Skip to main content
updated broken link, minor corrections
Source Link
Wolfgang
  • 13.4k
  • 5
  • 45
  • 102

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 itthat really so?

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

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

LetDenote by $a_n$ isthe terms of 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 an infinite family of polynomials with this property.

Added For referencereferences 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. here "11. Squarefree values of polynomials".

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.

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 that really so?

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

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

Denote by $a_n$ the terms of 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 an infinite family of polynomials with this property.

Added For references 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".

Added references per comments
Source Link
joro
  • 25.4k
  • 10
  • 66
  • 121

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$$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.

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 $f$ 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.

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.

Source Link
joro
  • 25.4k
  • 10
  • 66
  • 121

Is OEIS A007018 really a subsequence of squarefree numbers?

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 $f$ 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.