Identifying origin in convolution table I am taking the convolution of $x(n)=\left\{2,1,-1,-2,3 \right\}$ with $n=0$ at the third position with $h(n)=\left\{1,2,0,3 \right\}$ with $n=0$ at the second position. The answer is $y(n)=\left\{2,5,1,-10, -10, -3, 6, -9 \right\}$ with the $n=0$ at the fourth position. I studied convolution more than a decade ago and I used a table to do the convolution sum as follows:



There used to be a simple rule to obtain the location of the $n=0$ position of the output $y(n)$ from the table itself or perhaps from $x(n)$ and $y(n)$. But alas I cannot remember it. Grateful for any suggestions.
Given a signal $x[n]$ with the nonzero range of $N_1 \leq n \leq N_2$ and the signal $h[n]$ with the nonzero range $M_1 \leq n \leq M_2$, then the nonzero range of their comvolution $y[n] = x[n] \star h[n]$ is $N_1+M_1 \leq n \leq N_2 + M_2$.

Applying this to your case yields the following:
For the signal $x[n]$ with $n=0$ at third position the nonzero range is $-2 \leq n \leq 2$ and for $h[n]$ with $n=0$ at the second position the nonzero range is $-1 \leq n \leq 2$ which therefore yields the nonzero range of their convolution $y[n]$ is: 
$$ -3 \leq n \leq 4 $$

which means that the $n=0$ for $y[n]$ is at the fourth position.