It is well known that if $c(K)=2n+1$, then $u(K)$ is less than $n+1$. It can not be sharper because of the trefoil knot. On the other hand, if $c(K)=2n$, then similarly we have $u(K)$ is less than $n+1$. I think $u(K)=n$ is impossible in this case, i.e. there does not exist a knot $K$ with $c(K)=2n$ and $u(K)=n$. Maybe it is fairly easy, but I have no idea how to deduce it. Any hint is welcome :)
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$\begingroup$ Related question: mathoverflow.net/questions/108312/… $\endgroup$– Ian AgolOct 9 '12 at 14:03
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$\begingroup$ Thank you very much for your response. Given a knot with c(K)=2n, then u(K)\leq a(K)\leq n1. Although a(K) (the ascending number) is not easy to calculate in general, it works well in this question~ $\endgroup$ Oct 10 '12 at 8:25

$\begingroup$ Here is another proof inspired from Makoto Ozawa's ascending number. Assume there exists a knot diagram K with c(K)=2n and u(K)=n. By switching each crossing point one can get $2^n$ different knot diagrams totally. We say two diagrams are connected if one diagram can be obtained from the other one by switching one crossing point. If u(K)=n, it is not difficult to conclude that every unknot diagram is isolated. However this is impossible since each ascending diagram is connected to another ascending diagram, and both of them represent unknots. Hence that is a contradiction. $\endgroup$ Oct 10 '12 at 13:25

$\begingroup$ What a good structural proof! You gave a new view point. $\endgroup$ Oct 12 '12 at 0:58