Search Results

Let $\sigma(x)$ denote the classical sum-of-divisors function, and let $$I(x) = \frac{\sigma(x)}{x}$$ be the abundancy index of the positive integer $x$. My question is this: What proportion of th …

Let $$I(x) = \frac{\sigma(x)}{x}$$ be the abundancy index of the positive integer $x$. Note that $\sigma(x)$ is the classical sum-of-divisors function. For example, $$\sigma(12) = 1 + 2 + 3 + 4 + …

Hint: Since $q \mid \sigma(2^m)$, then $$q \leq 2^{m+1} - 1,$$ which implies that $$\frac{1}{q} \geq \frac{1}{2^{m+1} - 1}.$$ Can you finish?

I asked the following question in MSE four ($4$) days ago, but so far nobody has posted an answer. The gist of the question is as follows: Are all known $k$-multiperfect numbers (for $k > 2$) …

(Note: This question is an improved version of and has been cross-posted from this MSE post.) Let $\sigma(x)$ denote the sum of the divisors of $x$. If $\sigma(x) = 2x - 1$, then we call $x$ an almo …

This question was posted in MSE in early August 2020. It did garner several upvotes, but did not receive any responses. I have therefore cross-posted it here, hoping that it gets answered. Let $\sig …

Further to this recent corrigendum in NNTDM, which corrects an oversight in Some modular considerations regarding odd perfect numbers – Part II, we realized that we do in fact have the following bicon …

The following assertion appears in Theorem 3.3 (page 7, equations (5) to (6)) of Odd multiperfect numbers by Shi-Chao Chen and Hao Luo: Let $n=\pi^{\alpha} M^2$ be an odd $2$-perfect number, with $\p …

(Note: I asked this question in MSE this June 2018 but did not receive any responses there. I have therefore cross-posted it here, hoping that it gets answered.) Let $\sigma(z)$ denote the sum of t …

This answer is a direct response to Professor Pace Nielsen's suggestion to perform a "sanity check" and complements this other answer by providing a proof of the following biconditionals: If $p^k m^2 …

The title says it all. Question If $N = qn^2$ is an odd perfect number with Euler prime $q$ and $\gcd(q,n)=1$, is it possible to have $q + 1 = \sigma(n)$? Heuristic From the Descartes spoof, with qu …