\[ p(m, n) = \frac{n(n+1)}{2}+\frac{m(m-1)}{2}+(m-1)(n-1)\]

Too easy. It is just $p(m,n) = \binom{n+m}{2} - m + 1$.

https://artofproblemsolving.com/community/c6h1827482p12226300 General form of it

Baltic Way 1990.2. The squares of a squared paper are enumerated as shown on the picture. \[\begin{array}{|c|c|c|c|c|c} \ddots &&&&&\\ \hline 10&\ddots&&&&\\ \hline 6&9&\ddots&&&\\ \hline 3&5&8&12&\ddots&\\ \hline 1&2&4&7&11&\ddots\\ \hline \end{array}\]Devise a polynomial $p(m, n)$ in two variables such that for any $m, n\in\mathbb N$ the number written in the square with coordinates $(m, n)$ is equal to $p(m, n)$. Solution. Obviously, we can take $p(m, n)=\binom{m+n-1}2+n$.

obviously $P(m,n)=\sum_{i=1}^{n+m-2} i+n=\frac{(n+m-1)(n+m-2)}{2}+n$