Timeline for Estimating $\prod_{p\mid n}(1+1/p)$ in terms of n
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 8, 2013 at 21:41 | comment | added | user9072 | @GH: liminf phi(n) loglog n / n = e^(-gamma) is for example in Hardy&Wright (Theorem 328). | |
| Mar 8, 2013 at 20:55 | comment | added | Gerhard Paseman | GH: Schonfeld and Rosser, updated by Dusart and others. Gerhard "Ask Me About Prime Functions" Paseman, 2013.03.08 | |
| Mar 8, 2013 at 20:50 | comment | added | Ping Chen | Dear GH and Greg Martin, thank you very much for answering my questions. | |
| Mar 8, 2013 at 20:49 | vote | accept | Ping Chen | ||
| Mar 8, 2013 at 20:49 | |||||
| Mar 8, 2013 at 20:44 | comment | added | GH from MO | Thanks, Greg, I did not know the exact value of the constant. Do you know a reference off hand? | |
| Mar 8, 2013 at 20:25 | comment | added | Greg Martin | Indeed, for any constant $C>6e^\gamma/\pi^2$ (where $\gamma$ is Euler's constant), we have $$ \prod_{p\mid n}\bigg(1+\frac1p\bigg) < C\log\log n $$ for all $n>n_0(C)$. And this is best possible, in that there are infinitely many violations if $C<6e^\gamma/\pi^2$. One proof uses the identity $$ \prod_{p\mid n} \bigg(1+\frac1p \bigg) = \frac n{\phi(n)} \prod_{p\mid n} \bigg(1-\frac1{p^2} \bigg) $$ together with known lower bounds for $\phi(n)$. | |
| Mar 8, 2013 at 20:09 | history | answered | GH from MO | CC BY-SA 3.0 |