Let $f(n)$ be
$$
f(n)=\begin{cases}
1,&\small{\text{if there are digits 1 in the constant $\textit{e}$ $\textit{n}$ times in a row}}\\
0,&\small{\text{otherwise.}}\\
\end{cases}
$$
Is it true that $f(n)$ is computable?


I strongly believe that is not, but I can't even understand how to prove it. What to do?



Computable function
The function is computable. It is either the constant $1$ function, if there are arbitrarily long sequences of consecutive $1$s in the expansion of the number $e$ (and I guess you mean the decimal expansion or the binary expansion), or if there is not, then the function is a simple step function, with value $1$ up to some value $k$, the largest sequence occurring, and $0$ thereafter. All such functions are computable, and therefore, your function is computable. 


One shouldn't confuse the fact of the computability of your function with our lack of knowledge about which computable function it is. The argument exhibits a collection of functions, each of which is computable, and a proof that your function is one of those functions. So your function is computable, even if we don't know which one it is.


See also [my answer to a related question](https://mathoverflow.net/a/48031/1946), where a similar point is made. 


