1. Introduction

Any function whose domain of definition is the set of positive integers is said to be an arithmetic function. Let $f:\mathbb{N}\rightarrow\mathbb{N}$ be an arithmetic function with the property that for each $n\in\mathbb{N}$ there exists at least one $k\in\mathbb{N}$ such that $n|f(k)$. Let

$F_{f}(n)=\min\{k\in\mathbb{N}:n|f(k)\}$

(1.1)

This function generalizes some particular functions. If $f(k)=k!$, then one gets the well known Smarandache function, while for $f(k)=\frac{k(k+1)}{2}$ one has the Pseudo Smarandache function ^[1,5,6]. The dual of these two functions are defined by J. Sandor ^[5,7]. If $g$ is an arithmetic function having the property that for each $n\in\mathbb{N}$ , there exists at least one $k\in\mathbb{N}$ such that $g(k)|n$, then the dual of $F_{f}(n)$ is defined as

$G_{g}(n)=\max\{k\in\mathbb{N}:g(k)|n\}$

(1.2)

The dual Smarandache function is obtained for $g(k)=k!$ and for $g(k)=\frac{k(k+1)}{2}$ one gets the dual Pseudo-Smarandache function. The Euler minimum function has been first studied by P. Moree and H. Roskam ^[4] and it was independently studied by Sandor ^[11]. Sandor also studied the maximum and minimum functions for the various arithmetic functions like unitary toitent function $\varphi^{*}(n)$ ^[9], sum of divisors function $\sigma(n)$, product of divisors function $T(n)$ ^[10], the exponential totient function $\varphi^{e}(n)$ ^[8].

---

2. Preliminary

A positive integer $d$ is called a unitary divisor of $n$ if $d|n$ and gcd$\left(d, \frac{n}{d}\right)=1$ . The notion of unitary divisor related to arithmetical function was introduced by E.Cohen ^[3]. If the integer $n>1$ has the prime factorization $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}.....p_{r}^{\alpha_{r}}$, then $d$ is a unitary divisor of $n$ if and only if $d=p_{1}^{\beta_{1}}p_{2}^{\beta_{2}}.....p_{r}^{\beta_{r}}$ , where ${\beta_{i}}=0$ or ${\beta_{i}}={\alpha_{i}}$ for every $i\in\{{1, 2, 3....r}\}$ . The unitary divisor function, denoted by $\sigma^{*}(n)$ , is the sum of all positive unitary divisors of $n$. It is to noted that $\sigma^{*}(n)$ is a multiplicative function. Thus $\sigma^{*}(n)$ satisfies the functional condition $\sigma^{*}(nm)= \sigma^{*}(n) \sigma^{*}(m)$ for gcd$(m, n) =1$. If $n > 1$ has the prime factorization $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}.....p_{r}^{\alpha_{r}}$, then we have $\sigma^{*}(n)= \sigma^{*}(p_{1}^{\alpha_{1}}) \sigma^{*}(p_{2}^{\alpha_{2}})...\sigma^{*}(p_{r}^{\alpha_{r}})= (p_{1}^{\alpha_{1}}+1)(p_{2}^{\alpha_{2}}+1)....(p_{r}^{\alpha_{r}}+1)$ In this paper, we consider the case (1.2) for the unitary divisor function $\sigma^{*}(n)$ and investigate various characteristics of this function. In (1.2), taking $g(k)=\sigma^{*}(k)$ we define maximum function as follows

$U^{*}(n)=\max\{k\in\mathbb{N}:\sigma^{*}(k)|n\}$

First we discuss some preliminary results related to the function $\sigma^{*}(n)$.

Lemma 2.1. Let $n\geq2$ be a positive integer and let $r$ denote the number of distinct prime factors of $n$. Then

$\sigma^{*}(n)\geq(1+n^\frac{1}{r})^r\geq1+n$

Proof. Let $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}.....p_{r}^{\alpha_{r}}$ be the prime factorization of the natural number $n\geq 2$ , where $p_{i}$ are distinct primes and $\alpha_{i}\geq 0$ . For any positive numbers $x_{1}, x_{2}, x_{3}..., x_{r}$ by Huyggens inequality, we have $((1+x_{1})(1+x_{2})(1+x_{r}))^\frac{1}{r}\geq1+(x_{1}x_{2}..x_{r})^\frac{1}{r}$ For $i=1, 2, ..r$, putting $x_{i}=p_{i}^{\alpha_{i}}$ in the above inequality, we obtain $\sigma^{*}(n)^\frac{1}{r}\geq1+n^\frac{1}{r}$, giving $\sigma^{*}(n)\geq(1+n^\frac{1}{r})^r$ . Again for any numbers $a, b\geq0, r\geq1$, from binomial theory we have $(a+b)^r\geq a^r +b^r$. Therefore we obtain $\sigma^{*}(n)\geq(1+n^\frac{1}{r})^r\geq 1+n$. Thus for all $n\geq2$, we have $\sigma^{*}(n)\geq1+n$. The equality holds only when $n$ is a prime or $n$ is power of a prime.

Remark 2.1. From the lemma 2.1, for all $k\geq2$ we have $\sigma^{*}(k)\geq k+1$ and from $\sigma^{*}(k)| n$ it follows that $\sigma^{*}(k)\leq n$, so $k +1\leq n$ . Thus $U^{*}(n) \leq n -1$. Therefore the maximum function $U^{*}(n)$ is finite and well defined.

Lemma 2.2. Let $p$ be a prime. The equation $\sigma^{*}(x)=p$ has solution if and only if $p$ is a Fermat prime.

Proof. If $x$ is a composite number with at least two distinct prime factors, then $\sigma^{*}(x)$ is also a composite number. Therefore, for any composite number $x$ with at least two distinct prime factors, $\sigma^{*}(x)\neq p$, a prime. So $x$ must be of the form $x=q^{\alpha}$ for some prime $q$. Thus $x=q^{\alpha}$ gives $\sigma^{*}(x)=q^{\alpha}+1=p$ if and only if $q^{\alpha}=p-1$. If $p=2$, then $q=1$ and $\alpha=1$ , which is impossible , so $p$ must be an odd prime . If $p\geq3$, then $p-1$ is even, so we must have $q=2$, i.e., $p=2^{\alpha}+1$. It is clear that such prime exists when $\alpha$ is a power of $2$ giving thereby that $p$ is Fermat prime (see ^[2], page-236).

Lemma 2.3. Let $p$ be a prime. The equation $\sigma^{*}(x)= p^2$ only has the following two solutions: $x=3$, $p=2$ and $x=8$, $p=3$.

Proof. Let $x=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}.....p_{r}^{\alpha_{r}}$ be solution of $\sigma^{*}(x)=p^{2}$, then $(1+ p_{1}^{{\alpha}_{1}}) (1+ p_{2}^{{\alpha}_{2}})....(1+ p_{r}^{{\alpha}_{r}})=p^{2}$ if and only if

(a) $p_{1}^{\alpha_{1}}+1= p^2$

(b)$p_{1}^{\alpha_{1}}+1= 1$, $p_{2}^{\alpha_{2}}+1= p^2$

(c)$p_{1}^{\alpha_{1}}+1= 1$, $p_{2}^{\alpha_{2}}+1= p$ , $p_{3}^{\alpha_{3}}+1= p$

(d)$p_{1}^{\alpha_{1}}+1= p$, $p_{2}^{\alpha_{2}}+1= p$

Since $p_{i}$ are distinct primes, so the cases (b), (c) and (d) are impossible. Therefore only possible case is (a). If $x$ is odd, then from the case (a) we must have only $p=2$. In this case we have $p_{1}=3$, $\alpha_{1}=1$, so $x=3$. If $x$ is even then only possibility is $p_{1}=2$, so from the case (a), we have $2^{\alpha_{1}}=p^{2}-1$ , then $2^{\alpha_{1}}=(p-1)(p+1)$, giving the equations $2^{a}=p-1$, $2^{b}=p+1$, where $a+b=\alpha_{1}$. Solving we get $2^{b}=2(1+2^{a-1})$ and $p=2^{a-1}+2^{b-1}$. Since $2^{b}=2(1+2^{a-1})$ is possible only when $b=2$ and $a=1$, therefore $p=2^{a-1}+2^{b-1}$ gives $p=3$. Thus $\alpha_{1}=3$ and hence $x=8$.

Lemma 2.4. Let $p$ be a prime. The equation $\sigma^{*}(x)= p^3$ has a unique solution: $x=7$, $p=2$.

Proof. Proceeding as the lemma 2.3, we are to find the solution of the equation $p_{1}^{\alpha_{1}}+1=p^{3}$. If $p_{1}$ is odd, then only possible value of $p$ is $2$ and in that case the solution is $x=7$. If $p_{1}$ is even, then $p_{1}=2$ and $2^{\alpha_{1}}=p^{3}-1$ . In that case $p\neq2$ and hence $p$ is an odd prime. Since $p^{3}-1=(p-1)(p^{2}+p+1)$ and $p^{2}+p+1$ is odd for any prime $p$, hence $2^{\alpha_{1}}\neq p^{3}-1$ .

Lemma 2.5. Let $k>1$ be an integer. The equation $\sigma^{*}(x)=2^{k}$ is always solvable and its solutions are of the form $x =$ Mersenne prime or $x =$ a product of distinct Mersenne primes.

Proof. Let $x=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}.....p_{r}^{\alpha_{r}}$, then $(1+ p_{1}^{{\alpha}_{1}}) (1+ p_{2}^{{\alpha}_{2}})....(1+ p_{r}^{{\alpha}_{r}})=2^{k}$ , which gives $(1+ p_{1}^{{\alpha}_{1}})=2^{k_{1}}$ , $(1+ p_{2}^{{\alpha}_{2}})=2^{k_{2}}$ , ....$(1+ p_{r}^{{\alpha}_{r}})=2^{k_{r}}$, where $k_{1}+k_{2}+...+k_{r}=k$. Clearly each $p_{i}$ is odd. Now we consider the equation $p^{\alpha}=2^{a}-1$ , $(a>1)$ .

If $\alpha= 2m$ is an even and $p\geq3$, then $p$ must be of the form $4h\pm1$ and $p^{2}=16h\pm8h+1=8h(2h\pm1)+1=8j+1$. Therefore $p^{2m}+1=(8j+1)^{m}+1=(8r+1)+1=2(4r+1)\neq 2^{a}$. If $\alpha=2m+1$, $(m\geq0)$, then $p^{2m+1}+1 = (p+1)(p^{2m}-p^{2m-1}+...-p+1)$. Clearly the expression $p^{2m}-p^{2m-1}+...-p+1$ is odd. Thus $p^{\alpha}+1\neq2^{a}$, when $\alpha=2m+1$, $(m>0)$ . If $m=0$, then $p=2^{a}-1$, a prime. Any prime of the form $p=2^{a}-1$ is always a Mersenne prime. Thus each $p_{i}$ is Mersenne prime. Hence the lemma is proved.

Lemma 2.6. Let $p$ be a prime and $k>2$ be an integer. The equation $\sigma^{*}(x)=p^{k}$ has solution only for $p=2$.

Proof. Let $x=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}.....p_{r}^{\alpha_{r}}$ .Then proceeding as the lemma 2.5, we have to solve the equation of the form $q^{\alpha}+1=p^{k}$, where $q$ is a prime. If $q$ is odd , then $q^{\alpha}=p^{k}-1$ must be odd. This is possible only when $p=2$ and $\alpha=1$. In that case $\sigma^{*}(x)=2^{k}$ and by the lemma 2.5, this equation is solvable. If $q$ is even, then only possibility is $q=2$ and $2^{\alpha}=p^{k}-1$ .One can easily show that this equation has no solution for $k>2$ . It is clear that $p$ is an odd prime. If $k=2m+1, (m>0)$ is an odd, then $p^{2m+1}-1=(p-1)(p^{2m}+p^{2m-1}+...+p+1)$.Since the expression $p^{2m}+p^{2m-1}+...+p+1$ gives an odd number, so in that case $p^{2m+1}-1\neq2^{ \alpha}$ . If $k=2m+2, (m>0)$ is an even (since $k>2$), then $p^{2m+2}-1=2^{\alpha}$. This equation gives $p^{m+1}-1=2^{a}$ and $p^{m+1}+1=2^{b}$, where $a+b={\alpha}$. Solving we obtain $2^{b}-2^{a}=2$. The last equation has only solution $a=1$, $b=2$. Therefore we get ${\alpha}=3$. For ${\alpha}=3$, the equation $p^{2m+2}-1=2^{\alpha}$ strictly implies that $m=0$. But by our assumption $k>2$. Hence the lemma is proved.

---

3. Results

In this section we discuss our main results.

Following result follows from the definition of $U^{*}(n)$

Theorem 3.1. For all $n\geq1$, $\sigma^{*}(U^{*}(n))\leq n$.

Theorem 3.2. For all $n\geq2$, $U^{*}(n)\leq n-1$

Proof. From the lemma 2.1, for all $k\geq 2$ , we have $\sigma^{*}(k)\geq k+1$ . Putting $k=U^{*}(n)$, one can get $\sigma^{*}(U^{*}(n))\geq U^{*}(n)+1$ . Using the theorem 3.1, we obtain $n\geq \sigma^{*}(U^{*}(n)) \geq 1 + U^{*}(n)$, for all $n\geq2$ .

Theorem 3.3. If $p$ is a prime and $\alpha\geq 1$ , then $U^{*}(p^{\alpha}+1)= p^{\alpha}$

Proof. Since for any prime power $p^{\alpha}$ , we have $\sigma^{*}(p^{\alpha}) = p^{\alpha} + 1$ , so we can write $\sigma^{*}(p^{\alpha})|p^{\alpha}+1$. Therefore from the definition of $U^{*}(n)$, we get $p^{\alpha}\leq U^{*}(p^{\alpha}+1)$ , for all $\alpha\geq1$ .Putting $n= p^{\alpha}+1$ in the inequality of the theorem 3.2, we get $U^{*}(p^{\alpha}+1)\leq p^{\alpha}$ .

Theorem 3.4. For $i=1, 2, ....r$, let $p_{i}$ be distinct primes . If $n$ be a positive integer such that $(1+ p_{1}^{{\alpha}_{1}}) (1+ p_{2}^{{\alpha}_{2}})....(1+ p_{r}^{{\alpha}_{r}})|n$ , where $\alpha_{i}\geq1$ , then $U^{*}(n)\geq p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{r}^{\alpha_{r}}$

Proof. Since $\sigma^{*}(p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{r}^{\alpha_{r}})= (1+ p_{1}^{{\alpha}_{1}}) (1+ p_{2}^{{\alpha}_{2}})....(1+ p_{r}^{{\alpha}_{r}})|n$, so from the definition of $U^{*}(n)$, the result follows.

Theorem 3.5.

$U^{*}(p)= \left\{ \begin{array}{ll} 2^{m}, & \hbox{if $p=2^{m}+1$ is Fermat prime, } \\ 1, & \hbox{if $p=2$ or $p$ is not Fermat prime} \end{array} \right.$

Proof. We have $\sigma^{*}(k)|p$, when $\sigma^{*}(k)=p$ or $\sigma^{*}(k)=1$ . Thus from the lemma 2.2 and the definition of $U^{*}(n)$ the result follows.

Theorem 3.6.

$U^{*}(p^{2})= \left\{ \begin{array}{ll} 3 , & \hbox{if $p=2$, } \\ 8, &\hbox{if $p=3$}\\ 2^{m}, & \hbox{if $p=2^{m}+1>3$ is Fermat prime, } \\ 1, & \hbox{if $p$ is not Fermat prime} \end{array} \right.$

Proof. The result follows from the lemma 2.3 and the definition of $U^{*}(n)$.

Theorem 3.7.

$U^{*}(p^{3})= \left\{ \begin{array}{ll} 7 , & \hbox{if $p=2$, } \\ 8, &\hbox{if $p=3$}\\ 2^{m}, & \hbox{if $p=2^{m}+1>3$ is Fermat prime, } \\ 1, & \hbox{if $p$ is not Fermat prime} \end{array} \right.$

Proof. The result follows from the lemma 2.4 and the definition of $U^{*}(n)$.

Theorem 3.8. $U^{*}(2^{t})=g$, where $g$ is the greatest product $(2^{p_{1}}-1)(2^{p_{2}}-1)...(2^{p_{r}}-1)$ of Mersenne primes, where $p_{1}+p_{2}+...+p_{r}\leq t$ .

Proof. Let $\sigma^{*}(k)|2^{t}$, then $\sigma^{*}(k)=2^{a}$, where $0\leq a\leq t$ .From the definition of $U^{*}(n)$ and the lemma 2.5, the greatest value of such $k$ is $k=g$, where $g=(2^{p_{1}}-1)(2^{p_{2}}-1)...(2^{p_{r}}-1)$, with $p_{1}+p_{2}+...+p_{r}\leq t$ .

Example 3.1. For $n=2^{8}$, $p_{1}+p_{2}+...+p_{r}= 8$ , so we get $p_{1}=3$, $p_{2}=5$. Therefore $g=(2^{p_{1}}-1)(2^{p_{2}}-1)=217$, i.e. $U^{*}(2^{8})=217$ .

Theorem 3.9. For $k>3$,

$U^{*}(p^{k})= \left\{ \begin{array}{ll} g , & \hbox{if $p=2$, where g is given in the theorem 3.8, } \\ 8, & \hbox{if $p=3$}, \\ 2^{m}, & \hbox{if $p=2^{m}+1>3$ is Fermat prime, } \\ 1, & \hbox{if $p$ is not Fermat prime} \end{array} \right.$

Proof. The result follows from the lemma 2.6 and the definition of $U^{*}(n)$.

Corollary 3.10. For any $a\geq1$, $U^{*}(7^{a})=1$, $U^{*}(11^{a})=1$, $U^{*}(13^{a})=1$, $U^{*}(19^{a})=1$ etc.

Theorem 3.11. For $a\geq {1}$, any number of the form $n=(2^{m}+1)(2^{p}-1)^{a}$, $U^{*}(n)=2^{l}$ for some $l$, where $2^{m}+1$ is Fermat prime and $2^{p}-1$ is Mersenne prime.

Proof. Since $3$ is the only prime which is both Mersenne and Fermat prime, so in that case for $a\geq {1}$, $n=3^{a+1}$ from the theorem 3.9, it follows that $U^{*}(n)=2^{3}$. For $n\neq3^{a+1}$, if $\sigma^{*}(k)|n=(2^{m}+1)(2^{p}-1)^{a}$, then the only possibility is $\sigma^{*}(k)|2^{m}+1$. Therefore the result follows from the lemma 2.2.

Example 3.2. $U^{*}(35)=2^{2}$, $U^{*}(51)=2^{4}$, $U^{*}(7967)=2^{8}$.

---

4. Conclusion

We study the maximum function $U^{*}(n)$ in detail and determine the exact value of $U^{*}(n)$ if $n$ is prime power. There is also a scope for the study of the function $U^{*}(n)$ for other values of $n$.

---

Acknowledgement

We are grateful to the anonymous referee for reading the manuscript carefully and giving us many insightful comments.

---

Conflict of Interest

All authors declare no conflicts of interest in this paper.

---