Is this true that a prime divisor of a Mersenne number $M_n = 2^n-1$ when $n$ is odd, cannot be a Proth prime (i.e. a prime number of the form $2^mk+1$, where $k<2^m$)?
If yes, how is it demonstrated?
because, if I am not wrong, for any prime $p>3$ ($= r2^v +1$, where $r$ is odd and $v>1$),
either there exists a unique $j < v-1$ so that $p$ is a divisor of $2^J+1$, where $J=r2^j$,
or $p$ is divisor of $2^r-1$.
In the latter case, whenever I have checked so far, if I write $p$ as $k2^m+1$, ($k$ odd) $k$ is larger than $2^m$. Is this always like that?
In the former case, most of the times $j$ is $v-2$ or a slightly smaller. In particular, so far I have checked, it is rather rare that $v$ and $j$ differ by more than say $5$, like for instance when $p=65537(j=4,v=16)$, or $p=59393(j=2,v=11)$, or $p=25601 (j=3,v=10)$, or $p=2113 (j=1,v=6)$ or $p=6529 (j=0,v=7)$... I wonder how many such prime numbers exist ?