For $n\ge3$ let $a(n)$ be the minimum possible number of Hamiltonian paths in a strongly connected tournament on $n$ nodes.
What is a good (or at least nontrivial) lower bound for $a(n)$? What is a good reference for this question? Is $a(n)$ in the Encyclopedia of Integer Sequences?
My attempt: All I know is that $a(3)=3,a(4)=5,$ and $a(5)=9$ which I found by counting by hand. (In a strongly connected tournament on $5$ nodes, the number of Hamiltonian paths can be $9,\ 11,\ 13,$ or $15.$) Trivially $a(n)\ge n$ (because a strongly connected tournament has a Hamiltonian cycle), and $a(n)$ is odd by Rédei's theorem.
Update. As suggested in a comment by @Michael, for each $n\ge3$ there is a strongly connected tournament on $n$ nodes with exactly $2^{n-2}+1$ Hamiltonian paths; namely, take a transitive tournament and reverse the arc between the first and last nodes. This proves that $a(n)\le2^{n-2}+1.$ Is $a(n)=2^{n-2}+1$?
For $n\ge3,$ does every strongly connected tournament of $n$ nodes have at least $2^{n-2}+1$ Hamiltonian paths? (True for $n=3,4,5.$) [Nope — see the third update.]
Update. It's easy to see that $a(n+1)\ge a(n)+n-1,$ whence it follows by induction that $a(n)\ge\lfloor(n-1)^2/2\rfloor+1=$ OEIS sequence A099392, so I guess the trivial bounds are $$\lfloor(n-1)^2/2\rfloor+1\le a(n)\le2^{n-2}+1.$$
Update. Consider the transitive tournament with nodes $v_1,v_2,v_3,v_4,v_5,v_6$ and arcs $v_iv_j\ (1\le i\lt j\le6).$ Let $T$ be the strongly connected tournament obtained from this transitive tournament by reversing the arcs $v_1v_5$ and $v_4v_6.$ Then $T$ has exactly $15$ Hamiltonian paths, showing that $a(6)\le15.$ An obvious generalization of this construction shows that $$a(n)\le2^{n-2}-2^{n-4}+3.$$
I posted the same question at Math Overflow and Fedor Petrov answered there by referring to the paper "A Note on the Number of Hamiltonian Paths in Strong Tournaments" by Arthur H. Busch where it is shown that $a(n)$ is approximately $5^{\frac{n-1}3}.$