The topic of odd perfect numbers likely needs no introduction. Let $N$ be an odd perfect number given in the so-called Eulerian form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. It is known that $n^2 - q^k$ is not a square. (Note that $...

In Syracuse_function one of the described properties is this: $f^{p-1}(2^p \cdot h-1)=2 \cdot 3^{p-1}\cdot h -1 \quad$ with $\quad h$ odd. But if we consider $h=2\cdot k+1$ we have $f^{p-1}(2^{p+1} \cdot k+2^p-1)=2 \cdot 3^{p -1}\cdot (2\cdot k+1) -1$ so for $n=2^{p+1} \cdot k+2^p-1$ we have $f^{...