Timeline for Sum of reciprocals of integers minus primes
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 11, 2018 at 15:14 | vote | accept | Bogdan | ||
| Jun 8, 2018 at 21:51 | comment | added | GH from MO | @GregMartin: I think that $f(m)$ tends to infinity. Plausibly, any constant $c>0$ is admissible in my argument below, and $\liminf f(m)\geq\log(1/c)$ by that argument. The point is that the primes very close to $m$ give a bigger contribution than one would expect. | |
| Jun 8, 2018 at 19:22 | answer | added | GH from MO | timeline score: 4 | |
| Jun 8, 2018 at 17:08 | comment | added | Greg Martin | Yes, $f(m)$ is in fact bounded. The contribution from primes less than $m/2$ is trivially at most $1$ (even if all such integers were prime). For primes between $m/2$ and $m$, if all integers in that range were prime then we'd get a contribution of $\log(m/2)$ or so; but at most a proportion $1/log(m/2)$ of those numbers are prime, so this contribution will be bounded as well (one can make this precise with partial summation). | |
| Jun 8, 2018 at 15:57 | comment | added | Gerhard Paseman | Upon reflection (pun intended), there are fewer terms with small denominator, so it is possible there is an upper bound to all the sums. This is going to require more thought. Gerhard "Backward May Be Wrong Direction" Paseman, 2018.06.08. | |
| Jun 8, 2018 at 15:46 | comment | added | Gerhard Paseman | Likely yes. The sum seems not far from log(log m) - log (log (q)) where q is the distance between m and the next smallest prime. If you can prove it easily, it would establish a sub optimal but impressive upper bound on the size of prime gaps. Gerhard "Looking Backwards To Move Forward" Paseman, 2018.06.08. | |
| Jun 8, 2018 at 15:42 | history | edited | Liviu Nicolaescu | CC BY-SA 4.0 |
edited title
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| Jun 8, 2018 at 15:33 | history | asked | Bogdan | CC BY-SA 4.0 |