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
Jose Arnaldo Bebita's posts
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If $N=q^k n^2$ is an odd perfect number and $q = k$, why does this bound not imply $q > 5$?
Let $\mathbb{N}$ denote the set of natural numbers (i.e., positive integers). A number $N \in \mathbb{N}$ is said to be perfect if $\sigma(N)=2N$, where $\sigma=\sigma_{1}$ is the classical sum of divisors.  For example, $\sigma(6)=1+2+3+6=2\cdot{6}$, so th...﻿
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A question on the Euler prime of odd perfect numbers
(Note:  This post was copied verbatim from this MSE link .) A number $N \in \mathbb{N}$ is said to be perfect if $\sigma(N)=2N$, where $\sigma=\sigma_{1}$ is the classical sum-of-divisors function.  For example, $\sigma(6)=1+2+3+6=2\cdot{6}$, so that $6$ is...﻿
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Paolo Starni's "On Dris Conjecture about Odd Perfect Numbers"
Paolo Starni recently uploaded the preprint titled " On Dris Conjecture about Odd Perfect Numbers ", with details below: Abstract The Euler's form of odd perfect numbers, if any, is $n = {{\pi}^{\alpha}} N^2$, where $\pi$ is prime, $\gcd(\pi, N) = 1$ and $\...﻿ Post has attachment If$N = qn^2$is an odd perfect number, is it possible to have$q + 1 = \sigma(n)$? (Note: This blog post was pulled from this [ MO link ].) The title says it all. Question If$N = qn^2$is an odd perfect number, is it possible to have$q + 1 = \sigma(n)$? Heuristic From the Descartes spoof, with quasi-Euler prime$q_1\$: $$n_1 = 3003 < \s...﻿ Post has attachment On the condition n < \sigma(q^k) where N = q^k n^2 is an odd perfect number given in Eulerian form Let N = q^k n^2 be an odd perfect number with Euler prime q. Brown recently announced a proof for the inequality q < n, and a partial proof that q^k < n holds "in many cases". The inequality q^k < n was conjectured by Dris in his M. Sc. thesis. N...﻿ Post has attachment Verification of an equation from a recent preprint on odd perfect numbers (to appear in NNTDM) The paper titled "Conditions Equivalent to the Descartes-Frenicle-Sorli Conjecture on Odd Perfect Numbers" has finally been accepted and is set to appear in the journal Notes on Number Theory and Discrete Mathematics (Volume 23, Number 2) this June or Ju...﻿ Post has attachment Building on work from previous MSE question 2306650 (Re: Odd Perfect Numbers) (Note: This post builds on work from this previous MSE question .) Let \sigma(N) denote the sum of the divisors of the natural number N. If \sigma(N)=2N and N is odd, then N is called an odd perfect number . Denote the abundancy index I of ...﻿ Post has attachment If N=q^k n^2 is an odd perfect number, does q \leq 97 imply that I(q^k) + I(n^2) \leq 2.99? (Note: This blog post was copied verbatim from this MSE question .) Let \sigma(N) denote the sum of the divisors of the natural number N. If \sigma(N)=2N and N is odd, then N is called an odd perfect number . Denote the abundancy index I of ...﻿ Post has attachment If q^k n^2 is an odd perfect number, then k is bounded. If k is arbitrarily large in the equations$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$then$$\lim_{k \to \infty}{\frac{\sigma(n^2)}{q^k}}=\lim_{k \to \infty}{\frac{2n^2}{\sigma(q^k...﻿
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Paper Accepted in Notes on Number Theory and Discrete Mathematics
My paper titled "Conditions Equivalent to the Descartes-Frenicle-Sorli Conjecture on Odd Perfect Numbers" has finally been accepted and is set to appear in the journal Notes on Number Theory and Discrete Mathematics   (Volume 23, Number 2) this June or July...﻿