$A_{n+1}=f(A_n)$, were $f(x)=x^3-x+1=x(x_1)(x+1)+1.$ For $n>2$ $A_n$ is odd. Let p is odd prime and $a_n=A_n\mod p$. Then for $p=3$ $a_i=2,1,1,...$, for $p=5$ $2,2,2,...$, for $p=7$, $a_i=2,0,1,1,..$ For any $p$ it had some $i,t$, suth that $a_j=a_{j+t}, j+t<p \ (because \ f(0)=f(-1)=f(1)=1)$ for any $j\ge i$. $0$ not belong in period, because $f(0)=f(1)=1$. It mean $p\not| a_n$ for $n\ge p$.

thanks a lot!