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Notation: $$ \{x\}\ :=\ x-\lfloor x\rfloor $$


APF-functions $\ \tau(n)\ $ for $\ 2<n\in\mathbb N,\ $ and $\ \xi(n)\ $ for $\ 3<n\in\mathbb N,\ $ are defined as follows:

$$ \tau(n)\ :=\ \sum_{k=2}^{n-1}\,\left\{\frac nk\right\}\qquad\qquad\text{and}\qquad \qquad\xi(n)\ :=\ \sum_{k=2}^{\lfloor\sqrt n\rfloor}\ \left\{\frac nk\right\} $$

These functions are sensitive to their arguments being or not a prime. Locally, primes seem (how true is it?) dominate over their neighborhoods. In this spirit,

Question:   Do you already know or can you prove non-obvious results about the APF-functions $\ \tau\ $ and $\ \xi\,?$

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    $\begingroup$ What does APF stand for? I can't find it by a web search. $\endgroup$
    – YCor
    Commented Aug 27, 2022 at 20:59
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    $\begingroup$ I've removed my Q. about ipj-functions. I would remove present one too but the system protects the Answer to this Q. Oh, well... $\endgroup$
    – Wlod AA
    Commented Aug 27, 2022 at 22:51
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    $\begingroup$ "please, see the other thread for a partial answer." What other thread? And, can't you just change the title to something more informative? $\endgroup$ Commented Aug 28, 2022 at 1:02
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    $\begingroup$ Never mind, I took care of it. $\endgroup$ Commented Aug 28, 2022 at 12:32
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    $\begingroup$ I read the posts on this page and have no idea what “APF function” means. $\endgroup$
    – KConrad
    Commented Aug 28, 2022 at 13:13

1 Answer 1

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Your $\tau(n)$ and $\xi(n)$ are essentially the same as the divisor summatory function, often denoted by $\sigma(n)$. Indeed, we have $$\sigma(n)=\sum_{m=1}^n\sum_{k\mid m}1=\sum_{k=1}^n\left[\frac{n}{k}\right]=2\sum_{k=1}^{\left[\sqrt{n}\right]}\left[\frac{n}{k}\right]-\left[\sqrt{n}\right]^2.$$ That is, $$\sigma(n)=\tau(n)+n+1=2\xi(n)+2n-\left[\sqrt{n}\right]^2.$$ Note that we have rather precise estimates for $\sigma(n)$, this is what the Dirichlet divisor problem is about. See the above Wikipedia page for more details. For example, Huxley (2003) proved for any $\varepsilon>0$ that $$\sigma(n)=n\log n+(2\gamma-1)n+O_\varepsilon(n^{131/416+\varepsilon}).$$ In particular, these functions $\sigma$, $\tau$, $\xi$ are not sensitive to their arguments being prime or not prime.

Added. In my response, $[x]$ denotes the integral part of $x$, not the fractional part as in the OP's post. Sorry about that. At any rate, it is straightforward to relate the sum of the integral parts of $n/k$ and the sum of the fractional parts of $n/k$, because the sum of $n/k$ has a well-known asymptotic expansion.

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  • $\begingroup$ My earlier (removed by me) Q. had $\ \prod\ $ instead of $\ \sum.\ $ Possibly, IPJ-functions from my removed post are new or relatively new -- possibly, they have some research potential. $\endgroup$
    – Wlod AA
    Commented Aug 27, 2022 at 23:44
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    $\begingroup$ @WlodAA See my "Added" section. Sorry about the mismatch in notation. $\endgroup$
    – GH from MO
    Commented Aug 27, 2022 at 23:48
  • $\begingroup$ Sorry for (temporarily) forgetting about the standard fractional part notation. I'll correct it. #### The local domination mentioned in my Q. has a good chance to be at least partially true. $\endgroup$
    – Wlod AA
    Commented Aug 27, 2022 at 23:51
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    $\begingroup$ @WlodAA: The arithmetic nature of $n$ does not influence your functions $\tau$ and $\xi$ too much. For example, your $\tau(n)$ is really $\sum_{k=1}^n n/k-\sigma(n)$, which by my last display is $(1-\gamma)n+O_\varepsilon(n^{131/416+\varepsilon})$ for any $\varepsilon>0$. $\endgroup$
    – GH from MO
    Commented Aug 28, 2022 at 0:05
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    $\begingroup$ Well, my numerical experiment tells me that you, GH, are right. $\endgroup$
    – Wlod AA
    Commented Aug 28, 2022 at 3:05

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