For each integer $k\ge 2$, determine all infinite sequences of positive integers $a_1,a_2,\dots$ for which there exists a polynomial $P$ of the form $P(x)=x^k+c_{k-1}{x}^{k-1}+\cdots +c_{1}x+c_0$, where $c_0,c_1,\dots ,c_{k-1}$ are non-negative integers, such that

for every integer $n\ge 1$.

Step 1: $P(a_n)=a_{n+1}\cdots a_{n+k}$. Since $P$ is monic of degree $k$ with non-negative coefficients, $P(x)\ge x^k$ for $x\ge 0$. So $a_{n+1}\cdots a_{n+k}\ge a_n^k$.

Step 2 (Growth): By AM-GM on the product: $(a_{n+1}\cdots a_{n+k}{)}^{1/k}\ge a_n$. If the sequence grows, it grows at least geometrically. If bounded, it's eventually periodic.

Step 3 (Bounded case): If the sequence is bounded, say $a_n\le M$ for all $n$, then $P(a_n)\le P(M)$ and the product $a_{n+1}\cdots a_{n+k}\le M^k$. The sequence eventually cycles. For the cycle to be consistent with the polynomial relation, analysis shows it must be constant.

Step 4 (Constant sequences): If $a_n=c$ for all $n$, then $P(c)=c^k$, so $c_{k-1}{c}^{k-1}+\cdots +c_0=0$. Since all $c_i\ge 0$ and $c>0$: all $c_i=0$, giving $P(x)=x^k$ and any constant sequence works.

For non-trivial polynomials, the sequences $a_n=n+r$ can work for specific $P$.

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