3 improved formatting (a second time)

In the PhD dissertation titled "Algorithms in the Study of Multiperfect and Odd Perfect Numbers" (hyperlinked here) and completed in 2003, Ronald Sorli conjectured that the exponent k $k$ on the Euler prime $p$ for an odd perfect number $N = (p^k)(m^2)$ is one (i.e. we can drop $k$).

Assuming Sorli's conjecture is true, does anyone know if there exist (any) "effective" results (pardon my use of the term, I just could not think of a better word) in the literature, particularly with respect to relations between the Euler prime p, $p$, the exponent k $k$ and the number $\sqrt{\frac{N}{p^k}}$? I have, so far, only been able to get hold of Paolo Starni's article titled "Odd Perfect Numbers: A Divisor Related to the Euler′s Factor".

In particular, note that Sorli's conjecture implies the following relations:

$$I(p^k) = I(p) = \frac{p+1}{p}$$

$$I(m^2) = \frac{2}{I(p)} = \frac{2p}{p + 1}$$

which, in turn, gives the (trivial) algebraic identity:

$$\frac{1}{p} = \frac{1}{p+1} + \frac{1}{p}\left(\frac{1}{p+1}\right)$$

where $p$ is the Euler prime (i.e. $p^k$ is the Euler's Factor) and $I(x)$I(x) = \frac{\Sigma(x)}{x}$frac{\sigma(x)}{x}$$is the abundancy index of x. 2 Texified In the PhD dissertation titled "Algorithms in the Study of Multiperfect and Odd Perfect Numbers" (hyperlinked here) and completed in 2003, Ronald Sorli conjectured that the exponent k on the Euler prime p p for an odd perfect number N = (p^k)(m^2) p^k)(m^2) is one (i.e. we can drop k).k). Assuming Sorli's conjecture is true, does anyone know if there exist (any) "effective" results (pardon my use of the term, I just could not think of a better word) in the literature, particularly with respect to relations between the Euler prime p, the exponent k and the number \Sqrt{\Frac{N}{p^k}}? \sqrt{\frac{N}{p^k}}? I have, so far, only been able to get hold of Paolo Starni's article titled "Odd Perfect Numbers: A Divisor Related to the Euler′s Factor". In particular, note that Sorli's conjecture implies the following relations: I(p^k)$$I(p^k) = I(p) = \Frac{p+1}{p} I(m^2) frac{p+1}{p}I(m^2) = \Frac{2}{I(p)} frac{2}{I(p)} = \Frac{2p}{p frac{2p}{p + 1}1}$$which, in turn, gives the (trivial) algebraic identity: \Frac{1}{p}$$\frac{1}{p} = \Frac{1}{p+1} frac{1}{p+1} + (\Frac{1}{p})(\Frac{1}{p+1})\frac{1}{p}\left(\frac{1}{p+1}\right)$$where p$p$is the Euler prime (i.e. p^k$p^k$is the Euler's Factor) and I(x)$I(x) = \Frac{Sigma(x)}{x} frac{\Sigma(x)}{x}$is the abundancy index of x.$x\$.

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