Computation of busy beaver function

The busy beaver max shifts function, $S(n)$, has known values for $n\leq4$. Is there some basic, structural reason why it's inconceivable that we will ever find $S(n)$ for $n>4$? What is so different about $n=4$ than $n=5$? Or $n=6$? Somewhere along the way there must be some fundamental difference, otherwise $S(n)$ would be, in principle, computable for all $n$, so what exactly is this difference?

The reason that no program can compute $S(n)$ is that if you knew what $S(n)$ is you could decide the halting problem - you'd know when to stop waiting. On the other hand, for each $m$ there is a program that computes $S(n)$ for all $n \leq m$ - it just uses a table.

If it were possible to prove the value of $S(n)$ for all $n$ (that is, for all $n$ we could prove $S(n) = \alpha$ for some $\alpha$) then we could compute $S(n)$ by searching through all proofs (this assumes that our proof system is valid). So for each proof system there is a minimal value of $n$ for which you cannot prove that $S(n) = \alpha$ for any $\alpha$.

Finally, the reason that we know $S(4)$ is probably because $4$ is a really small number. The number $5$ is slightly bigger, and so things get more complicated. There's no deep reason why we know $S(4)$ but not $S(5)$, just like there is no deep reason why we know the Ramsey number $R(4)$ but not $R(5)$ (though Ramsey numbers are of course computable).

StackExchange Computer Science Q#12340, answer score: 6