<s>[INST] Could there be a formula of this? Say you have a function:  f (m,n)
f (m,n) =  m  if n = 1
f (m,n) =  n  if m = 1
otherwise f (m,n) = f (m - 1, n) + f (m, n - 1)

Pre-calculated value:
1 2  3  4  5  6 
2 4  7  11 16
3 7  14 25
4 11 25
5 16
6

Just wondering if there could be formula of f(m,n), instead of doing a dynamic progrmming or recursive calculation to get the value.
 [/INST] There is indeed. Let $g(m,n)=f(m-n+1,n)$, so that for instance $g(5,3)=f(3,3)=14$. The corresponding table for $g$ is:
$$\begin{array}{}
1\\
2&2\\
3&4&3\\
4&7&7&4\\
5&11&14&11&5\\
6&16&25&25&16&6
\end{array}$$
Now 
$$\begin{align*}
g(m,n)&=f(m-n,n)\\
&=f(m-n-1,n)+f(m-n,n-1)\\
&=f\big((m-1)-n,n\big)+f\big((m-1)-(n-1),n-1\big)\\
&=g(m-1,n)+g(m-1,n-1)\;,
\end{align*}$$
which is the recurrence that generates Pascal’s triangle of binomial coeffients. Moreover, $g$’s table looks a lot like Pascal’s triangle in overall form:
$$\begin{array}{}
1\\
1&1\\
1&2&1\\
1&3&3&1\\
1&4&6&4&1\\
1&5&10&10&5&1\\
1&6&15&20&15&6&1
\end{array}$$
Ignore that first column of Pascal’s triangle:
$$\begin{array}{}
1\\
2&1\\
3&3&1\\
4&6&4&1\\
5&10&10&5&1\\
6&15&20&15&6&1
\end{array}\tag{1}$$
Subtract this from $g$’s triangle:
$$\begin{array}{}
0\\
0&1\\
0&1&2\\
0&1&3&3\\
0&1&4&6&4\\
0&1&5&10&10&5
\end{array}\tag{2}$$
Ignore the first column of this, and put $1$’s along the diagonal, and Pascal’s triangle shows up again. Now the $(m,n)$-entry in $(1)$ is $\binom{m}n$, and if $(2)$ really is Pascal’s triangle, its $(m,n)$-entry is $\binom{m-1}{n-2}$, so we conjecture that $g(m,n)=\binom{m}n+\binom{m-1}{n-2}$ and hence that
$$f(m,n)=g(m+n-1,n)=\binom{m+n-1}n+\binom{m+n-2}{n-2}\;.$$
We first verify the recurrence:
$$\begin{align*}
\binom{m+n-1}n+\binom{m+n-2}{n-2}&=\binom{m+n-2}{n-1}+\color{red}{\binom{m+n-2}n}+\binom{m+n-3}{n-3}+\color{blue}{\binom{m+n-3}{n-2}}\\
&=\binom{m+(n-1)-1}{n-1}+\color{red}{\binom{m+(n-1)-2}{n-3}}\\
&\qquad\qquad+\binom{(m-1)+n-1}n+\color{blue}{\binom{(m-1)+n-2}{n-2}}\\
&=f(m,n-1)+f(m-1,n)\;,
\end{align*}$$
as desired. Finally,
$$f(1,n)=\binom{1+n-1}n+\binom{1+n-2}{n-2}=1+(n-1)=n$$
and
$$f(m,1)=\binom{m+1-1}1+\binom{m+1-2}{-1}=m+0=m\;,$$
so the initial conditions are also satisfied. To repeat,
$$f(m,n)=\binom{m+n-1}n+\binom{m+n-2}{n-2}\;,$$
which can easily be manipulated into a variety of other forms involving binomial coefficients, e.g.,
$$f(m,n)=g(m+n-1,n)=\binom{m+n-2}n+\binom{m+n-2}{n-1}+\binom{m+n-2}{n-2}\;.$$
</s>