Busy Beaver function

Remarks

There are $(4n + 4)^{2n}$ candidates, and at least one of them halts, so both maxima exist. Example: for $n = 1$, the candidate with $\delta(q_1, 0) = (H, 1, R)$ halts after one step with one $1$, so $\Sigma(1) \ge 1$. Every other $1$-state candidate either halts at once with at most one $1$ or never halts, so $\Sigma(1) = 1$ and $S(1) = 1$. Known values: $\Sigma(2) = 4$, $\Sigma(3) = 6$, $\Sigma(4) = 13$, $\Sigma(5) = 4098$, and $S(2) = 6$, $S(3) = 21$, $S(4) = 107$, $S(5) = 47{,}176{,}870$. Non-example: a machine that writes $1$s forever is not counted, because only halting runs enter the maximum. For every total computable function $g \colon \N \to \N$ there is $N$ with $\Sigma(n) > g(n)$ for all $n \ge N$. Hence $\Sigma$ and $S$ are not computable. If $S$ were computable, the halting problem on blank tape would be decidable: run an $n$-state candidate for $S(n)$ steps and see whether it has halted.