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\begin{document}

\begin{center}

{\large {\bf NEW NUMBERS ON EULER'S TOTIENT FUNCTION WITH
APPLICATIONS }}
\rule{0mm}{6mm}\renewcommand{\thefootnote}{}%Enter at least one, but not more than 3 MSCs.
% First entered MSC will be a primary one, others (at most 2) will be secondary.
\footnotetext{\scriptsize  Mathematics Subject Clasification (2000)
05C25, 11E04,20G15.

\rule{2.4mm}{0mm}Keywords and Phrases. Totient number, Super totient
number, Hyper totient number, Graph labeling. }

\vspace{1cc} {\large\it Shahbaz Ali and M. Khalid Mahmood}\\ {\it
shahbaz.math@gmail.com and khalid.math@pu.edu.pk}

\vspace{1cc}
\parbox{24cc}{{\small
 For any positive integer $m, ~\varphi(m)$ find out how many residues of $m$ thats are co-prime to
 $m$, where $\varphi$ is the
Euler's totient function. In this work, we introduce the notion of
totient, super totient and hyper totient numbers and
 discuss their relations.
  Many postulates and characterizations of these numbers have been proposed with straight forward
   proofs. Finally, applications of these
    numbers in graph labeling have been demonstrated over a family of well known graph.}}
\end{center}
\begin{center}
\section{Introduction}
\end{center}
A number is called perfect if the sum of its positive divisors is
twice of the number. That is $\sigma (n) = 2n,$ where $\sigma(n)$
denote the sum of positive divisors on $n$ [2]. A generalization of
the concept of perfect number, so called Zumkeller numbers, has been
investigated and published by Zumkeller in 2003. Zumkeller generated
a sequence of integers A08207 [15] in which the positive divisors of
any integer can be partitioned into two disjoint sets whose sums are
equal. The notion of Zumkeller numbers has been labeled and
investigated by Clark et al. in [12]. Many results regarding
Zumkeller numbers have been stated and proved in [12] by Clark et
al. Later on, Yuejian Peng, K.P.S. Bhaskara Rao proved several
results about Zumkeller numbers, and half Zumkeller numbers in [14].
In [1], B. J. Balamurugan, K. Thirusangu and  D. G. Thomas, gives
the Algorithms for Zumkeller labeling of bitartrate graph and wheel
graph. In [6], Hoque generalized the concept of perfect numbers
using arithmetic functions. McDaniel proved some results on
non-existence of odd perfect numbers [8]. In this piece of work, we
introduce and investigate a new class of numbers over Euler Totient
function $\varphi$ and demonstrate their
applications in graph labeling over Wheels graphs. We organize our paper as follows.\\

\noindent First we state some previous results without proof and few
important definitions from [2,13] to make this paper self readable.
In Section 2, we introduce super totient numbers and characterize
these completely.
 In Section 3,  we give the notion of hyper totient numbers and prove that a zumkeller number is
either a super totient or a hyper totient number. In Section 4, we
validate these numbers in graph labeling over Wheels graphs.\\

\begin{theorem}$[2]$ ``For $n>1$, the sum of positive integers less than $n$ and relatively prime to  $n$  is
$\frac{n\varphi(n)}{2}$."
\end{theorem}
\begin{definition}$[5]$ ``A  perfect
number is of Euclid type if it can be written in the form
$2^{k-1}(2^{k}-1),~  k>1$"
\end{definition}
\begin{definition}$[5,8]$
``A number $n$ is said to be $k$-multiperfect if $\sigma(n)= kn.$"
\end{definition}
\begin{definition}$[2,8]$
``An integer $n>0$ is near 3-perfect if $\sigma(n) = 3n+d,$
 where $\sigma(n)$ is the divisor function and $d$ is a
proper divisor of $n$.\\
\noindent  Every 3-perfect number will be of the form
$2^{\alpha}p_{1}^{t}p_{2},$ where $p_{1}<~p_{2}$ are distinct odd
primes provided $\alpha \geq 1,~ 1 \leq t\leq 2.$"
\end{definition}
\begin{definition}$[12]$ ``A positive integer $n$ is said to be a Zumkeller number if the
positive divisors of $n$ can be partitioned into two disjoint
subsets of equal sum."
\end{definition}
\begin{proposition}$ [14]$ ``If $n$ is a zumkeller number, then\\
 (a) $\sigma(n)$ is an even.\\
 (b)The prime factorization of n must include at least one odd prime to
an odd power."
\end{proposition}

\begin{definition}
 An integer $n>0$ is called totient, if the
  sum of co-prime residues of $n$ is $2^{k}n,~k\geq 1$. That is,
$$\sum\limits_{d<n,(d,n)=1}d = 2^{k}n ,~k\geq 1 $$  The numbers  5, 8, 10 ,12,
15, 16, 17, 20, 24,... are the examples of some totient numbers. In
the following proposition , we characterize the totient numbers to
view
  their postulates as simple consequences.
\end{definition}

\begin{proposition}
An integer $n>0$ is totient if and only if $\varphi(n)=2^{k+1},
k\geq 1.$
\end{proposition}
\begin{proof} By Theorem 1.1., $n$ is totient
\begin{eqnarray}
&\Leftrightarrow & \rho(n) = \frac{n\varphi(n)}{2} = 2^{k}n,~k\geq 1\nonumber\\
&\Leftrightarrow & \varphi(n) = 2^{k+1},k\geq 1
\nonumber\end{eqnarray}
\end{proof}
\noindent $\textbf{Consequences}$\\
\noindent \textbf{1.} Let $n_{i}$ for $i=1,2,3,...,r$ be pairwise
relatively
prime totient numbers. Then $\prod\limits_{i=1}^r n_{i}$  is a totient number.\\
\textbf{2.} If $2^{k}$ and $2^{k+1}$ are totient numbers, then so is
their sum.\\
\noindent\textbf{3.} An odd integer $n$ is a totient number if and
only if $2n$ is a totient number.\\

\noindent A  prime number of the form $2^{2^{n}}+1 ,n\geq 1$ is
called a Fermat prime $[11]$. Sine it is well-known that $p$ is
prime if and only if $ \varphi(p) = p-1.$ Thus in case of Fermat's
prime, we must get $ \varphi(p) = \varphi(2^{2^{n}}+1)= 2^{2^{n}},$
where $2^{n}>1.$ Thus in the light of Proposition 1.2, it is clear
that all Fermat's primes are totient numbers. This gives the
following theorem.
\begin{theorem}
 Every Fermat's prime is a totient numer.
 \end{theorem}
 \noindent By Proposition 1.2 and Theorem 1.2, we characterize
 totien number by prime factorization.
 \begin{theorem}
 A positive integer $n$ is totient number if and only if
  $n=2^{k}p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{m}^{k_{m}}$,
$k\geq 1$ and $p_{i},s$ are Fermat's prime number.
 \end{theorem}
 \noindent From Proposition 1.2 and Consequence (1), we conclude the
 following result.
 \begin{theorem}
The set of totient numbers is infinite.
 \end{theorem}
\begin{center}
\section{Super Totient Numbers}
\end{center}
Recall that an integer $n>0$ is super totient if the co-prime
residues of $n$ can be separated disjoint sets whose sums are equal.
For example if we take $n=5$, then the set of residues of 5 can be
partitioned as, $A=\{1,4\}$ and $B=\{2,3\}.$ Hence by definition, 5
is a Super totient number. Similarly 8, 10, 12, 13, 14, 15, 16, 17,
20, ... are all super totient numbers. In this section, we
completely characterize all super totient numbers. We further show
that the class of super totient numbers is a bigger class then to
class of totient numbers. For the characterization of super totient
numbers, the subsequent
lemma is of essential importance.\\

\begin{lemma}$[7]$ Let $n>0$ be an integer, if $\varphi(m)\equiv0~(mod~4)$
then $n$ is super totient.
 \end{lemma}
 \begin{proof}
 Let $t_{i}$ be the coprime residues of $n$ , we note that  $(t_{i}, n)=1\Leftrightarrow (n-t_{i},n)=1$.
 \par Let $k=\varphi (n)/4$, we can partition the set of co-prime
 residues
\begin{center}
$1=t_{1}<t_{2}<t_{3}<\cdots<t_{\varphi(n)}<n$
\end{center}
into disjoint sets $A$ and $B$ as:
\begin{center}
$A=~~\{~t_{1},~~ t_{2},~~ t_{3} \cdots, ~t_{k}\}\cup \{n-t_{1},~n-
t_{2},~ n-t_{3}\cdots,~n- t_{k}\}$\\
$B=\{t_{2k+1},~ t_{2k+2}, \cdots, ~t_{3k}\}\cup \{n-t_{2k+1},~
n-t_{2k+2}, \cdots, ~n-t_{3k}\} $
\end{center}
Then it is clear that
\begin{center}
 $ \sum\limits_{a\in
A}^{}a = \sum\limits_{i=1}^{k} t_{i}+(n-t_{i})= nk = \sum\limits_{j
= 2k+1}^{3k} t_{j}+(n-t_{j})=\sum\limits_{b\in B}^{}b$
\end{center}
\end{proof}
\noindent Converse of above lemma is not true in general, as 14 is
super totient but $4\nmid\varphi(14).$
\begin{theorem}
 A prime number $p$ is super totient if and only if $p= 1 +4t$ for some $t\in\mathbb{Z^{+}}$.
 \end{theorem}
 \begin{proof}
Let be $p$ a super totient prime. Then by definition, $1, 2, 3,...,
p-1$ can be divided into two sets of equal sums. This sum certainly
an integer, so by Theorem 1.1, $4 | p ~\varphi(p). $ But $p$ is
prime, so $\varphi(p) \equiv 0 ( mod~4)$ if and only if $p= 1 +4t$
for some $t\in\mathbb{Z^{+}}$
 \end{proof}
 \begin{corollary}
If a prime $p$ is a super totient then $p^{k} ,k\geq1$ is also super
totient.
\end{corollary}
\begin{proof}
Let $p$ be a super totient prime. Then by Theorem 2.1,
$p\equiv1(mod~4).$ That is, $p - 1 = 4t$ for some integer $t.$ Note
that,
$$\varphi(p^{k})= p^{k-1}(p-1)=p^{k-1}(4t)=4tp^{k-1}\equiv
0(mod~4)$$
 Thus by Lemma 2.1, $p^{k}$ is a super totient.
\end{proof}
\noindent Observe that, if $n$ is a totient number, then by
Proposition 1.2,  $\varphi(n)=2^{k+1},~~ k\geq 1$. Hence by Lemma
2.1, $n$ must be super totient. However, the converse is not true in
general, since 14 is a super totient number but $\varphi(14) = 6$
cannot be written as a power of 2. Hence, it is not a
 totient number. This confab leads to the  subsequent theorem.

\begin{theorem} Every
totient number is super totient. The converse is not asserted and
infect, is not true in general.
\end{theorem}

\noindent The proof of the subsequent theorem can be  regarded by
means of Lemma 2.1.
 \begin{theorem}~~~~~~\\
 $1$: Let $n_{i}$ for $i=1,2,3,...,r$ be pairwise
relatively
prime super totient numbers. Then $\prod\limits_{i=1}^r n_{i}$  is a super totient number.\\
 $2$: If $m>0$ has at least two odd prime factors
 then  $m$ is a super totient number.\\
$3$: If $m>0$ and $n$ is super totient number
then $mn$ is a super totient number.\\
$4$: If $n$ is super totient number, then $n^{2}$ is also super totient.\\
$5$: If $p|n$, where $p= 1 +4t$ for some $t\in\mathbb{Z^{+}}$ then
$n$ is super totient number.
\end{theorem}
\begin{theorem}
The positive integer $n\geq 30$ is super totient if and only if
$n\varphi (n)$ is a multiple of  $4$.
\end{theorem}
\begin{proof}
If $n$ is super totient, then $\{ r_{1},\cdots, r_{\varphi(n)}\}=
A\cup B $, where $A$ and $B$ are disjoint and $\sum\limits_{a\in
A}a=\sum\limits_{b\in B}b$. Letting $S$ denote the common value of
the above sum, we have
 $$2S=\sum\limits_{a\in A}a+\sum\limits_{b\in B}b= n\varphi (n)/2,$$
 so $S=n\varphi (n)/4$. Thus, $4|n\varphi (n)$. This proves the
 necessary condition.\\
 For the sufficiency, assume $4|n\varphi (n)$. If  $4|\varphi (n)$,
 then $n$ is super totient by Lemma 2.1. So assume that $4\nmid \varphi
 (n)$. Since $n\geq 30, \varphi (n)$ is even. Since $4\nmid \varphi
 (n)$, it follows that $2\parallel \varphi (n)$. Since $4\nmid \varphi
 (n)$ but $4|n\varphi (n)$, it follows that $2|n$. In particular,
 $n=  2p^{k}$ for some positive integer $k$ and prime $p$ with
 $p\equiv 3 (mod ~4)$.\\
 Consider the following numbers\\
 $$ r_{1}, r_{2},\cdots, r_{\varphi(n)/2},$$
 which are all the numbers smaller than $n/2$ and coprime to $n$.
 Let $s=\varphi(n)/2$. If $p>3$, the string of these numbers
 contains $ 1, 3, \cdots, n/2-4,$ which are all smaller than $n/2$,
 coprime to $n$, and distinct since $n/2-4>3$, which is equivalent
 to $ n>14$, which is satisfied for us. There are $s-3$ numbers
 left. Select $t$ of them say
 $$ r_{i1},\cdots, r_{it}$$
 where $t= (\varphi(n)- 2)/4$. The number $t$ is a positive integer
 since $\varphi(n)\equiv 2 (mod ~4)$ and it is possible to choose $t$
 numbers out of $s-3$ because $s-3\geq t$, an inequality equivalent
 to
$$\varphi(n)/2-3\geq (\varphi(n)- 2)/4 ,$$
which is equivalent to $\varphi(n)\geq 10$. To see that this is
satisfied for $n\geq 30$, recall that $n= 2p^{k}$. We want to show
that $p^{k-1}(p-1)\geq 10$. If $k=1$, then since $n\geq 30$, we get
that $p\geq 15$, so the above inequality is satisfied. If $k\geq 2$,
then either $p= 3$, in which case $k\geq 3$ and so $p^{k-1}(p-1)\geq
9\cdot2 >10$, or $p\geq 7$, in which case $p^{k-1}(p-1)\geq
7\cdot6>10.$ So, the inequality $\varphi (n)\geq 10$ is indeed
satisfied for $n\geq 30$ of the form $n= 2p^{k}$ with $p\equiv3
(mod~4)$.\\
Consider now
$$ A=\{ 1,~ 3,~ n/2-4,~ r_{i1},~\cdots,~ r_{it},~n- r_{i1},~\cdots,~ n- r_{it}\} $$
Then $A$ has $3+2t$ elements, the first $t+3\leq s$ being distinct
and smaller than $n/2$ and the last $t$ being distinct and larger
than $n/2$. The sum of elements of $A$ is
$$(1+3+  n/2-4) +\sum\limits_{j=1}^{t}(a_{ij}+n-a_{ij})=n/2+tn=n/2(1+2t)=n\varphi(n)/4= S.  $$
Thus, the complement $B$ of $A$ has sum $n\varphi(n)/2 -S=
n\varphi(n)/4= S$.\\
In case $p=3$, replace the beginning elements $ 1, 3, n/2-4$ by $ 1,
7, n/2-8$ which are also coprime to $n=2\cdot3^{k}$. We need
$n/2-8>7$, so $n>30$, which is satisfied for $n\geq 30$ of the form
$2\cdot3^{k}$.  Everything else stays the same ( namely the choice
of $t,$ etc.)
\end{proof}
 \noindent It is important to note that every even numbers $n$ must be any one of the
 two types namely $\varphi(n)\equiv 0 (mod
 ~4)$ or $\varphi(n)\equiv 2 (mod ~4).$ The numbers lying in first
 class have been determined as super totient numbers by means of Lemma
 2.1. However, the determination of all super totient numbers from
 the second type is difficult and challenging. Since there are
 many numbers from second class which do not follow the definition of
 super totient numbers. For example, $\varphi(18)\equiv 2 (mod ~4),$ but 18 is not super
 totient, whereas 14 is a super totient number and $\varphi(14)\equiv 2 (mod ~4)$ as
 well. After proving the following result, we characterize the
 supper totient numbers completely. Thus the following theorem is of
 vital importance.
\begin{theorem}
 An even integer not divisible by $4$ of kind $\varphi(n)\equiv 2 (mod ~4)$
 is super totient if and only if there exists residue
 $1<r_i<(n+2)/2,  ~ i=1,~2,~3,~\cdots~,~\varphi(\frac{n+2}{2})$, such that
 $(\frac{n+2}{2}  ,r_i ) = 1$ and~ $(\frac{n+2}{2} -r_{i}  ,~n ) ~=~1$.
 \end{theorem}
\begin{proof} Let $n$ be an even integer not divisible by $4$ satisfying the congruence
$\varphi(n)\equiv 2 (mod ~4).$
 Suppose there exists a residue $r_{i}, 1<r_i<(n+2)/2$, for
  $i = 1,~2,~3,~\cdots~,~\varphi(\frac{n+2}{2})$,
 such that  $(\frac{n+2}{2}  ,r_i )= 1 = (\frac{n+2}{2} -r_{i}  ,~n ).$ Without any loss,
 we take$ i = 2, $ and get, $(\frac{n+2}{2} ,r_2 ) = 1= (\frac{n+2}{2}
-r_{2}  ,~n ).$
 Let $k = (\varphi(n)-2)/4 > 0$ be any integer. If $k=1$ then $(\varphi(n)-2)/4 = 1 $ yields that
 $\varphi(n)=6$.  Thus in this case there are the six residues of $n$ which can be rearranged
 after renaming as: $$r_{1}=r_{1}, r_{i}=r_{2},  (n+2)/2-r_{2}=r_{3}, n-r_{3} = r_{4}
 , n-r_{2}=r_{5}, n-1= r_{6}.$$
These can be partitioned as under:
\begin{center}
$A=~~\{~r_{6},~r_{2},~r_{3}\}$$ ~~~~~~~~ $ $B=~~\{~r_{1},~r_{5},~ r_{4}\}$\\
\end{center}
Then it is clear that,
\begin{center}
 $ \sum\limits_{a\in
A}^{}a =n-1+\frac{n+2}{2}=\frac{3n}{2}=
1+n-r_{2}+n-(\frac{n+2}{2}-r_{2})=\sum\limits_{b\in B}^{}b$
 \end{center}That is, the case $k=1$ is true. Let $k>1$  and after renaming,
  we fix six residues as, $r_{1}=r_{1},~r_{i}=r_{2} ,~ (n+2)/2-r_{2}=r_{3},~ n-r_{3}=r_{4}
 ,~n-r_{2}=r_{5},~n-1= r_{6}.$ The rest of the residues can be
 rearranged as, $r_{7}<r_{8}<r_{9}<\cdots<r_{4k+1}<r_{4k+2}$. Then after, we can partition these
 co-prime residues of $n$ in the following two disjoint sets:\\
 $A=~~\{~r_{6},~r_{2},~r_{3}\}\cup
\{r_{7},~r_{8},~\cdots,~r_{k+5}\}\cup\{n-r_{7},~n-r_{8},~\cdots,~n-r_{k+5}\} $\\
$~~~~$\\
$B=~~\{~r_{1},~r_{5},~ r_{4}\}\cup
\{r_{k+6},~r_{k+7}, \cdots, r_{2k+4}\}\cup\{n-r_{k+6}, n-r_{k+7},\cdots,n-r_{2k+4}\}$\\

This gives,

\begin{eqnarray}
&&\sum\limits_{a\in A}^{}a=n-1+\frac{n+2}{2}+\sum\limits_{i=7}^{k+5}r_{i}+(n-r_{i})\nonumber \\
&&~~~~~~=n-1+\frac{n+2}{2}+(k-1)n\nonumber\\
&&~~~~~~=\frac{n(2k+1)}{2}\nonumber\\
&&~~~~~~=1+n-r_{2}+n-(\frac{n+2}{2}-r_{2})+(k-1)n\nonumber\\
&&~~~~~~=1+ n-r_{2}+n-r_{3}+\sum\limits_{j= k+6}^{2k+4}
r_{j}+(n-r_{j})=\sum\limits_{b\in B}^{}b\nonumber
\end{eqnarray}
Hence by definition, $n$ is super totient. \\

\noindent Conversely, suppose $n$ is a super totient number of the
type $\varphi(n)\equiv 2 (mod ~4).$ Then by Theorem 1.1, the sum of
co-prime residue of $n$ is $n(2k+1)$. But then the sum of residues
appearing in both the disjoint partitioned sets is
$\frac{n(2k+1)}{2}$.\\ That is,
\begin{eqnarray}
\frac{n(2k+1)}{2} &=& \sum\limits_{a\in A}^{}a\nonumber \\
&=&\sum\limits_{i=1}^{k-1}r_{i}+(n-r_{i})+n-1+s~~ \textmd{where~} ~
s =\sum_{j}r_{j}~ \textmd {for some~}~
 j.\nonumber\\
&=& n(k-1)+n-1+s\nonumber\\
&=& nk-1+s
\end{eqnarray}
Equation (1) is balanced only if $s=\frac{n+2}{2}$. Since $4\nmid
n,$ so $\frac{n+2}{4}$ must be an integer so, $s=\frac{n+2}{2}$ is
even integer. On contrary we suppose that there does not exist any
residue $r_{i},$ co-prime to $n$ provided $1<r_i<(n+2)/2,  ~
i=1,~2,~3,~\cdots~,~\varphi(\frac{n+2}{2})$, satisfying the
condition
 $(\frac{n+2}{2}  ,r_i ) = 1$ and  $(\frac{n+2}{2} -r_{i}  ,~n )
= 1.$ Then of course, it is impossible to find their sum as the
number
 $s.$ This further implies that $n$ is not super totient, a contradiction.
 This completes the proof.
\end{proof}
\noindent It is well-known that every even perfect number is of
Euclid type. That is, it can be expressed in the form
$2^{k-1}(2^{k}-1),~k>1$ . Let $n$ be an even perfect number  greater
then 6. Then it can be written as $2^{k-1}(2^{k}-1),~k>1$ for $k=2,~
n=6$, which is not super totient number. Let $k>2$ then we obtain,
$\varphi(n)=\varphi(2^{k-1}(2^{k}-1)=\varphi(2^{k-1})\varphi(2^{k}-1)=4k,$
for some positive integer $k$. Hence, $n$ is a super totient number.
This leads to the following result.
\begin{theorem}
 Every even perfect number   greater then $6$ is super totient.
 \end{theorem}
 Also, in view of Definition 2.1-2.2, following results are easy to prove.
\begin{theorem}~~\\
(1) Every $3$-perfect and $4$-perfect numbers  are super totient.\nonumber\\
(2)  Every near $3$-perfect is super totient.
 \end{theorem}
 \noindent From Lemma 2.1 and Theorem 2.3 (1), we conclude the
 following result.
 \begin{theorem}
The set of super totient numbers is infinite.
 \end{theorem}
\begin{center}
\section{Hyper Totient Numbers}
\end{center}
An integer $n>0$ is termed as hyper totient, if the co-prime
residues of $n$ including $n,$ can be divided into two separated
sets of equal sums. In this section, we state and prove results
regarding hyper totient numbers and show that the any Zumkeller
number is either a super totient number or hyper totient number. For
example the numbers 6 and 7 are not super totient since their
respective set of relatively prime residues are $\{1, 5\}$ and $\{1,
2, 3, 4, 5, 6\}.$ Both can never be divided into disjoint sets of
equal sums. However, if we add 6 and 7 to above sets then the
desired partitions are
possible. These are: $\{1,5\}$; $\{6\}$ and $\{1,6,7\}$ ;$\{2,3,4,5\}.$\\

\noindent The following theorems give few postulates and enumerate
the complete set of hyper totient numbers.\\

\begin{theorem} A positive integer $n$ is hyper totient  if $4|(\varphi(n)+2).$
However, converse is not asserted and, in fact, is false in general.
 \end{theorem}
\begin{proof}
 We note that,  $(r_{i}, n)=1$ if and only if $(n-r_{i},n)=1$.
 Let $k=(\varphi (n)+2)/4$. If $k=1,$ then $1=(\varphi (n)+2)/4$ gives that $\varphi (n)= 2.$
 This means that 1 and $n-1$ are the only co-prime residues of $n,$
 in this case, we find the desired partition as,
$A =\{1,~n-1\}$ and $B =~~\{n\}$. This clearly shows that $n$ is
hyper totient. Next we let $k>1.$ A partition of the set of co-prime
 residues of $n$ namely
$1=r_{1}<r_{2}<r_{3}<\cdots<r_{\varphi(n)}<n$ and including $n$, is
given as,
\begin{eqnarray}
 A &=& \{~r_{1},~~ r_{2},~~ r_{3} \cdots, ~r_{k}\}\cup
\{n-r_{1},~n-r_{2},~ n-r_{3}\cdots,~n- r_{k}\}\nonumber\\
  B &=& \{r_{k+1},~ r_{k+2}, \cdots, ~r_{2k-1}\}\cup \{n-r_{k+1},~
n-r_{k+2}, \cdots, ~n-r_{2k-1}\}\cup \{n\}\nonumber
\end{eqnarray}
Then it is clear that,
\begin{center}
 $ \sum\limits_{a\in
A}^{}a = \sum\limits_{i=1}^{k} r_{i}+(n-r_{i})= nk = \sum\limits_{j
= k+1}^{2k-1} r_{j}+(n-r_{j})+n =\sum\limits_{b\in B}^{}b$
\end{center}
 Conversely if we take $n=8,$ then it is  hyper totient
 since its set of co-prime residues can be
 partitioned as $\{1,3,8 \}$ and $\{5,7 \}$. However $\frac{\varphi(8)+2}{4}$ is not integer.
 \end{proof}
  \noindent The following results can be proved using Theorem 3.1.
 \begin{corollary}
 A prime number $p$ is hyper totient if and only if $p=3+4t$ for some $t\in\mathbb{Z^{+}}$.
 \end{corollary}
  \begin{corollary}
If a prime $p$ is a hyper totient then $p^{k} ,k\geq1$ is also hyper
totient.
\end{corollary}
\begin{theorem} Every integer divisible by $4$ is hyper totient.
 \end{theorem}
\begin{proof}
Let $n$ be any integer divisible by 4. In view of  Theorem 3.1, all
integers of the type $\varphi(n)\equiv 2(mod~4)$ divisible 4 are
hyper totient. Now we discuss the other case. That is, all integers
of kind $\varphi(n)\equiv 0(mod~4)$ and divisible by 4. So we, let
$k = \varphi(n)/4.$ It is easy to verify that $(n,~\frac{n}{2}-1)=1$
and $(n,~\frac{n}{2}+1)=1.$  If $k=1$, there are four co-prime
residues of $n.$ These can be rearranged as,
$r_{1}=r_{1},~r_{2}=\frac{n}{2}-1 ,~ r_{3}=\frac{n}{2}+1,~
r_{4}=n-1. $ Then, $ A = \{r_{1},~r_{2},~n\}$ and $B = \{~r_{3},~
r_{4}\}$ is the desired partition. Hence $n$ is hyper totient as,
\begin{center}
 $ \sum\limits_{a\in
A}^{}a =1+\frac{n}{2}-1+n=\frac{3n}{2}=
\frac{n}{2}+1+n-1=\sum\limits_{b\in B}^{}b$
 \end{center}
Let $k>1.$  We fix the above four co-prime residue of $n$ and the
remaining co-prime residues are given as,
 $r_{5}<r_{6}<r_{7}<\cdots<r_{4k}.$
Then we can find the desired partition as,
\begin{eqnarray}
A &=& \{~r_{1},~r_{2},~n\}\cup
\{r_{5},~r_{6},~\cdots,~r_{k+3}\}\cup\{n-r_{5},~n-r_{6},~\cdots,~n-r_{k+3}\}\nonumber\\
B &=& \{~r_{3},~r_{4}\}\cup
\{r_{k+4},~r_{k+5},~\cdots,~r_{2k+2}\}\cup
\{n-r_{k+4},~n-r_{k+5},~\cdots,~n-r_{2k+2}\}.\nonumber
\end{eqnarray} But then,
\begin{eqnarray}
\sum\limits_{a\in A}^{}a &=& 1+\frac{n}{2}-1+n+\sum\limits_{i=5}^{k+3}r_{i}+(n-r_{i})\nonumber \\
&=&\frac{3n}{2}+(k-1)n\nonumber\\
&=&\frac{n(2k+1)}{2}\nonumber\\
&=&\frac{n}{2}+1+n-1+(k-1)n\nonumber\\
&=& \frac{3n}{2}+\sum\limits_{j= k+4}^{2k+2}
r_{j}+(n-r_{j})+n=\sum\limits_{b\in B}^{}b\nonumber
\end{eqnarray}\end{proof}
\noindent Again we note that the number lying in the type
$\varphi(n)\equiv 2 (mod ~4)$ have been determined as hyper totient
numbers by means of Theorem
 3.1.  However, the determination of all super totient numbers from
 the  type $\varphi(n)\equiv 0 (mod ~4)$ was much difficult and challenging. Since there are
 many numbers from second class which do not follow the definition of
 super totient numbers. For example, $\varphi(30)\equiv 0 (mod ~4),$ but 30 is not
 hyper
 totient, whereas 26 is a hyper totient number and $\varphi(26)\equiv 0 (mod ~4)$ as
 well. After proving the following result, we characterize the
 hyper totient numbers completely. Thus the following theorem is of
 vital importance.

\begin{theorem}
 An even integer not divisible by $4$ of kind $\varphi(n)\equiv 0 (mod ~4)$
 is hyper totient if and only if there exists residue
 $1<r_i<(n+2)/2,  ~ i=1,~2,~3,~\cdots~,~\varphi(\frac{n+2}{2})$, such that
 $(\frac{n+2}{2}  ,r_i ) = 1$ and~ $(\frac{n+2}{2} -r_{i}  ,~n ) ~=~1$.
 \end{theorem}
 \begin{proof}
Let $n$ be an even integer not divisible by $4$ satisfying the
congruence $\varphi(n)\equiv 0 (mod ~4).$
 Suppose there exists a residue $r_{i}, 1<r_i<(n+2)/4$, for
  $i = 1,~2,~3,~\cdots~,~\varphi(\frac{n+2}{4})$,
 such that  $(\frac{n+2}{2}  ,r_i )= 1 = (\frac{n+2}{2} -r_{i}  ,~n ).$ Without any loss,
 we take$ i = 2, $ and get, $(\frac{n+2}{2} ,r_2 ) = 1= (\frac{n+2}{2}
-r_{2}  ,~n ).$
 Let $k = \varphi(n)/4 > 1$ be any integer. The six residues of $n$ which can be rearranged
 after renaming as: $$r_{1}=r_{1}, r_{i}=r_{2},  (n+2)/2-r_{2}=r_{3}, n-r_{3} = r_{4}
 , n-r_{2}=r_{5}, n-1= r_{6}.$$The rest of the residues can be
 rearranged as,
 $r_{7}<r_{8}<r_{9}<\cdots<r_{4k}$. we can partition the set of co-prime
 residues and including $n$ in the following two disjoint sets:\\
 $A=~~\{~r_{6},~r_{2},~r_{3}\}\cup
\{r_{7},~r_{8},~\cdots,~r_{k+5}\}\cup\{n-r_{7},~n-r_{8},~\cdots,~n-r_{k+5}\} $\\
$~~~~$\\
$B=~~\{~r_{1},~r_{5},~ r_{4}\}\cup
\{r_{k+6},~r_{k+7},~\cdots,~r_{2k+3}\}\cup\{n-r_{k+6},~n-r_{k+7},~\cdots,~n-r_{2k+3}\}\cup\{n\}$\\
Then it is clear that,
\begin{eqnarray}
&&\sum\limits_{a\in A}^{}a=n-1+\frac{n+2}{2}+\sum\limits_{i=7}^{k+5}r_{i}+(n-r_{i})\nonumber \\
&&~~~~~~=n-1+\frac{n+2}{2}+(k-1)n\nonumber\\
&&~~~~~~=\frac{n(2k+1)}{2}\nonumber\\
&&~~~~~~=1+n-r_{2}+n-(\frac{n+2}{2}-r_{2})+(k-2)n+n\nonumber\\
&&~~~~~~=1+ n-r_{2}+n-r_{3}+\sum\limits_{j= k+6}^{2k+3}
r_{j}+(n-r_{j})+n=\sum\limits_{b\in B}^{}b\nonumber
\end{eqnarray}
 Hence, $n$ is a hyper totient number.\\
\noindent Conversely, suppose $n$ is a hyper totient number of the
type $\varphi(n)\equiv 2 (mod ~4).$ Then by Theorem 1.1, the sum of
co-prime residue of $n$ is $n(2k+1)$. But then the sum of residues
appearing in both the disjoint partitioned sets is
$\frac{n(2k+1)}{2}$.\\ That is,
\begin{eqnarray}
\frac{n(2k+1)}{2} &=& \sum\limits_{a\in A}^{}a\nonumber \\
&=&\sum\limits_{i=1}^{k-1}r_{i}+(n-r_{i})+n-1+s~~ \textmd{where~} ~
s =\sum_{j}r_{j}~ \textmd {for some~}~
 j.\nonumber\\
&=& n(k-1)+n-1+s\nonumber\\
&=& nk-1+s
\end{eqnarray}
Equation (2) is balanced only if $s=\frac{n+2}{2}$. Since $4\nmid
n,$ so $\frac{n+2}{4}$ must be an integer so, $s=\frac{n+2}{2}$ is
even integer. On contrary we suppose that there does not exist any
residue $r_{i},$ co-prime to $n$ provided $1<r_i<(n+2)/2,  ~
i=1,~2,~3,~\cdots~,~\varphi(\frac{n+2}{2})$, satisfying the
condition
 $(\frac{n+2}{2}  ,r_i ) = 1$ and  $(\frac{n+2}{2} -r_{i}  ,~n )
= 1.$ Then of course, it is impossible to find their sum as the
number
 $s.$ This further implies that $n$ is not super totient, a contradiction.
 This completes the proof.
\end{proof}
\begin{theorem}
The positive integer $n\geq 32$ is hyper totient if and only if
$n(\varphi (n)+2)$ is a multiple of  $4$.
\end{theorem}
\begin{proof}
If $n$ is hyper totient, then $\{ r_{1},\cdots, r_{\varphi(n)},~n\}=
A\cup B $, where $A$ and $B$ are disjoint and $\sum\limits_{a\in
A}a=\sum\limits_{b\in B}b$. Letting $S$ denote the common value of
the above sum, we have
 $$2S=\sum\limits_{a\in A}a+\sum\limits_{b\in B}b= n(\varphi (n)+2)/2,$$
 so $S=n(\varphi (n)+2)/4$. Thus, $4|n(\varphi (n)+2)$. This proves the
 necessary condition.\\
 For the sufficiency, assume $4|n(\varphi (n)+2)$. If  $4|(\varphi (n)+2)$,
 then $n$ is super totient by Theorem 3.1. So assume that $4\nmid (\varphi
 (n)+2)$. Since $n\geq 32, \varphi (n)$ is even. Since $4\nmid (\varphi
 (n)+2)$, it follows that $2\parallel \varphi (n)$. Since $4\nmid \varphi
 (n)$ but $4|n\varphi (n)$, it follows that $2|n$. In particular,
 $n=  2p^{k}$ for some positive integer $k$ and prime $p$ with
 $p\equiv 1 (mod ~4)$.\\
 Consider the following numbers\\
 $$ r_{1}, r_{2},\cdots, r_{\varphi(n)/2},$$
 which are all the numbers smaller than $n/2$ and coprime to $n$.
 Let $s=(\varphi(n)+2)/2$. If $p\geq 5$, the string of these numbers
 contains $ 1, 3, \cdots, n/2-4,$ which are all smaller than $n/2$,
 coprime to $n$, and distinct since $n/2-4>5$, which is equivalent
 to $ n>18$, which is satisfied for us. There are $s-3$ numbers
 left. Select $t$ of them say
 $$ r_{i1},\cdots, r_{it}$$
 where $t= \varphi(n)/4$. The number $t$ is a positive integer
 since $\varphi(n)\equiv 2 (mod ~4)$ and $n\geq 32$, it is possible to choose $t$
 numbers out of $s-3$ because $s-3\geq t$, an inequality equivalent
 to
$$(\varphi(n)+2)/2-3\geq \varphi(n)/4 ,$$
which is equivalent to $\varphi(n)\geq 8$. To see that this is
satisfied for $n\geq 32$, recall that $n= 2p^{k}$. We want to show
that $p^{k-1}(p-1)\geq 8$. If $k=1$, then since $n\geq 32$, we get
that $p\geq 17$, so the above inequality is satisfied. If $k\geq 2$,
then either $p= 5$, in which case $k\geq 3$ and so $p^{k-1}(p-1)\geq
25\cdot4 >10$, or $p\geq 13$, in which case $p^{k-1}(p-1)\geq
13\cdot12>10.$  So, the inequality $\varphi (n)\geq 8$ is indeed
satisfied for $n\geq 32$ of the form $n= 2p^{k}$ with $p\equiv1
(mod~4)$.\\
Consider now
$$ A=\{ 1,~ 3,~ n/2-4,~ r_{i1},~\cdots,~ r_{it},~n- r_{i1},~\cdots,~ n- r_{it}\} $$
Then $A$ has $3+2t$ elements, the first $t+3\leq s$ being distinct
and smaller than $n/2$ and the last $t$ being distinct and larger
than $n/2$. The sum of elements of $A$ is
$$(1+3+  n/2-4) +\sum\limits_{j=1}^{t}(a_{ij}+n-a_{ij})= n/2(1+2\varphi(n)/4)= n(\varphi(n)+2)/4= S.  $$
Thus, the complement $B$ of $A$ has sum $n(\varphi(n)+2)/2 -S=
n(\varphi(n)+2)/4= S$.\\
Now if $n=2p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{m}^{k_{m}}$, then
clearly $n$ is even and not divisible by 4 and of the kind
$\varphi(n)\equiv0 (mod~4)$, so by Theorem 3.3,  $n$ is hyper
totient. If $n=2^{k}p_{1}^{k_{1}}p_{2}^{k_{2}}\cdots p_{m}^{k_{m}}$,
$k>1$ then $n$ is divisible by four, so by Theorem 3.2, $n$ is hyper
totient.
\end{proof}
\begin{remark} A hyper totient number my not be super totient. For example $4$ is
hyper totient but not a super totient number.
\end{remark}

\noindent Finally, we prove that the class of Zumkeller numbers is
either a sub class of super totient number or of hyper totient
numbers. The assertion can be entertained in the following theorem.
\begin{theorem}
Every zumkeller number is either a super totient or a hyper totient
number.
\end{theorem}
\begin{proof}
Let $n$ be a zumkeller number ( then $n\neq 2$ ), so by Proposition
1.1 the canonical representation of $n$ must contains an odd prime
number with odd exponent. This odd prime number must be of the form
$p\equiv1(mod ~4)$ or $p\equiv3(mod ~4)$. If $p\equiv1(mod ~4)$ then
by Theorem 2.1, $n$ is super totient number. However, if $p\equiv
3(mod~ 4)$, then by Corollary 3.1, $n$ is hyper totient number.
\end{proof}
\begin{conjecture}
The set of hyper totient numbers is infinite.
 \end{conjecture}
 \newpage
\noindent  In Table 1, we list first hundred numbers of each of the
classes for totient numbers, super totient numbers and hyper totient
numbers.
\begin{center}
\begin{tabular}{|c|c|c|}
 \hline
% after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
   Totient nubers& Super Totient numbers &   Hyper Totient numbers \\
   \hline
  5, 8, 10,  & 5, 8, 10, 12, 13, 14, 15, 16,    & 3, 4, 6, 7, 8, 9,11, 12,   \\
12, 15, 16,  & 17, 20, 21, 22, 24, 25, 26, 28,  & 14, 16, 18, 19,20, 22, 23, 24, \\
 17, 20, 24, & 29, 30, 32, 33, 34, 35, 36, 37,  &  26, 27,28, 31, 32, 34, 36, 38, \\
 30, 32, 34, & 38, 40, 41, 42, 44, 45, 46, 48,  & 40, 42, 43, 44, 46, 47, 48, 49,  \\
  40, 48, 51, & 50, 51, 52, 54, 55, 56, 57, 58,  & 50, 52, 54, 55, 56, 58, 59, 60, \\
 60, 64, 68, & 60, 61, 62, 63, 64, 65, 66, 68,  & 62, 64, 66,67, 68, 70, 71, 72,  \\
80, 85, 96  & 69, 70, 72, 73, 74, 75, 76, 77,  & 74,76, 78, 79, 80, 81, 82, 83,  \\
             & 78, 80, 82, 84, 85, 86, 87, 88,  & 84, 86, 87, 88, 90, 92, 94, 96, \\
             & 89, 90,91, 92, 93, 94, 95, 96,   & 98, 100 \\
             & 97, 98, 99, 100                                 & \\
\hline
\end{tabular}
\end{center}
\centerline{\textbf{Table 1}}
\begin{center}
\section{Applications of Super Totient Numbers}
\end{center}
In previous sections, we have introduced and investigated new
numbers by means of Euler's totient function. It would be more
interesting and of great worth if these numbers could be employed in
some well known mathematics. Initially, we are offering a new graph
labeling by revenues of super totient numbers. We have validated
these labeling over many graphs such as Wheels graphs, Bipartite
graphs, Friendship graphs and Cyclic graphs. In this paper, we
demonstrate these labeling only for Wheels graphs. The rest of the
labeling over
others classes can be validated in a similar technique.\\
\begin{definition}
 Let $V$ be the set of vertices and $E$ be the set of edges of given graph $G$.
  An one-one function $h:V\rightarrow $ $\mathbb{N}$ is call as super
  totient labeling of the graph $G$, if the
  induced function
$h^{*}:E\rightarrow$  $\mathbb{N}$ given by $h^{*} (xy)= kh(x)+
h(y)$ allocates a super totient number,
  $ \forall~ xy\in E$, where  $x, y\in V $ and $k\in \mathbb{Z^{+}}$.
  \end{definition}
 \begin{definition}
 We call a graph as super totient if it states a super totient
labeling.
\end{definition}
\begin{example}
 Let $V=\{3, 4, 5,6,7,8\}$ be the vertex set of graph $G$, then by the induced
 function $h^{*} (xy)= h(x)+h(y)$, we obtain the super totient graph in Fig.1.
\end{example}
\begin{center}
 \includegraphics[width=3 in, height=2.7 in]{figc.eps}
\centerline{\textbf{Fig.1: Super Totient Graph}}
  \end{center}
\begin{definition} $[4]:$
A graph $G$ with $n$ vertices is called a wheel graph if the
vertices of cyclic graph $v_{1}, v_{2}, v_{3},...,v_{n-1}$ are
adjacent with central vertex $v_{0}$.
\end{definition}
\begin{theorem}
The Wheel graph $W_{n}$ is super totient graph $i.e,$ $W_{n}$ states
a super totient labeling.
\end{theorem}
\begin{proof}
Let $v_{0}$ be the central  vertex of a given wheel graph and
$v_{1},v_{2},v_{3},\cdots ,v_{n}.$ be the remaining vertices of
wheel graph. Define the edge set $E$ of $W_{n}$  by:\\ \noindent
$E=\{e^{'}_{i}=v_{0}v_{i},~i = 1,2,...,n\}\cup
\{e_{i}=v_{i}v_{i+1},~i = 1,2,...,n-1 \}\cup\{v_{n}v_{1}=e_{n}\}.$\\
\noindent We deliberate the cases $n=2t$ and $n=2t+1$ for some $t$.
Let $n=2t,~t\in\mathbb{Z^{+}}$ and $p, q$ be distinct odd primes.\\
We establish  an one-one function $h:V\rightarrow\mathbb{ N}$ as:
\[
h(v_{i})= \left\{
\begin{array}{ll}
1,& {\rm~if~}i=0 \nonumber\\
p^{\frac{i+1}{2}},~~~ p\equiv 1(mod~4)& {\rm~if~}i {\rm~ is ~odd~}
\nonumber\\
q^{\frac{i}{2}},~~~~~ 3\neq q\equiv 3(mod~4)& {\rm~if~}i {\rm~ is
~even~}\nonumber
\end{array}
\right. \] Now, we define an induced function $h^{*}$ on $h$:
\begin{eqnarray}
h^{*}(e_{i})&=&h^{*}(v_{i}v_{i+1})=h(v_{i})+h(v_{i+1}), i\equiv1(mod~2)\\
h^{*}(e_{i})&=&h^{*}(v_{i}v_{i+1})=h(v_{i+1})+h(v_{i}), i=2t,~t\in\mathbb{O^{+}}\\
h^{*}(e_{i})&=&h^{*}(v_{i}v_{i+1})=3h(v_{i+1})+h(v_{i}), i=4t,~t\in\mathbb{Z^{+}}\\
h^{*}(e^{'}_{i})&=&h^{*}(v_{0}v_{i})=h(v_{0})+h(v_{i}),~t\in\mathbb{O^{+}}\\
h^{*}(e^{'}_{i})&=&h^{*}(v_{0}v_{i})=3h(v_{0})+h(v_{i}), i=4t,~t\in\mathbb{Z^{+}}\\
h^{*}(e^{'}_{i})&=&h^{*}(v_{0}v_{i})=3h(v_{0})+h(v_{i}),i\equiv1(mod~2)
\end{eqnarray}
By, applying definition of $h,$ we obtain,
\begin{eqnarray}
h^{*}(e_{i})&=&h^{*}(v_{i}v_{i+1})=h(v_{i})+h(v_{i+1})=p^{\frac{i+1}{2}}+q^{\frac{i+1}{2}}, i\equiv1(mod~2)\\
h^{*}(e_{i})&=&h^{*}(v_{i}v_{i+1})=h(v_{i+1})+h(v_{i})=q^{\frac{i}{2}}+p^{\frac{i+2}{2}}, i=2t,~t\in\mathbb{O^{+}}\\
h^{*}(e_{i})&=&h^{*}(v_{i}v_{i+1})=3h(v_{i+1})+h(v_{i})=3p^{\frac{i}{2}}+q^{\frac{i+2}{2}}, i=4t,~t\in\mathbb{Z^{+}}\\
h^{*}(e^{'}_{i})&=&h^{*}(v_{0}v_{i})=h(v_{0})+h(v_{i})=1+q^{\frac{i}{2}}, i=2t,~t\in\mathbb{O^{+}}\\
h^{*}(e^{'}_{i})&=&h^{*}(v_{0}v_{i})=3h(v_{0})+h(v_{i})=3+q^{\frac{i}{2}}, i=4t,~t\in\mathbb{Z^{+}}\\
h^{*}(e^{'}_{i})&=&h^{*}(v_{0}v_{i})=3h(v_{0})+h(v_{i})=3+p^{\frac{i+1}{2}},i\equiv1(mod~2)
\end{eqnarray}
Since, $p\equiv 1(mod~4)$ and $3\neq q\equiv 3(mod~4)$, thus
$p^{\frac{i+1}{2}}+q^{\frac{i+1}{2}},~~
q^{\frac{i}{2}}+p^{\frac{i+2}{2}},3p^{\frac{i}{2}}+q^{\frac{i+2}{2}},~~1+q^{\frac{i}{2}},~~3+q^{\frac{i}{2}}$
and $3+p^{\frac{i+1}{2}}$ all are multiple of 4 and greater then 4,
so
by Lemma 2.1, equation (9)-(14) are super totient number.\\
 Now if $n =2t,~~ t\in\mathbb{Z^{+}}$  then, we take distinct odd primes $p$ and
$q$. We define an one-to-one function $h$ by:
\[
h(v_{i})= \left\{
\begin{array}{ll}
1,& {\rm~if~}i=0 \nonumber\\
p^{\frac{i+1}{2}},~~~ p\equiv 1(mod~4)& {\rm~if~}i {\rm~ is ~odd~}
\nonumber\\
q^{\frac{i}{2}},~~~~~3\neq q\equiv 3(mod~4)& {\rm~if~}i {\rm~ is
~even~}
\nonumber\\
 3,& {\rm~if~}i=n, \nonumber
\end{array}
\right. \] Also we define an induced function $h^{*}$ to $h$ as
follows,
\begin{eqnarray}
h^{*}(e_{i})&=&h^{*}(v_{i}v_{i+1})=h(v_{i})+h(v_{i+1}), i\equiv1(mod~2)\\
h^{*}(e_{i})&=&h^{*}(v_{i}v_{i+1})=h(v_{i+1})+h(v_{i}), i=2t,~t\in\mathbb{O^{+}}\\
h^{*}(e_{i})&=&h^{*}(v_{i}v_{i+1})=3h(v_{i+1})+h(v_{i}), i=4t,~t\in\mathbb{Z^{+}}\\
h^{*}(e^{'}_{i})&=&h^{*}(v_{0}v_{i})=h(v_{0})+h(v_{i}),~t\in\mathbb{O^{+}}\\
h^{*}(e^{'}_{i})&=&h^{*}(v_{0}v_{i})=3h(v_{0})+h(v_{i}), i=4t,~t\in\mathbb{Z^{+}}\\
h^{*}(e^{'}_{i})&=&h^{*}(v_{0}v_{i})=3h(v_{0})+h(v_{i}),i\equiv1(mod~2)\\
h^{*}(e^{'}_{n})&=&h^{*}(v_{0}v_{n})=5h(v_{0})+h(v_{n})\\
h^{*}(e_{n-1})&=&h^{*}(v_{n-1}v_{n})=h(v_{n-1})+3h(v_{n}),~ n-1=2t,~ t\in\mathbb{O^{+}}\\
h^{*}(e_{n-1})&=&h^{*}(v_{n-1}v_{n})=h(v_{n-1})+h(v_{n}),~ n-1=4t,~ t\in\mathbb{Z^{+}}\\
h^{*}(e_{n})&=&h^{*}(v_{n}v_{1})=h(v_{n})+h(v_{1})
\end{eqnarray}
Equations (15)-(20) follow the previous case, so we only to prove
 that the equations (21)-(24) assign super totient numbers.
\begin{eqnarray}
 h^{*}(e^{'}_{n})&=&h^{*}(v_{0}v_{n})=5h(v_{0})+h(v_{n})=5+3=8\\
 h^{*}(e_{n-1})&=&h^{*}(v_{n-1}v_{n})=h(v_{n-1})+3h(v_{n})=q^{\frac{i}{2}}+9\\
h^{*}(e_{n-1})&=&h^{*}(v_{n-1}v_{n})=h(v_{n-1})+h(v_{n})=q^{\frac{i}{2}}+3\\
h^{*}(e_{n})&=&h^{*}(v_{n}v_{1})=h(v_{n})+h(v_{1})=3+p
\end{eqnarray}
Since, $p\equiv 1(mod~4)$ and $3\neq q\equiv 3(mod~4)$, thus
$q^{\frac{i}{2}}+9,~q^{\frac{i}{2}}+3$ and $3+p$ are multiple 4 and
greater then 4, so by Lemma 2.1, equation (25)- (28) assign super
totient number. In both cases wheel graph admits a super totient
labeling.
\end{proof}
\begin{example} For
$n=6$ the wheel graph $W_{6}$  with  $v_{0}=1,~ p=5$ and $q=7$  and
for $n=7$, the wheel graph $W_{7}$ with $v_{0}, p=5,~ q=7$ and
$v_{7}=3$ is super totient wheel graph is given in Fig.$2$.
 \end{example}
\begin{center}
 \includegraphics[width=2.3 in, height=2 in]{figa.eps}
 \includegraphics[width=2.3 in,
height=2 in]{figb.eps} \centerline{\textbf{Fig.2: Super Totient
Wheel Graphs}}
\end{center}
\noindent \textbf{Algorithm }\\
\noindent This algorithm gives the set of wheel graph $W_{n}$ in
such
a way that edges of wheel graph states a super totient number. \\
\textbf {Step (a).}
\par $W_{n},$ a wheel graph over $n$ vertices;
\par $V:$ Set of vertices of  $W_{n};$
\par $E:$ Set of edges of  $W_{n}$ and
$E=$ $\{e^{'}_{i}=v_{0}v_{i},~i = 1,2,...,n\}\cup
\{e_{i}=v_{i}v_{i+1},~i = 1,2,...,n-1 \}\cup\{v_{n}v_{1}=e_{n}\};$
\par $h: h$ is an injective function on $V;$
\par $p,~q:~p,~q$ are distinct odd prime;\\
$~~~~~~~~$ Set $h(v_{0}) = 1$;\\
\textbf {Step (b).}
 $\textbf { do}$ $\hspace{0.1 in}$ \textbf{\{}\\
$~~~~~$ $\textbf {if}$  $~~n=2t,~t\in\mathbb{Z^{+}}$ \textbf{then}\\
 \textbf{\{}\\
 \textbf{for} $i = 1, 3,..., n-1$ $\textbf { do}$

$\hspace{0.3 in}$ \textbf{\{}

$\hspace{0.3in}$ $h(v_{i})=p^{\frac{i+1}{2}}$

$\hspace{0.3 in}$ $h(v_{i+1})=q^{\frac{i+1}{2}}$

$\hspace{0.3 in}$ \textbf{\}}

\textbf{\}}

 $\textbf {else}$

\textbf{\{}

 \textbf{for} $i = 1, 3,..., n-2$ $\textbf { do }$

$\hspace{0.3 in}$ \textbf{\{}

$\hspace{0.3 in}$ $h(v_{i})=p^{\frac{i+1}{2}}$

$\hspace{0.3 in}$ $h(v_{i+1})=q^{\frac{i+1}{2}}$

$\hspace{0.3 in}$\textbf{\}}

\textbf{\}}\\
$~~~~~~~~~~~~~~~~~~~~$\textbf {if ~~ } $ i = n $ $\textbf{then}$
~~~ $h(v_{n})=3$\\
 \textbf{\}} \\
$\textbf {Step (c).}$ Output (super totient wheel graph $W_{n}$).

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\vspace{2cc}
\begin{center}

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\end{document}
