Number of integer functions satisfying three constraints I am trying to understand how many functions $\mathbb{Z^+}\to \mathbb{Z^+}$ which satistfy the three following constraints exist:


*

*For every $n \in \mathbb{Z^+}$
$$f(f(n))\leq\frac{n+f(n)}{2}$$

*For every $n \in \mathbb{Z^+}$
$$f(n)\leq f(n+1)$$

*For every $(n,m)\in \mathbb{Z}^{+}\times\mathbb{Z}^+ $ such that $n \neq m$ 
$$\text{GCD}(f(n),f(m))=1$$
One of these functions is indeed $f(n)=1$ for every $n \in \mathbb{Z}^+$, but any other constant function does not satisfy the third constraint, which is probably the strongest. 
My first intuition about other functions was something on the lines of $f(n)=\text{the n-th prime number}$, but this functions (or any of its multiples) does not satisfy the first constraint, already for $n=2$. 
Are there any other functions that satisfy these constraints? How can they be found?

The constant function $f(n)=1$ is indeed the only solution, even if we weaken the first condition to: $f(f(n)) \le C(n+f(n))$ for some fixed $C>0$ (no matter how large).
As a proof-of-concept, we prove: there is no such solution with $f(1)>1$. Note that if $f(1)>1$ then $f(n+1)>f(n)$ (strictly) for all $n$, because of the gcd condition. In particular, $f(n)>n$; so it suffices to show there is no solution even if we replace $f(f(n)) \le C(n+f(n))$ by $f(f(n)) \le 2Cf(n)$. To do so, it suffices to show that $\lim_{k\to\infty} \frac{f(k)}k = \infty$.
Each of the values $f(1),\dots,f(k)$ is divisible by at least one prime, and these primes are distinct by the gcd condition; therefore $f(k)$ is at least as large as the $k$th prime. In particular, $f(k) > k\ln k$ by Rosser's theorem, and so indeed $\lim_{k\to\infty} \frac{f(k)}k = \infty$.
This shows that there is no such solution with $f(1)>1$. To consider any solution with $f(1)=1$: If $f(n)$ is not constant, then choose $K$ such that $f(K+1)>1$, and redo the above proof with the values $f(K+1), \dots, f(K+k)$ instead of $f(1),\dots,f(k)$.
