First, we can easily find that $f(p^e t)=(t\mod p)$, which $t$ is not divisible by $p$. Lemma. If $n-1=\overline{c_k\cdots c_1c_0}_{(p)}$, $a_n=1+\sum_{i=0}^{k}{(c_i\mod 2)}$. pf) Induct on $n$. Assume that the statement holds for $n$. Denote $j$ be the minimal index such that $c_j \ne p-1$, then $n=\overline{c_k\cdots c_{j+1}(c_j+1)0\cdots0}_{(p)}$, $f(n)=c_j+1$. Since $((x+1) \mod 2) = (x\mod 2)+(-1)^{x+2}$, $a_{n+1}=a_n+(-1)^{f(n)+1}=1+\sum_{i=0}^{k}{(c_i\mod 2)}+(-1)^{c_j+2}=1+(c_j \mod 2)+\sum_{i=j+1}^{k}{(c_i\mod 2)}+(-1)^{c_j+2}=1+((c_j+1) \mod 2)+\sum_{i=j+1}^{k}{(c_i\mod 2)}$. Thus, also holds for $n+1$. By the lemma, now we can easily show that the answer is all positive integers.

We prove this using a modulus argument Assume WLOG that p is an odd prime. It can be see that adding (-1)^(f(n)+1) alternates between increase and decrease by 1 for all remainders of f(n) except for n congruent to p-1, where it adds two consecutive 1s after one move. Give that the sequence starts with 1, and that the increasing and decreasing 1s cancel out by n congruent to p-1 mod p (p-1 is even so half of it is decrease and half of it is increase). We have that through this increasing property all positive numbers should be able to be reached. And the answer is All positive integers