Does this recursive function have a function in terms of n? I am trying to convert the following recursive function to a non-recursive equation:
$$f(2) = 2$$
For $n>2$:
$$f(n)=nf(n-1)+n$$
I have calculated the results for n=2 through to n=9:
$$\begin{align}
f(2)&=2\\
f(3)&=9\\
f(4)&=40\\
f(5)&=205\\
f(6)&=1236\\
f(7)&=8659\\
f(8)&=69280\\
f(9)&=623529
\end{align}$$
I've tried graphing the function, but have got nowhere
Any help is appreciated!
 A: This is a linear, nonhomogenous recurrence relation. There are general methods to solve it, which you might find in the appropriate courses (I know my combinatorics text goes over it), but here's perhaps a more intuitive derivation.
Notice that you have
$$f(n) = nf(n-1) + n$$
Imagine iterating this several times: that is, we use the definition of $f$ for $f(n-1)$. We see:
$$\begin{align}
f(n) &= n+ nf(n-1)\\
&= n + n((n-1) + (n-1)f(n-2)) \\
&= n + n(n-1) + n(n-1)f(n-2)\\
&= n + n(n-1) + n(n-1)((n-2) + (n-2)f(n-3)) \\
&= n + n(n-1) + n(n-1)(n-2) + n(n-1)(n-2)f(n-3)\\ 
&= ...
\end{align}$$
If we keep iterating this until we get to our initial condition of $f(2)=2$, then we have
$$f(n) = n + n(n-1) + n(n-1)(n-2) + ... + n(n-1)(n-2)...(3)f(2)$$
Take note: since $f(2) = 2$, the last term is actually $n!$. So what we essentially have is a sum of all of the "falling factorials" of $n$ and $n!$ itself. (A falling factorial is something like $9\cdot 8 \cdot 7$ - it exhibits factorial-like behavior, but doesn't go all of the way to $2$ or $1$.)
Suppose we factor out $n!$ from each term. Then we see
$$f(n) = n! \left( \frac{1}{(n-1)!} + \frac{1}{(n-2)!} + ... + \frac{1}{2!} + \frac{1}{1!} \right)$$
This begs the summation notation:
$$f(n) = n! \left( \sum_{k=1}^{n-1} \frac{1}{k!} \right)$$
This is the summation noted in MachineLearner's answer. Then, leaning on  a past MSE post, we get the sequence OP mentioned wanting a derivation of in the comments of their question - an expression involving a floor function and $e$:
$$f(n) = n! \left( \sum_{k=1}^{n-1} \frac{1}{k!} \right) = \lfloor n! \cdot (e-1) \rfloor - 1$$
This is derived by simply noting that $e$ has the power series
$$e = \sum_{k=0}^\infty \frac{1}{k!}$$
If you start at $1$ instead, you get $e-1$ since $1/0!=1$. The summation in $f$ then becomes
$$\sum_{k=1}^{n-1} \frac{1}{k!} = \sum_{k=1}^\infty \frac{1}{k!} -  \sum_{k=n}^\infty \frac{1}{k!} = e-1 - \sum_{k=n}^\infty \frac{1}{k!}$$
The remaining summation is less than $1$, and thus invites the floor function and the resulting minus one.

And thus, we conclude:
$$f(n) = \lfloor n! \cdot (e-1) \rfloor - 1$$
the expression noted on the OEIS by Don Thousand in the comments.
A: The general term is given by
$$f(n)= n! \sum_{k=1}^{n-1} \dfrac{1}{k!}=n!\left[e-1+\sum_{k=n}^{\infty}\dfrac{1}{k!} \right]=n!(e-1)+n!\sum_{k=n}^{\infty}\dfrac{1}{k!} .$$
