If $x_{n(k)}=k*n(k)$, then ${x_{n(k)}+i}=k(n(k)+i)+2i,i<n(k)$ and $n(k+1)=2n(k)$. Therefore $x_{2^k}=(k+1)*2^k, k\ge 0$. and for $2^k<n<2^{k+1}$ we get $x_n=(k+1)*n+2(n-2^k)$.

Rust wrote: If $x_{n(k)}=k*n(k)$, then ${x_{n(k)}+i}=k(n(k)+i)+2i,i<n(k)$ and $n(k+1)=2n(k)$. Therefore $x_{2^k}=(k+1)*2^k, k\ge 0$. and for $2^k<n<2^{k+1}$ we get $x_n=(k+1)*n+2(n-2^k)$. I don't really understand your solution,can you clarify it? Thanks

I denote $n(k)$ suth index $n$, when $\frac{x_n}{n}=k\in N$. For example $x_1=1\to n(1)=1, x_2=4\to n(2)=2, x_4=12\to n(3)=4$. Then I proved $n(k)=2^{k-1}$ and for $n(k)=2^{k-1}<n<2^k=n(k+1)$ we get $x_n=k*n+2(n-n(k))$.