Let $n, d$ be integers with $1 \leq d \leq \left \lfloor \frac{ n-1}{ 2} \right \rfloor$, and set $h(n,d):={ n-d \choose 2} + d^2$ and $e(n,d):= \max\{ h(n,d),h(n, \left \lfloor \frac{ n-1}{ 2} \right \rfloor)\}$. Because $h(n,d)$ is quadratic in $d$, there exists a $d_0(n)=(n/6)+O(1)$ such that $e(n,1)> e(n, 2)> \dots >e(n,d_0)=e(n, d_0+1)=\dots = e(n,\left \lfloor \frac{ n-1}{ 2} \right \rfloor)$. A theorem by Erd\H{ o}s states that for $d\leq \left \lfloor \frac{ n-1}{ 2} \right \rfloor$, any $n$-vertex nonhamiltonian graph $G$ with minimum degree $\delta(G) \geq d$ has at most $e(n,d)$ edges, and for $d > d_0(n)$ the unique sharpness example is simply the graph $K_n-E(K_{ \lceil (n+1)/2\rceil})$. Erd\H{ o}s also presented a sharpness example $H_{ n,d}$ for each $1\leq d \leq d_0(n)$. We show that if $d< d_0(n)$ and a $2$-connected, nonhamiltonian $n$-vertex graph $G$ with $\delta(G) \geq d$ has more than $e(n,d+1)$ edges, then $G$ is a subgraph of $H_{ n,d}$. Note that $e(n,d) - e(n, d+1) = n - 3d - 2 \geq n/2$ whenever $d< d_0(n)-1$.
Categories: math.CO