Timeline for Sharp upper bounds for sums of the form $\sum_{p \mid k} \frac{1}{p+1}$
Current License: CC BY-SA 2.5
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 17, 2010 at 0:40 | vote | accept | user02138 | ||
| Nov 4, 2010 at 8:05 | vote | accept | user02138 | ||
| Nov 4, 2010 at 8:06 | |||||
| Nov 1, 2010 at 22:55 | comment | added | Gerry Myerson | OK, much better. $\vartheta=\theta^2$, no? | |
| Nov 1, 2010 at 15:06 | comment | added | user02138 | Hi Gerry. The formula is $\lfloor \vartheta^{2^{i-1}} - \frac{1}{2} \rfloor$, not $\lfloor \vartheta^{2^{i}} - \frac{1}{2} \rfloor$. Take $i = 3$, \begin{eqnarray} \lfloor \theta^{2^{3}} + \tfrac{1}{2} \rfloor - 1 = \lfloor \vartheta^{2^{2}} - \tfrac{1}{2} \rfloor = 6, \end{eqnarray} where $\theta$ and $\vartheta$ as above. | |
| Nov 1, 2010 at 12:00 | comment | added | Gerry Myerson | Never mind those websites, I'm asking YOU: does it make sense that $d_i$ is pretty much $a^{2^i}$ while $d_i-1$ is pretty much $b^{2^i}$ for some $b$ not equal to $a$? I mean, take $i=3$, say, and calculate those powers, and make your own judgement as to what's reasonable. | |
| Nov 1, 2010 at 6:13 | history | edited | user02138 | CC BY-SA 2.5 |
added 9 characters in body
|
| Nov 1, 2010 at 5:51 | comment | added | user02138 | As for the double exponential growth formulas that I quote above, see OEIS entries, along with Wikipedia and MathWorld articles for "Sylvester's Sequence". | |
| Nov 1, 2010 at 5:44 | history | edited | user02138 | CC BY-SA 2.5 |
deleted 12 characters in body; Post Made Community Wiki
|
| Nov 1, 2010 at 5:37 | history | edited | user02138 | CC BY-SA 2.5 |
deleted 2 characters in body
|
| Nov 1, 2010 at 5:31 | history | edited | user02138 | CC BY-SA 2.5 |
added 21 characters in body
|
| Nov 1, 2010 at 5:22 | history | edited | user02138 | CC BY-SA 2.5 |
added 37 characters in body; added 7 characters in body
|
| Nov 1, 2010 at 5:19 | comment | added | user02138 | Right, it's not as sharp as I'd like. For the case that you mention, 7/12 = 1/(2+1) + 1/(3+1) <= 1/2 + 1/6 = 2/3 isn't best possible but it'll do until I have a better understanding of the sequence 3,4,6,8,12,30,128,1932,309122... | |
| Nov 1, 2010 at 4:39 | comment | added | Gerry Myerson | Also, the bounds given in this answer can't be the sharp bounds you asked for since (for example) it's impossible to have $p+1=d_1=2$. For $\omega(k)=2$ it would seem that the sharp bound is $(1/3)+(1/4)=7/12$ obtained for $k=6$. | |
| Nov 1, 2010 at 4:33 | comment | added | Gerry Myerson | Can it really be that $d_i=\lfloor\theta^{2^i}+(1/2)\rfloor$ where $\theta\approx1.264$, while $d_i-1=\lfloor\vartheta^{2^i}-(1/2)\rfloor$ where $\vartheta\approx1.5979$? | |
| Nov 1, 2010 at 2:17 | vote | accept | user02138 | ||
| Nov 1, 2010 at 2:18 | |||||
| Nov 1, 2010 at 2:16 | history | edited | user02138 | CC BY-SA 2.5 |
deleted 2 characters in body
|
| Nov 1, 2010 at 2:11 | history | edited | user02138 | CC BY-SA 2.5 |
added 30 characters in body
|
| Nov 1, 2010 at 2:04 | history | edited | user02138 | CC BY-SA 2.5 |
added 229 characters in body; edited body; deleted 8 characters in body
|
| Nov 1, 2010 at 1:59 | history | edited | user02138 | CC BY-SA 2.5 |
added 121 characters in body
|
| Nov 1, 2010 at 1:53 | history | answered | user02138 | CC BY-SA 2.5 |