Timeline for Are all partial consecutive harmonic subsums distinct?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2017 at 22:07 | comment | added | Włodzimierz Holsztyński | @Gerhard, than you. Now, it has finally occurred to me that there should be plenty of examples because $\ H(a\ b)\ $ is a 2-parameter set while $\ l\ $ is only a 1-parmeter set, i.e. it is expected that $\ H(a\ b) \mapsto l\ $ is not injective. | |
| Feb 9, 2017 at 21:30 | comment | added | Gerhard Paseman | Let a=3, b=5, then H(a,b)=9/20. Similarly, H(19,20)=1/20. Of more interest is 3,7 and 19,21, with l=n=420. Gerhard "See How Easy It Is?" Paseman, 2017.02.09. | |
| Feb 9, 2017 at 21:09 | comment | added | Włodzimierz Holsztyński | Gerhard, would you write the d=20 example explicitly? (Perhaps, you could insert it into my "Answer", noting there that it is your subtext). One could consider the whole class of such examples as the first step toward your conjecture. | |
| Feb 9, 2017 at 20:18 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
inserting an ommission + format
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| Feb 9, 2017 at 15:27 | comment | added | Gerhard Paseman | For the latest,many counter examples exist, one with c=19 and d=20. There may even be counterexamples with d-c larger than 1, but I don't know of them. Yet. (How about d=21?) Gerhard "The Future Can Be Tricky" Paseman, 2017.02.09. | |
| Feb 9, 2017 at 9:27 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
A stronger conjecture
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| Feb 2, 2017 at 12:22 | comment | added | Włodzimierz Holsztyński | @MichaelStoll, thank you for bringing here more precise estimates of $\ H_n.\ $ Overall and ultimately you may be right that my proposition for helping to solve the conjecture may be not adequate. I only want to clarify (just in case) that I meant comparisons--i.e. estimations of the differences--between the two sides of the above inequalities, and not simply the fact that we (possibly) had inequalities. | |
| Feb 2, 2017 at 11:20 | comment | added | Michael Stoll | More precisely, $H_n = \log(n) + \gamma + 1/(2n) + O(n^{-2})$, so $H(a,b) \approx \log(b/a) - 1/2(1/a-1/b)$, which basically already implies that $d/c$ must be a bit smaller than $b/a$ if $H(a,b) = H(c,d)$ and $c > a$. So I don't think that proving $H(a,b) \neq H(ka,kb)$ for $k \ge 2$ is going to help a lot. | |
| Feb 2, 2017 at 7:29 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
TeX typo
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| Feb 2, 2017 at 5:50 | history | answered | Włodzimierz Holsztyński | CC BY-SA 3.0 |