Timeline for Are all partial consecutive harmonic subsums distinct?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3, 2017 at 23:07 | comment | added | Michael Stoll | @WłodzimierzHolsztyński You are welcome. | |
| Feb 3, 2017 at 12:54 | comment | added | Włodzimierz Holsztyński | @MichaelStoll, thank you for your answer, and for communicating with me. | |
| Jan 30, 2017 at 16:46 | comment | added | Michael Stoll | @Seva If $c \ge b$, then $p > b$, so the condition is sufficient. Of course, $d/2 > b$ implies $p > b$ as well. Neither of these will imply the other in general (which one is "better" depends on $d/c$). | |
| Jan 30, 2017 at 16:42 | comment | added | Seva | To conclude that the denominator of $H(a,b)$ is not divisible by $p$, we need to assume that $d/2>b$, not that $c\ge b$, correct? | |
| Jan 30, 2017 at 16:36 | history | edited | Michael Stoll | CC BY-SA 3.0 |
Added alternative argument.
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| Jan 30, 2017 at 16:19 | history | answered | Michael Stoll | CC BY-SA 3.0 |