Prove domain of partial computable function exists

Question

Prove that there is an n such that $W_n$ = {$2n, . . . , 2n + n^2$}

Now I don't know where to start with this question, how can I go about answering it? Would I construct a computable function that has that domain? What is that domain? I'm not sure I understand the question properly.

$W_n$ is the domain of a partial computable function with godel number n, is that right?

Ok, heres what I have so far with the recursion theorem:

define \begin{equation} g(x,y)=\begin{cases} 1, & \text{if $2x\le y \le 2x+x^2$}.\\ ↑, & \text{otherwise}. \end{cases} \end{equation}

Say g has an index e, such that $g = \varphi_e$

By S-M-N theorem we have a total computable $\varphi_{s(x)}(y) =\varphi_e(x,y)$.

Then we have $\varphi_{s(x)}$ = $\varphi_x$ fixed point by the recursion theorem

$\varphi_x$ has domain {$2x,...,2x+x^2$}, therefore such an n=x exists.

Answer 1

Hint: Use the recursion theorem.

Answer 2

Define

$\Psi(x,y) = \begin{cases} 0 & \quad 2x \leq y \leq 2x + x^2 \\ \uparrow & \quad \text{otherwise} \end{cases}$

It is clear that $\Psi$ is partial computable. By the s-m-n theorem, there exists a computable function $f$ such that

$\Psi(x,y) = \Phi_{f(x)}(y)$

By the recursion theorem, there exists a $n$ such that

$\Phi_{f(n)} = \Phi_n$

Therefore

$\Phi_n(y) = \Phi_{f(n)}(y) = \Psi(n,y) = \begin{cases} 0 & \quad 2n \leq y \leq 2n + n^2 \\ \uparrow & \quad \text{otherwise} \end{cases}$

Hence the domain of $W_n = \{2n, 2n + 1, ..., 2n + n^2\}$.