1$, then $N=p^l|b|_p$. Since $l>0$, Theorem \ref{th2Motose} implies that $p^2$ does not divide $\Phi_N(b)$. Therefore, we get the following corollary. \begin{corollary}\label{th7Trio} Let $N>1$ and $p$ the greatest prime divisor of $N$. Then $\gcd(N,\Phi_N(b))=1$ or $p$. \end{corollary} In the sequel, we prove that some known kinds of numbers are primovers to some base $b$. \begin{theorem}\label{fermatnumbers} A generalized Fermat number, $F_n(b)=b^{2^n}+1$, with $n$ a positive integer and $b$ even; is primover to base $b$. \end{theorem} \begin{proof} It is well known that if $p$ is prime, then $\Phi_{p^{r}}\left( x\right)=\dfrac{x^{p^{r}}-1}{x^{p^{r-1}}-1}$. Since $\gcd\left( 2^{n+1},\ \Phi_{2^{n+1}}\left( b\right) \right) =1$, we have $P_{2^{n+1}% }(b)=F_{n}\left( b\right) $ and the result follows from Theorem \ref{primover}. \end{proof} \begin{theorem}\label{mersennenumber} A generalized Mersenne number, $M_{p}\left( b\right) =\dfrac {b^{p}-1}{b-1}$, with $p$ a prime such that $\gcd(p,b-1)=1$, is primover to base $b$. \end{theorem} \begin{proof} Note that $\Phi_p(b)=M_{p}\left( b\right)$ and $\gcd(p,\Phi_p(b))=1$. So $P_p(b)=M_p(b)$ and the result follows from Theorem \ref{primover}. \end{proof} By Theorems \ref{t14} and \ref{mersennenumber}, once again, we can prove that the numbers $M_p{(b)}$ satisfy a similar property of the Mersenne numbers $M_p$. \begin{corollary}\label{cor4} If ~$\gcd(p,b-1)=1$, then for every pair of divisors $d_1 < d_2$ of $M_p{(b)}$, including trivial divisors $1$ and $M_p{(b)}$, we have \begin{equation}\label{22} p| d_2-d_1. \end{equation} \end{corollary} The following corollary give us an interesting property of $M_r(b)$. \begin{corollary} Let $r$ be a prime with $\gcd(r,b-1)=1$. Then $M_r(b)$ is prime if and only if the progression $(1+rx)_{x\geq0}$ contains just one prime $p$ such that $|b|_p=r$. \end{corollary} \begin{proof} Assume that $M_r(b)$ is prime. If there exists a prime $p$, such that $|b|_p=r$, then $p=M_r(b)$. Since $r|p-1$, i.e., $p$ is the unique prime in the progression $(1+rx)_{x\geq 0}$. Conversely, assume that there exists only one prime of the form $p=1+rx$, with $x\geq 0$, such that $|b|_p=r$. So $p$ divides $M_r(b)$. If $M_r(b)$ is composite, then it is overpseudoprime to base $b$ and thus to other prime divisor $q$ of $M_r(b)$ we obtain $|b|_q=r$. This contradicts our assumption. \end{proof} The next result shows that Fermat numbers to base $2$ are the only ones, of the form $2^m+1$, which are primover to base $2$. \begin{theorem} The following properties hold. \begin{enumerate} \item Assume that $b$ is even. Then $P_m(b)=b^{m}+1$ is primover to base $b$ if and only if $m$ is a power of $2$. \item Suppose that $\gcd(n,b-1)=1$. Then $M_n(b)=\dfrac{b^{n}-1}{b-1}$ is primover to base $b$ if and only if $n$ is prime. \end{enumerate} \end{theorem} \begin{proof} Sufficient conditions were proved in Theorems \ref{fermatnumbers} and \ref{mersennenumber}. Now assume that $m$ has an odd prime divisor. So $b+1$ is a divisor of $P_m(b)$ and thereby it is not a prime. Since, $\left\vert b\right\vert _{b+1}=2$ and $\left\vert b \right\vert _{b^m+1}=2m$; also it is not an overpseudoprime to base $b$. To prove the necessity of the second part, suppose that $n$ is not prime. Thus for a prime $p$ divisor of $n$, we have $M_{n}(b)$ is composite and $b^{p}-1$ is one of its proper divisors. As $\left\vert b\right\vert _{b^{p}-1}=p$ and $\left\vert b\right\vert _{M_{n}(b)}=n$, we get that $M_{n}(b)$ is not an overpseudoprime to base $b.$ \end{proof} We note that, for $p$ and $q$ primes with $q