Timeline for Inequality with Euler's totient function
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 7, 2013 at 11:34 | vote | accept | Werner Aumayr | ||
| Feb 7, 2013 at 23:43 | comment | added | user9072 | @Peter Mueller: thank you for this additional information. I'd guessed the situations are more similar. | |
| Feb 7, 2013 at 23:01 | comment | added | Gerhard Paseman | Ah. We are, and I mistook Emil's k for omega(n). My mistake. Indeed Emil's k can be two digits. My k is omega(n), and for every such counterexample, my gut says my k has more than two digits. Peter, I was hoping you would compute it mod p for a few million choice primes p, but it is not as important now. Gerhard "Not Quite The Omega Man" Paseman, 2013.02.07 | |
| Feb 7, 2013 at 22:40 | comment | added | Peter Mueller | @Gerhard: I believe there is no way to factor $3^n-2$, this number has more that $10^{26}$ decimal digits! @quid: Indeed, if one starts with $7$, things look quite differently. Up to $p<70000$, this greedy approach yields $425$ compatible primes, starting with $7, 17, 23, 47, 71, 167,\ldots$, yet the product of the $1-1/p$ for these primes is still bigger than $0.692$. | |
| Feb 7, 2013 at 21:38 | comment | added | user9072 | Thanks for this answer! After I initially saw the question I got quite curious what would be the outcome. Did you also try what happens if you start with 7 insted of 5? I would be quite curious how it compares. ps for Gerhard Paseman: regarding the k again, it occurs to me we were also talking about somewhat different things (which is my fault) but to me the k here is seventeen, if I counted right in any case the number of primes one had to take into account. But sure the 'rest' seems huge so there might be quite a few additional factors 'hiding'. | |
| Feb 7, 2013 at 20:51 | comment | added | Gerhard Paseman | I would like the answer more if I had a good lower bound on k, the number of distinct prime factors of the counterexample. Can you at least trial divide by candidates up to 3^20 or even perhaps 3^50 to get an idea? Gerhard "Would Be Ever So Grateful" Paseman, 2013.02.07 | |
| Feb 7, 2013 at 19:06 | history | edited | Peter Mueller | CC BY-SA 3.0 |
Added some more code
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| Feb 7, 2013 at 17:46 | history | answered | Peter Mueller | CC BY-SA 3.0 |