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Van der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers

In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$ satisfies the following recurrence relation ($n>1$): $$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\...
Perfect Number's user avatar
2 votes
1 answer
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If $q^k n^2$ is an odd perfect number with Euler prime $q$, is $\sigma(q^k)/n + \sigma(n)/q^k$ bounded from above?

STATEMENT OF THE PROBLEM If $q^k n^2$ is an odd perfect number with Euler prime $q$, is $\sigma(q^k)/n + \sigma(n)/q^k$ bounded from above? MOTIVATION Let $\sigma=\sigma_{1}$ denote the classical ...
Jose Arnaldo Bebita Dris's user avatar
1 vote
0 answers
69 views

Shapiro inequality for divisor sets

The Shapiro inequality is the statement that if $x_1, x_2, \dots, x_n$ are positive, with $x_{n+1}=x_1, x_{n+2}=x_2$, then $$\sum_{i=1}^{n} \frac{x_i}{x_{i+1}+x_{i+2}} \geq \frac{n}{2}.$$ This can be ...
JoshuaZ's user avatar
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1 vote
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305 views

About inequalities that involve the sum of divisors, the Euler's totient and the aliquot part $\sigma(n)-n$

In this post, for integers $n\geq 1$, I denote the sum of divisors $\sum_{1\leq d\mid n}d$ as $\sigma(n)$ and the Euler's totient function as $\varphi(n)$. It's easy to check* that if we assume that $...
user142929's user avatar
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0 answers
107 views

On improving the upper bound $I(m^2) \leq \frac{2p}{p+1}$, if $p^k m^2$ is an odd perfect number with special prime $p$

(Preamble: This question is an offshoot of this answer to an MSE question with the same title.) Denote the classical sum of the divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$ and the ...
Jose Arnaldo Bebita's user avatar
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On variants of the abc conjecture in terms of Lehmer means

In this post we denote the Lehmer mean of a tuple $\text{x}$ of positive real numbers as $$L_p(\text{x})={\sum_{k=1}^nx_k^p\over\sum_{k=1}^nx_k^{p-1}},$$ see the reference Wikipedia Lehmer mean. The ...
user142929's user avatar