A comment in A007018 a(n) = a(n-1)^2 + a(n-1), a(0)=1 claims

Subsequence of squarefree numbers (A005117). - Reinhard Zumkeller, Nov 15 2004

Is it really so?

As far as I know, it is open problem if a polynomial $f \in \mathbb{Z[x]}$ of degree $\ge 5$ can be squarefree infinitely often (some source require $f$ to be irreducible).

If the OEIS comment is correct, the sequence will give infinite family of (irreducible) polynomials which are squarefree infinitely often.

Let $a_n$ is OEIS A007018. Set $a_n = x$ and $$f(x)=a_{n+4}=x \cdot (x + 1) \cdot (x^{2} + x + 1) \cdot (x^{4} + 2 x^{3} + 2 x^{2} + x + 1) \\\\ \cdot (x^{8} + 4 x^{7} + 8 x^{6} + 10 x^{5} + 9 x^{4} + 6 x^{3} + 3 x^{2} + x + 1)$$

$f(a_n)=a_{n+4}$ will be squarefree infinitely often (including the irreducible degree 8 factor) and iterating $x \mapsto x^2+x$ will produce infinite family of polynomials with this property.

**Added** For reference of squarefree values of polynomials the search terms are *square free values of polynomials*. E.g. here p.1 and here 11. Squarefree values of polynomials.