Skip to main content
Update following Seva's explanations.
Source Link
Stefan Kohl
  • 19.6k
  • 21
  • 75
  • 137

Is there a set $A$ of positive integers such that

  • $\sum_{n \in A} \frac{1}{n} = \infty$, and

  • there is no polynomial $f \in \mathbb{Z}[x]$ of degree at least $2$ which takes infinitely many values in $A$?

Added on Feb 16, 2015: Seva gave a complete answer toanswered this question;question completely. though a little later, he claimedHe proved even a great dealgreat deal more -- his claim amounts to the following:

Claim: There is a partition of $\mathbb{N}$ into

  • a set $A$ of asymptotic density $1$ of 'non-values of non-linear polynomials', which has finite intersection with the image of any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$, and

  • a set $B$ of asymptotic density $0$ of 'values of non-linear polynomials', which contains all but finitely many positive values taken by any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$.

If Seva is rightnamely, can the construction of suchthat there is a partition of of $\mathbb{N}$ be made more explicit? into

  • a set $A$ of asymptotic density $1$ of 'non-values of non-linear polynomials', which has finite intersection with the image of any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$, and

  • a set $B$ of asymptotic density $0$ of 'values of non-linear polynomials', which contains all but finitely many positive values taken by any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$.

Is there a set $A$ of positive integers such that

  • $\sum_{n \in A} \frac{1}{n} = \infty$, and

  • there is no polynomial $f \in \mathbb{Z}[x]$ of degree at least $2$ which takes infinitely many values in $A$?

Added on Feb 16, 2015: Seva gave a complete answer to this question; though a little later, he claimed a great deal more -- his claim amounts to the following:

Claim: There is a partition of $\mathbb{N}$ into

  • a set $A$ of asymptotic density $1$ of 'non-values of non-linear polynomials', which has finite intersection with the image of any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$, and

  • a set $B$ of asymptotic density $0$ of 'values of non-linear polynomials', which contains all but finitely many positive values taken by any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$.

If Seva is right, can the construction of such partition of $\mathbb{N}$ be made more explicit?

Is there a set $A$ of positive integers such that

  • $\sum_{n \in A} \frac{1}{n} = \infty$, and

  • there is no polynomial $f \in \mathbb{Z}[x]$ of degree at least $2$ which takes infinitely many values in $A$?

Added on Feb 16, 2015: Seva answered this question completely. He proved even a great deal more -- namely, that there is a partition of $\mathbb{N}$ into

  • a set $A$ of asymptotic density $1$ of 'non-values of non-linear polynomials', which has finite intersection with the image of any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$, and

  • a set $B$ of asymptotic density $0$ of 'values of non-linear polynomials', which contains all but finitely many positive values taken by any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$.

Extended the question, asking for an explanation of Seva's claim.
Source Link
Stefan Kohl
  • 19.6k
  • 21
  • 75
  • 137

Is there a set $A$ of positive integers such that

  • $\sum_{n \in A} \frac{1}{n} = \infty$, and

  • there is no polynomial $f \in \mathbb{Z}[x]$ of degree at least $2$ which takes infinitely many values in $A$?

Added on Feb 16, 2015: Seva gave a complete answer to this question; though a little later, he claimed a great deal more -- his claim amounts to the following:

Claim: There is a partition of $\mathbb{N}$ into

  • a set $A$ of asymptotic density $1$ of 'non-values of non-linear polynomials', which has finite intersection with the image of any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$, and

  • a set $B$ of asymptotic density $0$ of 'values of non-linear polynomials', which contains all but finitely many positive values taken by any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$.

If Seva is right, can the construction of such partition of $\mathbb{N}$ be made more explicit?

Is there a set $A$ of positive integers such that

  • $\sum_{n \in A} \frac{1}{n} = \infty$, and

  • there is no polynomial $f \in \mathbb{Z}[x]$ of degree at least $2$ which takes infinitely many values in $A$?

Is there a set $A$ of positive integers such that

  • $\sum_{n \in A} \frac{1}{n} = \infty$, and

  • there is no polynomial $f \in \mathbb{Z}[x]$ of degree at least $2$ which takes infinitely many values in $A$?

Added on Feb 16, 2015: Seva gave a complete answer to this question; though a little later, he claimed a great deal more -- his claim amounts to the following:

Claim: There is a partition of $\mathbb{N}$ into

  • a set $A$ of asymptotic density $1$ of 'non-values of non-linear polynomials', which has finite intersection with the image of any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$, and

  • a set $B$ of asymptotic density $0$ of 'values of non-linear polynomials', which contains all but finitely many positive values taken by any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$.

If Seva is right, can the construction of such partition of $\mathbb{N}$ be made more explicit?

Source Link
Stefan Kohl
  • 19.6k
  • 21
  • 75
  • 137

Set of integers having finite intersection with the image of any polynomial of degree $\geq 2$

Is there a set $A$ of positive integers such that

  • $\sum_{n \in A} \frac{1}{n} = \infty$, and

  • there is no polynomial $f \in \mathbb{Z}[x]$ of degree at least $2$ which takes infinitely many values in $A$?