A non-recursively enumerable subset of $\mathbb{N}$ which has a short description in terms of arithmetic

Abstract

Let $F(x,n)$ denote the arithmetical formula which starts with $\exists ab ~\forall i \leqslant n ~\exists swpq ~\forall jv ~\exists eg$ and appeared in J. P. Jones' article from 1978. From the results of this article, it follows that the set $\{n \in \mathbb{N}: \neg F(n,n)\}$ is non-recursively enumerable and co-recursively enumerable. We prove that the set $$T=\{n \in \mathbb{N}: \exists p,q \in \mathbb{N} ~((2n=(p+q)(p+q+1)+2q) ~\wedge$$ $$\forall (x_0,\ldots,x_p) \in \mathbb{N}^{p+1} ~\exists (y_0,\ldots,y_p) \in \{0,\ldots,q\}^{p+1}$$ $$((\forall j,k \in \{0,\ldots,p\}~(x_j+1=x_k \Rightarrow y_j+1=y_k)) ~\wedge$$ $$(\forall i,j,k \in \{0,\ldots,p\} ~(x_i \cdot x_j=x_k \Rightarrow y_i \cdot y_j=y_k))))\}$$ is not recursively enumerable. We prove that the set $\mathbb{N} \setminus T$ is recursively enumerable. Let $β:\mathbb{N}^3 \to \mathbb{N}$ denote Gödel's $β$ function. For $x_1,x_2,x_3 \in \mathbb{N}$, $β(x_1,x_2,x_3)$ equals the remainder after integer division of $x_1$ by $1+(x_3+1) \cdot x_2$. We prove that the set $T$ consists of all $n \in \mathbb{N}$ such that $$\forall u,v \in \mathbb{N} ~\exists a,b,p,q \in \mathbb{N} ~((2n=(p+q)(p+q+1)+2q) \wedge \forall i,j,k \in \{0,\ldots,p\}$$ $$((β(a,b,i) \leqslant q) \wedge (β(u,v,j)+1=β(u,v,k) \Rightarrow β(a,b,j)+1=β(a,b,k)) ~\wedge$$ $$(β(u,v,i) \cdot β(u,v,j)=β(u,v,k) \Rightarrow β(a,b,i) \cdot β(a,b,j)=β(a,b,k))))$$ The above formula can be easily translated into a formula in Peano arithmetic.