What is the maximum number of edges in a connected graph without Hamiltonian path?
I've searched the Internet on a while, and read the question Maximum number of edges in a non-Hamiltonian graph here, which gives the answer and proof for Hamiltonian cycle. I've tried to apply the same method of the first answer, but it gives $\binom{n-1}2$ for this question. This is right for non-connected graph, where we can construct a $K_{n-1}$ and a single point. For connected graphs, I know the lower bound is $\binom{n-2}2+2$, which is to connect two single points with a same point in $K_{n-2}$, differ from the upper bound.
Which is the correct answer, the lower or upper bound? Or something between them? I'd appreciate it if you gave me any references or improved either lower or upper bound.
Update on 03-17-2023: The problem is solved as a special case of the paper Connected graphs without long paths. The counterexamples exist if and only if $n=6$ or $8$, which are:
o---o---o
\ /|\ /
X | X
/ \|/ \
o---o---o
o o o o o
XXXXXXX (All upper-level nodes are connected to all lower-level nodes)
o-o-o
_/
```
