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Jan
4
comment Is $\frac1\pi \arctan \frac\pi{\ln x}- \frac1{\ln x}$ related to the trivial solutions $\zeta(-2n)$?
Thanks again Raymond: I would be glad if you could have a look at this question on Two Representations of the Prime Counting Function as well. Cheers, draks...
Jan
4
accepted Is $\frac1\pi \arctan \frac\pi{\ln x}- \frac1{\ln x}$ related to the trivial solutions $\zeta(-2n)$?
Jan
4
revised Two Steps away from the Hamilton Cycle
added modified version of previous example and some more idea towards a general solution
Jan
3
revised Two Representations of the Prime Counting Function
gram added
Jan
3
asked Two Representations of the Prime Counting Function
Jan
3
revised Riesel and Gohl's Approximation of the Modified Prime Counting Function, $\pi_{0}$
fixed broken prime tex
Jan
3
comment Gram's series for integral equation
(i) What does $f(x)$ in $(1)$ correspond to in $(0)$, $\pi(e^t)$? (ii) Is $K(st)=1/(e^{st}-1)$? (iii) Why $g(s)/s$? Looks like you imply a logarithmic derivative somewhere...
Jan
3
revised Gram's series for integral equation
some typos, a little tex, two tags and a question mark
Jan
3
comment Aside from approximation functions, are there any proven functions that produce an exact $n$th prime?
@RaymondManzoni I love it...
Jan
3
comment Aside from approximation functions, are there any proven functions that produce an exact $n$th prime?
@RaymondManzoni I just realized that it was your answer back then...
Jan
3
revised Two Steps away from the Hamilton Cycle
extended
Jan
3
comment Two Steps away from the Hamilton Cycle
@Hagen Yes you miss cd and get ab twice. Do you have another subgraph? Sorry the late answer...
Jan
2
comment Two Steps away from the Hamilton Cycle
Where do you go from the second $y$? To $f$,$x$ or back to where you came from (call it $t$), but anyway you introduce more than two errors. Draw the picture by hand, make a photo and post that.
Jan
1
revised Two Steps away from the Hamilton Cycle
deleted 213 characters in body; edited title
Jan
1
revised Two Steps away from the Hamilton Cycle
typos fixed
Dec
31
comment Is this face inside or outside the Hamilton cycle?
Thanks for setteling the case. Would be glad if you could have a look at the linked one...
Dec
31
accepted Is this face inside or outside the Hamilton cycle?
Dec
30
comment Two Steps away from the Hamilton Cycle
(iv) How is Figure 3 false? When the original HC is $abcd..ef..yx$ and you have exactly these two 2 squares (here the subtitle is correct), you could introduce two errors if take the path $ad\dots e\color{red}{xy}f\dots xy$. Maybe a picture of yours would help. Thanks for your time so far... I will put some of my motivation in the next edit together with some more clarifications...
Dec
30
comment Two Steps away from the Hamilton Cycle
Brian, thanks for your answer, here are mine: Yes, the path should be closed (it was/is a cycle) and of the same length as the original HC. Not sure if this is called an edge path, but it sounds right. (i)But I don't want to remove/add edges to the graph. (ii)To admit the drawing are a little wrong since some of the vertices miss the cubic property (I will update it). (iii) Also "2 squares" doesn't really fit. I removed the $ex$/$fy$-edge(s) (see the post revisions), because you don't need them, to introduce two errors.
Dec
28
revised Two Steps away from the Hamilton Cycle
added ad