Find a binomial term/general formula for recurrence relation We know that Pascal's triangle obeys the recurrence relation $\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k} $
And we can simply $\binom{n}{k}$ by $\frac{n!}{k!\,(n - k)!}$
I have a recurrence relation where 
$$ f(n, k) = f(n - 1, k) + f(n - 2, k - 1) $$
How can I get a generel formula for that?
Thanks in advance!
Edited:
Base Case:
$f(n, 1) = n$ and $f(n, k) = 0$ when $n < k$

Let us compute $f(n, 2)$ for any integer $n\ge2$.
First see that $$f(2, 2) =  f(1, 2) + f(0, 1) = 0$$
$$f(3, 2) =  f(2, 2) + f(1, 1) = 1$$
$$f(4, 2) =  f(3, 2) + f(2, 1) = 3$$
$$f(5, 2) =  f(4, 2) + f(3, 1) = 6$$
We can conjecture that $f(n, 2) = {(n-2)(n-1)\over2} = \sum_{k=0}^{n-2}k$
Indeed $f(2,2)$ is our base step, and if $f(n, 2) = \sum_{k=0}^{n-2}k$, then $$f(n+1) = f(n, 2) + f(n-1, 1) = \sum_{k=0}^{n-2}k + n-1 = \sum_{k=0}^{n-1}k$$
Thus $f(n, 2) = \sum_{k=0}^{n-2}k$ for all $n$.

Now compute $f(n, 3)$ for any integer $n\ge 3$ :
$$f(3, 3) =  f(2, 3) + f(1, 2) = 0$$
$$f(4, 3) =  f(3, 3) + f(2, 2) = 0$$
$$f(5, 3) =  f(4, 3) + f(3, 2) = 1$$
$$f(6, 3) =  f(5, 3) + f(4, 2) = 4$$
$$f(7, 3) =  f(6, 3) + f(5, 2) = 10$$
We have $$f(n, 3) = \sum_{i=1}^{n-2} f(i, 2) = \sum_{i=1}^{n-2}\sum_{k=1}^{i-2}k = {(n-2)(n-3)(n-4)\over6}$$

What follows is that $$f(n, k) = \sum_{i=1}^{n-2} f(i, k-1) =\sum_{i_1=1}^{n-2} \sum_{i_2=1}^{i_1-2} {\cdot \cdot \cdot} \sum_{i_{k-1}=1}^{i_{k-2}-2} i_{k-1}$$
Computing this for different values of $k$, we can conjecture that for $k\ge 3$:
$$f(n,k) = (c_k + n(n-2k-1))(n-2){(n-k-3)!\over(n-2k)!k!}$$
Where $c_k$ is a sequence of integers. Now since :
$$f(n+2,k+1) = (c_{k+1} + (n+2)(n-2k-1))n{(n-k-4)!\over(n-2k)!(k+1)!}$$
and $$f(n+1,k+1) + f(n,k) = (c_{k+1} + (n+1)(n-2k-2))(n-1){(n-k-3)!\over(n-2k-1)!(k+1)!} + (c_k + n(n-2k-1))(n-2){(n-k-3)!\over(n-2k)!k!}$$
We get : $$(n-k-3)[(c_{k+1}+(n+1)(n-2k-2))(n-1)(n-2k) + (c_k + n(n-2k-1))(n-2)(k+1)] = (c_{k+1} + (n+2)(n-2k-1))n$$
Therefore $$-(k+3)[2k(c_{k+1}-2k-2)-2(k+1)c_k] = 0$$ and thus $$c_{k+1} = {k+1\over k}c_k + 2(k+1)$$
And $c_1 = 0$

We figured : $$f(n,k) = (c_k + n(n-2k-1))(n-2){(n-k-3)!\over(n-2k)!k!}$$ where $c_1 = 0$ and $c_{k+1} = {k+1\over k}c_k + 2(k+1)$ for any natural $k$.
