Let c=2011, C=[1,2,3,...,c], if f does exist
1) f(n+c) = f(n) + c
2) f(a) = f(b) <===> a = b
3) for any n in C, f(n) <= 2c
4) let A and B the subsets of C, so that
for all a in A, f(a) <= c)
for all b in B, c<f(b)<=2c
5) B can not be empty.
6) for each b in B, there is an unique a in A, with f(a)=b
That means A and B have the same number of elements, ...impossible since c is odd!
So, you can not construct any relation for C.
This is true for Cn=[nc+1,nc+2, ... nc+c], n=1,2,3 ......