Minimal DFA with more states than its equivalent NFA I understand that using a bad case for subset construction as provided through an example in the book -  Introduction to Automata Theory, Languages and Computation, we can definitely have an NFA with $n+1$ states, and get a corresponding minimal DFA with $2^n$ states due to the power set logic.

But, my question is, is there a case where you can have an NFA  (with or without epsilon moves), with $n$ states, having an equivalent minimal DFA of $p$ states, where $p \neq 2^n \ and \ p &gt; 2n$ ?
Consider the language $(0+1)^*0(0+1)^{n-1}$ over the alphabet $\{0,1,2\}$. It is accepted by an NFA having $n+1$ states, and its minimal DFA contains $2^n+1$ states.