The constant clearly does not satisfy the condition.

For example, the polynomial $P(x)=x-1$ works.

Note that the leading coefficient of the polynomial $P$ is positive (otherwise, $P(x)x$ for each $x \in N$. Suppose the sequence exists. Starting from some index, the terms less than $N$ will end, that is, there will be an $n$ such that $a_{k} \geq N$ for all $k>n$. Then $a_{n}=P\left(a_{n+1}\right)$ $>a_{n+1}, a_{n+1}=P\left(a_{n+2}\right)>a_{n+2}, \ldots$, that is, $a_{n}>a_{n+1}>a_{n+2}>\ldots-$ an infinite decreasing sequence of natural numbers, which is impossible.

## Answer

Only the first degree
$\underline{\text { Frankin B.R. }}$

Given a polynomial $P(x)$ with real coefficients. An infinite sequence of distinct natural numbers $a_{1}, a_{2}, a_{3}, \ldots$ is such that

$P\left(a_{1}\right)=0, P\left(a_{2}\right)=a_{1}, P\left(a_{3}\right)=a_{2}$, and so on. What degree can $P(x)$ have?