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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You can see the answer in Proposition 2.1 of link text 

