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Jose Arnaldo Bebita Dris
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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.

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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)
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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 $$\frac{\sigma(x)}{x} = \frac{p+2}{...
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Arnie Dris's Publications - 2nd Quarter, 2018
The Non-Euler Part of a Spoof Odd Perfect Number is not Almost Perfect (co-authored with Doli-Jane T. Lugatiman)
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Would like to get numerical (lower [and upper?]) bounds for $p$
(This post is copied verbatim from this MSE question .) This question is an offshoot of this earlier MSE question . Let $\sigma(z)$ denote the sum of divisors of $z \in \mathbb{N}$, the set of positive integers.  Denote the abundancy index of $z$ by $I(z) :...
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Can the following argument be pushed to a full proof that $(p + 2)/p$ is an outlaw if $p$ is an odd prime?
(The following post is extracted verbatim from this MSE question .) This is related to this earlier MSE question . Let $\sigma(x)$ be the sum of the divisors of $x$, and denote the abundancy index of $x$ by $I(x):=\sigma(x)/x$. If the equation $I(a) = b/c$ ...
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If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?
(Note:  This post was copied verbatim from this MSE question .) Let $\sigma(x)$ be the sum of the divisors of $x$.  Denote the deficiency of $x$ by $D(x) := 2x - \sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x) := \sigma(x) - x$. Here is my q...
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If $\frac{σ(x)}{x}=\frac{p+2}p$ where $p$ is an odd prime, does it follow that $x$ is an odd square?
(Note:  The following proof was copied verbatim from the answer of MSE user Alex Francisco .) First, note that for any coprime $a, b \in \mathbb{N}_+$, there is$$ I(ab) = I(a) I(b). $$ Suppose there is an even number $n = 2^k \cdot l$, where $k \geq 1$ and ...
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Arnie Dris's Publications - 3rd Quarter, 2017
The abundancy index of divisors of odd perfect numbers – Part III The Abundancy Index of Divisors of Spoof Odd Perfect Numbers
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OPN Research - October 2017
If there are infinitely many odd perfect numbers $q^k n^2$ with Euler prime $q$, then by a contrapositive to Dris and Luca's result (Dris, Luca 2017) , it follows that $\sigma(n^2)/q^k < K$ does not hold for any $K \in \mathbb{N}$.  This implies that $$\fra...
OPN Research - October 2017
OPN Research - October 2017
arnienumbers.blogspot.com
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$|\text{Odd Perfect Numbers }| < \infty \Longrightarrow |\text{Even Perfect Numbers}| < \infty$
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...
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