10
$\begingroup$

Given a positive integer $n$ in the standard form $$n=\prod_k p_k^{\alpha_k}$$ and the arithmetic function $$f(n)=\sum_k \alpha_k p_k$$ let's define the subset $F$ of positive integers $$F=\Big\{n\in N:f(n)\,|\,n,\;f(n)\lt n\Big\}=\Big\{16,27,30,60,70,72,84,105,150,\dots\Big\}$$ I ask if the density of this subset has ever been studied and, in particular, if it is possible to prove the convergence of the series $$\sum_{n\,\in\,F}\frac 1 n$$ Numerical experiments would show the convergence of such series towards a value quite close to the inverse of Euler's number $$\sum_{n\,\in\,F}\frac 1 n\sim\frac 1 e$$

Edit

My script is still running, but after $5\cdot 10^5$ terms ($n=584504910$) the sum of the series is $0.36652132586744884...\;(\frac 1 e = 0,36787944117144232...)$: the growth is extremely slow.

Work in progress

The most recent values obtained are the following:

$n=9928531324,\;\;3986000$-th term of the series$,\;\;$partial sum$\,=0.36776500537719703...$

$n=9931911561,\;\;3987000$-th term of the series$,\;\;$partial sum$\,=0.36776510608002266...$

$n=9935361024,\;\;3988000$-th term of the series$,\;\;$partial sum$\,=0.36776520674763440...$

$n=9938801814\,(\sim 10^{10}),\;\;3989000$-th term of the series$,\;\;$partial sum$\,=0.36776530738064540...$

I am cautiously optimistic about the convergence of the series.

$\endgroup$
8
  • 1
    $\begingroup$ Looks like this is A046346 on OEIS, except for the 4. (70 is in your set, right?) $\endgroup$ Feb 17 at 7:24
  • $\begingroup$ Many thanks for your suggestion. I have corrected the set. $\endgroup$ Feb 17 at 7:32
  • 2
    $\begingroup$ For every prime pair $p, p+2$, we have $2p(p+2) \in F$. Not sure if this suggests a difficulty related to the prime pair conjecture, or if it's not that important since there are other elements too. $\endgroup$
    – aschepler
    Feb 17 at 15:12
  • 1
    $\begingroup$ With the numbers with the desired property upto $2.9\cdot 10^9$ I got the sum $$0.36733612496219974942232403795926400423$$ but my guess would still be that the sum slowly diverges. $\endgroup$
    – Peter
    Feb 17 at 17:10
  • 2
    $\begingroup$ Nobody still hasn't verified this question??? If the sum converges, then the OP should publish it. No matter it converges to $\frac{1}{e}$ or not. This was one of the most interesting question I've seen in this site! $\endgroup$
    – user281
    Mar 12 at 16:23

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.