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Post Undeleted by A.Neves
Post Deleted by A.Neves
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Hi,

Here is what I found

$f(n)=v_{2}(n)=\displaystyle\sum\limits_{r=1}^{\infty}\frac{r}{2^{r+1}}\sum\limits_{k=0}^{2^{r+1}-1}e^{\frac{2k\pi i(n+2^{r})}{2^{r+1}}}$

This is a special case of $v_{m}(n)$ that differs a little from this one.

For the general case the formula is

$v_{m}(n)=\displaystyle\sum\limits_{r=1}^{\infty}\frac{r}{m^{r+1}} \sum_{j=1}^{(m-1)} \sum\limits_{k=0}^{(m^{r+1}-1)}e^{\frac{2k\pi i(n+(m-j)m^{r})}{m^{r+1}}}$

Now, with this we can put together some arithmetical formulas

the divisor functions

$\sigma_{k}(n)=1+\displaystyle\sum_{m=2}^{\infty}\sum_{r=1}^{\infty}\frac{m^k}{m^{r+1}} \sigma_{a}(n)=1+\displaystyle\sum_{m=2}^{\infty}\sum_{r=1}^{\infty}\frac{m^a}{m^{r+1}} \sum_{j=1}^{(m-1)} \sum\limits_{k=0}^{(m^{r+1}-1)}e^{\frac{2k\pi i(n+(m-j)m^{r})}{m^{r+1}}}$

and the divisor summatory function defined as

$\sigma_{0}=d(n)=\displaystyle\sum\limits_{k|n}1$

and

$D(x)=\displaystyle\sum\limits_{n \leq x}\sigma_{0}(n)$

so for the divisor summatory function we have this formula

$D(n)=\displaystyle\sum_{m=2}^{\infty}\sum_{r=1}^{\infty}\frac{r}{m^{r+1}} \sum_{j=1}^{(m-1)} \sum\limits_{k=0}^{(m^{r+1}-1)}e^{\frac{2k\pi i(p^n+(m-j)m^{r})}{m^{r+1}}}$ where $p$ is some arbitrary fixed ($2$ for example) prime number.

We can also express $\Omega(n)$ and $\omega(n)$ as sums over primes.

Hope this helps.

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