There is a typo in the problem. It should read "The function has repulsion degree $ k > 0$ if for every $ p$, $ f^{i}(p)\neq p$ for $ i = 1,2,...,\lfloor kp\rfloor$." Solution coming later today by yours truly (unless somebody's faster).

Obviously $ f(1) = 1$, and we shall assume that it is possible to find a circular function $ f$ with repulsion degree $ k\geq\frac{1}{2}$. Because $ \lfloor 2k\rfloor = 1$ by hypothesis, $ f(2)\neq2$, and because of circularity, $ m > 2$ exists such that $ f(2) = m$ and $ f(m) = 2$. If $ m\geq4$, then $ \lfloor km\rfloor\geq 2$, in contradiction with $ f(f(m)) = f(2) = m$, hence $ m = 3$ necessarily. For any positive integer $ q$, define $ C_{q}$ as the set of different numbers $ p$ such that $ f^{i}\left(2^{q}\right) = p$ for some $ i$. Because of circularity, $ C_{q}$ is also the set of integers $ p$ such that $ f^{i}(p) = 2^{q}$ for some $ i$, and the cardinal $ n_{q}$ of $ C_{q}$ is the smallest number such that $ f^{n_{q}}(p) = p$ for all $ p\in C_{q}$, being obviously $ 2^{q}\geq n_{q}$. Assume that $ p\in C_{q}$ exists such that $ p\geq2^{q+1}$. Then, $ \lfloor kp\rfloor\geq2^{q}$, or $ f^{i}(2^{q})\neq2^{q}$ for all $ i = 1,2,...,2^{q}$, which is absurd because of circularity. Therefore, all $ p\in C_{q}$ are less at most $ 2^{q+1}-1$. In particular, $ 2^{q}$ cannot belong in $ C_{q'}$ for any $ q' < q$, or all $ C_{q}$ are disjoint. We will now show by induction that $ C_{q}=\left\{2^{q},2^{q}+1,2^{q}+2,...,2^{q+1}-1\right\}$. The result is proved for $ q = 1$. Assume the result true for $ 1,2,...,q-1$. Then, any element in $ C_{q}$ is not lower than $ 2^{q}$, since all positive integers less than $ 2^{q}$ are either 1, or belong to some $ C_{q'}$ with $ q' < q$. The largest element in $ C_{q}$ is then at least $ 2^{q}+n_{q}-1$, or $ n_{q}-1\geq\lfloor\frac{2^{q}+n_{q}-1}{2}\rfloor\geq2^{q-1}+\frac{n_{q}}{2}-1$, and we can conclude that $ n_{q}\geq2^{q}$, yielding $ n_{q}= 2^{q}$. Since no element of $ C_{q}$ may be smaller than $ 2^{q}$ or larger than $ 2^{q+1}-1$, then $ C_{q}=\left\{2^{q},2^{q}+1,2^{q}+2,...,2^{q+1}-1\right\}$, qed. Since the largest element of $ C_{q}$ is $ 2^{q+1}-1$, and $ i = n_{q}= 2^{q}$ is the smallest $ i$ such that $ f^{i}\left(p\right) = p$ for any $ p\in C_{q}$, then $ 2^{q}-1\geq\lfloor k\left(2^{q+1}-1\right)\rfloor > k\left(2^{q+1}-1\right)-1$, or $ k <\frac{2^{q}}{2^{q+1}-1}=\frac{1}{2}+\frac{1}{2\left(2^{q+1}-1\right)}$. When $ q$ takes on increasing values of integers, the second term tends to zero, so $ k$ cannot be larger than $ \frac{1}{2}$. Note that a circular function with $ k =\frac{1}{2}$ may be constructed by setting $ f(p) = p+1$ if $ p+1$ is not an integral power of 2, and $ f(p) =\frac{p+1}{2}$ if $ p+1$ is an integral power of 2.

A function $f: N \rightarrow N$ is circular if for every $p \in N$ there exists $n\in N,\ n \le p$ such that $f^n(p)=p$ ($f$ composed with itself $n$ times) The function $f$ has repulsion degree $k>0$ if for every $p\in N$ $f^i(p)\neq{p}$ for every $i=1,2,\dots,\lfloor{kp}\rfloor$. Determine the maximum repulsion degree can have a circular function. Note: Here $\lfloor{x}\rfloor$ is the integer part of $x$.