A Strange Function.

It is easy to see $f(2)=f(1)+1$ and $f(4)=f(1)+3$ .Conclude that $f(3)=f(1)+2$ .By induction,get $f(n)=f(1)+n-1$ .

Suppose $f(1)=k>1$ .

Note that the numbers $(k+1)!+2,(k+1)!+3,....,(k+1)!+(k+1)$ are all composite.

Take the least prime $p$ exceeding $(k+1)!+(k+1)$ .

Set $n=p-k+1$ .Then $p=f(n)$ and hence $n$ is a prime.

But $n>(k+1)!+2$ and hence $p>n>(k+1)!+(k+1)$ .This contradicts the minimality of $p$ .

Conclude $f(1)=1$ and hence $f(n)=n$ for all natural numbers $n$ .

A simple solution is as follows: one such functions is the identity $f(n) = n$ . Since this function satisfies all the given properties, it must be that it is a solution. Hence $f(2014) = 2014$ .