Is function with finite amount of $\Phi_x(x+1)$ occurrences Turing computable? I'm studying for my exam from Logic and Computability and we have to face following kind of examples:

$$f(x) = \begin{cases}
         \ 1 &amp;\text{if }\Phi_x(x+1)\!\downarrow\text{ and }x\leq50\\
         \ 0 &amp;\text{otherwise.}
\end{cases}$$

Unfortunatelly I'm not really clear on how to tackle those. $\Phi_x$ is the $x$-th computable function, thus it's also Turing computable. This implies, there exists a TM (we have $N$ of them) which halts and returns the output.

The task is to determine whether it's computable. In order to prove it we either provide an informal algorithm to show it's computable or a formal proof that it's not. Can any of you solve this example? 

If $f(x)$ is computable, then $\Phi_x$ needs to halt for every $x \in\{1, \dots, 51\}$. However $\Phi_x$ could be undefined on one of the elements from $\{1, \dots, 51\}$ and keep looping forever.

At the other hand, there is an upper bound on $x$, which indicates its computability.

Please, can you show me the right direction?

Thanks!
You can compute this function using a table of size 51. More generally, if $f$ is a function which differs from a computable function $g$ on a finite number of inputs, then $f$ is computable as well. In our case, $g$ is the constant zero function, which is trivially computable.