Number of non-primes in the set of units modulo $n$

TL;DR

This paper investigates the properties of an arithmetic function counting non-prime units modulo n, providing insights into its behavior and distribution within number theory.

Contribution

It introduces and analyzes the function f, a novel arithmetic function counting non-prime units modulo n, expanding understanding of unit groups.

Findings

Characterization of f for various n

Distribution patterns of non-prime units

Relationships between f and classical arithmetic functions

Abstract

In this work, we studied various properties of arithmetic function $\tilde{\varphi}$, where $\tilde{\varphi}(n)=|\{m\in \mathbb{N} | 1\le m\le n, (m,n)=1, \mbox{$m$ is not a prime}\}|.$

Full text

Key Words: Greatest common divisor, Arithmetic function, Euler-totient function.

AMS(2010): 11A05,11A25.

1. Introduction

For a fixed positive integer $n$, let $U_{n}=\{k:1\leq k\leq n,\gcd(k,n)=1\}$. If $|S|$ denotes the cardinality of the set $S$, then $|U_{n}|=\varphi(n)$, the well known Euler-totient function. Let E_{n}=\{m\in{\mathbb{N}}|1\leq m\leq n,(m,n)=1,\mbox{m is not a prime}\}. That is E_{n}=\{m\in U_{n}|\mbox{m is not a prime}\}. It is known that $\varphi(n)=n-1$ if and only if $n$ is prime, hence $E_{n}=\{m\in U_{n}|\varphi(m)\neq m-1\}.$ Let $\tilde{\varphi}(n)=|E_{n}|.$ It is clear that $\tilde{\varphi}$ is an arithmetic function. First twenty values of $\tilde{\varphi}$ are given in the following tables.

$n$ $E_{n}$ $\tilde{\varphi}(n)$

1 $\{1\}$ 1

2 $\{1\}$ 1

3 $\{1\}$ 1

4 $\{1\}$ 1

5 $\{1,4\}$ 2

6 $\{1\}$ 1

7 $\{1,4,6\}$ 3

8 $\{1\}$ 1

9 $\{1,4,8\}$ 3

10 $\{1,9\}$ 2

$n$ $E_{n}$ $\tilde{\varphi}(n)$

11 $\{1,4,6,8,9,10\}$ 6

12 $\{1\}$ 1

13 $\{1,4,6,8,9,10,12\}$ 7

14 $\{1,9\}$ 2

15 $\{1,4,8,14\}$ 4

16 $\{1,9,15\}$ 3

17 $\{1,4,6,8,9,10,12,14,15,16\}$ 10

18 $\{1\}$ 1

19 $\{1,4,6,8,9,10,12,14,15,16,18\}$ 11

20 $\{1,9\}$ 2

The following results provides few immediate properties of $\tilde{\varphi}$ function. Recall that for $n\in{\mathbb{N}}$, $\pi(n),\omega(n)$ denotes the number of primes less than or equal to $n$, the number of distinct primes divisors of $n$ respectively. We denote $N_{j}=p_{1}p_{2}\cdots p_{j}$, product of first $j$ primes, for example $N_{3}=30.$ It would be interesting to note that $\frac{\sqrt{n}}{2}<\varphi(n)<n-\sqrt{n}$ and by prime number theorem, for a large enough $n$, $\pi(n)\approx\frac{n}{log(n)}$. But, $\omega(n)$ does not have any ordinary behaviour. In fact for any natural number $k$, one can find a subsequence $\{n_{i}\}$ of natural numbers such that $\omega(n_{i})=k$ for all $i$. To know more properties of $\pi(n),\omega(n),\varphi(n)$ and $U_{n}$ refer any one of [1, 2, 4, 5].

Proposition 0**.**

If $p_{k}$ denotes the $k^{th}$ prime, then $\tilde{\varphi}(p_{k})=p_{k}-k.$ In general $\tilde{\varphi}(n)=\varphi(n)-\pi(n)+\omega(n).$

Proof.

First part follows from the observation that $E_{p_{k}}=U_{p_{k}}\setminus\{p_{1},p_{2},\ldots,p_{k-1}\}.$

Let $n=q_{1}^{{\alpha}_{1}}q_{2}^{{\alpha}_{2}}\cdots q_{k}^{{\alpha}_{k}}.$ We claim $U_{n}\cup\{q_{1},q_{2},\ldots,q_{k}\}=E_{n}\cup\{p_{1},p_{2},\ldots,p_{\pi(n)}\}$ Suppose $x\in U_{n}\cup\{q_{1},q_{2},\ldots,q_{k}\}.$ Now if $x$ is prime, then then $x\in\{p_{1},p_{2},\ldots,p_{\pi(n)}\}$ and if $x$ is not prime, then $x\in E_{n}.$ Thus $x\in E_{n}\cup\{p_{1},p_{2},\ldots,p_{\pi(n)}\}.$

Suppose $y\in E_{n}\cup\{p_{1},p_{2},\ldots,p_{\pi(n)}\}.$ If $y\in E_{n}$, then $y\in U_{n}$ if $y\in\{p_{1},p_{2},\ldots,p_{\pi(n)}\}$ then $y\in U_{n}$ or $y\in\{q_{1},q_{2},\ldots,q_{k}\}$ depending on $y\nmid n$ or $y|n$ respectively. By definition, $E_{n}\cap\{p_{1},p_{2},\ldots,p_{\pi(n)}\}=U_{n}\cap\{q_{1},q_{2},\ldots,q_{k}\}=\emptyset.$ Hence $\tilde{\varphi}(n)+\pi(n)=|E_{n}|+|\{p_{1},p_{2},\ldots,p_{\pi(n)}\}|=|U_{n}|+|\{q_{1},q_{2},\ldots,q_{k}\}|=\varphi(n)+\omega(n).$ ∎

Proposition 1**.**

1. (1)

If $n_{0}$ is the square free part of $n$ then $\tilde{\varphi}(n)\geq\tilde{\varphi}(n_{0}).$ 2. (2)

*Let $n,k\in{\mathbb{N}}$ and $n\geq 3.$ Then $\tilde{\varphi}(n^{k})<\tilde{\varphi}(n^{k+1}).$ * 3. (3)

Let $n,p,q\in{\mathbb{N}}$ where $p,q$ primes with $p<q$ and $(p,n)=1$. Then $\tilde{\varphi}(np)\leq\tilde{\varphi}(nq).$ 4. (4)

If $a\in{\mathbb{N}}$ and $\omega(a)=i$, then $\tilde{\varphi}(N_{i})\leq\tilde{\varphi}(a).$

Proof.

Proof of Part 1: Since $n_{0}$ and $n$ have same set of prime factors, $E_{n_{0}}\subseteq E_{n}.$

Proof of Part 2: It is easy to see that $E_{n^{k}}\subseteq E_{n^{k+1}}.$ Sufficient to show that $E_{n^{k+1}}\setminus E_{n^{k}}\neq\emptyset.$

Since $n\geq 3$ there exists $\ell\in\{2,3,\ldots\}$ such that $(n-1)^{\ell-1}\leq n^{k}<(n-1)^{\ell}.$ Consequently $n^{k}<(n-1)^{\ell}<(n-1)^{\ell-1}\cdot n\leq n^{k+1}.$ Thus we produced an element $(n-1)^{\ell}$ in $E_{n^{k+1}}\setminus E_{n^{k}}.$

Proof of Part 3: Let $x\in E_{np}$ and $q^{m}||x$ for some $m\in{\mathbb{N}}\cup\{0\}.$ Then $x=q^{m}y$ for some $y\in{\mathbb{N}}.$ Further that $(p^{m}y,nq)=1$ as $(p^{m},n)=1,(y,q)=1.$ Now we define a map $f:E_{np}\to E_{nq}$ as $f(x)=p^{m}y.$ It is easy to see that $f$ is one-one. Hence the result follows.

Proof of Part 4 Let $q_{1},q_{2},\ldots,q_{i}$ be all distinct prime divisors of $a$ and $s=q_{1}q_{2}\cdots q_{i}.$ Then from the proof of Part 1 we have $\tilde{\varphi}(s)\leq\tilde{\varphi}(a).$ Hence the result follows from the facts that $N_{i}\leq s$ and $E_{N_{i}}\setminus\{q_{1},q_{2},\ldots,q_{i}\}\subseteq E_{s}\setminus\{p_{1},p_{2},\ldots,p_{i}\}.$ ∎

2. Main results

It is easy to see that $\tilde{\varphi}(n)=1$ if and only if $n\in\{1,2,3,4,6,8,12,18,24,30\}.$ That is $\tilde{\varphi}(\ell)>1$ whenever $\ell\geq 31.$ In general we prove.

Theorem 2**.**

For every $n\in{\mathbb{N}}$, there exists $N\in{\mathbb{N}}$ such that $\tilde{\varphi}(m)>n$, for all $m>N.$

Lemma 3**.**

Let $n\in{\mathbb{N}}.$ There exists an $M_{n}\in{\mathbb{N}}$ such that $\tilde{\varphi}(N_{k})>n$ for all $k\in{\mathbb{N}}$, $k\geq M_{n}.$

Proof.

Let $p_{i}$ denote the $i^{th}$ prime. For $i\geq 4,$ we define $P_{i}=\{m\in{\mathbb{N}}|\varphi(m)=m-1,m\in U_{N_{i}}\}$, $Q_{i}=\{r\in P_{i}|rp_{i+1}<N_{i}\}.$ From Bertrand’s Postulate $p_{i+1}\in Q_{i}.$ Suppose $Q_{i}=\{p_{i+1},p_{i+2},\ldots,p_{h}\}$ then $|Q_{i}|=h-i.$ Further $p_{i+2}<2p_{i+1}$ and $p_{h+2}<2p_{h+1}<4p_{h}.$ Thus $p_{i+2}p_{h+2}<8p_{i+1}p_{h}<8N_{i}<N_{i+1}$ as $p_{i+1}>8.$ Hence $\{p_{i+2},\ldots,p_{h+2}\}\subseteq Q_{i+1}$ consequently $|Q_{i}|<|Q_{i+1}|.$ Now $\{rp_{i+1}|r\in Q_{i}\}\subset E_{N_{i}}.$ Thus $\tilde{\varphi}(N_{i})>|Q_{i}|$. Hence the result follows from the fact that $|Q_{i}|$ increases with $i.$ ∎

Since $|Q_{4}|=4$, therefore for all $n\geq 4$, $|Q_{n}|\geq n$ and $\tilde{\varphi}(N_{k})>k$ when $k\geq 4$. Hence $M_{n}\leq\max\{4,n\}$. Further, it is easy to see that $\tilde{\varphi}$ is an increasing function on $\{N_{3},N_{4},N_{5},\ldots\}.$

Lemma 4**.**

Let $a,b\in{\mathbb{N}}$ and $A(a,b)=\{n\in{\mathbb{N}}|\tilde{\varphi}(n)=a,\omega(n)=b\}.$ Then $A(a,b)$ is finite.

Proof.

Let $p$ be the $(b+\ell)^{th}$ prime, where $a<\frac{\ell(\ell+1)}{2}.$ Let $n\in{\mathbb{N}}$ with $\omega(n)=b$ and $n>p^{2}.$ Since there are at least $\ell$ primes less than $p$,which are co-prime to $n$, we can choose two such primes $q_{1},q_{2}.$ Then $q_{1}q_{2}<p^{2}<n.$ Thus $q_{1}q_{2}\in E_{n}$ as a consequence $\tilde{\varphi}(n)>\frac{\ell(\ell+1)}{2}>a.$ Hence the result follows. ∎

Now we will prove Theorem 2.

Proof.

Let $n\in{\mathbb{N}}.$ There exists $k\in{\mathbb{N}}$ such that if $\omega(t)>k$, then $\tilde{\varphi}(t)\geq\tilde{\varphi}(N_{\omega(t)})>n.$ Now from Lemma 4 for each $i\in{\mathbb{N}}$, $i\leq k$ there exists $m_{i}\in{\mathbb{N}}$ such that for all $t\in{\mathbb{N}}$ with $\omega(t)=i$ and $t>m_{i}$, then $\tilde{\varphi}(t)>n.$ Thus if $N=\max\{m_{1},m_{2},\ldots,m_{n}\}$ and $\ell>N,$ then $\tilde{\varphi}(\ell)>n.$ ∎

3. Future work

The following question is natural.

Let $k\in{\mathbb{N}}$ be fixed. Find all $n$ such that $\tilde{\varphi}(n)=k.$

One can see the following result by calculating the appropriate bounds and checking accordingly.

Lemma 5**.**

Let $s(k)=\{n\in{\mathbb{N}}|\tilde{\varphi}(n)=k\}.$ Then the set of values $k\in\{1,2,\ldots,100\}$ such that $s(k)$ is empty set is $\{13,31,70\}.$

Conjecture: If $M=\{\tilde{\varphi}(n)|n\in{\mathbb{N}}\},$ then ${\mathbb{N}}\setminus M$ contains infinite number of elements.

Also recall the Carmichael conjecture [3]: if $N(m)=|\{n\in{\mathbb{N}}|\varphi(n)=m\}|$, then $N(m)\neq 1.$ This conjecture no longer true for $\tilde{\varphi}$ as $\tilde{\varphi}(n)=16$ only for $n=144.$ If $s(k)$ is as defined in the Lemma 5, then then the values of $k\in\{1,\ldots,100\}$ such that $s(k)$ contains a single element is $\{16,39,47,49,53,57,58,65,66,76,85,91,94\}.$

The following tables and observations may be useful in answering the above questions.

Given a positive integer $k$, the following tables provides the smallest value of $n$ such that $\tilde{\varphi}(n)=k.$ For example $\tilde{\varphi}(1)=\tilde{\varphi}(2)=\tilde{\varphi}(3)=\tilde{\varphi}(4)=1,$ hence the smallest value $n$ such that $\tilde{\varphi}(n)=2$ is $5.$

[TABLE]

[TABLE]

[TABLE]

[TABLE]

The following observations are useful for the question posed in the beginning of the section for the case $k\leq 20.$

1. (1)

$\tilde{\varphi}(n)=1\Leftrightarrow n\in\{2,3,4,6,8,12,18,24,30\}.$ 2. (2)

$\tilde{\varphi}(n)=2\Leftrightarrow n\in\{5,10,14,20,42,60\}.$ 3. (3)

$\tilde{\varphi}(n)=3\Leftrightarrow n\in\{7,9,16,36,48,90\}.$ 4. (4)

$\tilde{\varphi}(n)=4\Leftrightarrow n\in\{15,22,54,84\}.$ 5. (5)

$\tilde{\varphi}(n)=5\Leftrightarrow n\in\{26,28,66,120\}.$ 6. (6)

$\tilde{\varphi}(n)=6\Leftrightarrow n\in\{11,21,32,40,72,78,210\}.$ 7. (7)

$\tilde{\varphi}(n)=7\Leftrightarrow n\in\{13,34,50\}.$ 8. (8)

$\tilde{\varphi}(n)=8\Leftrightarrow n\in\{38,44,70,150\}.$ 9. (9)

$\tilde{\varphi}(n)=9\Leftrightarrow n\in\{102,114,126\}.$ 10. (10)

$\tilde{\varphi}(n)=10\Leftrightarrow n\in\{17,27,46,56,96,108,180\}.$ 11. (11)

$\tilde{\varphi}(n)=11\Leftrightarrow n\in\{19,33,52,132\}.$ 12. (12)

$\tilde{\varphi}(n)=12\Leftrightarrow n\in\{25,45,80,168\}.$ 13. (13)

$\tilde{\varphi}(n)=14\Leftrightarrow n\in\{23,39,58,62,110,138\}.$ 14. (14)

$\tilde{\varphi}(n)=15\Leftrightarrow n\in\{35,64,68,156,240\}.$ 15. (15)

$\tilde{\varphi}(n)=16\Leftrightarrow n=144.$ 16. (16)

$\tilde{\varphi}(n)=17\Leftrightarrow n\in\{74,76,100,140\}.$ 17. (17)

$\tilde{\varphi}(n)=18\Leftrightarrow n\in\{198,270,330\}.$ 18. (18)

$\tilde{\varphi}(n)=19\Leftrightarrow n\in\{29,51,88,98,162,174,420\}.$ 19. (19)

$\tilde{\varphi}(n)=20\Leftrightarrow n\in\{31,63,82,130\}.$

Acknowledgment: We would like to thank referee for valuable comments.