Numbers 101

Recall that we have from [Dris, Dagal (2021) - discussion of the proof of Theorem 2.1 from pp. 13 to 15] that $$I(m) \geq \left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)}$$ which is equivalent to $$I(m^2) \leq \left(I(m)\right)^{\ln(13/9)/\ln(4/3)},$$ from which we get $$\frac{I(m^2)}{I(m)} \leq \left(I(m)\right)^{\ln(13/12)/\ln(4/3)}.$$ This last inequality implies that $$1 > \frac{I(m)}{I(m^2)} \geq \left(I(m)\right)^{\ln(4/3)/\ln(13/12)}$$ $$> \Bigg(\left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)}\Bigg)^{\ln(4/3)/\ln(13/12)}= \left(I(m^2)\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))}.$$ But in general, we know that $I(m^2) > 8/5$, so that $$1>\frac{I(m)}{I(m^2)}> \left(I(m^2)\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))}$$ $$>\left(\frac{8}{5}\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))}> 3$$ which is a contradiction. This seems to prove that there are, in fact, no odd perfect numbers. We will stop here for the time be

I was the last presenter during the 2023 MSP-NCR Annual Convention, held via Zoom last May 27 (Saturday), and hosted by the University of Santo Tomas - Manila, Philippines. I was at home when I presented. Please do check out the Beamer slides in ResearchGate , if you are interested to know more about what ideas were communicated during my talk. For the record, no questions were asked after I was done.  (IIRC, s ome of the participants were graduate students from Polytechnic University of the Philippines and one of my former professors from De La Salle University .) I finished presenting in less than fifteen minutes.

Continuing from this earlier blog post : (In particular, if $m$ is almost perfect, then $m$ must be a square (since $m$ is odd). Consequently, if the equation $\sigma(p^k)=2m$ holds (which is true if and only if $G=H=I$), then $\sigma(p^k)/2$ is also a square. This implies that $k=1$, since $$p^k + 1 \leq \sigma(p^k)=2m<2p$$ holds under the assumption $m<p$.) Summarizing our results so far, we have the chain of implications: $$m < p \Rightarrow \Bigg((m < p^k) \land (k = 1) \land (D(m) = 1)\Bigg).$$  But this blog post proves that the condition $D(m) = 1$ is equivalent to $m < p$. Since $\sigma(m^2)/p^k$ is also a square if $\sigma(p^k)/2$ is a square, then under the assumption that $m < p$, we obtain $\sigma(m^2)/p^k = \sigma(p^k)/2 = m$ is a square. (Note that the difference $m^2 - p^k$ is also a square.) Additionally, we obtain $$\sigma(p^k) = p^k + 1$$ since $k=1$ follows from the assumption $m < p$.  Since $m^2 - p^k$ is a square if and only if $$m^2 - p^k

Let $N = p^k m^2$ be an odd perfect number with special prime $p$. Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x) = \sigma_1(x)$ is the classical sum of divisors of $x$. Additionally, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$. I claim that the estimate $m < p$ holds if and only if $D(m) = 1$. Recall the following material facts: $$\left(\frac{5}{3}\right)^{\ln(4/3)/\ln(13/9)} \approx 1.4912789897463723558$$ $$k = 1 \Rightarrow I(m) \geq \left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)} \geq \left(\frac{5}{3}\right)^{\ln(4/3)/\ln(13/9)}$$  Dagal, Dris (NNTDM, 06/2021) [ The Abundancy Index of Divisors of Odd Perfect Numbers - Part II ] We have the (generic/sharp [?]) upper bound $$\frac{I(m^2)}{I(m)} \leq \frac{\zeta(2)}{\zeta(3)}.$$  Reference: Comment by Erick Wong underneath an answer by Will Jagy to MSE question https://math.stackexchange.com/questions/1097803 . If $I(m) < 2$, then $$\frac{2m}{m+D(m)} \leq I(m) &

https://rasputinnumbers.blogspot.com/2023/05/some-new-results-on-odd-perfect-numbers.html https://arnienumbers.blogspot.com/2023/05/dris-conjecture-holds-if-and-only-if.html In what follows, we will denote the   classical sum of divisors   of the positive integer $x$ by $$\sigma(x)=\sigma_1(x).$$ We will also denote the   abundancy index  of $x$ by $I(x)=\sigma(x)/x$. We start with a minor technical lemma: Lemma 1:   If $N = p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then $I(m^2) < \zeta(2)$ if and only if $p = 5$ and $k \neq 1$. Proof: Suppose that $I(m^2) < \zeta(2)$.  Since $2(p - 1)/p < I(m^2)$, then it follows that $$\frac{2(p - 1)}{p} < \zeta(2).$$ Hence, we derive $$p < \frac{12}{12 - {\pi}^2} < 6,$$ from which we infer that $p = 5$ (since $p$ is a prime satisfying $p \equiv 1 \pmod 4$). Now, since we already have $p = 5$, then $$I(m^2) = \frac{2}{I(p^k)} = \frac{2}{I(5^k)} = \frac{2\cd

(The following is a summary of some new results on odd perfect numbers, obtained on May 05, 2023 .)   Let $N = p^k m^2$ be a hypothetical odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m) = 1$.  Note that, since $\gcd(p, m) = 1$ and $p$ is (the special) prime, then $p^k \neq m$.  (This also follows from the fact that prime powers are deficient, contradicting $N$ is perfect.)  By trichotomy, either $(p^k < m) \oplus (m < p^k)$ is true, where $\oplus$ denotes exclusive-OR.  (That is, $A \oplus B$ is true if and only if exactly one of $A$ or $B$ holds.)   We will denote the   classical sum of divisors  of the positive integer $z$ by $\sigma(z)=\sigma_1(z)$, and the   abundancy index  of $z$ by $I(z)=\sigma(z)/z$. Furthermore, we will denote the   deficiency  of $z$ by $D(z)=2z-\sigma(z)$, and the   aliquot sum  of $z$ by $s(z)=\sigma(z)-z$.   Here is a summary of some new results on odd perfect numbers, which were realized by the au