Solution 1

Let f(p) = 1 for every prime p.
Let f(pq) = 2 for every pair of primes p and q.
In general, let f(n) = k for n, the product of k primes.
We have to map 1 someplace, let f(1) = 1.

Solution 2

Let f(1) = 1.
Let f(2) = 1 and f(3) = 2.
Let f(4) = 1 and f(5) = 2 and f(6) = 3.
In general, let f(n(n+1)/2 + k) = k for n = 0, 1, 2, ..., and k = 1, 2, 3, ..., n+1.

Solution 3

Let f(2n + 1) = 1 for n = 0, 1, 2, ....
Let f(4n + 2) = 2 for n = 0, 1, 2, ....
Let f(8n + 4) = 3 for n = 0, 1, 2, ....
In general, let f(2^kn + 2^k-1) = k for n = 0, 1, 2, ..., and k = 1, 2, 3, ....