16. $\frac{n}{16}$.

From the problem, we know that $a_{n 3}$ is located in the $n$-th row and the 3rd column, and the common difference of the first column is $\frac{1}{4}$, with the common ratio of each row being $\frac{1}{2}$. By the general term formula of an arithmetic sequence, the first number in the $n$-th row is
$$
\frac{1}{4}+(n-1) \times \frac{1}{4}=\frac{n}{4} \text {. }
$$

Therefore, $a_{n 3}=\frac{n}{4} \times\left(\frac{1}{2}\right)^{2}=\frac{n}{16}$.
16. As shown in Figure 3, a right-angled triangular array of numbers is given, where the numbers in each column form an arithmetic sequence, and from the third row onwards, the numbers in each row form a geometric sequence with equal common ratios. Let the number in the $i$-th row and $j$-th column be $a_{i j}(i \geqslant j, i 、 j$ $\left.\in \mathbf{Z}_{+}\right)$. Then $a_{n 3}=$ $\qquad$