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. Many authors have studied the denseness of ratio sets of different subsets of
$\mathbb {N}$
in the positive real numbers. See, for example, [Reference Brown, Dairyko, Garcia, Lutz and Someck4–Reference Bukor and Tóth7, Reference Garcia, Poore, Selhorst-Jones and Simon12, Reference Hedman and Rose14–Reference Mišík18, Reference Šalát24, Reference Šalát25, Reference Starni28, Reference Strauch and Tóth29]. An analogous study has also been done for algebraic number fields, see for example [Reference Garcia9, Reference Sittinger27].
For a prime p, let
$\mathbb {Q}_p$
denote the field of p-adic numbers. In recent years, the denseness of ratio sets in
$\mathbb {Q}_p$
have been studied by several authors, see for example [Reference Antony and Barman1, Reference Antony, Barman and Miska3, Reference Donnay, Garcia and Rouse8, Reference Garcia, Hong, Luca, Pinsker, Sanna, Schechter and Starr10, Reference Garcia and Luca11, Reference Miska19–Reference Miska and Sanna21, Reference Sanna26]. 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 [Reference Garcia and Luca11], Garcia and Luca showed that the ratio set of Fibonacci numbers is dense in
$\mathbb {Q}_p$
for all primes p. Later, Sanna [Reference Sanna26, Theorem 1.2] showed that, for any
$k\geq 2$
and any prime p, the ratio set of the k-generalized Fibonacci numbers is dense in
$\mathbb {Q}_p$
and made the following open question.
Question 1.1 [Reference Sanna26, Question 1.3]
Let
$(S_n)_{n\geq 0}$
be a linear recurrence sequence of order
$k\geq 2$
satisfying
$$ \begin{align*} S_n=a_1S_{n-1}+a_2S_{n-2}+\dots+a_kS_{n-k}, \end{align*} $$
for all integers
$n\geq k$
, where
$a_1,\dots ,a_k,S_0,\dots ,S_{k-1}\in \mathbb {Z}$
, 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 [Reference Garcia, Hong, Luca, Pinsker, Sanna, Schechter and Starr10], Garcia et al. solved the problem partially for second-order recurrences. Later, in [Reference Antony and Barman2], we considered kth-order recurrence sequences for which
$a_k=1$
and initial values
$S_0=\cdots =S_{k-2}=0$
,
$S_{k-1}=1$
. We showed that if the characteristic polynomial of the recurrence sequence has a root
$\pm \alpha $
, where
$\alpha $
is a Pisot number and 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
$\mathbb {Q}_p$
. In this article, our objective is to study the denseness of quotient sets of linear recurrence sequences whose characteristic polynomials are reducible over
$\mathbb {Q}$
. Also, we extend [Reference Antony and Barman2, Theorem 1.9], which gives condition for the denseness of ratio sets of second order linear recurrence sequences
$(x_n)_{n\geq 0}$
whose characteristic polynomials are of the form
$(x-a)^2$
, to kth order linear recurrence sequences with characteristic polynomials of the form
$(x-a)^k$
in the case when the initial values are given as
$x_0=x_1=\dots =x_{k-2}=0, x_{k-1}=1$
.
In our first theorem, we consider recurrence sequences having characteristic polynomials whose roots are all distinct.
Theorem 1.2 Let
$(x_n)_{n\geq 0}$
be a linear recurrence of order
$k\geq 2$
satisfying
$$ \begin{align*} x_n=b_1x_{n-1}+b_2x_{n-2}+\dots+b_kx_{n-k}, \end{align*} $$
for all integers
$n\geq k$
, where
$b_1,\dots ,b_k,x_0,\dots ,x_{k-1}\in \mathbb {Z}$
, with
$b_k\neq 0$
and
$x_0,x_1,\dots , x_{k-1}$
not all zeros. Suppose that the characteristic polynomial of
$(x_n)_{n\geq 0}$
is given by
$$ \begin{align*}(x-a_1)(x-a_2)\dots (x-a_k),\end{align*} $$
where
$a_i\in \mathbb {Z}$
,
$ a_i\neq a_j $
for
$1\leq i\neq j\leq k$
, and
$\gcd (a_i, a_j)=1$
for all
$i\neq j$
. Let p be a prime such that
$p\nmid a_1a_2\cdots a_k$
. If
$x_0=0$
, then the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
.
Example 1.3 Suppose that
$p_1,p_2$
, and
$p_3$
are distinct primes. Let
$(x_n)_{n\geq 0}$
be a linear recurrence sequence defined by the recurrence relation
$$ \begin{align*}x_{n}=(p_1+p_2+p_3)x_{n-1}-(p_1p_2+p_1p_3+p_2p_3)x_{n-2}+(p_1p_2p_3)x_{n-3}\end{align*} $$
for
$n\geq 3$
, where
$x_0=0$
, and
$x_1$
and
$x_2$
are any integers not both zero. The characteristic polynomial is equal to
$(x-p_1)(x-p_2)(x-p_3)$
. Hence, by Theorem 1.2, the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
for all primes
$p\neq p_1,p_2, p_3$
.
In the following theorem, we consider kth order linear recurrence sequences whose characteristic polynomials have exactly two equal roots.
Theorem 1.4 Let
$(x_n)_{n\geq 0}$
be a linear recurrence of order
$k\geq 3$
satisfying
$$ \begin{align*} x_n=b_1x_{n-1}+b_2x_{n-2}+\dots+b_kx_{n-k}, \end{align*} $$
for all integers
$n\geq k$
, where
$b_1,\dots ,b_k,x_0,\dots ,x_{k-1}\in \mathbb {Z}$
, with
$b_k\neq 0$
. Suppose that the characteristic polynomial of
$(x_n)_{n\geq 0}$
is given by
$$ \begin{align*}(x-a_1)^2(x-a_2)(x-a_3)\dots (x-a_{k-1}),\end{align*} $$
where
$a_i\in \mathbb {Z}, a_i\neq a_j $
for
$1\leq i\neq j\leq k-1$
, and
$x_0=x_1=\dots =x_{k-2}=0,x_{k-1}=1$
. Let p be a prime such that
$p\nmid a_1a_2\cdots a_{k-1}$
. If
$a_i\not \equiv a_j\ \pmod {p}$
for all
$i\neq j$
, then the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
.
Example 1.5 Given an integer a, let
$(x_n)_{n\geq 0}$
be a linear recurrence sequence defined by the recurrence relation
$$ \begin{align*}x_{n}=4ax_{n-1}-5a^2x_{n-2}+2a^3x_{n-3}\end{align*} $$
for
$n\geq 3$
, where
$x_0=x_1=0$
and
$x_2=1$
. The characteristic polynomial is equal to
${(x-a)^2(x-2a)}$
. By Theorem 1.4, the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
for all primes
$p\nmid 2a$
.
Theorem 1.6 Let
$(x_n)_{n\geq 0}$
be a linear recurrence of order
$k\geq 2$
satisfying
$$ \begin{align*} x_n=b_1x_{n-1}+b_2x_{n-2}+\dots+b_kx_{n-k}, \end{align*} $$
for all integers
$n\geq k$
, where
$b_1,\dots ,b_k,x_0,\dots ,x_{k-1}\in \mathbb {Z}$
, with
$b_k\neq 0$
. Suppose that the characteristic polynomial of
$(x_n)_{n\geq 0}$
is given by
$(x-a)^k$
, where
$a\in \mathbb {Z}$
, and
$x_0=x_1=\dots =x_{k-2}=0,x_{k-1}=1$
. If p is a prime such that
$p\nmid a$
, then the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
.
Remark 1.7 Let
$a\in \mathbb {Z}$
. Consider the kth order linear recurrence sequence
$(x_n)_{n\geq 0}$
generated by the recurrence relation
$$ \begin{align*}x_{n}=\binom{k}{1}ax_{n-1}-\binom{k}{2}a^2x_{n-2}+\dots+(-1)^{k-1}\binom{k}{k}a^kx_{n-k}\end{align*} $$
for
$n\geq k$
, where
$x_0=\dots =x_{k-2}=0,x_{k-1}=1$
. Then, the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
for all primes p not dividing a. This generalizes [Reference Antony and Barman2, Theorem 1.9] for the case
$k=2$
.
Note that a linear recurrence sequence generated by a relation of the above form may not always have a dense quotient set in
$\mathbb {Q}_p$
. For example, consider the pth order linear recurrence sequence
$(x_n)$
generated by the recurrence relation
$$ \begin{align*}x_{n}=\binom{p}{1}ax_{n-1}-\binom{p}{2}a^2x_{n-2}+\dots+(-1)^{p-1}\binom{p}{p}a^px_{n-p}\end{align*} $$
for
$n\geq p$
, where the initial values
$x_0,\dots ,x_{p-1}\in \mathbb {Z}\backslash \{0\}$
have the same p-adic valuation. Then, the quotient set of
$(x_n)$
is not dense in
$\mathbb {Q}_p$
which follows from [Reference Antony and Barman2, Theorem 1.10].
In case of third- order recurrence sequences, we prove the following result where we do not need to fix all the initial values.
Theorem 1.8 Let
$(x_n)_{n\geq 0}$
be a third -order linear recurrence sequence given by
$$ \begin{align*} x_n=b_1x_{n-1}+b_2x_{n-2}+b_3x_{n-3}, \end{align*} $$
for all integers
$n\geq 3$
, where
$b_1,b_2, b_3,x_0,x_1, x_2\in \mathbb {Z}$
, with
$b_3\neq 0$
. Suppose that the characteristic polynomial of
$(x_n)_{n\geq 0}$
is given by
$(x-a)(x-b)(x-c)$
, where
$a,b,c\,{\in}\, \mathbb {Z}$
. Let p be a prime such that
$p\nmid abc$
. Then, the following hold.
-
(a) Suppose that
$a=b=c$
. If
$p|x_0 $
and
$p\nmid 4ax_1-x_2-3a^2x_0$
, then the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
. Moreover, if
$x_0=0,$
then the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
if and only if
$4ax_1\neq x_2$
. -
(b) Suppose that
$a=c\neq b$
. If
$p|x_0$
and
$p\nmid \left (a-b\right )\left (x_2-x_1(a+b)+x_0ab\right )$
, then the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
.
Example 1.9 Let
$a\in \mathbb {Z}$
be such that
$p\nmid a$
, and let
$(x_n)_{n\geq 0}$
be a linear recurrence sequence defined by the recurrence relation
$$ \begin{align*}x_{n+1}=3ax_n-3a^2x_{n-1}+a^3x_{n-2}\end{align*} $$
for
$n\geq 2$
, where
$x_0=0$
, and
$x_1$
and
$x_2$
are any integers satisfying
$\gcd (4a,x_2)=1$
. Then, by Theorem 1.8(a), the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
.
2 Preliminaries
Let r be a nonzero rational number. Given a prime number p, 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 defined as
$\nu _p(r)=k$
and its p-adic absolute value is defined as
$\|r\|_p=p^{-k}$
. 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
$\mathbb {K}$
over
$\mathbb {Q}_p$
of degree n. For
$\alpha \in \mathbb {K}$
, define
$\|\alpha \|_p$
as the nth root of the determinant of the matrix of linear transformation from the vector space
$\mathbb {K}$
over
$\mathbb {Q}_p$
to itself defined by
$x\mapsto \alpha x$
for all
$x\in \mathbb {K}$
. Also,
$\nu _p(\alpha )$
is the unique rational number satisfying
$\|\alpha \|_p=p^{-\nu _p(\alpha )}$
. The ring of integers of
$\mathbb {K}$
, denoted by
$\mathcal {O}$
, is defined as the set of all elements in
$\mathbb {K}$
with p-adic absolute value less than or equal to one. A function
$f: \mathcal {O}\rightarrow \mathcal {O}$
is called analytic if there exists a sequence
$(a_n)_{n\geq 0}$
in
$\mathcal {O}$
such that
$$ \begin{align*}f(z)=\sum_{n=0}^{\infty}a_nz^n\end{align*} $$
for all
$z\in \mathcal {O}$
.
We recall definitions of p-adic exponential and logarithmic function. For
$a\in \mathbb {K}$
and
$r>0$
, we denote
$\mathcal {D}(a,r):=\{z\in \mathbb {K}: \|z-a\|_p<r\}$
. Let
$\rho =p^{-1/(p-1)}$
.
For
$z\in \mathcal {D}(0,\rho )$
, the p-adic exponential function is defined as
$$ \begin{align*}\exp_p(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}.\end{align*} $$
The derivative is given by
$\exp _p'(z)=\exp _p(z)$
. For
$\mathcal {D}(1,\rho )$
, the p-adic logarithmic function is defined as
$$ \begin{align*}\log_p(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(z-1)^n}{n}.\end{align*} $$
For
$z\in \mathcal {D}(1,\rho )$
, we have
$\exp _p(\log _p(z))=z$
. If
$\mathbb {K}$
is unramified and
$p\neq 2$
, then
$ \mathcal {D}(0,\rho )= \mathcal {D}(0,1)$
and
$\mathcal {D}(1,\rho )= \mathcal {D}(1,1)$
. More properties of these functions can be found in [Reference Gouvêa13].
Next, we state a result for analytic functions which will be used in the proofs of our theorems.
Theorem 2.1 [Reference Gouvêa13, Hensel’s lemma]
Let
$f: \mathcal {O}\rightarrow \mathcal {O}$
be analytic. Let
$b_0\in \mathcal {O}$
be such that
$\|f(b_0)\|_p<1$
and
$\|f'(b_0)\|_p=1$
. Then there exists a unique
$b\in \mathcal {O}$
such that
$f(b)=0$
and
$\|b-b_0\|_p<\|f(b_0)\|_p$
.
Note that in [Reference Gouvêa13], Gouvêa states Hensel’s lemma for polynomials with coefficients in
$\mathcal {O}$
. However, Hensel’s lemma is also true and follows similarly for functions given by power series with coefficients in the ring
$\mathcal {O}$
. We will only be considering
$\mathbb {K}=\mathbb {Q}_p$
throughout this article. The following results are useful in proving denseness of quotient sets.
Theorem 2.2 [Reference Miska, Murru and Sanna20, Corollary 1.3]
Let
$f\colon \mathbb {Z}_p\rightarrow \mathbb {Q}_p$
be an analytic function with a simple zero in
$\mathbb {Z}_p$
. Then,
$R(f(\mathbb {N}))$
is dense in
$\mathbb {Q}_p$
.
Lemma 2.3 [Reference Garcia, Hong, Luca, Pinsker, Sanna, Schechter and Starr10, 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.
3 Proof of the theorems
In the proofs, we will use certain representation of the nth term of linear recurrence sequence in terms of the roots of the characteristic polynomial. More details on such representations can be found in [Reference Rosen23].
Proof of Theorem 1.2
For
$n\geq 0$
, the nth term of the sequence
$(x_n)$
is given by
$$ \begin{align*}x_n=c_0a_1^n+c_1a_2^n+\dots+c_{k-1}a_k^n,\end{align*} $$
where
$$\begin{align*}C=\begin{bmatrix} c_0&c_1&\dots & c_{k-1} \end{bmatrix}^{t}\end{align*}$$
is given by
$C=\frac {1}{\det (A)}\text {adj}(A)\cdot X_0$
, where
$$\begin{align*}X_0= \begin{bmatrix} x_0\\ x_1\\ \vdots \\ x_{k-1} \end{bmatrix}, A = \begin{bmatrix} 1&1&\dots &1\\ a_1 &a_2 &\dots &a_{k}\\ a_1^2&a_2^2 &\dots &a_k^2\\ \vdots &\vdots &\ddots &\vdots\\ a_1^{k-1}&a_2^{k-1} &\dots &a_k^{k-1} \end{bmatrix}.\end{align*}$$
We define a function f as
$$ \begin{align*}f(z):=\det(A)\left[c_0\exp_p{(z\log_p{a_1^{p-1}})}+\dots+c_{k-1}\exp_p{(z\log_p{a_k^{p-1}})}\right].\end{align*} $$
Since
$p\nmid a_1a_2\dots a_k$
, f is defined for all
$z\in \mathbb {Z}_p$
and
$f(n)=\det (A)x_{n(p-1)}$
for all
${n\in \mathbb {Z}_{\geq 0}}$
. Moreover,
$\mathbb {Z}_{\geq 0}$
is dense in
$\mathbb {Z}_p$
. Therefore, f is an analytic function from
$\mathbb {Z}_p$
to
$\mathbb {Z}_p$
. We have,
$$ \begin{align*}f(0)=\det(A)(c_0+c_1+\dots+c_{k-1})=\det(A)x_0=0\end{align*} $$
and
$$ \begin{align*}f'(0)=\det(A)(c_0\log_p{a_1^{p-1}}+c_1\log_p{a_2^{p-1}}+\dots+c_{k-1}\log_p{a_{k}^{p-1}}).\end{align*} $$
Suppose that
$f'(0)=0$
. Since
$\gcd (a_i,a_j)=1$
for all
$i\neq j$
, therefore,
$a_1^{p-1},\dots ,a_k^{p-1}$
are multiplicatively independent i.e,
$(a_1^{p-1})^{u_1}(a_2^{p-1})^{u_2}\dots (a_k^{p-1})^{u_k}=1$
for some integers
$u_1,u_2,\dots ,u_k$
only if
$u_1=u_2=\dots =u_k=0$
. Hence,
$$ \begin{align*}\log_p{a_1^{p-1}},\log_p{a_2^{p-1}},\dots,\log_p{a_{k}^{p-1}}\end{align*} $$
are linearly independent over
$\mathbb {Z}$
. Thus, if
$f'(0)=0$
then
$c_0=c_1=\dots =c_{k-1}=0$
which is not possible. Hence,
$f'(0)$
is nonzero. Therefore,
$0$
is a simple zero of f in
$\mathbb {Z}_p$
. By Theorem 2.2,
$R(f(\mathbb {N}))=R((x_{n(p-1)}))$
is dense in
$\mathbb {Q}_p$
. Hence, the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
.
Proof of Theorem 1.4
The nth term of the sequence is given by
$$ \begin{align*} x_n&=a_1^n(c_0+c_1n)+c_2a_2^n+c_3a_3^n+\dots+c_{k-1}a_{k-1}^n\\ &=a_1^n(c_0+c_1n+c_2(a_2a_1^{-1})^n+c_3(a_3a_1^{-1})^n+\dots+c_{k-1}(a_{k-1}a_1^{-1})^n), \end{align*} $$
where
$$\begin{align*}C=\begin{bmatrix} c_0&c_1&\dots &c_{k-1} \end{bmatrix}^{t}\end{align*}$$
is given by
$C=\frac {1}{\det (A)}\text {adj}(A)\cdot X_0$
, where
$$\begin{align*}X_0= \begin{bmatrix} 0\\ 0\\ \vdots \\ 0\\ 1 \end{bmatrix}, A = \begin{bmatrix} 1&0&1&\dots&1\\ a_1 & a_1 & a_2 & \dots & a_{k-1}\\ a_1^2 & 2a_1^2 & a_2^2 & \dots & a_{k-1}^2\\ \vdots& \vdots & \vdots & \ddots & \vdots\\ a_1^{k-1} & (k-1)a_1^{k-1} & a_2^{k-1} & \dots & a_{k-1}^{k-1} \end{bmatrix}.\end{align*}$$
We define an analytic function
$f:\mathbb {Z}_p\rightarrow \mathbb {Z}_p$
as
$$ \begin{align*} f(z)&:=\det(A)\exp_p{(z\log_p{(a_1)^{p-1}})}(c_0+c_1z(p-1)+c_2\exp_p{(z\log_p{(a_2a_1^{-1})^{p-1}})}\\ &\hspace{1.2cm}+\dots+ c_{k-1}\exp_p{(z\log_p{(a_{k-1}a_1^{-1})^{p-1}})}). \end{align*} $$
Then,
$f(n)=\det (A)x_{n(p-1)}$
for all
$n\in \mathbb {Z}_{\geq 0}$
. Also, we have
$$ \begin{align*} f(0)=\det(A)(c_0+c_2+\dots+c_{k-1})=\det(A)x_0=0 \end{align*} $$
and
$$ \begin{align*} f'(0)&=\det(A)(c_1(p-1)+c_2\log_p{(a_2a_1^{-1})^{p-1}}+\dots+c_{k-1}\log_p{(a_{k-1}a_1^{-1})^{p-1}}\\ &\hspace{1.2cm} +(c_0+c_2+\dots+c_{k-1})\log_p{(a_1)^{p-1}})\\ &=\det(A)(c_1(p-1)+c_2\log_p{(a_2a_1^{-1})^{p-1}}+\dots+c_{k-1}\log_p{(a_{k-1}a_1^{-1})^{p-1}}). \end{align*} $$
We find that
$\det (A)c_1=(-1)^{k+1}\prod _{1\leq i<j\leq (k-1)}(a_i-a_j)$
. By the hypothesis, we have
$p\nmid \det (A)c_1$
. Using the definition of
$\log _p(z)$
, we obtain that p divides
$\log _p{(a_ia_1^{-1})^{p-1}}$
for
$2\leq i \leq k-1$
. Therefore,
$p\nmid f'(0)$
which implies
$f'(0)$
is nonzero. Hence,
$0$
is a simple zero of f in
$\mathbb {Z}_p$
. By Theorem 2.2,
$R(f(\mathbb {N}))=R(x_{n(p-1)})$
is dense in
$\mathbb {Q}_p$
. Hence, the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
.
Proof of Theorem 1.6
The nth term of the sequence is given by
$$ \begin{align*}x_n=a^n(c_0+c_1n+\dots +c_{k-1}n^{k-1}),\end{align*} $$
where
$$ \begin{align*}C=\begin{bmatrix} c_0&c_1&\dots &c_{k-1} \end{bmatrix}^{t}\end{align*} $$
is given by
$C=\frac {1}{\det (A)}\text {adj}(A)\cdot X_0$
, where
$$\begin{align*}X_0= \begin{bmatrix} 0\\ 0\\ \vdots \\ 0\\ 1 \end{bmatrix}, A = \begin{bmatrix} 1 & 0 & 0 & \dots & 0\\ a & a & a & \dots & a\\ a^2 & 2a^2 & 2^2a^2 & \dots & 2^{k-1}a^2\\ \vdots& \vdots & \vdots & \ddots & \vdots\\ a^{k-1} & (k-1)a^{k-1} & (k-1)^2a^{k-1} & \dots & (k-1)^{k-1}a^{k-1} \end{bmatrix}.\end{align*}$$
We simplify
$C=\frac {1}{\det (A)}\text {adj}(A)\cdot X_0$
and obtain
$$ \begin{align*} \begin{bmatrix} c_0 \\ c_1 \\ c_2\\\vdots \\ c_{k-1} \end{bmatrix}=\begin{bmatrix} 1 & 0 & \dots & 0\\ 1 & 1 & \dots & 1\\ 1 & 2 &\dots & 2^{k-1}\\ \vdots& \vdots & \ddots & \vdots\\ 1 & (k-1) & \dots & (k-1)^{k-1} \end{bmatrix}^{-1}\begin{bmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 1/a^{k-1} \end{bmatrix}. \end{align*} $$
Next, we consider an analytic function
$f:\mathbb {Z}_p\rightarrow \mathbb {Z}_p$
defined as
$$ \begin{align*} f(z)&:=\det(A)\exp_p{(z\log_p{(a^{p-1})})}(c_0+c_1(p-1)z+c_2(p-1)^2z^2+\\ &\hspace{2.6cm}+\dots+c_{k-1}(p-1)^{k-1}z^{k-1}). \end{align*} $$
Let
$$ \begin{align*} h(z):=\exp_p{(z\log_p{(a^{p-1})})} \end{align*} $$
and
$$ \begin{align*} g(z):=\det(A)(c_0+c_1(p-1)z+c_2(p-1)^2z^2+\dots+c_{k-1}(p-1)^{k-1}z^{k-1}). \end{align*} $$
We have
$\|a^n\|_p=1$
and
$h(n)=a^{n(p-1)}$
for all positive integers n. Hence,
$\|h(z)\|_p=1$
for all
$z\in \mathbb {Z}_p$
. Therefore,
$f(z)=0$
if and only if
$g(z)=0$
for some
$z\in \mathbb {Z}_p$
. We have,
$g(0)=\det (A)c_0=\det (A)x_0=0$
and
$g'(0)=\det (A)c_1(p-1)$
. Using [Reference Rawashdeh22, Lemma 2.2], we find that
$$ \begin{align*} c_1=\frac{(-1)^{k}}{a^{k-1}(k-1)}. \end{align*} $$
Thus,
$c_1\neq 0$
for all
$k\geq 2$
. Therefore,
$0$
is a simple zero of f in
$\mathbb {Z}_p$
. By Theorem 2.2,
$R(f(\mathbb {N}))=R(x_{n(p-1)})$
is dense in
$\mathbb {Q}_p$
, which yields that the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
.
Proof of Theorem 1.8
We first prove part (a) of the theorem. For
$n\geq 0$
, the nth term of the sequence is given by the formula
$$ \begin{align*}x_n=a^n(c_0+c_1n+c_2n^2),\end{align*} $$
where
$$ \begin{align*} c_0&=x_0,\\ c_1&=\frac{4ax_1-x_2-3a^2x_0}{2a^2},\\ c_2&=\frac{x_2-2ax_1+a^2x_0}{2a^2}. \end{align*} $$
We define a function f as
$$ \begin{align*}f(z):= 2a^2\exp_p(z\log_p{a^{p-1}})(c_0+c_1(p-1)z+c_2(p-1)^2z^2).\end{align*} $$
Since
$p\nmid a$
, f is defined for all
$z\in \mathbb {Z}_p$
and
$f(n)=2a^2x_{n(p-1)}$
for all
$n\in \mathbb {Z}_{\geq 0}$
. Moreover,
$\mathbb {Z}_{\geq 0}$
is dense in
$\mathbb {Z}_p$
. Therefore, f is an analytic function from
$\mathbb {Z}_p$
to
$\mathbb {Z}_p$
. We have,
$$ \begin{align*}f(0)=2a^2c_0=2a^2x_0\equiv 0\quad \pmod{p}\end{align*} $$
and
$$ \begin{align*}f'(0)\equiv 2a^2c_1(p-1)=2a^2(4ax_1-x_2-3a^2x_0)(p-1)\not\equiv 0\quad \pmod{p}.\end{align*} $$
Therefore, by Hensel’s lemma, f has a zero
$z_0$
in
$\mathbb {Z}_p$
such that
$z_0\equiv 0\ \pmod {p}$
. Since f has a power series expansion with p-adic integral coefficients, we have
$f'(z_0)\equiv f'(0)\ \pmod {p}$
. Hence,
$z_0$
is a simple zero of f in
$\mathbb {Z}_p$
. Therefore, by Theorem 2.2,
$R(f(\mathbb {N}))=R((x_{n(p-1)}))$
is dense in
$\mathbb {Q}_p$
. Hence, the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
.
Next, if
$x_0=0$
, then
$c_0=0$
and
$c_1=\frac {4ax_1-x_2}{2a^2}$
. We have
$f(0)=0$
. Suppose that
$4ax_1\neq x_2$
. Then,
$f'(0)\neq 0$
which implies that
$0$
is a simple zero of f. Therefore, by Theorem 2.2, the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
. Conversely, suppose that
$4ax_1=x_2$
. This gives
$c_1=0$
, and hence
$x_n=a^nc_2n^2$
. If
$c_2= 0$
, then
$x_n=0$
for all n. If
$c_2\neq 0$
, then the quotient set of
$(x_n)_{n\geq 0}$
is equal to the quotient set of
$\{a^nn^2:n\in \mathbb {Z}_{\geq 0}\}$
. Since
$\nu _p(a^nn^2)=2\nu _p(n)$
, the p-adic valuation of these elements is even for all
$n\in \mathbb {Z}_{>0}$
. Therefore, by Lemma 2.3, the quotient set of
$(x_n)_{n\geq 0}$
is not dense in
$\mathbb {Q}_p$
. This completes the proof of part (a) of the theorem.
Next, we prove part (b) of the theorem. For
$n\geq 0$
, the nth term of the sequence is given by
$$ \begin{align*}x_n=c_0a^n+c_1na^n+c_2b^n=a^n(c_0+c_1n+c_2(ba^{-1})^n),\end{align*} $$
where
$$ \begin{align*} c_0&=\frac{b^2x_0-2abx_0-x_2+2ax_1}{(b-a)^2}, \\ c_1&=\frac{x_2-x_1(a+b)+x_0ab}{a(a-b)}, \\ c_2&=\frac{x_2-2ax_1+a^2x_0}{(b-a)^2}. \end{align*} $$
Since
$p\nmid ab(a-b)$
, we can define an analytic function
$f:\mathbb {Z}_p\rightarrow \mathbb {Z}_p$
as
$$ \begin{align*} f(z):=\exp_p(z\log_p{a^{p-1}})(c_0+c_1z(p-1)+c_2\exp_p(z\log_p({ba^{-1})^{p-1}})), \end{align*} $$
which satisfies the equation
$f(n)=x_{n(p-1)}$
for all
$n\geq 0$
. Now, we have
$$ \begin{align*} f(0)=c_0+c_2=x_0\equiv 0\quad \pmod{p} \end{align*} $$
and
$$ \begin{align*} f'(0)=c_1(p-1)+c_2\log_p(ba^{-1})^{p-1}+(c_0+c_2)\log_p{a^{p-1}}\not\equiv 0\quad \pmod{p}. \end{align*} $$
Therefore, by Hensel’s lemma, f has a zero
$z_0$
in
$\mathbb {Z}_p$
such that
$z_0\equiv 0\ \pmod {p}$
. Since f has a power series expansion with p-adic integral coefficients, we have
$f'(z_0)\equiv f'(0)\ \pmod {p}$
. Hence,
$z_0$
is a simple zero of f in
$\mathbb {Z}_p$
. Therefore, by Theorem 2.2,
$R(f(\mathbb {N}))=R(x_{n(p-1)})$
is dense in
$\mathbb {Q}_p$
. Hence, the quotient set of
$(x_n)_{n\geq 0}$
is dense in
$\mathbb {Q}_p$
.
Acknowledgements
We are grateful to the referee for going through the article and providing many helpful comments.
