I'm trying to find a tight lower bound (i.e. some function of $n\in \mathbb{N}$) for $$f(n)=\sum_{j=0}^n {n\choose j}\exp\left(\frac{j(j-1)}{2}+\frac{(n-j)(n-j-1)}{2}-j(n-j)\right)$$
My attempt: Define
$$g(x):= \left(\frac{x(x-1)}{2}+\frac{(n-x)(n-x-1)}{2}-x(n-x)\right)$$
$g'(x)=0$ only if $x=n/2$ and $g''(n/2)>0$. So $g(x)\geq g(n/2)$ for all $x$. In particular $g(j)\geq g(n/2)$ for all $0\leq j\leq n$. Thus
\begin{equation} f(n) \geq \exp(g(n/2))\sum_{j=0}^n {n\choose j} = \exp(-n/2)2^n \end{equation}
which is a crude lower bound and am hoping to find a better lower bound. The reason the bound I have is not good is because I have found a lower bound for each exponential term in the definition of $f(n)$ i.e. $g(n/2)$ which is exponentially decreasing in $n$ whereas in the definition of $f(n)$, there is at least one exponential term that is exponentially increasing in $n$ (For example, when $j=0$, the exponential term is $\exp(n(n-1)/2)$). Any ideas?