Combinatorial properties of sparsely totient numbers

TL;DR

This paper investigates the properties and distribution of sparsely totient numbers, revealing their divisibility patterns, constructing infinite families, and analyzing their additive and multiplicative structures.

Contribution

It provides new results on divisibility, explicit constructions, and the combinatorial structure of sparsely totient numbers, connecting them with prime distribution and progressions.

Findings

Squarefree integers divide all large sparsely totient numbers

Constructed infinite families of sparsely totient numbers

Proved sparsely totient numbers are multiplicatively piecewise syndetic

Abstract

Let $N_1(m)=\max\{n \colon \phi(n) \leq m\}$ and $N_1 = \{N_1(m) \colon m \in \phi(\mathbb{N})\}$ where $\phi(n)$ denotes the Euler's totient function. Masser and Shiu \cite{masser} call the elements of $N_1$ as `sparsely totient numbers' and initiated the study of these numbers. In this article, we establish several results for sparsely totient numbers. First, we show that a squarefree integer divides all sufficiently large sparsely totient numbers and a non-squarefree integer divides infinitely many sparsely totient numbers. Next, we construct explicit infinite families of sparsely totient numbers and describe their relationship with the distribution of consecutive primes. We also study the sparseness of $N_1$ and prove that it is multiplicatively piecewise syndetic but not additively piecewise syndetic. Finally, we investigate arithmetic/geometric progressions and other additive and multiplicative patterns like $\{x, y, x+y\}, \{x, y, xy\}, \{x+y, xy\}$ and their generalizations in the sparsely totient numbers.

Tables

Table 1. Table 1 :

Prime $q>$	$p$	$\frac{1}{2}\left(p+\sqrt{p^2+4{\left(p-1\right)}^2}\right)$	$p\left(p-1\right)$
$C_p\left(q\right)\le$	$2$	$1$	$0$

Equations

$$N_1(m)$$

$$\underset{n\in N_1}{\mathrm{liminf}}\frac{Q(n)}{\log n}\ge \sqrt{2}-1,$$

$$B{N}_1:=\{m\in V:N_1(m)=\max ({\varphi }^{-1}(m))\}.$$

$$X_{n,p}:=\left(n\prod _{q\mid n,q\in \mathbb{P}}{q}^{-1}\right)\prod _{q\le p,q\in \mathbb{P}}q,Y_{p,k}:=\prod _{\begin{array}{c}q\in [2,p_k]\cap \mathbb{P}\\ q\ne p\end{array}}q,D(p):=\frac{(p_1-p)(p_2-p)}{p(p-1)},$$

$$A(p)=\frac{p_1{p}_2}{p},E(2):=\{p\in \mathbb{P}:\left(A(p),A(p)+D(p)\right]\cap \mathbb{P}=\varnothing \},$$

$$E(3):=\left\{p\in \mathbb{P}:\left(\frac{p_1{p}_2{p}_3}{pq},1+\frac{(p_1-1)(p_2-1)(p_3-1)}{(p-1)(q-1)}\right]\cap {\mathcal{O}}_q=\varnothing \forall q\in {\mathcal{O}}_p\right\},$$

$$\mathcal{X}:=\left\{u:u=X_{n,p}\text{ for }n\in \mathbb{N},p\in \mathbb{P}\cap \left(\frac{n}{2},\infty \right)\right\},{\mathcal{Y}}_1:=\{Y_{p,1}:p\in \mathbb{P}\cap [5,\infty )\},$$

$${\mathcal{Y}}_2:=\{Y_{p,2}:p\in E(2)\text{ and }p\ge 11\}\text{and }{\mathcal{Y}}_3:=\{Y_{p,3}:p\in E(3)\text{ and }p\ge 11\}.$$

$$\phi(n)=n\displaystyle\prod_{p\mid n}\Big{(}1-\frac{1}{p}\Big{)}.$$

$$p\in \mathbb{P}$$

$$n:=q^r+k,I_1:$$

$$\varphi (y)\le q^{v_q(N)-r}\varphi (ST)\left(\frac{p-1}{p}\right)(q^r+p-1).$$

$$\frac{(q^r+p-1)(p-1)}{p}\le q^{r-1}(q-1)⟺\{\begin{array}{ll}(p-1{)}^2\le q^{r-1}(q-p)& \text{mbox}ifr>1,\\ q\ge (p-1{)}^2+p& \text{mbox}ifr=1.\end{array}$$

$$\prod _{q\in \mathbb{P},p<q<n}\frac{q}{q-1}=C\prod _{q\in \mathbb{P},q<n}\frac{q}{q-1}=C\prod _{q\in \mathbb{P},q<n}\left(\sum _{r=0}^{\infty }\frac{1}{q^r}\right)=\sum _{k\in I_n}\frac{C}{k}\ge \sum _{k=1}^{n-1}\frac{C}{k},$$

$$\varphi (A)<q^{-1}A,$$

$$B{N}_1=\{m\in V:N_1(m)=\max ({\varphi }^{-1}(m))\}.$$

$$D(A,B):=\left(\prod _{q\in A}\frac{q-1}{q}\right)\left(\prod _{q\in B}\frac{q}{q-1}\right).$$

$$D(B,A)\ge \prod _{x\in B}\frac{(x-1)f(x)}{x(f(x)-1)}=\prod _{x\in B}\frac{xf(x)-f(x)}{xf(x)-x}\ge 1.$$

$$L(n,y):=\left(\prod _{q|K(n,y)}{q}^{v_q(n)-v_q(y)}\right)\left(\prod _{q|n,q\nmid K(n,y)}{q}^{v_q(n)-1}\right).$$

$$L(n,y)$$

$$\varphi (y)$$

$$\varphi (X_{n,p})$$

$$\therefore \varphi (y)\le \varphi (X_{n,p})$$

$$\Rightarrow \prod _{q\in W(y)}(q-1)\le L(n,y)\prod _{q\in W(X_{n,p})}(q-1).$$

$$(r-1)\prod _{q\in W(y),q\ne r}(q-1)\le \frac{n}{2}\prod _{q\in W(X_{n,p})}(q-1).$$

$$B:=W(y)\setminus \{p\},C:=W(Y_{p,k})\setminus \{s\}.$$

$$D(W(y),W(Y_{p,k}))$$

$$y$$

$$y=\varphi (y)\left(\prod _{q\le p}\frac{q}{q-1}\right)\left(\prod _{i=1}^l\frac{q_i}{q_i-1}\right).$$

$$y\le \varphi (Y_{p,k})\left(\prod _{q\le p}\frac{q}{q-1}\right)\left(\prod _{i=1}^l\frac{q_i}{q_i-1}\right).$$

$$y\le Y_{p,k}\left(\frac{p(p_{k-1}-1)(p_k-1)}{(p-1)p_{k-1}{p}_k}\right).$$

Full text

**Keywords: Euler’s function, Sparsely totient numbers, IP Set, piecewise syndetic set **

**2010 Mathematics Subject Classification: Primary 11B75; Secondary 11A41, 11A25, 11B05, 11B25 **

1 Introduction

Euler’s totient function $\phi(n),$ which enumerates the number of positive integers which are co-prime to and less than or equal to $n,$ is a classical arithmetical function. It is a well known fact that the number of solutions to the equation $\phi(x)=m$ is finite for each $m\in\mathbb{N}$ ($\mathbb{N}$ is the set of positive integers). It is natural, then, to ask the following question:

Question 1**.**

For a given $m\in\mathbb{N},$ what is the largest integer $n$ such that $\phi(n)\leq m$?

We denote the set $\{x:\phi(x)=m\}$ by $\phi^{-1}(m)$ and the image of $\phi$ by $V,$ i.e. $V=\{\phi(m):m\in\mathbb{N}\}.$ The elements of $V$ are called totients. For $m\in V,$ we define the following quantities with the above question in view:

[TABLE]

Note that $N_{1}(m)$ can be defined on the whole of $\mathbb{N}.$ But this doesn’t contribute any new elements to the image $N_{1}$ of $N_{1}(m),$ since $N_{1}(m)=N_{1}(m-1)$ if $m\notin V.$ Hence, from here on, we study $N_{1}(m)$ only for $m\in V.$

Definition 1.1** (Sparsely totient number).**

An element of $N_{1}$ is called a sparsely totient number.

D. W. Masser and P. Shiu initiated the study of sparsely totient numbers and described several of their properties. Here we will focus on describing prime divisors of sparsely totient numbers, constructing infinite families of sparsely totient numbers, understanding the relationship of sparsely totient numbers with primes in short intervals and searching for interesting configurations involving addition as well as multiplication in the sparsely totient numbers.

In Section 2, we study some divisibility properties of sparsely totient numbers. It follows from the definition of $N_{1}(m)$ that $N_{1}$ is an infinite set. Masser and Shiu [3] showed that given a prime $p,$ all large enough elements of $N_{1}$ are divisible by $p.$ This leads to the first theorem in this paper where we give an alternate proof of this result.

Theorem 1.2**.**

Let $n\in\mathbb{N}.$ Then the following are equivalent:

1. (i)

n is a squarefree integer. 2. (ii)

There exists $M,$ depending on $n,$ such that $N_{1}(m)\equiv 0\pmod{n}$ for all $m>M.$

In their proof, Masser and Shiu studied the smallest prime not dividing $N_{1}(m)$ and proved that it tends to infinity as $m\rightarrow\infty.$ To achieve this, they showed the following using the prime number theorem:

[TABLE]

where $Q(n)$ is the smallest prime factor not dividing $n.$ We give an alternate proof of Theorem 1.2 without using the prime number theorem. This proof also gives some quantitative information about prime divisors of sparsely totient numbers in specific cases which is described below. We prove the following two propositions from which the ‘sufficient’ part of Theorem 1.2 follows as a consequence. For this, we need the concept of $q$-valuations of an integer $n,$ namely, $v_{q}(n)$(see Definition 2.2).

Proposition 1.3**.**

Suppose that $p$ and $q$ are primes such that $q>p.$ Then there exists $C_{p}(q)\in\mathbb{N}\cup\{0\}$ such that for each $N\in N_{1}$ with $N\not\equiv 0\pmod{p},$ we have $v_{q}(N)\leq C_{p}(q).$ In fact, we have the following upper bounds for $C_{p}(q)$:

Proposition 1.4**.**

Suppose that $p$ and $q$ are primes such that $q<p.$ Then there exists $D_{p}(q)\in\mathbb{N}\cup\{0\}$ such that for $N\in N_{1}$ with each prime factor less than $p,$ we have $v_{q}(N)\leq D_{p}(q).$

In particular, these propositions give upper bounds for the exponents of prime factors $q$ of those elements of $N_{1}$ which are not divisible by a particular prime $p.$ Table 1 shows that if an element of $N_{1}$ is not divisible by $p,$ then it cannot be divisible by any prime $q>p(p-1).$ In particular, this gives us the following corollary in the case when $p=3.$

Corollary 1.5**.**

All sparsely totient numbers other than $2,$ are divisible by $3.$ In other words, $N_{1}(m)$ is divisible by $3$ for $m>1.$

However, in Proposition 1.4 where we deal with primes $q<p,$ we have not been able to get explicit values of $D_{p}(q)$ as above. We get existence of $D_{p}(q)$ in Proposition 1.4 due to the fact that Riemann Zeta function $\zeta(s)$ has a simple pole at $s=1$ [see Question 3].

Next, we study equations of the type $N_{1}(x)=N_{1}(x+1)=\dots=N_{1}(x+k)$ with $N_{1}(x-1)<N_{1}(x)$ and $N_{1}(x+k)<N_{1}(x+k+1)$ as an application of the sufficient part of Theorem 1.2 and show that $k$ can be arbitrarily large. For this, we define a subset $BN_{1}$ of $V$ comprising the “first” elements of such blocks $[x,x+k]\cap\mathbb{N}.$

[TABLE]

If $m_{1}$ and $m_{2}$ are two consecutive elements in $BN_{1},$ then $N_{1}(m)=N_{1}(m_{1})$ for each $m\in[m_{1},m_{2})\cap\mathbb{N}.$ We get a nice pattern in the set $BN_{1}$ as follows:

Corollary 1.6**.**

For $n\in\mathbb{N},$ $n$ divides all sufficiently large elements of $BN_{1}.$

The ‘necessary’ part of Theorem 1.2 follows from the observation that $\prod_{q\leq p}q\in N_{1}.$ In Section 3, we generalize this observation to construct infinite families of sparsely totient numbers which are defined as follows:

Definition 1.7**.**

Let $\mathbb{P}$ be the set of primes and $\mathcal{O}_{p}=(p,\infty)\cap\mathbb{P}.$ If $p\in\mathbb{P},n,k\in\mathbb{N}$ and $p_{i}$ is the $i\text{th}$ smallest prime greater than $p,$ we define

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Theorem 1.8**.**

The sets $\mathcal{X}$ and $\mathcal{Y}_{1}$ are infinite subsets of $N_{1}.$ Also, $\mathcal{Y}_{2}$ and $\mathcal{Y}_{3}$ are non-empty subsets of $N_{1}.$

From Theorem 1.8, we have the following general divisibility result by observing that $X_{n,p}$ is a sparsely totient number for all $p>\frac{n}{2}$:

Corollary 1.9**.**

Every integer divides infinitely many sparsely totient numbers.

Elements of $\mathcal{Y}_{2}$ and $\mathcal{Y}_{3}$ depend on the sets $E(2)$ and $E(3)$ respectively and hence the study of elements of these sets is linked to the study of distribution of consecutive primes. The next theorem in this paper gives some explicit elements in the sets $\mathcal{Y}_{2}$ and $\mathcal{Y}_{3}.$

Theorem 1.10**.**

For $p\in\mathbb{P},\ p\geq 11,$ we have:

1. (i)

If $p,\ p+2i$ are consecutive primes for some $i\in\{1,2,3,4\},$ then $Y_{p,2}\in N_{1}.$ 2. (ii)

If $p,\ p+2$ and $p+6$ are consecutive primes, then $Y_{p,3}\in N_{1}.$ 3. (iii)

Let $a,b$ be distinct positive even integers. If $p\geq 2ab$ and $p,\ p+a,\ p+b$ are consecutive primes, then $Y_{p,2}\in N_{1}.$

By Theorem 1.10(i), $\mathcal{Y}_{2}$ will be an infinite set if there is an affirmative answer to the twin prime conjecture which is the case $k=2$ of the following conjecture.

Conjecture 1.11** (Polignac’s conjecture).**

For every even natural number k, there are infinitely many pairs of primes that differ by $k.$

Our next theorem guarantees the infinitude of $\mathcal{Y}_{2}$ if there exist infinitely many primes $p$ such that the fractional part of $\frac{p_{1}p_{2}}{p}$ is bounded above by an absolute constant $\delta_{0}<1.$

Theorem 1.12**.**

Let $p_{i}$ be the $i\text{th}$ smallest prime greater than $p$ and $0<\delta_{0}<1.$ If $A:=\left\{p\in\mathbb{P}\colon\frac{p_{1}p_{2}}{p}-\left\lfloor\frac{p_{1}p_{2}}{p}\right\rfloor<\delta_{0}\right\}$ is an infinite set of primes, then $Y_{p,2}\in N_{1}$ for all but finitely many elements $p$ in $A.$

In 2014, Yitang Zhang [6] proved the weaker version of the twin prime conjecture.

Proposition 1.13** ([6], Yitang Zhang).**

Let $p_{1}$ be the next consecutive prime to a prime $p.$ Then there are infinitely many primes $p$ such that $p_{1}-p<7\times 10^{7}.$

Using this, we get that $\mathcal{Y}_{2}$ is an infinite set as a corollary of Theorem $\ref{t4}.$

Corollary 1.14**.**

$\mathcal{Y}_{2}$* is an infinite set.*

On the basis of computations for the fractional part of $\frac{p_{1}p_{2}}{p}$ for $p\in\mathbb{P}$ in Figure 1, one can hope that the fractional part of $\frac{p_{1}p_{2}}{p}$ may tend to [math] as $p\rightarrow\infty$ and the fractional part of $\frac{p_{1}p_{2}}{p}$ is less than $0.983607$ for each $p\in\mathbb{P}$ [see Question 6]. If this expectation is true, then $Y_{p,2}$ is a sparsely totient number for all sufficiently large primes $p.$ In view of this observation we can ask the following question.

Question 2**.**

Is $\mathcal{Y}_{2}=\{Y_{p,2}\colon p\in\mathbb{P},p\geq 11\}$?

In Section 4, we study certain additive and multiplicative patterns in the sparsely totient numbers. More precisely, we investigate patterns like $\{x,y,x+y\},\{x,y,xy\}$ and generalizations of them. $N_{1}$ contains the set of finite sums of arbitrarily long sequences (an $IP_{0}$ set) but does not contain the set of finite sums of any infinite sequence (an $IP$ set). However, in the multiplicative scenario, it does contain the set of finite products of an infinite sequence (see Definition 4.3 and 4.4). One can also ask whether both sum and product occur simultaneously in $N_{1}.$ This has been answered in the negative for the subset $\mathcal{X}$ of $N_{1}$ [see Question 8]. We also look at the presence of arithmetic/geometric progressions in $N_{1}.$ We describe sparseness of the set $N_{1}$ using the notion of piecewise syndeticity [see Definition 4.1]. The following theorem summarizes these properties:

Theorem 1.15**.**

1. (i)

The set of sparsely totient numbers is multiplicatively piecewise syndetic but not additively piecewise syndetic. 2. (ii)

The set of sparsely totient numbers is an additive $IP_{0}$ and multiplicative $IP$ set but not an additive $IP$ set. 3. (iii)

There do not exist $x,y\in\mathbb{N}$ such that $x+y,xy\in\mathcal{X}.$ 4. (iv)

The set of sparsely totient numbers contains arbitrarily long arithmetic and geometric progressions.

In Section 5, we pose some open problems arising from the present work.

Notation**.**

Let $\mathbb{N},\mathbb{P},\mathbb{Z}$ denote, respectively, the set of positive integers, the set of prime numbers and the set of integers. $p,q$ will always represent prime numbers unless otherwise mentioned. $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x,$ $[a,b]$ denotes the set $\{x\in\mathbb{N}:a\leq x\leq b\}$ and similarly for the sets $(a,b],[a,b),(a,b)$ and $W(x)$ denotes the set of prime divisors of $x.$ By convention, we assume empty products and empty sums to take the values 1 and 0 respectively.

2 The prime divisors of sparsely totient numbers

Euler’s totient function $\phi$ is a well studied arithmetical function in number theory. For a positive integer $n,$ $\phi(n)$ counts the number of positive integers less than or equal to $n$ which are co-prime to $n.$ It has a closed formula in terms of the prime factors dividing $n,$ namely,

[TABLE]

From this formula, we immediately see the following relations for $\phi$ which are frequently used in the sequel.

Lemma 2.1**.**

[1]** Let $k,u,n\in\mathbb{N}$ and $p,p_{1},p_{2},\dots p_{k}$ be prime numbers such that $p_{i}\mid n$ for $i\in[1,k].$ Then

(i) $\phi(n)\leq n\displaystyle\prod_{i=1}^{k}\displaystyle\left(1-\frac{1}{p_{i}}\right),$ (ii) $\phi(pu)=\begin{cases}p\phi(u)&\mbox{if }p\mid u,\\ (p-1)\phi(u)&\mbox{if }p\nmid u.\end{cases}$

From the definition of $N_{1}(m),$ it follows that $N_{1}(m)\rightarrow\infty$ as $m\rightarrow\infty.$ It also follows from the definition that if $N\in\mathbb{N}$ is not in $N_{1},$ then there exists a positive integer $y$ greater than $N$ with $\phi(y)\leq\phi(N)$ and our goal is to try to investigate such elements in this section. To prove the ** ‘sufficient’** part of Theorem 1.2, it is enough to prove the equivalent statement:

[TABLE]

We make the following definition which will be used frequently in the sequel.

Definition 2.2** (Valuation).**

Let $p$ be a prime number. Then the $p$-valuation on the integers $\mathbb{Z}$ is the map $v_{p}\colon\mathbb{Z}\rightarrow\mathbb{N}\cup\{0,\infty\}$ defined by $v_{p}(0)=\infty$ and $v_{p}(n)=r$ for $n\neq 0,$ where $r$ is the largest non-negative integer such that $p^{r}\mid n.$

Therefore, for a given prime $p$, we need to find $c(p)\in\mathbb{N}$ and $p_{0}(p)\in\mathbb{P}$ to prove 1 such that if $N\in N_{1}$ and $p\nmid N,$ then

1. (A)

$v_{q}(N)=0\ \forall q\geq p_{0}(p)$ and 2. (B)

$v_{q}(N)<c(p)\ \forall q<p_{0}(p).$

Proposition 1.3 deals with the value of $v_{q}(N)$ in case (A) and later on, Proposition 1.4 deals with the value of $v_{q}(N)$ in case (B). In the proof of Proposition 1.3 below, the element $y$ satisfying $y>N$ and $\phi(y)\leq\phi(N)$ is obtained by using the smallest multiple of $p$ greater than $q^{r}.$

We provide the statement of Proposition 1.3 again in full detail.

Proposition 2.3**.**

Suppose $N$ is a positive integer and $p,q$ are primes such that $p<q,\ p\nmid N$ and $q^{r}\mid N$ for some $r\in\mathbb{N}.$

If $r>2,$ then  $\exists\ y>N$ such that $\phi(y)\leq\phi(N).$ 2. 2.

If $r=2$ and $q\geq\frac{p+\sqrt{p^{2}+4(p-1)^{2}}}{2},$ then $\exists\ y>N$ such that $\phi(y)\leq\phi(N).$ 3. 3.

If $r=1$ and $q\geq(p-1)^{2}+p,$ then  $\exists\ y>N$ such that $\phi(y)\leq\phi(N).$

Proof.

For $p,q\in\mathbb{P}$ with $p<q$ and $r\in\mathbb{N},$ choose $k\in[1,p-1]$ such that $q^{r}+k\equiv 0\pmod{p}.$ Since $k<p,$ it follows that $\gcd(q,q^{r}+k)=1.$ Now define

[TABLE]

Put $S=\prod_{z\in I_{2}}z^{v_{z}(N)}\text{ and }T=\prod_{z\in I_{1}}z^{v_{z}(N)}.$ Using the fact that $q^{r}\mid N,$ we get $v_{q}(N)\geq r.$ We set $y:=STnq^{v_{q}(N)-r}>N.$

Observe that $\phi(y)\leq q^{v_{q}(N)-r}\phi(STn)=q^{v_{q}(N)-r}\phi(S)\phi(Tn),$ since $(Tn,S)=1.$ Using $I_{1}\subset W(n)$ along with $p\mid n$ and $p\notin I_{1},$ we get that $\phi(Tn)\leq\phi(T)\frac{p-1}{p}n$ by Lemma 2.1. Since $n\leq q^{r}+p-1,$ we have $\phi(Tn)\leq\phi(T)\frac{p-1}{p}(q^{r}+p-1).$ Inserting this in the above estimate for $\phi(y),$ we find that

[TABLE]

[TABLE]

The right hand side of the equivalence in (3) is satisfied in the following cases.

(I) r>2,$$q>p (II) r=2,$$q\geq\frac{p+\sqrt{p^{2}+4(p-1)^{2}}}{2} (III) r=1,$$q>p(p-1).

Therefore, the inequality in (2) and the equivalence in (3) together imply that $\phi(y)\leq\phi(STq^{v_{q}(N)})=\phi(N)$ in the above cases I, II and III. This completes the theorem. ∎

We now proceed to prove Proposition 1.4 in which the lemma below will be crucial.

Lemma 2.4**.**

For $p\in\mathbb{P},$ the sequence $\displaystyle\prod_{q\in\mathbb{P},p<q<n}\displaystyle\frac{q}{q-1}\longrightarrow\infty\text{ as }~{}n\longrightarrow\infty.$

Proof.

Let $I_{n}=\{x\in\mathbb{N}\colon q\mid x,q\in\mathbb{P}\Rightarrow q<n\}$ for $n\in\mathbb{N}$ and $C=\prod_{q\in\mathbb{P}\cap[2,p]}\frac{p-1}{p}.$

[TABLE]

which tends to $\infty$ as $n\longrightarrow\infty.$ ∎

It can be rephrased in the following manner which is used to show Proposition 1.4.

Corollary 2.5**.**

Let $p\in\mathbb{P}$ and $n\in\mathbb{N}.$ Then there exists a natural number $\beta(p,n)$ such that for each prime $t>\beta(p,n),$ $\phi(B_{t})\leq n^{-1}B_{t}$ with $B_{t}=\prod_{q\in[p,t]\cap\mathbb{P}}q.$

Proof of Proposition 1.4.

Corollary 2.5 gives $\beta(p,q)\in\mathbb{P}$ such that

[TABLE]

where $A=\prod_{u\in[p,\beta(p,q)]\cap\mathbb{P}}u.$ Define $D_{p}(q):=\alpha$ where $\alpha$ is the unique non-negative integer satisfying $q^{\alpha}<A<q^{\alpha+1}.$ If $v_{q}(N)>\alpha,$ then choose $y:=Sq^{v_{q}(N)-\alpha}A,$ where $S=Nq^{-v_{q}(N)}$ satisfies $\gcd(S,q)=1.$ Observe that $y>N$ as $A>q^{\alpha}.$ Since the prime factors of $A$ are greater than or equal to $p,$ each prime factor of $N$ is less than $p$ and $\gcd(S,q)=1,$ it follows that the elements of $\{S,A,q\}$ are pairwise co-prime. Then $\phi(y)=\phi(S)\phi(q^{v_{q}(N)-\alpha})\phi(A).$ Using equation (4) along with $A<q^{\alpha+1}$ and Lemma 2.1(ii), we get that $\phi(y)<\phi(N).$ This completes the proof. ∎

Proof of Theorem 1.2.

To prove the sufficient part of the theorem, it is enough to assume that $n$ is a prime. Let $s$ be a prime and $s_{0}$ be the smallest prime greater than $(s-1)^{2}+s.$ If $N\in N_{1}$ such that $s\nmid N,$ then by Proposition 2.3, $v_{q}(N)=0\ \forall\ q\geq s_{0}.$ Since $s<s_{0},$ applying Proposition 1.4 with $p=s_{0}$ gives us $v_{q}(N)<D_{s}(q)$(depending only on $s$ and $q$) and hence the ‘sufficient’ part of the theorem follows.

To prove the converse, it is enough to show that $Z_{p}=\prod_{q\leq p}q\in N_{1}$ $\forall$ $p\in\mathbb{P}.$ Suppose for some $p,$ $Z_{p}\notin N_{1}.$ Then  $\exists\ y>Z_{p}$ with $\phi(y)\leq\phi(Z_{p}).$ So, $\prod_{q\in W(y)}q^{v_{q}(y)-1}(q-1)\leq\prod_{q\leq p}(q-1).$ Therefore $\left|W(y)\right|\leq\left|W(Z_{p})\right|.$ Then the quantity $Q:=\left(\prod_{q\in W(y)}\frac{q-1}{q}\right)\left(\prod_{q\leq p}\frac{q}{q-1}\right)\geq 1,$ because $x>p$ $\forall$ $x\in W(y)\setminus W(Z_{p}).$ It gives that $\frac{\phi(y)}{\phi(Z_{p})}=\frac{yQ}{Z_{p}}\geq\frac{y}{Z_{p}}.$ Then $y>Z_{p}$ implies $\phi(y)>\phi(Z_{p}).$ But, this is a contradiction to $\phi(y)\leq\phi(Z_{p})$ and hence the theorem follows. ∎

Proof of Corollary 1.5.

By Proposition 2.3, if $3\nmid N_{1}(m)$ for some $m\in V,$ then $N_{1}(m)=2^{a}\cdot 5^{b},$ with $a\geq 1,b\in\{0,1\}.$ If $a>1,$ then $\phi(2^{a-1}\cdot 3\cdot 5^{b})=\phi(2^{a}\cdot 5^{b}),$ but $2^{a-1}\cdot 3\cdot 5^{b}>2^{a}\cdot 5^{b}.$ Therefore, $N_{1}(m)=2,2\times 5.$ But $\phi(10)=\phi(12)$ and so $N_{1}(m)\neq 10.$ Hence, $N_{1}(m)=2,$ which implies that $m=1.$ ∎

There is a nice application of Theorem 1.2 to equations of the type $N_{1}(m_{0})=N_{1}(m_{0}+1)=\dots=N_{1}(m_{0}+k)$ with $N_{1}(m_{0}-1)<N_{1}(m_{0})$ and $N_{1}(m_{0}+k)<N_{1}(m_{0}+k+1).$ Recall the definition of the set $BN_{1}$:

[TABLE]

We observe that $m_{0}\in BN_{1}.$ Note that the image of $BN_{1}$ under the function $h\colon x\mapsto N_{1}(x)$ is the whole of $N_{1}.$ We enumerate the elements of $BN_{1}$ as $m_{1},m_{2},\cdots,m_{t},..$ and so on. We show that given any $n\in\mathbb{N},$ all large enough elements of $BN_{1}$ are divisible by $n.$ To show this, we make use of the following result:

Lemma 2.6**.**

( [5, Theorem 2.17, page 88]). For any integer $n>1,$ there exists a prime $p\equiv 1\pmod{n}.$

Proof of Corollary 1.6.

Consider the arithmetic progression $\{1+tn\}_{t\in\mathbb{N}}.$ By Lemma 2.6, there is a prime in this sequence, say $p,$ i.e., $1+k_{0}n=p$ for some $k_{0}\in\mathbb{N}.$ So, $p-1=k_{0}n\Rightarrow n\mid(p-1).$ Using Theorem 1.2, $\exists~{}k_{1}$ such that $p\mid N_{1}(m_{t})\ \forall\ t\geq k_{1}.$ So, $(p-1)\mid\phi(N_{1}(m_{t}))=m_{t}\ \forall\ t\geq k_{1}.$ Hence $n\mid m_{t}$ for all $t\geq k_{1}.$ ∎

Suppose $N\in N_{1}$ and there exists a prime $p,\ p\nmid N.$ If there is a prime factor of $N$ greater than $p,$ then we have seen in Proposition 2.3 that $v_{q}(N)\leq 2.$ But the existence of such an $N$ in $N_{1},$ which has holes in its prime factorization, is not guaranteed. In the next section, we show that there are infinitely many elements of this type in $N_{1}.$

3 Explicit construction of sparsely totient numbers

In this section, we explicitly construct several infinite families of elements in $N_{1}.$ As a corollary of these constructions, we prove a divisibility result analogous to Theorem 1.2 for non-squarefree integers. From the previous section, we can expect two types of elements, say $x$ and $y$, in $N_{1}.$ Here $x$ is divisible by all the primes smaller than some prime $p$ and $y$ has some “holes” in its prime factorization. The main tool in these proofs will be a generalization of the technique used in the proof of the ‘necessary’ part of Theorem 1.2 (i.e., $\prod_{q\in[2,p]\cap\mathbb{P}}q\in N_{1}\ \forall\ p\in\mathbb{P}).$

3.1 Proof of Theorem 1.8

Definition 3.1** (D(A,B)).**

Let $A$ and $B$ be two finite subsets of $\mathbb{P}.$ Then

[TABLE]

If $x\notin N_{1},$ then $\exists\ y$ such that $y>x,\ \phi(y)\leq\phi(x).$ The following lemma gives the value of $D(W(y),W(x))$ in this case.

Lemma 3.2**.**

If $\phi(y)\leq\phi(x)$ and $y>x$ for $y,x\in\mathbb{N},$ then $D(W(y),W(x))$ $<1.$

Proof.

Using Euler product formula for $x$ and $y,$ we have $\frac{\phi(y)}{\phi(x)}=\frac{yD(W(y),W(x))}{x}.$ Applying $\phi(y)\leq\phi(x)$ and $y>x,$ we get $D(W(y),W(x))<1.$ ∎

For an $x\in\mathbb{N}\setminus\{1\},$ if we are able to show that $D(W(y),W(x))\geq 1$ for all $y$ satisfying $y>x$ and $\phi(y)\leq\phi(x),$ then it means that $x\in N_{1}.$ The two lemmas below help us in this regard.

Lemma 3.3**.**

Let $A$ and $B$ be two finite subsets of $\mathbb{P}$ such that $|B|\leq|A|.$ If $\min(B\setminus A)>\max(A)$ or $B\subset A,$ then $D(B,A)\geq 1.$

Proof.

For the case $B\subset A,$ $D(B,A)=\left(\prod_{q\in A\setminus B}\frac{q}{q-1}\right)\geq 1.$ In the case when $B\not\subset A,$ one can define an injective map $f\colon B\rightarrow A$ such that $f(x)=x$ for $x\in A\cap B,$ since $|B|\leq|A|.$ Using $\min(B\setminus A)>\max(A),$ it follows that $f(x)\leq x~{}\forall~{}x\in B.$ Therefore,

[TABLE]

∎

Lemma 3.4**.**

Let $A,B$ be two finite subsets of $\mathbb{P}$ such that $|A|\leq|B|$ and $A\neq B.$ If $\min(B\setminus A)>\max(A),$ then $\prod_{q\in B}(q-1)>\prod_{q\in A}(q-1).$

Proof.

If $A$ is a proper subset of $B,$ then the result is trivially true. On the other hand, if $A\setminus B$ is non-empty, define an injective map $f\colon A\rightarrow B$ such that $f(x)=x$ for $x\in A\cap B.$ This can be done as $|A|\leq|B|.$ Then $f(x)>x~{}\forall~{}x\in A\setminus B$ as $\min(B\setminus A)>\max(A).$ Hence $\prod_{q\in B}(q-1)\geq\prod_{q\in f(A)}(q-1)=\prod_{q\prime\in A}(f(q\prime)-1)>\prod_{q\in A}(q-1).$ ∎

We will now proceed to construct elements of the type $X_{n,p}$ in $N_{1}.$ For this purpose, we use two parameters $K(n,y)$ and $L(n,y).$

Definition 3.5**.**

Let $n,y\in\mathbb{N}.$ Then, the quantity $K(n,y)$ is defined by $q$-valuations as $v_{q}(K(n,y)):=v_{q}(y),$ if $v_{q}(n)>0,$ and $v_{q}(K(n,y)):=0,$ otherwise. The quantity $L(n,y)$ is defined as follows:

[TABLE]

Lemma 3.6**.**

If $n\in\mathbb{N}\setminus\{1\},$ then $L(n,y)\leq\displaystyle\frac{n}{2}$ for each $y\in\mathbb{N}.$

Proof.

For $n,y\in\mathbb{N},$ define $A_{1}:=\{q\in\mathbb{P}\colon q\mid n\},$ $A_{2}:=\{q\in\mathbb{P}\colon q\mid K(n,y)\}$ and $A_{3}:=\{q\in\mathbb{P}\colon q\mid n\text{ and }q\nmid K(n,y)\}.$ Observe that $A_{1}=A_{2}\cup A_{3}.$ Since $n\geq 2,$ it follows that $A_{1}\neq\varnothing$ and hence $A_{2}\neq\varnothing$ or $A_{3}\neq\varnothing.$ If $A_{2}\neq\varnothing,$ there exists a $q_{0}\in\mathbb{P}$ such that $v_{q_{0}}(K(n,y))\geq 1.$ It follows that $v_{q_{0}}(y)\geq 1.$ If $A_{3}\neq\varnothing,$ there exists $q_{1}\in A_{3}.$ Hence,

[TABLE]

Since $q_{0},q_{1}\geq 2$ and $n=\displaystyle\prod_{q\in A_{1}}q^{v_{q}(n)},$ we get $L(n,y)\leq\displaystyle\frac{n}{2}.$ ∎

Lemma 3.7**.**

Suppose that $p\in\mathbb{P},$ $y\in\mathbb{N}$ and $n\in\mathbb{N}\setminus\{1\}$ satisfy $p>\displaystyle\frac{n}{2}.$ Then $\phi(y)\leq\phi(X_{n,p})\Rightarrow D(W(y),W(X_{n,p}))\geq 1.$

Proof.

Note that since $p>\frac{n}{2}$, we have

[TABLE]

[TABLE]

Since $n\in\mathbb{N}\setminus\{1\},$ by Lemma 3.6, we have $\displaystyle\prod_{q\in W(y)}(q-1)\leq\frac{n}{2}\prod_{q\in W(X_{n,p})}(q-1).$ If $|W(y)|>|W(X_{n,p})|,$ then $\exists$ $r\in\mathbb{P}$ such that $r>p$ and $r\in W(y).$ This gives

[TABLE]

Applying Lemma 3.4, we get $r-1\leq\displaystyle\frac{n}{2}$ which implies that $p\leq\displaystyle\frac{n}{2},$ a contradiction. Therefore $|W(y)|\leq|W(X_{n,p})|.$ Applying Lemma 3.3, we get the desired result. ∎

From the above lemma and Lemma 3.2, we get the following proposition:

Proposition 3.8**.**

$X_{n,p}\in N_{1}$* for $n\in\mathbb{N}$ and $p\in\mathbb{P}$ with $p>\frac{n}{2}.$*

Now, we shall investigate elements of the type $Y_{p,k},$ which have “holes” in their prime factorization. This is achieved by the series of lemmas below.

Lemma 3.9**.**

If $y,k\in\mathbb{N},p\in\mathbb{P}\cap[3,\infty)$ and $\phi(y)\leq\phi(Y_{p,k}),$ then $|W(y)|\leq|W(Y_{p,k})|.$

Proof.

Let $B=W(y)\setminus\{p\}.$ If $B\not\subset W(Y_{p,k}),$ then $\min(B\setminus W(Y_{p,k}))>\max(W(Y_{p,k})).$ Since $\phi(y)\leq\phi(Y_{p,k}),$ we have $\prod_{q\in B}(q-1)\leq\prod_{q\in W(Y_{p,k})}(q-1).$ By Lemma 3.4, it follows that $|B|<|W(Y_{p,k})|$ and hence $|W(y)|\leq|W(Y_{p,k})|.$ If $B$ is a proper subset of $W(Y_{p,k})$ or $B=W(Y_{p,k})\text{ with }$ $p\not\in W(y),$ then $|W(y)|\leq|W(Y_{p,k})|.$

If $B=W(Y_{p,k})$ and $p\in W(y),$ then $\phi(y)=\prod_{q\in W(y)}q^{v_{q}(y)-1}(q-1)$ $\geq$ $(p-1)\phi(Y_{p,k})>\phi(Y_{p,k})$ as $p\geq 3,$ a contradiction. Hence, $|W(y)|\leq|W(Y_{p,k})|.$ ∎

Lemma 3.10**.**

Suppose $y,k\in\mathbb{N}$ and $p\in\mathbb{P}$ satisfy $|W(y)|\leq|W(Y_{p,k})|.$ If there exists a prime $s\in[2,p]$ such that $s\notin W(y),$ then $D(W(y),W(Y_{p,k}))\geq 1.$

Proof.

If $s=p\notin W(y),$ then $\min(W(y)\setminus W(Y_{p,k}))>\max(W(Y_{p,k}))$ or $W(y)\subset W(Y_{p,k}).$ Then Lemma 3.3 along with $|W(y)|\leq|W(Y_{p,k})|$ ensure the result in this case. Suppose $p\in W(y)$ and $s\in[2,p)$ be a prime such that $s\notin W(y).$ Define

[TABLE]

Observe that $|B|\leq|C|$ as $|W(y)|\leq|W(Y_{p,k})|.$ Since $s\not\in W(y),$ it follows that $\min(B\setminus C)>\max(C)$ or $B\subset C.$  So, $D(B,C)\geq$ 1 by Lemma 3.3. Therefore,

[TABLE]

∎

Now we shall state an important quantitative result due to Nagura on primes in short intervals to characterize elements of the type $Y_{p,k}$ in $N_{1}.$

Proposition 3.11** (Nagura [4]).**

Let $n$ be a positive integer greater than $25.$ Then the interval $(n,1.2n)$ contains a prime.

Lemma 3.12**.**

Let $p_{i}$ be the $i\text{th}$ smallest prime greater than $p.$ Then, $p_{3}<\frac{9}{5}p\ \forall\ p\geq 11$ and $p_{1}<\frac{9}{5}p\ \forall\ p\geq 5.$

Proof.

Let $p>25$ be a prime. Then by Proposition 3.11, we have $p_{1}<1.2p,\ p_{2}<(1.2)^{2}p$ and $p_{3}<(1.2)^{3}p.$ Hence $p_{3}<\frac{9}{5}p.$ For primes less than twenty-five, one can easily verify the truth of the lemma. ∎

From Lemma 3.9, $y$ has at most $k-1$ prime factors greater than $p$ in the case when $W(y)\supset[2,p]\cap\mathbb{P}.$ The following lemma describes the situation when $y$ has exactly $k-1$ prime factors greater than $p.$

Lemma 3.13**.**

*Let $k\in\mathbb{N},\ k\leq 3.$ If $p\in\mathbb{P}$ and $y\in\mathbb{N}$ satisfy

(a) $W(y)\supset[2,p]\cap\mathbb{P},$ (b) $|W(y)\cap(p,\infty)|=k-1$and (c) $y$is not squarefree,

then $\phi(y)>\phi(Y_{p,1})\ \forall\ p\geq 5$ and $\phi(y)>\phi(Y_{p,k})\ \forall\ p\geq 11,k\in\{2,3\}.$*

Proof.

Let $p\in\mathbb{P}.$ Let $p_{i}$ be the $i\text{th}$ smallest prime greater than $p$ for $i=1,2,\dots,k$ and $(q_{i})_{i=1}^{k-1}$ be $k-1$ primes greater than $p$ in increasing order such that

[TABLE]

If possible, let $\phi(y)\leq\phi(Y_{p,k}).$ Since $y$ is not squarefree, $\ \exists\ r\in W(y)$ such that $v_{r}(y)>1.$ Then $\phi(y)\leq\phi(Y_{p,k})\Rightarrow r(p-1)\prod_{i=1}^{k-1}(q_{i}-1)\leq\prod_{i=1}^{k}(p_{i}-1).$ This implies that $2(p-1)\leq(p_{k}-1)$ i.e. $p_{k}\geq 2p-1$ using the facts that $r\geq 2$ and $q_{i}\geq p_{i}$ for $i\in[1,k-1].$ Then $p_{k}\geq\frac{9}{5}p$ $\forall\ p\geq 5$ which contradicts Lemma 3.12. Therefore, $\phi(y)>\phi(Y_{p,1})$ $\forall$ $p\geq 5$ and $\phi(y)>\phi(Y_{p,k})$ $\forall$ $p\geq 11,~{}k\in\{2,3\}.$

∎

Proposition 3.14**.**

For any prime $p\geq 5,\ Y_{p,1}\in N_{1}.$

Proof.

Let $Y_{p,1}\notin N_{1}$ for some $p\geq 5.$ Then, there exists $y\in\mathbb{N}$ such that $\phi(y)\leq\phi(Y_{p,1})$ and $y>Y_{p,1}.$ Then Lemma 3.2 gives $D(W(y),W(Y_{p,1}))<1$ and Lemma 3.9 gives $|W(y)|\leq|W(Y_{p,1})|.$ If there exists a prime $s\in[2,p]$ such that $s\notin W(y),$ then by Lemma 3.10, we have $D(W(y),W(Y_{p,1}))\geq 1,$ a contradiction. On the other hand, if $W(y)\supset[2,p]\cap\mathbb{P},$ then Lemma 3.13 together with $\phi(y)\leq\phi(Y_{p,1})$ and $|W(y)|\leq|W(Y_{p,1})|$ implies that $y=\prod_{q\leq p}q<Y_{p,1},$ a contradiction. Hence, $Y_{p,1}\in N_{1}$ $\forall$ $p\geq 5.$ ∎

In the next two lemmas, for $k=2,3,$ we analyze the situation where $y$ contains at most $k-2$ prime factors greater than $p.$

Lemma 3.15**.**

For $k=2$ or $3$ and $p\geq 11,$ $(p-1)p_{k-1}p_{k}>p(p_{k-1}-1)(p_{k}-1),$ where $p_{i}$ is the $i\text{th}$ smallest prime greater than $p.$

Proof.

For $k\in\{2,3\},$ Lemma 3.12 gives $p_{k}<\frac{9}{5}p$ $\forall$ $p\geq 11.$ This along with $p_{k}>p_{k-1}+1,$ gives $\frac{p_{k-1}p_{k}}{p_{k}+p_{k-1}-1}<\frac{p_{k}}{2}<p,\text{ i.e., }(p-1)p_{k-1}p_{k}>p(p_{k-1}-1)(p_{k}-1).$ ∎

Lemma 3.16**.**

*Let $k=2$ or $3.$ If $p\in\mathbb{P},\ p\geq 11$ and $y\in\mathbb{N}$ satisfy

(i) $W(y)\supset[2,p]\cap\mathbb{P}$and

(ii) $|W(y)\cap(p,\infty)|\leq k-2.$

Then we have $\phi(y)\leq\phi(Y_{p,k})\Rightarrow y\leq Y_{p,k}.$*

Proof.

Let $p_{i}$ be the $i\text{th}$ smallest prime greater than $p$ and $q_{1}<q_{2}<\dots<q_{l}$ be the $l\ (\leq k-2)$ prime factors of $y$ greater than $p.$ (In the case when $l=0,$ the rightmost product in the equation below is an empty product with value $1.$) Then,

[TABLE]

Inserting $\phi(y)\leq\phi(Y_{p,k})$ in the above equation, we get

[TABLE]

By the definition of $q_{i}$ and $p_{i},$ we have $q_{i}\geq p_{i}$ for $i\in[1,l].$ Using this along with the Euler product formula for $Y_{p,k}$ in the above equation, we have

[TABLE]

Using Lemma 3.15, we obtain $y\leq Y_{p,k}.$ ∎

We shall see, for $k=2,$ which of the $Y_{p,k}$ are in $N_{1}.$ Recall $E(2),D(p)$ and $A(p)$ from Definition 1.7.

Proposition 3.17**.**

Let $p\geq 11$ be a prime and $p\in E(2),$ then $Y_{p,2}\in N_{1}.$

Proof.

To show $Y_{p,2}\in N_{1},$ it is enough to show that $D(W(y),W(Y_{p,2}))\geq 1$ for $y>Y_{p,2}$ with $\phi(y)\leq\phi(Y_{p,2})$ by Lemma 3.2.

Consider $y\in\mathbb{N}$ and $p\geq 11$ satisfying $y>Y_{p,2}$ with $\phi(y)\leq\phi(Y_{p,2}).$ Then by Lemma 3.9, we have $\left|W(y)\right|\leq\left|W(Y_{p,2})\right|.$

Suppose that $[2,p]\cap\mathbb{P}\subset W(y).$ Since $\left|W(y)\right|\leq\left|W(Y_{p,2})\right|,$ it follows that $W(y)\cap(p,\infty)$ contains at most one element. If $W(y)\cap(p,\infty)=\varnothing,$ then Lemma 3.16 gives us $y\leq Y_{p,2},$ a contradiction. On the other hand, if $W(y)\cap(p,\infty)=\{q_{1}\}$ for some prime $q_{1},$ then we have $y=\left(\prod_{q\leq p}q\right)q_{1}$ by Lemma 3.13 and the fact that $\phi(y)\leq\phi(Y_{p,2}).$ Since $y>Y_{p,2}$ and $\phi(y)\leq\phi(Y_{p,2}),$ it follows that $q_{1}\in(A(p),A(p)+D(p)]\cap\mathbb{P}$ and hence $p\not\in E(2),$ which is a contradiction.

Therefore, $[2,p]\cap\mathbb{P}\not\subset W(y).$ Then by Lemma 3.10, $D(W(y),W(Y_{p,2}))\geq 1.$ ∎

Now, we move on to discuss elements of the type $Y_{p,3}$ in $N_{1}.$ Recall $E(3)$ from Definition 1.7.

Proposition 3.18**.**

Let $p\geq 11$ be a prime and $p\in E(3),$ then $Y_{p,3}\in N_{1}.$

Proof.

To show $Y_{p,3}\in N_{1},$ it is enough to show that $D(W(y),W(Y_{p,3}))\geq 1$ for $y>Y_{p,3}$ with $\phi(y)\leq\phi(Y_{p,3}),$ by Lemma 3.2.

Consider $y\in\mathbb{N}$ and $p\geq 11$ satisfying $y>Y_{p,3}$ with $\phi(y)\leq\phi(Y_{p,3}).$ Then by Lemma 3.9, we have $\left|W(y)\right|\leq\left|W(Y_{p,3})\right|.$

Suppose that $[2,p]\cap\mathbb{P}\subset W(y).$ Since $\left|W(y)\right|\leq\left|W(Y_{p,3})\right|,$ it follows that $W(y)\cap(p,\infty)$ contains at most two elements. If $|W(y)\cap(p,\infty)|\leq 1,$ then Lemma 3.16 and $\phi(y)\leq\phi(Y_{p,3})$ imply $y\leq Y_{p,3},$ a contradiction. On the other hand, if $W(y)\cap(p,\infty)=\{q_{1},q_{2}\}$ for some distinct primes $q_{1},\ q_{2},$ we have $y=\left(\prod_{q\leq p}q\right)q_{1}q_{2}$ by Lemma 3.13 and the fact that $\phi(y)\leq\phi(Y_{p,3}).$ Since $y>Y_{p,3}$ and $\phi(y)\leq\phi(Y_{p,3}),$ we get $p\not\in E(3),$ a contradiction.

Therefore, $[2,p]\cap\mathbb{P}\not\subset W(y).$ Then by Lemma 3.10, $D(W(y),W(Y_{p,3}))\geq 1.$ ∎

Recall $\mathcal{X},\mathcal{Y}_{k}$ for $k\in[1,3]$ from Definition 1.7. One can easily check that $11\in E(2)\cap E(3).$ This shows that $\mathcal{Y}_{2}$ and $\mathcal{Y}_{3}$ are non-empty families. Therefore combining Propositions 3.8, 3.14, 3.17 and 3.18, we get the proof of Theorem 1.8.

Remark 3.19**.**

If $a=X_{n,p}\in\mathcal{X},$ then we can rewrite this as $a=b\prod_{q\leq p}q$ where $b=n\prod_{q\in W(n)}q^{-1}$ depends only on the value of $X_{n,p}.$ Infact, $\mathcal{X}$ can be expressed as $\mathcal{X}=\left\{b\prod_{q\leq p}q\in\mathbb{N}\colon\ b\prod_{q\in W(b)}q<2p\right\}.$

3.2 Proof of Theorem 1.10, Theorem 1.12 and Corollary 1.14

In this section, we will proceed to find explicit elements of $\mathcal{Y}_{2}$ and $\mathcal{Y}_{3}$ and then study the infiniteness of these sets. Due to previous discussions, we turn our attention to the set $E(2)$ and $E(3).$ If $p\in E(2),$ then $(A(p),A(p)+D(p)]\cap\mathbb{P}=\varnothing.$ We shall see below that, as a consequence of the prime number theorem (Lemma 3.20), the length of the interval $(A(p),A(p)+D(p)]$ becomes smaller and smaller as $p$ increases, which reduces the chances of there being a prime in it. Thus, one may expect that $Y_{p,2}\in N_{1}$ for infinitely many $p.$

Lemma 3.20**.**

For $\epsilon>0,$ there is an $n_{0}\in\mathbb{N}$ such that $(n,n(1+\epsilon))\cap\mathbb{P}\neq\varnothing\ \forall\ n>n_{0}.$

Lemma 3.21**.**

$D(p)\rightarrow 0$* as $p\rightarrow\infty.$*

Proof.

Let $\epsilon>0.$ Lemma 3.20 gives $p_{0}(\epsilon)\in\mathbb{N}$ such that $p_{1}<(1+\epsilon)p$ and $p_{2}<(1+\epsilon)^{2}p$ for all primes $p>p_{0}(\epsilon).$ Hence, $D(p)<2\epsilon(\epsilon^{2}+2\epsilon)<6\epsilon^{2}$ and the result follows. ∎

We now give another type of sufficient conditions for $Y_{p,2}$ to be in $N_{1}.$ These conditions are in terms of the distance between successive primes.

Proof of Theorem 1.10(i).

To show that $Y_{p,2}\in N_{1},$ by Proposition 3.17, it is enough to show that $\left(\frac{p_{1}p_{2}}{p},\frac{p_{1}p_{2}}{p}+D(p)\right]\cap\mathbb{P}=\varnothing,$ where $p_{i}$ is the $i\text{th}$ smallest prime greater than $p.$ If possible, let there be a prime $q$ in $\left(\frac{p_{1}p_{2}}{p},\frac{p_{1}p_{2}}{p}+D(p)\right].$ Then, $pq=p_{1}p_{2}+l\ \text{ where }l\in\left[2,\frac{(p_{1}-p)(p_{2}-p)}{p-1}\right]\cap 2\mathbb{N}.$

We now consider several cases according to the value of $p_{1}-p.$ Define $d_{p}:=\displaystyle\frac{p_{1}-p}{2}$ and $\Delta(d_{p}):=\displaystyle\frac{(p_{1}-p)(p_{2}-p)}{p-1}.$ For $p>25,$ we get the following upper bounds for $\Delta(d_{p})$ for various values of $d_{p},$ by using Proposition 3.11.

[TABLE]

Let $d_{p}\in\{1,2\}.$ If $p>25$ or $(p,d_{p})=(11,1),$ $(17,1),$ then $\Delta(d_{p})<2.$ It follows that there does not exist any even $l\in[2,\Delta(d_{p})]$ such that $pq=p_{1}p_{2}+l,$ a contradiction. On the other hand, if $(p,d_{p})=(13,2)$, $(19,2)$, then $\Delta(d_{p})<3$ and thus $pq=p_{1}p_{2}+2.$ We can see by simple calculation that $q$ isn’t a prime here, again giving a contradiction.

Now we come to the case when $d_{p}=3,4.$ For $p>25,$ we again see that $l=2$ is the only possible value in $[2,\Delta(d_{p})].$ So, $pq=p_{1}p_{2}+2.$ Inserting the value $p_{1}=p+2d_{p}$ in this, we get $q=p_{2}+2\frac{d_{p}p_{2}+1}{p}.$ This implies that $k_{d_{p}}:=\frac{d_{p}p_{2}+1}{p}$ is an integer. Since $p<p_{2},$ we observe that $k_{d_{p}}>d_{p}.$ Applying Proposition 3.11, we get $p_{2}<1.2p_{1}=1.2(p+2d_{p}).$ Hence,

[TABLE]

Inequality (5) gives us $k_{3}<4\text{ for }p>53$ and $k_{4}<5\text{ for }p>197,$ which results in a contradiction in each case since $k_{d_{p}}$ is an integer greater than $d_{p}.$ So the only remaining cases are $p=23,31,47,53$ for $d_{p}=3$ and $p=89$ for $d_{p}=4.$ For $(p,d_{p})=(47,3),(53,3)$ and $(89,4),\Delta(d_{p})<2,$ which gives a contradiction to the existence of $q.$ For $(p,d_{p})=(23,3),(31,3),$ we get $\Delta(d_{p})<3$ and so $pq=p_{1}p_{2}+2.$ We can see by simple calculation that $q$ isn’t a prime here.

Therefore, in each of the above cases, we get a contradiction to the existence of a prime in $(\frac{p_{1}p_{2}}{p},\frac{p_{1}p_{2}}{p}+D(p)]$ and hence $Y_{p,2}\in N_{1}.$ ∎

Generalizing the observations in Theorem 1.10(i) to three successive primes, we give certain qualitative results for primes $p$ such that $Y_{p,2}$ is in $N_{1}$ by means of Theorem 1.10(iii).

Proof of Theorem 1.10(iii).

For given $a,b\in 2\mathbb{N},$ let $p\geq 2ab$ and $p\in\mathbb{P}.$ Suppose that $p,p+a$ and $p+b$ are consecutive primes. Clearly $p\geq 11$ and $\frac{ab}{p}\leq\frac{1}{2}.$ It follows that $p+a+b<\frac{(p+a)(p+b)}{p}\leq p+a+b+\frac{1}{2}\text{ and }D(p)\leq\frac{1}{2(p-1)}.$ Therefore $\left(\frac{(p+a)(p+b)}{p},\frac{(p+a)(p+b)}{p}+D(p)\right]\cap\mathbb{N}=\varnothing$, since $p\geq 11.$ So $p\in E(2)$ and hence, by Proposition 3.17, $Y_{p,2}\in N_{1}.$ ∎

The Theorem 1.10(ii) gives concrete examples of elements of the type $Y_{p,3}$ in $N_{1}.$

Proof of Theorem 1.10(ii).

Let $p_{i}$ be the $i\text{th}$ smallest prime greater than $p.$ To prove that $Y_{p,3}\in N_{1},$ it is enough to show that $p\in E(3)$ by Proposition 3.18.

Suppose $p\not\in E(3),$ then there exist primes $q_{1},q_{2}\in\mathcal{O}_{p}$ with $q_{1}<q_{2}$ such that $q_{2}\in\left(\frac{p_{1}p_{2}p_{3}}{pq_{1}},1+\frac{(p_{1}-1)(p_{2}-1)(p_{3}-1)}{(p-1)(q_{1}-1)}\right].$ Consider the cases on the values of $q_{1}.$

  $\underline{\textbf{Case I:}}$ ($q_{1}\geq p_{3}$)

Since $q_{2}>q_{1}$ and $q_{1}\geq p_{3},$ we have $q_{2}\geq p_{3}+2.$ It follows that $p_{3}+1\leq\frac{(p_{1}-1)(p_{2}-1)}{p-1}.$ Setting $p_{1}=p+2,$ $p_{2}=p+6$ and $p_{3}\geq p_{2}+2$ in the above inequality, we get $(p+9)(p-1)\leq(p+1)(p+5),$ i.e., $p\leq 7,$ which is a contradiction.

  $\underline{\textbf{Case II:}}$ ($q_{1}=p_{2}$)

In this case, $q_{2}\in\left(\frac{p_{1}p_{3}}{p},\frac{p_{1}p_{3}}{p}+\frac{(p_{1}-p)(p_{3}-p)}{p(p-1)}\right].$ By Lemma 3.12, we have $p_{3}<\frac{9p}{5}$ for $p\geq 11.$ Using this along with $p_{1}=p+2,$ we get $q_{2}p\in\left(p_{1}p_{3},p_{1}p_{3}+\frac{8p}{5(p-1)}\right).$ Since $p\geq 11,$ we have $\frac{8p}{5(p-1)}<2.$ Inserting this in the above expression, we have $q_{2}p=p_{1}p_{3}+1$ which is not possible because $q_{2}p$ is odd and $p_{1}p_{3}+1$ is even.

  $\underline{\textbf{Case III:}}$ ($q_{1}=p_{1}$)

In this case, $q_{2}\in\left(\frac{p_{2}p_{3}}{p},\frac{p_{2}p_{3}}{p}+\frac{(p_{2}-p)(p_{3}-p)}{p(p-1)}\right].$ By Lemma 3.12, we have $p_{3}<\frac{9p}{5}$ for $p\geq 11.$ Along with $p_{2}=p+6,$ this gives $q_{2}p\in\left(p_{2}p_{3},p_{2}p_{3}+\frac{24p}{5(p-1)}\right).$ Since $\frac{24p}{5(p-1)}<6$ for $p\geq 11,$ $q_{2}p$ can be either $p_{2}p_{3}+2$ or $p_{2}p_{3}+4.$ Since $p_{2}=p+6,$ we have $q_{2}=p_{3}+2d,$ where $d$ is an integer equal to either $\frac{3p_{3}+1}{p}$ or $\frac{3p_{3}+2}{p}.$ Since $p<p_{3}<\frac{9p}{5}$ by Lemma 3.12, we get $3<d<6.$ It follows that either $\frac{3p_{3}+1}{p}=4$ or $\frac{3p_{3}+2}{p}=5$ as $3p_{3}+1$ is an even integer. In both cases, $p\equiv 1\pmod{3}$ and hence $p+2\equiv 0\pmod{3}.$ This contradicts the fact that $p+2$ is a prime.

Therefore, in each case, we get a contradiction to the existence of the prime $q_{1}.$ Hence $p\in E(3).$ ∎

Now we shall see a criteria for the infiniteness of the set $\mathcal{Y}_{2}$ in Theorem 1.12 by taking a condition on the fractional part of $\frac{p_{1}p_{2}}{p}.$

Proof of Theorem 1.12.

By Lemma 3.21, $D(p)\rightarrow 0$ as $p\rightarrow\infty.$ So there exists a prime $p_{0}(\delta_{0})$ such that $D(p)<1-\delta_{0}\ \forall\ p>p_{0}(\delta_{0}).$ Therefore, for $p\in A\cap\left[p_{0}(\delta_{0}),\infty\right),$ we have

[TABLE]

Hence, $\left(\frac{p_{1}p_{2}}{p},\frac{p_{1}p_{2}}{p}+D(p)\right]$ doesn’t contain any integers for $p\in A\cap[p_{0}(\delta_{0}),\infty).$ Therefore $p\in E(2)$ for each $p$ in $A\cap[p_{0}(\delta_{0}),\infty).$ Hence, by Proposition 3.17, $Y_{p,2}\in N_{1}$ for $p\in A\cap[p_{0}^{\prime}(\delta_{0}),\infty),$ where $p_{0}^{\prime}(\delta_{0})=\max\{11,p_{0}(\delta_{0})\}.$ ∎

We are now going to show that $\mathcal{Y}_{2}$ is an infinite set using the result on Bounded gaps between consecutive primes by Yitang Zhang (see Proposition 1.13).

Proof of Corollary 1.14.

Let $\epsilon<1$ be a positive real number and $C=7\times 10^{7}.$ Let $p,p_{1}$ and $p_{2}$ be consecutive primes for each $p\in\mathbb{P}.$ By prime number theorem, there exists $p_{0}(\epsilon)\in\mathbb{N}$ such that $p_{2}-p<\frac{\epsilon p}{C}$ $\forall$ $p\geq p_{0}(\epsilon).$ Consider the set

[TABLE]

By Proposition 1.13, $R$ is an infinite set.

Let $p\in R.$ Then $p_{1}=p+N$ for some $N<C.$ It gives us $\frac{p_{1}p_{2}}{p}=p_{2}+N+\frac{N(p_{2}-p)}{p}$ where $\frac{N(p_{2}-p)}{p}<\frac{N\epsilon}{C}<\epsilon$ as $p_{2}-p<\frac{\epsilon p}{C}$ for $p\geq p_{0}(\epsilon)$ and $N<C.$ Hence the fractional part of $\frac{p_{1}p_{2}}{p}$ is less than $\epsilon$ for each $p\in R.$ Then Theorem 1.12 ensures that $Y_{p,2}$ is a sparsely totient number for all sufficiently large primes $p\in R.$ Since $R$ is an infinite set, $\mathcal{Y}_{2}$ is an infinite set. ∎

4 Additive and multiplicative patterns in sparsely totient numbers

We now turn our attention to additive and multiplicative configurations like sets of finite sums, sets of finite products and arithmetic and geometric progressions inside $N_{1}.$ Below, we give the relevant definitions.

Let $G$ be a non-empty set and let $P_{f}(G)$ be the collection of all non-empty finite subsets of $G.$

Definition 4.1**.**

Let $(G,+)$ be a commutative semigroup and $A\subset G.$ Then

1. (a)

$A$* is syndetic iff there exists $S\in P_{f}(G)$ such that $G=\displaystyle\cup_{t\in S}(-t+A).$* 2. (b)

$A$* is thick iff whenever $F\in P_{f}(G),$ there exists $x\in G$ such that $F+x\subset A.$* 3. (c)

$A$* is piecewise syndetic iff $\exists$ $S\in P_{f}(G)$ such that $\displaystyle\cup_{t\in S}(-t+A)\text{ is thick}.$*

If $G=\mathbb{N}$ with binary operation addition $+(\text{or }\cdot),$ then syndetic, thick and piecewise syndetic sets are called, respectively, additively (or multiplicatively) syndetic, additively (or multiplicatively) thick and additively (or multiplicatively) piecewise syndetic.

Proposition 4.2**.**

$N_{1}$* is multiplicatively piecewise syndetic but not additively piecewise syndetic.*

Proof.

Suppose $S\in P_{f}(\mathbb{N}).$ Let $u=\max(S).$ Suppose $p$ is a prime with $p>u.$ Then by Theorem 1.2,  $\exists\ n_{0}(p)\in N_{1}$ such that  $\forall\ n_{1}>n_{2}>n_{0},\ n_{1},n_{2}\in N_{1},$ we have $p\mid(n_{1}-n_{2}).$ In particular, $n_{1}-n_{2}\geq p.$ Observe that the set $\left(\cup_{t=1}^{u}(-t+N_{1})\right)\cap[n_{0}(p),\infty)$ does not contain any interval of length greater than $p.$ This means that $\left(\cup_{t=1}^{u}(-t+N_{1})\right)$ is not additively thick and therefore its subset $\left(\cup_{t\in S}(-t+N_{1})\right)$ is also not additively thick. Therefore, $N_{1}$ is not additively piecewise syndetic.

We will prove that $N_{1}$ is multiplicatively piecewise syndetic by showing that its subset $\mathcal{X}$ itself has this property. It is enough to show that $2^{-1}\mathcal{X}=\{n\in\mathbb{N}\colon 2n\in\mathcal{X}\}$ is multiplicatively thick. Let $F$ be a finite set. Define $r=\max\{\lambda_{u}\colon u\in F\},$ where $\lambda_{u}=2u\prod_{q\in W(2u)}q.$

Let $p\in\mathbb{P}$ and $p>r.$ Then, $X_{\lambda_{u},p}=2u\prod_{q\leq p,q\in\mathbb{P}}q\ \forall\ u\in F.$ Since $\lambda_{u}\leq r<p,$ we have $X_{\lambda_{u},p}\in\mathcal{X}.$ Hence, $F\cdot\displaystyle\prod_{q\leq p,q\in\mathbb{P}}q\subset 2^{-1}\mathcal{X}$ and so $2^{-1}\mathcal{X}$ is a thick set. ∎

Definition 4.3** (Finite sums and Finite products).**

Let $J$ be a subset of $\mathbb{N}$ and suppose $x=(x_{n})_{n\in J}$ is a sequence indexed by this subset. Then we define the set of Finite sums $FS(x)$ and the set of Finite products $FP(x)$ as follows:

[TABLE]

Definition 4.4**.**

Let $A\subset\mathbb{N}$ and $r\in\mathbb{N}.$ Then

1. (a)

$A$* is called an additive $IP_{r}$ set (respectively, multiplicative $IP_{r}$ set) if there is a sequence $x=(x_{n})_{n\in J}$ with $|J|=r$ such that $FS(x)\subset A$ $(FP(x)\subset A).$* 2. (b)

$A$* is called an additive $IP_{0}$ set (respectively, multiplicative $IP_{0}$ set) if $A$ is an additive $IP_{r}$ set (multiplicative $IP_{r}$ set) for each $r\in\mathbb{N}.$* 3. (c)

$A$* is called an additive $IP$ set (respectively, multiplicative $IP$ set) if there exists an infinite sequence $x=(x_{n})_{n\in\mathbb{N}}$ such that $FS(x)\subset A$ $(FP(x)\subset A).$*

First, we study the existence of an additive $IP_{r}$ set in $N_{1}.$ For example, if $r=2,$ by an additive $IP_{2}$ set, we mean a set which contains a triplet of the form $\{x,y,x+y\}.$ In the following proposition, we show that $\mathcal{X}$ contains the set of finite sums of arbitrarily long finite sequences.

Proposition 4.5**.**

$\mathcal{X}$* is an additive $IP_{0}$ set.*

Proof.

It is enough to show that $\mathcal{X}$ is an additive $IP_{r}$ set for each $r\in\mathbb{N}.$ Let $r\in\mathbb{N}$ and $\{n_{i}\colon i\in[1,r]\}\subset\mathbb{N}.$ Define $m_{i}:={n_{i}}{\prod_{q\in W(n_{i})}q^{-1}}$ and $u_{I}=\sum_{i\in I}m_{i}$ $\forall$ $i\in[1,r]$ and $I\subset[1,r].$

Let $p$ be a prime such that $p>\left(\sum_{i=1}^{r}m_{i}\right)^{2}$ and consider the finite sequence $E=\{X_{n_{i},p}\}_{i=1}^{r}.$ We will show that $\mathcal{X}$ is an additive $IP_{r}$ set by showing $FS(E)\subset\mathcal{X}.$

Let $P$ be the product of primes less than or equal to $p.$ Then $X_{n_{i},p}=m_{i}P.$ It gives $\sum_{i\in I}X_{n_{i},p}=\left(\sum_{i\in I}m_{i}\right)P=u_{I}P$ for $I\subset[1,r].$ So, $\sum_{i\in I}X_{n_{i},p}=X_{b_{I},p},$ where $b_{I}=u_{I}\prod_{q\in W(u_{I})}q\leq u_{I}^{2}<p,\text{ as }I\subset[1,r].$ Thus, $X_{b_{I},p}\in\mathcal{X}.$ Hence, $FS(E)\subset\mathcal{X}.$ ∎

However, as the next proposition shows, we see that $N_{1}$ does not contain the set of finite sums of any infinite sequence via an application of Theorem 1.2.

Proposition 4.6**.**

$N_{1}$* is not an additive IP set.*

Proof.

If possible, suppose $N_{1}$ is an additive $IP$ set. Then there exists an increasing infinite sequence $x=(x_{n})_{n\in\mathbb{N}}$ in $N_{1}$ such that $FS(x)\subset N_{1}.$ Set $y_{n}=\sum_{i=1}^{n}x_{i}$ and $z_{n}=\sum_{i=2}^{n}x_{i}$ for $n\geq 2.$ Since $FS(x)\subset N_{1},$ it follows that $y_{n},z_{n}\in N_{1}$ and $y_{n}-z_{n}=x_{1}$ $\forall$ $n\geq 2.$

Suppose $p$ is a prime greater than $x_{1}.$ Then by Theorem 1.2, there exists $u_{0}\in\mathbb{N}$ such that $u\geq u_{0}$ and $u\in N_{1}\Rightarrow u\equiv 0\pmod{p}.$ Since $(y_{n})$ and $(z_{n})$ are strictly increasing sequences, there exists $n_{0}\in\mathbb{N}$ such that $y_{n}\geq u_{0}$ and $z_{n}\geq u_{0}$ for all $n\geq n_{0}.$ Since $y_{n},z_{n}\in N_{1}$ for all $n\geq\max\{n_{0},2\},$ it follows that $x_{1}=y_{n}-z_{n}\equiv 0\pmod{p},$ a contradiction to $0<x_{1}<p.$ Therefore $N_{1}$ is not an additive $IP$ set. ∎

But, $N_{1}$ does contain the set of finite products of an infinite sequence as shown below.

Proposition 4.7**.**

$\mathcal{X}$* is a multiplicative $IP$ set.*

Proof.

Define an increasing sequence of primes $(p_{n})_{n\in\mathbb{N}}$ and an increasing sequence $(x_{n})_{n\in\mathbb{N}}$ in $\mathcal{X}$ such that

[TABLE]

Let $E=FP((x_{n})_{n\in\mathbb{N}}).$ To prove that $\mathcal{X}$ is a multiplicative $IP$ set, it suffices to show that $E\subset\mathcal{X}.$ Suppose $\prod_{i=1}^{l}x_{k_{i}}\in E,$ where $k_{i}<k_{i+1}$ $\forall$ $i\in[1,l).$ If $k_{l}=1,$ then $l=1$ and $\prod_{i=1}^{l}x_{k_{i}}=x_{1}=X_{2,2}\in\mathcal{X}.$ On the other hand, if $k_{l}>1,$ we have $\displaystyle\prod_{i=1}^{l}x_{k_{i}}=X_{b,p_{k_{l}}}$ with $b=X_{2,p_{k_{1}}}^{l}\displaystyle\prod_{i=2}^{l-1}\left(\displaystyle\frac{X_{2,p_{k_{i}}}}{X_{2,p_{k_{i-1}}}}\right)^{l+1-i}\leq X_{2,p_{k_{l-1}}}^{l}=x_{k_{l-1}}^{l}.$ Since $k_{i}<k_{i+1}$ $\forall$ $i\in[1,l),$ we get that $k_{l-1}\leq k_{l}-1$ and $k_{l}\geq l.$ Then $b\leq x_{k_{l}-1}^{k_{l}}$ as $(x_{n})_{n\in\mathbb{N}}$ is an increasing sequence. Since $k_{l}>1,$ we get $b<p_{k_{l}}$ and hence $\displaystyle\prod_{i=1}^{l}x_{k_{i}}=X_{b,p_{k_{l}}}\in\mathcal{X}.$ This completes the proof. ∎

Next, we would like to know whether additive and multiplicative configurations can occur together in $N_{1}.$ Even for the simplest configuration $\{x+y,xy\}$ for $x,y\in\mathbb{N},$ we do not know the answer to this. However, we prove that $\mathcal{X}$ doesn’t contain such a pair.

Proposition 4.8**.**

The set $\mathcal{X}$ does not contain a configuration of the type $\{x+y,xy\}$ for any positive integers $x,y.$

Proof.

If possible assume that $x+y,xy\in\mathcal{X}$ for some $x,y\in\mathbb{N}.$ Let $x+y=X_{n_{1},p_{1}}$ and $xy=X_{n_{2},p_{2}}$ for some $p_{1},p_{2}\in\mathbb{P}$ and $n_{1},n_{2}\in\mathbb{N}$ satisfying $p_{1}>\frac{n_{1}}{2}$ and $p_{2}>\frac{n_{2}}{2}.$

Since $xy=X_{n_{2},p_{2}}$ with $p_{2}>\frac{n_{2}}{2},$ we have $v_{p_{2}}(xy)=1.$ This means that $p_{2}$ divides exactly one of $x$ and $y.$ Then $v_{p_{2}}(x+y)=0.$ Therefore $p_{2}>p_{1}$ as $x+y=X_{n_{1},p_{1}}.$

Since $p_{1}<p_{2},$ we have $v_{q}(x+y)\geq 1$ and $v_{q}(xy)\geq 1$ for each prime $q\leq p_{1}.$ This gives $v_{q}(x)\geq 1$ and $v_{q}(y)\geq 1$ for each prime $q\leq p_{1}.$ Then there exists $T(x,y)\in\mathbb{N}$ such that

[TABLE]

[TABLE]

Since $xy=X_{n_{2},p_{2}},$ we have $p_{2}\mid x$ or $p_{2}\mid y$ and hence $T(x,y)\geq p_{2}$, as $p_{2}>p_{1}.$

Using the fact that $x+y=X_{n_{1},p_{1}}\in\mathcal{X}$ along with the definition of $\mathcal{X}$ as stated in Remark 3.19, eq.(6) gives us

[TABLE]

Since $T(x,y)>1,$ there exists a prime $q_{0}$ such that $q_{0}\mid T(x,y).$ So, the inequality in (7) gives $2T(x,y)\leq q_{0}T(x,y)<2p_{1}$ i.e. $T(x,y)<p_{1}<p_{2},$ a contradiction to the fact that $T(x,y)\geq p_{2}.$

Hence, there exist no $x,y\in\mathbb{N}$ such that $x+y,xy\in\mathcal{X}.$ ∎

Now, we look for arithmetic and geometric progressions in $N_{1}.$ By Propopsition 4.2 and a result in [2] stated below, we conclude that $N_{1}$ contains arbitrarily long arithmetic and geometric progressions.

Proposition 4.9** (Hindman [2], p. 361).**

Let $A$ be a multiplicatively piecewise syndetic subset of $\mathbb{N}$ and $k\in\mathbb{N}.$ Then there exists $a,d\in\mathbb{N}$ and $r\in\mathbb{N}\setminus\{1\}$ such that {$r^{j}(a+id)\colon i,j\in\{0,1,\dots,k\}\}\cup\{dr^{j}\colon j\in\{0,1,\dots,k\}\}\subset A.$

Proposition 4.10**.**

$N_{1}$* contains arbitrarily long arithmetic, geometric and geo-arithmetic progressions.*

Using the ideas of Proposition 4.5, we will now give explicit examples of infinite families of arithmetic and geometric progressions in $N_{1}$ via $\mathcal{X}.$ Suppose $\{m_{1},m_{2},\dots,m_{n}\}$ is an arithmetic (or geometric) progression of length $n$ in $\mathbb{N}.$ Let $b_{i}=m_{i}\prod_{q\in W(m_{i})}q$ for each $1\leq i\leq n$ and let $p>\max\{\frac{b_{i}}{2}\colon 1\leq i\leq n\}.$ Then $\{X_{b_{i},p}\}_{i=1}^{n}$ is an arithmetic (or geometric) progression in $\mathcal{X}$ of length $n.$

Combining Propositions 4.2, 4.5, 4.6, 4.7, 4.8 and 4.10, we get Theorem 1.15.

5 Questions

In this section, we state some problems on sparsely totient numbers arising from the present work.

For a prime $p,$ define $TN_{1}(p):=\inf\{m\in V\colon k>m\Rightarrow p\mid N_{1}(k)\}.$ For example, $TN_{1}(2)=1$ and Corollary 1.5 gives $TN_{1}(3)=2.$ Theorem 1.2 ensures the existence of $TN_{1}(p)$ for each $p$. One can observe that $TN_{1}(p)=\max\{m\in V\colon p\nmid N_{1}(m)\}.$ Proposition 1.3 and Proposition 1.4 give us that if $p\nmid N_{1}(k),$ then $v_{q}(N_{1}(k))=0$ for $q>p(p-1)$, $v_{q}(N_{1}(k))\leq 2$ for $p<q\leq p(p-1),$ and $v_{q}(N_{1}(k))\leq D_{p}(q)$ for $q<p$, where $D_{p}(q)$ is independent of $k.$ Here we do not know quantitative information about $D_{p}(q)$. A knowledge of this will give an explicit upper bound of $TN_{1}(p)$. These observations give rise to the following problems:

Question 3**.**

For a prime $p,$ what is the value of $TN_{1}(p)$?

Question 4**.**

Let $p$ and $q$ be primes such that $q<p$. If $N$ is a sparsely totient number such that each prime factor of $N$ is less than $p,$ then find an explicit upper bound for $v_{q}(N)$.

Let $m_{1},m_{2},\dots,m_{k},\dots$ be the enumeration of the elements of the set $BN_{1},$ where $BN_{1}=\{m\in V:N_{1}(m)=\max(\phi^{-1}(m))\}$. From the table of values of $N_{1}(m)$ below, it seems that $N_{1}(m_{k})>k^{2}$ for each $k.$ Can we prove $N_{1}(m_{k})>k^{1+\epsilon},$ where $\epsilon$ is a small positive real number. This would also mean that $\sum_{m\in BN_{1}}(N_{1}(m))^{-1}$ is convergent.

[TABLE]

Hence we ask the following question.

Question 5**.**

Does there exist $\epsilon>0$ such that $N_{1}(m_{k})>k^{1+\epsilon}$ for each $k\in\mathbb{N}$? Does $\sum_{m\in BN_{1}}(N_{1}(m))^{-1}$ converge?

As discussed in Section 1 via observations from Figure 1, we raise the following two questions:

Question 6**.**

Let $p,p_{1}$ and $p_{2}$ be consecutive primes for each $p\in\mathbb{P}.$ Does the fractional part of $\frac{p_{1}p_{2}}{p}$ tend to [math] as $p\rightarrow\infty$?

Question 7**.**

Is $\mathcal{Y}_{2}=\{Y_{p,2}\colon p\in\mathbb{P},p\geq 11\}$? Is the set $\mathcal{Y}_{3}$ infinite?

As we have seen in Proposition 4.8, $\mathcal{X}$ doesn’t contain a configuration of the type $\{x+y,xy\}.$ So, the question arises whether such a configuration is possible in other families of subsets of $N_{1}.$ More generally,

Question 8**.**

Does $N_{1}$ contain the set $\{x+y,xy\}$ for some $x,y\in\mathbb{N}$?

To the best of the authors’ knowledge, all the above problems are open.

Acknowledgement

We would like to thank the Department of Atomic Energy, Government of India for the financial support. In addition, the research of authors Bhuwanesh Rao Patil and Pramod Eyyunni was also supported by the ‘INFOSYS scholarship for senior students’. We would also like to thank Harish-Chandra Research Institute for the excellent facilities. We sincerely thank Samrat Kadge for help with the computation and programming. We thank the referee for a careful reading of this paper and suggestions.