How can you compute the asymptotics of
$$T=\sum_{k=1}^{n} k^{k-1} (n-k)^{2n-k} \binom{n}{k}\;?$$
This is related to Asymptotics of sum of binomials .
I attempted to simply use Stirling's approximation. This gives you
$$T \approx U =\frac{n^{n+1/2}}{\sqrt{2\pi}}\sum_{k=1}^n \frac{(n-k)^{n-1/2}}{k^{3/2}}$$
This approximation $U$ seems to be some constant times $n^{2n}$ asymptotically and also for $T$ itself.
By experimentation it looks like
$$T \sim \frac{n^{2n}}{2\pi}.$$
Is this correct?
Update. No it isn't.
The summand is decreasing with $k$:
$$ \sum_{k=1}^n k^{k-1} (n-k)^{2n-k} \binom{n}{k} = n^{2n} \sum_{k=1}^n \left(\frac{k}{n}\right)^{k-1} \left(1-\frac{k}{n}\right)^{2n-k} \frac{1}{n} \binom{n}{k} $$ We can now use the fact that $$ \lim_{n \to \infty} \left(\frac{k}{n}\right)^{k-1} \left(1-\frac{k}{n}\right)^{2n-k} \frac{1}{n} \binom{n}{k} = \frac{k^{k-1}}{k!} \left( \frac{1}{\mathrm{e}^2} \right)^k $$ Thus $$ \sum_{k=1}^n k^{k-1} (n-k)^{2n-k} \binom{n}{k} \rightarrow_{n \to \infty} n^{2n} \sum_{k=1}^\infty \frac{k^{k-1}}{k!} \left( \frac{1}{\mathrm{e}^2} \right)^k = n^{2n} \left( -W\left(-\frac{1}{\mathrm{e}^2}\right)\right) = n^{2n} \exp\left( W\left(-\mathrm{e}^{-2}\right) - 2 \right) $$ where $W(z)$ denotes the Lambert $W$-function. Note that $$ \left( -W\left(-\frac{1}{\mathrm{e}^2}\right)\right) \approx 0.158594 < 0.159155 \approx \frac{1}{2\pi} $$ Numerically it seems that estimate is from above: