All Questions
9 questions
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 ...
- 6,969
1
vote
0
answers
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 $...
0
votes
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 ...
5
votes
1
answer
357
views
Can the Lagarias inequality be written as a "kernel inequality"?
The Lagarias inequality, which is equivalent to the Riemann hypothesis, is:
$$\sigma(n) \le H_n + \exp(H_n) \log(H_n) =:L(n)$$
for all natural numbers $n$, where $\sigma=$ sum of divisors, $H_n=n$-th ...
- 3,144
2
votes
2
answers
643
views
On the nearest-square function and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number
This question has been cross-posted from this MSE question and is an offshoot of this other MSE question.
(Note that MSE user mathlove has posted an answer in MSE, which I could not completely ...
2
votes
1
answer
631
views
$\frac{\sigma(n)}{n} < e \ln \ln (n)$ is true?
In Guy Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187–213 (pdf)
we find the following result:
If the Riemann hypothesis is true ...
- 53
9
votes
0
answers
695
views
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 ($n>1$):
$$n^2(n-1) = \frac{6}{\sigma(n)} \sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\...
- 213
0
votes
0
answers
264
views
On variants of the abc conjecture in terms of Lehmer means
In this post we denote the Lehmer mean of a tuple $\text{x}$ of positive real numbers as $$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 ...
2
votes
1
answer
482
views
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 ...
