\left( 0,\dots,0,\underbrace{1}_{j\quad th \quad  component},0,\dots,0,\underbrace{1}_{k\quad th \quad  component},0,\dots,0\right)\] are different $\Re-$ classes in $Z(s)$. This, in particular, would mean that there is a factorization $\left( s_1,\dots,s_p\right)$ of $s$ different from
 \[\left( 0,\dots,0,\underbrace{1}_{j\quad th \quad  component},0,\dots,0,\underbrace{1}_{k\quad th \quad  component},0,\dots,0\right)\] such
that \[supp(\left( s_1,\dots,s_p\right))\bigcap  supp( \left( 0,\dots,0,\underbrace{1}_{j\quad th \quad  component},0,\dots,0,\underbrace{1}_{k\quad th \quad  component},0,\dots,0 \right))=\emptyset.\] This contradicts our
acceptance.	
\end{enumerate}
\paragraph{} It is known that every element in the semigroup involved in one of its minimal presentations has a set of factorizations with at least two $\Re-$ classes. According to the above, $Z(s)$ has at least two $\Re-$ classes. Namely, for every $ x=(x_1,\dots,x_p ),y=(y_1,\dots,y_p )$ in $Z(s)$ we can write $supp(x)\bigcap  supp(y )=\emptyset$. Thus, $ gcd(x,y) =(0,\dots,0)$. This in particular implies that $ dist(x,y) =\max\left\lbrace |x|,|y |\right\rbrace $. Therefore, the catenary degree of $S$ is the maximum of the lengths of these factorizations. 
\paragraph{} Again according to the above, one easily deduces that the largest length of a factorization in $Z(s)$ is reached when $s=a_j+a_k\equiv0 \pmod p$ for $j\neq k$. Since $ p< c< c+1<\dots< c+p-i-1< c+p-i+1<\dots<c+p-1 $, the smallest $s$ that meets these conditions $\min(s)=\min(a_j+a_k)=(c+p-i-1)+(c+p-i+1)=2c+2p-2i$. But can we find another element $s$ larger than $2c+2p-2i$. Namely, there is an element $s$ in $Z(s)$  with $2c+2p-2i<s <2c+2(p-1)$? Since $s=a_j+a_k\equiv0 \pmod p$  for $j\neq k$, if there is, then $s=a_j+a_k=2c+2p-2i+pk<2c+2(p-1)$ for some possitive integers $k$. When we make the necessary cancellations, the inequality $i>\frac{pk+2}{2}$. Where $k=0$ or $k=1$ due to the values of $i$. Thus, we have two cases:
\begin{itemize}
	\item[Case 1.] If $i>\frac{pk+2}{2}$, then $\max(s)=\max(a_j+a_k)=2c+2p-2i+p=2hp+2i+3p-2i=2h+3$. And the factorization of $s$ is $\left(2h+3, 0,\dots,0\right)$. Since the catenary degree is the length of this factorization,  $C(S)=2h+3$ from the Corollary \ref{corollary 2}.
	\item[Case 2.] If other cases, namely  $i<\frac{p+2}{2}$, then $\max(s)=\max(a_j+a_k)=2c+2p-2i=2hp+2i+2p-2i=2h+2$. And the factorization of $s$ is $\left(2h+2, 0,\dots,0\right)$. Since the catenary degree is the length of this factorization,  $C(S)=2h+2$ from the Corollary \ref{corollary 2}.
\end{itemize}	
\end{proof}	

\begin{example}
Let $S$ be the saturated numerical with multiplicity  $5$ and conductor  $33$. Then $S=<5,33,34,36,37>=\left\lbrace 0,5,10,15,20,25,30,33,\rightarrow \right\rbrace $. Where $p=5$, $i=3$ and $h=6$. The set of
elements of the form $w+n$ with $n\in \left\lbrace 33,34,36,37\right\rbrace $ and $w\in Ap(S,5)\setminus\{0\}=\left\lbrace 33,34,36,37\right\rbrace$ is $\left\lbrace 66,67,68,69,70,71,72,73,74\right\rbrace$. The factorizations of these elements are the
following:
$$ Z(66)=\left\lbrace \left(0, 2,0,0,0\right), \left(6,0,0,1,0\right)\right\rbrace $$
$$ Z(67)=\left\lbrace \left(0, 1,1,0,0\right), \left(6,0,0,0,1\right)\right\rbrace $$
$$ Z(68)=\left\lbrace \left(0, 0,2,0,0\right), \left(7,1,0,0,0\right)\right\rbrace $$
$$ Z(69)=\left\lbrace \left(0, 1,0,1,0\right), \left(7,0,1,0,0\right)\right\rbrace $$
$$ Z(70)=\left\lbrace \left(14, 0,0,0,0\right), \left(0,1,0,0,1\right),\left(0,0,1,1,0\right) \right\rbrace $$
$$ Z(71)=\left\lbrace \left(0, 0,1,0,1\right), \left(7,0,0,1,0\right)\right\rbrace $$
$$ Z(72)=\left\lbrace \left(0, 0,0,2,0\right), \left(7,0,0,0,1\right)\right\rbrace $$
$$ Z(73)=\left\lbrace \left(0, 0,0,1,1\right), \left(8,1,0,0,0\right)\right\rbrace $$
$$ Z(74)=\left\lbrace \left(0, 0,0,0,2\right), \left(8,0,1,0,0\right)\right\rbrace $$

Each element of $Z(s)$ is in the different $\Re-$ classes. The catenary degree of $S$ is reached for in
$70$. The catenary degree of $S$ is $14$. Moreover, since $i=3<\frac{5+2}{2}=\frac{p+2}{2}$ and $h=6$, it can
easily be found that $C(S)=2h+2=(2\cdot6)+2=14$ from the Theorem \ref{thm3}
\end{example}
Claire had a big fit at the pool yesterday regarding going off the diving board at free swim. She went off the diving board during her lesson and felt I should ok it during free swim too. I told her it wasn’t fully up to me, that she had to pass the pool’s deep water swim test before she would be allowed to be in the deep end at free swim. She was mad because she felt I should override that and wouldn’t. One dive and one swim back wasn’t enough for me to feel confident she would be safe in the deep end for two hours. So she pitched a crying fit and was very angry, both at my assessment and the pool’s “stupid” test.
The swim test consists of one full lap of swimming, non-stop, any stroke you choose. I’ve watched Claire in lessons and knew that she wasn’t quite there yet; she can go a half lap at most right now. But I told her I had no problem with her taking the pool’s swim test to see how she does. Better to let certain results speak for themselves than to argue with a crying, unreasonable child.
Then today at lessons, Claire’s teacher pushed them a bit, told them to swim a half lap without stopping. This was the most the kids had to swim by themselves so far. My confident Claire had a little scare about 2/3 of the way down. She kind of panicked and started bobbing in the water, touching bottom and then coming up for air.