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When I do the substitution $x=1+\frac{1}{d}$ in the explicit formula related with the Riemann Zeta function, for divisors $1<d\mid n$ and take the sum over all those divisors of such integer $n>1$ (notice that then LHS is simplified, because is equal to zero) one can define, combining some summands from the identity, for example the following arithmetic function

$$f(n)=-\log (2\pi)\sigma_0(n)-\frac{1}{2}\sum_{1<d\mid n}\log\left(\frac{1+2d}{(d+1)^2}\right),$$ for integers $n>1$, where $\sigma_0(n)=\sum_{d\mid n}1$ is the number of divisors.

After some experiments I saw that the graph of this arithmetic function seems curious because has twists. At least I've detected one, but I believe that it is possible that there are more. I don't know if it is due to a bad implementation or the precision of my program. I've detected also a twist of this second arithmetic function $(1-\log (2\pi))\sigma_0(n)-\frac{1}{2}\sum_{1<d\mid n}\log\left(\frac{1+2d}{(d+1)^2}\right)$, but I don't know if has more.

Question. Can you deduce some relevant facts about the asymptotic behaviour of some of previous arithmetic functions? Thanks in advance.

I am asking if it is possible determine something about an approximation of the size of the first or of the second arithmetic functions; if there are twists or were mistakes.

A picture of the graph of the first arithmetic function:

enter image description here

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  • $\begingroup$ If some user is interested in the full identity that one gets from the recipe of the first paragraph is welcome to ask about it, and I add it as a comment here. $\endgroup$ – user243301 Aug 16 '16 at 8:04
  • $\begingroup$ All users, now I believe that only has mathematical meaning a twist. $\endgroup$ – user243301 Aug 16 '16 at 13:48
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Your functions have essentially the behavior of the $\sigma_{0}\left(n\right)=d\left(n\right) $ function. Let $$f_{1}\left(n\right)=-\log\left(2\pi\right)d\left(n\right)-\frac{1}{2}\sum_{\underset{{\scriptstyle d>1}}{d\mid n}}\log\left(\frac{1+2d}{\left(d+1\right)^{2}}\right) $$ now note that $$\frac{\log\left(\frac{5}{9}\right)}{2}\left(d\left(n\right)-1\right)\leq\frac{1}{2}\sum_{\underset{{\scriptstyle d>1}}{d\mid n}}\log\left(\frac{1+2d}{\left(d+1\right)^{2}}\right)\leq0\tag{1} $$ since every addends $\log\left(\frac{1+2d}{\left(d+1\right)^{2}}\right) $ is negative and $d\geq2 $, so $$-\log\left(2\pi\right)d\left(n\right)\leq f_{1}\left(n\right)\leq d\left(n\right)\left(-\log\left(2\pi\right)-\frac{\log\left(\frac{5}{9}\right)}{2}\right)+\frac{\log\left(\frac{5}{9}\right)}{2} $$ and if we define $$f_{2}\left(n\right)=\left(1-\log\left(2\pi\right)\right)d\left(n\right)-\frac{1}{2}\sum_{\underset{{\scriptstyle d>1}}{d\mid n}}\log\left(\frac{1+2d}{\left(d+1\right)^{2}}\right) $$ we can again use $(1)$ and get $$-\log\left(2\pi\right)d\left(n\right)\leq f_{2}\left(n\right)\leq d\left(n\right)\left(1-\log\left(2\pi\right)-\frac{\log\left(\frac{5}{9}\right)}{2}\right)+\frac{\log\left(\frac{5}{9}\right)}{2} $$ so we can use the well known informations about $d\left(n\right) $, for example $$\forall\epsilon>0\,\, f_{n}\left(n\right)=o\left(n^{\epsilon}\right),\, n=1,2.$$

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  • $\begingroup$ Many thanks for your answer. $\endgroup$ – user243301 Aug 16 '16 at 13:04

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