MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Do you have a reference for that question, about some polynomials of high degree being squarefree infinitely often? – Per Alexandersson Jan 4 '13 at 13:00
As noticed here:, it is not known whether all of them are square-free. – Ilya Bogdanov Jan 4 '13 at 13:15
Sylvester’s sequence is your $a(n)+1$. Since $a(n)=a(n-1)(a(n-1)+1)$, your sequence consists of squarefree numbers if and only if Sylvester’s sequence does. – Emil Jeřábek Jan 4 '13 at 13:35
For the first 100 primes, the sequence becomes periodic modulo $p^2$ without passing through $0$. – David Speyer Jan 4 '13 at 13:39
Hm, Mathworld only says that the sequence is squarefree for the first 10e15... – Per Alexandersson Jan 4 '13 at 14:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.