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On odd perfect numbers and a GCD - Part III

Let $m = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. It is known that $$\gcd(\sigma(q^k),\sigma(n^2)) = \frac{(\gcd(n,\sigma(n^...
Jose Arnaldo Bebita's user avatar
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)\...
Perfect Number's user avatar
8 votes
2 answers
722 views

Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1}k^2 \sigma(k)\sigma(n-k)$?

This question is related to the last question about van der Pol's identity for the sum of divisors. In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$ satisfies the following ...
mathoverflowUser's user avatar
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 ...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
345 views

On odd perfect numbers $q^k n^2$ satisfying $n^2 - q^k = 2^r t$

Let $N = q^k n^2$ be an odd perfect number with special prime $q$, satisfying $$n^2 - q^k = 2^r t$$ where $r \geq 2$ and $\gcd(2,t)=1$. We could prove that: (1) $2^r t > 2n$. (We can modestly ...
Jose Arnaldo Bebita's user avatar
2 votes
0 answers
751 views

Does $n \mid \sigma(n^2)$, if $q^k n^2$ is an odd perfect number?

Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$. It is known that $$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\gcd(n^2,\sigma(n^2))...
Jose Arnaldo Bebita's user avatar
5 votes
0 answers
171 views

Touchard / van der Pol's identity for the sum of divisors and an elliptic curve for perfect numbers

In Touchard (1953) it is mentioned that the sum of divisors $\sigma(n)$, satisifies 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
3 votes
0 answers
180 views

Does this Theorem 2 from Dandapat et al. imply that $\gcd(\sigma(p^k),\sigma(a^2)) > 1$?

Write the odd perfect number $m=p^k a^2$ as a product of primes $$m = p^k {p_1}^{2a_1} \cdots {p_v}^{2a_v}.$$ (Note that it is known that $v \geq 9$ by work of Nielsen.) Let $N(m)$ be the number of ...
Jose Arnaldo Bebita's user avatar
3 votes
4 answers
1k views

A conjecture regarding odd perfect numbers

(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 ...
Jose Arnaldo Bebita's user avatar
2 votes
1 answer
326 views

Does there exist an integer that is both solitary and almost perfect?

This question is an offshoot from the following MSE post. I hope that it is appropriate for this site. Let $\sigma(x)$ be the sum of the divisors of $x$. An integer $a$ is said to be solitary if ...
Jose Arnaldo Bebita's user avatar
1 vote
0 answers
167 views

On "Euclidean" odd perfect numbers

In what follows, we let $N = r^s u^2$ be an odd perfect number given in Eulerian form, i.e. $r$ is the special prime satisfying $r \equiv s \equiv 1 \pmod 4$ and $\gcd(r,u)=1$. In this preprint, ...
Jose Arnaldo Bebita's user avatar
1 vote
2 answers
387 views

Improving the lower bound $I(n^2) > \frac{2(q-1)}{q}$ when $q^k n^2$ is an odd perfect number

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Define the abundancy index $$I(x)=\frac{\sigma(x)}{x}$$ where $\sigma(x)$ ...
Jose Arnaldo Bebita's user avatar