

If A is a geometric progression in $\mathbb {Z}$ , then $R(A)$ is not dense in any $\mathbb {Q}_p$ .

Proof of Theorem 1.8.

By Theorem 3.3, $(x_n)_{n\geq 1}$ forms a geometric progression whose nth term is $ax_0(a+r)^{n-1}$ for $n\geq 1$ . Hence, by Lemma 3.4, $R((x_n)_{n\geq 0})$ is not dense in $\mathbb {Q}_p$ for any prime p.

To prove Theorem 1.9 we need some results on the uniform distribution of sequences of integers. Recall that a sequence $(x_n)_{n\geq 0}$ is said to be uniformly distributed modulo m if each residue occurs equally often, that is,

$$ \begin{align*}\lim_{N\rightarrow\infty}\frac{\# \{n\leq N \mid x _n\equiv t \pmod{m}\}}{N}=\frac{1}{m} \quad\mbox{for all } t\in\mathbb{Z}.\end{align*} $$

Proposition 3.5 [8, Proposition 1].

Suppose $(G_n)_{n\ge 0}$ is the sequence of integers determined by the recurrence relation $G_{n+1}=AG_n-BG_{n-1}$ with initial values $G_0,G_1$ where $A,B,G_0,G_1\in \mathbb {Z}$ . If $A=2a,B=a^2$ , then $(G_n)_{n\ge 0}$ is uniformly distributed modulo a prime p if and only if $p\nmid a(G_1-aG_0)$ .

Theorem 3.6 [8, Theorem].

Suppose $(G_n)_{n\ge 0}$ is the sequence of integers determined by the recurrence relation $G_{n+1}=AG_n-BG_{n-1}$ with initial values $G_0,G_1$ where $A,B,G_0,G_1\in \mathbb {Z}$ . If $(G_n)_{n\ge 0}$ is uniformly distributed modulo p, then $(G_n)_{n\ge 0}$ is uniformly distributed modulo $p^h$ with $h>1$ if and only if

1. (1) $p>3$ ; or

2. (2) $p=3$ and $A^2\not \equiv B\pmod {9}$ ; or

3. (3) $p=2, A\equiv 2\pmod {4},B\equiv 1\pmod {4}$ .

Proof of Theorem 1.9.

Let p be a prime. The given recurrence sequence $(x_n)_{n\geq 0}$ satisfies the hypotheses of Proposition 3.5, and hence $(x_n)_{n\geq 0}$ is uniformly distributed modulo p. If $p>3$ , then by Theorem 3.6(1), $(x_n)_{n\geq 0}$ is uniformly distributed modulo $p^k$ with $k>1$ , that is,

$$ \begin{align*}\lim_{N\rightarrow\infty}\frac{\# \{n\leq N\mid x _n\equiv t \pmod{p^k}\}}{N}=\frac{1}{p^k}>0.\end{align*} $$

Therefore, for all $t\in \mathbb {N}$ and for all $k>1$ , there exists $x_n$ such that $\| x_n-t\|_p\leq p^{-k}$ . Hence, $R((x_n)_{n\geq 0})$ is p-adically dense in $\mathbb {N}$ . Therefore, by Lemma 2.2, $R((x_n)_{n\geq 0})$ is dense in $\mathbb {Q}_p$ .

We next consider the remaining primes $p=2,3$ . Since $p\nmid a(x_1-ax_0)$ , we have $p\nmid a$ . It is easy to check that $p=3$ satisfies the condition given in Theorem 3.6(2) and $p=2$ satisfies the condition given in Theorem 3.6(3). The rest of the proof follows similarly as shown in the case $p>3$ . This completes the proof of the theorem.

We need the following lemma to prove Theorem 1.10.

Lemma 3.7 [26, Lemma 3.3].

Let $(r_n)_{n\geq 0}$ be a linearly recurring sequence of order $k\geq 2$ given by $r_n=a_1r_{n-1}+\cdots +a_k r_{n-k}$ for each integer $n\geq k$ , where $r_0,\ldots ,r_{k-1}$ and $a_1,\ldots ,a_k$ are all integers. Suppose that there exists a prime number p such that $p\nmid a_k$ and $\min \{\nu _p(a_j):1\leq j<k\}>\max \{\nu _p(r_m)-\nu _p(r_n):0\leq m,n<k\}$ . Then $\nu _p(r_n)=\nu _p(r_{n\pmod {k}})$ for each nonnegative integer n.

Proof of Theorem 1.10.

By Lemma 3.7,

$$ \begin{align*} \nu_p(x_n/x_m)=\nu_p(x_{n\pmod{k}})-\nu_p(x_{m\pmod{k}})\leq M \end{align*} $$

for all $n,m\in \mathbb {N}\cup \{0\}$ , where $M=\max \{\nu _p(x_i):i=0,1,\ldots ,k-1\}$ . Therefore, by Lemma 2.1, $R((x_n)_{n\geq 0})$ is not dense in $\mathbb {Q}_p$ .