Timeline for Certain totient chain growth assuming Carmichael's Totient Function Conjecture
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 8, 2016 at 2:41 | comment | added | Gerhard Paseman | OK, as long as you have $m/4\log\log m$ monotonic increasing where you use the inequality. Gerhard "Not Valid For Small Naturals" Paseman, 2016.06.07. | |
| Jun 8, 2016 at 2:16 | history | edited | Greg Martin | CC BY-SA 3.0 |
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| Jun 8, 2016 at 2:16 | comment | added | Greg Martin | Yes, extra squares, my fault. But I don't need $\phi(m^2)=m\phi(m)$; I'm literally plugging in $n=\phi^{-1}(n_j)^2$ in the lower bound. | |
| Jun 7, 2016 at 22:17 | comment | added | Gerhard Paseman | Uh, I don't think so Greg. You have an extra exponent. However, phi(m^2)=m*phi(m), so you can still show super exponential growth independent of Carmichael's conjecture. Gerhard "Since We Are Being Frank" Paseman, 2016.06.07. | |
| Jun 7, 2016 at 21:43 | comment | added | Greg Martin | clarity shmarity, I just messed up! fixed now | |
| Jun 7, 2016 at 21:43 | history | edited | Greg Martin | CC BY-SA 3.0 |
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| Jun 7, 2016 at 20:48 | comment | added | Gerhard Paseman | You should note for clarity that the $n$'s you are using are different from those in the post. Gerhard "I Would Use M's Instead" Paseman, 2016.06.07. | |
| Jun 7, 2016 at 20:18 | comment | added | user76479 | If $k\neq2$ should be $2^{2^{i\log k}}$? | |
| Jun 7, 2016 at 20:16 | comment | added | user76479 | I think $2$ is also true. | |
| Jun 7, 2016 at 19:21 | history | answered | Greg Martin | CC BY-SA 3.0 |