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Questions tagged [divisor-counting-function]

For questions that involve the divisor counting function, also known as $\sigma_0$, $\tau$, or $d$.

For questions that involve the divisor counting function, also known as $\sigma_0$, $\tau$, or $d$.

The divisor counting function counts the number of divisors of $n$, so for example, $6$ has the divisors $1,2,3,6$, and so $\sigma_0(6)=4$.

It can be defined as $\sigma_0(n)=\sum\limits_{d|n} 1$.

258 questions
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Exact formula of divisor problem

given the sum of divisors of a function $$ D(x)= \sum_{n=1}^{x}d (n) = x\log x + x(2\gamma-1) + \Delta(x)\ $$ then what is the EXACT formula of $ \Delta (x) $ not the O-notation but the EXACT power series correction to the Dirichlet's divisor…
Jose Garcia
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The number of divisors of 2700 including 1 and 2700 equals

I don't really know how to approach this kind of problems, is there any trick or formula for this?
Iti Shree
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Are there any integers $n$ such that $\sigma_0(n) = n$?

Where $\sigma_0(n)$ is the divisor function, are there any integers $n$ such that $\sigma_0(n) = n$? I already know that $n=1, 2$ work, but are there any others?
fwoosh
  • 107
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Asymptotics of the sum $\sum_{n\leq x}\frac{\omega(n)}{n}$ as $x \to \infty$.

If $\omega(n)$ denotes the number of distinct prime factors of $n$, it is known ( e.g. wikipedia) that $$ \sum_{n\leq x}\omega(n)=x \log \log x+O(x) $$ as $x \to \infty$. Is it possible to use this formula to estimate the sum in question? Thanks
user385459
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explicit exact formula for divisor functions and sums

from basic number theory we know $$ \sum_{n\le x}\sigma _{0}(n)= \sum_{n=1}^{\infty}[x/n] $$ where $ [x] $ is the floor function then for the divison function of any order can we evaluate exactly the sums $ \sum_{n\le x}\sigma _{k}(n) $ for an…
Jose Garcia
  • 8,506