Analytical formula for a summation I am looking for an "analytical" formula for the following summation; that is, write the following function without a summation sign:
$$ f(n) = \sum_{k=0}^{n}(-1)^k \cdot {n \choose k} \cdot (n-k)^{3n-1} $$
defined for all natural numbers $ n \in \mathbb{N} $. A disclaimer in beforehand: I actually do not know whether there is a solution to my problem, but I am grateful for any idea (also for asymptotically accurate results). My intuition was to somehow use the binomial expansion here, but I did not get anywhere...
 A: Rather surprisingly, I can tell you the answer and give a combinatorial example:
This can be expressed as  $$n! \, S_2(3n-1,n)$$
where $S_2(\cdot,\cdot)$ is a Stirling number of the second kind.  The sequence starts $1, 30, 5796, 3498000, 4809004200, \ldots$ when $n=1,2,3,4,5,\ldots$
It is the number of different ways of putting $3n-1$ labelled balls into $n$ labelled boxes so that each box has at least one ball.
So it is less than $n^{3n-1}$ (which would be the number if there were no requirement to have each box have at least one ball) but not an enormous amount less. That sequence starts $1, 32, 6561, 4194304, 6103515625, \ldots$
Empirically if you multiplied $n^{3n-1}$ by something like $0.981561^{3n-2}$ you might get closer and a sequence starting approximately $0.98, 29.7, 5759.6, 3482033, 4791868460,\ldots$ which is not that far away, though I have no theoretical justification for that choice.  It seems to work reasonably well for larger $n$ too.
A: An asymptotics may be obtained as follows. Using the generating function of the Stirling numbers of the second kind and Cauchy's formula, we find
$$
f(n)=n!\left\{{3n-1 \atop n}\right\}= \frac{{(3n - 1)!}}{{2\pi i}}\oint_{(0 + )} {\frac{{(e^t  - 1)^n }}{{t^{3n} }}dt}  = \frac{{(3n - 1)!}}{{2\pi i}}\oint_{(0 + )} {e^{ - n\varphi (t)} dt} 
$$
where
$$
\varphi (t) = \log \left( {\frac{{t^3 }}{{e^t  - 1}}} \right).
$$
There is a relevant saddle point at $t_0  = 3 + W( - 3e^{ - 3} ) = 2.82143937214277 \ldots$, where $W$ denotes the principal branch of the Lambert $W$-function. Following the steps in N. Temme's paper Asymptotic Estimates of Stirling Numbers (https://doi.org/10.1002/sapm1993893233) and applying Stirling's formula for the factorial, we deduce
$$
f(n) \sim \frac{1}{{n\sqrt {3\left| {\varphi ''(t_0 )} \right|} }}\left( {\frac{{27n^3 }}{{e^3 }}\frac{{e^{t_0 }  - 1}}{{t_0^3 }}} \right)^n 
$$
as $n\to +\infty$, with
\begin{align*}
\left| {\varphi ''(t_0 )} \right| & = 0.309567474978694 \ldots ,
\\
\frac{{e^{t_0 }  - 1}}{{t_0^3 }} & = 0.703514172238102 \ldots\, .
\end{align*}
For example, $f(5)=4809004200$, whereas the asymptotic formula gives $4790795320$.
Note that you can write the asymptotic formula in the form
$$
f(n) \sim An^{3n - 1} B^n ,
$$
with
\begin{align*}
& A = \frac{1}{{\sqrt {3\left| {\varphi ''(t_0 )} \right|} }} = 1.037675851879 \ldots, \\ &
B = \frac{{27}}{{e^3 }}\frac{{e^{t_0 }  - 1}}{{t_0^3 }} = 0.9456995211564 \ldots ,
\end{align*}
showing why @Henry's empirical formula $f(n) \approx n^{3n - 1} C^{3n - 2}$ with $C = B^{1/3}  = 0.98156196 \ldots$ works so well.
A: A connection with the binomial theorem is given via the exponential series expansion. Using the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series we can write
\begin{align*}
[z^k]e^{nz}=[z^k]\left(1+nz+\frac{(nz)^2}{2!}+\frac{(nz)^3}{3!}+\cdots\right)=\frac{n^k}{k!}\tag{1}
\end{align*}

We obtain using (1)
\begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k\binom{n}{k}(n-k)^{3n-1}}\\
&=\sum_{k=0}^n(-1)^k\binom{n}{k}(3n-1)![z^{3n-1}]e^{z(n-k)}\\
&=(3n-1)![z^{3n-1}]e^{nz}\sum_{k=0}^n\binom{n}{k}\left(-e^{-z}\right)^k\\
&=(3n-1)![z^{3n-1}]e^{nz}\left(1-e^{-z}\right)^n\\
&\,\,\color{blue}{=(3n-1)![z^{3n-1}]\left(e^{z}-1\right)^n}
\end{align*}

