Sum of Stirling numbers with exponents I have a trouble with the following sum
$\sum_{i=0}^n\binom{n}{i}S(i,m)3^i$, where $S(i,m)$ is the Stirling number of the second kind (the number of all partitions of $i$ elements into $m$ nonempty sets).
Below it was obtained the sum
$f(n,m)=\frac{1}{m!}\sum_{k=0}^m(-1)^{m-k}\binom{m}{k}(1+3k)^n$. Is there a further simplification of this expression?

here is a table of $f(m,n)$ for small values of $m$ and $n\geq m$ (note that $f(m,n)=0$ for $n<m$):
$f(n,1)=-1+4^n$
$f(n,2)=\tfrac{1}{2}\left(1-2\cdot 4^n+7^n\right)$
$f(n,3)=\tfrac{1}{6}\left(-1+3\cdot 4^n-3\cdot 7^n+10^n\right)$
$f(n,4)=\tfrac{1}{24}\left(1-4\cdot 4^n+6\cdot 7^n-4\cdot 10^n+13^n\right)$
$f(n,5)=\tfrac{1}{120}\left(-1+5\cdot 4^n-10\cdot 7^n+10\cdot 10^n-5\cdot 13^n+16^n\right)$
$f(n,6)=\tfrac{1}{720}\left(1-6\cdot 4^n +15\cdot 7^n-20\cdot 10^n+15\cdot 13^n-6\cdot 16^n+19^n\right)$
there is a pattern, which as Todd Trimble points out is
$$f(n,m)=\frac1{m!} \sum_{k=0}^m (-1)^{m-k} \binom{m}{k} (1 + 3k)^n$$

note: Mathematica gets $f(n,2)$ wrong... [SE post]
