There exists a non-recursively enumerable set {n∈N: φ(n)} such that the formula φ(n) is short and can be easily translated into a first-order formula which uses only + and \cdot

We prove that the set T={n∈N: ∃p,q∈N ((2n=(p+q)(p+q+1)+2q) ∧ ∀(x_0,...,x_p)∈N^{p+1} ∃(y_0,...,y_p)∈{0,...,q}^{p+1} ((∀k∈{0,...,p} (1=x_k ⇒ 1=y_k)) ∧ (∀i,j,k∈{0,...,p} (x_i+x_j=x_k ⇒ y_i+y_j=y_k)) ∧ (∀i,j,k∈{0,...,p} (x_i \cdot x_j=x_k ⇒ y_i \cdot y_j=y_k))))} is not recursively enumerable. By using Gödel's β function, we prove that the formula that defines the set T can be easily translated into a first-order formula which uses only + and \cdot. The same properties has the set {n∈N: ∃p,q∈N ((2n=(p+q)(p+q+1)+2q) ∧ ∀(x_0,...,x_p)∈N^{p+1} ∃(y_0,...,y_p)∈{0,...,q}^{p+1} ((∀j,k∈{0,...,p} (x_j+1=x_k ⇒ y_j+1=y_k)) ∧ (∀i,j,k∈{0,...,p} (x_i \cdot x_j=x_k ⇒ y_i \cdot y_j=y_k))))}.