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Divisor function as in the function that counts the number of positive integers dividing the input n.

I would think that it would be $ln(n)+\gamma$ since $$\sum_{f=1}^n \frac{1}{f} \approx ln(n) + \gamma$$ The logic is that any number $f$ has a $\frac{1}{f}$ chance of dividing a random $n>f$, but I can't find anything online that says this, only approximate growth rates of the divisor function among other things.

metholympian
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    this is not clear. Which "divisor function" are you thinking of? Why would some other limit have to be the same? – lulu Mar 02 '23 at 18:39
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Mar 02 '23 at 18:40
  • should be more clear now – metholympian Mar 02 '23 at 18:50
  • So, if $d(n)$ is the number of positive factors of $n$, you think $d(n)$ should approach $\ln (n)$? If $n=p$ is prime, then $d(p)=2$ so, maybe you mean $\limsup$ or something like that? – lulu Mar 02 '23 at 18:51
  • How can this be "approximate"? There are infinitely many n (namely the primes) for which the number of divisors is 2. For the partial sums of this sequence, we have $n \log(n) + (2 \gamma - 1) n - 4 \sqrt{n} - 1 \le a(n) \le n \log(n) + (2 \gamma - 1) n + 4 \sqrt{n}$ (see OEIS sequence A006218) – Robert Israel Mar 02 '23 at 18:53
  • See this post for an upper bound on $d(n)$. – lulu Mar 02 '23 at 18:55
  • ok as n goes to infinity, what function f yields the lowest sum of $|f(x)-d(x)| $ for all positive x less than n? – metholympian Mar 02 '23 at 19:00
  • That's not the question you mean. $f(x)=d(x)$ clearly gives the minimum for that sum. here is a question which addresses the asymptotics of $\sum \frac {d(n)}n$. And this is a discussion of "average order" which is possibly what you intend. – lulu Mar 02 '23 at 19:22
  • See this post as well. – TravorLZH Mar 02 '23 at 23:05

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