Jose Arnaldo Bebita Dris
15 followers -
I know nothing. Therefore, you can ask me anything, but I cannot promise that I will be able to answer everything.
I know nothing. Therefore, you can ask me anything, but I cannot promise that I will be able to answer everything.

15 followers
Posts
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Arnie Dris's Publications - 3rd Quarter, 2018
Conditions equivalent to the Descartes–Frenicle–Sorli Conjecture on odd perfect numbers – Part II (co-authored with Doli-Jane Uvales Tejada)
On the abundancy index/outlaw status of the fraction $\frac{p+2}{p}$, where $p$ is an odd prime
As before, we consider the equation $$\frac{\sigma(x)}{x} = \frac{p+2}{p}$$ where $p$ is an odd prime. We consider several cases.  (Note that the list of cases presented here is not exhaustive.) Case 1:   $3 \mid x$ Since $$\frac{\sigma(x)}{x} = \frac{p+2}{... Add a comment... Post has attachment Arnie Dris's Publications - 2nd Quarter, 2018 The Non-Euler Part of a Spoof Odd Perfect Number is not Almost Perfect (co-authored with Doli-Jane T. Lugatiman) Add a comment... Post has attachment Would like to get numerical (lower [and upper?]) bounds for p (This post is copied verbatim from this MSE question .) This question is an offshoot of this earlier MSE question . Let \sigma(z) denote the sum of divisors of z \in \mathbb{N}, the set of positive integers. Denote the abundancy index of z by I(z) :... Add a comment... Post has attachment Can the following argument be pushed to a full proof that (p + 2)/p is an outlaw if p is an odd prime? (The following post is extracted verbatim from this MSE question .) This is related to this earlier MSE question . Let \sigma(x) be the sum of the divisors of x, and denote the abundancy index of x by I(x):=\sigma(x)/x. If the equation I(a) = b/c ... Add a comment... Post has attachment If q^k n^2 is an odd perfect number with Euler prime q, does this equation imply that k=1? (Note: This post was copied verbatim from this MSE question .) Let \sigma(x) be the sum of the divisors of x. Denote the deficiency of x by D(x) := 2x - \sigma(x), and the sum of the aliquot divisors of x by s(x) := \sigma(x) - x. Here is my q... Add a comment... Post has attachment If \frac{σ(x)}{x}=\frac{p+2}p where p is an odd prime, does it follow that x is an odd square? (Note: The following proof was copied verbatim from the answer of MSE user Alex Francisco .) First, note that for any coprime a, b \in \mathbb{N}_+, there is$$ I(ab) = I(a) I(b). $$Suppose there is an even number n = 2^k \cdot l, where k \geq 1 and ... Add a comment... Post has attachment Arnie Dris's Publications - 3rd Quarter, 2017 The abundancy index of divisors of odd perfect numbers – Part III The Abundancy Index of Divisors of Spoof Odd Perfect Numbers Add a comment... Post has attachment OPN Research - October 2017 If there are infinitely many odd perfect numbers q^k n^2 with Euler prime q, then by a contrapositive to Dris and Luca's result (Dris, Luca 2017) , it follows that \sigma(n^2)/q^k < K does not hold for any K \in \mathbb{N}. This implies that$$\fra...
$|\text{Odd Perfect Numbers }| < \infty \Longrightarrow |\text{Even Perfect Numbers}| < \infty$
Let $q^k n^2$ be an odd perfect number given in Eulerian form, and let $(2^p - 1){2^{p-1}}$ be an even perfect number given in Euclidean form.  Denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$, where $\sigma(x)$ is the sum of the divis...