Is every busy beaver strictly monotonic asymptotically?
Take an $n$-state Turing machine $M$ which outputs $k$ one symbols. Define the new Turing machine $M&#39;$ with $n+1$ states as follows. Every transition in $M$ to the ACCEPT state instead goes into a new state $q&#39;$. The state $q&#39;$ has the following behavior. If the head currently contains a ZERO, we overwrite with a ONE and ACCEPT. If the head currently contains a ONE, we shift to the right and remain in state $q&#39;$.

As machine $M$ halts, so does machine $M&#39;$. Furthermore, machine $M&#39;$ always puts one $k+1$ ONES onto the tape.

So $BB(n+1) \geq k+1 &gt; k = BB(n)$