Answer by fgrieu for Majority-based feedback shift register

I'm reading the question as asking for $b(n)$, the largest possible, such that we can exhibit distinct tap points (in odd number), and $n$-bit state, leading to a periodic sequence of (shortest) period at least $b(n)$ steps.

I have a construction with $b(n)=2n-2$ for $n\ge3$. The taps are the next three bits that will drop out of the MVFSR. The state is all-zeroes, except for the next bit that will drop out of the MVFSR. Here is an illustration with $n=8$: time elapses from left to right, the MVSFR shifts down, the next bit enters on top, the three taps are marked with *.

   0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0   0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0   0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0   0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0   0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0*  0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0*  0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1*  1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1

If we allow a single tap, we get to $b(n)=2n$.

I hope that $b(n)$ can be improved (increased), perhaps to become exponential, but it's already more than $1$ of this answer, and $2$ of this comment.