Finding a formula for a triangle similar to Pascal's. I need to find a formula expressed in binomial coefficients of the following triangle:
$$D_n^k=\begin{cases}
n \quad\quad\quad\quad\quad\text{ if } n=k \text{ or } k=0 \\
D_{n-1}^k+D_{n-1}^{k-1} \text{ otherwise}
\end{cases}$$
This triangle is the same as the Pascal's triangle, except for the base value ($n$ instead of $1$).

As darij grinberg has pointed out, the formula is $$D_n^k=\binom{n+2}{k+1}-2\binom{n}{k}$$ Here is a simple proof that it fits the definition:
$$D_n^n=\binom{n+2}{n+1}-2\binom{n}{n}=n+2-2=n$$
$$D_n^0=\binom{n+2}{1}-2\binom{n}{0}=n+2-2=n$$
$$D_n^k=\binom{n+2}{k+1}-2\binom{n}{k}\stackrel{(*)}{=}\binom{n+1}{k+1}+\binom{n+1}{k}-2(\binom{n-1}{k}+\binom{n-1}{k-1})\\=\binom{n+1}{k+1}-2\binom{n-1}{k}+\binom{n+1}{k}-2\binom{n-1}{k-1}=D_{n-1}^k+D_{n-1}^{k-1}$$
