Find formula for this sequence of numbers. Does anybody know how to compute the cell (i,j) of this table, using just i and j?
In practice, given a cell (i,j): 
- the left-top number is computing summing the top row of cell (i,j-1)
- the right-top number is computing summing the bottom row of cell (i,j-1)
- the left-bottom number is computing summing the left column of cell (i-1,j)
- the right-bottom number is computing summing the right column of cell (i-1,j)
This said, I want a formula based on i and j that allows me to compute a cell, without having to compute the whole table.

Thanks
 A: Your table does not agree with your rules. Let $c(i,j)$ be the entry in cell $\langle i,j\rangle$, considered as a $2\times 2$ matrix; then your table is constructed according to the rule $c(i,j)=c(i-1,j)+c(i,j-1)$, with initial conditions
$$c(1,j)=\begin{pmatrix}1&j-1\\0&1\end{pmatrix}\qquad\text{and}\qquad c(i,1)=\begin{pmatrix}1&0\\i-1&1\end{pmatrix}\;.$$
Assuming that the table is right, so that this is what you actually want, the general formula is
$$c(i,j)=\begin{pmatrix}\binom{i+j-2}{j-1}&\binom{i+j-2}{j-2}\\\binom{i+j-2}{j}&\binom{i+j-2}{j-1}\end{pmatrix}\;.$$
(Note that $\binom{n}k=0$ if $k<0$.) This is easily verified by induction on $i+j$. Suppose that it holds true for $i+j<k$, and let $m+n=k$. Then
$$\begin{align*}
c(m,n)&=c(m-1,n)+c(m,n-1)\\\\
&=\begin{pmatrix}\binom{m+n-3}{n-1}+\binom{m+n-3}{n-2}&\binom{m+n-3}{n-2}+\binom{m+n-3}{n-3}\\\binom{m+n-3}{n}+\binom{m+n-3}{n-1}&\binom{m+n-3}{n-1}+\binom{m+n-3}{n-2}\end{pmatrix}\\\\
&=\begin{pmatrix}\binom{m+n-2}{n-1}&\binom{m+n-2}{n-2}\\\binom{m+n-2}{n}&\binom{m+n-2}{n-1}\end{pmatrix}\;,
\end{align*}$$
as desired.
A: Have a closer look at the rules as these don't agree with the top rows of the cells in the table and cell $(4,4)$ may not be correct. Also, strictly speaking since all four rules are inductive you need a base case - this is the content of cell $(1,1)$. 
The first thing you will notice, if you are familiar with Pascal's triangle, is that the numbers in any row of a cell are consecutive numbers somwhere in Pascal's triangle. Writing an "extended" form of Pascal's triangle that is zero-padded for numbers outside the usual triangle we get something like:
$$\begin{matrix}
 1 &  0 &  0 &  0 &  0 &  0 &  0 &  0 & \ldots \\
 1 &  1 &  0 &  0 &  0 &  0 &  0 &  0 & \ldots \\
 1 &  2 &  1 &  0 &  0 &  0 &  0 &  0 & \ldots \\
 1 &  3 &  3 &  1 &  0 &  0 &  0 &  0 & \ldots \\
 1 &  4 &  6 &  4 &  1 &  0 &  0 &  0 & \ldots \\
 1 &  5 & 10 & 10 &  5 &  1 &  0 &  0 & \ldots \\
 1 &  6 & 15 & 20 & 15 &  6 &  1 &  0 & \ldots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots 
\end{matrix}$$
where the rows and columns are numbered starting at $0$.
In the top row of cell $i,j=1$ in your table, you have the $1, 0$ elements starting on the diagonal of the $(i-2)-th$ row in Pascal's triangle.
To relate the top row of cells in your table to Pascal's triangle (PT), observe that:


*

*moving down a column in your table equates to moving down the diagonal in PT

*moving along a row in your table equates to moving down a column in PT


So then for the top left cell in row $i$, column $j$ in your table:
$$t_{i,j} = p_{i+j-2,j-1} = \binom{i+j-2}{j-1}$$
You will get the same formulae as given by Brian Scott in his answer.   
