The Left Most Number

If we look at the first of left-most number of the first four rows, we note that they are 1, 2, 4, and 8, indicating that the first number of row $r$ is given by $n_{r,1} = 2^{r-1}$ . Therefore, $n_{12,1} = 2^{12-1} = 2^{11} = \boxed{2048}$ .

To prove that the claim $n_{r,1} = 2^{r-1}$ is true for all $r\ge 1$ . We consider as follows. Let the number of elements (or numbers) in $r$ th row be $N_r$ . We note that $N_1=1$ and $N_r = 2N_{r-1}=2^{r-1}$ . The total number of elements for the first $r$ rows is then $S_r = \displaystyle \sum_{k=1}^r N_k = \sum_{k=1}^r 2^{k-1} = 2^r - 1$ . We also that $S_r = n_{r,l}$ the last number of row $r$ . Therefore, the first number of $(r+1)$ th row is $n_{r+1.1} = 2^r - 1+1 = 2^r$ , $\implies \boxed{n_{r,1} = 2^{r-1}} \blacksquare$ .