# Questions tagged [divisor-sum]

For questions on the divisor sum function and its generalizations.

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### How to write this sum $\sum_{m=0}^\infty (2m+1)^{-2k-1} \sum_{r=1}^\infty (-1)^{r-1} e^{-2(2m+1)r a}$ as a sum over single index?

So I want to write the sum $$\sum_{m=0}^\infty (2m+1)^{-2k-1} \sum_{r=1}^\infty (-1)^{r-1} e^{-2(2m+1)r a}$$ where $a>0$ and $k\in \mathbb{N}$, as a sum over single index which probably uses odd ...
1answer
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### Resources for asymptotic analysis in analytic number theory?

Currently, I'm looking for some good resources on the methods of asymptotic analysis that can be applied in analytic number theory. So far, I've found books on asymptotic analysis (e.g. the book by de ...
1answer
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0answers
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### On odd perfect numbers $p^k m^2$ with special prime $p$, satisfying $m^2 - p^k = 2^r t$

In what follows, we will denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. A positive integer $y$ ...
1answer
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### Partial summation of $d(n)/(n-1)$

In the answers to this question, it is established that $$\sum_{n\leq x}\frac{d(n)}{n}=\frac{1}{2}(\log(x))^{2}+2\gamma\log (x)+\gamma^{2}-2\gamma_{1}+O\left(x^{-1/2}\right).$$ A related result can be ...
1answer
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1answer
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### Why does an odd perfect number seemingly “violate” basic inequality rules?

Let $p^k m^2$ be an odd perfect number (OPN) with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. My question is as is in the title: Why does an OPN seemingly "...
1answer
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1answer
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### Estimate of $\prod_{d\mid n}(d+1)$.

I would like to ask if there is any cool estimate of $\prod_{d\mid n}(d+1)$. I know that $\prod_{d\mid n}d=n^{\tau(n)/2}$ ($\tau(n)$ is the number of divisors of $n$) so we have the trivial estimate \$\...